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```PRACTICE EXAMS
PRACTICE EXAM 1
PE-1
ACTEX EXAM C/4 - PRACTICE EXAM 1
1. ? has a Weibull distribution with parameters and . Find the density function ²'³ of the
random variable A ~ c c²?°³ .
A) for  '  B) for  '  C) c' for '  D) c'° for '  E) for  '  2. A portfolio of risks models the annual loss of an individual risk as having an exponential
distribution with a mean of \$. For a randomly selected risk from the portfolio, the value of \$ has
an inverse gamma distribution with a mean of 40 and a standard deviation of 20. For a randomly
chosen risk, find the probability that the annual loss for that risk is greater than 20.
A) .524
B) .544
C) .564
D) .584
E) .604
3. You are given the following:
- Losses follow a distribution (prior to the application of any deductible) with mean 2000.
- The loss elimination ratio (LER) at a deductible of 1000 is 0.3.
- 60 percent of the losses (in number) are less than the deductible of 1000.
Determine the average size of a loss that is less than the deductible of 1000.
A) Less than 300
B) At least 300 nut less than 320
C) At least 320 but less than 340
D) At least 340 but less than 360
E) At least 360
4. A casino has a game that makes payouts at a Poisson rate of 5 per hour and the payout
amounts are 1,2,3,... without limit. The probability that any given payout is equal to is .
Payouts are independent. Calculate the probability that there are no payouts of 1, 2, or 3 in a
given 20 minute period.
A) 0.08
B) 0.13
C) 0.18
D) 0.23
E) 0.28
5. Zoom Buy Tire Store, a nationwide chain of retail tire stores, sells 2,000,000 tires per year of
various sizes and models. Zoom Buy offers the following road hazard warranty:
"If a tire sold by us is irreparably damaged in the first year after purchase, we'll replace it free,
regardless of the cause."
The average annual cost of honoring this warranty is \$10,000,000, with a standard deviation of
\$40,000. Individual claim counts follow a binomial distribution, and the average cost to replace a
tire is \$100. All tires are equally likely to fail in the first year, and tire failures are independent.
Calculate the standard deviation of the placement cost per tire.
A) Less than \$60
B) At least \$60, but less than \$65
C) At least \$65, but less than \$70
D) At least \$70, but less than \$75
E) At least \$75
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-2
PRACTICE EXAM 1
6. You are given the following random sample:
7 , 12 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53
Determine the method of percentile matching estimate of T Ò\  \$!Ó using the 25-th and
"
75-th smoothed empirical percentiles for a distribution with cdf J ÐBÑ œ "  "Ð B Ñα .
)
A) Less than .30
B) At least .30 but less than .32
D) At least .34 but less than .36
E) At least .36
C) At least .32 but less than .34
7. Suppose a 3-year data set is divided into a year-by-year count of new entrants, deaths and
right-censored observations:
.! œ "!!! ß B! œ #! ß ?! œ \$! ß ." œ #!! ß B" œ "! ß ?" œ #! ß
.# œ #!! ß B# œ "& ß ?# œ \$! Þ
E is the estimate of WÐ#Ñ using the approximation for large data sets if α œ " and " œ ",
and F is the estimate of WÐ#Ñ using the approximation for large data sets if α œ Þ& and " œ !.
Find EÎF .
A) Less than 1.000
B) At least 1.000 but less than 1.025
C) At least 1.025 but less than 1.050
D) At least 1.050 but less than 1.075
E) At least 1.075
8. Claim sizes of 10 or greater are described by a single parameter Pareto distribution, with
parameter α. A sample of claim sizes is as follows: 10 12 14 18 21 25
Calculate the method of moments estimate for α for this sample.
A) Less than 2.0
B) At least 2.0, but less than 2.1
C) At least 2.1, but less than 2.2
D) At least 2.2, but less than 2.3
E) At least 2.3
9. Let \1 , \2 , \3 be independent Poisson random variables with means ), 2), and 3)
respectively. What is the maximum likelihood estimator of ) based on sample values B" ß B# ß and
B\$ from the distributions of \1 , \2 and \3 , respectively,
_
_
3B 2B B
6B 3B 2B3
B 2B 3B3
A) 12 B
B) B
C) 1 62
D) 1 6 2 3
E) 1 112
10. You are given the following random sample of 12 data points from a population distribution
\:
7 , 15 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53
Suppose that the distribution variance is 100.
Determine the bias in the biased form of the sample variance as an estimator of the distribution
variance.
A) Less than  &
B) At least  & but less than !
C) At least ! but less than &
D) At least & but less than "!
E) At least "!
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 1
PE-3
11. The following table was obtained by fitting both a Poisson distribution and a binomial
distribution to a data set of 100,000 integer-valued observations.
V ~ À
V ~ Á V ~ À
Fitted Poisson
Fitted Binomial
expected
expected
,
,À
À
Á À
À
,
Á À
À
Á À
À
À
À
À
À
À
À
À
À
 Totals Á Á À
Á À
Degrees of freedom
cc~
cc~
 À
À   À
-value
You are given that the negative loglikelihood of the fitted Binomial model is 36,787 .
Use Schwarz Bayesian Criterion to choose between the Poisson and Binomial models.
A) Choose Poisson
B) Choose Binomial
C) They are equally preferable
D) Not enough information is available to determine the Poisson likelihood function.
E) None of A, B, C are correct
12. You are given the following random sample of 8 data points from a population distribution
?:
1,2,2,2,2,3,4,8
It is assumed that ? has an exponential distribution with parameter , and the prior distribution of
is discrete with 7 ´# ~ µ ~ ÀÁ 7 ´# ~ µ ~ ÀÁ 7 ´# ~ µ ~ À
Find the mean of the posterior distribution.
A) 2.5
B) 2.6
C) 2.7
D) 2.8
E) 2.9
13. Prior to tossing a coin, it is believed that the chance of tossing a head is equally likely to be
or . The coin is tossed twice, and both tosses result in a head. Determine the posterior
A) B) C) D) E) 14. In a portfolio of risks, each risk has an exponential claim amount distribution. The mean of
the claim amount distribution for a randomly chosen risk is , where has a Gamma distribution
with parameters ~ À and ~ . A single claim amount of 2 is observed for a randomly
chosen risk. Find the Buhlmann credibility premium for the next claim amount for the same risk.
A) Less than 1.0
B) At least 1.0, but less than 1.25
C) At least 1.25, but less than 1.5
D) At least 1.5, but less than 1.75
E) At least 1.75
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-4
PRACTICE EXAM 1
15. A risk class is made up of three equally sized groups of individuals. Groups are classified as
Type A, Type B and Type C. Any individual of any type has probability of .5 of having no claim
in the coming year and has a probability of .5 of having exactly 1 claim in the coming year. Each
claim is for amount 1 or 2 when a claim occurs. Suppose that the claim distributions given that a
claim occurs, for the three types of individuals are
° % ~ 7 ²claim of amount %OType A and a claim occurs³ ~ D
Á
° % ~ ° % ~ 7 ²claim of amount %OType B and a claim occurs³ ~ D
Á
° % ~ °
% ~ 7 ²claim of amount %OType C and a claim occurs³ ~ D
À
°
% ~ If an individual is chosen at random from the risk population and ~ observation ? is
available for that individual, find the credibility premium for the next exposure period for this
individual.
A) À? b À
B) À? b À
C) À
? b À
D) À
? b À
E) À? b À
16. A scientist perform experiments, each with a 60% success rate. Let ? represent the number
of trials until the first success. Use the inverse transform method to simulate the random variable,
?, and the following random numbers (where low numbers correspond to a high number of
trials): 0.15 , 0.62 , 0.37, 0.78 . Generate the total number of trials until three successes result.
A) 3
B) 4
C) 5
D) 6
E) 7
17. You are given the following information on towing losses for two classes of insureds, adults
and youths:
Exposures
Year
Youth
Total
1996
2000
450
2450
1997
1000
250
1250
1998
1000
175
1175
1999
1000
125
1125
Total
5000
1000
6000
Year
1996
1997
1998
1999
Weighted Average
0
5
6
4
3
Youth
15
2
15
1
10
Total
2.755
4.400
7.340
3.667
4.167
You are also given that the estimated variance of the hypothetical means is 17.125.
Determine the nonparametric empirical Bayes credibility premium for the youth class.
A) Less than 5
B) At least 5, but less than 6
C) At least 6, but less than 7
D) At least 7, but less than 8
E) At least 8
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 1
PE-5
18. \ and ] are independent random variables with IÒ\Ó œ Z +<Ò\Ó œ " ß
IÒ] Ó œ Z +<Ò] Ó œ # . Two random variables ^" and ^# are defined as follows.
^" œ "# \  "# ] , and ^# is a mixture of \ and ] where J^# ÐDÑ œ "# J\ ÐDÑ  "# J] ÐDÑ .
Find Z +<Ò^" Ó  Z +<Ò^# Ó .
A) #
B) "
C) !
D)  "
E)  #
19. The XYZ Insurance Company sells property insurance policies with a deductible of \$5,000,
policy limit of \$500,000, and a coinsurance factor of 80%. Let \3 be the individual loss amount
of the 3th claim and ]3 be the claim payment of the 3th claim. Which of the following represents
the relationship between \3 and ]3 ?
!
\3 Ÿ &ß !!!
!Þ)!Ð\

&ß
!!!Ñ
&ß
!!!  \3 Ÿ '#&ß !!!
A) ]3 œ H
3
&!!ß !!!
\3  '#&ß !!!
!
\3 Ÿ &ß !!!
B) ]3 œ H !Þ)!Ð\3  %ß !!!Ñ %ß !!!  \3 Ÿ &!!ß !!!
&!!ß !!!
\3  &!!ß !!!
!
\3 Ÿ &ß !!!
C) ]3 œ H !Þ)!Ð\3  &ß !!!Ñ &ß !!!  \3 Ÿ '\$!ß !!!
&!!ß !!!
\3  '\$!ß !!!
!
\3 Ÿ 'ß #&!
D) ]3 œ H !Þ)!Ð\3  'ß #&!Ñ 'ß #&!  \3 Ÿ '\$"ß &!!
&!!ß !!!
\3  '\$"ß &!!
!
\3 Ÿ &ß !!!
E) ]3 œ H !Þ)!Ð\3  &ß !!!Ñ &ß !!!  \3 Ÿ &!&ß !!!
&!!ß !!!
\3  &!&ß !!!
20. A Poisson distribution has mean 1 and probability function
"
05 œ /5x for 5 œ !ß "ß #ß ÞÞÞ.
":
A geometric distribution has mean : , where !  :  ", and probability function
15 œ :Ð"  :Ñ5 for 5 œ !ß "ß #ß ÞÞÞ
0
A comparison of the two distributions is made by summing the ratios of the probabilities 15 .
5
5œ!
Find the value of : that minimizes the sum.
A) Less than .300
B) At least .300, but less than .325
D) At least .350, but less than .375
E) At least .375
∞
C) At least .325, but less than .350
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-6
PRACTICE EXAM 1
21. Actuaries have modeled auto windshield claim frequencies. They have concluded that the
number of windshield claims filed per year per driver follows the Poisson distribution with
parameter , where follows the gamma distribution with mean 3 and variance 3. Calculate the
probability that a driver selected at random will file no more than 1 windshield claim next year.
A) 0.15
B) 0.19
C) 0.20
D) 0.24
E) 0.31
22. A compound Poisson claim distribution has ~ and individual claims amounts distributed
as follows:
%
? ²%³
À
À
Determine the expected cost of an aggregate stop-loss insurance subject to a deductible of 6.
A) Less than 15.0
B) At least 15.0 but less than 15.3
C) At least 15.3 but less than 15.6
D) At least 15.6 but less than 15.9
E) At least 15.9
23. A mortality study is conducted on 50 lives observed from time zero. You are given:
(i)
Time Number of Deaths
Number Censored
!
!
!
15
2
0
17
0
3
25
4
0
30
0
30
32
8
0
40
2
0
(ii) : (35) is the Product-Limit estimate of : (35).
(iii) = V
([: (35)] is the estimate of the variance of : (35) using Greenwood’s formula.
=V
([: (35)]
(iv)
~ 0.011467
[:(35)]2
Determine 30 , the number censored at time 30.
A) 3
B) 6
C) 7
D) 8
E) 11
24. You are given the following random sample of 12 data points from a population
distribution ? :
7 , 12 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53
Using the uniform kernel with bandwidth 5 , find V ²³ .
A) .03
B) .04
C) .05
D) .06
E) .07
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 1
PE-7
25. Loss random variable \ has a uniform distribution on Ð!ß )Ñ.
A sample is taken of 8 insurance payments from policies with a limit of 100.
Eight of the sample values are limit payments of 100.
The maximum likelihood estimate of ) is s).
Another sample is taken, also of 8 insurance payments, but from policies with a limit of 150.
Three of the sample values are limit payments of 150.
The maximum likelihood estimate of ) is %\$s).
Determine 8.
A) 40
B) 42
C) 44
D) 46
E) 48
26. For a group of policies, you are given:
(i) Losses follow a uniform distribution on the interval Ð!ß )Ñ , where )  #& .
(ii) A sample of 20 losses resulted in the following:
Interval
Number of Losses
B Ÿ "!
8"
"!  B Ÿ #&
8#
B  #&
8\$
The maximum likelihood estimate of ) can be written in the form #&  C. Determine C.
#&8
#&8"
#&8#
"
#
A) 8#&8
B) 8#&8
C) 8 8\$
D) 8 8
E) 8 8
8
8
8
8
#
\$
"
\$
"
#
"
#
\$
"
#
\$
27. You are given:
(i) Loss payments for a group health policy follow a gamma distribution with unknown mean.
but with α known to be equal to 2.
(ii) A sample of losses is: 100 200 400 800 1400 3100
Find the estimated asymptotic variance of the maximum likelihood estimate of ).
A) Less than 20,000
B) At least 20,000 but less than 21,000
C) At least 21,000 but less than 22,000
D) At least 22,000 but less than 23,000
E) At least 23,000
28. The following table displays the number of policyholders by territory by number of claims:
Number
of Claims
Territory 1
Territory 2
Territory 3
Territory 4
Total
0
97
188
392
293
970
1
2
10
4
4
20
2
1
2
4
3
10
Total
100
200
400
300
1000
You are testing the hypothesis that the claim count distributions are the same in each territory
using the Chi-Square goodness of fit test with a significance level of 5%. Calculate the absolute
value of the difference between the test statistic and the critical value. Do not adjust for small
numbers of expected observations that may occur in some cells.
A) Less than 0.30
B) At least 0.30, but less than 0.40
C) At least 0.40, but less than 0.50
D) At least 0.50, but less than 0.60
E) At least 0.60
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-8
PRACTICE EXAM 1
29. An analysis of credibility premiums is being done for a particular compound Poisson claims
distribution, where the criterion is that the total cost of claims is within 5% of the expected cost of
claims with a probability of 90%. It is found that with ~ exposures (periods) and
c
? ~ À, the credibility premium is À . After 20 more exposures (for a total of 80) and
c
revised ? ~ , the credibility premium is À . After 20 more exposures (for a total of
c
100) the revised ? is À. Assuming that the manual premium remains unchanged in all cases,
and assuming that full credibility has not been reached in any of the cases, find the credibility
premium for the 100 exposure case.
A) 191.5
B) 192.5
C) 193.5
D) 194.5
E) 196.5
30. A random sample of ~ 8 values from the distribution of ? is given:
Á Á Á Á Á Á Á The distribution of ? is assumed to be exponential with parameter , so that ²%O³ ~ c%° À
The prior distribution of # is assumed to be uniform on the interval ´Á µ , so that ²³ ~ À
for   À Find the Buhlmann credibility premium.
A) Less than 14.0
B) At least 14.0, but less than 14.3
D) At least 14.6, but less than 14.8
E) At least 14.8
C) At least 14.3, but less than 14.6
31. You are given the following information about six coins:
-
0.5
5
0.25
6
0.75
A coin is selected at random and then flipped repeatedly. ? denotes the outcome of the th flip,
where “1” indicates heads and “0” indicates tails. The following sequence is obtained:
: ~ ¸? Á ? Á ? Á ? ¹ ~ ¸Á Á Á ¹
Determine the Buhlmann estimate of ? ..
A) Less than .50
B) At least .50 but less than .52
C) At least .52 but less than .54
D) At least .54 but less than .56
E) At least .56
32. An insurance company has two group policies. The aggregate claim amounts (in millions of
dollars) for the first three policy years are summarized in the table below. Assume that the two
groups have the same number of insureds.
Aggregate Claim Amounts
Group
Policy Year
Policy
5
8
11
2
8
4
Determine the estimated Buhlmann credibility premium for group 1 for the fourth year.
A) 8.0
B) 8.2
C) 8.4
D) 8.6
E) 8.8
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 1
PE-9
33. An estimate is being made of the probability of flipping a head with a particular coin. The
coin will be flipped until the estimated standard deviation of the estimated value of is less than
.075. In the first 40 coin flips there are 25 heads and 15 tails. The next 10 coin flips are
T H H H T H H T H H . With which flip is the stopping criterion reached?
A) 41
B) 42
C) 43
D) 44
E) 45
34. 5 is a non-negative integer valued random variable with expected value 9.
5 ²³ is a zero-modified distribution of 5 with 7 ´5 ²³ ~ µ ~ À and with mean 7.
5 ²³ is a zero-truncated distribution of 5 . Determine ,´5 ²³ µ .
A) 6
B) 8
C) 10
D) 12
E) 14
35. You are given:
(i) Claim amounts follow a mixture of two exponential distributions with probability density
function:
²%³ ~ c%° b c%° , %  (ii) A random sample of claim amounts 5 7 7 9 10 10 14 16 18 24
is estimated using the method of moments. Find the median of the estimated distribution.
A) Less than 8 B) At least 8 but less than 9
C) At least 9 but less than 10
D) At least 10 but less than 11
E) At least 11
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-10
PRACTICE EXAM 1
ACTEX EXAM C/4 - PRACTICE EXAM 1 SOLUTIONS
1. A ~ ²?³ ~ c c²?°³ S ? ~ ´ c ² c A³µ° ~ ²A³ .
A ²'³ ~ ? ²²'³³ h Z ²'³
~
° °³
²´c²c'³µ° ³ c c²´c²c'³µ
~ for  '  .
h ´ c ² c '³µ c h c'
A has a uniform distribution on ²Á ³À
2. Given \$ ~ , the annual loss ? has an exponential distribution with mean and \$ has an
inverse gamma distribution. ? is a continuous mixture distribution of an "exponential over an
inverse gamma".
Suppose that the inverse gamma distribution of \$ has parameters and . We are given that the
mean and standard deviation of \$ are 20 and 10. Therefore, the variance of \$ is 100 and the 2nd
moment of \$ is ,´\$ µ ~ = ´\$µ b ²,´\$µ³ ~ b ~ .
The mean of an inverse gamma is c
. and the 2nd moment is ²c³²
c³ À
From the two equations c
~ and ²c³²
c³ ~ ,
c
we get ~ ²c³²c³ ,² c ³ ~ c . Then solving for results in ~ .
Substituting back into c
~ , we get that ~ .
When we have a continuous mixture distribution for ? over \$, the pdf, expected values and
probabilities for the marginal distribution of ? can be found by conditioning over \$.
The conditional pdf of ? given \$ ~ is ²%O\$ ~ ³ ~ c%° and the pdf of the inverse
c°
gamma distribution of \$ is \$ ²³ ~ bh!²³ .
c°
From the calculated parameter values, we have \$ ²³ ~ h!²
³ .
We can find the probability 7 ²?  ³ by conditioning over :
c°
B
B
7 ²?  ³ ~ 7 ²?  O³ h \$ ²³ ~ c° h ~
!²
³
h
c°
B h!²
³
.
We know that the inverse gamma pdf must integrate to 1 (as any pdf must), so that
c°
!²³
B c°
B b
~ . It follows that b ~ .
h!²³
c°
B
Therefore, !²
³
B c°
!²
³
= b ~ , and 7 ²?  ³ ~ !
²
³ h ~ À
.
It is possible to show in general that if the conditional distribution of ? given \$ is exponential
with mean \$, and if \$ has an inverse gamma distribution with parameters and , then the
marginal (unconditional) distribution of ? is (two-parameter) Pareto with parameters and .
In this example, the marginal distribution of ? would be Pareto with parameters ~ and
~ , and 7 ²?  %³ ~ ² b%
³ , so that 7 ²?  ³ ~ ² Answer: C
³ .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 1
PE-11
²%³
% h - ²³ % ~ À
% ²%³ % ,
since we are given - ²³ ~ À
and since the conditional density function of ? given that
3. We are to find ,´?O?  µ ~ ²%³
?  is - ²³ . We are also given ,´?µ ~ , and 3,9 ~
Therefore, ,´? w µ ~ .
,´?wµ
,´?µ
~ À .
But ,´? w µ ~ % ²%³ % b ´ c - ²³µ ~ % ²%³ % b ,
so that % ²%³ % ~ . Then ,´?O?  µ ~ À
²³ ~ À .
4. When a payout occurs, it is 1, 2 or 3 with probability b b ~ .
The number of payouts that are 1, 2 or 3 follows a Poisson process with an hourly rate of
d ~ . The expected number of payouts that are 1, 2 or 3 in 20 minutes, say 5 , has a
°
Poisson distribution with mean ~ . The probability that there are no payouts of 1, 2, or
3 in a given 20 minute period is the probability that 5 ~ , which is c° ~ À .
5. We denote by ? the warranty claim that arises from the sale of a tire with a mean of ,´?µ and
ÁÁ
variance = ´?µ. The total claim for the year is : ~ ? , where ? is the warranty claim
~
that arises from the sale of the -th tire (we are told that 2,000,000 tires are sold).
,´:µ ~ Á Á ,´?µ , and since tire failures are independent of one another,
= ´:µ ~ Á Á = ´?µ . We are given that ,´:µ ~ Á Á , so that ,´?µ ~ , and
Á
we are given = ´:µ ~ Á so that = ´?µ ~ ÁÁ ~ .
When a tire is replaced, the cost of replacement is the random variable @ . We are given
,´@ µ ~ , and we are asked to find the standard deviation of @ . We are told that individual
claim counts follow a binomial distribution. We interpret this as saying that the tire will either
fail during the year with probability or the tire will not fail, with probability c .
@ prob. For each tire sold, the warranty cost ? has distribution ? ~ F
.
prob. c Then ,´?µ ~ ~ h ,´@ µ ~ , so that ~ À .
Above we found that = ´?µ ~ , so that ,´? µ c ~ and then ,´? µ ~ .
It is also true that ,´? µ ~ ~ h ,´@ µ ~ À ,´@ µ , so that ,´@ µ ~ Á .
Then, = ´@ µ ~ ,´@ µ c ²,´@ µ³ ~ Á c ~ Á .
Finally, the standard deviation of @ is j
Á ~ À
.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-12
PRACTICE EXAM 1
6. The smoothed empirical estimate of the 25-th percentile is 16, and of the 75-th percentile is
32.25. The cdf is - ²%³ ~ c b²% ³ . Using the empirical estimates, we get two equations:
À ~ - ²
³ ~ c b²
³ Á À ~ - ²À³ ~ c b² À
.
³
À Then, b ² ³ ~ À ~ and b ² ³ ~ À ~ , so that
²³
À ² À
³ ,² ³ ~ °² ³ ~ . Then ² ³ ~ , so that ~ ²À°
³ ~ À ,
and ~ À . The estimated value of 7 ´?  µ is c - ²³ ~ b² ³À ~ À .
À
7. With ~ and ~ , the numbers at risk are
~ c " ~ Á ~ ² b ³ c ²% b " ³ c " ~ .
The estimate of the survival probability to time 2 is ( ~ 4 c 54 c 5 ~ À .
With ~ À and ~ , we have
7 ~ Á 7 ~ b c c ~ Á 7 ~ b c c ~ Á
and ~ À c ²³" ~ ²À³²³ ~ Á and
~ c ²% b " ³ b À c ²³" ~ b ²À³²³ ~ .
The estimate of the survival probability to time 2 is ) ~ 4 c 54 c 5 ~ À .
Then (°) ~ À .
8. "Claim sizes are 10 or greater" indicates that ~ . The mean of the single parameter Pareto
with
~ is  c ~ c À The sample mean is . According to the method of moments,
we set
c ~ and solve for . The resulting value is ~ À .
9. The likelihood function is
3 ~ 1 ²%1 ³ h 2 ²%2 ³ h 3 ²%3 ³ ~ c %1 %1 [ h c2 (2)%2 %2 [ h c3 (3)%3 %3 [
%
%
~ c6 %1 b%2 b%3 h2 2 h3 3 %1 [ %2 [ %3 [ À
3 ~ c 6 b ²%1 b%2 b%3 ³ h b (where does not depend on ).
_
% b% b%
Setting 3 ~ 0 results in ~ 1 62 3 ~ %2 . Answer: A
10. The bias in the estimator is ,´ ²% c c
%³ µ c . We know that
~
,´ c
²% c c
%³ µ ~ , since c
²% c c
%³ is an unbiased estimator of = ´?µ.
~
~
c
Therefore, ,´ ²% c c
%³ µ c ~ ,´ c
h c ²% c %³ µ c ~
~
~ ²³ c ~ c À À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 1
PE-13
11. 3 ~ ² ³Á ² ³Á ² ³ ² c c c ³
3 ~ Á b Á b b ² c c c ³
V
V³
²
V ~ c ~ cÀ ~ À
Á V ~ c h [ ~ À Á V ~ c h [ ~ À À
The estimate of c 3 is c Á À
The SBC measure for the estimated Poisson model is ( ~ parameter)
3 c h ~ c Á c Á ~ c Á .
The SBC measure for the estimated Binomial model is ( ~ parameters)
3 c h ~ c Á 87 c Á ~ c Á .
According to the SBC, choose the fitted Binomial model over the fitted Poisson model.
V
V
V
12. The model distribution is ²%O³ ~ c% ° ~ c'% ° ~ c° .
~
The joint distribution of ? and # is
À d c'%
?Á# ²%Á ³ ~ ²%O# ~ ³ h 7 ´# ~ µ ~ H À d c c'% °
À d c c'% °
The marginal distribution of ? is
? ²%³ ~ ?Á# ²%Á ³ b ?Á# ²%Á ³ b ?Á# ²%Á ³
~ ²À³c'% b ²À³c c'% ° b ²À³c c'% ° .
The posterior distribution of # is
#~
#~ .
#~
²À³c'%
#O? ²O%³ ~ ²À³c'% b²À³c c'%° b²À³c c'%° ,
²À³c c'% °
#O? ²O%³ ~ ²À³c'% b²À³c c'%° b²À³c c'%° ,
²À³c c'% °
#O? ²O%³ ~ ²À³c'% b²À³c c'%° b²À³c c'%° À
For the given vector of ? values, this becomes
#O? ²O%³ ~ À Á #O? ²O%³ ~ À Á #O? ²O%³ ~ À
À
The posterior mean is ²À³²³ b ²À³²³ b ²À
³²³ ~ À .
7 ´3HO1H and 2HO ~
13. Posterior probability of H is
7 ´3H,1H,2Hµ
7 ´1H,2Hµ .
We will denote by " H" the event that the probability of tossing a head with the coin is , with a
similar definition for " H".
7 ´1H,2Hµ ~ 7 ´1H,2HO Hµ h 7 ´ Hµ b 7 ´1H,2HO Hµ h 7 ´ Hµ ~ ² ³ ² ³ b ² ³ ² ³ ~ .
7 ´3H,1H,2Hµ ~ 7 ´3H,1H,2HO Hµ h 7 ´ Hµ b 7 ´3H,1H,2HO Hµ h 7 ´ Hµ
~ ² ³ ² ³ b ² ³ ² ³ ~ ~ .
°
Then 7 ´3HO1H and 2HO ~ ° ~ . Answer: C
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-14
PRACTICE EXAM 1
14. Hypothetical mean ~ ²³ ~ ,´?Oµ ~ Á Process variance ~ #²³ ~ = ´?Oµ ~ .
~ ,/4 ~ ,´²³µ ~ ,´µ ~ ~ À
# ~ ,7 = ~ ,´ #²³µ ~ ,´ µ ~ ² b ³ ~ À .
~ = /4 ~ = ´²³µ ~ = ´µ ~ ~ À .
A ~ b
# ~
À ~ S the Buhlmann credibility premium is
b
À
c
A? b ² c A³ ~ h b ² c ³²À³ ~ À
.
15. # ~ Á Á each with probability
, and
²³ ~ ,´?O# ~ µ : ²³ ~ ,´?O# ~ µ ~ ²À³´ h b h µ ~ Á
²³ ~ ,´?O# ~ µ ~ Á ²³ ~ ,´?O# ~ µ ~ À
Therefore ~ ,´²#³µ ~ ² ³´ b b µ ~ À
#²³ ~ = ´?O# ~ µ .
%~
?O# ²%O ~ ³ ~ F h ~ h ~ %~
%~
S #²³ ~ ²³ ² ³ b ²³ ² ³ b ²³ ² ³ c ²,´?O# ~ µ³ ~ c ² ³ ~ ,
%~
?O# ²%O ~ ³ ~ F ?O# ²%O ~ ³ ~ F
% ~ Á S #²³ ~ c ² ³ ~ , and
%~
%~
% ~ S #²³ ~ c ² ³ ~ .
%~
Then # ~ ,´#²#³µ ~ ² ³² ³ b ² ³² ³ b ² ³² ³ ~ À
Also, ~ = ´²#³µ ~ ,´²#³ µ c ²,´²#³µ³
~ ² ³ ² ³ b ² ³ ² ³ b ² ³ ² ³ c ² ³ ~ À
Then ~ # ~ ~ . c
The credibility premium is A? b ² c A³ À
In this case, ~ and ~ so that A ~ b
~ À , and
c
? ~ ? . From Problem 6, we have ~ . The credibility premium is
²À³? b ²À³² ³ ~ À? b À
À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 1
PE-15
16. To say that low numbers correspond to a low number of trials is the standard form of the
inverse transform method: given a uniform random number we find the integer such that
-? ² c ³   -? ²³ , and the simulated value of ? is . If we apply the inverse transform
method in the form where low numbers correspond to a high number of trials, then given a
uniform random number , we find the integer such that -? ² c ³  c  -? ²³ .
The random variable ? is the number of success until the first trial. This is a geometric
distribution with probability function ²%³ and distribution function - ²%³ as follows:
?~
ÀÀÀ
²³
À
²À³²À
³ ~ À
²À³ ²À
³ ~ À
²À³ ²À
³ ~ À ÀÀÀ
- ²³
À
À
À
À
Each uniform number simulates the number of trials until the next success. We must simulate
? three times to get ? (the simulated number of trials until the first success, ? (the additional
number of trials until the second success) and ? (the additional number of trials until the third
success). Then the total number of trials until the third success is ? b ? b ? .
The first random number is ~ À, so that c ~ À . We see that
- ²³ ~ À  À  À
~ - ²³ so that the simulated value of ? is 3. This is ? , the
simulated number of trials until the first success. The second random number is ~ À
, with
c ~ À . We see that À  À
~ - ²³ , so that the simulated value of ? is 1.
~ À , so that c ~ À
. We see that - ²³ ~ À
 À
 À ~ - ²³, so that the
simulated value of ? is 2. Then ? b ? b ? ~ b b ~ . Answer: D
17. Since the numbers of exposures differs from one exposure period to the next, we use
non-parametric empirical Bayes estimation for the Buhlmann-Straub model.
Under the Buhlmann-Straub model, there are exposure periods for policy holder ,
~ Á Á ÀÀÀÁ . For policyholder and exposure period , there are exposure units (2000
adults for 1996 adult exposure, etc.), and ? represents the observed average claim per exposure
unit (for "cell" Á ) (0 for 1996 adult exposure).
In this case, ~ policyholders (adult and youth), and ~ ~ exposure periods for each
c
policyholder. The usual unbiased estimates that are used for the structural parameters are V~?
c
h ²? c ? ³ (estimated mean of the process variances), and
V# ~ ² c³
~
~
V
~ ~
c
c
h ´ ²? c ? ³ c V#² c ³ µ (estimated variance of the hypothetical
~
c
~
c
means). We are given that V
~ À and ? ~ À
V ~ ~ V# .
The estimated credibility factor for group is A
V
b V
b
c
c
We are given ? ~ (adult) and ? ~ (youth) , so that
h ´² c ³ b ² c ³ b ²
c ³ b ² c ³
V# ~ ²c³
~
b ² c ³ b ² c ³ b ² c ³ b ² c ³ µ ~ Á .
For adults, ~ , for youths, ~ .
1
V 2 ~ 2 V# ~
A
Á ~ À .
2 b V
1b À
Credibility premium for the youth class is
c
V ?
V ³
A
V ~ ²À³²³ b ²À³²À
³ ~ À
.
b ² c A
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-16
PRACTICE EXAM 1
18. = ´A µ ~ ² ³ = ´?µ b ² ³ = ´@ µ ~ .
= ´A µ ~ ,´A µ c ²,´A µ³ .
= ´?µ ~ ~ ,´? µ c ²,´?µ³ ~ ,´? µ c S ,´? µ ~ Á and
,´@ µ ~ = ´@ µ b ²,´@ µ³ ~ b ~ .
,´A µ ~ ,´?µ b ,´@ µ ~ . ,´A2 µ ~ ,´? µ b ,´@ µ ~ .
= ´A µ ~ c ² ³ ~ . = ´A µ c = ´A µ ~ c .
19 . With coinsurance factor , deductible , policy limit ²" c ³ , the amount paid per loss is
?
(we are assuming in inflation rate of ~ )
@ ~ H ²? c ³  ?  " .
²" c ³ ?  "
In this problem, the coinsurance factor is ~ À , the deductible is ~ Á , and the policy
limit is À²" c Á ³ ~ Á , so that the maximum covered loss is " ~ Á .
The amount paid per loss becomes
@ ~ H À²? c Á ³ Á  ?  Á .
Á ?  Á c
c
20. ~ ² [ ³,²² c ³ ³ ~ h²c³
h[
B
B
c
c
°²c³
°²c³
~ h
~ h °²c³ ~ .
h[
~
~
°²c³
It can be seen that approaches B as Sb (from above 0)
and as Sc (from below 1).
°²c³
°²c³
To find where is minimized, we take ´ µ , and minimize that.
°²c³
°²c³
´ µ ~ c c , and ´ µ ~ [ c c ]
c²c³
~ ²c³
c ~ ²c³ h .
The critical points occur where c ² c ³ ~ c b c ~ ,
so that ~
fj
. We ignore the root  , and ~
cj
~ À .
21. ? , the number of claims filed by a driver in a year is Poisson with mean \$, where \$ has a
gamma distribution with mean ~ , and variance ~ . The gamma distribution must
have parameters ~ and ~ . ? is an example of a distribution created by continuous
mixing. The combination of the conditional distribution of ?O\$ being Poisson and the
distribution of \$ being gamma results in a marginal distribution of ? that is negative binomial
with ~ and ~ . Thus, ? has a negative binomial distribution with ~ and ~ .
7 ´?  µ ~ 7 ´? ~ Á µ ~ ²b ³ b ²b ³b ~ b ~ .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 1
PE-17
22. The minimum claim amount is 5 if a claim occurs. W must be 0 or a multiple of 5.
The stop-loss insurance with deductible 6 pays ÐW  'Ñ œ W  ÐW • 'Ñ ,
! Wœ!
where W • ' œ H & W œ &
.
' W "!
IÒWÓ œ IÒR Ó † IÒ\Ó œ Ð\$ÑÒÐ&ÑÐÞ'Ñ  Ð"!ÑÐÞ%ÑÓ œ #" .
IÒW • 'Ó œ & † T ÐW œ &Ñ  'Ò"  T ÐW œ !ß &ÑÓ.
T ÐW œ !Ñ œ T ÐR œ !Ñ œ /\$ and
T ÐW œ &Ñ œ T ÐR œ "Ñ ‚ T Ð\ œ &Ñ œ /\$ † \$ ‚ ÐÞ'Ñ œ "Þ)/\$ .
IÒW • 'Ó œ &Ð"Þ)/\$ Ñ  'Ò"  T ÐW œ !ß &ÑÓ œ 'Ò"  #Þ)/\$ Ó œ &Þ'" .
Then IÒÐW  'Ñ Ó œ IÒWÓ  IÒW • 'Ó œ #"  &Þ'" œ "&Þ\$* . Answer: C
23. Greenwood's formula for the variance is
#
%
)
#
s
s
s ÒWÐ\$&ÑÓ
Z
œ ÒWÐ\$&ÑÓ
Ò Ð&!ÑÐ%)Ñ  Ð%&ÑÐ%"Ñ  Ð%"- ÑÐ\$\$- Ñ Ó .
\$!
\$!
#
%
)
Therefore, Þ!""%'( œ Ð&!ÑÐ%)Ñ  Ð%&ÑÐ%"Ñ  Ð%"- ÑÐ\$\$- Ñ , so that
\$!
\$!
)
Ð%"-\$! ÑÐ\$\$-\$! Ñ œ Þ!!)%'&'& , and then Ð%"  -\$! ÑÐ\$\$  -\$! Ñ œ *%& .
We can solve the quadratic equation, or substitute in the possible answers. -! œ ' .
24. For the point B œ \$! , there are five C4 values within the band from \$!  & œ #& to
\$!  & œ #& . These are the data values C& œ #' ß C' œ #( ß C( œ #* (repeated twice) ß
"
C) œ \$! and C* œ \$\$ . Therefore, 5C& Ð\$!Ñ œ â œ 5C* Ð\$!Ñ œ "!
and 5C4 Ð\$!Ñ œ ! for all other
C4 's since B œ \$! is outside the interval C4  & ß C4  & for the other C4 's .
Then s0 Ð\$!Ñ œ :ÐC4 Ñ † 5C4 Ð#!Ñ œ :ÐC& Ñ † 5C& Ð\$!Ñ  â  :ÐC* Ñ † 5C* Ð\$!Ñ
""
4œ"
œ :Ð#'Ñ † 5#' Ð\$!Ñ  :Ð#(Ñ † 5#( Ð\$!Ñ  :Ð#*Ñ † 5#* Ð\$!Ñ  :Ð\$!Ñ † 5\$! Ð\$!Ñ  :Ð\$\$Ñ † 5\$\$ Ð\$!Ñ
"
"
"
"
#
"
"
"
"
"
"
œ Ð "#
ÑÐ "!
Ñ  Ð "#
ÑÐ "!
Ñ  Ð "#
ÑÐ "!
Ñ  Ð "#
ÑÐ "!
Ñ  Ð "#
ÑÐ "!
Ñ œ #!
Þ
#
Note that :Ð#*Ñ œ "#
since two of the twelve data points are 29.
25. Suppose that 7 of the sample values are limit values.
7
The likelihood function for the first estimation is Ð ") Ñ87 Ð )"!!
) Ñ œ
The log of the likelihood is jÐ)Ñ œ 7 68Ð)  "!!Ñ  8 68 ) .
Ð)"!!Ñ7
)8
.
For the first sample we have
`
`)
)
jÐ)Ñ œ )"!!
 8) œ ! , so that the mle of ) is s) œ "!!8
8) .
For the second sample we have
`
`)
\$
jÐ)Ñ œ )"&!
 8) œ ! , so that the mle of ) is "&!8
8\$ .
"&!8
% "!!8
We are given that 8\$ œ \$ † 8) .
Solving for 8 results in 8 œ %) .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-18
PRACTICE EXAM 1
26. The likelihood function is the product of probabilities for each interval.
The probability for the interval %  is 7 ²?  ³ ~ - ²³ ~ .
The probability for the interval  %  is - ²³ c - ²³ ~ c ~ .
c
The probability for the interval %  is c - ²³ ~ c ~
.
The likelihood function is
²c³
c 3 ~ ´- ²³µ h ´- ²³ c - ²³µ h ´ c - ²³µ ~ ² ³ ² ³ ² ³ ~ h b b ,
where ~ h .
The log of 3 is 3 ~ b ² c ³ c ² b b ³ , and the derivative is
b b
. Setting this equal to 0 and solving for results in the
3 ~ c c
mle ~
² b b ³
b
~ b b . Answer: C
c%°
27. The pdf of the gamma with ~ is ²%³ ~ % .
Then ²%³ ~ % c % c , and ²%³ ~ % c ,
and ²%³ ~ c %
b .
The asymptotic variance of the mle of is 0²³ , where 0²³ ~ c ,´ ²³µ ,
and ²³ ~ ' ²% ³ is the loglikelihood function..
Therefore, ²³ ~ ' ²% ³ ~ '´ c %
b µ ~ c h '% b where is the number of data points. Then,
,´ ²³µ ~ ,´ c h '% b µ ~ c h ,´'% µ b À
Since each % has a gamma distribution with ~ , we have ,´%µ ~ , so that
,´ ²³µ ~ c h b ~ c b ~ c .
The asymptotic variance is 0²³ ~ .
In this problem we have ~ . Also, for the gamma distribution with known , the mle of c
V
is V ~ % . In this case we have V ~ . The estimated asymptotic variance of is
V
~
²°³
²
³
~ Á .
c
c
An alternative solution would be to consider the mle V ~ % ~ % , and then
c
c . For any random variable and random sample, we have
= ´Vµ ~ = ´ % µ ~ h = ´%µ
c ~ = ´?µ . In the case of a gamma distribution with ~ , the variance is
= ´%µ
c ~ = ´?µ ~ , and then
= ´?µ ~ ~ , so that = ´%µ
c
V
= ´µ ~ h = ´%µ ~ h
~
.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 1
PE-19
28. Based on the total data set, the probabilities of 0 , 1 and 2 claims are .97 , .02 , and .01.
The expected numbers for each territory and claim amount are
Number
of Claims
Territory 1
Territory 2
Territory 3
Territory 4
Total
0
97
194
388
291
970
1
2
4
8
6
20
2
1
2
4
3
10
²6 c, ³
The Chi-Square statistic is 8 ~ ' , , over all cells, where , is the expected number of
observations for a cell and 6 is the observed number. There are 12 cells, and summing columnby-column, we get
²c³
²c³
²c³
²c³
²c
³
²c³
8~
b b b
b Ä b b ~ À .
The number of degrees of freedom is 6. There are 12 categories (cells), and we lose a degree of
freedom for each cell in the bottom row, because the ~ values are known once the ~ and
~ values are known for each Territory (for instance, with 100 observations for Territory 1,
with 97 at ~ and 2 at ~ , there must be at ~ ).
We also lose a degree of freedom for Territory 4 at ~ and ~ , since once we know the
number of ~ observations for Territories 1,2, and 3 out of 970 observations of ~ , the rest
must be for Territory 4. The 5% significance critical value for the Chi-Square distribution with 6
degrees of freedom is 12.59. OÀ c ÀO ~ À
.
29. For the 60 exposure case, the credibility premium is
À ~ A
b 4 ² c A
³ , and for the 80 exposure case,
À ~ A b 4 ² c A ³ . We wish to find ÀA b 4 ² c A ³ .
In going from 60 to 80 exposures, the credibility factor changes from
A
~
m
= ´@ µ
h6b ²,´@ µ³ 7
to A ~
m
= ´@ µ
h6b ²,´@ µ³ 7
(where @ is the severity distribution).
A
Thus, A ~ k ~ À , and the two credibility premium equations become
À ~ A
b 4 ² c A
³ , À ~ À
A6 b 4 ² c ÀA
³ .
ÀcA
cA
Juggling these equations results in ÀcÀ
A
~ cÀA
Á
which results in the quadratic equation
ÀA
c ÀA
b À ~ S A
~ À Á À
.
Using A
~ À Á and substituting into the equations above, we get 4 ~ ,
and using A
~ À
, we get 4 ~ À
.
With A
~ À , we get A ~ A
h k ~ À
, and the new credibility premium is
²À
³²À³ b ² c À
³²³ ~ À À
With A
~ À
, we get A ~ A
h k ~ À
, and the new credibility premium is
²À
³²À³ b ² c À
³²À
³ ~ À À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-20
PRACTICE EXAM 1
30. \l) has an exponential distribution, and @ is uniformly distributed on Ò"!ß #!Ó.
.Ð)Ñ œ IÒ\l)Ó œ ) ß @Ð)Ñ œ Z +<Ò\l)Ó œ )# .
Ð#!"!Ñ#
. œ IÒ.Ð)ÑÓ œ IÒ)Ó œ "!#!
œ
"&
,
+
œ
Z
+<Ò
.
Ð
)
ÑÓ
œ
Z
+<Ò
)
Ó
œ
œ )Þ\$\$\$ ,
#
"#
#! #
#
@ œ IÒ@Ð)ÑÓ œ IÒ) Ó œ '"! ) ÐÞ"Ñ . ) œ #\$\$Þ\$\$ .
)
For 8 œ ) data points, the Buhlmann credibility factor is ^ œ )) @ œ ) #\$\$Þ\$\$
œ Þ### .
+
)Þ\$\$\$

For the 8 œ ) data points \$ ß % ß ) ß "! ß "# ß ") ß ## ß \$& , with \ œ "% , the Buhlmann
31. ) œ Mß MMß MM with 1ÐMÑ œ #\$ , 1ÐMMÑ œ "' , and 1ÐMMMÑ œ "' .
The hypothetical means are IÒ\l) œ MÓ œ "# ß IÒ\l) œ MMÓ œ "% and IÒ\l) œ MMMÓ œ \$% .
\$
The process variances are Z +<Ò\l) œ MÓ œ "% ß Z +<Ò\l) œ MMÓ œ "'
\$
and Z +<Ò\l) œ MMMÓ œ "'
The expected hypothetical mean is "# ‚ #\$  "% ‚ "'  \$% ‚ "' œ Þ& œ . .
The variance of the hypothetical means is
"
*
"
Ð "% ‚ #\$  "'
‚ "'  "'
‚ "' Ñ  Ð "# Ñ# œ %)
œ+.
\$
\$
""
The expected process variance is #\$ ‚ %"  '" ‚ "'
 '" ‚ "'
œ %)
œ@.
There are 8 œ % observations, so the Buhlmann credibility factor is

8
%
\$
^ œ 8
@ œ
""Î%) œ Þ\$'\$' . From the given information we have \ œ % ,
% "Î%)
+
so the Buhlmann estimate is ÐÞ\$'\$'ÑÐ \$% Ñ  ÐÞ'\$'%ÑÐÞ&Ñ œ Þ&* .



32. \ " œ &)""
œ ) , \ # œ #)"%
œ ) , \ œ ))
s,
\$
\$
# œ)œ.
"
[email protected] œ Ð#ÑÐ#Ñ ŠÒÐ&  )Ñ#  Ð)  )Ñ#  Ð""  )Ñ# Ó  ÒÐ#  )Ñ#  Ð)  )Ñ#  Ð"%  )Ñ#Ó‹
œ ##Þ& , and
<


"
##Þ&
+ œ <"
Ð\ 3  \ Ñ#  [email protected] œ "" ÒÐ)  )Ñ#  Ð)  )Ñ# Ó  ##Þ&
s
\$ œ  \$ !
3œ"
s " . The credibility premium is
Since s
+  !, the we assign a value of 0 to ^

s
s
^ " \ "  Ð"  ^ " Ñ.
sœ.
sœ).
33. \3 œ š
"
!
.
if tail
number of heads in 5 flips

.
\5 œ
5
\ Ð"\ Ñ
The stopping criterion is É 5 5 5  Þ!(& .



É \ %! Ð"\ %! Ñ œ Þ!('& Þ
After the first 40 flips, we have \ %! œ #&
œ
Þ'#&
ß
so
that
%!
%!


\ %" Ð"\ %" Ñ

#&
É
5 œ %" p \ %" œ %" œ Þ'"! p
œ Þ!('# ß
%"



É \ %# Ð"\ %# Ñ œ Þ!(%* Þ
5 œ %# p \ %# œ #'
%# œ Þ'"* p
%#


SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 1
c
PE-21
²³
À
34. ,´5 ²³ µ ~ c d ,´5 µ S ~ c
d À S ~ À .
,´5 ²³ µ ~ c d ,´5 µ ~ À d ~ À
35. The mixture components are exponential random variables with means of and , and the
mixing weights are for each component. The distribution mean is d b d ~ .
The sample mean of the data set is 12. Applying the method of moments, we have ~ ,
so that the estimate of is 8. The cdf of the estimated distribution is
c%°
c%°
- ²%³ ~ d ² c c%° ³ b d ² c c%° ³ ~ c c .
c°
c°
The median satisfies the equation - ²³ ~ ~ c c .
c°
c°
This can be written in the form b c ~ .
Letting " ~ c° , this is the quadratic equation " b " c ~ .
cfj
. We ignore the negative
j
cb
c°
~
, so that ~ À .
The roots are " ~
Then, " ~
root, because " ~ c°  .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-22
PRACTICE EXAM 1
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 2
PE-23
ACTEX EXAM C/4 - PRACTICE EXAM 2
1. A loss distribution is a two-component spliced model using a Weibull distribution with
~ Á and ~ for losses up to \$4000, and a Pareto distribution with ~ Á and
~ for losses \$4000 and greater. The probability that losses are less than \$4000 is 0.60.
Calculate the probability that losses are less than \$25,000.
A) Less than 0.900
B) At least 0.900, but less than 0.925
C) At least 0.925, but less than 0.950
D) At least 0.950, but less than 0.975
E) At least 0.975
2. Low Risk Insurance Company provides liability coverage to a population of 1,000 private
passenger automobile drivers. The number of claims during a given year from this population is
Poisson distributed. If a driver is selected at random from this population, his expected number of
claims per year is a random variable with a Gamma distribution such that ~ and ~ .
Calculate the probability that a driver selected at random will not have a claim during the year.
A) 11.1%
B) 13.5%
C) 25.0%
D) 33.3%
E) 50.0%
3. Annual loss follows a uniform distribution between 0 and 1000. An insurer provides a policy
with a maximum covered loss of 750 and a policy deductible of 100 (with deductible applied
after maximum covered loss). The insurer collects a premium which is 25% above the expected
cost for the annual loss. If the insurer's cost for the annual loss is less than the premium, the
insurer's risk manager receives a bonus of 10% of the amount by which the insurer's cost is below
the premium collected. Find the expected bonus to be received by the risk manager.
A) Less than 5
B) At least 5 but less than 10
C) At least 10 but less than 15
D) At least 15 but less than 20
E) At least 20
4. An actuary determines that claim counts follow a negative binomial distribution with
unknown and . It is also determined that individual claim amounts are independent and
identically distributed with mean 700 and variance 1,300. Aggregate losses have a mean of
48,000 and variance 80 million. Calculate the values for and .
A) ~ ÀÁ ~ À
B) ~ ÀÁ ~ À
C) ~ ÀÁ ~ À
D) ~ Á ÀÁ ~ À
B) ~ Á ÀÁ ~ À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-24
PRACTICE EXAM 2
5. 40 observed losses have been recorded in thousands of dollars and are grouped as follows:
Interval
Number of
Total Losses
Losses
(\$000)
(\$000)
² Á µ
² Á µ
²Á µ
²Á B³
Determine the empirical limited expected value with limit 2000 (answer in \$000).
A) Less than .5
B) At least .5, but less than 1.0
C) At least 1.0, but less than 1.5
D) At least 1.5, but less than 2.0
E) At least 2.0
6. A mortality study of a large number of heart attack survivors shows that during the first year
after the heart attack occurs 20% of patients withdraw from the study. Most of those who
withdraw from the study do so because they believe their heart attack was minor and don't feel
that they should be included. A standard Product-Limit estimate, V , is found of the mortality
probability for death within one year after heart attack (assuming the withdrawers are rightcensored). A diligent researcher tracks down all 20% who withdrew and calculates a new
Product-Limit estimate of the mortality probability for death within one year after heart attack,
and finds that this new estimate is 95% of the original estimate. The diligent researcher also
calculates a Product-Limit estimate of the mortality probability for death within one year after
heart attack for the 20% who withdrew, V Z . Find the ratio V Z ° V .
A) .50
B) .625
C) .75
D) .875
E) 1.00
7. The number of claims follows a negative binomial distribution with parameters and ,
where is unknown and is known. You wish to estimate based on observations,
where c
% is the mean of these observations. Determine the maximum likelihood estimate of .
c
c
c
A) %
B) %
C) c
%
D) %
E) c
%
8. The following observations are assumed to come from the continuous distribution with pdf
²%Â ³ ~ % c% : Á Á Á Á Á . Find the mle of .
A) .25
B) .50
C) .75
D) 1.00
E) .125
9. The following grouped data set for times of death of 100 individuals is given:
Interval:
²Á µ
²Á µ
²Á µ
²Á µ
Number:
12
31
44
13
The probability of surviving to time 2.5 is : ²À³, and is estimated using linear interpolation
between survival to time 2 and survival to time 3. Find the estimated variance of : ²À³ .
A) .0006
B) .0008
C) .0010
D) .0012
E) .0014
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 2
PE-25
10. You are given the following random sample of 12 data points from a population distribution
?
7 , 12 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53
The estimated distribution is exponential with estimated parameter V ~ .
In constructing a - plot of the data, find the maximum absolute deviation of the plot from the
line & ~ % on the interval ´Á µ (find %O- i ²% ³ c O).
A) .185
B) .195
C) .205
D) .215
E) .225
11. ? has an exponential distribution with mean and has a uniform distribution on ²Á ³.
@ has a uniform distribution on ²Á ³ and has an exponential distribution with mean 100.
= ´?µ
Find = ´@ µ .
A) 0
B) C) 1
D) 2
E) B
12. The parameter has a prior distribution with pdf ²³ ~ for   .
The conditional distribution of ? given is uniform on the interval ²Á ³ .
Find the Bayesian premium ,´? O? ~ %µ .
A) b % B) b%
C) b%
D) b%
E) b%
13. Survival times are available for four insureds, two from Class A and two from Class B. The
two from Class A died at times ! ~ 1 and ! ~ 9. The two from Class B died at times ! ~ 2 and
! ~ 4. Nonparametric Empirical Bayes estimation is used to estimate the mean survival time for
each class. Unbiased estimators of the expected value of the process variance and the variance of
the hypothetical means are used. Estimate A , the Buhlmann credibility factor.
A) 0
B) 2/19
C) 4/21
D) 8/25
E) 1
14. You are to use the inverse transform method to generate two random observations from the
%
c%
distribution with probability density function ²%³ ~ F .
elsewhere
You are to use the following random numbers from the uniform distribution on ´Á µ:
À Á À
Calculate the sum of the resulting random observations.
A) c À
B) c À
C) c À
D) c À
E) c À
15. ? is a mixture of a uniform distribution on ´Á µ and a uniform distribution on ´Á µ,
with mixing weights of .5 for each mixture component. Find the 90% Conditional Tail
Expectation for ? .
A) 140
B) 150
C) 160
D) 170
E) 180
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-26
PRACTICE EXAM 2
16. Annual loss follows a distribution with density function
c%°
²%³ ~ , %B .
A premium of 1200 is charged to cover losses. If losses are below the premium, the risk manager
receives a bonus of 25% of the amount by which losses are below the premium. Find the
expected bonus that the risk manager will receive.
A) 100
B) 125
C) 150
D) 175
E) 200
17. An automobile insurer classifies drivers as good or bad. The number of claims per year for a
good driver has a Poisson distribution with a mean of .2 and the number of claims per year for a
driver randomly chosen from the insurer's population of bad drivers has
a Poisson distribution with mean \$, where \$ is uniformly distributed between 1 and 2.
The insurer's portfolio consists of 75% good drivers and 25% bad drivers. A randomly selected
driver from the insurer's portfolio is found to have 0 claims during the past year. Find the
probability that this driver will have 0 claims this year.
A) .76
B) .77
C) .78
D) .79
E) .80
18. For a group of lives aged 30, containing an equal number of smokers and non-smokers, you
are given the following information about ; , the time until death.
(i) For non-smokers, ; has a constant hazard rate of .08 .
(ii) For smokers, ; has a constant hazard rate of .16 .
Find the hazard rate ; ²³ for a randomly selected survivor from this group at age 40,
where ; is the time until death random variable for the randomly chosen individual at age 40.
A) Less than 0.08
B) At least 0.08, but less than 0.10
C) At least 0.10, but less than 0.12
D) At least 0.12, but less than 0.14
E) At least 0.14
19. A loss ? is partially insured. The insurance policy has an ordinary deductible of 100.
The insurance pays of the loss in excess of 100 up to a loss (not payment) amount of 1000.
For a loss ? above 1000, the insurance pays ? c .
You are given the following limited expected values related to the loss variable ? :
,²?³ ~ Á ,²? w ³ ~ Á ,²? w ³ ~ Á
,²? w ³ ~ Á ,²? w ³ ~ Á ,²? w ³ ~ .
Find the expected amount paid by the insurance when a loss occurs.
A) Less than 1500
B) At least 1500 but less than 1520
C) At least 1520 but less than 1540
D) At least 1540 but less than 1560
E) At least 1560
20. The time elapsed between claims processed is modeled such that = represent the time
elapsed between processing the c ! and the ! claim. (= ~ time until the first claim is
processed).
You are given:
(i) = Á = Á ÀÀÀ are mutually independent.
(ii) The pdf of each = is ²!³ ~ ÀcÀ! , !  , where ! is measured in minutes.
Calculate the probability of at least two claims being processed in a ten minute period.
A) 0.2
B) 0.3
C) 0.4
D) 0.5
E) 0.6
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 2
PE-27
21. A compound distribution : has frequency distribution 5 which is Poisson with a mean of .
The severity distribution ? is a discrete random variable, with
7 ´? ~ µ ~ for ~ Á Á Á ÀÀÀ
Find 7 ´:  µ
A) Less than .24
B) At least .24 but less than .26
C) At least .26 but less than .28
D) At least .28 but less than .3
E) At least .30
22. Calculate the empirical estimate of the skewness from the following random sample.
A) .30
B) .35
C) .40
D) .45
E) .50
23. For a complete study of 10 lives, we are given that the times of death are
3 , 4 , 4 , 6 , 7 , 8 , 8 , 9 , 11 , 12 .
The underlying survival distribution is of the form :²!³ ~ !b
.
Two attempts are made to estimate the parameter :
(i) the method of moments , and
(ii) the method of percentile matching based on the 50th percentile.
Which of the following statements are true?
I. The method of moments results in an estimate of of .120 .
II. The method of percentile matching based on the median results in an estimate of of .
A) Both are false
B) I is false, II is true
C) I is true, II is false
D) Both are true
E) The correct answer is not given by A, B, C, or D
24. A distribution has density function ²%³ ~ 2%
2 for 0  %  , where  .
You are given the following information about a four-point data set:
% ~ Á % ~ Á % ~ Á %  What is the maximum likelihood estimator of ?
A) j
B) j
C) j
D) j
E) 25. You are given the following:
- The random variable ? has the density function ²%³ ~ c%° Á  %  B Á  .
~
- is estimated by an estimator based on a large random sample of size .
- is the proportion of the observations in the sample that are greater than 1.
~
- The probability that ? is greater than 1 is estimated by the estimator c° .
Using the delta method, determine the approximate variance of the estimator for the probability
~ c
that ? is greater than 1 if is ? .
A) B) c°
C) c°
D) c°
E) c° ² c c° ³
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-28
PRACTICE EXAM 2
26. You are given the following random sample of 6 observations:
8.0 , 1.8 , 6.9 , .5 , 4.1 , 2.2
The data set is drawn from an exponential distribution.
A likelihood ratio test is applied to test the null hypothesis that the mean of the distribution is )! .
For how many of the following values of )! does the test reject the null hypothesis at the 10%
level of significance?
)! : 2 , 4 , 6 , 8
A) None
B) 1 only
C) 2 only
D) 3 only
E) All 4
27. You are given the following random sample of 12 data points from a population distribution
\
7 , 12 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53
The estimated distribution is exponential with estimated parameter s) œ \$!.
Find the Anderson-Darling statistic for this data (based on the estimated value of s) œ \$!).
A) Less than .2
B) At least .2, but less than .4
C) At least .4, but less than .6
D) At least .6, but less than .8
E) Greater than .8
28. Total claim amount per period W follows a compound Poisson claims distribution. The
standard for full credibility for total claims in a period W based on number of claims is 1500
claims. It is then discovered that an incorrect value of the coefficient of variation for the severity
distribution ] was used to determine the full credibility standard. The original coefficient of
variation used was Þ'#"" , but the corrected coefficient of variation for ] is Þ&#!! .
Find the corrected standard for full credibility for W based on number of claims.
A) 1300
B) 1325
C) 1350
D) 1375
E) 1400
29. A portfolio of risks is divided into three classes. The characteristics of the annual claim
distributions for the three risk classes is as follows:
Class I
Class II
Class III
Annual Claim
Poisson
Poisson
Poisson
Number Distribution mean 1
mean 2
mean 5
50% of the risks are in Class I, 30% are in Class II, and 20% are in Class III.
A risk is chosen at random from the portfolio and is observed to have 2 claims in the year. Find
the expected number of claims for the risk next year.
A) Less than 1.0
B) At least 1.0 but less than 1.3
C) At least 1.3 but less than 1.6
D) At least 1.6 but less than 1.9
E) At least 1.9
30. You are given the following:
- Claim size \ for a given risk follows a distribution with density function
1 /Ð -B ) , 0  B  ∞, -  0Þ
0 (B) œ The prior distribution of - is assumed to follow a distribution with mean 50 and density function
100
1(-) œ 500,000
/Ð - ) , 0  -  ∞
-4
Find the Buhlmann credibility factor for a single observation of \ .
A) "'
B) "\$
C) "#
D) #\$
E) &'
31. The distribution of \ in three consecutive periods has the following characteristics:
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 2
PE-29
31. The distribution of ? in three consecutive periods has the following characteristics:
,´? µ ~ Á = ´? µ ~ Á ,´? µ ~ Á = ´? µ ~ Á ,´? µ ~ Á
*#´? Á ? µ ~ Á *#´? Á ? µ ~ Á *#´? Á ? µ ~ À
Find the credibility premium for period 3 in terms of ? and ? using Buhlmann's credibility
approach.
A) b ? b ?
B) b ? b ?
C) b ? b ?
D) b ? b ?
E) b ? b ?
32. For a large sample of insureds, the observed relative frequency of claims during an
observation period is as follows:
Number
Relative Frequency
of Claims
of Claims
0
61.9%
1
28.4%
2
7.8%
3
1.6%
4
.3%
5 or more
0
Assume that for a randomly chosen insured, the underlying conditional distribution of number of
claims per period given the parameter # is Poisson with parameter #.
Given an individual who had claims in the observation period, use semiparametric empirical
Bayesian estimation to find expected number claims that the individual will have in the next
period.
A) À b À
B) À b À
C) À b À
D) À b À E) À b À
33. The random variable ? has distribution function - ²%³. You are given:
- ²³ ~ Á - Z ²%³ ~ for  %  , 7 ´? ~ µ ~ , - Z ²%³ ~ for  %  , - ²³ ~ You simulate ? by using a function of < , the uniform distribution on the interval ´Á µ, and
obtain the following sequence of values from < : Á Á Á .
c
Determine ? , the sample mean.
A) B) 1
C) 2
D) E) 3
34. Losses have an Inverse Exponential distribution. The mode is 10,000.
Calculate the median.
A) Less than 10,000
B) At least 10,000, but less than 15,000
C) At least 15,000, but less than 20,000
D) At least 20,000, but less than 25,000
E) At least 25,000
35. Suppose that the ground up loss has a uniform distribution on the interval ²Á ³ and the
following data set of insurance payments is available
7 , 12 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 30 , 30 , 30
Assume the data set represents insurance payments after a maximum covered loss of " ~ and
an ordinary deductible of ~ is applied (so the policy limit is " c ~ c ~ ).
Assume that the data is conditional given that ?  . Find the mle of .
A) 40
B) 45
C) 50
D) 55
E) 60
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-30
PRACTICE EXAM 2
ACTEX EXAM C/4 - PRACTICE EXAM 2 SOLUTIONS
c c²%°³
c%°
1. The Weibull density is ²%³ ~ % ~ (actually, exponential with mean
²Á³
1500). The Pareto density is ²%³ ~ ²%b³b ~ ²%bÁ³
²%³ %  The pdf of the spliced distribution is ²%³ ~ D À
²%³ %  B
In order to be a valid pdf, we must have ²%³ % ~ .
B
This implies that ²%³ % b ²%³ % ~ . This equation becomes
Á
- ²³ b ´ c - ²³µ ~ ´ c c° µ b ² ÁbÁ ³ ~ , which is
À b À
~ .
We are also told that the probability that losses are less than \$4000 is 0.6.
Therefore, ²%³ % ~ À
, from which we get À ~ À
, and ~ À
.
Then À
~ À, so that ~ À .
We wish to find 7 ²?  Á ³ ~ 7 ²?  ³ b 7 ²  ?  Á ³
~ À
b 7 ²  ?  Á ³ À
Á
7 ²  ?  Á ³ ~ ²%³ % ~ À´- ²Á ³ c - ²³µ
Á
Á
~ À´² ÁbÁ ³ c ² ÁbÁ ³ µ ~ À .
Finally, 7 ²?  Á ³ ~ À
b À ~ À .
2. Given , the distribution of the annual number of claims for a driver is Poisson with mean .
Therefore, the conditional distribution of 5 given is Poisson with mean .
has a Gamma distribution with parameters and . This is an example of a continuous mixture
in which we are know the conditional distribution of 5 given , and we know the distribution of
also ( is the "mixing" random variable, and it has a continuous distribution in this case). For
the continuous mixture in which 5 given is Poisson with mean and has a gamma
distribution with parameters and , the unconditional (or marginal) distribution of 5 for a
randomly chosen individual is negative binomial with ~ and ~ . The probability
7 ´5 ~ µ for the negative binomial is ²b ³b . In this case, ~ ~ and ~ ~ , so
that 7 ´5 ~ µ ~
²b³b
3. The ground up loss random variable is uniform between 0 and 1000. The cdf of the ground up
loss random variable is - ²%³ ~ À% for  %  . With maximum covered loss 750 and
policy deductible 100, the expected cost for the annual loss is ´ c À%µ % ~ À .
The premium charged is ²À³²À³ ~ À .
²À³²
À³
@3 ~ The bonus received is ) ~ H ²À³²
À c @3 ³  @3  À ,
@3  À
?  where @3 is the cost per loss random variable, @3 ~ ) ~ H ? c  ?  .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 2
PE-31
3. continued
²À³²
À³
?  The bonus can be formulated as ) ~ H ²À³²
À c ²? c ³³  ?  À .
?  À
À
The expected bonus is ,´)µ ~ ²
À³ h 7 ´?  µ b ²
À c À%³ ²%³ %
À
~ ²
À³²À³ b ²
À c À%³²À³ % ~ À .
4. The aggregate loss : has a compound distribution with negative binomial frequency 5 and
severity @ . ,´5 µ ~ , = ´5 µ ~ ² b ³ , ,´@ µ ~ , = ´@ µ ~ Á À
We are also given that ,´:µ ~ Á and = ´:µ ~ Á Á .
For a compound distribution, we have ,´:µ ~ ,´5 µ h ,´@ µ and
= ´@ µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ . These equations become
Á ~ and Á Á ~ b Á ² b ³ À
The second equation then becomes
Á
Á
Á Á ~ ² ³ b Á ² ³² b ³ Á
from which we get ~ À and ~ À À
5. In general, with grouped data, and limit " equal to the endpoint of one of the data groupings,
the empirical limited expected value is
V w "µ ~ ²total amount of claims"³b²number of claims "³h" À
,´?
total number of claims
V w µ ~ bb²³²³ ~ À .
,´?
6. Suppose there are heart attack patients. There are À right-censored observations. Suppose
that there are deaths observed. Then V ~ cÀ ~ À .
Suppose that there are Z deaths from the À withdrawers.
b Z
b Z
Then the new estimate is . We are given that
~ ²À³² À ³ from which it follows
Z
that ~ À . The Product-Limit estimate based on the 20% withdrawers is
Z
V Z ~ À ~
À
À
À ~ ² À ³²À³² À ³ ~ À V .
7. The mle equations for the negative binomial distribution are
%
V ~c
V ³ ~ ² b c
% and ² b ³ . The second of these equations would usually require a
V
V
numerical solution for V. Since we are assuming that is known, we can use the first equation to
%
V~ c
as the mle of .
8. The likelihood function is ²% Â ³ ~ ² h c ³² h c ³Ä² h c ³
~
c
²Á Á ³ S 3²³ ~
S M²³ ~ 3²³ ~ c b ²Á Á ³ b c V
S MZ ²³ ~ Answer: C
c ~ S ~ ~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-32
PRACTICE EXAM 2
"
"
"
&(
"\$
9. W"!! Ð#Þ&Ñ œ # W"!! Ð#Ñ  # W"!! Ð\$Ñ œ Ð # ÑÐ "!!  "!! Ñ œ Þ\$& .
The estimated variance is found from
Z +<ÒW 8 ÐBÑÓ œ
Ð-4 -4" Ñ# Z +<Ò] ÓÐB-4" Ñ# Z +<Ò^Ó#Ð-4 -4" ÑÐB-4" Ñ[email protected]Ò] ß^Ó
,
Ò8Ð-4 -4" ÑÓ#
where Z +<Ò] Ó œ 8WÐ-4" ÑÒ"  WÐ-4" ÑÓ ,
Z +<Ò^Ó œ 8 † ÒWÐ-4" Ñ  WÐ-4 ÑÓ † Ò"  WÐ-4" Ñ  WÐ-4 ÑÓ , and
[email protected]Ò] ß ^Ó œ  8 † Ò"  WÐ-4" ÑÓ † ÒWÐ-4" Ñ  WÐ-4 ÑÓ .
&(
&(
Therefore, Z s
+<Ò] Ó œ "!!Ð "!! ÑÐ"  "!! Ñ œ #%Þ&" ß
&(
"\$
&(
"\$
Zs
+<Ò^Ó œ "!!Ð "!!  "!! ÑÐ"  "!!  "!! Ñ œ #%Þ'% , and
&(
&(
"\$
[email protected]Ò]
ß ^Ó œ  "!!Ð"  "!! ÑÐ "!!  "!! Ñ œ  ")Þ*# , and
s
Zs
+<ÒW"!! Ð#Þ&ÑÓ œ
œ Þ!!""(& .
Ð\$#Ñ# Ð#%Þ&"ÑÐ#Þ&#Ñ# Ð#%Þ'%Ñ#Ð\$#ÑÐ#Þ&#ÑÐ")Þ*#Ñ
Ò"!!Ð\$#ÑÓ#
10. The model cdf is J ‡ ÐBÑ œ "  /BÎ) . With ) œ \$!, we have the following table
4
B4
J ‡ ÐB4 Ñ
4Î"\$
lJ ‡ ÐB4 Ñ  4Î"\$l
"
(
Þ#!)"
Þ!('*
Þ"\$"#
#
"#
Þ\$#*(
Þ"&\$)
Þ"(&*
\$
"&
Þ\$*\$&
Þ#\$!)
Þ"'#(
%
"*
Þ%'*#
Þ\$!((
Þ"'"&
&
#'
Þ&(*'
Þ\$)%'
Þ"*&!
'
#(
Þ&*\$%
Þ%'"&
Þ"\$"*
(
#*
Þ'"*(
Þ&\$)&
Þ!)"#
)
#*
Þ'"*(
Þ'"&%
Þ!!%\$
*
\$!
Þ'\$#"
Þ'*#\$
Þ!'!#
"!
\$\$
Þ''("
Þ('*#
Þ"!#"
""
\$)
Þ(")#
Þ)%'#
Þ"#)!
"#
&\$
Þ)#*"
Þ*#\$"
Þ!*%!
The maximum absolute difference is ."*&.
11. IÒ\Ó œ IÒ IÒ\l)Ó Ó œ IÒ)Ó œ &! and
"!!
#!ß!!!
"
IÒ\ # Ó œ IÒ IÒ\ # l)Ó Ó œ IÒ#)# Ó œ '! #)# † Ð "!!
Ñ .) œ \$ Þ
Z +<Ò\Ó œ
#!ß!!!
\$
 &!# œ
"#ß&!!
\$
Þ
IÒYÓ œ IÒ IÒYlαÓ Ó œ IÒ α# Ó œ "# † IÒαÓ œ &! and
#
∞ #
#!ß!!!
" αÎ"!!
IÒ] # Ó œ IÒ IÒ] # lαÓ Ó œ IÒ α\$ Ó œ '! α\$ † "!!
/
.) œ \$ Þ
#!ß!!!
"#ß&!!
Z +<Ò] Ó œ \$  &!# œ \$ Þ
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 2
PE-33
12. The model density is ²%O³ ~ for  %  .
The joint density is ²%Á ³ ~ ²%O³ h ²³ ~ h ~ on the (triangular) region
 %   .
The marginal density of ? is ? ²%³ ~ % ²%Á ³ ~ % ~ ² c %³ for  %  .
²%Á³
The posterior density is ²O%³ ~ ²%³ ~ ²c%³
~ c%
for %   ?
(this is uniform on the interval ²%Á ³ ).
c%
,´? O? ~ %µ ~ % ,´? Oµ h ²O%³ ~ % h c%
~ ²²c%³³
~ b%
.
13. For empirical Bayes estimation in the equal sample size case, the estimated credibility factor
V ~ V# À In this formulation, ~ is the number of observations for each
for each is A
b V
c
group (or policyholder, or sample), and V# ~ ²c³ ²? c ? ³ ~ V# ,
~ ~
~
c
where V# ~ c ²? c ? ³ ) , and is the number of policyholders.
~
c
c
V#
Also, V
~ c ²? c ? ³ c .
~
In this example ~ , ~ , ? ~ Á ? ~ (Class A death times) ,
c
c
c
? ~ Á ? ~ (Class B death times) , ? ~ Á ? ~ Á ? ~ À
V# ~ ²³ h ´² c ³ b ² c ³ b ² c ³ b ² c ³ µ ~ Á
~ h ´² c ³ b ² c ³ µ c ~ c À  .
V
V is set equal to 0.
When V
 , the credibility factor A
%
14. -? ²%³ ~ c ²!³ ! ~ % b c  %  . We solve " ~ -? ²%³ for %.
% b
% b
À ~ S % ~ À , À ~ S % ~ c À . % b % ~ c À . Answer: A
15. The 90-th percentile of ? is 8À where
- ²8À ³ ~ ²À³´- ²8À ³ b - ²8À ³µ
Since - ²³ ~ ²À³´- ²³ b - ²³µ ~ ²À³´ b Àµ ~ À , it follows that 8À  .
8À
Therefore, - ²8À ³ ~ ²À³´- ²8À ³ b - ²8À ³µ ~ ²À³´1 b µ ~ À .
Solving for 8À results in 8À ~ .
*; ,À ~ ,´?O?  8À µ ~ ,´?O?  µ .
²%³ ~ ²À³²À b À³ ~ À for  %  and
²%³ ~ ²À³²À³ ~ À for  %  .
7 ²?  ³ ~ ²À³²³ ~ À .
²%³
²%O?  ³ ~ 7 ²?
³ ~ À
À ~ À for  %  .
,´?O?  µ ~ % h ²À³ % ~ .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-34
PRACTICE EXAM 2
²À³² c ?³ ?  16. The bonus received is F
~ ²À³´ c ²? w ³µ .
?  The expected bonus is ,´²À³´ c ²? w ³µµ ~ ²À³² c ,´? w µ³ .
,´? w µ ~ ´ c - ²%³µ % ~ cÀ% % ~ ´ c cÀ µ ~ À
The expected bonus is ²À³´ c µ ~ À
17. We are to find 7 ´5 ~ O5 ~ µ ~
7 ´²5 ~³q²5 ~³µ
7 ²5 ~³
.
7 ´5 ~ µ and 7 ´²5 ~ ³ q ²5 ~ ³µ are found by conditioning over driver type.
7 ´5 ~ µ ~ 7 ´5 ~ O.µ h 7 ´.µ b 7 ´5 ~ O)µ h 7 ´)µ , where . and ) denote the event that
the randomly selected driver is good or bad, and 7 ´.µ ~ À Á 7 ´)µ ~ À were given. Since
good drivers have Poisson parameter .2, we get 7 ´5 ~ O.µ ~ cÀ .
The Poisson parameter for bad drivers is distributed uniformly from 1 to 2, so 7 ´5 ~ O)µ is
found by conditioning over the Poisson parameter:
7 ´5 ~ O)µ ~ 7 ´5 ~ O)Á µ h ²³ ~ c h ²³ ~ c c c .
Then 7 ´5 ~ µ ~ ²cÀ ³²À³ b ²c c c ³²À³ ~ À
.
In a similar way we can find 7 ´²5 ~ ³ q ²5 ~ ³µ .
7 ´²5 ~ ³ q ²5 ~ ³µ
~ 7 ´²5 ~ ³ q ²5 ~ ³O.µ h 7 ´.µ b 7 ´²5 ~ ³ q ²5 ~ ³O)µ h 7 ´)µ .
As usual in Bayesian analysis, it is implicitly assumed that for a randomly chosen driver with a
particular value of \$, the numbers of claims in separate years are independent of one another.
Then 7 ´²5 ~ ³ q ²5 ~ ³O.µ ~ 7 ´5 ~ O.µ h 7 ´5 ~ O.µ ~ ²cÀ ³²cÀ ³ ~ À
and
7 ´²5 ~ ³ q ²5 ~ ³O)µ ~ 7 ´5 ~ O)Á µ h 7 ´5 ~ O)Á µ h ²³ ~ c h ²³ ~ ²c c c ³ ~ À .
Then, 7 ´²5 ~ ³ q ²5 ~ ³µ ~ ²À
³²À³ b ²À³²À³ ~ À
Finally, 7 ´5 ~ O5 ~ µ ~
7 ´²5 ~³q²5 ~³µ
7 ´5 ~µ
~ À
À
~ À
18. With a constant hazard rate of , the random variable ; is exponential with mean .
For a non-smoker, the probability of surviving to age 40 is
7 ²;  ³ ~ c ~ c²³²À³ ~ À ,
and for a smoker, it is 7 ²;  ³ ~ c ~ c²³²À
³ ~ À .
Since the group starts out evenly split into non-smokers and smokers, 10 years later, at age 40
cÀ
the proportion of survivors that are non-smokers is cÀbcÀ
~ À
and the proportion that
cÀ
are smokers is cÀbcÀ
~ À . The future lifetime of a random chosen individual from the
survivors at age 40 is a mixture of non-smoker lifetime, ; , and smoker lifetime, ; , with mixing
weights .69 for non-smokers and .31 for smokers. The hazard rate for the randomly chosen
individual is
so that ²³ ~
²!³
À
²!³bÀ ²!³
²!³ ~ :²!³ ~ À
: ²!³bÀ: ²!³ ~
À
²À³bÀ²À
³
À
bÀ
~ À .
À
²ÀcÀ! ³bÀ²À
cÀ
! ³
À
cÀ! bÀcÀ
!
,
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 2
PE-35
19. For an insurance with ordinary deductible and maximum covered loss ,
the insurance payment is ²? w ³ c ²? w ³ .
For a deductible 100 and maximum covered loss of 1000, the insurance in question pays
´²? w ³ c ²? w ³µ for a loss up to 1000.
For a loss just a 1000, this insurance would pay ² c ³ ~ .
If we do not modify this expression, ´²? w ³ c ²? w ³µ would
pay 450 for any loss at or above 1000. In order to have an insurance payment
of ? c for a loss above 1000, we must add ? c c ~ ? c to ´²? w ³ c ²? w ³µ . Therefore, we can represent the insurance payment as
´²?
w ³ c ²? w ³µ b ²? c ³b ~ ? c ²? w ³ c ²? w ³ À
The expected amount paid by insurance when a loss occurs is
,²?³ c ,²? w ³ c ,²? w ³ ~ c ²³ c ²³ ~ .
20. Any 10-minute period can be considered. In order for there to be at least two claims in a 10
minute period, there must be a first claim, say at time !  , and then a second claim within
c ! minutes after the first claim. ( ~ event that there are 2 claims within the 10 min. period.
7 ²(³ ~ 7 ²1st claim is at time ! q 2nd claim is within c ! minutes after 1st claim) !
~ 7 ²2nd claim within c ! minutes after 1st claimO1st claim at !³ h ²!³ !
~ 7 ²;  c !³ h ; ²!³ ! ~ ´ c cÀ²c!³ µ²ÀcÀ! ³ !
We use the impendence of successive claim processing times, to get that
7 ²2nd claim within c ! minutes after 1st claimO1st claim at !³
~ 7 ²2nd claim within c ! minutes after 1st claim³ ~ - ² c !³ ~ cÀ²c!³ .
The integral is ´ÀcÀ! c Àc µ ! ~ c c ~ À .
Alternatively, because the time between successive claims processed is exponential with mean
À ~ , the number of claims processed follows a Poisson process with rate ~ À per minute.
The number of claims processed in a 10 minute period has a Poisson distribution with a mean of
2. The probability of at least 2 claims being processed in a 10 minute period is the complement of
0 or 1 claims processed, which is c 7 ²5 ~ ³ c 7 ²5 ~ ³ ~ c c c c ~ c c .
21. 7 ´:  µ ~ c 7 ´: ~ Á Á µ
7 ´: ~ µ ~ 7 ´5 ~ µ ~ c , 7 ´: ~ µ ~ 7 ´5 ~ µ h 7 ´? ~ µ ~ c h 7 ´: ~ µ ~ 7 ´5 ~ µ h 7 ´? ~ µ b 7 ´5 ~ µ h ²7 ´? ~ µ³
c
~ c h b h ² ³ ~ c .
c
c
Then 7 ´: ~ Á Á µ ~ and 7 ´:  µ ~ c ~ À . Answer: E
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-36
PRACTICE EXAM 2
,´²?c³ µ
22. The skewness of random variable ? is
, where ~ ,´²? c ³ µ .
c
The empirical estimate of is V ~ ? ~ .
The empirical estimate of is V ~ ´² c ³ b Ä b ² c ³ µ ~ .
The empirical estimate of ,´²? c ³ µ is ´² c ³ b Ä b ² c ³ µ ~ .
The empirical estimate of the skewness is ²³
° ~ À .
23. :²!³ ~ !b
¦ ,´; µ ~ :²!³ ! ~ B .
B
Therefore the method of moments cannot be applied since the mean of the underlying distribution
is infinite. I is false.
If ! denotes the median of the underlying distribution, then :²! ³ ~ ¦ ! b ~ ¦ ~ ! .
But the median time of death of the given data set is ~! ~ À ¦ ~ À
~ . II is true.
24. The survival function is :²%³ ~ c % , so the likelihood function is
3²³ ~ ²³ d ²³ d ²³ d :²
³ ~ d d d ² c ³ ~ d ² c ³ .
The loglikelihood is M²³ ~ c b ² c ³ so we solve for from
M²³ ~ c b ~ . This results in ~ j . Answer: D
c
25. ? has an exponential distribution. The mle is the same as the moment estimator of , which
c
is ? . The estimate of 7 ´?  µ is the estimate of c° , which is found using the mle of .
~
c
The variance of a function of the mle of
The estimate of 7 ´?  µ is of c° ~ c°? .
Z
V
V
parameter is = ´²³µ ~ ´ ²³µ h = ´µ . In this case, ²³ ~ 7 ´?  µ ~ c° S
c
V~?
Z ²³ ~ c° h . Since the mle of in an exponential distribution is ,
=
´?µ
c
Vµ ~ = ´?µ ~
= ´
~ . The variance of the estimate of 7 ´?  µ is
c°
²c° h ³ h ~ . Answer: D
26. The likelihood ratio test of / ¢ ~ versus / ¢ £ has test statistic ; ~ ² 3
3 ³ ~ ´ 3 c 3 µ ,
where 3 is the loglikelihood based on the maximum likelihood estimate of , and 3
is the loglikelihood based on the null hypothesis value.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 2
PE-37
26. continued
For the exponential distribution with one parameter, the test statistic will have a chi-squared
distribution with 1 degree of freedom. The critical value found for a 10% level of significance
found from the chi-square table is À ²³ ~ 2.706 . The null hypothesis will be rejected at the
10% level if ´ 3 c 3 µ  .
.
The maximum likelihood estimate of the mean of the exponential distribution is 3.92 (it is the
sample mean). Since ²%³ ~ c%° is the pdf, the log of the pdf is
²%³ ~ c c % , and the loglikelihood is
'%
3 ~ c c .
Using the sample values and the mle estimate of , we get
3 ~ c À c À
À ~ c À .
The null hypothesis will be rejected if ´ 3 c 3 µ  .
, which is equivalent to
´ c À c 3 µ  .
, which is equivalent to 3  c À .
For a hypothesized value of , the loglikelihood is
3 ~ c c À
.
The loglikelihood for each of the stated values of and the test results are:
3
Test Result
c À Reject /
c À Don't reject /
c À Don't reject /
c À Don't reject /
27. The data is not truncated or censored, so the Anderson-Darling statistic is
c
i
i
c- ²& ³
- ²& ³
( ~ c b > ² c - ²& ³³ h ² c- i ²& ³ ³ b - ²& ³ h ² - i ²&b³ ³?.
~
i
The model cdf is - ²%³ ~ c b
c%°
c- i ²& ³
h ² c- i ²& ³ ³
b
~
. There are 11 distinct %-values.
- i ²&
³
, ~ - ²& ³ h ² - i ²&b³ ³
&
- ²& ³
- i ²& ³
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
The Anderson-Darling statistic is
c b ´À b Ä b À b b À b Ä b Àµ ~ À À Answer: E
~ ² c - ²& ³³
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-38
PRACTICE EXAM 2
28. For the compound Poisson distribution with Poisson parameter (frequency distribution or
number of claims per period) and claim amount distribution @ (severity distribution or amount
per claim), the standard for full credibility for expected number of claims is
= ´@ µ
´ b ²,´@ µ³ µ ~ ´ b ² @ ³ µ . Thus, with @ ~ À
Á
@
@
´ b ² @ ³ µ ~ À ~ . With the coefficient of variation of @ changed to
@
@
@
~ À5200 , we have ´ b ² @ ³ µ ~ À ~ .
@
29. 7 ´\$ ~ O? ~ µ ~
7 ´\$ ~ O? ~ µ ~
7 ´\$ ~ O? ~ µ ~
7 ´?~O\$~µh7 ´\$~µ
7 ´?~µ
~
c h
[ h²À³
c h
c h
c h
h²À³b
[
[ h²À³b [ h²À³
c h
[ h²À³
c
h
c h
c h
[ h²À³b [ h²À³b [ h²À³
c h
[ h²À³
c h
c h
c h
[ h²À³b [ h²À³b [ h²À³
~ À ,
~ À , and
~ À .
Then, since ,´? O\$ ~ O ~ Á ,´? O\$ ~ O ~ and ,´? O\$ ~ O ~ , we have
,´? O? ~ µ ~ ²³²À³ b ²³²À³ b ²³²À³ ~ À . Answer: D
30. ?O has an exponential distribution, and has an inverse gamma distribution
with ~ Á ~ . ²³ ~ ,´?Oµ ~ and #²³ ~ = ´?Oµ ~ .
# ~ ,´#²³µ ~ ,´ µ ~ ²c³²
c³ ~ .
~ = ´²³µ ~ = ´µ ~ ,´ µ c ²,´µ³ ~ c ² c
³ ~ .
A ~ b # ~ b ~ .
~ -coefficients:
31. According to Buhlmann's approach, we solve the normal equations for the ~
~
~
,´? µ ~ b ,´? µ b ,´? µ
~ = ´? µ b ~ *#´? Á ? µ
*#´? Á ? µ ~ ~
~ = ´? µ .
*#´? Á ? µ ~ *#´? Á ? µ b Substituting the given values, these equation become
~ b ~ b ~
~
~ b ~
~
~
~
~ b ~
~
~ ~ . The credibility premium is
with solution ~ Á ~ Á ~ b ~ ? ~ b ? b ? , with ~ .
~
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 2
PE-39
32. The average number of claims per insured is
c
? ~ ²³²À
³ b ²³²À³ b ²³²À³ b ²³²À
³ b ²³²À³ ~ À À
This is the estimate of ,´?µ ~ . Since the conditional distribution of ? given # is Poisson
with parameter #, we have ,´?O#µ ~ = ´?O#µ ~ # . Then, since
,´,´?O#µµ ~ ,´?µ , our estimate for # ~ ,´= ´?O#µµ is also À (since
= ´?O#µ ~ ,´?O#µ ). We estimate = ´?µ, the variance of the relative claim frequency per
insured; the relative frequency of claims forms the estimated probability distribution of ? .
=V
´?µ ~ ² c À³ ²À
³ b ² c À³ ²À³
b ² c À³ ²À³ b ² c À³ ²À
³ b ² c À³ ²À³ ~ À À
Since = ´?µ ~ # b , we use the estimated variance of ? along with the estimate of # to get
an estimate of ; V
~ =V
´?µ c V# ~ À c À ~ À À
À
V
The estimate of is ~ VV# ~ À
~ À , and the estimated credibility factor for one
V
individual is A ~
~ À À The credibility premium for the next period for an individual
bÀ
V b ² c A³
V who had claims in the current period is A
V ~ À b À
.
33. ? has a mixed distribution with cdf - ²%³ ~
H
%
%b
%
%
%~
.
%
%
According to the inversion method of simulation, given a uniform random number " from ´Á µ,
the simulated value of ? is
% ~ " if  "  , % ~ if  "  , % ~ " c if  "  .
Then, " ~ S % ~ Á " ~ S % ~ Á " ~ S % ~ Á " ~ S % ~ .
The sample mean is bbbÀ
.
34. For an Inverse Exponential distribution with parameter , the mode is . We are given that
the mode is 10,000, so that ~ Á . The median, say , is the point for which
7 ´?  µ ~ - ²³ ~ À . For the Inverse Exponential, - ²%³ ~ c°% . We solve for from
cÁ
cÁ° ~ À . Taking the natural log of both sides results in
~ ²À³ , so that
~ Á .
35. The conditional density function at the uncensored ground up loss % is
²%O?  ³ ~ c ~ c
À The conditional probability at a censored loss is
c 7 ´?  O?  µ ~ c ~ c
c À The likelihood function for the given data is
²c³
c 3²³ ~ ² c
³ ² c ³ ~ ²c³ À The loglikelihood is
3²³ ~ ² c ³ c ² c ³ , and setting the derivative equal to 0 results in
V
3²³ ~ c c c ~ , from which we get ~ . Answer: C
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-40
PRACTICE EXAM 2
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 3
PE-41
ACTEX EXAM C/4 - PRACTICE EXAM 3
1. A Mars probe has two batteries. Once a battery is activated, its future lifetime is exponential
with mean 1 year. The first battery is activated when the probe lands on Mars. The second battery
is activated when the first fails. Battery lifetimes after activation are independent. The probe
transmits data until both batteries have failed. Calculate the probability the probe is transmitting
data three years after landing.
A) 0.05
B) 0.10
C) 0.15
D) 0.20
E) 0.25
2. An insurance company sells hospitalization reimbursement insurance. You are given:
• Benefit payment for a standard hospital stay follows a lognormal distribution with ~ and
~ .
• Benefit payment for a hospital stay due to an accident is twice as much as a standard benefit.
• 25% of all hospitalizations are for accidentally causes.
Calculate the probability that a benefit payment will exceed \$15,000.
A) Less than 0.12
B) At least 0.12, but less than 0.14
C) At least 0.14, but less than 0.16
D) At least 0.16, but less than 0.18
E) At least 0.18
3. Losses during the current year follow a Pareto distribution with ~ and ~ Á .
Annual inflation is 10%. Calculate the ratio of the expected proportion of claims that will exceed
\$750,000 next year to the proportion of claims that exceed \$750,000 this year.
A) Less than 1.105
B) At least 1.105, but less than 1.115
C) At least 1.115, but less than 1.125
D) At least 1.125, but less than 1.135
E) At least 1.135
4. Customers arrive at a bank according to a Poisson process at the rate of 100 per hour. 20% of
them make only a deposit, 30% make only a withdrawal and the remaining 50% are there only to
complain. Deposit amounts are distributed with mean 8000 and standard deviation 1000.
Withdrawal amounts have mean 5000 and standard deviation 2000. The number of customers and
their activities are mutually independent. Using the normal approximation, calculate the
probability that for an 8-hour day the total withdrawals of the bank will exceed the total deposits.
A) 0.27
B) 0.30
C) 0.33
D) 0.36
E) 0.39
5. An insurance company increases the per claim deductible of all automobile policies from \$300
to \$500. The mean payment and standard deviation of claim severity are shown below:
Mean Payment Standard Deviation
Deductible
\$300
1,000
256
\$500
1,500
678
The claims frequency is Poisson distributed both before and after the change of deductible. The
probability of no claims increases by 30%, and the probability of having exactly one claim
decreases by 10%. Calculate the percentage increase in the variance of the aggregate claims.
A) Less than 30%
B) At least 30%, but less than 50%
C) At least 50%, but less than 70%
D) At least 70%, but less than 90%
E) 90% or more
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-42
PRACTICE EXAM 3
6. You are given the following for a sample of five observations from a bivariate distribution:
%
&
1
4
2
2
4
3
5
6
6
4
(ii)
% ~ 3.6, & ~ 3.8
( is the covariance of the empirical distribution - as defined by these five observations. ) is
the maximum possible covariance of an empirical distribution with identical marginal distributions to - . Determine ) c (.
A) 0.9
B) 1.0
C) 1.1
D) 1.2
E) 1.3
(i)
7. When data is in interval grouped form, the usual assumption that is made is that within each
interval the data points for that interval are uniformly distributed on the interval. This means that
for interval ²Á µ , a data point within that interval would have pdf ²%³ ~ c
.
For an interval grouping with interval endpoints ~    Ä  c  ,
and with data points in interval ²c Á µ , this results in an empirical estimate of the first
b
moment being h c , where ~ (total number of data points).
~
~
²%c³
Suppose that the following pdf is assumed for data in the interval ²Á µ : ²%³ ~ ²c³ .
Find the estimate of the first moment based on the grouped data set.
b
A) h c ~
c
D) h c ~
b
B) h c ~
c
E) h c
~
b
C) h c ~
8. An inverse gamma distribution is fit to a data set using maximum likelihood estimation.
The estimates of and are V ~ À and V ~ À .
À
À
The information matrix that results from the estimation of and is >
.
À À ?
Apply the delta method to find a 95% confidence interval for the mean of the distribution.
What is the upper limit of the interval?
A) 8.0
B) 8.1
C) 8.2
D) 8.3
E) 8.4
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 3
PE-43
9. 1000 workers insured under a workers compensation policy were observed for one year. The
number of work days missed is given below:
Number of Days of Work
Number of Workers
Missed
0
818
1
153
2
25
3 or more
4
Total
1000
Total Number of Days Missed
230
The chi-square goodness-of-fit test is used to test the hypothesis that the number of work
days missed follows a geometric distribution where:
(i) The geometric parameter is estimated by the average number of work days missed.
(ii) Any interval in which the expected number is less than one is combined with the previous
interval.
Determine the results of the test (use at least one decimal accuracy in all calculations).
A) The hypothesis is not rejected at the 0.10 significance level.
B) The hypothesis is rejected at the 0.10 significance level, but is not rejected at the 0.05
significance level.
C) The hypothesis is rejected at the 0.05 significance level, but is not rejected at the 0.025
significance level.
D) The hypothesis is rejected at the 0.025 significance level, but is not rejected at the 0.01
significance level.
E) The hypothesis is rejected at the 0.01 significance level.
10. A loss distribution is being analyzed using the Bayesian credibility approach.
The parameter has a prior gamma distribution with ~ and ~ .
The model distribution ? is Poisson with a mean of .
A sample of 6 observations of ? results in a Bayesian premium of 37.48 .
A 7-th observation of X is obtained and the Bayesian premium is recalculated to be 37.84.
Find the value of the 7-th observation.
A) 40
B) 41
C) 42
D) 43
E) 44
11. You are given:
(i) An individual automobile insured has annual claim frequencies that follow a Poisson
distribution with mean .
(ii) An actuary’s prior distribution for the parameter has probability density function:
() = (0.5)5c b ²À³ c° À
(iii) In the first policy year, no claims were observed for the insured.
Determine the expected number of claims in the second policy year.
A) 0.3
B) 0.4
C) 0.5
D) 0.6
E) 0.7
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-44
PRACTICE EXAM 3
12. Type A risks have each year's losses uniformly distributed on the interval ²Á ³ .
Type B risks have each year's losses uniformly distributed on the interval ²Á ³ .
A risk is selected at random, with each type being equally likely. The first year's losses equal 3.
Find the Buhlmann credibility premium for the second year's losses in terms of 3.
A) À3 b À B) À3 b À C) À3 b À D) À3 b À E) À3 b À
13. You are given the following table of data for three policyholders over a three year period.
Policy Year S
Policyholder
¨
1
1
2
3
Number of Claims
Average Claim Size
40
200
50
220
2
Number of Claims
Average Claim Size
100
200
120
200
3
Number of Claims
Average Claim Size
50
200
60
250
120
150
Apply the nonparametric empirical Bayes credibility method to find the credibility premium per
claim in the 4-th year for Policyholder 2.
A) Less than 165
B) At least 165 but less than 175
C) At least 175 but less than 185
D) At least 185 but less than 195
E) At least 195
14. A random sample of 18 data points has a sample mean of 8 and an unbiased sample variance
of 4. ? ~ and ? ~ are added to the sample. Find the updated unbiased sample
variance based on all 20 data points.
A) 4.0
B) 4.1
C) 4.2
D) 4.3
E) 4.4
15. You are given a random sample of 3 values from a distribution - : 4 , 5 , 9
You estimate the median of ? using the estimator ²smallest ? b largest ? ³ .
Determine the bootstrap approximation to the mean square error.
A) 2.2
B) 2.4
C) 2.6
D) 2.8
E) 3.0
16. For a particular data set, the product limit estimator results in the following estimates:
(i) : ²³ ~ .94
(ii) With data truncated at 1, the estimated conditional probability of surviving to time 2 is .88 .
(iii) With data truncated at 2, the estimated conditional probability of surviving to time 3 is .75 .
Find : ²³ .
A) .58
B) .62
C) .66
D) .70
E) .74
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 3
PE-45
17. For an insurance portfolio, you are given:
(i) For each individual insured, the number of claims follows a Poisson distribution.
(ii) The mean claim count varies by insured, and the distribution of mean claim counts follows a
gamma distribution.
(iii) For a random sample of 1000 insureds, the observed claim counts are as follows:
Number of Claims, 0
1
2
3
4 5
Number of Insureds, 512 307 123 41 11 6
~ 750 2 ~ 1494
(iv) Claim sizes follow a Pareto distribution with mean 1500 and variance 6,750,000.
(v) Claim sizes and claim counts are independent.
(vi) The full credibility standard is to be within 5% of the expected aggregate loss 95% of the
time.
Determine the minimum number of insureds needed for the aggregate loss to be fully credible.
A) Less than 8300
B) At least 8300, but less than 8400
C) At least 8400, but less than 8500
D) At least 8500, but less than 8600
E) At least 8600
18. For a group of lives you are given that the time until death random variable for each member
of the group has a constant hazard rate that is drawn from the uniform distribution on
´ À Á À µ . For someone selected at random from this group, calculate the probability
that the individual dies between one and 3 years from now.
A) .026
B) .029
C) .032
D) .035
E) .038
19. Losses follow and exponential distribution with parameter . For a deductible of 100, the
expected payment per loss is 2,000. Which of the following represents the expected payment per
loss for a deductible of 500?
A) B) ² c c° ³
C) Á c°
c°
D) Á E) Á ² c c° ³² c c° ³
20. 5 has a geometric distribution with a mean of 2. Determine the mean of the zero-modified
distribution with 4 ~ .
A) 2.1
B) 2.2
C) 2.3
D) 2.4
E) 2.5
21. For a portfolio of 2,500 policies, claim frequency is 10% per year, and severity is distributed
uniformly between 0 and 1,000. Each policy is independent and has no deductible. Calculate the
reduction in expected annual aggregate payments, if a deductible of \$200 per claim is imposed on
the portfolio of policies.
A) Less than \$46,000
B) At least \$46,000, but less than \$47,000
C) At least \$47,000, but less than \$48,000
D) At least \$48,000, but less than \$49,000
E) \$49,000 or more
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-46
PRACTICE EXAM 3
22. In a given week, the number of projects that require you to work overtime has a Poisson
distribution with a mean of 2. For each project, the distribution of the number of overtime hours
in the week is the following:
%
²%³
5
0.2
10
0.3
20
0.5
The number of projects and the number of overtime hours are independent. You will get paid for
overtime hours in excess of 15 hours in the week. Calculate the expected number of overtime
hours for which you will get paid in the week.
A) 15
B) 16
C) 17
D) 18
E) 19
23. A study of the time until failure, ? , of an electronic device is based on observing 20 of the
devices. One failure and one right-censoring is observed at each of the integer time
points 1, 2, . . , 10. The probability 7 ´  ?  O ?  µ is to be estimated.
Find the absolute difference between the Kaplan-Meier product limit estimate and the
Nelson-Aalen estimate.
A) .0020
B) .0022
C) .0024
D) .0026
E) .0028
24. You are given the following random sample of 12 data points from a population
distribution ? :
7 , 12 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53
V ²³.
Using the triangle kernel with bandwidth 3, find A) B) C) D) E) 25. The following sample is taken from the distribution ²%Á ³ ~ 2 3c%°
Observation
1
2
3
4
5
6
7
%
0.49
1.00
0.47
0.91
2.47
5.03
16.09
Determine the maximum likelihood estimator of , where 7 ²?  ³ ~ À .
A) Less than 1.0
B) At least 1.0, but less than 1.2
C) At least 1.2, but less than 1.4
D) At least 1.4, but less than 1.6
E) 1.6 or more
26. Call center response times are described by the cumulative distribution function
- ²%³ ~ %b . where  %  and  c .
A random sample of response times is as follows:
0.56
0.83
0.74
0.68
0.75
Calculate the maximum likelihood estimate of .
A) Less than 1.4
B) At least 1.4, but less than 1.6
C) At least 1.6, but less than 1.8
D) At least 1.8, but less than 2.0
E) At least 2.0
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 3
PE-47
27. The mean of a distribution is being estimated, using the sample mean of a random sample as
an estimator. A sample of size 2 is drawn: % ~ Á % ~ . 4 :,²- ³ is the mean square error
of the estimator when the empirical distribution is used. Find the exact value of 4 :,²- ³.
A) .125
B) .250
C) .375
D) .500
E) .625
28. You are given the following sample of five claims: 43 , 145 , 233 , 396 , 775
The distribution with density function ²%³ ~ ²b%³
Á %  is being tested with the
Kolmogorov-Smirnov test. Find the value of the Kolmogorov-Smirnov statistic.
A) Less than .05
B) At least .05, but less than .140
C) At least .140, but less than .230
D) At least .230, but less than .320
E) At least .320
29. Total claims per period : follows a compound Poisson distribution and claim severity has
the pdf ²&³ ~ &c
, for &  . A full credibility standard based on number of exposures of :
needed has been determined so that the total cost of claims per period is within 5% of the
expected cost with a probability of 90%. If the same number of exposures for full credibility of
total cost is applied to the number of exposures needed for the frequency variable 5 , the actual
number of claims per exposure period would be within 100% of the expected number of claims
per exposure period with probability 95%. Find .
A) .054
B) .058
C) .062
D) .066
E) .070
30. An individual insured has a frequency distribution per year that follows a Poisson
distribution with mean . The prior distribution for is a mixture of two distributions.
Distribution 1 is constant with value 1, and distribution 2 is exponential with a mean of 3, and the
mixing weights are both .5. An individual is observed to have 0 claims in a year. Find the
Buhlmann credibility premium for the same individual for the following year.
A) Less than .50
B) At least .50 but less than .52
C) At least .52 but less than .54
D) At least .54 but less than .56
E) At least .56
31. For a portfolio of independent risks, the number of claims per period for a randomly chosen
risk has a Poisson distribution with a mean of #, where # has pdf ²³ ~ ² b ³ ,   .
Two risks are chosen at random and observed for one period, and it is found that Risk 1 has no
claims for the period and Risk 2 has 2 claims for the period. 7 is the Buhlmann credibility
premium for Risk 1 for the next period and 7 is the Buhlmann credibility premium for Risk 2 for
7
the next period. Find lim 7 .
¦
C) D) E) B
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-48
PRACTICE EXAM 3
32. Annual aggregate claims for a particular policy are modeled as a compound Poisson
distribution with Poisson parameter for the frequency (number of claims per year), and a severity
(individual claim size) @ that is either 1 or 2 with 7 ²@ ~ ³ ~ 7 ²@ ~ ³ ~ À . An insurer has a large
portfolio of policies, and each policy has its own value of . For a randomly chosen policy from the
portfolio, the distribution of is exponential with a mean of 1.
The claim sizes and the numbers of claims are independent of one another given .
A policy is chosen at random from the portfolio, and : denotes the aggregate claim for that policy for one
year. The policy is observed for three years and the observed aggregate losses for the 3 years are
: ~ Á : ~ and : ~ . Find the Buhlmann credibility premium for the 4-th year for this policy.
A) Less than 0.7
B) At least 0.7 but less than 0.8
C) At least 0.8 but less than 0.9
D) At least 0.9 but less than 1.0
E) At least 1.0
33. The prior distribution of is a gamma distribution with parameters ~ and ~ .
The conditional distribution of ? given is Poisson with a mean of .
The unconditional distribution of ? is to be simulated in two steps.
Step 1: simulate a value of Step 2: simulate a value of ? given the value of simulated in Step 1.
To apply step 1, we use the fact that the gamma distribution with parameters ~ and is the
sum of three independent exponential random variables each with mean , and simulate three
independent exponentials using the inverse transformation method and add the simulated values to
get the simulated gamma distribution value. To apply step 2 we use the product algorithm for the
Poisson.
The sequence of uniform ²Á ³ numbers to be used in the overall simulation are
À Á À
Á À Á À , À Á À Á À
These numbers are used in the order given, with the first three used to simulate the gamma
distribution and the remaining numbers used in step 2. Each number is used once until the
simulation is complete.
Determine the value of ? simulated.
A) 0
B) 1
C) 2
D) 3
E) 4
34. According to the Loss Models book, which of the following is/are true, based on the
existence of moments test?
I. The Loglogistic Distribution has a heavier tail than the Gamma Distribution.
II. The Paralogistic Distribution has a heavier tail than the Lognormal Distribution.
III. The Inverse Exponential has a heavier tail than the Exponential Distribution.
A) I only
B) II only
C) I and III only
D) II and III only
E) I, II and III
V ~ À
, and
35. For these data, the maximum likelihood estimate for the Poisson distribution is for the negative binomial distribution, it is V
~ À and ~ À. The Poisson has a negative
loglikelihood value of 385.9, and the negative binomial has a negative loglikelihood value of
382.4. Determine the likelihood ratio test statistic, treating the Poisson distribution as the null
hypothesis.
A) c 1
B) 1
C) 3
D) 5
E) 7
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 3
PE-49
ACTEX EXAM C/4 - PRACTICE EXAM 3 SOLUTIONS
1. There are a few ways to approach this problem. One approach is the convolution approach to
finding the distribution function of the sum of random variables ? and @ . If ? and @ are
continuous independent non-negative random variables, the
!
[email protected] ²!³ ~ ? ²%³ h [email protected] ²! c %³ % . In this case, ? and @ are both exponential with mean 1,
and ! ~ , this becomes c% h ´ c c²c%³ µ % ~ c% % c c %
~ c c c c ~ À . This is the probability that the total time until failure of both batteries
is  . The probability that total time until failure is  is À .
An alternative solution is based on the observation that if ? and @ are independent exponential
random variables both with mean 1, then ? b @ has a gamma distribution with ~ and
~ , and the pdf of ? b @ is [email protected] ²!³!c! .
B
B
Then 7 ´? b @  µ ~ !c! ! ~ c !c! c c! c ~ c ~ À .
!~
Yet another approach is to note that with exponential inter-event time with mean 1 year, the
number of failures forms a Poisson process with a rate of 1 per year. The probe will be
transmitting in 3 years if there is at most one battery failure in the 3 year period. The number of
failures in a 3-year period, 5 ²³, is Poisson with mean 3, so the probability is
7 ´5 ²³  µ ~ c b c h ~ c . Answer: D
2. The benefit payment ? is a mixture of standard benefit payment ? with weight .75 and
accident benefit payment ? with weight .25.
7 ²?  Á ³ ~ À7 ²?  Á ³ b À7 ²?  Á ³ .
? has a lognormal distribution, so
Á c 7 ²?  Á ³ ~ c )²
³ ~ c )²À³ ~ c À ~ À
.
? ~ ? , so
Á c 7 ²?  Á ³ ~ 7 ²?  Á ³ ~ 7 ²?  Á ³ ~ c )²
³
~ c )²À
³ ~ c À ~ À
.
Then 7 ²?  Á ³ ~ À²À
³ b À²À
³ ~ À .
3. Let ? denote the claim random variable for this year. ? has a Pareto distribution with ~ and ~ Á . The expected proportion of claims that exceed 750,000 is the same as
7 ´?  Á µ, which is the probability that a claim exceeds 750,000. This probability is (from
Á
the distribution table) c - ²Á ³ ~ ² ÁbÁ ³ ~ À .
Next year after inflation of 10%, the claim random variable will be @ ~ À? .
The Pareto distribution is a "scale distribution" with scale parameter . This means that if ? is
Pareto with parameters and , and if @ ~ ? (  ), then @ also has a Pareto distribution
with the same , and with Z ~ . In this case, @ ~ À? will have a Pareto distribution with
Á
~ and ~ Á , and 7 ´@  Á µ ~ ² ÁbÁ ³ ~ À
.
The ratio is À
À ~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-50
PRACTICE EXAM 3
4. Because of independence, we can separate the deposit process and the withdrawal process as
two independent processes. The rate per hour at which depositors arrive is ²À³ ~ , and the
rate per hour of withdrawers arriving is 30. The number of depositors arriving in an 8-hour day
has a Poisson distribution with a mean of ²³ ~ , and the number of withdrawers arriving
in an 8-hour day has a Poisson distribution with a mean of ²³ ~ .
The total amount deposited in a day has a compound Poisson distribution with Poisson parameter
160, and individual deposit amount (severity distribution) with mean 8000 and standard deviation
1000. The mean and variance of the total deposit in an 8-hour day is
,´:+ µ ~ ,´5+ µ h ,´?+ µ ~ ²
³²³ ~ Á Á and
= ´:+ µ ~ ,´5+ µ h = ´?+ µ b = ´5+ µ h ²,´?+ µ³ ~ ²
³² ³ b ²
³²³
~ Á Á Á .
In a similar way, we get the mean and variance of the total withdrawals in an 8-hour day:
,´:> µ ~ ,´5> µ h ,´?> µ ~ ²³²³ ~ Á Á and
= ´:> µ ~ ,´5> µ h = ´?> µ b = ´5> µ h ²,´?> µ³ ~ ²³² ³ b ²³²³
~ Á Á Á .
We wish to find 7 ´:>  :+ µ using the normal approximation. ,´:> c :+ µ ~ c Á , and
since :> and :+ are independent,
= ´:> c :+ µ ~ Á Á Á b Á Á Á ~ Á Á Á .
Then, using the normal approximation,
: c: c²cÁ³
c²cÁ³
+
7 ´:>  :+ µ ~ 7 ´:> c :+  µ ~ 7 ´ >
jÁ
ÁÁ  jÁ
ÁÁ µ
~ c )²À
³ ~ c À ~ À .
5. Let ,´5 µ ~ be the Poisson expected claim frequency before the change in deductible and let
,´5 Z µ ~ Z be the Poisson expected claim frequency after the change in deductible. We are given
Z
that 7 ´5 Z ~ µ ~ c ~ Àc ~ À7 ´5 ~ µ and
Z
7 ´5 Z ~ µ ~ Z c ~ Àc ~ À7 ´5 ~ µ . Dividing the second equation by the first, we get
Z ~ À The second moment of the claim severity @ before the change in deductible is
,´@ µ ~ = ´@ µ b ²,´@ µ³ ~ b ~ Á Á .
The second moment of the claim severity @ after the change in deductible is
,´²@ Z ³ µ ~ = ´@ Z µ b ²,´@ Z µ³ ~ b ~ Á Á .
Since aggregate claims follow a compound Poisson distribution, the variance of : , aggregate
claims before deductible is = ´:µ ~ ,´@ µ ~ Á Á , and the variance of : Z , aggregate
claims before deductible is = ´: Z µ ~ Z ,´²@ Z ³ µ ~ Á Á ² ³ .
The proportional increase in the variance of aggregate claims is
Á Á ² ³°Á Á ~ À
. This is a 76% increase. Answer: D
6. The empirical distribution assigns probability of ~ À for each ²%Á &³ pair.
The covariance of the empirical distribution is
,´²? c ? ³²@ c @ ³µ ~ ²À³´² c À
³² c À³ b ² c À
³² c À³
b ² c À
³² c À³ b ² c À
³²
c À³ b ²
c À
³² c À³µ ~ À ,
or ,´[email protected] µ c ? h @ ~ bbbb
c ²À
³²À³ ~ À ~ ( .
Using the same marginal distributions (the same ? 's and the same @ 's, perhaps in different
pairings), the covariance will be maximized if ,´[email protected] µ is maximized.
Maximization will occur if the largest @ 's are paired with the largest ? 's. The following bivariate
distribution has the same marginal distributions as the original bivariate distribution:
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 3
PE-51
6. continued
?
@
The covariance of ? and @ in this bivariate distribution is
b
b
bb
c ²À
³²À³ ~ À ~ ) . This is the maximum covariance with identical
marginal distributions. ) c ( ~ À c À ~ À . Answer: D
7. An underlying relationship upon which the estimate is based is
,´?µ ~ ,´? O c  ?  µ h 7 ²c  ?  ³ .
~
When the uniform distribution is assumed for each interval, this results in
b
,´? O c  ?  µ ~ c . Also, the estimate of 7 ²c  ?  ³
is (fraction of total set of data points that lie in the interval).
²%c³
With the new pdf, we get ,´? O  ?  µ ~ % h ²c³ % .
A somewhat simplified approach is to find
²%c³
²c³
²%c³
,´? c O  ?  µ ~ ²% c ³ h ²c³ % ~ ²c³ % ~ ,
so that ,´? O  ?  µ ~ ,´? c O  ?  µ b ~
Then, ,´? O c  ?  µ ~
c b
²c³
b ~ b
.
b
, and the estimated mean of ? is h c ~
8. The covariance matrix is the inverse of the information matrix. This will be
c
À
À
À
~ ²À³²À
³c²À³²À³
h >
*#²
VÁ V³ ~ >
?
À À
c À
À
c À
=V
²
V³ *#²
V VÁ V³
.
~ >
~ @
?
c À
À
*#²
=V
²V³ A
V VÁ V³
c À
À
?
À
The mean of the inverse gamma is c
. The estimate of this is Àc
~ À .
According to the delta method, the variance of the mle estimate of ²Á ³ ~ c
is
C
C
C
C
²
²Á ³³ h = ²
V³ b ² ²Á ³³² ²Á ³³ h *#²
VÁ V³ b ² ²Á ³³ h = ²V ³
C
C
evaluated at the estimate values. This is
C
C
² c ²c³
³ ²À³ b ² c ²c³ ³² c ³² c À³ b ² c ³ ²À
³
À
À
~ ² c ²Àc³
³ ²À³ b ² c ²Àc³ ³² Àc ³² c À³ b ² Àc ³ ²À
³ ~ À
À
The 95% confidence interval for the mean is À f À
jÀ
~ ²ÀÁ À³ .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-52
PRACTICE EXAM 3
9. The estimate of the geometric mean is ~ À .
Of the 1000 workers, the expected number who miss 0 days of work is h À
~ À,
À
the expected number who miss 1 day of work is h ²À³ ~ À,
²À³
the expected number who miss 2 days of work is h ²À³ ~ À,
and the expected number who miss 3 or more days of work is
c ²À b À b À³ ~ À
. Each "interval", "0", "1", "2" and "3 or more" has an
expected number of at least 1, so we keep all four intervals.
The chi-squared statistic is
²Àc³
²Àc³
²Àc³
²
À
c³
8~
b
b
b À
~ À .
À
À
À
The test has 2 degrees of freedom (there are 4 intervals and one parameter estimated).
From the chi-square table with 2 degrees of freedom we see that the 90-th percentile is 4.605.
Therefore, the hypothesis is not rejected at the 10 per cent level of significance. Answer: A
~
~
% ³² ³ ~ À
10. The original Bayesian premium is ² b % ³² b
³ ~ ² b
from which it follows that % ~ .
~
~
~
% ³² ³ ~ À
The updated Bayesian premium is ² b % ³² b
³ ~ ² b
from which it follows that % ~ .
~
~
~
Therefore, % ~ % c % ~ . Answer: A
11. The problem involves determining the Bayesian premium. The prior distribution of the
parameter has density ²³ ~ ²À³c b ²À³² ³c° (a mixture of two exponential
distributions, one with mean and on with mean 5, which can also be regarded as a mixture of
two gamma distributions, the first with ~ Á ~ Á the second with ~ Á ~ ).
The model distribution, ? given , is Poisson with mean .
We are given one observation of ? , which is ? ~ in the first policy year.
We wish to find ,´? O? ~ µ , which is the Bayesian premium.
B
We can write the expectation as ,´? Oµ h ²O? ~ ³ .
Since ? given has a Poisson distribution with mean , this integral becomes
B h ²O? ~ ³ Á which is the mean of the posterior distribution.
Therefore, if we can identify the posterior distribution, it may be easy to determine its mean.
The joint density of % and at % ~ is
%
²Á ³ ~ ²O³ h ²³ ~ c h %[ h ´²À³c b ²À³² ³c° µ
~ ²À³´c
b ² ³cÀ µ , and the marginal probability that ? ~ is
B
²³ ~ ²Á ³ ~ ²À³´ b µ ~ À .
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PRACTICE EXAM 3
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11. continued
The posterior density of given ? ~ is then
²Á³
²À³´c
b² ³cÀ µ
²O? ~ ³ ~ 7 ´?~µ ~
~ c
b ² ³cÀ .
À
This can be written in the form ²O? ~ ³ ~ ² ³²
c
³ b ² ³²ÀcÀ ³ .
Therefore the posterior distribution is a mixture of two exponential distributions, with mixing
weight for the exponential with mean , and mixing weight for the exponential with mean
À . The mean of the posterior distribution is then ² ³² ³ b ² ³² À ³ ~ À . Answer: A
12. Credibility is being applied to aggregate losses : , which has a compound distribution.
The frequency 5 depends on the parameter , and the severity @ depends on the parameter ,
so : depends on both parameters. In general, for a compound distribution, the mean is
,´:µ ~ ,´5 µ h ,´@ µ and the variance is = ´:µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ .
The hypothetical mean in this example is
,´:OÁ µ ~ ,´5 Oµ h ,´@ Oµ ~ h and the process variance is
= ´:OÁ µ ~ ,´5 Oµ h = ´@ Oµ b = ´5 Oµ h ²,´@ Oµ³ ~ b ~ .
The variance of the hypothetical mean is
~ = ´ ,´:OÁ µ µ ~ = ´µ ~ ,´²³ µ c ²,´µ³ .
Since and are independent, we have ,´µ ~ ,´µ h ,´µ ~ d ,
and ,´²³ µ ~ ,´ µ ~ ,´ µ h ,´ µ ~ ²³²³ ~ (since has an exponential distribution with mean 1, the second moment of is ²mean³ ~ ,
and since has a Poisson distribution with mean 1,
~ = ´µ ~ ,´ µ c ²,´µ³ ~ ,´ µ c ²³ , so that ,´ µ ~ ).
Therefore, ~ c ~ . The expected process variance is
# ~ ,´ = ´:OÁ µ µ ~ ,´ µ ~ ,´µ h ,´ µ ~ ²³²³ ~ . Then ~ # ~ .
²³²³b²³²³
c
13. ? ~
~ À Á ~ b
²³²³b²³²³b²³²³
c
? ~
~ À Á ~ bb
c
²³²³b²
³²³
? ~
~ À Á ~ b
²³²³b²³²³b²³²³b²³²³b²³²³b²³²³b²
³²³
c
~ À
V~?~
bb
~ .
V# ~
V# ~
V# ~
c
c
c
h ´² c À³ b ² c À³ µ ~ À Á
h ´² c À³ b ² c À³ b ² c À³ µ ~ Á À Á
h ´² c À³ b ² c À³ µ ~ Á À
V# b#
V b#
V
~ Á À .
V# ~ bb
c
c
~
h ´ ²? c ? ³ c V#² c ³ µ
V
c ~
~
~ c ²b b ³
d ´ ´²À c À³ b ²À c À³ b ²À c À³ µ c Á À²³µ
~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 3
13. continued
The credibility premium for policyholder 2 is
V c b ² c A³
V A?
V~²
ÁÀ ³ h ²À³ b ² c
b
À
³²
À³
b ÁÀ
À
~ À .
c
c
? b?
14. The sample mean based on the first 19 points is ? ~
~ ,
c
c
? b?
and the sample mean based on all 20 points is ? ~
~ À
We use the relationship
b
c
c
c :b
~ ²? c ? b ³ ~ ² c
³: b ² b ³²? b c ? ³ to get
~
c
c :
~ ² ³: b ²³²? c ? ³ , and then
c
c :
~ ² ³: b ²³²? c ? ³
c
c c
c ~ ² ³: b ²³²? c ? ³ b ²³²? c ? ³
~ ² ³²³ b ²³² c ³ b ²³² c ³ ~ À .
15. 4 :, is approximated (estimated) by using the corresponding quantity in the empirical
distributionÀ The empirical distribution consists of the three data points, so the median of the
empirical distribution is . The mean square error of the estimator is ,´²V c ³ µ . Since the
empirical distribution consists of three points, there are ~ possible samples of size 3 that
can be drawn from the empirical distribution, each with probability . The samples and values
of the estimator V are
Sample #
1
2
3
4
5
6
7
8
9
Sample
4,4,4 4,4,5 4,4,9 4,5,4 4,5,5 4,5,9 4,9,4 4,9,5 4,9,9
V ~ Sample median
4
4.5
6.5
4.5
4.5
6.5
6.5
6.5
6.5
Sample #
Sample
Sample median
10
5,4,4
4.5
11
5,4,5
4.5
12
5,4,9
6.5
13
5,5,4
4.5
14
5,5,5
5
15
5,5,9
7
16
5,9,4
6.5
17
5,9,5
7
18
5,9,9
7
Sample #
Sample
Sample median
19
9,4,4
6.5
20
9,4,5
6.5
21
9,4,9
6.5
22
9,5,4
6.5
23
9,5,5
7
24
9,5,9
7
25
9,9,4
6.5
26
9,9,5
7
27
9,9,9
9
4 :, ~ ² c ³ b ²À c ³ b ² c ³
b ²
À c ³ b ² c ³ b ² c ³ ~ À .
: ²³
: ²³
16. We are given : ²³ ~ .94 , : ²³ ~ À and : ²³ ~ À .
: ²³
: ²³
: ²³ ~ : ²³ h : ²³ h : ²³ ~ À
.
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17. The general standard for the number of exposures (or insureds in this case) needed for full
= ´:µ
credibility for the random variable : is
²,´:µ³ , since : is the aggregate loss per insured.
The distribution of : is a compound distribution (but not compound Poisson, since the claim
count parameter has a gamma distribution). We use the more general relationships for compound
distributions for which claim frequency is 5 and claim severity is @ .
,´:µ ~ ,´5 µ h ,´@ µ and = ´:µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ .
In this case, we are given that @ has a Pareto distribution with mean ,´@ µ ~ , and
variance = ´@ µ ~ Á Á . Since the full credibility standard is to be within 5% of the
expected aggregate loss 95% of the time, we have ~ ² À
À ³ ~ À
.
From the given observations about the claim counts of 1000 insureds we can estimate the mean of
c
V µ~5
5 , ,´5
~ h ' ~ ~ À , and we can estimate the variance of 5 ,
c c
=V
´5 µ ~ c h '² c 5 ³ ~ h ´' c ²5 ³ µ ~ À .
The estimated mean of : is ²À³²³ ~ , and the estimated variance is
²À³²
Á Á ³ b ²À³²³ ~ Á Á .
= ´:µ
Á
Á
The full credibility standard is then ²,´:µ³ ~ ²
À
³´ ²³ µ ~ ,
which is the number of insureds needed for full credibility.
Note that for a compound aggregate loss distribution : , with frequency 5 and severity @ , there
are three alternative equivalent standards for full credibility. This is reviewed in Section 1 of the
notes on credibility in Volume 1 of this study guide. The standard for full credibility used above
is the standard for the number of insureds (or exposures) needed. The other two standards are:
= ´:µ
(i) standard based on aggregate losses; this is ,´:µ , which in this example would be
9,780,750 in aggregate losses needed for full credibility; and
(ii) standard based on total number of claims; this is would be d ²À³ ~ À .
= ´:µh,´5 µ
²,´:µ³ ,
which in this example
18. Constant hazard rate means that time until death has an exponential distribution
with mean . The time until death for a randomly chosen individual has a continuous mixture
distribution. The probability 7 ²  ;  ³ is found by conditioning over .
The pdf of is ²³ ~ À
for À   À.
À
À
7 ²  ;  ³ ~ À 7 ²  ;  O³ h ²³ ~ À ´c c c µ h ~ ´²
cÀ
c
cÀ
³c
c²À³ cc²À³
µ
À
~ À .
19. With deductible , the expected payment per loss is ,´²? c ³b µ ~ ,´?µ c ,´? w µ .
For the exponential with parameter , from the table of distributions, we have ,´?µ ~ and
,´? w µ ~ ² c c° ³ , so that ,´²? c ³b µ ~ c° .
We are given that ,´²? c ³b µ ~ Á , and therefore, c° ~ Á .
We are asked to find ,´²? c ³b µ ~ c° .
,´²?c³ µ
c°
We see that ,´²?c³b µ ~ c° ~ c° , and therefore
b
,´²? c ³b µ ~ ,´²? c ³b µ h c° ~ Á c° .
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PRACTICE EXAM 3
20. The probability function of the geometric distribution with mean ,´5 µ ~ ~ is of the
form ~ ²b ³b ~ ²³b ~ ² ³² ³ , so that ~ .
The zero-modified distribution has probability function
c4
c 4 ~ c h ~ °
h ~ for ~ Á Á Á ÀÀÀ
c4
The mean of the zero-modified distribution is ,´54 µ ~ c h ,´5 µ ~ h ~ .
Note that since the probability at 0 is reduced by an amount of ²reduced from to ), the total
°
probability  1 must be increased from to , which is a proportional increase of ° ~ À.
It follows that the mean (or any moment of distribution) is increased proportionally by 1.25 from
21. With no deductible the aggregate annual payment : has a compound distribution with
frequency 5 that is binomial with ~ and ~ À, and severity @ that is uniform on the
interval ´Á µ . The expected aggregate payment with no deductible is
,´:µ ~ ,´5 µ h ,´@ µ ~ ²³²À³²³ ~ Á .
If a deductible of 200 is applied to each claim, the severity becomes @ Z ~ ²@ c ³b
and the expected severity is ,´@ Z µ ~ ,´²@ c ³b µ ~ ²& c ³²À³ & ~ .
With deductible of 200 per claim, the aggregate annual payment : Z has a compound distribution
with frequency 5 that is still binomial with ~ and ~ À, and severity @ Z .
The expected aggregate payment with a deductible of 200 per claim is
,´: Z µ ~ ,´5 µ h ,´@ Z µ ~ ²³²À³²³ ~ Á .
The reduction in expected annual claims is Á c Á ~ Á .
Alternatively, the expected reduction per claim when the deductible of 200 is imposed is
,´? w µ ~ &²À³ & b ´ c [email protected] ²³µ ~ b ´ c Àµ ~ .
The expected number of claims is ²À³ ~ , so the expected reduction in aggregate claims
is ²³²³ ~ Á .
22. This is a stop-loss problem where : is the aggregate number of overtime hours worked in the
week and the deductible is 15. : has a compound distribution with frequency 5 that is Poisson
with mean 2 and severity ? that is 5 (prob. .2), 10 (prob. .3) or 20 (prob. .5). We wish to find
,´²: c ³b µ ~ ,´:µ c ,´: w µ . The mean of : is
,´:µ ~ ,´5 µ h ,´?µ ~ ´²À³ b ²À³ b ²À³µ ~ .
Note that : must be a multiple of 5, with
7 ²: ~ ³ ~ 7 ´5 ~ µ ~ c ~ À (the only way that : ~ is if 5 ~ ),
7 ´: ~ µ ~ 7 ´5 ~ µ h 7 ´? ~ µ ~ ²À³²À³ ~ À , and
7 ´: ~ µ ~ ²7 ´5 ~ µ h 7 ´? ~ µ b 7 ´5 ~ µ h ²7 ´? ~ µ³ ³
~ ²À³²À³ b ²À³²À³ ~ À .
Then, 7 ²:  ³ ~ c 7 ²: ~ Á Á ³ ~ À .
: ~ , prob. À
: ~ , prob. À
,
: w ~
: ~ , prob. À
:  , prob. .
so ,´: w µ ~ ²À³ b ²À³ b ²À³ ~ À .
Then, ,´²: c ³b µ ~ c À ~ À . Answer: B
H
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 3
PE-57
7 ´?
µ
23. 7 ´  ?  O ?  µ ~ 7 ´?µ
Since the observed failure times are all integers, the estimate of 7 ´  ?  µ is the same as
the estimate of 7 ´  ?  µ ~ :²³ c :²³ .
:²³c:²³
Therefore, we wish to estimate c:²³ .
The numbers at risk and the numbers of failures at each failure time are:
& ¢
¢
¢
The product limit estimate of :²³ is ² c
³²
c
³²
c
³
~ À
.
The product limit estimate of :²³ is ² c ³² c ³² c ³² c ³² c ³ ~ À .
The product limit estimate of :²³ is
² c ³² c ³² c ³² c ³² c ³² c ³² c ³² c ³² c ³ ~ À .
The product limit estimate of
:²³c:²³
c:²³
is À
cÀ
~ À À
cÀ
V
The Nelson-Aalen estimate of /²³ is /²³
~ b
cÀ
so the N-A estimate of :²³ is ~ À
.
b
~ À
,
V
The Nelson-Aalen estimate of /²³ is /²³
~ b
cÀ
so the N-A estimate of :²³ is ~ À
.
b
b
b
~ À ,
The Nelson-Aalen estimate of /²³ is
V
/²³
~ b b b b b b b b ~ À
,
so the N-A estimate of :²³ is cÀ
~ À .
The Nelson-Aalen estimate of
:²³c:²³
c:²³
is À
cÀ
~ À À
cÀ
The absolute difference between the two estimates is .0026 . Answer: D
24. Triangle kernel with bandwidth ~ .
For the point % ~ there is one & value within the band from c ~ to 0c ~ ;
this is the data value & ~ . .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 3
24. continued
The intervals centered at points & ~ Á & ~ Á and & ~ all lie completely to the left of
% ~ , so that 2 ²³ ~ 2 ²³ ~ 2 ²³ ~ . For the &'s from 26 to 53, the intervals all lie
completely to the right of 20, so 2& ²³ ~ for each of them. From the triangle diagram above,
it can be seen that the area in the triangle to the right of 20 is d d ~ , so that the area to
the left of 20 in that triangle is 2 ²³ ~ c ~ À Therefore,
V ²³ ~ ² ³²³ b ² ³²³ b ² ³²³ b ² ³² ³ ~ . Answer: D
V
25. The estimate of 7 ´?  µ is c° , where V is the mle of .
Since ? has an exponential distribution with mean , the mle of is the sample mean of the data
set, which is 3.78 . The estimate of 7 ´?  µ is c°À , which we are given as 0.75 .
Therefore c À
~ ²À³ and ~ À .
26. The pdf is ²%³ ~ ² b ³% and ²%³ ~ ² b ³ b %
and ²%³ ~ b
b % . Setting the derivative of the loglikelihood function to 0, we get
3 ~ ' ²% ³ ~ b
b ' % ~ S ~ 'c
% c ~ À .
27. The parameter being estimated is the distribution mean, and the mean of the empirical
distribution is .5 . The estimator being used is the sample mean, % b%
. Thus, for the 4 pairs
²% Á % ³, we have estimator values Á À Á À Á , and the estimate of the MSE is
´² c À³ b ²À c À³ b ²À c À³ b ² c À³ µ ~ À .
28. The distribution function of the hypothesized distribution is
%
%
- ²%³ ~ ²!³ ! ~ ²b!³
! ~ c b! À
Alternatively, it can be noted that ? has a Pareto distribution with parameters ~ and
c
~ , so that - ²%³ ~ c ² %b
~ c %b
À
³
The empirical and model distribution function values are
%
- ²%³
- ²%c ³
- ²%³
+
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
The Kolmogorov-Smirnov statistic is the maximum from column +, .3404. Answer: E
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 3
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B
29. The severity distribution has mean ,´@ µ ~ & h &c
& ~ Á and
B
,´@ µ ~ & h &c
& ~ Á so that = ´@ µ ~ c ² ³ ~ À
With ~ À and ~ À , we have ~ ² À
À ³ ~ À À The full credibility standard for
= ´@ µ
number of exposures needed for the compound Poisson distribution is ´ b ²,´@ µ³ µ .
The full credibility standard for the number of exposures needed for the Poisson frequency
Z
distribution only is  . Since we are considering the same Poisson frequency distribution,
the value of (which is not known) stays the same. If the same value of for full credibility
from the aggregate compound Poisson distribution is applied to the Poisson frequency
Z
= ´@ µ
distribution alone, then we set ~ ´ b ²,´@ µ³ µ and the " " for the Poisson frequency
credibility standard must change, which is why it has been denoted Z .
Z
= ´@ µ
°
Then ~ ´ b ²,´@ µ³ µ ~ À
´ b ²°³ µ S Z ~ À.
With 7 ~ À , & ~ À
, and then in order for this to be the proper Z for 7 ~ À , we must
have À ~ ² À
³ S ~ À .
30. ,´?Oµ ~ Á = ´?Oµ ~ . Since is a mixture, ,´µ ~ ²À³²³ b ²À³²³ ~ and ,´ µ ~ ²À³²³ b ²À³² d ³ ~ À (the first component is constant at 1). Then
~ ,´²³µ ~ ,´µ ~ Á # ~ ,´#²³µ ~ ,´µ ~ Á
~ = ´²³µ ~ = ´µ ~ ,´ µ c ²,´µ³ ~ À .
For a single observation of ? , ~ , the Buhlmann credibility factor is A ~ b ~ À .
With ? ~ , the Buhlmann credibility estimate for the next year is
c
A? b ² c A³ ~ b ²À
³²³ ~ À .
À
31. Prior distribution is 7 ²(³ ~ 7 ²)³ ~ À
Hypothetical means are ²(³ ~ ,´?O(µ ~ Á ²)³ ~ ,´?O)µ ~ .
Process variances are #²(³ ~ = ´?O(µ ~ ~ ~ Á #²)³ ~ .
~ ,´?µ ~ expected hypothetical mean ~ ²³² ³ b ²À³² ³ ~ .
# ~ expected process variance ~ ² ³² ³ b ² ³² ³ ~ .
~ variance of hypothetical mean ~ ² c À³ ² ³² ³ ~ .
A ~ b
~ À .
# ~
b °
°
A3 b ² c A³ ~ À3 b À
² ³ ~ À3 b À
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PRACTICE EXAM 3
32. Hypothetical mean is ,²:O³ ~ h ,²@ ³ ~ À À
Process variance is = ²:O³ ~ h ,²@ ³ ~ À .
Expected hypothetical mean is ~ ,²,²:O³³ ~ ,²À³ ~ À.
Expected process variance is # ~ ,²= ²:O³³ ~ ,²À³ ~ À .
Variance of hypothetical mean is ~ = ²,²:O³³ ~ = ²À³ ~ À .
A ~ b # ~ bÀ ~ À À
À
c
Buhlmann credibility premium is A: b ² c A³ ~ ²À³² bb
³ b ²À³²À³ ~ À.
33. The simulation of an exponential variable @ with mean using the inverse transform
method is & ~ c ² c "³ . In this case, ~ , so the three simulated exponential values are
@ ~ c ² c À³ ~ À
Á @ ~ c ² c À
³ ~ À Á and
@ = c ² c À³ ~ À .
The simulated value of the gamma is ~ @ b @ b @ ~ À .
This completes Step 1.
According to the product algorithm we multiply the successive uniform ²Á ³ values until the
product is first less then cÀ ~ À . The remaining uniform ²Á ³ numbers are
.1 , .7 , .7 , .1 . We see that À d À  À  À d À d À .
Since the product of the first two uniform numbers was greater than cÀ , but the product of
the first three was less than cÀ , the simulated value of ? is 2. Answer: C
34. According to the existence of moments test, a distribution has a light right tail if all positive
moments exists; this means that ,´? µ  B if  . If not all positive moments exist then
the distribution has a heavy right tail. From the Exam C Tables we see that all positive moments
exist for the Gamma, Lognormal and Exponential distributions and we also see that not all
positive moments exist for the LogLogistic, Paralogistic and Inverse Exponential distributions.
For instance, for the Loglogisitic distribution with parameters and , ,´? µ exists only for
c   (not for all ). Therefore, according to the existence of moments test, the
LogLogistic, Paralogistic and Inverse Exponential distributions all have heavier tails than the
Gamma, Lognormal and Exponential distributions. All three statements are true.
35. The likelihood ratio test is an approximate test of the null hypothesis that model B is
preferable to model A, where model B has more parameters than model A. The test statistic is
²) c ( ³ , where represents the log of the likelihood at the estimated parameter values. This
statistic is approximately chi-square with degrees of freedom equal to
#parameters in model ) c #parameters in model (.
In this case the Poisson is model A (one parameter) and the negative binomial is model B (two
parameters). The test statistic is ² c À c ² c À³³ ~ .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 4
PE-61
ACTEX EXAM C/4 - PRACTICE EXAM 4
1. ? has a Pareto distribution with parameters and .
The random variable @ defined as follows: @ ~ ² ?b
³.
Identify the type of distribution that @ has.
A) Exponential with mean B) Exponential with mean C) Gamma with parameters and E) Inverse Pareto with parameters ~
D) Gamma with parameters and and 2. ? has a normal distribution with a mean of \$ and variance of 1. \$ has a normal distribution
with a mean of 1 and variance of 1. Find the 95-th percentile of ? .
A) 3.00
B) 3.33
C) 3.67
D) 4.00
E) 4.33
3. A loss random variable ? has the following characteristics. There is a 90% chance of no loss
occurring, ? ~ , and there is a 10% chance that a positive loss occurs. If a positive loss occurs,
it is uniformly distributed between 1000 and 5000. An insurance policy on this loss has an
ordinary deductible of 2000 applied. Find the expected cost per loss (including when ? ~ ) and
the expected cost per payment for this policy. Find the variance of the cost per payment random
variable.
A) 600,000
B) 650,000
C) 700,000
D) 750,000
E) 800,000
4. Insurance losses are a compound Poisson process where:
(i) The approvals of insurance applications arise in accordance with a Poisson process at a rate
of 1000 per day.
(ii) Each approved application has a 20% chance of being from a smoker and an 80% chance of
being from a non-smoker.
(iii) The insurances are priced so that the expected loss on each approval is c 100.
(iv) The variance of the loss amount is 5000 for a smoker and is 8000 for a non-smoker.
Calculate the variance for the total losses on one day's approvals.
A) 13,000,000
B) 14,100,000
C) 15,200,000
D) 16,300,000
E) 17,400,000
5. The random variable ? has an exponential distribution with mean .
The loss random variable @ is defined to be a mixture of ? with mixing weight , and 0 with
mixing weight c , where   .
A random sample of losses is observed: @ Á À À À Á @ .
Suppose that of the losses are 0 and the remaining losses are  .
Find the maximum likelihood estimators of and in terms of Á and @ Á À À À Á @ .
'@
'@
'@
'@
A) B) c
C) D) b
E) None of A, B, C or D is correct
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-62
PRACTICE EXAM 4
6. XYZ Re provides reinsurance to Bigskew Insurance Company. XYZ agrees to pay Bigskew
for all losses resulting from "events", subject to:
a \$500 deductible per event and
a \$100 annual aggregate deductible
For providing this coverage, XYZ receives a premium of \$150. Use a Poisson distribution with
mean equal to 0.15 for the frequency of events. Event severity is from the following distribution.
Probability
Loss
250
0.10
500
0.25
800
0.30
1,000
0.25
1,250
0.05
1,500
0.05
~ %
What is the actual probability that XYZ will pay out more than it receives?
A) 8.9%
B) 9.0%
C) 9.1%
D) 9.2%
E) 9.3%
7. You are given the following grouped data set of 80 random data points taken from the
distribution of ? :
Interval
Number of Data Points
²Á µ
²Á µ
²Á µ
²Á µ
²Á µ
²Á µ
²Á µ
Find the empirical estimate of the -th percentile of ? .
A) 640
B) 644
C) 648
D) 652
E) 656
8. Loss data for 925 policies with deductibles of 300 and 500 and policy limits of 5,000 and
10,000 were collected. The results are given below:
Deductible
Range
300
500
Total
²Á µ
c
²Á µ
²Á µ
²Á Á µ
At At Á Total
Using the Kaplan-Meier approximation for large data sets with ~ À and ~ À,
estimate - ²³.
A) 0.25
B) 0.36
C) 0.47
D) 0.58
E) 0.69
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 4
PE-63
Questions 9 and 10 are based on the following random sample of 12 data points from a random
variable ?
7 , 12 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53
9. Assuming a uniform distribution on interval ²Á ³ , maximum likelihood estimation is applied
to estimate . Find the estimated variance of the parameter estimate.
A) 10
B) 12
C) 14
D) 16
E) 18
10. The mle found in question 5 is used with the delta method to construct an approximate 95%
confidence interval for the probability 7 ´?  µ . Find the upper limit of the interval.
A) .47
B) .49
C) .51
D) .53
E) .55
11. The random variable ? has pdf ²%³ ~ % for  %  .
? Á ÀÀÀÁ ? is a random sample from the distribution of the continuous random variable ? .
V ~ ? is taken as an estimate of the distribution mean . Find 4 :, ²V³ .
c
~
A) b
B) ²c³
C) b
D) ²c³
b
E) ²c³
12. You are given the following random sample of size 4: 1 , 2 , 5 , 9
An exponential distribution with a mean of is used as a model for the data.
In the -plot for this data set and model distribution, it is found that the right tails of the
exponential distribution are thinner than the right tails of the (smoothed) empirical distribution at
each of the data points. Determine the maximum value of (rounded to the nearest .1) that is
consistent with this -plot.
A) 3.5
B) 3.9
C) 4.5
D) 4.9
E) 5.5
13. The prior distribution of the parameter has pdf ²³ ~ for  .
The model distribution has a uniform distribution on the interval ´ Á µ .
Find the posterior density of given % if %  , and indicate the region of density for the
posterior.
A) %
B) %
C) %
D) %
E) %
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-64
PRACTICE EXAM 4
Questions 14 and 15 relate to the following information. A portfolio of independent risks is
divided into two classes of equal size. All of the risks in Class 1 have identical claim count and
claim size distributions as follows:
Class 1
Class 1
Number of Claims Probability
Claim Size Probability
1
1/2
50
2/3
2
1/2
100
1/3
Class 2
Number of Claims
1
2
Probability
2/3
1/3
Class 2
Claim Size Probability
50
1/2
100
1/2
The number of claims and claim size(s) for each risk are independent. A risk is selected at
random from the portfolio, and a pure premium of 100 is observed for the first exposure period.
14. Determine the Bayesian estimate of the expected number of claims for this same risk for the
second exposure period.
A) 1.0
B) 1.1
C) 1.2
D) 1.3
E) 1.4
15. An aggregate claim of 150 is observed for this risk for the second exposure period.
Determine the Buhlmann credibility estimate of the expected aggregate claim for this same risk
for the third exposure period.
A) 100
B) 125
C) D) 150
E) 16. An actuary applies empirical Bayes credibility analysis to a portfolio consisting of ~ policyholders. Each policyholder has ~ exposure observations. The data summary is as
c
c
c
follows: ? ~ ? ~ À Á ? ~ À Á
c c c ²? c ?
³ ~ À Á ²? c ? ³ ~ À Á ²? c ? ³ ~ À .
~
~
~
The estimated process variance V
# is found based on this data.
Another actuary combines policyholders 1 and 2 together as a single policy holder, and applies
empirical Bayes credibility analysis to the newly grouped portfolio of 2 policyholders (previous
policyholders 1 and 2 combined are now "policyholder 1" and previous policyholder 3 is now
"policyholder 2"). The estimated process variance V# is found by the second actuary.
Find V# c V# .
A) c À B) c À
C) D) À
E) À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 4
PE-65
17. You are given a sample of losses from an exponential distribution. However, if a loss is
1000 or greater, it is reported as 1000. The summarized sample is:
Reported Loss
Number
Total Amount
Less than 1000
62
28,140
38,000
1000
38
Total
100
66,140
Determine the maximum likelihood estimate of , the mean of the exponential distribution.
A) Less than 650
B) At least 650, but less than 850
C) At least 850, but less than 1050
D) At least 1050, but less than 1250
E) At least 1250
18. A portfolio of risks is divided into three classes. The characteristics of the annual claim
distributions for the three risk classes is as follows:
Class I
Class II
Class III
Annual Claim
Poisson
Poisson
Poisson
Number Distribution mean 1
mean 2
mean 5
50% of the risks are in Class I, 30% are in Class II, and 20% are in Class III.
A risk is chosen at random from the portfolio and is observed to have 2 claims in the first year
and 2 claims in the second year. Find the probability that the risk will have 2 claims in the third
year.
A) Less than .20
B) At least .20 but less than .21
C) At least .21 but less than .22
D) At least .22 but less than .23
E) At least .23
19. You are given:
(i) claim count 5 has a binomial distribution with parameters 4 and ~ À .
(ii) 4 has a discrete uniform distribution on the integers Á Á Á ÀÀÀÁ .
Find = ´5 µ .
A) 1.21
B) 1.24
C) 1.27
D) 1.30
E) 1.33
20. For a particular loss random variable ? , if an ordinary deductible of 1000 is applied, the
mean excess loss is 3500. If the deductible of 1000 is applied to ? as a franchise deductible, then
the expected cost per loss is 3600. Find the expected cost per loss if the deductible of 1000 is
applied as an ordinary deductible.
A) 2000
B) 2200
C) 2400
D) 2600
E) 2800
21. A compound distribution : has frequency 5 and severity ? , both of which are members of
the ²Á Á ³ class. You are given the following:
,²5 ³ ~ À Á = ²5 ³ ~ À Á ,²:³ ~ À Á = ²:³ ~ À
Find 7 ²: ~ ³ .
A) Less than .025
B) At least .025 but less than .050
D) At least .075 but less than .100
E) At least .100
C) At least .050 but less than .075
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-66
PRACTICE EXAM 4
22. For a collective risk model:
(i) The number of losses has a geometric distribution with a mean of 2.
(ii) The common distribution of the individual losses is:
%
? ²%³
À
À
An insurance covers aggregate losses subject to a deductible of 3. Calculate the expected
aggregate payments of the insurance.
A) Less than 1.0
B) At least 1.0 but less than 1.2
C) At least 1.2 but less than 1.4
D) At least 1.4 but less than 1.6
E) At least 1.6
23. PQR Re provides reinsurance to Telecom Insurance Company. PQR agrees to pay Telecom
for all losses resulting from "events", subject to a \$500 per event deductible. For providing this
coverage, PQR receives a premium of \$250. Use a Poisson distribution with mean equal to 0.15
for the frequency of events. Event severity is from the following distribution:
Probability
Loss
250
0.10
500
0.25
750
0.30
1,000
0.25
1,250
0.05
1,500
0.05
~ %
Using the normal approximation to PQR's annual aggregate losses on this contract, what is the
probability that PQR will pay out more than it receives?
A) Less than 12%
B) At least 12%, but less than 13%
C) At least 13%, but less than 14%
D) At least 14%, but less than 15%
E) 15% or more
24. The "redistribute to the right" procedure is a method of dealing with censored data when
estimating :²!³. All data points start out with "probability mass" of . When an
observation(s) is (are) censored, its (their) probability mass(es) at the time of censoring is (are)
divided among the remaining individuals still alive at that time. The redistribute to the right
method is applied to a group of 10 individuals, one of whom had a censored observation. The
other 9 deaths were observed to occur at all different times. We are given that : ²! ³ ~ À (!
is the time of the 8-th observed death). The censored observation occurred
A) before the first death
B) between the first and third deaths
C) between the third and fifth deaths
D) between the fifth and seventh deaths
E) between the seventh and ninth deaths
25. The random variable ? has the density function
²%³ ~ ²% b ³cc Á  %  B Á  Assuming  , determine the method of moments estimator of .
c
c
c
c
c
?
?
?b
c
c
c
c
A) ?
B) ?c
C) ?b
D) ?c
E)
?
?
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 4
PE-67
26. Which of the following statements is true?
A) For a null hypothesis that the population follows a particular distribution, using sample data to
estimate the parameters of the distribution tends to decrease the probability of a Type II error.
B) The Kolmogorov-Smirnov test can be used on individual or grouped data.
C) The Anderson-Darling test tends to place more emphasis on a good fit in the middle rather
than in the tails of the distribution.
D) For a given number of cells, the critical value for a chi-square goodness-of-fit test becomes
larger with increased sample size.
E) None of A, B, C or D are true.
27. Annual claim counts follow a Negative Binomial distribution. The following claim count
observations are available:
Year:
2005
2004
2003
Claim Count:
0
3
5
Assuming each year is independent, calculate the likelihood function of this sample.
A) ² b
³ ² b ³
B) ² b
³ ² b ³
C) ² b
³ ² b ³
D) ² b
³ ² b ³
E) ² b
³ ² b ³
²b³ ²b³
[ [
²b³ ²b³
[ [
²b³ ²b³ ²b3³
[ [
²b³ ²b³ ²b3³²b³
[ [
²b³ ²b³ ²b3³²b³
[ [
28. 40 observed losses have been recorded in thousands of dollars and are grouped as follows:
Interval
Number of
Total Losses
Losses
(\$000)
(\$000)
² Á µ
² Á µ
²Á µ
²Á B³
The null hypothesis, / , is that the random variable ? underlying the observed losses, in
thousands, has the density function
²%³ ~ % , %  .
Calculate the value of the chi-square goodness-of-fit statistic used to test the null hypothesis.
A) 7.0
B) 7.2
C) 7.4
D) 7.6
E) 7.8
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-68
PRACTICE EXAM 4
29. The ABC Insurance Company has decided to establish its full credibility requirements for an
individual state rate filing using a standard under which the observed total cost of claims should
be within 5% of the true value with probability .95 . The claim frequency follows a Poisson
distribution and the claim severity is distributed according to the distribution ²%³ ~ Á
for  %  Á . What is the expected number of claims necessary to obtain full
credibility?
A) Less than 1500
B) At least 1500, but less than 1800
C) At least 1800, but less than 2100
D) At least 2100, but less than 2400
E) At least 2400
30. ? has the following spliced distribution: ²%³ ~ H The mean of ? is 1.25. Find the variance of ? .
A) B) C) D) E) %
%
otherwise
31. A portfolio of insurance policies consists of two types of policies. Policies of type 1 each
have a Poisson claim number per month with mean 2 per period and policies of type 2 each have
a Poisson claim number with mean 4 per period. of the policies are of type 1 and are of type
2. A policy is chosen at random from the portfolio and the number of claims generated by that
policy in the following is the random variable ? . Suppose that a policy is chosen at random and
the number of claims is observed to be 1 for that month. The same policy is observed the
following month and the number of claims is ? (assumed to be independent of the first month's
claims for that policy). Find 7 ´? ~ O? ~ µ.
A) .15
B) .20
C) .25
D) .30
E) .35
32. You are given the following:
- Partial Credibility Formula A is based on the methods of limited fluctuation credibility, with
1600 expected claims needed for full credibility.
- Partial Credibility Formula B is based on Buhlmann's credibility formula with a of 391.
- One claim is expected during each period of observation.
Determine the largest number of periods of observation for which Partial Credibility Formula B
yields a larger credibility factor than Partial Credibility Formula A.
A) 524
B) 526
C) 528
D) 530
E 532
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 4
PE-69
33. Semi-parametric empirical Bayesian credibility is being applied in the following situation.
The distribution of annual losses ? on an insurance policy has pdf ²%O³ ~ %
for  %  , where
has an unknown distribution. A sample of annual losses for 100 separate insurance policies is available.
~
~
It is found that ? ~ and ? ~ .
For a particular insurance policy, it is found that the total losses over a 3 year period is 4.
Find the semi-parametric estimate of the losses in the 4-th year for this policy.
A) Less than 1.1
B) At least 1.1 but less than 1.3
C) At least 1.3 but less than 1.5
D) At least 1.5 but less than 1.7
E) At least 1.7
34. The following random sample of size 7 is drawn from a distribution:
1 , 1 , 2 , 4 , 5 , 6 , 10
The distribution median is estimated, and the estimator used is the sample median.
The following five bootstrap samples, each of size 7 have been simulated from the empirical
distribution of the original random sample.
Sample 1 : 1 , 1 , 1 , 5 , 6 , 6 , 10
Sample 2 : 1 , 2 , 2 , 4 , 6 , 6 , 10
Sample 3 : 2 , 4 , 4 , 5 , 5 , 6 , 6
Sample 4 : 1 , 1 , 4 , 4 , 4 , 10 , 10
Sample 5 : 1 , 2 , 5 , 6 , 6 , 6 , 10
Use these five bootstrap samples to determine the bootstrap approximation to the mean square
error of the sample median estimator.
A) 1.0
B) 1.2
C) 1.4
D) 1.6
E) 1.8
35. The parameter has prior distribution ²³ ~ c b ² c° ³
(mixture of two exponentials). The model distribution ? has a conditional distribution given that is
Poisson with mean . Find the Buhlmann credibility premium if there is a single observation of 0.
A) B) C) D) E) © ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-70
PRACTICE EXAM 4
ACTEX EXAM C/4 - PRACTICE EXAM 4 SOLUTIONS
1. ? ²%³ ~ ²%b³b for %  .
@ ~ ² ?b
S ? ~ ²@ c ³ ~ ²@ ³.
³ ~ ²?³
&
&
c&
@ ²&³ ~ ? ²²&³³ h Z ²&³ ~ ²²&³b
.
³b h ~ ²& ³b h ~ This is the pdf of the exponential distribution with mean .
2. The conditional distribution of ? given \$ has ,´?O\$µ ~ \$ and = ´?O\$µ ~ .
We are given that ,´\$µ ~ = ´\$µ ~ .
? is a continuous mixture with mean ,´?µ ~ ,´ ,´?O\$µ µ ~ ,´\$µ ~ and
= ´?µ ~ = ´ ,´?O\$µ µ b ,´ = ´?O\$µ µ ~ = ´\$µ b ,´µ ~ b ~ .
The continuous mixture of a normal "over" a normal distribution is also normal, so the
unconditional distribution of ? is normal with mean 1 and variance 2.
c
c
The 95-th percentile of ? is , where 7 ²?  ³ ~ 7 ² ?c
j  j ³ ~ )² j ³ ~ À .
From the standard normal table we get c
j ~ À
, so that ~ À .
3. The distribution of ? is a mixture of a discrete point at ? ~ , with mixing weight
7 ²? ~ ³ ~ À, and a continuous uniform distribution on then interval ´Á µ with mixing
weight .1. The pdf of ? on ´Á µ is ? ²%³ ~ ²À³²À³ ~ À
for  %  .
if ?  .
? c if ?  ,´@3 µ ~ ²% c ³ h ? ²%³ % ~ ²% c ³ h ²À³ % ~ À À
,´@3 µ ~ ²% c ³ h ²À³ % ~ Á À
The cost per loss variable is @3 ~ ²? c ³b ~ D
,´@ µ
3
The cost per payment variable @7 has expected value ,´@7 µ ~ 7 ²?³
.
7 ²?  ³ ~ ? ²%³ % ~ À % ~ À .
,´@ µ
3
,´@7 µ ~ 7 ²?³
~ À
À ~ .
,´@ µ
Á
3
,´@7 µ ~ 7 ²?³
~ À ~ Á Á .
= ´@7 µ ~ Á Á c ~ Á .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 4
PE-71
4. The total losses for the day can be formulated as : ~ : b : , where
: is the total losses from smokers and : is the total losses from non-smokers.
The number of approvals in a day has a Poisson mean of 1000. Since 20% are smokers and 80%
non-smokers, the number of approvals of smokers per day is Poisson with mean 200 and for nonsmokers it is Poisson with mean 800. We must assume that application approvals arise
independently, and : is independent of : , so that = ´:µ ~ = ´: µ b = ´: µ .
: has a compound Poisson distribution with ~ (frequency) and severity @ with
,´@ µ ~ c and = ´@ µ ~ . Similarly, : has a compound Poisson distribution with
~ (frequency) and severity @ with ,´@ µ ~ c and = ´@ µ ~ .
Then using the general form of variance of a compound distribution, we have
= ´: µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ ~ ²³ b ² c ³ ~ Á Á ,
and
= ´: µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ ~ ²³ b ² c ³ ~ Á Á .
= ´:µ ~ Á Á b Á Á ~ Á Á .
Note that since : has a compound Poisson distribution, it is also true that
= ´: µ ~ h ,´@ µ ~ h ´= ²@ ³ b ²,´@ µ³ µ ~ ²³´ b ² c ³ µ ~ Á Á (and a similar comment applies to : ).
5. @ has a mixed distribution with a probability mass at @ ~ which has probability
7 ²@ ~ ³ ~ and density function @ ²&³ ~ ² c ³ h c&° for &  .
The likelihood function will be the product of a factor of for each loss that is 0, and a factor of
² c ³ h c&° for each loss that is  . The loglikelihood will be a sum of factors of for
&
each loss that is 0, and ² c ³ c c for each loss that is  .
There are ~ c losses that are  . Let us denote them ' Á ÀÀÀÁ ' (there are the @
values that are  ). The loglikelihood function is
''
3 ~ b ² c ³ c c .
C
C
The mle's of and are found by solving the two equations C
3 ~ and C
3 ~ .
Because of the additive separation of and , it can be seen that the mle of is V ~ b
~ ''
'' (the usual binomial parameter mle) and V ~ ~
. Since the 'Z s are the non-zero @ 's, it
'@
is true that '' ~ '@ , so that V ~ c .
c
6. We denote by : the aggregate amount of payment made by XYZ before the aggregate
deductible of 100 is applied, but after the individual claim deductible of 500 has been applied to
each individual claim for the year. : has a compound Poisson distribution with mean frequency
0.15 (per year) and severity distribution @ (after the individual claim deductible of 500):
Probability
Individual Claim Payment @
0
0.35 (if original loss is 250 or 500)
300
0.30
500
0.25
750
0.05
1,000
0.05
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-72
PRACTICE EXAM 4
6. continued
XYZ receives 150. With an aggregate deductible of 100, XYZ will pay out more than it receives
if :  . This will happen if there is at least one individual loss ? of 800 or more, which is
equivalent to an individual claim @ (after deductible of 500) of 300 or more. Alternatively,
7 ´:  µ ~ c 7 ´:  µ , and :  in one of two cases:
(i) there are 5 ~ claims, or
(ii) there are more than 0 claims, but each one is of amount 0 (after 500 deductible).
B
7 ´:  µ ~ 7 ´:  O5 ~ µ h 7 ´5 ~ µ .
~
We note that 7 ´:  O5 ~ µ ~ and 7 ´:  O5 ~ µ ~ ²À³ (since there is a .35
probability of each loss being  500 and thus having a claim payment of 0).
B
B
²À³
7 ´:  µ ~ ²À³ h 7 ´5 ~ µ ~ ²À³ h cÀ h
B
~ cÀ h ~
~
²ÀdÀ³
[
[
~
~ cÀ h ²ÀdÀ³ ~ cÀ ~ À .
Then, 7 ´:  µ ~ c À ~ À .
7. The empirical cdf has - ²³ ~ À Á - ²³ ~ À . The estimated 60-th percentile
À
cÀ
VÀ
c
~ ÀcÀ
(linear interpolation in the ²c Á µ for which
VÀ
is found from c
- ²c ³  À
 - ² ³ ). Answer: C
VÀ
~ À .
8. Using the notation associated with the Kaplan-Meier approximation for large data sets, we
have ~ Á ~ Á ~ Á ~ Á ~ Á .
~ (there are 400 observations with deductible of 300) , ~ 25 (observations with
deductible of 500) , ~ ~ ~ (there are no observations with deductible higher than
5000). % ~ , % ~ Á % ~ Á % ~ Á " ~ Á " ~ À
c
c
~
~
With ~ À and ~ À, the number at risk at is ~ b À c ²% b " ³ c À" ,
V ² ³ ~ c 4 c
and we use the estimator -
%c
%
5Ä4 c c 5
.
Then with ~ , we have ~ À c À" ~ À²³ c À²³ ~ Á
~ ² b À ³ c ²% b " ³ c À" ~ ´ b À²³µ c ² b ³ c À²³ ~ À Á
~ ² b b À ³ c ²% b % b " b " ³ c À"
~ ² b b ³ c ² b b b ³ c À²³ ~ .
V ²³ ~ c 4 c % 54 c % 54 c % 5
Then ~ c 4 c
54 c À 54 c 5
~ À
.
Note that " ~ is the number of censorings in the interval from ~ to ~ .
There is censoring at 5000, but that is counted in the interval from 5000 to 10,000.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 4
PE-73
9. The pdf of the uniform distribution is , so the likelihood function is 3´µ ~ .
This is maximized at the minimum feasible value for . Since each % must be in the interval
²Á ³ it must be true that  % for each % . Therefore  %¸% ¹ . The minimum
feasible value for is the largest of the %'s, which is V ~ 4 %¸? ¹ . To find the variance of V
we must find the distribution of @ ~ 4 %¸? ¹ and then the variance of V is = ´@ µ . Since
there are 12 %'s we have the cdf for @ is
[email protected] ²&³ ~ 7 ´@  &µ ~ 7 ´4 %¸? Á ÀÀÀÁ ? ¹  &µ
~ 7 ´²?  &³ q ²?  &³ q Ä q ²?  &³µ
&
~ 7 ´?  &µ h 7 ´?  &µÄ7 ´?  &µ ~ ² ³ .
&
The pdf of @ is @ ²&³ ~ [email protected] ²&³ ~ for  &  .
The first and second moments of @ are
&
,´@ µ ~ & h & ²&³ & ~ & ~ ,
&
,´@ µ ~ & h & ²&³ & ~ & ~ .
Then, = ´@ µ ~ ,´@ µ c ²,´@ µ³ ~ c ² ³
~
.
The estimated variance is found using the mle of , which is 53.
²³
The estimate variance is ~ À .
10. For the uniform distribution on the interval ²Á ³ , we have
7 ´?  µ ~ c ~ ² ³ À Using the maximum likelihood estimate for , the estimated
variance of the estimated probability is
=V
´²V³µ ~ ´Z ²V³µ h = ´Vµ ~ ² ³ h = ´Vµ . Using the mle V ~ and = V
´Vµ ~ À
V
from question 5, we get the estimated variance of the estimate of the probability 7 ´?  µ to be
² ³ h ²À³ ~ À
. The estimate of the probability using the mle is
7 ´?
 µ ~ c V
V ~ c ~ À . The approximate 95% confidence interval for
7 ´?  µ is À f À
jÀ
~ ²À Á À³ .
11. ~ ,´?µ ~ %²%³% ~ ,
= ´?µ ~ ,´²? c ³ µ ~ ²% c ³ ²%³% ~ 4 :, ²V³ ~ = ´Vµ b ´ ²V³µ .
°
= ´Vµ ~ = ´ c
? µ ~ ²c³
= ´? µ ~ ²c³ Á
~
~
²V³ ~ ,´Vµ c ~ ,´ c
? µ c ~ ²c³
c ~ ²c³
~
°
b
4 :, ²V³ ~ ²c³ b 4 ²c³
5 ~ ²c³
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-74
PRACTICE EXAM 4
12. The -plot uses the following smoothed empirical distribution function values, and model
probabilities:
%
Empirical
À
À
À
À
Dist. Fn. (left tail)
Model
c c°
c c°
c c°
c c°
Dist. Fn. (left tail)
In order for the right tails of the exponential model to be thinner than those of the (smoothed)
empirical distribution at each data point, we must have
c°  À , and c2°  À
, and and c°  À , and and c°  À .
This translates into the following required inequalities for that must all be satisfied:
 À Á  À Á  À Á  À .
Maximum value of consistent with all inequalities is 3.9 . Answer: B
13. ²%O³ ~ for  %  .
²%Á ³ ~ ²%O³ h ²³ ~ h ~ (on the appropriately defined region for ? and ) .
%
If  %  , then   % , and ? ²%³ ~ ~ c % ~ %%c
.
% %
If %  , then
  % , and ? ²%³ ~ %° ~ c ~ .
Since %  , ? ²%³ ~
%
and ²O%³ ~
%
%
,´O%µ ~ %° h %
²%Á³
? ²%³
~
%
°
°%
~
%
%
%
on the region %   % .
14. We wish to find ,´number of 2nd period claimO1st period claim ~ 100µ . The Bayesian
approach shows that this expectation can be "mixed" or "factored" through the two classes:
,´number of 2nd period claimsO1st period claim ~ 100µ
~ ,´number of 2nd period claimsOfrom class 1µ h 7 ´from class 1O1st period claim ~ 100µ
b ,´number of 2nd period claim 2Ofrom class 2µ h 7 ´from class 2O1st period claim ~ 100µ
From the description of classes 1 and 2, we have
,´number of 2nd period claimsOfrom class 1µ ~ ² ³ b ² ³ ~ Á and
,´number of 2nd period claimsOfrom class 2µ ~ ² ³ b ² ³ ~ À
Using Bayes theorem, we have
7 ´Oclass 1µh7 ´class 1µ
7 ´from class 1O1st period claim ~ 100µ ~ 7 ´Oclass 1µh7 ´class 1µb7 ´Oclass 2µh7 ´class 2µ
~
´² ³² ³b² ³² ³ µ² ³
´² ³² ³b² ³² ³ µ² ³b´² ³² ³b² ³² ³ µ² ³
~ (in the numerator, 7 ´Oclass 1µ is 7 ´1 claimµ h 7 ´µ b 7 ´ claimsµ h ²7 ´µ³ , similarly for
class 2 in the denominator). Also,
7 ´Oclass 2µh7 ´class 2µ
7 ´from class 2O1st period claim ~ 100µ ~ 7 ´Oclass 1µh7 ´class 1µb7 ´Oclass 2µh7 ´class 2µ
~
´² ³² ³b² ³² ³ µ² ³
~ À
´² ³² ³b² ³² ³ µ² ³b´² ³² ³b² ³² ³ µ² ³
Then,
,´number of 2nd period claimsO1st period claim ~ 100µ
³ b ² ³² ³ ~ ~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 4
PE-75
15. We are given ? ~ and ? ~ for the chosen risk.
c
The Buhlmann credibility estimate is A? b ² c A³ , where
~ ,´²#³µ and A ~ b # . In this case, # takes on two "values", class 1 and class 2.
²class 1³ ~ (expected frequency)(expected severity) ~ ² ³² ³ ~ , and
²class 2³ ~ ² ³²³ ~ . Since ²³ is a constant for all , we have ~ = ´²#³µ ~ ,
and this case A ~ . Thus, the Buhlmann credibility estimate is ~ ,´²#³µ ~ .
16. Applying empirical Bayes in the equal sample size case with ~ Á ~ ,
c
V# ~ ²c³ ²? c ? ³ ~ ²³²³ h ´À b À b Àµ ~ À
.
~ ~
When policyholders 1 and 2 are combined, the resulting model has ~ Á ~ ,
~ for ~ Á Á ÀÀÀÁ (combined old policyholders 1 and 2),
c
c
~ Á ~ for ~ Á Á ÀÀÀÁ . ? ~ À as before, and ? ~ À .
c c
²? c ?
²? c ? ³ ~ À b À ~ À Á
³ ~
~
~
~
~
c c
²? c ?
²? c ? ³ ~ À .
³ ~
In this unequal sample size case,
V# ~
c
h ²? c ? ³ ~ b
h ´À b Àµ ~ À
.
² c³ ~ ~
~
V# c V# ~ À
c À
~ c À .
17. This problem involves maximum likelihood estimation with right-censored data.
The likelihood function to be maximized is 3 ~ ²% ³ h ´ c - ²"³µ ,
~
where % Á ÀÀÀÁ % are the non-censored observations, and is the number of right-censored
observations, censored at limit ". In this case, the limit is " ~ , and there are ~ right-censored observations (observations that are 1000 or greater).
For the exponential distribution with mean , the density function is ²%³ ~ h c%°
and the distribution function is - ²%³ ~ c c%° . The likelihood function becomes
3 ~ ² h c% ° ³ h ´c° µ ~ h c²'% ³° h cÁ°
~
~ h cÁ° h cÁ° ~ h c
Á° .
It is generally easier to maximize the log of the likelihood. The resulting value of is still the
maximum likelihood estimate. The natural log of the likelihood is
Á
Á
3 ~ c c Á and 3 ~ c b .
Setting this equal to 0 results in ~ . This is the maximum likelihood estimate of .
The general (and simple) rule for maximum likelihood estimation for the exponential distribution
is that given a data set that may be include left-truncated (after deductible) payments, and rightcensored (limit) payments, the mle of the ground up loss mean is
total amount paid
Á
V ~
number of non-censored observations . in This example that will be ~ .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-76
PRACTICE EXAM 4
18. We are to find 7 ´? ~ O? ~ Á ? ~ µ and ,´? O? ~ Á ? ~ µ .
7 ´? ~ O? ~ Á ? ~ µ ~
7 ´²? ~³q²? ~³q²? ~³µ
7 ´²? ~³q²? ~³µ
.
In part (a) we found 7 ´²? ~ ³ q ²? ~ ³µ ~ À .
We find 7 ´²? ~ ³ q ²? ~ ³ q ²? ~ ³µ in a similar way.
7 ´²? ~ ³ q ²? ~ ³ q ²? ~ ³µ
c c c ~ 2 h 3 h ²À³ b 2 h 3 h ²À³ b 2 h 3 h ²À³ ~ À .
[
[
Then 7 ´? ~ O? ~ Á ? ~ µ ~
[
À
À ~
À .
19. = ´5 µ ~ = ´,´5 O4 µµ b ,´= ´5 O4 µµ
,´5 O4 µ ~ 4 ~ À4 , = ´5 O4 µ ~ 4 ² c ³ ~ À
4 .
,´4 µ ~ ²À³² b Ä b ³ ~ À Á
= ´4 µ ~ ,´4 µc²,´4 µ³ ~ ²À³² b Ä b ³ c ²À³ ~ À .
= ´,´5 O4 µµ ~ = ´À4 µ ~ À= ´4 µ ~ À ,
,´= ´5 O4 µµ ~ ,´À
4 µ ~ À
,´4 µ ~ À À
= ´5 µ ~ À b À ~ À .
20. For an ordinary deductible of 1000, the mean excess loss is
,´²?c³b µ
:²³
~ ,
where :²³ ~ 7 ²?  ³ .
For a franchise deductible, the expected cost per loss is
,´²? c ³b µ b :²³ ~ .
Therefore, ,´²? c ³b µ ~ :²³ , and it follows that
:²³ b :²³ ~ , from which we get :²³ ~ À .
Then, ,´²? c ³b µ ~ ²À³ ~ is the expected cost per loss if the deductible of
1000 is applied as an ordinary deductible. Answer: E
21. The probability generating functions of :Á 5 and ? satisfy the relationship
7: ²!³ ~ 75 ²7? ²!³³ . Then, 7 ²: ~ ³ ~ 7: ²³ ~ 75 ²7? ²³³ .
An ²Á Á ³ distribution must be either Poisson, Negative Binomial or Binomial.
Binomial is the only one of the three with expected value greater than variance.
Therefore, 5 is binomial, say with parameters and .
,´5 µ ~ ~ À and = ´5 µ ~ ² c ³ ~ À ,
and it follows that ~ À and ~ . The probability generating function of 5 is
75 ²'³ ~ ´ b ²' c ³µ ~ ´ b ²À³²' c ³µ .
The mean of : is ,´:µ ~ ,´5 µ h ,´?µ , so that À ~ À,´?µ
and we get ,´?µ ~ À .
The variance of : is = ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ,
so that À ~ À= ´?µ b À²
À³ , and we get = ´?µ ~ .
Since ? is an ²Á Á ³ distribution, it is either Poisson, Negative Binomial or Binomial.
The Negative Binomial distribution is the only one of the three whose mean is less than its
variance. Therefore ? has a negative binomial distribution, say with parameters and .
,´?µ ~ ~ and = ´?µ ~ ² b ³ ~ , and it follows that ~ and ~ .
The probability generating function of ? is 7? ²!³ ~ ´c ²!c³µ
~ ´c²!c³µ .
Then, 7? ²³ ~ ´c²c³µ
~ , and
7 ²: ~ ³ ~ 7: ²³ ~ 75 ²7? ²³³ ~ 75 ² ³ ~ ´ b ²À³² c ³µ ~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 4
PE-77
22. The amount paid is a stop-loss insurance with a deductible of 3 applied to aggregate losses.
The aggregate loss has a compound distribution with an integer-valued severity. Therefore, : is
integer-valued. ,´:µ ~ ,´5 µ h ,´?µ ~ ´²³²À
³ b ²³²À³µ ~ À .
,´²:c³b µ ~ ,´:µ c ,´: w µ .
The geometric distribution with mean ~ has probability function
7 ²5 ~ ³ ~ ²b ³b ~ b .
: is integer valued with probability values 7 ²: ~ ³ ~ 7 ²5 ~ ³ ~ Á
7 ²: ~ ³ ~ 7 ²5 ~ ³ h 7 ²? ~ ³ ~ d ²À
³ ~ Á
7 ²: ~ ³ ~ 7 ²5 ~ ³ h 7 ²? ~ ³ b 7 ²5 ~ ³ h ²7 ²? ~ ³³
~ d ²À³ b d ²À
³ ~ ,
and 7 ²:  ³ ~ c 7 ²: ~ Á Á ³ ~ c ² b b ³ ~ .
prob. Then,
: w~
prob. prob. H
,
prob. so that ,´: w µ ~ ²³² ³ b ²³² ³ b ²³² ³ ~ À
.
,´²: c ³b µ ~ À c À
~ À .
23. The amount paid out by PQR, say > (after deductible) has a compound Poisson distribution
with a mean frequency of ,´5 µ ~ À and the following severity distribution (after deductible):
Individual Claim Payment @
Probability
0
0.35 (if original loss is 250 or 500)
250
0.30
500
0.25
750
0.05
1,000
0.05
The first and second moments of the claim payment @ are
,´@ µ ~ ²³²À³ b ²³²À³ b ²³²À³ b ²³²À³ ~ À , and
,´@ µ ~ ´² ³²À³ b ² ³²À³ b ² ³²À³ b ² ³²À³µ ~ Á À
Then ,´> µ ~ ,´5 µ h ,´@ µ ~ À and = ´> µ ~ ,´5 µ h ,´@ µ ~ Á À (this is a
form of the variance for a compound Poisson distribution). We wish to find 7 ´>  µ using
the normal approximation with continuity correction. This is
7 ´>  ´ ~ c 7 ´>  Àµ ~ c )² ÀcÀ
³ ~ c )²À³ ~ c À
j
~ À .
Á
À
24. If the withdrawal is before the first death, then the amount of failure mass redistributed to
À
each survivor is ~ À , and the estimated : values will decrease by steps of size
À b À ~ À so that : ²! ³ ~ À . If the withdrawal is between the first and third
À
deaths then the amount of failure mass redistributed is either ~ À with : ²! ³ ~ À or
À
V ³ ~ .1149 . If the withdrawal is between the third and fifth deaths then the
~ À with :²!
À
À
amount of failure mass redistributed is either ~ À
with : ²! ³ ~ À
or ~ À
with : ²! ³ ~ .1 . Answer: C
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-78
PRACTICE EXAM 4
25. This is a Pareto distribution with ~ . From the Exam C distribution tables, we have
,´? µ ~
!²b³!²c³
!²³
!²³!²c³
~ c
.
!²³
c
c
?b
c
set c ~ ? S ~ ?
Á so that ,´?µ ~
According to the method of moments, we
26. A) False. See the bottom of page 427 of the Loss Models book.
B) False. The K-S test is applied to individual data.
C) False. See page 430 of the Loss Models text.
D) False. The critical value depends on the number of cells, the number of estimated parameters,
and the level of significance.
27. The likelihood function is ²³ h ²³ h ²³ .
For the Negative Binomial, ²³ ~ ²b ³ , ²³ ~
²b³²b³²b³²b³
h
[
²b ³b .
²b³ ²b³ ²b³²b³
likelihood function is
[ [
²b³²b³
[
h ²b ³b ,
and ²³ ~
The
h ²b ³b .
²6 c, ³
, where there interval groupings for the dataÀ
,
~
28. The chi-square statistic is ~ is the observed number of occurrences in interval grouping . , ~ ´- ² ³ c - ²c ³µ is the
expected number of occurrences of a total sample of size in interval grouping given that the
hypothesized distribution is correct ²- is the distribution function of the hypothesized
distribution, and the 's are the interval endpoints).
In this case, for the ~ interval groupings, we have ~ , with 6 ~ Á 6 ~ Á
6 ~ Á 6 ~ . The distribution function of the hypothesized distribution is
%
% - ²%³ ~ cB ²!³ ! ~ ! ! ~ c % for %  .
Then , ~ ´- ² ³ c - ²³µ ~ ´ c µ ~ Á
, ~ ´- ²³ c - ² ³µ ~ ´ c µ ~ ,
, ~ ´- ²³ c - ²³µ ~ ´ c µ ~ , and
, ~ ´- ²B³ c - ²³µ ~ ´ c µ ~ .
The chi-square statistic is ~
²
c³
²c³
²c³
²c³
b
b
b
~ À.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 4
PE-79
29. For a compound Poisson total claims distribution, the standard for number of exposures for
&
= ´@ µ
full credibility for total claims is ² ³ h < b ²,´@ µ³ = ,
and the standard for expected number of claims for full credibility for total claims is
&
² ³ h < b
= ´@ µ
²,´@ µ³ =.
In this question, ~ À so that & ~ À
is the 97.5 percentile of the
standard normal distribution, and ~ À . Also, @ has a uniform distribution on the interval
Á
À
°
À
² À ³ h < b À
²Á³ =
´ Á Á µ, so that ,´@ µ ~ Á Á = ´@ µ ~
The standard for full credibility for total claims is
~ À À
30. ²%³ % ~ S b ~ Á % ²%³ % ~ À S À b À ~ À À
Solving for and results in ~ À Á ~ À .
Then ,´? µ ~ % ²%³ % ~ , and = ´?µ ~ c ² ³ ~ . Answer: B
7 ´? ~q? ~µ
7 ´? ~µ
7 ´? ~q? ~OType 1µh7 ´Type 1µb7 ´? ~q? ~OType µh7 ´Type µ
7 ´? ~µ
31. 7 ´? ~ O? ~ µ ~
~
~
²c h [ ³ h² ³b²c h [ ³ h² ³
²c h [ ³h² ³b²c h [ ³h² ³
~ À .
Alternatively, 7 ´? ~ O? ~ µ
~ 7 ´? ~ OType 1µ h 7 ´Type 1O? ~ µ b 7 ´? ~ OType µ h 7 ´Type 2O? ~ µ
~ ²c h [ ³²À³ b ²c h [ ³²À³ ~ À
.
32. The "credibility factor" is the value of A . In the partial credibility context,
= ´?µ
A ~ ¸j ( Á ¹ , where ( ~ h ²,´?µ³ . We are told that full credibility, A ~ , is
j
reached with ~ À Therefore, ( ~ , and the partial credibility A is A ~ if
 . The Buhlmann credibility factor is A ~ b
~ b
. In order to have A  A ,
j
we must have b
 , or equivalently, j  b , or equivalently,
c b Á  . The roots of the quadratic equation c b Á ~ are
~ and . Any between 289 and 529 will result in A  A .
Since is an integer, the maximum is 528. Answer: C
33. Hypothetical mean is ,²?O³ ~ % h %
% ~ .
Process variance is = ²?O³ ~ ,²? O³ c ´,²?O³µ À
,²? O³ ~ % h % % ~ S process variance ~ = ²?O³ ~ c ² ³ ~ .
Expected hypothetical mean is ~ ,´?µ ~ ,´,²?O³µ ~ , ² ³ ~ 23 ,´µ ,
Expected process variance ~ # ~ ,´= ²?O³µ ~ ,´ 8 µ ~ 8 ,´ µ .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-80
PRACTICE EXAM 4
33. continued
Variance of hypothetical mean ~ ~ = ´,²?O³µ ~ = ² ³
~ = ²³ ~ ´,² ³ c ²,²³³ µ .
c
From the sample, we can estimate ,²?³ as ? ~ , so this is also the estimate of ,´µ.
The estimate of ,´µ is .
From the sample we can estimate = ²?³ using the unbiased sample estimate,
c
´'? c ? µ ~ ´
c ² ³µ ~ À .
But = ²?³ ~ b # ~ ,´ µ b ´,² ³ c ²,²³³ µ ~ ,´ µ c ²,² ³³ .
Using the estimated variance of ? and the estimated mean of , we have
À ~ ,´ µ c ² ³ , so that the estimate of ,´ µ is À .
Then, # ~ ,´ µ is estimated to be À
, and
~ ´,² ³ c ²,²³³ µ is estimated to be À .
V c b ² c A³
V The estimate of losses in the 4-th year is [email protected]
V
c
V ~ V# ~ À
~ À , and where A
V~?~,
b V
so that
b À
V c b ² c A³
V [email protected]
V ~ ²À³² ³ b ²À³²³ ~ À .
34. The bootstrap estimate of the mean square error of an estimator is found from the bootstrap
samples as follows. We find the parameter in the empirical distribution corresponding to the
parameter being estimated in the actual distribution. Since the original random sample has 7 (an
odd number) points, the median is the middle (4-th) point, ~ . The bootstrap procedure then
generates a number of random samples from the empirical distribution. There are 5 of these
bootstrap samples given. For each of these five samples, we find
´(sample median of bootstrap sample) c µ . The estimated mean square error of the sample
median estimator is the average of these values over all the bootstrap samples.
The estimate of the MSE of the sample median estimator is
´² c ³ b ² c ³ b ² c ³ b ² c ³ b ²
c ³ µ ~ À
35. Hypothetical mean ~ ,´?Oµ ~ and process variance ~ = ´?Oµ ~ .
~ Expected hypothetical mean ~ ,´µ ~ ²³ b ²³ ~ .
~ Variance of hypothetical mean ~ = ´µ ~ ,´ µ c ²,´µ³ À
,´ µ ~ ² d ³ b ² d ³ ~ S ~ c ² ³ ~ .
# ~ Expected process variance ~ ,´µ ~ .
° ~ .
b °
² ³²³ b ² ³² ³
There is ~ observed value, so A ~
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 5
PE-81
ACTEX EXAM C/4 - PRACTICE EXAM 5
1. A large retailer of personal computers issues a warranty contract with each computer that it
sells. The warranty covers any cost to repair or replace a defective computer within the first 30
days of purchase. 40% of all claims are easily resolved with minor technical help and do not
involve any cost to replace or repair. If a claim involves some cost to replace or repair, the claim
size is distributed as a Weibull with parameters ~ and ~ .
Which of the following statements are true?
I. The expected cost of a claim is \$60.
II. The survival function at \$60 is 0.243.
III. The hazard rate at \$60 is 0.012.
A) I only
B) II only
C) III only
D) I and II only
E) II and III only
2. In 2006, annual claim frequency follows a negative binomial distribution with parameters and . follows a uniform distribution on the interval ²Á ³ and ~ À Calculate the probability
that there is at least 1 claim in 2006.
A) Less than 0.85
B) At least 0.85, but less than 0.88
C) At least 0.88, but less than 0.91
D) At least 0.91, but less than 0.94
E) At least 0.94
3. An insurance company offers two policies. Policy R has no deductible and no limit. Policy S
has a deductible of \$500 and a limit of \$3000; this is, the company will pay the loss amount
between \$500 and \$3000. In year !, severity follows a Pareto distribution with parameters ~ and ~ . The annual inflation rate is 6%. Calculate the difference in expected cost per loss
between policies R and S in year ! b .
A) Less than \$500
B) At least \$500, but less than \$550
C) At least \$550, but less than\$600
D) At least \$600, but less than \$650
E) At least \$650
Questions 4 and 5 are based on the following information.
An insurance policy on the loss ? has an ordinary deductible of 40. The policy also has the
following adjustments. If the loss is between 40 and 60, the insurance policy pays the amount of
the loss above 40. If the loss is between 60 and 80, the insurance pays 20 plus 75% of the loss
above 60. If the loss is above 80, the insurance pays 35.
4. If the distribution of ? is uniform on the interval ²Á ³ , find the expected cost per loss .
A) 13.0
B) 13.5
C) 14.0
D) 14.5
E) 15.0
5. Express the cost per loss random variable as a combination of ? and ? w " factors for
appropriate values of ".
A) ²? w ³ b ²? w ³ c ²? w ³
B) À²? w ³ b ²? w ³ c ²? w ³
C) ²? w ³ b À²? w ³ c ²? w ³ D) À²? w ³ b À²? w ³ c ²? w ³
E) ²? w ³ b ²? w ³ c ²? w ³
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-82
PRACTICE EXAM 5
6. The number of calls arriving at a customer service center follows a Poisson distribution with
~ per hour. The length of each call follows an exponential distribution with an expected
length of 4 minutes. There is a \$3 charge for the first minute or an fraction thereof and a charge of
\$1 per minute for each additional minute or fraction thereof. Determine the expected charges in a
single hour.
A) Less than \$375
B) At least \$375, but less than \$500
C) At least \$500, but less than \$625
D) At least \$625, but less than \$750
E) At least \$750
7. A mortality study has right-censored data. The first time at which deaths occur is ! and you
are given the variance of the product-limit estimate at ! and variance of the Nelson-Aalen
estimate of the cumulative hazard rate at time ! are = V
´: ²! ³µ ~ À and
V
=V
´/²! ³µ ~ À. Find the product-limit estimate of the survival probability to time ! .
A) .63
B) .68
C) .73
D) .78
E) .83
8. A company is analyzing failures among it fleet of delivery vehicles. Over the course of 15
weeks, the company observes the following number of breakdowns per week:
Number of Weeks
Number of Breakdowns
0
1
1
2
2
3
3
3
4
3
5
2
6
1
The company is using a binomial distribution with ~ and to model the number of
breakdowns per week. Determine the moment estimate of .
A) Less than .25
B) At least .25, but less than .45
C) At least .45, but less than 65
D) At least .65, but less than .85
E) At least .85
9. Maximum likelihood estimation is being applied to estimate the mean of an exponential loss
distribution. The sample data is based on insurance payments with no deductible, but with a
common policy limit. The data set consists of 50 losses that are below the policy limit.
The MLE is 3.00 . If an additional limit payment is added to the data set the MLE becomes 3.08.
Suppose that an additional insurance payment of 3.5 is added to the original data. Find the new
MLE in this case.
A) Less than 3.1
B) At least 3.1 but less than 3.3
D) At least 3.5 but less than 3.7
E) At least 3.7
C) At least 3.3 but less than 3.5
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 5
PE-83
10. Losses come from an equally weighted mixture of an exponential distribution with
mean , and an exponential distribution with mean .
Determine the least upper bound for the coefficient of variation of this distribution.
A) B) j
C) j
D) E) j
11. Which of the following statements about hypothesis testing are true?
1. A Type I error occurs if / is rejected when it is true.
2. A Type II error occurs if / is rejected when it is true.
3. Type I errors are always worse than Type II errors.
A) 1. only
B) 2. only
C) 3. only
1. and 3. only
E) 2. and 3. only
12. The number of claims in one year for an individual chosen from a portfolio of insurance
policies has a Poisson distribution with mean . The prior distribution for is a Gamma
distribution with mean 0.14 and variance 0.0004 . During the past year a total of 50 claims has
been observed from an insurance portfolio of 310 policies. Determine the variance of the
posterior distribution of .
A) .0001
B) .0002
C) .0003
D) .0004
E) .0005
Questions 13 and 14 relate to the following information. You are given
- a portfolio of independent risks is divided into three classes of equal size
- for each risk in Classes 1 and 2, the probability of exactly one claim during one exposure
period is 1/3, while the probability of no claim is 2/3
- for each risk in Class 3, the probability of exactly one claim during one exposure period
is 2/3, while the probability of no claim is 1/3
- A risk is selected at random from the portfolio. During the first two exposure periods, two
claims are observed for this risk (one in each exposure period).
13. Determine the posterior probability that the risk was selected from Class 3.
A) B) C) D) E) 14. Determine the Buhlmann credibility estimate of the probability that a claim will be observed
for the same risk during the third exposure period.
A) B) C) D) E) 15. You are given that the number of claims for Risk 1 during a single exposure period follows a
Bernoulli distribution with mean . The prior distribution of is uniform on the interval ´ Á µ .
The number of claims for Risk 2 during a single exposure period follows a Poisson distribution
with mean . The prior distribution for has the density function
²³ ~ %² c ³ Á   B Á  .
The loss experience of both risks is observed for an equal number of exposure periods.
Determine all the values of for which the Buhlmann credibility factor A for Risk 2 will be
greater than the corresponding factor for Risk 1.
A) All B)  only
C)  only
D)  only
E)  only
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 5
16. You are given the following table of data for three policyholders over a three year period.
Policy Year S
Policyholder
¨
1
Number of Claims
Average Claim Size
1
2
40
200
50
220
2
Number of Claims
Average Claim Size
100
200
120
200
3
Number of Claims
Average Claim Size
50
200
60
250
3
120
150
Apply the nonparametric empirical Bayes credibility method to find the credibility premium per
claim in the 4th year for Policyholder 2.
A) Less than 185
B) At least 185 but less than 188
C) At least 188 but less than 191
D) At least 191 but less than 194
E) At least 194
17. A binomial distribution with n ~ 3 and p ~ .4 is simulated by the inverse transform method
with the uniform random numbers .31 , .71 , .66 , .48 , .19 . How many of the generated random
variables are equal to 2?
A) 1
B) 2
C) 3
D) 4
E) 5
18. You are given a random sample of two values from a distribution function - :
1
3
2
?2 .
You estimate = (? ) using the estimator (?1 , ?2 ) ~ (? c ? )2 , where ? ~ ?1 b
2
~1
Determine the bootstrap approximation to the mean square error.
(A) 0.0
(B) 0.5
(C) 1.0
(D) 2.0
(E) 2.5
19. You have observed the following claim severities:
11.0
15.2
18.0
21.0
25.8
You fit the following probability density function to the data:
(%) ~ 1 %4c 21% (% c )2 5, %  0,  0
j 2 %
Determine the maximum likelihood estimate of .
A) Less than 17
B) At least 17, but less than 18
C) At least 18, but less than 19
D) At least 19, but less than 20
E) At least 20
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 5
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20. For a certain insurance company, 60% of claims have a normal distribution with mean 5,000
and variance 1,000,000. The remaining 40% have a normal distribution with mean 4,000 and
variance 1,000,000. Calculate the probability that a random selected claim exceeds 6,000.
A) Less than 0.10
B) At least 0.10, but less than 0.15
C) At least 0.15, but less than 0.20
D) At least 0.20, but less than 0.25
E) At least 0.25
21. Lucky Tom deposits the coins he finds on the way to work according to a Poisson process
with a mean of 22 deposits per month.
5% of the time, Tom deposits coins worth a total of 10.
15% of the time, Tom deposits coins worth a total of 5.
80% of the time, Tom deposits coins worth a total of 1.
The amounts deposited are independent, and are independent of the number of deposits.
Calculate the variance of the monthly deposits.
A) 180
B) 210
C) 240
D) 270
E) 300
22. You are given the following random sample of 20 losses from the random variable ? :
20 , 22 , 31 , 38 , 42 , , 57 , 58 , 61 , 67 , 68 , 72 , 72 , 77 , 83 , 84 , 85 , 91 , 96 , 97
You are given that   .
The smoothed empirical estimate of the 25-th percentile is found, say (.
The data is now grouped into the following intervals:
² Á µ Á ² Á µ Á ² Á µ Á ²
Á µ Á ² Á µ .
The empirical estimate of the cdf of ? is also found based on the ogive of the grouped data set,
and the estimate of the 25-th percentile is ) .
Find the value of for which ( ~ ) .
A) 45
B) 48
C) 51
D) 54
E) 57
23. The mortality of a group of 20 people with a certain genetic defect is being compared to the
mortality of a reference population. The following data is for the 20 people. There is no
censoring the set of data.
Month
Number of Deaths During Month
1
2
2
1
11
1
14
1
22
2
24
1
The reference population has constant hazard rate of per year. Find the value of that results
in the same cumulative hazard function for the reference population and the group under study
over the two year period. Estimation for the group of 20 is based on the Nelson-Aalen method.
A) .20
B) .22
C) .24
D) .26
E) .28
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 5
24. A uniform kernel with bandwidth is used to estimate ²%³, the pdf of ? , based on a
random sample of size from the distribution of ? . The kernel density estimator is
V ²%³ . A new sample value becomes available, and the kernel density estimator is
reformulated based on b sample values. The new kernel density estimator
is V b ²%³. Find the minimum possible value of V b ²%³ c V ²%³ over all values of %.
A) c ²b³
B) c ²b³
C) D) ²b³
E) ²b³
25. When Mr. Jones visits his local racetrack, he places three independent bets. In his last 20
visits, he lost all of his bets 10 times, won one bet 7 times, and won two bets 3 times. He has
never won all three of his bets. Calculate the maximum likelihood estimate of the probability that
Mr. Jones wins an individual bet.
A) 13/60
B) 4/15
C) 19/60
D) 11/30
E) 5/12
26. The random variable ? has the density function
²%³ ~ %² c %°³ Á  %  B Á  V based on a large sample of data.
is estimated by the maximum likelihood estimator V³ .
The probability that ? is greater than is estimated by the estimator %² c °
Find the approximate variance of the estimator for the probability that ? is greater than .
V³
V³
V³
A) c° h ² ³ h = ²
B) c° h ² ³ h = ²
C) c° h ² ³ h = ²
V³
D) c° h ² ³ h = ²
V³
E) c° h ² ³ h = ²
27. Summary statistics are given for a sample of 100 losses:
Interval
Number of Losses Sum
Sum of Squares
(0 , 2000]
39
38,065
52,170,078
(2000 , 4000]
22
63,816
194,241,387
(4000 , 8000]
17
96,447
572,753,313
(8000 , 15000] 12
137,595 1,628,670,023
(15000 , B)
10
331,831 17,906,839,238
Total
100
667,754 20,354,674,039
A study was conducted on a different data set, and the data was fit to a Pareto distribution. The
estimated parameters were V ~ À Á V ~ Á . Determine the chi-square statistic for a test
(using 5 groups) to assess the acceptability of fit of this data to those parameter values.
A) .70
B) .75
C) .80
D) .85
E) .90
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28. Aggregate claims per period are assumed to follow a compound Poisson distribution with
Poisson frequency parameter ~ and a per claim severity distribution that is gamma with
parameters ~ and ~ . The standard for full credibility of aggregate claims per period
based on number of exposure periods is , where the standard has been selected so that aggregate
claims per period will be within 10% of expected aggregate claims per period 99% of the time. A
similar situation is considered but in which , , are all doubled. Find the standard for full
credibility of aggregate claims per period based on number of exposure periods in the new
situation.
A) B) C) D) E) 29. You are given:
A portfolio consists of 150 independent risks, 100 of the risks each have a policy with a
\$100,000 per claim policy limit, and 50 of the risks each have a policy with a \$1,000,000 per
claim policy limit. The risks have identical claim count distributions. Prior to censoring by
policy limits, the claim size distribution for each risk is as follows:
Claim Size Probability
\$10,000
1/2
\$50,000
1/4
\$100,000
1/5
\$1,000,000 1/20
A claims report is available that shows actual claim sizes incurred for each policy after censoring
by policy limits, but does not identify the policy limit associated with each policy. The claims
report shows exactly three claims for a policy selected at random. Two of the claims are
\$100,000, but the amount of the third is illegible. Determine the expected value of this illegible
number (nearest 100).
A) 53,000
B) 53,100
C) 53,200
D) 53,300
E) 53,400
30. A portfolio of Risks is equally divided in two types of risks, Type A and Type B. For a risk
of Type A, the claim frequency per period is Poisson with a mean of 2. For a risk of Type B, the
claim frequency per period is Poisson with mean where is uniformly distributed between 0
and 2. Find the Buhlmann factor for this portfolio of risks.
A) B) C) D) E) 31. The number of losses arising from b individual insureds over a single period of
observation is distributed as follows:
Number of Insureds
Number of Losses
0
1
3
2
1
3 or more
0
The number of losses for each insured follows a Poisson distribution, but the mean of each such
distribution may be different for individual insureds. You estimate the variance of the
hypothetical means using Empirical Bayes semiparametric estimation. Determine all values of for which the estimate of the variance of the hypothetical means will be greater than 0.
A) All B)  only
C)  only
D)  only
E)  only
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 5
32. A “mixed exponential" distribution has P[ X ~ 0 ] ~ p (0  p  1) and density function
f(x) ~ (1 c p) h h ecx for x  0. The distribution can be simulated by the inverse transform
method as follows:
if U is a random uniform (0,1) value and U  p , then X ~ 0 ; if p  U  1 then X ~
1cp
1cp
U
A) c h ln(1 c U) B) c 1 h ln(1 c U) C) h ln4 1cU 5 D) 1 h ln4 1cU 5 E) h ln4 11c
cp 5
33. A biologist is monitoring the number of cells generated by a single mutant cell over the
course of an hour. The single cell divides into two cells, then each of those two cells divide into
two, etc. After 5 subdivisions in one hour, there are 5 cells in the population. The mutant cell is
obtained from a sample of mutant cells, each of which has its own rate of subdivision. The
biologist's model for cell division behavior from this sample is that for a randomly chosen cell,
the distribution of 5 (number of subdivisions in one hour) has a Poisson distribution with mean
\$, where \$ has a gamma distribution with mean 10 and variance 5. Find the expected number of
cells in a population generated in one hour by a randomly chosen mutant cell.
A) Less than 1 million
B) At least 1 million but less than 1.25 million
C) At least 1.25 million but less than 1.5 million
D) At least 1.5 million but less than 1.75 million
E) At least 1.75 million
34. You fit an exponential distribution to the following data:
1000
1400
5300
7400
7600
Determine the coefficient of variation of the maximum likelihood estimate of the medianÀ
A) 0.33
B) 0.45
C) 0.70
D) 1.00
E) 1.21
35. For a risk, you are given:
(i) The number of claims during a single week has a Poisson distribution with mean \$.
(ii) The prior distribution of \$ is gamma with parameters ~ Á ~ .
(iii) Claims experience is available for 6 weeks.
(iv) Based on the claims experience for the 6 weeks, the Bayesian predictive expected
number of claims per week is 4.
An additional 4 weeks of claims experience becomes available. Based on the claims experience
for all 10 weeks, the Bayesian predictive expected number of claims per week is 6. Find the
Bayesian predictive expected number of claims per week if only the additional 4 weeks of
experience is used with the original prior distribution.
A) less than 4
B) at least 4 but less than 5
C) at least 5 but less than 6
D) at least 6 but less than 7
E) at least 7
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 5
PE-89
ACTEX EXAM C/4 - PRACTICE EXAM 5 SOLUTIONS
1. The claim amount ? is a mixture of @ =0 with weight .4, and a Weibull distribution @
( ~ Á ~ ³ with weight .6.
I. The expected value of the Weibull is h !² b ³ . For ~ and ~ , we have the
expected value of the Weibull is h !²³ ~ . Since only 60% of claims result in a Weibull
cost, the expected cost of a claim is ²À
³²
³ ~ .
I is false .
II. The survival function for claim amount at 60 is
°
7 ´?  µ ~ ²À³7 ´@  µ b ²À
³7 ´@  µ ~ b ²À
³c²
°³ ~ À
. False.
²À
³@ ²
³
²
³
III. The hazard rate at \$60 is :²
³ ~ ²À
³: ²
³ ~
@
III is true.
²
³ c c²
°³ °
c²
°³
~
² ³²
c° ³
°
~ À .
2. 5 is a continuous mixture distribution. 7 ²5  ³ ~ c 7 ²5 ~ ³ .
7 ²5 ~ ³ ~ 7 ²5 ~ O ³ ² ³ ~ ²b ³ h ~ ²b ³ h ( ² ³ is the pdf of the uniform distribution on ²Á ³, which is ).
h ~ ~ h h ´ c µ ~ À
.
²b ³
²b ³
Then, 7 ²5  ³ ~ c À
~ À .
3. Inflation will have increased cost by a factor of ²À
³ ~ À
by year ! b .
The Pareto distribution is a scale distribution with scale parameter , so the loss random variable
@ in year ! b is Pareto with ~ and ~ À
²³ ~ . The expected cost per
loss in year ! b for Policy R is just the expected loss ,´@ µ ~ c ~ . Policy S has a
deductible of 500 and a maximum covered loss of 3000, so the amount paid under Policy S is
²@ w ³ c ²@ w ³ . The expected cost per loss under Policy S is
c
c
,´@ w µ c ,´@ w µ ~ µ c µ ~ .
c ´ c ² b ³
c ´ c ² b ³
The difference in cost per loss between Policy R and Policy S is c ~ .
?  ? c  ?  4. Cost per loss random variable is @3 ~ H
b À²? c ³  ?  ?  ,´@3 µ ~ ²% c ³²À³ % b ´ b À²% c ³µ²À³ % b ²À³ %
~ b À b ~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 5
?  ? c  ?  5. Amount paid by insurer (cost per loss) is @3 ~ H
b À²? c ³  ?  ?  ?  ²? c ³b ~ H
is valid for @3 up to ? ~ .
? c ?  If we subtract À²? c ³b , we get ²? c ³b c À²? c ³b
?   ?  ~ H ? c ²? c ³ c À²? c ³ ~ b À²? c ³
?  If we subtract À²? c ³b , we get
²? c ³b c À²? c ³b c À²? c ³b
?  ? c  ?  ~H
b À²? c ³
 ?  b À²? c ³ c À²? c ³ ~ ?  This is the cost per payment made by the insurance.
This is ²? c ³b c À²? c ³b c À²? c ³b
~ ? c ²? w ³ c À´? c ²? w ³µ c À´? c ²? w ³µ
~ À²? w ³ b À²? w ³ c ²? w ³
6. The charge for a call can be described as follows:
Call time ; (min.)
; 1
; 
; 
Call charge \$
3
cÀ
cÀ
c²À³ c²À³
c²À³
Prob.
c
c
c
...
...
In general, for a call which lasts for ; , with  ;  b , the charge is b , and the
probability is c²À³ c c²b³²À³ . The expected charge for a call is
B
B
² b ³²c²À³ c c²b³²À³ ³ ~ b ² c cÀ ³ h h , where ~ cÀ .
~
~
B
This is true since
B
²c²À³ c c²b³²À³ ³ ~ .
~
cÀ
Then, h ~ b b b Ä ~ ²c³ ~ ²c
cÀ ³ , so that the expected charge per
~
cÀ
call is b ² c cÀ ³ h ²c
cÀ ³ ~ À .
The expected charge in one hour with an expected number of 100 calls is 652. Answer: D
7. = V
´: ²! ³µ ~ ² c ³ h ² c ³ ~
=V
´: ²! ³µ
² c ³
from Greenwood's formula and
À
V ³µ ~ , so that
=V
´/²!
V ³µ ~ c ~ : ²! ³ ~ À ~ À . Answer: D
=V
´/²!
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 5
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8. There are 15 observed values, one '0' , two '1's, etc. The sample mean of the 15 observations is
²³²³b²³²³b²³²³b²³²³b²³²³b²³²³b²³²
³
~.
We set this equal to the expected number of breakdowns per week using the model distribution,
~ S V ~ À .
9. For maximum likelihood estimation applied to the exponential distribution with rightcensoring at ", if a data set has uncensored data points (say % Á ÀÀÀÁ % ) and data points
% b "
censored at ", the MLE of the exponential mean is ~ uncensored observations). We are given the following:
% b "
~
(the denominator is the number of
% b ²b³"
~ and ~ ~ À
From this we see that " ~ .
Therefore, the insurance payment of 3.5 is below the policy limit, so if it is added to the original
% b "
data set, the new MLE is
~
~
% b À b "
~
.
% b "
From
~
~ we know that % b " ~ .
~
% b À b "
Therefore,
~
~ bÀ
~ À is the new MLE.
10. The mean will be ² ³² b ³ and the variance will be
² ³² b ³ c ´² ³² b ³µ .
The coefficient of variation is the ratio of standard deviation to mean.
The square of the coefficient of variation is
² ³² b ³c´² ³² b ³µ
c b
~
² b ³
´² ³² b ³µ
~
b
b c ~ c ² b
.
² b ³
³
The maximum square of the coefficient of variation is 3, and it occurs at the minimum of
² b ³ , which is 0 (if or is 0).
The least upper bound of the coefficient of variation is j .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 5
11. I. True. This is the definition of Type I error.
II. False.
III. False. It is generally the case that a small probability of Type I error is associated with a large
probability of Type II error, and vice-versa.
12. This is the Gamma prior-Poisson model pair, with ~ observations, and
% ~ observed claims (we are not given individual % 's). The prior gamma distribution has
~
mean ~ À and variance ~ À , so that ~ À and ~ .
The posterior distribution is also a gamma distribution with Z ~ b '% ~ b ~ À
and Z ~ b ~ ²À³b
~ À .
The variance of the posterior gamma distribution is ²Z ³ h Z ~ À . Answer: B
7 ´ÁO*µh7 ´*µ
13. 7 ´*OÁ µ ~ 7 ´ÁO*µh7 ´*µb7 ´ÁO*µh7 ´*µb7 ´ÁO*µh7 ´*µ
~
²°³ ²°³
²°³ ²°³b²°³ ²°³b²°³ ²°³
~ À
14. Each Class has the same probability of of being chosen.
The hypothetical (mean) claim occurrence probabilities from the three classes are
/4 ~ /4 ~ Á /4 ~ . Then the collective premium is
~ ,´/4 µ ~ ´ b b µ ~ À The process variances of the claim probability in each of
the classes are Bernoulli distribution variances:
7 = ~ 7 = ~ ² ³² ³ ~ Á 7 = ~ ² ³² ³ ~ , so that
# ~ ,´7 = µ ~ ´ b b µ ~ À
~ = ´/4 µ ~ ,´/4 µ c ²,´/4 µ³ ~ ´ b b µ c ² ³ ~ .
°
With ~ and ~ # ~ ° ~ , the Buhlmann credibility factor is A ~ b
~ .
A claim has occurred in each of the first two periods, so ? ~ ? ~ (claim count for each
c
exposure period), and ? ~ . The Buhlmann estimate of the probability that a claim will occur
c
in the third exposure period is A? b ² c A³ ~ ² ³²³ b ² ³² ³ ~ . Answer: C
15. The Buhlmann credibility (factor) is A ~ b
~ b
# (based on exposure periods).
For risk 1, has a uniform distribution on ´Á µ, so ²³ ~ , and ? has the Bernoulli
distribution with mean , so the hypothetical mean is ²³ ~ ,´?Oµ ~ (and the collective
(pure) premium is ~ ,´²³µ ~ ).
Then #²³ ~ = ´?Oq ~ µ ~ ² c ³ Á
# ~ ,´#²³µ ~ #²³ h ²³ ~ ² c ³ ~ ,
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 5
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15. continued
~ = ´²³µ ~ = ´µ ~ (variance of the uniform distribution on ´Á µ).
Then A ~ b
# ~
b .
For Risk 2, ²³ ~ c for  is the prior density ( has an exponential distribution with
mean ). ? has a Poisson with mean , so that ²³ ~ ,´?Oµ ~ (and
~ ,´µ ~ c ~ , the mean of the exponential distribution).
Then #²³ ~ = ´?O# ~ µ ~ S # ~ ,´#²³µ ~ ,´µ ~ .
B
~ = ´²#³µ ~ = ´#µ ~ c c ² ³ ~ (variance of the exponential
distribution). Then A ~ b
# ~
b .
B
Then A  A is equivalent to b
 b , which is equivalent to  .
²³²³b²³²³
c
16. ? ~
~ À Á ~ b
²³²³b²³²³b²³²³
c
? ~
~ À Á ~ bb
²³²³b²
³²³
c
? ~
~ À Á ~ b
²³²³b²³²³b²³²³b²³²³b²³²³b²³²³b²
³²³
c
~ À
V~?~
bb
~ .
V# ~ c h ´² c À³ b ² c À³ µ ~ À Á
V# ~ c h ´² c À³ b ² c À³ b ² c À³ µ ~ Á À Á
V# ~ c h ´² c À³ b ² c À³ µ ~ Á À
V# b#
V b#
V
~ Á À .
V# ~ bb
c
c
~
h ´ ²? c ? ³ c V#² c ³ µ
V
c ~
~
~ c ²b b ³
d ´ ´²À c À³ b ²À c À³ b ²À c À³ µ c Á À²³µ
~ À .
The credibility premium for policyholder 2 is
V c b ² c A³
V A?
V~²
ÁÀ ³ h ²À³ b ² c
b
À
³²
À³
b ÁÀ
À
~ À .
17.
s:
0
1
2
3
(s)
.216
.432
.288
.064
- (s)
.216
.648
.936
1.00
A binomial value of 2 will be simulated by a uniform (0,1) value that is both greater than or equal
to .648 and less than .936 . Thus, the uniform numbers .71 and .66 result in simulated binomial
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 5
18. Since the sample consists of only 2 sample points, there are only 4 possible, equally likely
bootstrap samples: (1) Á (2) Á (3) Á (4) Á .
We are trying to estimate = ²?³ , so we find the variance of the empirical distribution.
The empirical distribution is a 2-point random variable and its variance is
´² c ³ b ² c ³ µ ~ .
The bootstrap estimate of the MSE of the estimator V is ,´²V c ³ µ .
2
To find this expectation we first find V ~ (? c ? )2 for each of the samples:
~1
(1) Á S V ~ (2) Á S V ~ (3) Á S V ~ (4) Á S V ~ .
The bootstrap estimate is the average of the values of ²V c ³ for the samples:
´² c ³ b ² c ³ b ² c ³ b ² c ³ µ ~ . Answer: C
19. The log of the pdf is ²%³ ~ j c j% c
and the derivative of ²%³ is
²%³
~
²%c³
%
%c
% ~c % .
The derivative of the loglikelihood function with data points is
3²³ ~ ' ²% ³ ~ '² c % ³ ~ c ´ % b Ä b % µ
~ c ´ À
b À
b À
b À
b À
µ.
Setting 3²³ ~ and solving for results in V ~ À .
Notice that the loglikelihood function can be written as
²% c³
3²³ ~ c '´ µ , where ~ c h j c ²& % ³ .
%
²% c³
Since ~ , it follows that we only need to maximize c '´ % µ .
In general, suppose that a density function involves a parameter , and we are applying maximum
likelihood estimation to estimate . If the density function is of the form
²%³ ~ ²%³ h ²%Â ³ , where ²%³ does not involve the parameter , then for the data set
% Á % Á ÀÀÀÁ % , the likelihood function is 3²³ ~ ´²% ³µ h ´²% Â ³µ , which is proportional
to ²% Â ³ . To find the mle of , we need only maximize ²% Â ³ ~ ' ´²% Â ³µ .
20. Let ? denote the normal claim random variable with mean 4,000 and variance 1,000,000, and
let @ denote the normal claim random variable with mean 5,000 and variance 1,000,000.
A randomly selected claim will have claim size A that is a mixture of ? and @ with mixing
weight .6 for ? and .4 for @ . Then
7 ´A  Á µ ~ ²À
³7 ´?  Á µ b ²À³7 ´@  Á µ .
?cÁ
ÁcÁ
7 ´?  Á µ ~ 7 ´ j
 j
µ ~ c )²³ ~ À .
7 ´@  Á µ ~
ÁÁ
@ cÁ
7´j
ÁÁ

ÁÁ
ÁcÁ
jÁÁ µ
~ c )²³ ~ À .
Then 7 ´A  Á µ ~ ²À
³²À³ b ²À³²À³ ~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 5
PE-95
21. The number of deposits per month (frequency), 5 , is Poisson with mean 22. The amount of
each deposit (severity), ? , has a 3-point distribution, 7 ´? ~ µ ~ À Á 7 ´? ~ µ ~ À Á
7 ´? ~ µ ~ À . Aggregate monthly deposits, : , has a compound Poisson distribution with
variance = ´:µ ~ h ,´? µ ~ ²³´²³²À³ b ²³²À³ b ²³²À³µ ~ . Answer: B
22. With ~ data points, the smoothed percentiles of the given data points are
%
Smoothed Percentile
~ À
~ À
Å
Å
~
À
~ À
Å
Å
Since À is between ~ À and ~ À , the smoothed empirical estimate of
the 25-th percentile will be between % ~ and % ~ . The smoothed empirical 25-th
percentile ( is found by linear interpolation. The proportion of the way that ( is between
% ~ and % ~ is the same proportion of the way that À is between ~ À and
~
À
,
which
is
the
same
proportion
of
the
way
that
À
is
between
and (we get
these by multiplying À , À and À by ). We see that À is of the way
from to , and therefore ( is of the way from % ~ to % ~ . Therefore, ( ~ b c
is the smoothed empirical estimate of the 25-th percentile.
Based on the grouping of the data, there is 1 observation in ² Á µ so the empirical estimate of
the cdf at % ~ is ~ À . There are 3 observations in the interval ² Á µ , for a total
of 4 observations in the interval ² Á µ , so the empirical estimate of the cdf at % ~ is
~ À À There are 4 observations in the interval ² Á µ , for a total of 8 observations in the
interval ² Á µ , so the empirical estimate of the cdf at % ~ is ~ À À
The ogive is based on interpolating between interval endpoints. The empirical estimate of the
25-th percentile ) is the linearly interpolated point between 40 and 60, since the empirical
estimate of - ²³ is .20 and the empirical estimate of - ²³ is .4 . Therefore, since .25 is of
the way from .20 to .40, ) must be of the way from 40 to 60, so ) ~ .
In order to have ) ~ ( , we must have b c
~ , so that ~ . Answer: D
23. The estimated CHF to time 24 months of the group under study is
b b b b b ~ À .
With reference hazard rate per year for 2 years, (24 months), the reference CHF over the 2
years is . Therefore, ~ À S À .
24. The empirical distribution for the sample of size assigns a probability of
to each
of the original sample values, and
to each sample value in the random sample of size
c. The new estimator is V b ²%³ ~ b
h V ²%³ b b
h %b ²%³ .
b
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-96
PRACTICE EXAM 5
24. continued
Then
V b ²%³ c V ²%³ ~ h V ²%³ b h %b ²%³ c V ²%³
b
b
~ b
h ´%b ²%³ c V ²%³µ .
The maximum possible value of V ²%³ is and the minimum possible value of %b ²%³
is 0. The minimum value of the difference will occur if %b ²%³ ~ and V ²%³ ~ ,
and the difference is b
h ´ c µ ~ c ²b³
in that case.
25. Suppose that is the probability of winning an individual bet. This is a binomial success
probability. When the number of trials of binomial distribution is given, say , the maximum
likelihood estimate of the binomial parameter is the same as the moment estimator of , which
is the proportion of successes in the trials. In this example ~ , since Mr. Jones places 60
bets in total (20 visits to the race track, 3 bets on each visit). Mr. Jones wins 13 of the bets (he
won one bet on 7 visits, and 2 bets on 3 visits, for a total of ²³ b ²³ ~ ).
The mle of is Answer: A
.
26. Given a function ²³, if V is the maximum likelihood estimate of then the variance of ²V³
is ´Z ²V³µ h = ²V³ . In this case, ²³ ~ c° , so that Z ²³ ~ c° h ² ³ .
V³µ ~ ´c° h ² ³µ h = ²
V³ ~ c° h ² ³ h = ²
V³ .
Then = ´²
Although not required in this question, it is possible to find the variance of the mle estimate of the
V . For an exponential distribution, the mle based on a random sample is
exponential parameter V ~c
V in this case is the variance of the sample mean,
the sample mean, % . The variance of V
c ~ = ²?³ ~ . This would be estimated as which is = ²%³
.
²O c, ³
27. The chi-square statistic is 8 ~ ,
~
, where is the observed number of
losses in loss interval ²c Á µ and , is the expected number of losses in the interval,
, ~ ´- ² ³ c - ²c ³µ . The distribution function for the Pareto distribution with parameters
and is - ²%³ ~ c ² %b
³ . In this problem, ~ , so that
, ~ ´ c ² b
³ c ² c ² cb ³ ³µ ~ ´² cb ³ c ² b ³ µ.
With ~ Á ~ Á ~ Á ~ Á ~ Á Á ~ B and V ~ À and
V ~ Á , we have , ~ ´² Á ³À c ² Á ³À µ ~ À
Á
bÁ
bÁ
Á
Á
Á
Á
, ~ ´² bÁ ³À c ² bÁ ³À µ ~ À Á
, ~ ´² bÁ ³À c ² bÁ ³À µ ~ À Á
Á
Á
, ~ ´² bÁ ³À c ² ÁbÁ ³À µ ~ À Á
Á
, ~ ´² ÁbÁ ³À c ²³À µ ~ À À
²c
À
³
8~
À
b
²cÀ³
À
b
²cÀ³
À
b
²cÀ³
À
b
²cÀ³
À
~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 5
PE-97
28. The standard based on number of exposures for the compound Poisson distribution : is
= ´:µ
= ´@ µ
h ²,´:µ³ ~ h < b ²,´@ µ³ = .
For the gamma distribution @ , we have ,´@ µ ~ and = ´@ µ ~ .
The standard becomes h < b ²
³ = ~ h < b = ~ h < b = ~ ~ .
If , , and are doubled, then the standard becomes
h < b = ~ h < b = ~ ~ .
29. For claims with a 100,000 per claim limit, the claim distribution is
Claim Size after limit 10,000
50,000
100,000
Prob.
1/2
1/4
1/4
For claims with a 1,000,000 per claim limit, the claim distribution is as stated in the problem. We
wish to find the expected claim (after limit) of the third claim on a policy given that the first two
claims (after limit) are for amount 100,000 . This is
,´? O? ~ ? ~ Á µ
~ ,´? O2 limitµ h 7 ´2 limitO? ~ ? ~ 2µ
b ,´? O4 limitµ h 7 ´4 limitO? ~ ? ~ 2µ .
From the original distribution of claim amount ? , we have
,´? O2 limitµ ~ 2 ² ³ b 2 ² ³ b 2 ² ³ ~ Á Á and
,´? O4 limitµ ~ 2 ² ³ b 2 ² ³ b 2 ² ³ b 4 ² ³ ~ Á .
Using Bayesian probability relationships, we have
7 ´2 limitO? ~ ? ~ 2µ
7 ´? ~? ~2O2 limit´h7 ´2 limitµ
~ 7 ´? ~? ~2O2 limit
µh7 ´2 limitµb7 ´? ~? ~2O4 limit´h7 ´4 limitµ
~
² ³ ² ³
~ À
² ³ ² ³b² ³ ² ³
Then,
7 ´4 limitO? ~ ? ~ 2µ ~ c 7 ´2 limitO? ~ ? ~ 2µ ~ (or we could have found this probability by the same conditioning relationships).
The expectation in question becomes
,´? O? ~ ? ~ Á µ ~ ²Á ³² ³ b ²Á ³² ³ ~ Á . Answer: E
30. The hypothetical means are
²(³ ~ ,´?O(µ ~ , ²)³ ~ ,´?O)µ ~ ,´,´?O)µOµ ~ ,´µ ~ .
The process variances are
#²(³ ~ = ´?O(µ ~ Á
#²)³ ~ = ´?O)µ ~ ,´= ´?O)µOµ b = ´,´?O)µOµ ~ ,´µ b = ´µ
²c³
~ c
b ~ .
,7 = ~ # ~ ²³² ³ b ² ³² ³ ~ , = /4 ~ ~ ²³ ² ³ b ²³ ² ³ c ² ³ ~ À
°
~ # ~ ° ~ Answer: B
À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-98
PRACTICE EXAM 5
31. The variance of the hypothetical means is ~ = ´²#³µ . When estimation is based on
Empirical Bayes semiparametric estimation, we have the following relationships.
²#³ ~ ,´?O#µ is the hypothetical mean, #²#³ ~ = ´?O#µ is the process variance and
# ~ ,´#²#³µ . We then use the fact that = ´?µ ~ ,´#²#³µ b = ´²#³µ ~ # b to get ~ = ´?µ c # . With semiparametric estimation, we do not know the distribution of #,
so we cannot find # ~ ,´#²#³µ directly. In order to estimate , we estimate = ´?µ and #, so
that the estimate of is V
~ =V
´?µ c V# .
= ´?µ is usually estimated using the unbiased form of the sample variance.
The estimate V# depends on the nature of the conditional distribution of ? given #. In the case
that ? is Poisson with mean and variance #, #²#³ ~ ,´?O#µ ~ # , so that the estimate of # is
the estimate of #, which is the mean of ? . So we use the sample mean of the ? 's to estimate #:
c
#
~ =V
´?µ c V# À
V ~ ? ~ b . The estimate of is then V
Using the unbiased estimate of = ´?µ, we get
=V
´?µ ~ b
´² c b
³ b ² c b
³ b ² c b
³µ
b²c³ b²b³
= and
²b³
b²c³ b²b³
=V
´?µ c V# ~ b
<
= c b
²b³
~ b
<
~
V
, which is positive if  .
32. For x  0 , F(x) = P[ X  x ] = P[ X = 0 ] + P[ 0  X  x] = p + (1 c p) h h e–t dt
x
–x
= p + (1 c p)(1 c e ) . Thus, F(x) = H b ² c ³² c c% ³
Thus, for U  p , the inversion method implies that
1cp
U = F(x) = p + (1 c p)(1 c e–x ) S x = 1 h ln4 1cU 5 . Answer:
if %  if % ~ if %  D.
33. We are asked to find ,´5 µ , where 5 has a Poisson distribution with mean \$, and \$ has a
gamma distribution with mean ~ and variance ~ . Therefore, ~ and
~ À. We use the double expectation rule, ,´5 µ ~ ,´ ,´5 O\$µ µ .
Since 5 O\$ is Poisson, we see that ,´5 O\$µ is the probability generating function of 5 , which is
75 O\$ ²³ ~ ,´5 O\$µ ~ \$ . Then, ,´5 µ ~ ,´\$ µ , which is the moment generating function of
\$, 4\$ ²³ ~ ,´\$ µ ~ ²c³ ~ ²cÀ³
~ Á Á . This is the expected number of
~ cells after one hour. Note that the pgf of the Poisson distribution and the mgf of the gamma
distribution are in the Exam C table. Answer: B
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 5
PE-99
34. Given a random sample of date, the mle of the mean of the exponential distribution is the
c
sample mean of the random sample, V ~ ? . The mean of the mle is
c
,´Vµ ~ ,´?µ ~ , and variance of the mle in this case is
= ´?µ
c
= ´Vµ ~ = ´?µ ~
~
.
The median of the distribution is ~ , so the mle of the median is V .
The coefficient of variation of the median mle V V is
k= ´V µ
,´V µ
~
j² °³² ³
in this formulation..
~ k ~ À À Note that the actual sample values are irrelevant
35. If 5 is Poisson with parameter \$, and \$ has a gamma distribution with parameters and ,
and data values are Á Á ÀÀÀÁ are available, then the posterior distribution of \$ is also
gamma, but with parameters b ' and b , and the predictive expectation of 5 is
² b ' ³² b ³ . We are given ~ Á ~ .
Based on the first ~ weeks of experience, we are given that
² b ³²
~
² ³b
³ ~ , from which we get ~ .
~
Based on the first all 10 weeks of experience, we are given that
² b ³²
~
³
² ³b
~ , from which we get ~ .
~
Therefore, ~ , and using only the final 4 weeks of experience and the original prior, the
~
predictive expectation of 5 is ² b ³²
² ³b
³~
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-100
PRACTICE EXAM 5
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 6
PE-101
ACTEX EXAM C/4 - PRACTICE EXAM 6
1. ? has an exponential distribution with mean 1.
@ is a transformation of ? based on the increasing one-to-one transformation @ ~ ²?³.
The distribution of @ is Weibull with parameters and .
Find the transformation function ²%³.
B) %
C) %°
D) %°
A) %
E) %°
2. A portfolio of insurance policies is made up of non-smokers and smokers. Non-smokers make
up 75% of the policyholders. The model being used for mortality is that non-smokers have a
constant hazard rate of .01 and smokers have a constant hazard of .02. A policyholder is chosen at
random from the group, and it is found that the policyholder is still alive 20 years after the policy
is issued. Find the probability that the policyholder is a smoker.
A) Less than .200
B) At least .200 but less than .210
C) At least .210 but less than .220
D) At least .220 but less than .230
E) At least .230
3. The ground up loss random variable ? for a health insurance policy in 2006 is modeled with
an exponential distribution with mean 1000. An insurance policy pays the loss above an ordinary
deductible of 100, with a maximum annual payment of 500. The ground up loss random variable
is expected to be 5% larger in 2007, but the insurance in 2007 has the same deductible and
maximum payment as in 2006. Find the percentage increase in the expected cost per payment
from 2006 to 2007.
A) 1.005
B) 1.008
C) 1.011
D) 1.014
E) 1.017
4. A risk class is made up of three equally sized groups of individuals. Groups are classified as
Type A, Type B and Type C. Any individual of any type has probability of .5 of having no claim
in the coming year and has a probability of .5 of having exactly 1 claim in the coming year. Each
claim is for amount 1 or 2 when a claim occurs. Suppose that the claim distributions given that a
claim occurs, for the three types of individuals are
° % ~ 7 ²claim of amount %OType A and a claim occurs³ ~ D
Á
° % ~ ° % ~ 7 ²claim of amount %OType B and a claim occurs³ ~ D
Á
° % ~ °
% ~ 7 ²claim of amount %OType C and a claim occurs³ ~ D
À
°
% ~ An insured is chosen at random from the risk class and is found to have a claim of amount 2.
Find the probability that the insured is Type A.
A) B) C) D) E) © ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-102
PRACTICE EXAM 6
5. The following grouped data set is given
Interval
Number of Claims
² Á µ
² Á µ
² Á µ
² Á µ
For a policy limit " with  "  , the empirical estimate of the limited expected value is
²"³ . Find Z ²"³ .
c"
A) "c
B) c"
C) "c
D) 80
E) "
Questions 6 and 7 relate to the following situation. The number of employees leaving a company
for all reasons is tallied by the number of months since hire. The following data was collected for
a group of 50 employees hired one year ago:
Number of Months Since Hire Number Leaving the Company
1
1
2
1
3
2
5
2
7
1
10
1
12
1
6. Determine Greenwood's approximation to the variance of the Product-Limit estimate : ²³.
A) .0015
B) .0017
C) .0019
D) .0021
E) .0023
7. Determine the Nelson-Aalen estimate of the cumulative hazard at the sixth month since hire,
assuming employees always leave the company after a whole number of months.
A) .106
B) .116
C) .126
D) .136
E) .146
8. A group consisting of ten independent lives has a health policy with an ordinary deductible of
250, coinsurance of 80%, and a maximum covered loss of 1,000 (before application of the
deductible and coinsurance; the policy limit will be À² c ³ ~ ). In the past year, the
following individual payments were made to members of the group:
40 , 120 , 160 , 280 , 600 , 600 (two limit payments) ,
Determine the likelihood function for estimating parameters of the ground-up loss distribution
using ²%³ to represent the probability density function and - ²%³ to represent the cumulative
distribution function.
A) ²
À³ h ²
À³ h ²À³ h ²
À³ h ´ c - ²³µ
B) ²³ h ²³ h ²³ h ²
³ h ´ c - ²³µ
C)
D)
E)
²
À³h ²
À³h ²À³h ²
À³h´c- ²³µ
´c- ²³µ
²³h ²³h ²³h ²
³h´c- ²³µ
´c- ²³µ
²
À³h ²
À³h ²À³h ²
À³h´c- ²³µ
´c- ²À³µ
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 6
PE-103
9. The following sample was taken from a distribution with probability density function
²%³ ~ %c , where  %  and  .
0.21
0.43
0.56
0.67
0.72
Let R and S are the estimators of using the maximum likelihood and method of moments,
respectively. Calculate the value of R c S.
A) Less than 0.3
B) At least 0.3, but less than 0.4
C) At least 0.4, but less than 0.5
D) At least 0.5, but less than 0.6
E) At least 0.6
10. Blue Sky Insurance Company insures a portfolio of 100 automobiles against physical damage.
The annual number of claims follows a binomial distribution with ~ .
For the last 5 years, the number of claims is each year has been:
Year 1:
5
Year 2:
4
Year 3:
4
Year 4:
9
Year 5:
3
Two methods for estimating the variance in the annual claim count are:
Method 1: Unbiased Sample Variance
Method 2: Maximum Likelihood Estimation
Use each method to calculate an estimate of the variance. What is the difference between the two
estimates?
A) Less than 0.50
B) At least 0.50, but less than 0.60
C) At least 0.60, but less than 0.70
D) At least 0.70, but less than 0.80
E) 0.80 or more
11. The number of claims for an insured in a year has a Bernoulli distribution with parameter ,
which is the probability of no claim occurring in the year. Insureds are classified into two groups
of equal size. For an insured from Group A, the parameter is uniformly distributed on the
interval ²Á ³. For an insured from Group B, has density function ) ²³ ~ on the interval
²Á ³. In insured is chosen at random from a randomly chosen Group, and is observed to have no
claims in the year. Find the posterior density function of the parameter .
A) ´ b µ
B) ´ b µ
C) ´ b µ
D) ´ b µ
E) ´ b µ
12. Given a first observation with a value of 2, the Buhlmann credibility estimate for the
expected value of the second observation is . Given the first 2 observations each have value 2,
the Buhlmann credibility estimate for the expected value of the third observation is . Find the
Buhlmann credibility estimate for the expected value of the fourth observation if the first three
observations are each 2.
A) B) C) D) E) © ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-104
PRACTICE EXAM 6
Use the following information for Questions 13 and 14.
Two dice, A and B, are used to determine the number of claims. The faces of each die are
marked with either a 0 or a 1, where 0 represents 0 claims and 1 represents 1 claim. The
probabilities for each die are
Probability of 0 Claims
Probability of 1 Claim
Die
A
2/3
1/3
B
1/3
2/3
In addition, there are two spinners, X and Y, which are used to determine claim size. Spinner X
has two areas marked 2 and 8. Spinner Y has only one area marked 2. The probabilities for each
spinner are
Probability That Claim Size = 8
Spinner Probability That Claim Size = 2
X
1/3
2/3
Y
1
0
For the first trial, a die is randomly selected from A and B and rolled. If a claim occurs, a spinner
is randomly selected from X and Y and spun.
13. Determine the expected amount of total losses on the first trial.
A) Less than 1.4
B) At least 1.4, but less than 1.8
C) At least 1.8, but less than 2.2
D) At least 2.2, but less than 2.6
E) At least 2.6
14. For each subsequent trial, the same die selected in the first trial is rolled again. If a claim
occurs, a spinner is again randomly selected from X and Y and spun. Determine the limit of the
Bayesian analysis estimate of the expected amount of total losses for the -th trial as goes to
infinity if the first c trials each yielded total losses of 2.
A) Less than 1.4
B) At least 1.4, but less than 1.8
C) At least 1.8, but less than 2.2
D) At least 2.2, but less than 2.6
E) At least 2.6
15. An insurance company has two group policies. The aggregate claim amounts (in millions of
dollars) for the first three policy years are summarized in the table below. Assume that the two
groups have the same number of insureds. Use Buhlmann's model with empirical Bayesian
estimation to estimate the credibility premium group 1 for the next (4-th) policy year.
Aggregate Claim Amounts
Group
Policy Year
Policy
8
8
11
15
5
A) 9.0
B) 9.2
C) 9.4
D) 9.6
E) 9.8
16. The inverse transform method of simulation is applied to simulate a value of % from a
mixture of two exponential distributions, one with a mean of 1 and one with a mean of 2. The
uniform number . results in a simulated value of % ~ À. Find the mixing weight for the
exponential with mean 1.
A) .1
B) .2
C) .3
D) .4
E) .5
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 6
PE-105
Questions 17 and 18 are based on the following information. You are given the following times
of first claim for five randomly selected auto insurance policies observed from time ! ~ :
1 , 2 , 3 , 4 , 5
17. Calculate the empirical estimate of the kurtosis of this sample.
A) 0.0
B) 0.5
C) 1.7
D) 3.4
E) 6.8
18. You are later told that one of the five times given is actually the time of policy lapse, but you
are not told which one. The smallest Product-Limit estimate of : (4), the probability that the first
claim occurs after time 4, would result if which of the given times arose from the lapsed policy?
A) 1
B) 2
C) 3
D) 4
E) 5
19. You are given four classes of insureds, each of whom may have zero or one claim, with the
following probabilities:
Number of Claims
0
1
Class
I
.9
.1
II
.8
.2
III
.5
.5
IV
.1
.9
A class is selected at random (with probability ), and four insureds are selected at random from
the class. The total number of claims is two. If five insureds are selected at random from the
same class, estimate the total number of claims using Buhlmann-Straub credibility.
A) 2.0
B) 2.2
C) 2.4
() 2.6
E) 2.8
20. A continuous loss random variable ? is uniformly distributed on the interval ²Á ³.
If ?  , a risk manager is paid a bonus equal to 50% of the difference between ? and 50.
Find the variance of the bonus received by the risk manager.
A) 63.1
B) 64.1
C) 65.1
D) 66.1
E) 67.1
21. An aggregate claims random variable : has a compound distribution for which the frequency
prob. has a geometric distribution, and the severity distribution is ? ~ F
.
2 prob. c The mean of : is 2.55 and the stop loss premium with a deductible of 1 is 1.95 .
Find the stop loss premium with a deductible of 2.
A) 1.3
B) 1.4
C) 1.5
D) 1.6
E) 1.7
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 6
22. Suppose that    . An insurance on a loss of amount ? makes the following
payment:
À d ²? w ³ b À
d ²? c ³b c À²? c ³b
Which of the following is the piecewise definition of amount of insurance paid?
À?
?
À
?
A) H
À b À
²? c ³
?
À b À²? c ³ c À
²? c ³
?
À?
?
À
?
B) H
À b À
²? c ³
?
À b À
²? c ³ c À²? c ³
?
À
?
?
À
?
C) H
À
b À²? c ³
?
À
b À²? c ³ c À
²? c ³
?
À?
?
À
?
D) H
À b À²? c ³
?
À b À²? c ³ c À²? c ³
?
À?
?
À
?
E) H
À b À²? c ³
?
À b À²? c ³ c À²? c ³
?
23. A compound claim distribution is analyzed and it is decided to model the distribution of :
by using the geometric distribution for 5 (claim number) and the gamma distribution for ?
(claim amount). With this model it is found that ,´:µ ~ , = ´:µ ~ and
= ´5 µ ~ . An alternative model is proposed in which the distribution of 5 is Poisson with
the same mean as the geometric distribution for 5 in the first model. The claim amount
distribution in the new model is the same gamma distribution as in the old model, and the mean
aggregate claims in the new model is the same as in the old model (16). What is the variance of
the aggregate claims random variable in the new model?
A) 18
B) 36
C) 54
D) 72
E) 90
24. A study of claim numbers for 1000 similar policies results in the following data.
Number of Claims:
0
1
2
3
4
5
6
7
# of Policies:
700
200
60
20
10
8
2
0
The method of moments is applied to estimate the parameters and of the negative binomial
distribution. Find the probability that a policy has 0 claims using the estimated negative binomial
distribution.
A) .70
B) .71
C) .72
D) .73
E) .74
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 6
PE-107
25. The following 10 claim numbers were observed and are being used to obtain the maximum
likelihood estimate of the Poisson parameter for a Poisson claim number distribution.
1,1,1,1,1,2,2,3,3,5.
V³ at the maximum likelihood estimate V.
Find the value of the loglikelihood function M²
A) c B) c C) c D) c E) c 26. A random sample of size is taken from the distribution of ? . The mean and variance of ?
c
c
are and , respectively. The sample mean ? is calculated, and ? is used as an estimator for
À What is the bias of the estimator (the bias of the estimator V is ,´Vµ c , where is the
actual parameter value).
A) B) C) c D)
b
E) c
b
27. You are given the negative loglikelihoods associated with five models:
Model
Number of Parameters
Negative Loglikelihood
Generalized Pareto
3
219.1
Burr
3
219.2
Pareto
2
221.2
Lognormal
2
221.4
Inverse Exponential
1
224.2
Using the Likelihood Ratio Test at a 5% level of significance, how many models are less
preferable than the Generalized Pareto?
A) 0
B) 1
C) 2
D) 3
E) 4
28. Claim frequency follows a Poisson process with a rate of 10 per year. Claim severity is
exponentially distributed with mean 2,000. The method of moments is used to estimate the
parameters of a lognormal distribution for the aggregate losses. Using the lognormal
approximation, calculate the probability that annual aggregate losses exceed 105% of the
expected annual losses.
A) Less than 34.5%
B) At least 34.5%, but less than 35.5%
C) At least 35.5%, but less than 36.5%
D) At least 36.5%, but less than 37.5%
E) At least 37.5%
29. In a portfolio of risks, each risk has a Poisson claim amount distribution. The mean of the
claim amount distribution for a randomly chosen risk is , where has a Pareto distribution with
parameters ~ À and ~ . A single claim amount of 1 is observed for a randomly chosen
risk. Find the Buhlmann credibility premium for the next claim amount for the same risk.
A) 1
B) 1.5
C) 2
D) 2.5
E) 3
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-108
PRACTICE EXAM 6
30. You are given the following information for 10,000 risks grouped by number of claims. A
Poisson distribution was fit to the grouped risks, and minimum chi-square estimation has been
used to estimate the Poisson distribution parameter. The results are as follows.
Number of
Actual Number
Estimated Number of Risks
Claims
of Risks
Using Estimated Poisson
0
7,673
7,788
1
2,035
1,947
2
262
243
3 or more
30
22
Total
10,000
10,000
You are to use the chi-square goodness-of-fit statistic to test the hypothesis / that the Poisson
provides an acceptable fit. Which of the following is true?
A) Reject / at significance level .005
B) Accept / at significance level .005, but reject at significance level .010
C) Accept / at significance level .010, but reject at significance level .025
D) Accept / at significance level .025, but reject at significance level .050
E) Accept / at significance level .050
31. The number of claims per exposure follows a Poisson distribution with a mean of 10.
Claim sizes follow a lognormal distribution with parameters ~ and ~ . The number of
claims per exposure and claim sizes are independent. The method of limited fluctuation
credibility is used, and the full credibility standard has been selected so that total claim dollars per
exposure will be within 10% of expected total claim dollars per exposure 95% of the time.
The partial credibility factor A is calculated for each of the following three approaches:
(a) based on 60 exposures,
(b) based on 700 claims
(c) based on a total claim amount of 3,000.
Which of the following is the correct ranking of the three factors A Á A Á A ?
A) A  A  A
B) A  A  A
C) A  A  A
D) A  A  A
E) A  A  A
32. Type A risks have each period's losses exponentially distributed with mean 1.
Type B risks have each period's losses exponentially distributed with mean .
A risk is selected at random with each risk being equally likely. The firs! period's losses equal 1.
The Buhlmann credibility estimate for the expected second period losses depend on the value of
, say ²³. Find lim ²³ .
¦
C) D) E) SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 6
PE-109
33. From a random sample from a distribution, successive sample mean and sample standard
deviation estimates are made adding one new sample point at a time and recalculating the sample
mean and sample standard deviation. It appears that the sample standard deviation is stabilizing
in the interval ´À Á Àµ . Assuming that estimate is accurate, find the minimum number of
sample points needed in order to ensure that the 99% confidence interval for the distribution
mean has width less than .1 .
A) Less than 30
B) At least 30, but less than 35
C) At least 35, but less than 40
D) At least 45, but less than 50 E) At least 50
34. A Poisson claims process has two types of claims, Type I and Type II.
(i)
The expected number of claims is 3000.
(ii) The probability that a claim is Type I is 1/3.
(iii) Type I claim amounts are exactly 10 each.
(iv) The variance of aggregate claims is 2,100,000.
Calculate the variance of aggregate claims with Type I claims excluded.
A) 1,700,000
B) 1,800,000
C) 1,900,000
D) 2,000,000
E) 2,100,000
35. Two actuaries are simulating the number of automobile claims for a book of business. For
the population that are studying:
i) The claim frequency for each individual driver has a Poisson distribution.
ii) The means of the Poisson distributions are distributed as a random variable, \$.
iii) \$ has a gamma distribution.
In the first actuary's simulation, a driver is selected and one year's experience is generated. This
process of selecting a driver and simulating one year is repeated 5 times. In the second actuary's
simulation, a driver is selected an 5 years of experience are generated for that driver.
Which of the following is/are true?
I. The ratio of the number of claims the first actuary simulates to the number of claims the
second actuary simulates should tend towards 1 as 5 tends to infinity.
II. The ratio of the number of claims the first actuary simulates to the number of claims the
second actuary simulates will equal 1, provided that the same uniform random numbers are used.
III. When the variances of the two sequences of claim counts are compared the first actuary's
sequence will have a smaller variance because more random numbers are used in computing it.
A) I only
B) I and II only
C) I and III only
D) II and II only
E) None of I, II, or III is true
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-110
PRACTICE EXAM 6
ACTEX EXAM C/4 - PRACTICE EXAM 6 SOLUTIONS
1. Since the transformation is one-to-one and increasing, we have
c
[email protected] ²&³ ~ c c²&°³ ~ 7 ²@  &³ ~ 7 ²²?³  &³ ~ 7 ²?  c ²&³³ ~ c c ²&³ ,
where c ²&³ is the inverse function of .
&
It follows that % ~ c ²&³ ~ ² ³ , from which we get & ~ %° ~ ²%³ .
Alternatively, let us define ²&³ ~ c ²&³ .
The mechanical transformation approach gives us @ ²&³ ~ ? ²²&³³ h O Z ²&³O .
Therefore, &b c²&°³ ~ c²&³ h O Z ²&³O .
&
It appears from this relationship that % ~ ²&³ ~ ² ³ , and the & ~ %° ~ ²%³ .
Trying this transformation results in the correct distribution for @ . Answer: E
2. Let ? represent the exact time until death of the randomly chosen policyholder.
We wish to find 7 ´SmokerO;  µ ~
7 ´Smokerq²; ³µ
7 ´; µ
.
7 ´Smoker q ²;  ³µ ~ 7 ´;  OSmokerµ h 7 ´Smokerµ ~ c²À³ h ²À³ ~ À
.
7 ¸;  µ ~ 7 ´²;  ³ q Smokerµ b 7 ¸²;  ³ q Non-smokerµ
~ 7 ´;  OSmokerµ h 7 ´Smokerµ b 7 ´;  ONon-smokerµ h 7 ´Non-smokerµ
~ c²À³ h ²À³ b c²À³ h ²À³ ~ À
.
Then 7 ´SmokerO;  µ ~
7 ´Smokerq²; ³µ
7 ´; µ
~ À
À
~ À .
3. In 2006 the deductible is ~ and the maximum covered loss is " ~ (policy limit of
" c ~ ). The expected cost per loss in 2006 is ,´? w "µ c ,´? w µ ,
and the expected cost per payment is
,´?w"µc,´?wµ
c-? ²³
.
For the exponential distribution with mean , we have
-? ²%³ ~ c c%° and
,´? w %µ ~ ² c c%° ³ .
The expected cost per payment in 2006 is
,´?w
µc,´?wµ
c-? ²³
~
²cc
° ³c²cc° ³
c²cc° ³
~ ² c c° ³ À
In 2007 the loss random variable is @ ~ À? . The exponential distribution is a scale
distribution, which means that a constant multiple is also exponential with a scaled mean.
Therefore, @ has an exponential distribution with mean 1050.
The expected cost per payment in 2007 (same deductible and policy limit as 2006)
,´@ w
µc,´@ wµ
[email protected] ²³
~
²cc
° ³c²cc° ³
c²cc° ³
~ ² c c° ³ À
The ratio of expected cost per payment in 2007 to that of 2006 is
²cc° ³
²cc° ³
~ À , an increase of 1.1% from 2006 to 2007.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 6
PE-111
3. continued
An alternative way to find the expected cost per payment in 2007 is as follows.
If @ ~ ² b ³? and a maximum covered loss of " is applied to @ , then it is possible to
"
formulate the expected cost per loss for @ in terms of ? : ,´@ w "µ ~ ² b ³,´? w b
µ.
&
Also, the distribution function of @ is [email protected] ²&³ ~ -? ² b ³ .
If there is a policy deductible of and a maximum covered loss of " applied to @ , then
the expected cost per loss is
"
,´@ w "µ c ,´@ w µ ~ ² b ³²,´? w b
µ c ,´? w b
µ³ .
The expected cost per payment is
,´@ w"µc,´@ wµ
[email protected] ²³
~
"
²b³²,´?w b
µc,´?w b
µ³
c-? ² b ³
.
Applying this to exponential random variable ? with ~ ÀÁ ~ Á ~ and
" ~ , we get an expected cost per payment in 2007 of
²À³²,´?w À
µc,´?w À
µ³
c-? ² À ³
~
²À³´²cc
° ³c²cc° ³µ
c²cc° ³
This reduces to the same expression as the first approach.
4. 7 ´Type AO? ~ µ ~
~
7 ´?~OAµh7 ´(µ
7 ´?~µ
.
7 ´?~OAµh7 ´(µ
~ 7 ´?~O(µh7 ´(µb7 ´?~O)µh7 ´)µb7 ´?~O*µh7 ´*µ
´²À³² ³µh² ³
´²À³² ³µh² ³b´²À³² ³µh² ³b´²À³² ³µh² ³
~ .
5. The empirical estimate of the limited expected value with limit " in the interval from 20 to 50
"²c"³
b
" c
b
"c"
is ²"³ ~ ´² b
.
³ b ² ³ b ² ²c³ ³ b ² c ³ b "µ ~
Z ²"³ ~ 80c"
.
6. = V
´: ²³µ ~ : ²³ h ² c
Á ! ~ .
³
!  : ²³ ~ ² c ³² c ³² c ³² c ³ ~ ~ À
~ Á ~ Á ~ Á ~ Á ~ Á ~ Á ~ Á ~ =V
´: ²³µ ~ ²À³ ´ ²³²³
b ²³²³
b ²³²
³
b ²
³²³
µ ~ À .
V ²!³ ~ ,
7. The Nelson-Aalen estimate of the cumulative hazard function to time ! is /
! ! where is the number of deaths at death point ! and is the size of the group at risk just prior
to the deaths at time ! . In this problem,
! ~ Á ~ Á ~ Â ! ~ Á ~ Á ~ Â ! ~ Á ~ Á ~ Â
! ~ Á ~ Á @ ~ , and since ! ~  , we do not make use of the data from time !
V ²!³ ~ ~ b b b ~ À
À
onward. /
! 
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-112
PRACTICE EXAM 6
8. The likelihood function is a product of "likelihood factors", one for each loss recorded. The
likelihood factor is the density or probability of that loss occurring. There are 6 losses recorded
so the likelihood function will be a product of 6 factors. We use the relationship between the
payment and the ground up loss to reconstruct the ground-up loss amount for each payment
recorded. If the ground up loss is %, with  %  , then the payment is ²À³²% c ³ ,
and if the ground up loss is  , then the payment is ²À³² c ³ ~ . For a
payment of 40, the ground up loss is % where
~ ²À³²% c ³ , so that % ~ . In the same way, we get the following:
Payment
40
120
160
280
600
Ground up loss 300
400
450
600
 1000
Since only ground up losses above 250 will result in payments being made, all payments are
conditional on the ground up loss being over 250. Thus, the likelihood factors for the first 4
²³
payments are c- ²³ (this the conditional density of the ground up loss being 300 given that it
²³
²³
²
³
is at least 250), and c- ²³ , c- ²³ , c- ²³ . The likelihood factors for the last two
c- ²³
payments are both 7 ´?  O?  µ ~ c- ²³ À The overall likelihood function for the
6 payments listed is
²³h ²³h ²³h ²
³h´c- ²³µ
´c- ²³µ
.
9. Maximum likelihood estimation: The log of the density is ²%³ ~ b ² c ³ % .
The loglikelihood function is
M ~ ²% ³ ~ b ² c ³ %
~ b
~ b
Then setting
~
² c ³´ À b À b À
b À
b À µ
² c ³² c À³ .
C
C M ~ results in c À ~ , so that the mle of is
R ~ À
~ À
.
Moment estimation: The first moment of ? is ,´?µ ~ % ²%³ % ~ % % ~ b
.
According to the method of moments, this is set equal to the empirical estimate of the mean,
which is the sample mean c
% ~ À . Solving for in the equation b
~ À results in the
moment estimate for of S ~ À .
Then R c S ~ À . Answer: A
10. For a binomial random variable ? with given and unknown, and sample values
'%
% Á ÀÀÀÁ % , the mle of is V ~ . In this case, ~ and there are ~ sample values,
so V ~ bbbb
~ À . The mle estimate of is the same as the moment estimate of .
²³
The estimate of the variance of ? based on the mle estimate of is
²
V c V³ ~ ²À³²À³ ~ À .
The unbiased sample variance based on the 5 sample values is
²%
~
cc
%³ ~ ´² c ³ b ² c ³ b ² c ³ b ² c ³ b ² c ³ µ ~ À .
The difference between the two estimates is À c À ~ À . Answer: D
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 6
PE-113
11. The prior distribution of is a mixed distribution with pdf ²³ ~ ²À³´ b µ for
²Á³
  . The posterior density of given ? ~ is ²O³ ~ 7 ´?~µ .
²Á ³ ~ 7 ´? ~ Oµ h ²³ ~ h ²À³´ b µ ~ ²À³´ b µ .
Then 7 ´? ~ µ ~ ²Á ³ ~ ²À³´ b µ ~ .
Then, ²O³ ~
²À³´b µ
°
~ ´ b µ .
12. With ~ and ? ~ Á the credibility premium estimate is
² b
³²³ b ² b
³ ~ b
b ~ S b ~ ² b ³² ³ À
c
With ~ and ? ~ Á the credibility premium estimate is
² b
³²³ b ² b
³ ~ b
b ~ S b ~ ² b ³² ³ .
c
With ~ and ? ~ Á the credibility premium estimate is ² b
³²³ b ² b
³ ~ b
b .
From the first two premiums, we have ~ b S ~ and then ~ À
b c
The credibility premium with ~ and ? ~ is then b ~ À
13. The probability of no claim occurring is
7 ´no claimµ ~ 7 ´no claimOAµ h 7 ´Aµ b 7 ´no claimOBµ h 7 ´Bµ ~ ² ³² ³ b ² ³² ³ ~ .
If a claim occurs, the probability it is for amount 2 is
7 ²Oclaim³ ~ 7 ²O?³ h 7 ²?³ b 7 ²[email protected] ³ h 7 ²@ ³ ~ ² ³² ³ b ²³² ³ ~ and the probability that a claim is for amount 8 given that a claim occurs is 7 ´Oclaimµ ~ .
The unconditional probabilities of claim of 2 and 8 are
7 ²³ ~ 7 ²Oclaim³ h 7 ²claim³ ~ ² ³² ³ ~ , and similarly 7 ²³ ~ ² ³² ³ ~ .
The expected total losses for the first trial are ,´; µ ~ ²³² ³ b ²³² ³ b ²³² ³ ~ À
This is a frequency/severity distribution where frequency (number of claims 5 generated by the
roll of the die) and severity (claim amount * generated by the spinner) are independent. An
alternative way to find ,´; µ is as follows. The expectation of total losses is
,´; µ ~ ,´5 µ h ,´*µ . 5 and * are compound distributions.
,´5 µ ~ ,´5 O(µ h 7 ²(³ b ,´5 O)µ h 7 ²)³ ~ ² ³² ³ b ² ³² ³ ~ , and
,´*µ ~ ,´*O?µ h 7 ²?³ b ,´*[email protected] µ h 7 ²@ ³ ~ ²
³² ³ b ²³² ³ ~ À
Then ,´; µ ~ ,´5 µ h ,´*µ ~ ²³² ³ ~ À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-114
PRACTICE EXAM 6
14. The fact that the first c claims each yielded total losses of 2 is relevant only in that it tells
us that there was a claim on each of the first c trials, since a spinner will always be chosen at
random. The Bayesian estimate of total losses on the -th trial is the
conditional expectation
,´; on -th trialO c claimsµ
~ ,´; on -th trialO(µ h 7 ´(O c claimsµ b ,´; on -th trialO)µ h 7 ´(O c claimsµ À
,´; on -th trialO(µ ~ ² ³ ,´*µ ~ Á ,´; on -th trialO)µ ~ ² ³ ,´*µ ~ À
7 ²c claimsO(³h7 ²(³
7 ²(O c claims³ ~ 7 ²c claimsO(³h7 ²(³b7 ²c claimsO)³h7 ²)³
²°³c ²°³
~ ²°³c ²°³b²°³c ²°³ ~ bc Á and
c
7 ²)O c claims³ ~ c 7 ²(O c claims³ ~ b
c .
c
Then, ,´; on -th trialO c claimsµ ~ ² ³² bc ³ b ² ³² b
c ³ À
c
lim ² ³² bc ³ b ² ³² b
c ³ ~ ² ³²³ b ² ³²³ ~ À Answer: E
¦B
15. In this case, ~ (groups) , and ~ 2 (exposure periods per group,
and ~ exposure unit per group/year combination), and ~ ~ ~ 3, and
c
c
c
~ ~ 6. Then ? ~ ² b b ³ ~ Á ? ~ 4 Á ? ~ ² b 4³ ~ .5 .
c
The estimates of the structural parameters are V ~ ? ~ .5 Á
V# ~ ²³²³ ´² c ³ b ² c ³ b ² c ³
b ² c ³ b ² c ³ b ² c ³ µ ~ Á
and V
~ ´² c À³ b ² c À³ µ c ~ À À
Since ~ , the credibility factors for the two groups are equal:
V ~ ~ V# ~ ~ À À
A
V
b À
b V
b
c
V 1?
V 1 ³
The credibility premium for the 4-th year for group 1 is A
V.
1 b ² c A
For group 1 the credibility premium is (À)²³ b ²À³²À³ ~ À .
16. The cdf of the mixed distribution is - ²%³ ~ ´ c c% µ b ² c ³´ c c%° µ .
We are given that À ~ - ²À³ ~ ´ c cÀ µ b ² c ³´ c cÀ
µ ~ À b À
.
Solving for results in ~ À .
,´²?c³ µ
17. The kurtosis of a random variable ? is defined to be
, where ,´?µ ~ and
= ´?µ ~ . We use the empirical distribution to estimate the parameters. The mean of the
c
empirical distribution for the five sample points it the sample mean V ~ ? ~ . The variance
of the empirical distribution is V ~ h ²? c ³ ~ (notice that this is the biased form of
~
the sample variance). The kurtosis of this sample is the kurtosis of the empirical distribution.
This is h ´ h ²? c ³ µ ~ À .
V
~
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 6
PE-115
18. If ! ~ is a policy lapse time, then the at risk group at death time 2 is 4, at death time 3 is 3
and at death time 4 is 2, so that : ²³ ~ ² c ³² c ³² c ³ ~ .
If ! ~ 2 is a policy lapse time, then the at risk group at death time 1 is 5, at death time 3 is 3 and
at death time 4 is 2, so that : ²³ ~ ² c ³² c ³² c ³ ~ .
If ! ~ is a policy lapse time, then the at risk group at death time 1 is 5, at death time 2 is 4 and
at death time 4 is 2, so that : ²³ ~ ² c ³² c ³² c ³ ~ .
If ! ~ is a policy lapse time, then the at risk group at death time 1 is 5, at death time 2 is 4 and
at death time 3 is 3, so that : ²³ ~ : ²³ ~ ² c ³² c ³² c ³ ~ .
If ! ~ is a policy lapse time, then the at risk group at death time 1 is 5, at death time 2 is 4 and
at death time 3 is 3, and at death time 4 is 2 so that
: ²³ ~ ² c ³² c ³² c ³² c ³ ~ .
The minimum value of : ²³ occurs if ! ~ is the lapse time.
19. We use the usual Buhlmann credibility method to solve the problem. The parameter # is the
"class" variable: 7 ´# ~ 0µ ~ 7 ´# ~ 00µ ~ 7 ´# ~ 000µ ~ 7 ´# ~ 0= µ ~ ~ À ,
and the conditional distribution of ? for each class is a 0,1 random variable.
We are told that there is a sample of size ~ from a randomly chosen class, and the total
c
number of claims is ? ~ . Therefore ? ~ h ²³ ~ À ,
~
The hypothetical means are
²0³ ~ ²³²À³ b ²³²À³ ~ À Á ²00³ ~ À Á ²000³ ~ À and ²0= ³ ~ À .
The process variances are #²0³ ~ #´?O# ~ 0µ ~ h ²À³²À³ ~ À Á
#²00³ ~ h ²À³²À³ ~ À
Á #²000³ ~ h ²À³²À³ ~ À and #²0= ³ ~ h ²À³²À³ ~ À
(to find the process variances,we have used the rule for a two-point discrete random variable
prob. > ~F
, the variance of > is ² c ³ h ² c ³ ).
prob. c Then the expected hypothetical mean is ~ ,´²#µµ ~ ²À³²À b À b À b À³ ~ À .
The expected process variance is # ~ ,´#²#³µ ~ ²À³²À b À
b À b À³ ~ À .
The variance of the hypothetical means is
~ = ´²#³µ ~ ²À³²À b À b À b À ³ c ²À³ ~ À
.
The Buhlmann credibility premium for the next claim from the same group is
c
A? b ² c A³ , where A ~ b
# ~
À ~ À .
b À
The Buhlmann credibility premium for the next claim from the same group is
²À³²À³ b ²À³²À³ ~ À .
The credibility premium for 5 claims from the same group is d À ~ À À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 6
c ? ?  ~ À´ c ²? w ³µ
?  ,´Bonusµ ~ À´ c ,´? w µµ ~ À´ c ´ %²À³ % b 7 ´?  µµµ
~ À´ c ´À b µµ ~ À or
,´Bonusµ ~ À d ² c %³²À³ % ~ À
,´Bonus µ ~ ²À³ d ² c %³ ²À³ % ~ À
Variance of bonus ~ À c ²
À³ ~ À .
20. Bonus ~ À d F
21. Let us denote the geometric distribution parameter with the usual notation .
Then ,´5 µ ~ and ,´?µ ~ b ² c ³ ~ c ,
and ,´:µ ~ ,´5 µ d ,´?µ ~ d ² c ³ ~ À .
,´: w µ ~ d 7 ²:  ³ ~ c 7 ²: ~ ³ ~ c 7 ²5 ~ ³ ~ c b
À
We are given that ,´²: c ³b µ ~ À , so that
,´²: c ³b µ ~ ,´:µ c ,´: w µ ~ d ² c ³ c ´ c b
µ ~ À .
It follows that À b b
~ À from which we get ~ À ,
and then from d ² c ³ ~ À we get ~ À .
The stop loss premium with a deductible of 2 is
,´²: c ³b µ ~ ,´:µ c ,´: w µ ~ À c ´ d : ²³ b d 7 ²:  ³µ .
À
: ²³ ~ 7 ²: ~ ³ ~ 7 ²5 ~ ³ d 7 ²? ~ ³ ~ ²b ³ d ~ ²À³
d ²À³ ~ À ,
and 7 ²?  ³ ~ c : ²³ c : ²³ ~ c À c À ~ À .
Then ,´²: c ³b µ ~ À c ´À b ²À³µ ~ À . Answer: B
? ?
?
²? c ³b ~ F
?c
²? c ³b ~ F
?c
À?
À d ²? w ³ ~ F
À
22. ? w ~ F
À d ²? c ³b ~ F

, this is insurance with a limit of 
?
, this is insurance with a deductible of ?
?
, this is insurance with a deductible of ?
?
?
,
À
d ²? c ³b ~ F
,
?
.6(? c ) ?  . (? c )
?
?
Combining these insurance payments results is
À d ²? w ³ b À
d ²? c ³b c À d ²? c ³b
À?
À
~H
À b À
²? c ³
À b À
²? c ³ c À²? c ³ ~ À c À
b À b À?
?
?
?
?
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 6
PE-117
23. With a geometric distribution for 5 , = ´5 µ ~ ² b ³ ~ ¦ ~ 4
(ignore the negative root c ). Then ,´5 µ ~ ~ .
Since ,´:µ ~ ,´5 µ h ,´?µ it follows from the parameters in the old model that ,´?µ ~ .
Since = ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ , it follows from the parameters of the
old model that = ´?µ ~ . Then, ,´? µ ~ = ´?µ b ²,´?µ³ ~ , and
= ´:µ ~ h ,´? µ ~ .
24. The method of moments equations for the negative binomial distribution are
~ h and ² b ³ ~ h c 4 h 5 .
~
~
~
The equations become ~ À , ² b ³ ~ À À
V ~ À ~ À
, so that V ~ À
and then V ~ À ~ À
.
Then, b À
À
7 ´? ~ µ ~ ² b ³c , so the estimated probability is ²À
³cÀ
V ~c
25. The mle of for the Poisson is % ~ in this case.
%
'%
The likelihood function is 3²³ ~ ²c h % [ ³ ~ c h ~
% [
~
The loglikelihood function is M²³ ~ 3²³ ~ c b ² ³²'% ³ c '²% [³ .
In this case, we have ~ and
V³ ~ M²³ ~ c ²³ b ² ³² b b Ä b ³ c ´ [ b [ b Ä b [µ
M²
~ c À .
c
26. The bias is ,´? µ c . We know that
c c
? c ?
µ is an unbiased estimator for .
c ²? c ?³ ~ c ´
~
~
c
Thus, ,´ ? µ c ,´? µ ~ ² c ³ À But, ,´ ? µ ~ ,´? µ ~ ´ b µ À
~
~
c
Thus, ,´? µ ~ <,´ ? µ c ² c ³ = ~ <² b ³ c ² c ³ µ ~ b À
~
c
The bias is ,´? µ c ~ b c ~ .
27. The test statistic is ² 3( c 3) ³ . For the 5% test, the critical values are
3.84 with 1 degree of freedom and 5.99 with 2 degrees of freedom.
The generalized Pareto is automatically preferable to the Burr since it has the same number of
parameters but has a higher loglikelihood value.
With model A being the generalized Pareto, the test statistic values are
Test Statistic
Degrees of Freedom
Model B
Pareto
4.2
1
Lognormal
4.6
1
Inverse Exponential
10.2
2
The likelihood ratio test prefers the Generalized Pareto to all other models, since the test statistic
is greater than the critical value in all cases.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-118
PRACTICE EXAM 6
28. The mean and variance of the frequency 5 are ,´5 µ ~ = ´5 µ ~ .
The mean and variance of the severity ? are ,´?µ ~ Á and = ´?µ ~ Á .
The mean of the compound distribution : is ,´:µ ~ ,´5 µ h ,´?µ ~ Á , and the variance
of : is = ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ Á Á .
Using the lognormal approximation to : and the method of moments, we have, : has lognormal
parameters and that satisfy the relationships ,´:µ ~ b ~ Á and
,´: µ ~ b ~ Á Á (note that ,´: µ ~ = ´:µ b ²,´:µ³ ).
Then, b ~ ²Á ³ ~ À and b ~ ²Á Á ³ ~ À .
We can solve for and from these equations: ~ À and ~ À (and ~ À
).
Then, using the estimated lognormal distribution,
7 ´:  À,²:³µ ~ 7 ´:  Á µ ~ c 7 ´:  Á µ (we now use the cdf of the
lognormal given in the table of distributions)
Á c À
~ c )4
5 ~ c )²À³ ~ c )²À³ ~ c À
~ À .
À
29. Hypothetical mean ~ ²³ ~ ,´?Oµ ~ Á Process variance ~ #²³ ~ = ´?Oµ ~ .
# ~ ,7 = ~ ,´ #²³µ ~ ,´µ ~ c
~ .
~ = /4 ~ = ´²³µ ~ = ´µ ~ B since  .
c
c
A ~ b
# ~
~ S the Buhlmann credibility premium is A? b ² c A³ ~ ? ~ ,
b B
since there is only one observation, and it is of amount 1.
30. The chi-square statistic is
?~
~
²6 c, ³
,
~
²
c³
b
²c³
b
²
c³
b
²c³
~ À .
The degrees of freedom c c , where is the number of data intervals and is the
number of parameters estimated in the model. In this Poisson case there is one parameter, so
~ and there are ~ intervals, so that ? has 2 degrees of freedom. From the chi-square
table we see that the 99-th percentile is 9.21 and the 99.5-th percentile is 10.60. We reject / at
the c À ~ À significance level since À  À , but we do not reject / at the
c À ~ À significance level since À  À
31. The partial credibility factor is A ~ ¸k (
Á ¹ , where ( is the standard for full
credibility and is the observed value for that standard.
For the compound Poisson distribution, the three versions of the standard for full credibility are
= ´@ µ
(a) h < b ²,´@ µ³ = , standard based on exposure periods observed ,
= ´@ µ
(b) h < b ²,´@ µ³ = , standard based on number of claims observed ,
= ´@ µ
(c) h <,´@ µ b ,´@ µ = , standard based on total amount of claims observed .
b In this case, ~ ² À
~ À Á
À ³ ~ À
, ~ Á ,´@ µ ~ b b = ´@ µ ~ ,´@ µ c ²,´@ µ³ ~ c ²
³ ~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 6
PE-119
31. continued
The standards are (a) À , (b) , (c) .
With 60 exposures, we get A ~ k À
~ À .
With 700 claims we get A ~ k ~ À .
With total claims of 3000 we get A ~ k ~ À .
A  A  A .
32. The hypothetical means are ,´?O(µ ~ ²(³ ~ Á ²)³ ~ À
The collective mean is ~ ,´/4 µ ~ ² b ³ À
The process variances are = ´?O(µ ~ #²(³ ~ Á #²)³ ~ (variance of an exponential
distribution is the square of the mean).
The expected process variance is # ~ ,´7 = µ ~ b À
The variance of the hypothetical means is ~ = ´/4 µ ~ ² b ³ c ´ ² b ³µ ~ ² c
³
With ~ exposure, the Buhlmann credibility factor is A ~ b # ~
À
² b³
b
²c³
Then the Buhlmann credibility premium is
c
A? b ² c A³ ~
h b ² c
² b³
b
²c³
lim ²³ ~ b
b ² c b
³ ~ .
b³ ³² ³² b ³
b ²²c³
~ ²³ À
¦
:
33. 99% confidence interval has width ²À³ j , where is the number of sample points
(2.58 is the 99.5 percentile of the standard normal distribution). Since À  :  À . we must
²À³
ensure that ²À³ j  À S  À
34. For a compound Poisson claims distribution : , = ´:µ ~ ,´5 µ h ,´? µ .
Therefore, Á Á ~ ²³,´? µ S ,´? µ ~ .
Claim amount is a mixture of two distributions, therefore,
~ ,´? µ ~ ² ³,´? µ b ² ³,´? µ ~ ² ³² ³ b ² ³,´? µ S ,´? µ ~ .
If type 1 claims are eliminated, then the expected number of claims is 2000 , and the variance of
the remaining aggregate claim is ,´5 Z µ h ,´? µ ~ ²³²³ ~ Á Á . Answer: D
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-120
PRACTICE EXAM 6
35. I. The combination of a Poisson claim count with mean \$ and a gamma distribution for \$
results in a negative binomial distribution being simulated by actuary 1. The average number of
claims simulated by actuary 1 in 5 trials is 5 h ,´\$µ . The second actuary selects a driver with
Poisson parameter and the average number of claims in 5 years for that driver will be 5 .
The ratio 55,´\$µ tends to 1 only if the second actuary's driver's is equal to ,´\$µ. False
II. This is false for the same reason as I. False
III. For actuary 1, the variance of the sequence generated is the variance of a negative binomial
distribution. For actuary 2, the variance of the sequence generated is the variance of the Poisson
distribution with parameter (the for the driver chosen by actuary 2). Either variance could be
larger than the other depending on the value of for actuary 2's driver. False
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 7
PE-121
ACTEX EXAM C/4 - PRACTICE EXAM 7
1. Subway trains arrive at your station at a Poisson rate of 20 per hour. 25% of the trains are
express and 75% are local. The types and number of trains arriving are independent. An express
gets you to work in 16 minutes and a local gets you there in 28 minutes. You always take the first
train to arrive. Your co-worker always takes the first express. You are both waiting at the same
station. Calculate the conditional probability that you arrive at work before your co-worker,
given that a local arrives first.
A) 37%
B) 40%
C) 43%
D) 46%
E) 49%
2. A frailty model has a base age-at-death distribution that follows DeMoivre's Law with
~ (:²%³ ~ % for  !  ), and associated hazard rate function ²%³. The conditional
hazard rate for the age-at-death random variable ? for an individual with parameter is
?O ²%O³ ~ ²%³. For a new-born individual in the frailty model group, the value of is
uniformly distributed between .8 and 1.5. Find the probability that a randomly selected new-born
from the frailty group will survive to at least age 80.
A) .133
B) .144
C) .155
D) .166
E) .177
3. ? is a random variable for a loss.
Losses in the year 2000 have a distribution such that:
, [? w ] ~ c0.025 2 b 1.475 c 2.25, ~ 10, 11, 12 Ã , 26
Losses are uniformly 10% higher in 2001.
An insurance policy reimburses 100% of losses subject to a deductible of 11 up to a maximum
reimbursement of 11. Calculate the ratio of expected reimbursements in 2001 over expected
reimbursements in the year 2000.
A) 110.0%
B) 110.5%
C) 111.0%
D) 111.5%
E) 112.0%
4. You are given:
Number of claims
Individual Losses
Mean
8
10,000
Standard Deviation
3
3,937
Using the normal approximation, determine the probability that the aggregate loss will exceed
150% of the expected loss.
A) )(1.25)
B) )(1.5)
C) 1 c )(1.25)
D) 1 c )(1.5)
E) 1.5)(1)
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-122
PRACTICE EXAM 7
5. An insurance company sold 300 fire insurance policies as follows:
Policy
Probability of
Number of
Policies
Maximum Claim Per Policy
100
400
.05
200
300
.06
You are given:
(i) The claim amount for each policy is uniformly distributed between 0 and the policy maximum
(ii) The probability of more than one claim per policy is 0.
(iii) Claim occurrences are independent.
Calculate the variance of the aggregate claims.
A) 150,000
B) 300,000
C) 450,000
D) 600,000
E) 750,000
6. An analysis of losses results in an empirical distribution for which the following is known.
- ²Á ³ ~ À Á - ²Á ³ ~ À Á - ²Á ³ ~ À Á - ²Á ³ ~ À .
Find the empirical estimate of the mean residual lifetime ²Á ³.
A) 6000
B) 6250
C) 6500
D) 6750
E) 7000
7. The product limit estimate of a survival distribution results in the following estimates.
!
: ²! ³
À À À À À À
=V
´: ²! ³µ
À À À À
À À
c
A 99% linear confidence interval of :²! ³ is constructed at each death point ! . How many of the
points 1,2,4,5,8,9 will have intervals containing .5? The 99.5 percentile of the standard normal
distribution is 2.58 .
A) 0
B) 1
C) 2
D) 3
E) 4
8. The following grouped data set is given
Interval
Number of Claims
² Á µ
² Á µ
² Á µ
² Á µ
The method of percentile matching is used to estimate the parameter in an exponential model fit
to the data. ! denotes the estimated parameter if the !-th empirical percentile is used. Which of
the following is a correct expression for ! for À  !  À?
!
c!
!
c!
A) !
B) ²c!³
C) ²c!³
D) ²c!³
E) ²c!³
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 7
PE-123
9. A random variable ? has the density function ²%³ ~ ² c %³ for  %  À
A random sample of two observations of ? yields the values .50 and .90 .
Determine V , the maximum likelihood estimator of .
A) Less than .45
B) At least .45, but less than .95
C) At least .95, but less than 1.45
D) At least 1.45, but less than 1.95
E) At least 1.95
10. You are given the following
- three losses have been recorded as follows: 200 , 300 , 500
- losses below 100 have not been recorded
- the random variable ? underlying the losses has the density function ²%Â ³ and the
distribution function - ²%Â ³
Which of the following must be maximized to find the maximum likelihood estimate of ?
A) ´Â µ h ´Â µ h ´Â µ
C)
´Âµh ´Âµh ´Âµ
´1- ´Âµµ
E)
´Âµh ´Âµh ´Âµ
´1-- ´Âµµ
D)
B)
´Âµh ´Âµh ´Âµ
´ ´Âµµ
´Âµh ´Âµh ´Â µ
´- ´Âµµ
11. You are given the following distribution:
²%³ ~ %c Á  %  .
A single observation % is used to estimate the parameter by maximum likelihood estimation.
The probability 7 ´?  Àµ is estimated using that maximum likelihood estimator. Determine the
approximate variance of the estimator for that probability if % ~ À.
A) ²À³= ´Vµ
B) ²À³= ´Vµ
C) ²À³= ´Vµ
D) ²À³= ´Vµ
E) ²À³= ´Vµ
12. Two claims distributions are observed for the same number of exposure periods. Which of
the following statements is true regarding the Buhlmann credibility factors A and A for claim
distributions 1 and 2.
A) If the two distributions have the same hypothetical mean distribution then A ~ A .
B) If the two distributions have the same process variance distribution then A ~ A .
C) If the two distributions have the same hypothetical mean distribution and the same process
variance distribution then A ~ A .
D) If A ~ A then the two distributions must have the same collective mean .
E) If A ~ A then the two distributions have the same expected process variance, #.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-124
PRACTICE EXAM 7
Use the following information for Questions 13 and 14.
An insurance portfolio consists of two classes, A and B. The frequency distributions for the two
classes is
Probability of Number of Claims
Class
0
1
2
3
A
.7
.1
.1
.1
B
.5
.2
.1
.2
Class A has three times as many insureds as Class B.
A randomly selected risk from the portfolio generates 1 claim over the most recent policy period.
13. Determine the Bayesian analysis estimate of the claims frequency rate for the observed risk.
A) Less than .72
B) At least .72, but less than .78
C) At least .78, but less than .84
D) At least .84, but less than .90
E) At least .90
14. Determine the Buhlmann credibility estimate of the claims frequency rate for the observed
risk.
A) Less than .72
B) At least .72, but less than .78
C) At least .78, but less than .84
D) At least .84, but less than .90
E) At least .90
15. The number of claims per year for a given risk follows a Poisson distribution with mean .
The prior distribution of is assumed to be a gamma distribution with coefficient of variation À
Determine the coefficient of variation of the posterior distribution of after 160 claims have been
observed for this risk in one year.
A) Less than .05
B) At least .05, but less than .10
C) At least .10, but less than .15
D) At least .15, but less than .20
E) At least .20
16. Semi-parametric empirical Bayesian credibility is being applied in the following situation.
The distribution of annual losses ? on an insurance policy is uniform on the interval ²Á ³ , where has
an unknown distribution. A sample of annual losses for 100 separate insurance policies is available. It is
~
~
found that ? ~ and ? ~ .
For a particular insurance policy, it is found that the total losses over a 3 year period is 4.
Find the semi-parametric estimate of the losses in the 4-th year for this policy.
A) 1.1
B) 1.4
C) 1.7
D) 2.0
E) 2.3
17. The prior distribution of the parameter is exponential with a mean of 1, and the distribution
of the number of claims one 1 year is Poisson with a mean of . The following series of
simulations is performed, applying the inverse transform method where small random numbers
correspond to small values of the parameter or small numbers of claims. The random number .5 is
used to simulate a value of . Then the random number .9 is used to simulate the number of
claims for the Poisson distribution mean . Find the number of claims simulated.
A) 0
B) 1
C) 2
D) 3
E) 4
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 7
PE-125
18. Beginning with the first full moon in October deer are hit by cars at a Poisson rate of 20 per
day. The time between when a deer is hit and when it is discovered by highway maintenance has
an exponential distribution with a mean of 7 days. The number hit and the times until they are
discovered are independent. Calculate the expected number of deer that will be discovered in the
first 10 days following the first full moon in October.
A) 78
B) 82
C) 86
D) 90
E) 94
19. A group health insurance plan has the following characteristics:
Probability
Claim Distribution
Number
Subgroup
in Subgroup
of Claim
Given a Claim Occurs
1
200
.02
mean=, variance= 2
300
.01
mean=, variance= À 3
100
.01
mean=, variance=À The insurer determines that a premium of À will provide a 95% probability of covering
aggregate claims and a premium of À
will provide a 99% probability of covering
aggregate claims (using the normal approximation). You are given 7 ´A  À
µ ~ À , and
7 ´A  À
µ ~ À , where A is a standard normal random variable. What is ?
A) 0
B) 200
C) 400
D) 600
E) 800
20. A portfolio of risks models the annual loss of an individual risk as having an exponential
distribution with a mean of \$. For a randomly selected risk from the portfolio, the value of \$ has
an inverse gamma distribution with a mean of 20 and a standard deviation of 10. For a randomly
chosen risk, find the probability that the annual loss for that risk is greater than 20.
A) Less than .30
B) At least .30 but less than .31
C) At least .31 but less than .32
D) At least .32 but less than .33
E) At least .33
21. Payfast Auto insures sub-standard drivers.
• Each driver has the same non-zero probability of having an accident.
• Each accident does damage that is exponentially distributed with ~ .
• There is a \$100 per accident deductible and insureds only "report" claims that are larger
than the deductible.
• Next year each individual accident will cost 20% more.
• Next year Payfast will insure 10% more drivers.
• What will be the percentage increase in the number of "reported" claims next year?
A) Less than 15%
B) At least 15%, but less than 20%
C) At least 20%, but less than 25%
D) At least 25%, but less than 30%
E) At least 30%
22. You are given the following information and probability generating function for a discrete
distribution:
7 Z ²³ ~ 7 ZZ ²³ ~ Calculate the variance of the distribution.
A) Less than 1.5
B) At least 1.5, but less than 2.5
C) At least 2.5, but less than 3.5
D) At least 3.5, but less than 4.5
E) At least 4.5
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-126
PRACTICE EXAM 7
23. Taxicabs leave a hotel with a group of passengers at a Poisson rate ~ 10 per hour. The
number of people in each group taking a cab is independent and has the following probabilities:
Number of People
1
2
3
Probability
.60
.30
.10
Using the normal approximation, calculate the probability that at least 1050 people leave the hotel
in a cab during a 72-hour period.
A) 0.60
B) 0.65
C) 0.70
D) 0.75
E) 0.80
24. The Nelson-Aalen estimate of the cumulative hazard rate is given at the first 4 death points.
!
!
!
!
V ³
/²!
À
À
À
À
Find : ²! ³ .
A) .81
B) .83
C) .85
D) .87
E) .89
25. An insurance company collected the following data on the payment pattern for a group of 15
claims. Payments occur at the end of the month
Month:
1
2
3
5
7
10
12
Number Paid:
5
3
2
2
1
1
1
Determine the kernel-smoothed estimate of the cumulative distribution function for the time until
payment random variable at the end of month six using a bandwidth of three months and the
%&c
%c&b
uniform kernel
2& ²%³ ~ H
&c %&b .
&
A) B) C) D) E) 26. Forty observed losses from a loss distribution have been recorded. 10 of the losses are less
than amount 1 and 30 of the losses are at least amount 1. A Pareto distribution with parameters
~ and (unknown) is being fit to the data ( ²%³ ~ ²%b³ ). Find the maximum likelihood
estimator of .
A) Less than .675
B) At least .675, but less than .725
C) At least .725, but less than .775
D) At least .775, but less than .825
E) At least .825
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 7
PE-127
is a vector of parameters being estimated by maximum likelihood. You are given
?
that the current value of the vector of estimates is V ~ > ? as well as the estimated information
À
c À
matrix 0 ²V
³ ~ >
c À
À ?
Determine the approximate variance of V .
A) 1.70
B) 17.0
C) 27.4
D) 46.6
E) 274
27. ~ >
28. Given the distribution ²%³ ~ %c ,  %  , ~ , and the sample
.7 , .75 , .8 , .5 , .65 ,
what is the value of the Kolmogorov-Smirnov statistic for the fit of the distribution to the data?
A) Less than .1
B) At least .1, but less than .2
C) At least .2, but less than .3
D) At least .3, but less than .4
E) .4 or more
29. You are given that the total claims distribution per period follows a compound Poisson
distribution. The claim size follows a Pareto distribution with parameters ~ and
~ (distribution function - ²%³ ~ c ² %b
³ ).
A full credibility standard is established so that the actual number of claims will be within 5% of
the expected number of claims 95% of the time. Determine the number of expected claims
needed for 60% partial credibility for the distribution of number of claims.
A) Less than 500
B) At least 500, but less than 600
C) At least 600, but less than 700
D) At least 700, but less than 800
E) At least 800
30. An individual automobile insured has a claim count distribution per policy period that
follows a Poisson distribution with parameter . For the overall population of automobile
insureds, the parameter follows a distribution with density function ²³ ~ %² c ³ Á
 . One insured is selected at random from the population and is observed to have a total of claims during policy periods. Determine the expected number of claims that this same insured
will have during the b -st policy period.
b1
b
b
A) b
B) b
C) b
D) b
E) b
31. The random variable ? has distribution function - ²%³. You are given:
- ²³ ~ Á - Z ²%³ ~ for  %  , 7 ´? ~ µ ~ , - Z ²%³ ~ for  %  ,
and - ²³ ~ . You simulate ? by using the inverse transformation method. The first four
simulations are based on the uniform numbers Á Á Á .
c
Determine ? , the sample mean of the simulated values.
A) 0.5 B) 1.0
C) 1.5
D) 1.75
E) 2
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PRACTICE EXAM 7
32. An insurance company has three group policies. The aggregate claim amounts (in millions)
for the first two policy years are summarized in the following table:
Total Claim Amounts
Policy Year
Group
1
2
1
8
10
2
5
6
3
6
10
V for year three for group 1.
Find the estimated Buhlmann credibility factor A
A) B) C) () E) 33. You are given a random sample of three values from a distribution function - :
Á Á You estimate the median of ? using the estimator V ~ sample median of ? Á ? Á ? .
Determine the bootstrap approximation to the mean square error of the estimate.
A) 1.0
B) 1.1
C) 1.2
D) 1.3
E) 1.4
34. A loss random variable has a lognormal distribution with ~ and ~ . Find the
percentage change in the loss elimination ratio if the deductible is changed from 100 to 200.
A) less than 20%
B) at least 20% but less than 40%
C) at least 40% but less than 60%
D) at least 60% but less than 80%
E) at least 80%
35. You have modeled eight loss ratios as @! ~ b ! b ! , ! ~ Á Á Ã Á , where @! is the loss
ratio for year ! and ! is an error term. You have determined:
V
V
0.50
À c À
and
= 8> V ?9 ~ >
> V ? ~ > 0.02 ?
c À
À ?
V h , using the delta
Estimate the standard deviation of the forecast for year 10, @V ~ Vb
method.
A) Less than 0.01
B) At least 0.01, but less than 0.02
C) At least 0.02, but less than 0.03
D) At least 0.03, but less than 0.04
E) At least 0.04
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 7
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ACTEX EXAM C/4 - PRACTICE EXAM 7 SOLUTIONS
1. Given that a local train arrives first, you will get to work 28 minutes after that local train
arrives, since you will take it. Your co-worker will wait for first express train. You will get to
work before your co-worker if the next express train (after the local) arrives more than 12 minutes
after the local. We expect 5 express trains per hour, so the time between express trains is
exponentially distributed with a mean of of an hour, or 12 minutes. Because of the lack of
memory property of the exponential distribution, since we are given that the next train is local,
the time until the next express train after that is exponential with a mean of 12 minutes.
Therefore, the probability that after the local, the next express arrives in more than 12 minutes is
7 ´;  µ , where ; has an exponential distribution with a mean of 12. This probability is
c° ~ c ~ À
(37%).
2. The base hazard rate function is ²%³ ~ c%
, because DeMoivre's Law with upper age
limit has hazard rate function (force of mortality) c%
at age %.
The conditional survival probability to age 80 for an individual with parameter is
:²O³ ~ c ²%³ % ~ ´:²³µ ~ ² c
³ ~ ²À³ .
The pdf of the parameter is \$ ²³ ~ À
(uniform distribution on ²ÀÁ À³).
The survival probability to age 80 for a randomly chosen individual is
²À³À c²À³À
À
À
:²³ ~ À :²O³ \$ ²³ ~ À ²À³ h À
~ ² À³²À³ ~ À
3. The policy limit is 22, with maximum reimbursement of 11 (after deductible of 11) occurring
if the loss is above 22. The expected reimbursement in 2000 is
,´? w µ c ,´? w µ ~ À c À ~ À .
In 2001, the loss is @ ~ À? . The deductible of 11 is reached when @ is 11, or equivalently,
when ? is À
~ , and in a similar way, the limit is reached when @ ~ (or ? ~ ). Since
@ ~ À? , we have expected reimbursement in 2001
²À³²,´? w µ c ,´? w µ³ ~ À²À c À³ ~ À .
The ratio is À
À ~ À .
4. It must be assumed that number of claims 5 and individual loss amounts ? are mutually
independent. The aggregate loss : has mean and variance
,´:µ ~ ,´5 µ h ,´?µ ~ ²³²Á ³ ~ Á and
= ´:µ ~ = ´5 µ h ²,´?µ³ b ,´5 µ h = ´?µ ~ Á Á Á .
Using the normal approximation,
:c,´:µ
= ´:µ
7 ´:  À,´:µµ ~ 7 ´ j

À,´:µc,´:µ
j= ´:µ µ
~ 7 ´A  Àµ ~ c )²À³ .
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PRACTICE EXAM 7
5. Each of the two groups has policies of the following type. There is a probability of a claim
occurring, and given that a claim occurs, there is a claim amount random variable ) . Under these
circumstances, the expected loss is ,´)µ, and the variance of the loss can be found by mixing a
loss of 0 with probability c and a loss of ) with probability À This will be
h ,´) µ c ² h ,´)µ³ ~ ²,´)µ³ ² c ³ b = ´)µ h . For a policy from the first group of
100, the variance is ²³ ²À³²À³ b h ²À³ ~ . For a policy from the second group
of 200, the variance is ²³ ²À
³²À³ b h ²À
³ ~ . The variance of aggregate
claims is ²³² ³ b ²³²³ ~ Á .
The question can also be analyzed from the compound distribution point of view.
Each policy in the first group of 100 policies has a .05 probability of claim, so the number of
claims 5 that arise from the first group of 100 policies has a binomial distribution with ~ and ~ À, so that ,´5 µ ~ ²³²À³ ~ and = ´5 µ ~ ²³²À³²À³ ~ À.
The amount of a claim ? (from the first group of 100) is uniformly distributed on ´Á µ, and
has mean ,´?µ ~ and variance = ´?µ ~ . The variance of the aggregate of all
claims from the first group of 100 policies is
,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ ²³² ³ b ²À³²³ ~ Á .
In a similar way, for the second group of 200 policies. ,´5 µ ~ ²³²À
³ ~ Á
= ´5 µ ~ ²³²À
³²À³ ~ À Á ,´?µ ~ Á = ´?µ ~ .
The variance of aggregate claims from the second group of 200 policies is
²³² ³ b ²À³²³ ~ Á . The total variance is of aggregate claims from the two
groups of policies is Á b Á ~ Á . Answer: D
6. ²Á ³ ~
,´?µc,´?wÁµ
V ²Á³
c-
.
When data is in interval form with data points in interval ²c Á µ the empirical estimate of
V ~ h
,´?µ is found as ,´?µ
~
bc
( is the total number of data points).
The intervals implied by the data are ²Á µ ~ ² Á µ Á ²Á µ ~ ² Á µ Á
²Á Á µ ~ ² Á µ and ²Á Á µ ~ ² Á µ .
V ²Á ³ ~ À Á
From the given data we have ~ V ²Á ³ c V ²Á ³ ~ À Á ~ V ²Á ³ c V ²Á ³ ~ À Á and
~
V ²Á ³ c V ²Á ³ ~ À À Then
~
V ~ h
,´?µ
~
bc
~ ²À³²Á ³ b ²À³²Á ³ b ²À³²Á ³ b ²À³²
Á ³
~ Á À .
When " ~ is an interval endpoint, the empirical estimate of the limited expected value with
V w "µ ~ h
limit " is ,´?
~
bc
b " h .
~b
With " ~ Á ~ this becomes
V w Á µ ~ ²À³²Á ³ b ²À³²Á ³ b ²À b À³²Á ³ ~ Á .
,´?
Then V²Á ³ ~
V
V
,´?µc,´?wÁµ
V
c- ²Á³
~
Á
ÀcÁ
cÀ
~ Á .
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PRACTICE EXAM 7
PE-131
7. The linear 99% confidence interval of :²!³ is : ²!³ f Àj= V
´: ²!³µ .
The sample median is the point ! for which : ²! ³ ~ À . We construct a confidence interval at
each death point, and we wish to find the number of intervals which contain .5. The intervals are
Interval
!
²À Á À
³
²À Á À³
²À Á À³
²À
Á À³
²À Á À³
² c À
Á À
³
Three of the points, 5 , 8 and 9 have intervals containing .5.
8. If À  !  À , then the empirical !-th percentile is between 10 and 20, with value
! . Taking this as the !-th percentile of the exponential distribution, we have
!
c c!°! ~ ! . Solving for ! results in ! ~ c ²c!³
9. 3²³ ~ ²% À³ h ²% À³ ~ ² c À³ h ² c À³ ~ ² c À³² c À³ À
3²³ ~ c b ² c À³ b ² c À³ À
c²cÀ³²cÀ³b²cÀ³b² cÀ³
3²³ ~ c b cÀ
b cÀ
~S
~ .
²cÀ³²cÀ³
The numerator becomes c b À c À ~ , which has roots ~ À
Á À .
Since the distribution of ? requires that  %  , the value of ~ À
is inconsistent with the
data point % ~ À . Thus the mle of is 1.5 .
10. The likelihood function is the product of density function values when individual loss
amounts are know, and probability values when intervals in which loss amounts occurred are
known. When there is truncation from below - no loss below are recorded, then the conditional
density and probability factors (conditional given that the loss is above ) are included in the
likelihood function. That is the case in this question. As no losses below 100 are recorded, there
is truncation from the left at 100. Since actual loss amount are given, we use the conditional
density of 200, 300 and 500 given that the loss is above 100.
3²³ ~ ²Â Oloss is  ³ h ²Â Oloss is  ³ h ²Â Oloss is  ³ h
´Âµ
´Âµ
´Âµ
~ ´1-- ´Âµµ h ´1-- ´Âµµ h ´1-- ´Âµµ .
11. Given a function ²³, if V is the maximum likelihood estimate of then the variance of ²V³
is ´Z ²V³µ h = ²V³ . In this case, ²³ ~ 7 ´?  Àµ ~ ²À³ , so that Z ²³ ~ ²À³ .4 À
With one sample point %, the likelihood function is 3²³ ~ %c , so that
V
3²³ ~ b % ~ S ~ c % À
With % ~ À , the mle of is V ~ À À The estimate of = ´²³µ is then
V
´Z ²V³µ h = ²V³ ~ ´²À³ Àµ h = ²V ³ ~ ´²À³À Àµ h = ²V ³ ~ À
h = ²V ³ .
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PRACTICE EXAM 7
12. A ~ A is equivalent to # ~ ~ ~ # . If ²³ ~ ²³ have the same distribution
and if # ²³ ~ # ²³ have the same distribution, then
# ~ ,´# ²#³µ ~ ,´# ²#³µ ~ # and ~ = ´ ²#³µ ~ = ´ ²#³µ ~ ,
so that ~ .
13. Expected claims frequency rate given that there is one claim in the first period is
,´5 Oµ ~ ,´5 O(µ h 7 ²(O³ b ,´5 O)µ h 7 ²)O³
Where ,´5 O(µ ~ ²³²À³ b ²³²À³ b ²³²À³ b ²³²À³ ~ À
is the expected frequency rate for a
risk from Class A, and ,´5 O)µ ~ ²³²À³ b ²³²À³ b ²³²À³ b ²³²À³ ~ is the expected
frequency rate for a risk from Class B.
7 ²(O³ is the conditional probability that the risk is from Class A given that there was one claim
in the first period. We note that since Class A has 3 times as many insureds as Class B,
7 ²(³ ~ À Á 7 ²)³ ~ À are the probabilities randomly choosing a risk from Class A or B,
respectively (7 ²(³ ~ 7 ²)³ and 7 ²(³ b 7 ²)³ ~ ). The conditional probability 7 ´(Oµ is
found using the usual Bayesian approach:
7 ²O(³h7 ²(³
²À³²À³
7 ²O)³h7 ²)³
²À³²À³
7 ²(O³ ~ 7 ²O(³h7 ²(³b7 ²O)³h7 ²)³ ~ ²À³²À³b²À³²À³ ~ .6 Á and in a similar way
7 ²)O³ ~ 7 ²O(³h7 ²(³b7 ²O)³h7 ²)³ ~ ²À³²À³b²À³²À³ ~ . (alternatively, once we have
7 ²(O³ , we get 7 ²)O³ ~ c 7 ²(O³ ). Then ,´AOµ ~ ²À
³²À
³ b ²³²À³ ~ À
14. The hypothetical means are /4( ~ ,´5 O(µ ~ À
Á /4) ~ ,´5 O)µ ~ À
The collective mean is ~ ,´/4 µ ~ ²À
³²À³ b ²³²À³ ~ À À
The process variances are
7 =( ~ = ´5 O(µ ~ ,´5 O(µ c ²,´5 O(µ³
~ ²À³ b ²À³ b ²À³ b ²À³ c ²À
³ ~ À Á
7 =) ~ = ´5 O)µ ~ ,´5 O)µ c ²,´5 O)µ³
~ ²À³ b ²À³ b ²À³ b ²À³ c ²³ ~ À ,
and # ~ ,´7 = µ ~ ²À³²À³ b ²À³²À³ ~ À À
~ = ´/4 µ ~ ,´/4 µ c ²,´/4 µ³ ~ ²À
³ ²À³ b ²³ ²À³ c ²À³ ~ À .
The Buhlmann credibility factor is A ~ b
~ b
~ À , when
# ~
b À
À
there is only ~ observation. The Buhlmann credibility estimate of the expected number of
c
claims in the second period is A? b ² c A³ , and since there is only one ? , this becomes
²À³²³ b ²À³²À³ ~ À À
15. This problem involves Bayesian estimation based on the combination of a gamma prior
distribution with a Poisson model distribution. If the prior variable is gamma with parameters and and if the model is Poisson with mean , and if observations are available % Á ÀÀÀÁ % for a
particular policy, then the posterior distribution is also gamma with parameters Z ~ b '%
and Z ~ b . We are told that the coefficient of variation of the prior gamma distribution is .
j= ´³
,´µ
j
~ j ~ À Therefore,
~ . There is ~ observation, and it is % ~ . The posterior distribution is gamma
with Z ~ b ~ . The coefficient of variation of the posterior is
The coefficient of variation of the prior gamma is
jZ ²Z ³
Z Z
~ j Z ~ ~ .0714 .
~
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 7
PE-133
16. Hypothetical mean is ,²?O³ ~ .
Process variance is = ²?O³ ~ .
Expected hypothetical mean is ~ ,´?µ ~ ,´,²?O³µ ~ , ´ µ ~ ,´µ ,
Expected process variance ~ # ~ ,´= ²?O³µ ~ , ´ µ ~ ,´ µ .
Variance of hypothetical mean ~ ~ = ´,²?O³µ ~ = ² ³
~ = ²³ ~ ´,² ³ c ²,²³³ µ .
c
From the sample, we can estimate ,²?³ as ? ~ , so this is also the estimate of ,´µ.
The estimate of ,´µ is 4.
From the sample we can estimate = ²?³ using the unbiased sample estimate,
c
´'? c ? µ ~ ´
c ² ³µ ~ À .
But = ²?³ ~ b # ~ ,´ µ b ´,² ³ c ²,²³³ µ ~ ,´ µ c ²,² ³³ .
Using the estimated variance of ? and the estimated mean of , we have
À ~ ,´ µ c ² ³ , so that the estimate of ,´ µ is À
.
Then, # ~ ,´ µ is estimated to be À , and
~ ´,² ³ c ²,²³³ µ is estimated to be .515 .
V c b ² c A³
V The estimate of losses in the 4-th year is [email protected]
V
c
V
where A ~
V~?~,
À ~ À , and V# ~
b V
so that
b
À
V c b ² c A³
V [email protected]
V ~ ²À³² ³ b ²À³²³ ~ À
.
17. The cdf of the exponential distribution with mean 1 is - ²³ ~ c c . The simulated value
of is the solution of À ~ c c , so that ~ À
. Then the probability function and cdf
cÀ
²À
³%
of the Poisson distribution with mean .693 is ²%³ ~
%[
%
²%³
À
À
À
- ²%³
À
À
À
Since À  À  À
, the simulated number of claims is 2.
.
À
À
ÀÀÀ
18. We expect a deer to be hit by a car every ~ À days. For the deer that is expected to be
hit at À days, the chance of being discovered within the first 10 days is the probability of
being discovered within c À days after being hit. Since time of discovery after being hit
has an exponential distribution with mean 7 days, this probability is c c²cÀ³° (the prob.
7 ´;  c Àµ , where ; is exponential with mean 7). The expected number of deer
discovered within 10 days following the first full moon in October is ´ c c²cÀ³° µ ,
~
since each term in the sum is the expected number of deer discovered for the one deer hit at time
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 7
18. continued
. The sum goes to 199 since the 200-th deer is expected to be hit just at time 10, and cannot be
discovered before time 10.
´ c c²cÀ³° µ ~ c c° h À° ´ b À° b ²À° ³ b Ä b ²À° ³ µ
~
~ c cÀ° h
²À° ³ c
À° c
~ À (round up to the next integer value 94) . Answer: E
19. À ~ ,´:µ b À
h j= ´:µ , and À
~ ,´:µ b À
h j = ´:µ so that
Àc,´:µ
Á
c,´:µ
~ À
À
~ À and thus, ,´:µ ~ , and = ´:µ ~ Á .
But, ,´:µ ~ h ²À³ h b h ²À³ h b h ²À³ h ~ ~ ¦ ~ ,
and = ´:µ ~ h ´²À³²À³ h b ²À³ h µ
b h ´²À³²À³ h b ²À³ h À µ
b h ´²À³²À³ h b ²À³ h À µ ~ Á ¦ ~ .
20. Given \$ ~ , the annual loss ? has an exponential distribution with mean and \$ has an
inverse gamma distribution. ? is a continuous mixture distribution of an "exponential over an
inverse gamma".
Suppose that the inverse gamma distribution of \$ has parameters and . We are given that the
mean and standard deviation of \$ are 20 and 10. Therefore, the variance of \$ is 100 and the 2nd
moment of \$ is ,´\$ µ ~ = ´\$µ b ²,´\$µ³ ~ b ~ .
The mean of an inverse gamma is c
. and the 2nd moment is ²c³²
c³ À
From the two equations c
~ and ²c³²
c³ ~ ,
c
we get ~ ²c³²c³ ,² c ³ ~ c . Then solving for results in ~ .
Substituting back into c
~ , we get that ~ .
When we have a continuous mixture distribution for ? over \$, the pdf, expected values and
probabilities for the marginal distribution of ? can be found by conditioning over \$.
The conditional pdf of ? given \$ ~ is ²%O\$ ~ ³ ~ c%° and the pdf of the inverse
c°
gamma distribution of \$ is \$ ²³ ~ bh!²³ . From the calculated parameter values, we have
c°
\$ ²³ ~ h!²
³ .
We can find the probability 7 ²?  ³ by conditioning over :
B
7 ²?  ³ ~ 7 ²?  O³ h \$ ²³ c°
B
B c°
~ c° h ~ !
²
³ h h!²
³
We know that the inverse gamma pdf must integrate to 1 (as any pdf must), so that
c°
!²³
B c°
B b
~ . It follows that b ~ .
h!²³
Therefore,
c°
B 7 ²?  ³ ~ !²
³ h
B c°
= b
!²
³
!²
³
~ , and
~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 7
PE-135
20. continued
It is possible to show in general that if the conditional distribution of ? given \$ is exponential
with mean \$, and if \$ has an inverse gamma distribution with parameters and , then the
marginal (unconditional) distribution of ? is (two-parameter) Pareto with parameters and .
In this example, the marginal distribution of ? would be Pareto with parameters ~ and
~ , and 7 ²?  %³ ~ ² b%
³ , so that 7 ²?  ³ ~ ² ³ ~ À . Answer: E
21. We denote by ? an exponential random variable with ~ .
This year, a claim will be reported if the loss damage is over 100. The probability of this
occurring is 7 ´?  µ ~ c° ~ cÀ À This probability can also be interpreted as the
proportion of claims that are reported this year. If there are a total if * drivers this year, then the
number of reported claims this year will be cÀ * .
Next year the loss amount is À? . The probability that a loss next year is reported (with the
c² À ³° ~ c° À
same deductible of 100 as this year) is 7 ´À?  µ ~ 7 ´?  À µ ~ À* are insured next year, so the number of reported losses next year will be À*c° À
The percentage increase in the number of reported claims is
À*c°
*cÀ
c ~ À° c ~ À
22. For the integer random variable 5 with probability function 7 ²5 ~ ³ ~ , the
probability generating function is 75 ²!³ ~ b !b ! b Ä .
7 Z ²³ ~ ,´5 µ and 7 ZZ ²³ ~ ,´5 ²5 c ³µ ~ ,´5 µ c ,´5 µ .
We are given 7 Z ²³ ~ ~ ,´5 µ and 7 ZZ ²³ ~ ,´5 µ c ,´5 µ ~ ,´5 µ c ~ , so that
,´5 µ ~ . Then, = ´5 µ ~ ,´5 µ c ²,´5 µ³ ~ c ~ .
23. : ~ number leaving by in a one hour period has compound Poisson distribution, with
~ (average number of cabs (claims) per hour). The number per cab ("claim amount") ? is
1, 2 or 3, with probabilities .6, .3 and .1, respectively. The expected number of people leaving
per hour (expected aggregate claims per period) is
,´:µ ~ ,´?µ ~ ´²³²À
³ b ²³²À³ b ²³²À³µ ~ , and the variance of the number leaving
per hour is = ´:µ ~ ,´? µ ~ ´²³ ²À
³ b ²³ ²À³ b ²³ ²À³µ ~ .
In 72 (independent) hours, > , the number of people leaving by cab has a mean of
²³ ~ and variance ²³ ~ . If > is assumed to be approximately normal,
then since > is an integer 7 ´>  µ ~ 7 ´>  Àµ ~ 7 ´ >jc  Àc
µ
j
~ 7 ´A  c À
µ ~ 7 ´A  À
µ ~ À (A has a standard normal distribution).
V ³ ~ ~ À Á /²!
V ³ ~ b ~ À S ~ À Á
24. /²!
V ³ ~ b b ~ À S ~ À Á
/²!
V ³ ~ b b b ~ À S ~ À .
/²!
: ²! ³ ~ ² c ³² c ³² c ³² c ³ ~ ²À³²À
³²À
³²À³ ~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 7
25. The kernel density estimator of the cumulative distribution function is
V ²%³ ~ ²& ³ h 2& ²%³ , and the uniform kernel with bandwidth is
~
2& ²%³ ~ H
%c&b
%&c
&c %&b .
%&b
There are 15 data points and the empirical probability function is
²³ ~ Á ²³ ~ Á ²³ ~ Á ²³ ~ Á ²³ ~ Á
²³ ~ Á ²³ ~ .
For % ~ , we have 2& ²
³ ~ H
c&
&
& .
&
V
- ²
³ ~ ²³2 ²
³ b ²³ ²
³ b ²³2 ²
³ b ²³2 ²
³ b ²³2 ²
³
b ²³2 ²
³ b ²³2 ²
³
c
~ ² ³²³ b ² ³²³ b ² ³²³ b ² ³² c
³ b ² ³² ³ b ² ³²³ b ² ³²³
~ .
26. Since the data are grouped, the likelihood function is
3²³ ~ - ²³ ´ c - ²³µ ~ ´ c b
µ ´ b µ ~ ²b³
.
3²³ ~ c ² b ³ À
V
3²³ ~ c b ~ S ~ À
27. The covariance matrix of the estimates is the matrix inverse ´0 ²³µc (recall that in the
covariance matrix, the diagonal entries are the variances of the estimates of the 's, and the
entries off the diagonal are the covariances between the estimates of the 's).
c
The inverse of >
is c
.
> c
?
?
À À
À À
Therefore, [0 ²V
³µc ~ ²À³²À³c²cÀ
³
~
h>
À
À ? > À À ?
and the estimated variance of V is 27.4 .
28. ²%³ ~ % ,  %  S - ²%³ ~ % ,  %  .
% - ²%c ³
- ²%³
- ²%³
O- ²%³ c - ²%c ³O O- ²%³ c - ²%c ³O
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À
À À
À
À
À
À
À À
À
À
À
À
+ ~ the maximum of the final two columns ~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 7
PE-137
29. The claim amount distribution is irrelevant in establishing the standard for full credibility for
expected number of claims. With a Poisson claim number distribution, the standard for full
&
credibility of expected number of claims is ~ ² ³ .
In this question, with ~ À and ~ À , ~ ² À
À ³ ~ À
À
The number of expected claims needed for 60% credibility is where k À
~ À
,
so that ~ À .
30. The prior distribution is exponential (gamma) with ~ . With a gamma prior distribution
of (exponential is a special case of the gamma distribution) and a Poisson model distribution
whose mean is , the posterior distribution is a gamma distribution as well, with Z ~ b %
and Z ~
b
~
, where is the number of observations (exposure periods), and % is the
~
total number of claims observed during those periods. We are given that % ~ .
~
b
~ b
. For a
Poisson model distribution the predictive mean is the same as the expected value of the posterior.
Therefore the predicted number of claims in the b -st period is
b
Z Z ~ ² b ³² b
³ ~ b .
Z
Z
Therefore, the posterior distribution is gamma with ~ b and ~
31. ? is continuous on  %  , with - ²%³ ~ À% , which rises from 0 to .25.
? has a probability mass at % ~ , with probability ²³ ~ À . Then ? is continuous on
 %  , with - ²%³ ~ À b À²% c ³ (this is just the straight line with slope .25 on the
interval  %  and has - ²³ ~ À). Then ? has a probability mass at ? ~ , with
²³ ~ À (this is true because as %S from the left, - ²%³SÀ, but then - ²³ ~ , so there is
a jump in - ²%³ at % ~ ).
The inverse transform simulation method is applied as follows:
- if  "  À , then " ~ - ²%³ ~ À%, so % ~ " is the simulated value,
- if À  "  À , then the simulated value of ? is % ~ ,
- if À  "  À , then " ~ - ²%³ ~ À b À²% c ³ ~ À b À%, so that % ~ " c ,
- if À  "  , then % ~ .
From the given uniform values, the simulated values are % ~ Á % ~ Á % ~ Á % ~ .
The sample mean is 1.5 .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-138
PRACTICE EXAM 7
32. ~ groups, and ~ ~ ~ ~ exposure periods. Since no mention is made of
different numbers of policies from group to group, we can regard this situation as one exposure
unit per exposure period, which is the basic Buhlmann empirical Bayesian credibility approach.
V ~ ~ V# , where ~ ~ (one exposure unit per exposure period for two
A
V
b
b V
c
c
c
c
exposure periods). ? ~ Á ? ~ À Á ? ~ Á ? ~ À À
c
V# ~ ²c³ ²? c ? ³
~ ~
~ ²c³
<² c ³ b ² c ³ b ² c À³ b ²
c À³ b ²
c ³ b ² c ³ = ~ ÀÀ
c
V#
c
~ c
²? c ? ³ c ~ ´² c À³ b ²À c À³ b ² c À³ µ c À
V
~ À À
V ~
A
~
b À
À
~ .
33. In the empirical distribution based on the three sample values, the median is the middle
value, so the value of in the empirical distribution is . We wish to find ,´2V c ³ µ .
From the original sample of three points, Á Á , there are 27 possible equally likely bootstrap
samples ²? Á ? Á ? ³ with the following medians:
Sample : Median
(1,1,1):1 , (1,1,2):1 , (1,1,4):1 , (1,2,1):1 , (1,2,2):2 , (1,2,4):2 , (1,4,1):1 , (1,4,2):2 , (1,4,4):4 ,
(2,1,1):1 , (2,1,2):2 , (2,1,4):2 , (2,2,1):2 , (2,2,2):2 , (2,2,4):2 , (2,4,1):2 ,
(2,4,2):2 , (2,4,4):4 , (4,1,1):1 , (4,1,2):2 , (4,1,4):4 , (4,2,1):2 , (4,2,2):2 , (4,2,4):4
(4,4,1):4 , (4,4,2):4 , (4,4,4):4
Of the 27 possible bootstrap samples, 7 sample medians are equal to 1, 13 sample medians are
equal to 2 and 7 sample medians are equal to 4. The bootstrap estimate of the mean square error
is the estimate of ,´²V c ³ µ , which is
´² c ³ b ² c ³ b ² c ³ µ ~ À
.
,´?wµ
34. 3,9 ~ ,´?µ . ,´?µ ~ %² b ³ ~ b h ~ À .
For the lognormal distribution,
c c ,´? w µ ~ ´%² b ³µ h )²
³ b ´ c )² c ³µ .
c
,´? w µ ~ ²À³)² c c ³ b ´ c )² ³µ
~ ²À³)² c À³ b ´ c )²À³µ
~ ²À³²À
³ b ´ c À
µ ~ À .
c
,´? w µ ~ ²À³)² c c ³ b ´ c )² ³µ
~ ²À³)² c À³ b ´ c )²À
³µ
~ ²À³²À³ b ´ c Àµ ~ À .
3,9 ~
3,9
3,9
~
,´?wµ
À
~ À
~ À
,´?µ
À
À ~ À
. 56% increase
, 3,9 ~
in LER.
,´?wµ
,´?µ
À
~ À
~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 7
PE-139
35. According to the delta method suppose that ²%Á &³ is a function of two variables, with partial
C
C
derivatives C%
~ ²³ and C&
~ ²³ . The approximate variance of ²V , V ³ is
²²³ ²V , V ³³ h = ´ V µ b ²³ ²V , V ³ h ²³ ²V , V ³ h *#´ V , V µ
b ²²³ ²V , V ³³ h = ´ V µ .
V , and ²
V³ ~ V . Then,
In this case the estimators are denoted V and VÁ V b C
C
V ³ ~ and
V ³ ~ 10 . According to the delta method, we get
²
²
VÁ VÁ C
V
V
C
V µ ~ ²³ h = ´
V µ b ²³ h = ´
Vµ
= ´
V b Vµ b ²³²³ h *#´
VÁ ~ À b ²³² c À³ b ²³²À³ ~ À .
Standard deviation is jÀ ~ À .
Note that this can also be found from basic probability rules
V µ ~ = ´
V µ b ²³ h *#´
Vµ .
= ´
V b Vµ b ²³ h = ´
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-140
PRACTICE EXAM 7
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 8
PE-141
ACTEX EXAM C/4 - PRACTICE EXAM 8
1. An actuary has determined that the number of claims follows a negative binomial distribution
with mean 3 and variance 12. Calculate the probability that the number of claims is at least 3 but
less than 6.
A) Less than 0.20
B) At least 0.20, but less than 0.25
C) At least 0.25, but less than 0.30
D) At least 0.30, but less than 0.35
E) At least 0.35
2. You are given:
(i) The claim count 5 has a Poisson distribution with mean \$.
(ii) \$ has a gamma distribution with mean 1 and variance 2.
Calculate the probability that 5 ~ 1.
A) 0.19
B) 0.24
C) 0.31
D) 0.34
E) 0.37
3. An insurance agent will receive a bonus if his loss ratio is less than 70%. You are given:
(i) His loss ratio is calculated as incurred losses divided by earned premium on his block of
(ii) The agent will receive a percentage of earned premium equal to 1/3 of the difference between
70% and his loss ratio.
(iii) The agent receives no bonus if his loss ratio is greater than 70%.
(iv) His earned premium is 500,000.
(v) His incurred losses are distributed according to the Pareto distribution:
Á
- (%) ~ 1 c ² %b
Á ³ , %  0
Calculate the expected value of his bonus.
A) 16,700
B) 31,500
C) 48,300
D) 50,000
E) 56,600
4. A compound Poisson claim distribution : has Poisson parameter ~ and severity
prob. .4
distribution ? ~ H 2 prob. .2
3 prob. .4
A deductible of 1 is applied to each individual claim ? . The aggregate insurance payment
(after deductibles are applied) is : Z . Stop loss insurance with a deductible of is applied to the
original distribution of : . Find the deductible for which the expected stop loss insurance
payment is the same as ,´: Z µ.
A) Less than 2.00
B) At least 2.00 but less than 2.25
C) At least 2.25 but less than 2.50
D) At least 2.50 but less than 2.75
E) At least 2.75
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-142
PRACTICE EXAM 8
5. Given:
(i) denotes the probability that the number of claims equals for ~ Á Á Á ÀÀÀ
(ii) ~ [
[ ,  ,  Using the corresponding zero-modified claim count distribution with 4 ~ À, calculate 4 .
A) 0.1
B) 0.3
C) 0.5
D) 0.7
E) 0.9
6. A random sample of 1000 observations from a loss distribution has been grouped into five
intervals as follows:
Number of
Observations
Interval
´ Á À³
´À Á À³
´À Á À³
´À Á À³
´À Á B³
Find the empirical limited expected value of the loss distribution with a policy limit of 15.0 .
A) Less than 9.9
B) At least 9.9, but less than 10.0
C) At least 10.0, but less than 10.1
D) At least 10.1, but less than 10.2
E) At least 10.2
7. The 95% linear confidence interval for :²! ³ is ²À
Á À
³ À
Determine the 95% log-transformed confidence interval for :²! ³ .
A) ²À Á À
³
B) ²À
Á À³ C) ²À
Á À³
D) ²À Á À
³ E) ²À Á À
³
%
%~
8. The random variable whose distribution function is - ²%³ ~ H c c c% %  is being estimated in two ways:
(i) the method of moments, and (ii) the method of percentiles.
A sample has been collected, and the following is known:
- the sample mean is 2.8 and the sample second moment is 29
- the sample 60-th percentile is 2.3 and the sample 80-th percentile is 5.2 .
It is known that  À .
Find the absolute difference in the estimates of by the two methods.
A) Less than .01 (B) At least .01, but less than .02
C) At least .02 but less than .03
D) At least .03, but less than .04
E) At least .04
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 8
PE-143
Use the following information for Questions 9 and 10.
The observed number of claims for a group of 1000 risks has been recorded as follows:
Number of Risks
Number of Claims
0
729
1
242
2
29
3 or more
0
9. A Poisson distribution is fit to the data by the method of maximum likelihood estimation.
Find the estimated probability of a risk incurring no claims.
A) .72
B) .73
C) .74
D) .75
E) .76
10. The minimum chi-square estimate of the mean of the Poisson distribution is .3055 .
Determine the chi-square goodness-of-fit statistic to test the hypothesis that the Poisson with
parameter .3055 is a good fit to the data. The chi-square statistic is based on three groupings,
those risks with 0 claims, those risks with 1 claim, and those risks with 2 or more claims.
A) Less than 3.8
B) At least 3.8, but less than 4.0
C) At least 4.0, but less than 4.2
D) At least 4.2, but less than 4.4
E) At least 4.4
11. A Pareto distribution is know to have a value of ~ . A random sample of 100
observations is taken from the distribution. 30 of the observations were right-censored at 50 (all
30 values are over 50, but the actual values are not known). The maximum likelihood estimate of
is 3.50 . Suppose that of the 30 data points censored at 50, 25 of them are greater than 60, and
the others are known to be 52 , 55 , 56 , 56 and 58 . Using this information, find the revised mle
of .
A) Less than 3.0
B) At least 3.0 but less than 3.1
C) At least 3.1 but less than 3.2
D) At least 3.2 but less than 3.3
E) At least 3.3
12. Which of the following statements regarding selecting and validating a model is false?
A) When performing a chi-square goodness-of-fit test, a common convention is to have at least 5
expected observations in each grouping cell.
B) The likelihood ratio test is an appropriate hypothesis test to compare any two models.
C) The Schwartz Bayesian Criterion depends on sample size.
D) The critical values for the Kolmogorov-Smirnov test depend on whether or not parameters for
the model were estimated.
E) A null hypothesis is accepted if the significance level is less than the -value of the test
statistic.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-144
PRACTICE EXAM 8
13. The number of claims for an individual insured in a one year period has a Poisson
distribution with mean . The parameter has an exponential distribution with mean 1.
For a particular insured, the numbers of claims in different years are independent. Over a two
year period, the total number of claims for a particular insured is found to be 3. The posterior
distribution of is found and has pdf ²³. Later, it is found that the individual had 3 claims in
the first year and 0 claims in the second year. Based on this new information, the posterior
²³
distribution of is calculated again and is found to have pdf ²³. Find ²³ .
A) 3
B) 1
C) D) E) 14. Claim sizes follow a gamma distribution with pdf ²&³ ~
²&°³ c&°
&!²³
. You are given that
the value of the parameter is .5 , and that the prior distribution of is uniform on the interval
²Á ³. Determine the value of in Buhlmann's credibility approach.
A) B) C) D) E) 15. Simulation is to be applied to a compound aggregate claim distribution : by first simulating
the frequency and the simulating the severity for the appropriate number of claims. The inverse
transform method is used for simulation of both the frequency and the severity. The simulation is
performed so that small random uniform numbers correspond to small claim frequencies and
small severities. The claim frequency is binomial with parameters ~ , ~ À , and the
severity distribution is Pareto with mean 2,000 and variance 20,000,000 . One simulation of the
compound aggregate claim distribution is performed, with uniform random number .25 used to
simulate the frequency, and uniform random numbers .2 , .8 , .6 , .4 used (as many as needed, in
the order given) to simulate severity amounts. Find the simulated value of : .
A) 2950
B) 2970
C) 2990
D) 3010
E) 3030
²³
²³
16. TVaR and TVaR denote the Conditional Tail Expectation for exponential random
²³
variables with means 1 and 2, respectively. Find lim TVaR²³ .
A) 0
B) C) 1
D) 2
E) B
¦ TVaR
17. You are modeling a claim process as a mixture of two independent distributions ( and ) .
You are given:
(i) Distribution ( is exponential with mean 1.
(ii) Distribution ) is exponential with mean 10.
(iii) Positive weight is assigned to distribution (.
(iv) The standard deviation of the mixture is 2.
Determine using the method of moments.
A) 0.960
B) 0.968
C) 0.972
D) 0.979
E) 0.983
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 8
PE-145
18. The future lifetimes of a certain population can be modeled as follows:
(i)
Each individual’s future lifetime is exponentially distributed with constant hazard rate .
(ii)
Over the population, is uniformly distributed over (1, 11).
Calculate the probability of surviving to time 0.5, for an individual randomly selected at time 0.
A) 0.05
B) 0.06
C) 0.09
D) 0.11
E) 0.12
19. ? is a continuous non-negative random variable with a finite mean and infinite support
(defined on ²Á B³) and with pdf ²%³ and cdf - ²%³. Two new random variables, @ and A are
defined related to ? .
@ has cdf [email protected] ²&³ ~
& ´c- ²%³µ %
,´?µ
and A has cdf -A ²'³ ~
' %h ²%³ %
,´?µ
.
You are given that @ and A have similar (proportional) right tails.
Which of the following could be the distribution of ? ?
I. Exponential distribution
II. Pareto distribution
A) I only
B) II only
C) Both I and II
D) Neither I nor II
E) The correct answer is not given by A, B, C, or D
20. ? is a loss random variable.
An insurance policy pays nothing for losses below an ordinary deductible of 10.
The insurance pays half of the loss amount in excess of 10 up to a loss of amount 50.
The insurance pays the full loss amount for any loss in excess of 50.
Express the expected cost per loss in terms of factors of the form ,´? w %µ Á -? ²%³
and constants.
A) ,´?µ c ²,´? w µ c ,´? w µ³ b ²³´ c -? ²³µ
B) ,´?µ c ²,´? w µ b ,´? w µ³ b ²³´ c -? ²³µ
C) ,´?µ c ²,´? w µ c ,´? w µ³ b ²³´ c -? ²³µ
D) ,´?µ c ²,´? w µ b ,´? w µ³ b ²³´ c -? ²³µ
A) ,´?µ c ²,´? w µ c ,´? w µ³ b ²³´ c -? ²³µ
E) ,´?µ c ²,´? w µ b ,´? w µ³ c ²³´ c -? ²³µ
21. Aggregate claims : follow a compound distribution with a Poisson frequency distribution 5
with mean 1, and a geometric severity distribution ? with probability function
7 ²? ~ ³ ~ ²b ³b , for ~ Á Á Á ÀÀÀ
It is found that ,´:O:  µ ~ c c° . Determine ,²?³ .
A) .5
B) .6
C) .7
D) .8
E) . 9
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-146
PRACTICE EXAM 8
22. Survival is being estimated using the Product-Limit estimator. It is found that
: ²! ³ ~ À Á : ²! ³ ~ À , and the estimated (Greenwood approximation) variances of
these estimates is = V
´: ²! ³µ ~ À Á = V
´: ²! ³µ ~ À À Determine the number at risk
just before the deaths that take place at death point ! .
A) 35
B) 40
C) 45
D) 50
E) 55
23. A company provides insurance to a concert hall for losses due to power failure. You are
given:
(i) The number of power failures in a year has a Poisson distribution with mean 1.
(ii) The distribution of ground up losses due to a single power failure is:
%
10
20
50
Probability of %
.30
.30
.40
(iii) The number of power failures and the amounts of losses are independent.
(iv) There is an annual deductible of 30.
Calculate the expected amount of claims paid by the insurer in one year.
A) 5
B) 8
C) 10
D) 12
E) 14
24. You are given the following information. Losses follow a log-normal distribution with
parameters and . The following five losses have been observed:
500 , 1000 , 1500 , 2500 , 4500
Using the method of moments, determine the probability that a loss will exceed 4500.
A) Less than .04 B) At least .04, but less than .08 C) At least .08, but less than .12
D) At least .12, but less than .16
E) At least .16
25. The random variable ? has a uniform distribution on the interval ´Á µ . A random sample of
three observations of ? has been recorded and grouped as follows:
Number of
Observations
Interval
´Á ³
´Á ³
´Á ³
Determine the maximum likelihood estimator of .
A) 5
B) 7.5
C) 10
D) 5 b E) 10 c © ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 8
PE-147
26. Losses are assumed to follow a Pareto distribution with parameters ~ and and density
function ²%³ ~ ²%b
³ for %  . 300 losses have been observed and the parameter is
estimated using the method of moments. Find the variance of that estimator in terms of the true
parameter value.
A) À
B) À
C) À
D) À
E) À
27. : has a compound distribution in which claim frequency 5 follows a Poisson distribution
and claim severity ? is independent of claim frequency and has the following distribution:
²%³ ~ %c° Á %  .
A full credibility standard for the claim frequency random variable 5 is determined so that the
total number of claims is within 5% of the expected number with probability 98%. If that same
number of claims is used as a full credibility standard for the number of claims needed for the
total cost of claims : , the standard would be that of the total claim amount being within 100% of
the expected cost with probability 95%. Using a normal approximation to the aggregate loss
distribution, determine .
A) Less than .04
B) At least .04 , but less than .05
C) At least .05, but less than .06
D) At least .06, but less than .07
E) At least .07
28. : ~ state of the world , for ~ Á Á .
The probability of each state ~ .
In any state, the probability of a claim ~ .
The claim size is either 1 or 2 units.
Given that a claim has occurred, the following are conditional probabilities of claim size (in units)
for each possible state:
:
:
:
7 ²³ ~ 7 ²³ ~ 7 ²³ ~ 7 ²³ ~ 7 ²³ ~ 7 ²³ ~ Use the data given above and Bayes' Theorem. If you observe a single claim of size 2 units, in
which range is your estimate of the pure premium for that risk?
A) Less than .65
B) At least .65 but less than .67
C) At least .67 but less than .69
D) At least .69 but less than .71
D) .71 or more
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-148
PRACTICE EXAM 8
29. A portfolio of 200 independent insureds is subdivided into two classes as follows:
Expected
Variance
Number
Numberof Number
Expected
Variance of
of
of Claims
of Claims
Severity
Severity
Per Insured
Per Insured
Per Claim
Per Claim
Class Insureds
1
50
.25
.75
4
20
2
150
.50
.75
8
36
Claim count and severity for each insured are independent. An insured is selected at random
from this portfolio, and its claim amount ? for one exposure period is observed.
Find the Buhlmann credibility premium for the same insured for the next exposure period.
A) À
B) À? b À
C) À? b À D) À? b À
E) À? b À
30. Ten urns each contain 5 balls, numbered as follows:
Urns 1 to 5: 1,2,3,4,5
Urn 6: 1,1,1,1,1 Urn 7: 2,2,2,2,2 Urn 8: 3,3,3,3,3 Urn 9: 4,4,4,4,4 Urn 10: 5,5,5,5,5
An urn is selected at random and a ball selected at random from that urn is found to have the
number 2 on it (the ball is then returned to the urn). Another ball is selected at random from the
same urn and found to have the number 3 on it (the ball is then returned to the urn). A third ball
is selected from the same urn. Let ) denotes the Bayesian analysis estimate of the expected
value of the number on the third ball, and let )" denote the Buhlmann credibility estimate of the
expected value of the number on the third ball. Find the ratio ) °)" .
A) 5/4
B) 9/8
C) 1
D) 8/9
E) 4/5
%
%°
%
°
%~
31. A claim amount distribution has cdf - ²%³ ~
.
²% b ³°  %  %
The inversion method is applied to simulate 5 values of the claim amount distribution.
The five uniform random numbers to be used in the simulation are
À Á À Á À Á À Á À
Find the sample mean of the five claims amounts simulated.
A) .90
B) .92
C) .94
D) .96
E) .98
H
32. Random variables ? and @ have a joint distribution with density function ²%Á &³ ~ % b &
for  %  and  &  . The following method is used to simulate an ²?Á @ ³ pair from
random uniform ²Á ³ numbers " and #.
• The value of % is simulated from the distribution of ? using " and the inverse
transform method.
• The value of & is simulated from the conditional distribution of @ given % using # and
the inverse transform method.
Find the pair ²% Á & ³ if " ~ and # ~ À
A) ²ÀÁ À³ B) ²ÀÁ À³ C) ²ÀÁ À³ D) ²ÀÁ À³
E) ²ÀÁ À³
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 8
PE-149
33. Speedy Delivery Company makes deliveries 6 days a week. Accidents involving Speedy
vehicles occur according to a Poisson process with a rate of 3 per day and are independent. In
each accident, damage to the contents of Speedy's vehicle is distributed as follows:
Amount of damage
Probability
\$ 0
1/4
\$2,000
1/2
\$8,000
1/4
Using the normal approximation, calculate the probability that Speedy's weekly aggregate
damages will not exceed \$63,000.
A) 0.24
B) 0.31
C) 0.54
D) 0.69
E) 0.76
34. The parameter has a prior distribution with pdf ²³ ~ for   .
The conditional distribution of ? given is uniform on the interval ²Á ³ .
Find the posterior density ²O%³ .
A) Uniform on the interval ²Á %³
B) Uniform on the interval ²Á c %³
C) Uniform on the interval ²%Á ³
D) Uniform on the interval ² c %Á ³
E) Uniform on the interval ² c %Á %³
35. Using the Product-Limit estimator, estimate the mean time until death given the following
sample of ten policies.
2
3
3
5
5+
6
6
7+
9
9
(+ indicates that the loss exceeded the policy limit)
A) Less than 5.0
B) At least 5.0 but less than 5.4
C) At least 5.4 but less than 5.8
D) At least 5.8 but less than 6.2
E) At least 6.2
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 8
ACTEX EXAM C/4 - PRACTICE EXAM 8 SOLUTIONS
1. We want
7 ²5 ~ Á Á ³ ~
²b³²b³ [ ²b ³b
b
²b³²b³²b³ [ ²b ³b
b
²b³²b³²b³²b³ [ ²b ³b
.
,´5 µ ~ ~ , = ´5 µ ~ ² b ³ ~ S ~ Á ~ .
We see that since ~ , the distribution is geometric with probability function
~ ²b ³b ~ b À The probability is b b b ~ À .
2. If the conditional distribution of 5 given \$ ~ is Poisson with parameter , and if \$ has a
gamma distribution with parameters and , then the unconditional distribution of 5 is negative
binomial with parameters ~ and ~ . This is a relationship that has come up repeatedly
over the years. In this problem, \$ has a gamma distribution with mean ~ and variance
~ , so that ~ , ~ . Therefore, the unconditional distribution of 5 is negative
binomial with ~ and ~ , so that
7 ´5 ~ µ ~ [²b ³b ~ °
~ À .
´À c loss ratioµ h ´earned premiumµ ~ ´Á c ?µ if loss ratio  À
3. Bonus ~ F if loss ratio  À
~ ´Á c ²?Á Á ³µ .
The expected bonus is ´Á c ,´²?Á Á ³µµ .
Á
,´²?Á Á ³µ ~ % ²%³ % b Á ´ c - ²Á ³µ
Á
~ Á
´ c - ²%³µ % ~ Á ² %b
Á
³ % ~ Á .
Expected bonus is ´Á c Á µ ~ Á Answer: E
4. ,´²? c ³b µ ~ ²À³ b ²À³ ~ S ,´: Z µ ~ ,´5 µ h ,´²? c ³b µ ~ ²³²³ ~ With no deductible, ,´:µ ~ ²³²³ ~ À
With a deductible of 1 for stop loss insurance, ,´²: c ³b µ ~ ,´:µ c ,´: w µ ~ c ,´: w µ
where ,´: w µ ~ 7 ²:  ³ ~ c 7 ²: ~ ³ ~ c c .
Therefore, ,´²: c ³b µ ~ c ² c c ³ ~ b c .
With a deductible of 2 for stop loss insurance, ,´²: c ³b µ ~ ,´:µ c ,´: w µ ~ c ,´: w µ
where ,´: w µ ~ 7 ²: ~ ³ b 7 ²:  ³ ~ c h ²À³ b ´ c 7 ²: ~ or 1³µ .
7 ²: ~ or ³ ~ 7 ²: ~ ³ b 7 ²: ~ ³ ~ c b c h ²À³ ~ Àc .
Then ,´: w µ ~ Àc b ´ c Àc µ ~ c Àc , and
,´²: c ³b µ ~ c ² c Àc ³ ~ b Àc ~ À . This is still larger than 2.
With a deductible of 3 for stop loss insurance, ,´²: c ³b µ ~ ,´:µ c ,´: w µ ~ c ,´: w µ
where ,´: w µ ~ 7 ²: ~ ³ b 7 ²: ~ ³ b 7 ²:  ³ .
We have seen that 7 ²: ~ ³ ~ c .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 8
PE-151
4. continued
7 ²: ~ ³ ~ 7 ²5 ~ ³ h 7 ²? ~ ³ b 7 ²5 ~ ³ h ´7 ²? ~ ³µ
c
~ c h ²À³ b h ²À³ ~ Àc .
7 ²:  ³ ~ c 7 ²: ~ Á Á ³ ~ c ´c b Àc b Àc µ ~ c Àc À
Then, ,´²: c ³b µ ~ c Àc c ²Àc ³ c ´ c Àc µ ~ b Àc ~ À  .
Since : is integer valued, in order to have an expected stop loss payment of 2, the deductible
must be between 2 and 3. ,´²: c ³b µ is a linear function for between 2 and 3.
Applying linear interpolation for   , we see that
,´²: c ³b µ ~ ,´²: c ³b µ c ² c ³ h ´ ,´²: c ³b µ c ,´²: c ³b µ µ
~ À c ² c ³ h ²À c À³ .
In order for this to be 2, we get À c ² c ³ h ²À c À³ ~ , so that ~ À .
B
B
~
~
5. From (ii), ~ °[ S ~ ~ h [ ~ h S ~ c Á
~ c h [ S the distribution is Poisson with mean 1.
The zero-modified distribution with 4 ~ À has probability function
À
À
4
c
4 ~ c
h ~ c
³ ~ À .
c h ~ À S ~ ²À³²
6. We use the grouped data form for LEV, which uses average claim amount within each
claim interval below the policy limit
V w µ ~ (.001)´²À³ b ²À³ b ²À³ b ²³µ ~ À À Answer: B
,´?
7. The linear confidence interval for :²! ³ is of the form : ²! ³ f À
h j= V
´: ²! ³ .
Since this is a symmetric interval, it follows that : ²! ³ ~ À (the midpoint of the interval),
and À
h j= V
´: ²! ³ ~ À (the half-width of the interval).
The log-transformed confidence interval has lower limit
À
hj= V
´: ²! ³
V ³µ =.
: ²! ³h´:²!
À
%´ ²À³ ²À³
µ~
´: ²!³µ°< and upper limit ´: ²!³µ< , where < ~ %<
For ! , the value of < is < ~ %<
À
hj= V
´: ²! ³
V ³µ =
: ²! ³h´:²!
~
À
.
The lower and upper limits are ´: ²! ³µ°< ~ ²À³°À
~ À , and
´: ²! ³µ< ~ ²À³À
~ À
.
8. This distribution is a mixture of the constant at zero, with weight c , and the exponential
distribution with mean , with weight . The first and second moments of ? are ,´?µ ~ h and ,´? µ ~ h since the moments of a mixed distribution are the weighted averages of the
moments of the individual distributions (the moments of the constant at 0 are all 0, and the first
and second moment of an exponential are and )
Applying the method of moments, we have h ~ À and h ~ , from which we get
²³²À³
V ~ ~ À (and V ~ À).
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 8
8. continued
Since we are told that  À , we know that the 60-th and 80-th distribution percentiles are  ,
with - ²%³ ~ c c% . Applying the method of percentiles, we have
À
~ c cÀ and À ~ c cÀ . Then, cÀ ~ À , and cÀ ~ À ,
so that À ~ , and V ~ À (and V ~ À
) .
The absolute difference between the two estimates of is OÀ c ÀO ~ À
9. The mle of the Poisson parameter is the same as the moment estimate, which is
V ~ Total number of observed claims ~ b²³ ~ À À
Total number of exposures
V
Estimated probability of no claims is 7 ´5 ~ µ ~ c ~ cÀ ~ À . Answer: C
²6 c, ³
10. ~ ~
,
, where 6 ~ (0 claims)Á 6 ~ (1 claim)Á 6 ~ (2 or more
claims). , ~ cÀ ~ À Á , ~ , ~ c ², b , ³ ~ À .
~
²c
À³
À
b
²cÀ³
À
b
²cÀ³
À
cÀ ²À³
[
~ À Á
~ À
.
11. Based on the original data, the mle of is V~ Answer: A
% b
' ² ³b
² "b ³
,
where ~ is the number of non-censored data points, the % 's are the uncensored sample
values, ~ and " ~ . Therefore À ~ ,
' ²
% b
³b ² ³
from which we get ' ² %b ³ ~ À
.
Using the revised data, there are now 75 uncensored observations, and the mle is
V ~ ~
' ²
% b
³b ² ³
~ À .
À
b² ³b²
³b²
³b² ³b² ³b ² ³
12. A. True.
B. False.
C. True. D. True.
E. True.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 8
PE-153
13. Given a gamma prior with parameters and and Poisson data values Á ÀÀÀÁ , the
posterior is also gamma with Z ~ b ' and Z ~ b . Therefore, the posterior depends
only on ' , not on the individual values, so and are the same.
The algebraic verification of this is as follows. The prior pdf of is ²³ ~ c (exponential
with mean 1). The pf of number of claims in one year given is ²O³ ~ c h [ (Poisson
with mean ). Since number of claims for a particular individual are independent from one year
to the next, given the total number of claims for the individual in two years has a Poisson
distribution with mean (the sum of independent Poisson random variables is also Poisson and
²³
we add the means, b ), and pf ²O³ ~ c h [ .
²Á³
Given claims in two years, the posterior density of is ²³ ~ ²³ , where
²Á ³ ~ ²O³ h ²³ is the joint density of and (²³ ~ is the prior density of )
B
and ²³ ~ ²Á ³ is the marginal probability function of .
Therefore, with ~ ,
²³
²Á ³ ~ ²O³ h ²³ ~ c h [ h c ~ c h , and
c h
²Á³
B
²³ ~ c h , and then ²³ ~ ²³ ~ B c c
~ Bchh .
h This is the posterior density of if the total number of claims in the first two years is 3.
Now we assume that the number of claims in the first year is 3 and the number in the second year
is 0. Then, ²O³ ~ c h [ ( is the pf for number claims in one year), and
²O³ ~ c h [ ~ c , and ² ~ Á ~ O³ ~ c h [ h c ~ c h [ is the joint pf
of number of claims in first and second year being 3 and 0, respectively. The posterior density of
in this case is found in much the same way is the first case:
² Á Á³
² Á O³h²³
c h hc
c
²³ ~ ² Á ³ ~ B ² Á O³h²³ ~ B c [ c ~ Bchh ,
h h [
which is equal to ²³. Therefore, the posterior distribution of is the same in both cases.
14. ~ # , where # ~ ,´#²#³µ ~ ,´process varianceµ , and
~ = ´²#³µ ~ = ´hypothetical meanµ .
In this case, # is the parameter which has uniform distribution on ²Á ³ .
The conditional distribution of ? is gamma with parameters and ~ .5 .
Then, the hypothetical mean is ²³ ~ ,´?Oµ ~ ~ À (the mean of a gamma distribution),
and the process variance is #²³ ~ = ´?Oµ ~ ~ À .
(The variance of a gamma distribution; with reference to the distribution table made available
with the exam, for a gamma distribution with parameters and , the first moment is
!²b³
!²³
~ , since the gamma function ! satisfies the relation !²³ ~ ² c ³!² c ³ , and
the second moment is
!²b³
!²³
~ ² b ³ , so that the variance is
² b ³ c ²³ ~ ).
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 8
14. continued
Since is uniformly distributed on ²Á ³, it has mean 2, and therefore,
# ~ ,´#²³µ ~ ,´Àµ ~ ²À³²³ ~ À . Also, the variance of is ~ (variance of the
uniform distribution on ²Á ³), so that
~ = ´²³µ ~ = ´Àµ ~ ²À³= ´µ ~ ²À³² ³ ~ .
À
Then ~ # ~ °
~ À
15. The binomial distribution with ~ and ~ À has pf and cdf
5¢
¢
²À³ ~ À 2 3²À³ ²À³ ~ À
À
À
À
- ¢
À
À
À
À
With random uniform number .25, since - ²³ ~ À  À  À ~ - ²³Á
the simulated value of frequency is 2 claims, so we must simulate 2 claims (severity) amounts.
The mean of the Pareto with parameters and is ,´?µ ~ c
, and the second moment is
,´? µ ~ ²c³²
c³ , so that the variance is
= ´?µ ~ ,´? µ c ²,´?µ³ ~ ²c³²
c³ c ² c ³ ~ ²c³ ²c³ .
We are given ,´?µ ~ c
~ and = ´?µ ~ ²c³ ²c³ ~ Á Á .
= ´?µ
ÁÁ
Therefore, ²,´?µ³ ~ ~ S ~ À S ~ .
c ~
À
The cdf of the severity (Pareto) will be - ²%³ ~ c ² %b
³ ~ c ² %b ³ .
For a given random uniform number " the simulated Pareto claim amount is %, where
À
À
" ~ - ²%³ S " ~ c ² %b
³ . For " ~ À we have À ~ c ² %b
³ ,
so that % ~ is the first claim amount. For " ~ À we have À ~ c ² %b ³À ,
so that % ~ is the second claim amount. The total : of the two simulated claims is 2990 .
16. For the exponential distribution with mean , TVaR ~ b VaR ~ ´ c ² c ³µ .
²³
²³
Therefore, TVaR ~ c ² c ³ and TVaR ~ ´ c ² c ³µ ,
²³
so that TVaR²³ ~ .
TVaR
17. The mean of the mixed distribution is
h ,´?( µ b ² c ³ h ,´?) µ ~ b ² c ³ ~ c .
For an exponential random variable with mean , the variance is and the second moment is
. The second moment of the mixed distribution is
h ,´?( µ b ² c ³ h ,´?) µ ~ ²³ b ² c ³²³ ~ c .
The variance of the mixed distribution is
c c ² c ³ ~ c c .
We set this equal to the given variance of the mixture, which is 4 (square of the standard
deviation), so that c c ~ . Solving the quadratic equation for results in
~ À Á c À , and we ignore the negative root. Alternatively, we can substitute the given
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 8
PE-155
18. Suppose that # is a continuous random variable with pdf # ²³, and suppose that the
distribution of @ depends upon (or is conditional upon) the parameter , with conditional pdf (or
pf for the discrete case of @ ³ @ O# ²&O³ . Then the unconditional (marginal) distribution of @ has
pdf @ ²&³ ~ @ O# ²&O³ h # ²³ , and distribution function
7 ´@  &µ ~ [email protected] ²&³ ~ [email protected] O# ²&O³ h # ²³ (and 7 ´@  &µ ~ c [email protected] ²&³ ~ ´ c [email protected] O# ²&O ³µ h # ²³ ).
This is the situation in this problem. # has density # ²³ ~ À for   (uniform on an
interval of length 10) and the conditional distribution of @ given # is exponential with
distribution function [email protected] O# ²&O³ ~ c c& (constant hazard rate ).
Then, 7 ´@  Àµ ~ ´ c [email protected] O# ²ÀO³µ h # ²³ ~ cÀ h ²À³ ~ À .
: ²!³
19. We are given that lim :@ ²!³ ~ , where   B
!¦B A
(this is the definition @ and A having similar right tails).
:@ ²!³
:A ²!³
~
[email protected] ²!³
c-A ²!³
~
! ´c- ²%³µ %
,´?µ
! %h ²%³ %
c ,´?µ
c
~
!
,´?µc ´c- ²%³µ %
.
!
,´?µc %h ²%³ %
When limit is taken as !SB, the numerator and denominator both approach 0, so we apply
l'Hospital's
rule to take the limit. According to l'Hospital's
rule, we differentiate with respect to !
V
V
both the numerator and denominator, and then take the limit of the ratio. The ratio is
c- ²!³
!h ²!³
:²!³
~ !h ²!³ ~ !h²!³
, where ²!³ is the hazard function of ? .
Therefore lim !h²!³ ~ .
!¦B
c%°
Suppose that ? has an exponential distribution. Then ²%³ ~ , and :²%³ ~ c%° ,
²!³
so that ²!³ ~ :²!³ ~ . Then, lim !h²!³
~ ! ~ !¦B
Therefore, ? cannot have an exponential distribution.
Suppose that ? has a Pareto distribution with parameters and . Then ²%³ ~ ²%b³b
²!³
and :²%³ ~ ² %b
³ , so that ²!³ ~ :²!³ ~ !b .
Then, lim !h²!³
~ lim !b
! ~ À
!¦B
!¦B
For the Pareto distribution,   B , and it follows that @ and A have similar right tails.
? can have a Pareto distribution.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 8
20. The amount paid by the insurance is @ ~ H
²?
?  c ³  ?  .
?  ?
?  ´²? w ³ c ²? w ³µ ~ H ²? c ³  ?  .
?  ?  .
²? c ³b ~ ? c ²? w ³ ~ F
? c ?  ?  Therefore, ´²? w ³ c ²? w ³µ b ? c ²? w ³ ~ H ²? c ³  ?  .
? c ?  Therefore @ ~ ´²? w ³ c ²? w ³µ b ? c ²? w ³
b a payment of 30 if ?  ,
so that the expected cost per loss is
,´@ µ ~ ,´ ´²? w ³ c ²? w ³µ b ? c ²? w ³µ b ²³´ c -? ²³µ .
~ ,´?µ c ²,´? w µ b ,´? w µ³ b ²³´ c -? ²³µ .
21. ,²?³ ~ . : is a non-negative integer-valued random variable with probability function
B
B
7 ²:~³
,²:³
7 ²: ~ ³ ~ . ,´:O:  µ ~ h 7 ²: ~ O:  ³ ~ h c7 ²:~³ ~ c7 ²:~³ À
~
~
The severity distribution is geometric with 7 ²? ~ ³ ~ b
.
,²:³ ~ ,²5 ³ h ,²?³ ~ ²³² ³ ~ .
B
7 ²: ~ ³ ~ ~ 7 ²: ~ O5 ~ ³ h 7 ²5 ~ ³
~
B
B
~ 7 ²? ~ q ? ~ q Ä q ? ~ O5 ~ ³ h 7 ²5 ~ ³ ~ ~
c
~
c
² b
³ ²³
~
h °²b ³ ~ c °²b ³ .
,²:³
[
À
,´:O:  µ ~ c7 ²:~³ ~ cc°²b³ ~ c c° ~ ccÀ°²bÀ³
.
We see that ~ À .
22. À ~ : ²! ³ ~ : ²! ³´ c µ ~ ²À³´ c µ S ~ À , where is the number of deaths
at time and is the exposure or number at risk just prior to ! .
À ~ = V
´: ²! ³µ ~ : ²! ³ h ² c
³ ~ ²À³ h
² c ³ S
² c ³ ~ À
! ! À ~ = V
´: ²! ³µ ~ : ²! ³ h
! ! ! ! ² c ³
! ! ~ ²À
³ h ² c
~ ² c
b ² c ³ ~ À À
³
³
! ! ! ! It follows that
S
² À
² c ³
~ À , so that
c ³ ~ S ~ .
² c ³
~ ²
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 8
PE-157
23. The overall expected aggregate loss is ,´:µ ~ ,´5 µ h ,´?µ , where 5 is the frequency and
? is the severity (ground up loss). ,´5 µ ~ and
,´?µ ~ ²À³ b ²À³ b ²À³ ~ , so that ,´:µ ~ .
The amount paid by the insurer is ,´²: c ³b µ ~ ,´:µ c ,´: w µ .
: must be a multiple of 10, and the probability function of : is
:
Event(s)
5 ~
5 ~ Á ? ~ 5 ~ Á ? ~ or
5 ~ Á ? ~ ? ~ Probability
c ~ À
c ²À³ ~ À
c ²À³ b
c
h ²À³ ~ À
,´: w µ ~ ²³²À³ b ²³²À
³ b ²³² c À
b À b À
³ ~ À .
The expected amount paid by the insurer is c À ~ À . Answer: E
24. For the log-normal distribution with parameters and , the first and second moments are
,´?µ ~ %² b ³ , and ,´? µ ~ % ² b ³ .
The method of moments matches the sample moments with the theoretical distribution moments,
from which the parameters are found. The first and second sample moments are
c
? ~ '? ~ , '? ~ Á Á .
The method of moments equations are
%² b ³ ~ Á and % ² b ³ ~ Á Á .
Taking natural logs we get b ~ Á and b ~ Á Á .
Solving for and results in V ~ À Á V ~ À .
Since ? is normal with mean and variance , we have
?cÀ
7 ´?  µ ~ 7 ´ ?  µ ~ 7 ´ j

À
² ³cÀ
µ
jÀ
~ 7 ´A  Àµ
where A  5 ²Á ³. From the standard normal table, 7 ´A  Àµ ~ À
c
25. The likelihood function is 3²³ ~ - ²³ h ´- ²³ c - ²³µ h ´ c - ²³µ ~ h c
h .
The loglikelihood is 3²³ ~ b ² c ³ b ² c ³ c The maximum likelihood equation is 3²³ ~ c
c ~ S ~ À À Answer: B
26. The first moment of the Pareto distribution is c
~ , since ~ is given.
According to the method of moments when one parameter is being estimated, we set the
V
c
c
distribution mean equal to the sample mean: ~ ? S V ~ ? . Then the variance of
= ´?µ
= ´?µ
c
c
this estimate will be = ´Vµ ~ = ´?µ ~ = ´?µ ~ ~ (since is given as
300 observations). The variance of the Pareto distribution is
= ´?µ ~ ,´? µ c ²,´?µ³ ~ ²c³²
c³ c ² c ³ ~ ²³²³ c ² ³ ~ .
= ´?µ
°
Then = ´Vµ ~ ~ ~ À À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 8
27. Since 5 has a Poisson distribution, the standard for full credibility for number of claims 5 is
² À
À ³ ~ (2.327 is the 99-th percentile of the standard normal distribution). If this
number is used as the number of claims needed for the standard for full credibility for total claim
= ´?µ
&
costs : , then ² ³ < b ²,´?µ³ = ~ , where ? is the severity distribution.
B
B
,´?µ ~ & h h &c° & ~ Á ,´? µ ~ & h h &c° & ~ S = ´?µ ~ c ² ³ ~ À
Since we want total claims to be within 100% of total expected claims 95% of the time, we have
² À
³ < b
°
²°³ =
~ S ~ À
.
28. We use the following notation: ? ~ size of first claim from the selected risk
? ~ size of second claim from the selected risk.
We wish to find ,´? O? ~ µ . We condition the expectation over the states of the world:
,´? O? ~ µ ~ ,´? O: µ h 7 ´: O? ~ µ
b ,´? O: µ h 7 ´: O? ~ µ b ,´? O: µ h 7 ´: O? ~ µ .
In each state of the world there is a chance of a claim occurring, and we are given the
distribution of claim amount in each state given that a claim occurs. To find ,´? O: µ we
condition over whether or not a claim occurs:
,´? O: µ ~ ,´? Oclaim occurs while in : µ h 7 ´claim occurs while in : µ
b ,´? Ono claim occurs while in : µ h 7 ´no claim occurs while in : µ
~ ´²³² ³ b ²³² ³µ h ² ³ b ²³ h ² ³ ~ .
In a similar way, we get ,´? O: µ ~ ´²³² ³ b ²³² ³µ h ² ³ b ²³ h ² ³ ~ , and
,´? O: µ ~ ´²³² ³ b ²³² ³µ h ² ³ b ²³ h ² ³ ~ .
We use conditional probability rules (Bayes' Theorem) to find 7 ´: O? ~ µ:
7 ²: O? ~ ³ ~
7 ²: q²? ~³³
7 ²? ~³
~
7 ²? ~O: ³h7 ²: ³
7 ²? ~³
.
From the distribution of ? in state 1, we have
7 ²? ~ O: ³
~ 7 ²a claim occurs in : ³ h 7 ²claim amount is 2 given that a claim occurs in : ³
~ ² ³² ³ ~ . We are given that 7 ²: ³ ~ 7 ²: ³ ~ 7 ²: ³ ~ , so that
7 ²: q ²? ~ ³³ ~ 7 ²? ~ O: ³ h 7 ²: ³ ~ ² ³² ³ ~ .
Since the state has to be one of 1, 2 or 3, we have
7 ²? ~ ³ ~ 7 ²: q ²? ~ ³³ b 7 ²: q ²? ~ ³³ b 7 ²: q ²? ~ ³³ , and the reasoning
used to find 7 ²: q ²? ~ ³³ can be used to find
7 ²:2 q ²? ~ ³³ ~ 7 ²? ~ O: ³ h 7 ²: ³ ~ ² ³² ³² ³ ~ and
7 ²: q ²? ~ ³³ ~ 7 ²? ~ O: ³ h 7 ²: ³ ~ ² ³² ³² ³ ~ .
We then get 7 ²? ~ ³ ~ ² ³ b ² ³ b ² ³ ~ .
°
Then, 7 ²: O? ~ ³ ~ °
~ .
°
In a similar way, we find 7 ²: O? ~ ³ ~ °
~ , and
°
7 ²:3 O? ~ ³ ~ °
~ .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 8
PE-159
28. continued
Finally, we get the expectation we are looking for:
,´? O? ~ µ ~ ,´? O: µ h 7 ²: O? ~ ³ b ,´? O: µ h 7 ²: O? ~ ³
b ,´? O: µ h 7 ²: O? ~ ³
~ ² ³² ³ b ² ³² ³ b ² ³² ³ ~ Answer: D
~ À
.
29. The credibility premium is A? b ² c A³ .
The collective premium is ~ ,´hypothetical meanµ and
A ~ b
# , where ~ (data from one exposure period) , # ~ ,´process varianceµ , and
~ = ´hypothetical meanµ . Information within each class is given in form of frequency 5 and
severity @ .
For an insured from Class 1 the hypothetical mean (expected claim for an insured from Class 1) is
²Class 1³ ~ ,´5 µ h ,´@ µ ~ ²À³²³ ~ ,
and from Class 2, ²Class 2³ ~ ²À³²³ ~ .
~ ,´/4 µ ~ ²Class 1³²Probability of choosing Class 1³
b ²Class ³²Probability of choosing Class ³ ~ ²³² ³ b ²³² ³ ~ À .
The variance of the hypothetical mean is
~ = ´/4 µ ~ ,´/4 µ c ²,´/4 µ³ ~ ² ³² ³ b ² ³² ³ c ²À³ ~ À
For an insured from Class 1, the process variance (variance of a claim for an insured from
Class 1) is #²Class 1³ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ ~ ²À³²³ b ²À³² ³ ~ ,
and for Class 2 it is #²Class 1³ ~ ²À³²
³ b ²À³² ³ ~ .
The expected value of the process variance is
# ~ ,´7 = µ ~ ²³² ³ b ²
³² ³ ~ À À
#
À
Then ~ ~ À
~ À , and A ~ bÀ
~ À04 .
The credibility premium is A? b ² c A³ ~ À? b À²À³ ~ À? b À . Answer: B
30. Since the urn must contain a 2 and a 3, the urn must be one of urns 1 to 5. All of urns 1 to 5
have the same content. Thus the Bayesian analysis estimate of the third ball number is 3, the
expected number on the balls in urns 1 to 5.
This could have been found in a more formal algebraic way, identifying events and using
conditional probabilities as follows:
,´third ballO2,3µ ~ ,´third ballOurn 1µ h 7 ²urn 1O2,3³ b
Ä b ,´third ballOurn 10µ h 7 ²urn 10O2,3³ . But then for urns ~ 6Á ÀÀÀÁ 10, 7 ²urn µ2,3³ ~ .
7 ²ÁOurn 1³h7 ²urn 1³
7 ²urn 1O2,3³ ~ 7 ²ÁOurn 1³h7 ²urn 1³bÄb7 ²ÁOurn 5³h7 ²urn 5³
~
² ³² ³² ³
~ and the same for 7 ²urn O2,3³ for ~ Á Á Á .
² ³² ³² ³bÄb² ³² ³² ³
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 8
30. continued
Thus, ,´third ballO2,3µ
~ ² ³ b ² ³ b ² ³ b ² ³ b ² ³ b ²³ b ²³ b ²³ b ²³ b ²³ ~ ~ ) .
We find the Buhlmann credibility estimate of the expected value by first finding Á #Á and .
The hypothetical means for the urns are the expected ball number for each of the urns
/4 ~ ,´ballOurn 1µ ~ ~ /4 ~ /4 ~ /4 ~ /4 ,
and /4
~ Á /4 ~ Á /4 ~ Á /4 ~ Á /4 ~ À
Since each urn is equally likely to be picked,
~ ,´/4 µ ~ ² ³´ b b b b b b b b b µ ~ .
The process variance for each of the urns is the variance of the ball number for each urn.
For urns ~ 1 to 5, the process variance is 7 = ~ ´² c ³ b Ä b ² c ³ µ ~ ,
and for urns ~ 6 to 10, the process variance is 7 = ~ , since there is no variation within any
of those 5 urns. Then # ~ ,´7 = µ ~ ² ³´ b b b b b b b b b µ ~ À
The variance of the hypothetical means is ~ ,´/4 µ c ²,´/4 µ³ , where
,´/4 µ ~ ² ³´ b b b b b b b b b µ ~ ,
and ,´/4 µ ~ was found earlier. Then ~ = ´/4 µ ~ c ~ .
c
The Buhlmann credibility premium estimate is )" ~ A? b ² c A³ , where
c
A ~ b
# ~
~ (there were ~ picks). Since ? ~ À , we have
)" ~
b ² ³²À³ b ² ³²³
~
. Then ) °)" ~ ° . Answer: B
31. The simulated value is % , where - ²%³ ~ " if  "  À or À  "  , since ? is
continuous there, and the simulated value is 1 if À  "  À, since ? has a probability mass
there. " ~ À S % ~ Á " ~ À S % ~ Á " ~ À S - ²%³ ~ %° ~ À S % ~ À Á
" ~ À S - ²%³ ~ %° ~ À S % ~ À Á " ~ À S - ²%³ ~ ²% b ³° ~ À S % ~ À
.
Sample mean of the %'s is bbÀbÀbÀ
~ À
.
32. The marginal distribution of ? has pdf ? ²%³ ~ ²% b &³ & ~ % b À
%
The cdf of ? is -? ²%³ ~ ²! b ³ ! ~ % b% for  %  .
Applying the inverse transform method to simulate ? with " ~ À results in
% b%
~ S % b % c ~ S % ~ À Á c À . We ignore the negative root,
so that % ~ À . The density function of the conditional distribution of @ given ? ~ % is
%&b &
²%Á&³
&
%b&
@ O? ²&O%³ ~ ²%³ ~
, and the cdf is [email protected] O? ²&O%³ ~ %b! ! ~
?
%b %b %b for  &  . The cdf of the conditional distribution of @ given ? ~ À is
À&bÀ&
. Applying the inverse transform method
À
À&bÀ&
~ S & b & c ~ S & ~ À Á c .
À
[email protected] O? ²&OÀ³ ~
to simulate @ using # ~
results in
Again, we ignore the negative root for &. The ²?Á @ ³ pair is ²ÀÁ À³ .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 8
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33. In 6 days the expected number of accidents is 18 (6 days and an average of 3 accidents per
day). The aggregate damages for one week, say : , has a compound Poisson distribution with
~ (expected number of claims per week). The severity distribution ? has mean
,´?µ ~ ²³² ³ b ²Á ³² ³ b ²Á ³² ³ ~ Á , and the second moment is
,´? µ ~ ² ³² ³ b ²Á ³² ³ b ²Á ³² ³ ~ Á Á .
The mean and variance of : are ,´:µ ~ ,´?µ ~ Á and
= ´:µ ~ ,´? µ ~ Á Á . ? is an integer, so we apply the normal approximation
with continuity correction. We get
:cÁ
ÁÀcÁ
ÁÀcÁ
7 ´:  Á Àµ ~ 7 ´ j
 j
µ ~ )² j
³
~ )²À³ ~ À
.
ÁÁ
ÁÁ
ÁÁ
34. The model density is ²%O³ ~ for  %  .
The joint density is ²%Á ³ ~ ²%O³ h ²³ ~ h ~ on the triangular region  %   .
The marginal density of ? is
? ²%³ ~ % ²%Á ³ ~ % ~ ² c %³ for  %  .
The posterior density is
²%Á³
²O%³ ~ ²%³ ~ ²c%³
~ c%
for %   (this is uniform on the interval ²%Á ³ ).
?
B
35. We wish to estimate ,´?µ . We can use the rule ,´?µ ~ :²!³ ! .
Since we are using a discrete estimate of survival, the product-limit estimate of :²!³ is a step
function. The death points are at 2, 3, 5, 6, and 9À The numbers at risk ( at the -th death point)
and the number of deaths ( at the -th death point) are
Death Point
! ~ ! ~ ! ~ ! ~ ! ~ Risk Group ~ ~ ~ ~ ~ Number of
~
~
~
~
~
Deaths Note that at the 3rd death point (time 5) there is also a censored observation. That censored
observation is considered to be censored after the death at time 5 so that group at risk includes
that individual (but that individual will not be included in the risk group at the next death point).
A similar comment applies to the tie at time 9. The estimate of :²!³ is
: ²³ ~ c ~ À , : ²³ ~ ² c ³² c ³ ~ À ,
: ²³ ~ ² c ³² c ³² c ³ ~ À
, : ²
³ ~ ² c ³² c ³² c ³² c ³ ~ À
,
and : ²³ ~ .
!
À  !  À  !  : ²!³ ~
.
À
 !  À
 !  !
B
Then, ,´?µ ~ :²!³ ! ~ ²³²³ b ²À³²³ b ²À³²³ b ²À
³²³ b ²À
³²³ ~ À .
H
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 8
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 9
PE-163
ACTEX EXAM C/4 - PRACTICE EXAM 9
1. You are given:
• ? has density ²%³, where ²%³ ~ Á °% , for %  (single parameter Pareto with
~ )
• @ has density ²&³, where ²&³ ~ &c&° °Á (gamma with ~ and ~ )
Which of the following are true?
1. ? has an increasing mean residual life function.
2. @ has an increasing hazard rate.
3. ? has a heavier tail than @ based on the hazard rate test.
A) 1 only
B) 2 only
C) 3 only
D) 2 and 3 only
E) All of 1, 2 and 3
2. A population of auto insurance policies consists of three types of policies. Low risk policies
make up 60% of the population, medium risk policies make up 30% of the population, and the
other 10% are high risk. The number of claims per year for a low risk policy has a Poisson
distribution with a mean of .2. The number of claims per year for a medium risk policy has a
Poisson distribution with a mean of 1. The number of claims per year for a high risk policy has a
Poisson distribution with a mean of . For a randomly chosen policy from the population, the
variance of the number of claims in a year is 1.1701. Find the expected number of claims per year
for a high risk policy.
A) 1.5
B) 2.0
C) 2.5
D) 3.0
E) 3.5
Use the following information for Questions 3 and 4.
• losses follow a distribution with density function
c%°
²%³ ~ , %B
• there is a deductible of 500
• 10 losses are expected to exceed the deductible each year
3. Determine the amount to which the deductible would have to be raised to double the loss
elimination ratio.
A) Less than 550
B) At least 550, but less than 850
C) At least 850, but less than 1150
D) At least 1150, but less than 1450
E) At least 1450
4. Determine the expected number of losses that would exceed the deductible each year if all loss
amounts doubled, but the deductible remained at 500.
A) Less than 10
B) At least 10, but less than 12
C) At least 12, but less than 14
D) At least 14, but less than 16
E) At least 16
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 9
5. A company has 1,000 employees who are partly covered under a disability insurance plan.
The plan pays full salary for up to 4 weeks of disability (after which a government plan takes
over). Occurrences of disability among employees are independent of one another, and an
employee is covered for only one occurrence per year. The following information is known:
Disability
Employee Weekly
Number of
Category
Salary
Employees
Probability
1
400
300
.025
2
600
500
.02
3
800
200
.01
Length of
Disability
Probability
1 week
.8
2 weeks
.1
3 weeks
.05
 4 weeks
.05
The company actuary calculates the annual premium required to ensure (using the normal
approximation) with 95% probability that the premium will exceed disability claims. What
percentage of total weekly payroll is that premium (nearest %)?
A) 2%
B) 2.5%
C) 3%
D) 3.5%
E) 4%
6. Your are given the following.
- a sample of 2000 claims contains 1700 that are no greater than \$6000, 30 that are greater
than \$6000 but no greater than \$7000, and 270 that are greater than \$7000
- the total amount of the 30 claims that are greater than \$6000 but no greater than \$7000
is \$200,000
- the empirical limited expected value for this sample with a policy limit of \$6000 is \$1810
Determine the empirical limited expected value for this sample with a policy limit of \$7000.
A) Less than \$1900
B) At least \$1900, but less than \$1925
C) At least \$1925, but less than \$1950
D) At least \$1950, but less than \$1975
E) At least \$1975
7. A mortality study without any censored observations begins with individuals under
observation. There are deaths at the first death point ! and deaths at the second death
point ! . You are given the following:
(i) the product limit estimate of :²! ³ is .903614
(ii) the Nelson-Aalen estimate of :²! ³ is .906065 (both values rounded to 6 decimal places)
(iii)  Find the Nelson-Aalen estimate of :²! ³.
A) À
B) À
C) À
D) À
E) À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 9
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8. 40 observed losses have been recorded in thousands of dollars and are grouped as follows:
Interval
Number of
Total Losses
Losses
(\$000)
(\$000)
² Á µ
² Á µ
²Á µ
²Á B³
A deductible of 1 is applied to the loss data. The empirical distribution function is constructed
from the numbers of losses in each interval of the resulting data set after the deductible is applied.
The method of percentile matching is applied using the after-deductible empirical distribution
function to estimate the parameters and of the Weibull distribution (the distribution function
&
is - ²&³ ~ c %´ c ² ³ µ ). The percentiles to be matched are the 40-th and the 90-th
percentiles. Determine the estimated value of .
A) .4
B) .5
C) .6
D) .7
E) .8
Use the following information for Questions 9 and 10.
You are given the following:
- the random variable ? has the density function
²%³ ~ ²% b ³cc ,  %  B Á  - a random sample of size is taken of the random variable ?
~ , the method of moments estimator of .
9. Assuming  , determine c
c
c
c
c
?
?
?b
c
c
c
c
A) ?
B) ?c
C) ?b
D) ?c
E)
?
?
10. Determine the limit of V as the sample mean goes to infinity, where V is the maximum
likelihood estimator of .
A) 0
B) 1/2
C) 1
D) 2
E) B
11. You are given the following claims settlement activity for a book of automobile claims as of
the end of 1999:
Number of Claims Settled
Year
Year Settled
Reported
1997
1998 1999
1997
Unknown
3
1
1998
5
2
1999
4
3 ~ ²Year Settled c Year Reported) is a random variable describing the time lag in settling a
claim. The probability function of 3 is 3 ²³ ~ ²c³ , for
~ Á Á Á Ã .
Determine the maximum likelihood estimate of the parameter .
A) 3/11
B) 7/22
C) 1/3
D) 3/8
E) 7/15
Use the following information for Questions 12 and 13.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 9
A portfolio of independent risks is divided into two classes. Each class contains the same number
of risks.
For each risk in Class 1, the number of claims for a single exposure period follows a Poisson
distribution with mean 1.
For each risk in Class 2, the number of claims for a single exposure period follows a Poisson
distribution with mean 2.
A risk is selected at random from the portfolio. During the first exposure period, 2 claims are
observed for this risk. During the second exposure period, 0 claims are observed for this same
risk.
12. Determine the posterior probability that the risk selected came from Class 1.
A) Less than .53
B) At least .53, but less than .58
C) At least .58, but less than .63
D) At least .63, but less than .68
E) At least .68
13. Determine the Buhlmann credibility estimate of the expected number of claims for this same
risk for the third exposure.
A) Less than 1.32
B) At least 1.32, but less than 1.34
C) At least 1.34, but less than 1.36
D) At least 1.36, but less than 1.38
E) At least 1.38
14. The prior distribution of the parameter is exponential with a mean of 1. The conditional
distribution of ? , the number of claims for an insured in one year, given , is a mixture of two
Poisson random variables with probability function
c %
c ²³%
²%O³ ~ ²À³< %[ = b ²À³<
= Á % ~ Á Á Á ÀÀÀ
%[
An insured is chosen at random and observed to have no claims in the first year.
Find the Bayesian estimate of the expected number of claims next year for the same insured.
A) 0.5
B) 0.55
C) 0.60
D) 0.65
E) 0.70
15. The number of claims per month for a given risk is assumed to be Poisson distributed with an
unknown mean that varies by risk. It is found that for a risk that has reported no claims for the
past month, the semiparametric empirical Bayes estimate of the expected number of claims next
month is , and it is found that for a risk that has reported no claims for the past two months,
the semiparametric empirical Bayes estimate of the expected number of claims next month is .
Find the semiparametric empirical Bayes estimate of the expected number of claims next month
for a risk that has reported no claims for the past three months.
A) B) C) D) E) © ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 9
PE-167
Use the following information for Questions 16 and 17.
- a large portfolio of automobile risks consists solely of youthful drivers
- the number of claims for one driver during one exposure period follows a Poisson
distribution with mean c , where is the grade point average of the driver
- the distribution of within the portfolio is uniform on the interval ´Á µ
A driver is selected at random form the portfolio. During one exposure period, no claims are
observed for this driver.
16. Determine the posterior probability that the selected driver has a grade point average greater
than 3.
A) Less than .15
B) At least .15, but less than .35
C) At least .35, but less than .55
D) At least .55, but less than .75
E) At least .75
17. A second driver is selected at random from the portfolio. During five exposure periods, no
claims are observed for this second selected driver. Determine the Buhlmann credibility estimate
of the expected number of claims for this second driver during the next exposure period.
A) Less than .375
B) At least .375, but less than .425
C) At least .425, but less than .475
D) At least .475, but less than .525
E) At least .525
18. A sample of ten observations comes from a parametric family (%, &; 1 , 2 ) with loglikeli
hood function 3(1 , 2 ) ~ (% , & ; 1 , 2 ) ~ c2.512 c 31 2 c 22 b 51 b 22 b ,
where is a constant.
~
Determine the estimated covariance matrix of the maximum likelihood estimator, @
.5
A) >
.3
.3
20
B) >
c30
.2 ?
c30
.2
C) >
50 ?
.3
.3
.5 ?
5
D) >
3
3
2?
2
E) >
c3
V1
.
V2 A
c3
5 ?
19. Semi-parametric empirical Bayesian credibility is being applied in the following situation.
The distribution of annual losses on an insurance policy is uniform on the interval ²Á ³ ,
where has an unknown distribution. A sample of annual losses for 100 separate insurance
~
~
policies is available. It is found that ? ~ and ? ~ .
For a particular insurance policy, it is found that the total losses over a 3 year period is 3.
Find the semi-parametric estimate of the losses in the 4-th year for this policy.
A) Less than 1.5
B) At least 1.5, but less than 1.7
C) At least 1.7, but less than 1.9
D) At least 1.9, but less than 2.1
E) At least 2.1
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 9
Questions 20 and 21 relate to the following situation. Claims arrive for processing according to a
Poisson process with mean rate ~ per hour. Claim processing
takes either or hour, with any given claim having a .5 probability of taking hour.
20. Use the inverse transform method to simulate the number of claims in each of the first two
hours. The uniform random numbers to be used in sequence to simulate the numbers of claims in
hours 1 and 2 are: À Á À5 . Find the number of claims simulated in each hour.
A) 1 in the first hour and 3 in the second hour
B) 2 in the first hour and 2 in the second hour
C) 3 in the first hour and 2 in the second hour
D) 3 in the first hour and 1 in the second hour
E) None of A, B, C or D is correct
21. The simulated arrival times (in hours) of the claims during the first 2 hours are
.2 , .8 , 1.1 , 1.3 , 1.7
There is only one claims processor. The claim processing times of the successive claims are
simulated using the inversion method, using the following uniform random numbers, where small
random numbers correspond to small processing times:
.8 , .6 , .1 , .1 , .7 .
Determine the state of the claims processing system at the end of 2 hours.
A) No claims are being processed, no claims are waiting
B) A claim is being processed and no claims are waiting to be processed.
C) A claim is being processed and one claim are waiting to be processed.
D) A claim is being processed and two claims are waiting to be processed.
E) A claim is being processed and three claims are waiting to be processed.
22. ? has a uniform distribution on the interval ´Á j\$µ , and \$ has a uniform distribution on the
interval ´Á µ . Find the mean of the unconditional distribution of ? .
A) B) C) D) E) 23. A loss random variable has a continuous uniform distribution between 0 and \$100 .
An insurer will insure the loss amount above a deductible . The variance of the amount that the
insurer will pay is 69.75 . Find .
A) 65
B) 70
C) 75
D) 80
E) 85
24. The times of death in a mortality study are ! Á ! Á ! Á ÀÀÀ The following information is
given. There was one death at time ! , two deaths at time ! and one death at time ! .
The Product-Limit estimate of survival probability for those times are
: ²! ³ ~ À Á : ²! ³ ~ À
and : ²! ³ ~ À .
Determine the number of right-censorings that took place in the interval ´! Á ! ³ .
A) 0
B) 1
C) 2
D) 3
E) 4
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 9
PE-169
25. On Time Shuttle Service has one plane that travels from Appleton to Zebrashire and back and
each day. Flights are delayed at a Poisson rate of two per month. Each passenger on a delayed
flight is compensated \$100. The numbers of passengers on each flight are independent and
distributed with mean 30 and standard deviation 50. (You may assume that all months are 30
days long and that years are 360 days long). Calculate the standard deviation of the annual
compensation for the delayed flights.
A) Less than \$25,000
B) At least \$25,000, but less than \$50,000
C) At least \$50,000, but less than \$75,000
D) At least \$75,000, but less than \$100,000
E) At least \$100,000
26. A farmer develops a model for his seeding season for a particular crop. The number of days
in which crops can be seeded during the season has a Poisson distribution with a mean of 20. On
a day suitable for seeding, the number of acres than can be seeded is either 1 or 2, each with
probability .5. The farmer wishes to insure against a poor seeding season. The farmer purchases
insurance which will pay if the number of acres seeded during the season is under 20.
For each acre under 20 that is not seeded the insurance will 5000. : represents that number of
acres that will be seeded in the season. The farmer has determined ,´²: c ³b µ ~ 1À .
Find the expected insurance payment.
A) 4000
B) 5000
C) 6000
D) 7000
E) 8000
27. You are given the following random sample of 6 observations from the distribution of the
random variable ? :
2 , 4 , 4 , 5 , 7 , 10
Kernel smoothing is applied to estimate the density function of ? . The kernel function used for
the data point & is the pdf of the normal distribution with mean & and variance 1. Use kernel
V ²³ .
smoothing to estimate the distribution function of ? at the point % ~ , A) Less then .06
B) At least .06, but less than .12
C) At least .12 but less than .18
D) At least .18, but less than .24
E) At least .24
28. You are given:
(i) A sample of losses is:
600 700  900
(the third loss is known only to be at least 900)
(ii) No information is available about losses of 500 or less.
(iii) Losses are assumed to follow an exponential distribution with mean .
Determine the maximum likelihood estimate of .
A) Less than 500
B) At least 500 but less than 600
C) At least 600 but less than 700
D) At least 700 but less than 800
E) At least 800
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 9
29. If the proposed model is appropriate, which of the following tends to zero as the sample size
goes to infinity?
A) Kolmogorov-Smirnov test statistic
B) Anderson-Darling test statistic
C) Chi-square goodness-of-fit test statistic
E) None of A), B), C), or D)
30. When applying the method of limited fluctuation credibility to a certain compound Poisson
distribution of total claims cost so that total claims cost will be within % of expected total claims
cost % of the time, the full credibility standard is 1200 expected claims. We also know that the
coefficient of variation of the severity distribution is 2.
Suppose the following changes are made in our assumptions:
(i) the coefficient of variation of the severity distribution is doubled to 4, and
(ii) the standard for full credibility is based on total claims cost being within % of expected
claims cost % of the time ² is unchanged).
Find the new standard for full credibility.
A) 500
B) 1000
C) 1020
D) 1200
E) 2040
31. In a portfolio of insureds, each insured will have either 0 or 1 claim in a year, with
independence from one year to another. The probability that an individual insured will have a
claim in a given year is %. The portfolio of insureds is such that for a randomly chosen individual
from the portfolio, the probability % is uniformly distributed on ²Á ³ . A randomly chosen
individual is found to have no claims in consecutive years, where  . Determine the
expected number of claims that the individual will have in the b -st year.
A) c
B) c
C) D) b
E) b
32. Four machines are in a shop. The number needing repair in each week has a binomial
distribution with = 0.5. For each machine, the repair time, in hours, is uniformly distributed
on [0,10]. You are to estimate ? , the total repair time (in hours) for a three-week period, using
the inverse transformation method of simulation. Use the following numbers from the uniform
distribution on [0Á 1] to simulate the number of machines needing repair during each of three
weeks: 0.3, 0.6, 0.7. Use the following numbers from the uniform distribution on [0Á 1] to
simulate repair times: 0.3
0.1 0.7 0.6 0.5 0.8 0.1 0.3
Determine ? .
A) 17
B) 22
C) 23
D) 30
E) 34
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 9
PE-171
33. ? has the following distribution : 7 ´? ~ µ ~ À Á 7 ´? ~ µ ~ À
.
The distribution of @ is conditional on the value of ? :
if ? ~ then the distribution of @ is 7 ´@ ~ µ ~ À
Á 7 ´@ ~ µ ~ À Á 7 ´@ ~ µ ~ À , and
if ? ~ then the distribution of @ is 7 ´@ ~ µ ~ ÀÁ 7 ´@ ~ µ ~ À Á 7 ´@ ~ µ ~ À .
A is the sum of @ independent normal random variables, each with mean and variance 2.
What is = ´Aµ ?
A) 5.0
B) 6.0
C) 7.0
D) 8.0
E) 9.0
34. Random sampling from the distribution of ? results in the three sample values 1, 2 and 4.
A uniform distribution on the interval ´Á µ is fitted to the data set by finding the value of that
minimizes the Kolmogorov-Smirnov goodness-of-fit statistic. Find .
A) 4.4
B) 4.5
C) 4.6
D) 4.7
E) 4.8
35. It is known that there are two groups of drivers in an insured population. One group has a 20
percent accident probability per year and the other group has a 40 percent accident probability per
year. Two or more accidents per year per insured are not possible. The two groups comprise
equal proportions of the population and each has the following accident severity distribution:
Probability
Size of Loss
.80
100
.10
200
.10
400
A merit rating plan is based on the pure premium experience of individual insureds for the prior
year. Calculate the credibility of an insured's experience.
A) Less than .01
B) At least .01, but less than .02
C) At least .02, but less than .03
D) At least .03, but less than .04
E) At least .04
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 9
ACTEX EXAM C/4 - PRACTICE EXAM 9 SOLUTIONS
1. I. The mean residual lifetime with deductible is
B
For the single parameter Pareto, :²!³ ~ ! ²%³ %
B
B Á
Á
Then :²!³ ! ~ ! ~
.
!
Á°
Á° ~
B :²!³ !
.
:²³
B Á
~ !
%
% ~
Á
!
.
, which is increasing. I is true.
Alternatively, if ? has an decreasing hazard rate, then it has an increasing mean residual lifetime.
The hazard rate for ? is
c: Z ²%³
:²%³
²%³
Á°%
~ :²%³ ~ Á°% ~ % , which is a decreasing function of %.
II. It can be shown that the gamma distribution with  has an increasing hazard rate.
B
B &c&°
We would have to find :²!³ ~ ! ²&³ & ~ !
& .
Using integration by parts,
The hazard rate is
²!³
:²!³
~
Á
!c!°
c!°
this is :²!³ ~ Ád b Ád
.
c!°
!
°Á
~ ! ! , which is an
!c!°
c!°
b
b
Ád
Ád
increasing function.
II is true.
III. According to the hazard rate test, a random variable with a decreasing hazard rate has a heavy
right tail, and a if the hazard rate is increasing the right tail is light. ? has a decreasing hazard
rate so it has a heavy right tail, and @ has an increasing hazard rate so it has a light right tail. III
is true.
2. The randomly chosen policy is a mixture of the three policy types.
The expected number of claims per year for a randomly chosen policy is
²À
³²À³ b ²À³²³ b ²À³ ~ À b À .
The second moment of a Poisson random variable is ,´5 µ ~ = ´5 µ b ²,´5 µ³ ~ b À
The second moment of number of claims for a low risk policy is À b ²À³ ~ À.
The second moment of number of claims for a medium risk policy is b ²³ ~ .
The second moment of number of claims for a high risk policy is b ²³ .
The second moment of the number of claims per year for a randomly chosen policy is
²À
³²À³ b ²À³²³ b ²À³² b ³ ~ À b À² b ³ .
The variance of the number of claims in a year for a randomly chosen policy is
À b À² b ³ c ²À b À³ ~ À
b À
b À .
We are given that this is 1. , so that À
b À
b À ~ À .
This is the quadratic equation À b À
c À
~ .
The two roots of the equation are 2.5 and c 2.67. We ignore the negative root. ~ 2.5 .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 9
PE-173
3. The loss elimination ration with a deductible of amount is
,´?wµ
,´?µ
À
The distribution function of the loss distribution is - ²%³ ~ c c%° Á and ,´?µ ~ À
With a deductible of 500, the limited expected value is
,´? w µ ~ % ²%³ % b ´ c - ²³µ
c%°
~ % h % b h c°
~ ² c %c%° c c%° ³c
%~
%~
b c° ~ À À
The LER is À
~ À À
With a deductible of amount , the limited expected value is
,´? w µ ~ % ²%³ % b ´ c - ²³µ
c%°
~ % h % b h c°
~ ² c %c%° c c%° ³c
%~
%~
b c° ~ ² c c° ³ .
²cc° ³
The LER is
~ c c° .
In order for this to be twice as large as .39347, we must have
c c° ~ À
S ~ À
4. The probability of a given loss exceeding 500 is c° ~ c° ~ À
.
If there are exposures, then the expected number of losses exceeding the deductible
will be c° ~ À
. We are told that this is 10, so that ~ ° .
If all loss amounts doubled, the loss distribution will be exponential with mean 2000, so that
- ²%³ ~ c c%° , and the expected number of losses exceeding 500 will be
´ c -\$ ²³µ ~ ° c° ~ ° ~ À À
5. Consider an employee with salary * and probability of becoming disabled . Let 0 be the
indicator random variable indicating whether or not disability will take place. Let ? denote the
disability claim random variable for this employee. Then ,´?µ ~ ,´,´?  0µµ . If 0 ~ then
there is no claim so that ,´?  0 ~ µ ~ . If 0 ~ (probability ) then the expected number
of weeks of disability is h ²À³ b h ²À³ b h ²À³ b h ²À³ ~ À , so the expected
disability claims will be À* . Thus, ,´?µ ~ h ²À*³ . We can find the variance of ?
from = ´?µ ~ = ´,´?  0µµ b ,´= ´?  0µµ ~ ² c ³²À*³ b ,´= ´?  0µµ .
If 0 ~ then = ´?  0 ~ µ ~ , and if 0 ~ then
= ´?  0 ~ µ ~ * h ²À³ b ²*³ h ²À³ b ²*³ h ²À³ b ²*³ h ²À³ c ²À*³
~ À
* , and thus,
,´= ´?  0µµ ~ ² c ³ h b h ²À
* ³ .
Then, = ´?µ ~ ² c ³²À*³ b h ²À
* ³ .
The disability claim variance for each of the category 1 employees is
²À³²À³² ³ b ²À³²Á ³ ~ Á À ,
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 9
5. continued
for category 2 employees it is ²À³²À³² ³ b ²À³²Á ³ ~ Á À
, and
for category 3 employees it is ²À³²À³² ³ b ²À³²Á ³ ~ Á À
.
The expected aggregate claim for disability is
h ²À³²³ b h ²À³²³ b h ²À³²³ ~ Á .
The variance for the aggregate disability claim random variable for all 1000 employees is
²³²Á À³ b ²³²Á À
³ b ²³²Á À
³ ~ Á Á .
The premium required to ensure with 95% probability that claims will be covered is 7 where
7 cÁ
Á
Á
7 ´7  :µ ~ 7 ´ j
:c,´:µ
7 cÁ
 j
µ ~ À ¦ j
~ À
¦ 7 ~ Á .
Á
Á
= ´:µ
Total weekly payroll is h b h b h ~ Á . The fraction of weekly
Á
payroll is Á ~ À, or À% .
6. With the policy limit raised to 7000, the 30 policies between 6000 and 7000 will add
Á c Á ~ Á in losses (total of those 30 claims minus what was paid under the
policy limit of 6000). The 270 claims above 7000 will add 270,000 (and additional 1000 is paid
on each of these 270 policies). So the total amount added to losses paid is 290,000, or an
Á
average of ~ , which brings the LEV to b ~ À
An alternative approach is the following. With a policy limit of 6000, the empirical LEV is
V w µ ~ (Total claim amount from claims  6000)b²³²
³ ~ .
,´?
It follows that the total of the 1700 claims that were each no greater than 6000 is
²³²³ c ²³²
³ ~ Á Á À
V w µ ~ (Total claim amount from claims  7000)b²³²³ À
Then ,´?
Total claim amount from claims  7000 is equal to
Total claim amt from claims  6000 b Total claim amt from claims between 6000 and 7000
~ Á Á b Á ~ Á Á .
V w µ ~ ÁÁb²³²³ ~ À
Then ,´?
7. À
~ c c ~ c c S ~ À
c ~ À
c À
V
V ³ ~ b ~ c ²À
³ ~ À
À
À
~ c/²! ³ S /²!
c
Since
°
b c
~ b c²
°³
~ ~ À
or À
À
~ b À
c
~ À
Á we solve the quadratic for c
But if ~ À
Á then ~ À  À Thus,
V ³ Á and the Nelson-Aalen estimate of :²! ³ is cÀ
~ À
À
~ ~ À
~ /²!
and get
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 9
PE-175
8. After the deduction of 1 from the losses, the data grouping is
Interval
Number of
Losses
(\$000)
² Á µ
² Á µ
²Á µ
²Á B³
The 40-th and 90-th percentiles from the empirical distribution are and 3, since
~ À of the after-deductible payments are  , and ~ À of the after-deductible
payments are  . We use and 3 as the 40-th and 90-th percentiles in the Weibull
&
distribution. With Weibull distribution function - ²&³ ~ c %´ c ² ³ µ , we have the
following percentile matching equations:
°
c %´ c ² ³ µ ~ À Á c %´ c ² ³ µ ~ À .
°
Then, ² ³ ~ c À
~ À , and ² ³ ~ c À ~ À .
²°³
² °³
À
~ À ~ À ,
which simplifies to ~ À , so that V ~ À
(and then V ~ À). Answer: D
Dividing the second equation by the first, we have
9. According to the method of moments for a one parameter distribution, we set the sample mean
equal to the distribution mean and solve for the parameter. The distribution mean is
B % h ²% b ³cc % . This can be simplified using integration by parts
B % h ²% b ³cc % ~ B % ´ c ²% b ³c µ
%~B
B
~ c %²% b ³c c
c ´ c ²% b ³c µ % ~ c b b À
c
%~
Integration by parts can be avoided if it is noticed that the distribution of ? is Pareto with
parameters and ~ , and density ²%³ ~ ²%b³b ~ ²% b ³cc .
Then the mean of the Pareto is ,´?µ ~ c
~ c
.
c
c
~ ~ ?b
c .
Applying the method of moments, we have ? ~ c , so that ?
We could also avoid integration by parts by using the substitution " ~ % b in the integral.
10. Given the sample values % Á ÀÀÀÁ % the likelihood function is
~
~
~
cc
3²% Á ÀÀÀÁ % O³ ~ ²% Á ³ ~ ´²% b ³cc µ ~ < ²% b ³ =
The loglikelihood is
M²³ ~ 3 ~ c ² b ³ < ²% b ³ = ~ c ² b ³ ²% b ³
~
The maximum likelihood equation is
V~
²% b³
~
M²³ ~ c ²% b ³ ~ , so that the mle is
~
. If the sample mean goes to infinity, then some of the sample elements
~
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 9
10. continued
approach infinity (we are not necessarily assuming that the number of sample points goes to
infinity). In that case, the denominator in the expression for V goes to infinity and V goes to 0.
The significance of this is as follows, if the true value of is 1 or less, then distribution has an
infinite mean, so that sample means could be extremely large.
11. For claims reported in 1997, only those settled in 1 or 2 years can be observed. The
likelihood function will be formulated using conditional probabilities related to those events. The
likelihood factor for a claim reported in 1997 and settled in 1 year is
3 ²³
3 ²³b3 ²³
²c³
~ ²c³b²c³ ~ b
and the likelihood factor for a claim reported in 1997 and
²c³
²³
3
settled in 2 years ²³b
~ ²c³b²c³ ~ b .
3
3 ²³
The denominator 3 ²³ b 3 ²³ is the probability that a claim was observed in years 1 or 2, so
²³
3
7 ´claim observed in year 1Oclaim observed in year 1 or year 2µ ~ ²³b
, and
3
3 ²³
²³
3
7 ´claim observed in year 2Oclaim observed in year 1 or year 2µ ~ ²³b
is how the
3
3 ²³
conditional probabilities above are formulated.
For claims reported in 1998, only those settled in 0 or 1 years are observed. The likelihood factor
²³
²c³
3
for a claim reported in 1998 and settled in 0 years is ²³b
~ ²c³b²c³ ~ b
, and
3
3 ²³
the likelihood factor for a claim reported in 1998 and settled in 1 year is
3 ²³
3 ²³b3 ²³
²c³
~ ²c³b²c³ ~ b .
For claims reported in 1999, only those settled in 0 years are observed, and the likelihood factor
²³
is 3 ²³ ~ .
3
The likelihood function for the data set given is
3 ~ ´ b
µ h ´ b µ h ´ b
µ h ´ b µ h ´µ ~ ²b³ .
The log of the likelihood function is 3 ~ c ² b ³ . The maximum is found
where 3 ~ S c b
~ S ~ is the maximum likelihood estimate.
12. The two classes have the same number of risks, and so are equally likely to be chosen.
7 ²ÁOClass 1³h7 ²Class 1³
7 ²Class 1OÁ ³ ~ 7 ²ÁOClass 1³h7 ²Class 1³b7 ²ÁOClass 2³h7 ²Class 2³
~
c h c h [ ³² [ ³² ³
c c c h c h ² [ ³² [ ³² ³b² [h ³² [h ³² ³
²
~
c2
c2
c
b
~ b
c ~ À
À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 9
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13. The hypothetical means are the expected number of claims from each class:
/4 ~ Á /4 ~ .
Then ~ ,´/4 µ ~ ²³² ³ b ²³² ³ ~ , and
~ = ´/4 µ ~ ² ³´² c ³ b ² c ³ µ ~ .
The process variances are the variance of the number of claims from each class:
7 = ~ Á 7 = ~ À
Then # ~ ,´7 = µ ~ ²³² ³ b ²³² ³ ~ .
With ~ exposures, the Buhlmann credibility factor is
A ~ b
~ b
# ~
° ~ . The Buhlmann credibility premium is
b °
c
A? b ² c A³ ~ ² ³² b
³ b ² ³² ³ ~ ~ À À
14. The prior density is ²³ ~ c ,  . The joint density of ? and at ? ~ is
²Á ³ ~ ²O³ h ²³ ~ ²À³´c b c µ h c ~ ²À³´c b c µ .
The marginal probability that ? ~ is
B
B
7 ´? ~ µ ~ ²Á ³ ~ ²À³´c b c µ ~ ²À³´ b µ ~ .
²Á³
²À³´c bc µ
The posterior density of is ²O³ ~ 7 ´?~µ ~
~ ² ³´c b c µ .
°
The Bayesian estimate for the expected number of claims next year is
B
,´? O? ~ µ ~ ,´?Oµ h ²O³ .
Since the conditional distribution of ? given is a mixture of a Poisson with mean and a
Poisson with mean 2, with mixing weights of .5 for each part of the mixture, it follows that
,´?Oµ ~ ²À³´ b µ ~ À . Then
B ,´?Oµ h ²O³ ~ B À h ² ³´c b c µ ~ ² ³B ´c b c µ ~ ² ³´ b µ ~ .
c
15. For a risk with sample mean @ of claims for the past months, the semiparametric empirical
Bayes estimate of the expected number of claims next month is
c
V c b ² c A³
V V ~ . We are given that with ~ and @
[email protected]
~ , we have
V , where A
V
b
V
c
V h ²³ b ² c A³
V A
V ~ ² c V ³
V~
V ~ , and with ~ and @ ~ , we have
V h
b
b
V
V h ²³ b ² c A³
V A
V ~ ² c V ³
V~
V ~ . It follows that
V h
b
b
V
V
V
c
b
V
² V h
V³,² V h V³ ~
V ~ , so that ~ À . Then, with ~ and @ ~ , we have
b
b
b
V
V
V
b
À V h ²³ b ² c A³
V A
h
V ~ ² c V ³
V~
V ~ ² V ³² V h V³ ~ ² À ³² ³ ~ .
V
b
b
b b
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 9
16. 7 ´  O ~ µ ~
7 ´²~³ q² ³µ
7 ´~µ
À
c
7 ´² ~ ³ q ²  ³µ ~ 7 ´ ~ Oµ ²³ ~ c²c³ h ~ c
~ À
c
7 ´ ~ µ ~ 7 ´ ~ Oµ ²³ ~ c²c³ h ~ c
~ À À
7 ´  O ~ µ ~ À
À ~ À
.
17. The hypothetical mean for a driver with gradepoint is /4 ~ c , so that
~ ,´/4 µ ~ ,´ c µ ~ c ~ , since is uniformly distributed on ´Á µ .
The process variance of number of claims for a driver with gradepoint is also 7 = ~ c (since Poisson mean is equal to variance), and therefore, # ~ ,´7 = µ ~ . The variance of the
hypothetical mean is ~ = ´/4 µ ~ = ´ c µ ~ = ´µ ~ ~ (this is the variance of
the uniform distribution on ´Á µ ). The Buhlmann factor is ~ # ~ °
~ , and the
credibility factor is A ~ b
~ bÀ
~ for the driver with five years of no claims. Then
the Buhlmann credibility estimate for expected number of claims for this driver is
c
A? b ² c A³ ~ ² Answer: C
³²³ b ² ³²³ ~ ~ À
À
18. If maximum likelihood estimation is applied to a distribution with parameters and , then
C
v C M²³
M² ³ y
C
C
C x
{
the information matrix 0 ²³ is the following d matrix c ,
C
C
M²³ z
w C C M²³
C where M²³ is 3² Á ³. The covariance matrix of the estimates is the matrix inverse
´0 ²³µc (recall that in the covariance matrix, the diagonal entries are the variances of the
estimates of the 's, and the entries off the diagonal are the covariances between the estimates of
the 's). The 1,1 entry of 0 ²³ is c ,´ CC 3² Á ³µ ~ c ,² c ³µ ~ À
The 1,2 entry and the 2,1 entry are the same (0 ²³ is always a symmetric matrix) and they are
c ,´ C C C 3² Á ³µ ~ c ,´² c ³µ ~ .
The 2,2, entry is c ,´ CC M²³µ ~ c ,´ c µ ~ À Then 0 ²³ ~ >
matrix is ´ 0 ²³µc ~ hch
h>
c
c
c
~
.
? > c
?
Note that the inverse of the d matrix >
is hch
h>
?
c
, and the covariance
?
c
.
?
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 9
PE-179
19. Hypothetical mean is ,´?Oµ ~ .
Process variance is = ´?Oµ ~ .
Expected hypothetical mean is ~ ,´?µ ~ ,´ ,´?Oµ µ ~ , ´ µ ~ ,´µ .
Expected process variance is # ~ ,´ = ´?Oµ µ ~ , ´ µ ~ ,´ µ .
Variance of hypothetical mean is ~ = ´ ,´?Oµ µ ~ = ´ µ ~ = ´µ
~ ´ ,´ µ c ²,´µ³ µ À
c
From the sample, we can estimate ,´?µ as ? ~ , so this is also the estimate ,´µ .
The estimate of ,´µ is 4.
From the sample we can estimate = ´?µ using the unbiased sample variance,
c
´ '? c ? µ ~ ´ c ²³ µ ~ À .
But = ´?µ ~ b # ~ ,´ µ b ´ ,´ µ c ²,´µ³ µ ~ ,´ µ c ²,´µ³ .
Using the estimated variance of ? and the estimated mean of , we have
À ~ ,´ µ c ²³ , so that the estimate of ,´ µ is 18.06 .
Then, # ~ ,´ µ is estimated to be 1.505, and
~ ´ ,´ µ c ²,´µ³ µ is estimated to be .515 .
V c b ² c A³
V The estimate of losses in the 4th year is [email protected]
V,
c
V
where A ~
~
~ À, and V ~ ? ~ ,
b À
b VV#
À
c
V b ² c A³
V so that [email protected]
V ~ ²À³²³ b ²À³²³ ~ À .
20. The Poisson probability function for ~ is ²³ ~ c ~ À Á
c
c
c
²³ ~ [h ~ À
Á ²³ ~ [h ~ À
Á ²³ ~ [h ~ À Á ...
The probability function and distribution function are:
%
ÀÀÀ
²%³
À
À
À
À
ÀÀÀ
- ²%³
À
À
À
À
ÀÀÀ
The uniform number .7 results in 3 simulated claims in the first hour, since À
 À  À
, and
the uniform number .5 results in 2 simulated claims in the second hour. Answer: C
21. Processing time is hour if uniform random number is  À , and is hour if  .5 .
The simulated processing times are Á Á Á Á .
Claim 1 arrives at .2 and is processed at .7 . Claim 2 arrives at .8 and is processed at 1.3, and
there are 2 claims waiting (claims 3 and 4 have arrived). Claim 3 is processed at time 1.55, and
there is 1 claim waiting. Claim 4 is processed at time 1.8 and there is one claim waiting (Claim 5
has arrived). Claim 5 is processed at time 2.3 . At time 2, a claim is being processed and no
claims are waiting to be processed.
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22. The uniform distribution on ´Á j\$µ has pdf ?O\$ ²%O³ ~ j
or equivalently,  % . Also, the pdf of \$ is \$ ²³ ~ .
for  %  j ,
Then ? ²%³ ~ ?O\$ ²%O³ h \$ ²³ ~ % j h 1 d ~ ² c %³ for  %  .
The mean of % is ,´?µ ~ % h ² c %³ % ~ .
j\$
Alternatively, since ,´?O\$µ ~ , we have
j\$
j
j
,´?µ ~ ,´ ,´?O\$µ µ ~ ,´ µ ~ h \$ ²³ ~ ~ .
, if %
23. The insurer pays @ ~ D
%c
Then ,´@ µ ~ %
²c³
%c , if %
with constant density .01 .
²c³
²c³
²%c³
and ,´@ µ ~ % ~
²c³ c ´ µ . Substituting the possible answers, we
~
so that = ´@ µ ~
~ , the variance of @ is 69.75 .
see that with
24. À ~ : ²! ³ ~ : ²! ³´ c µ ~ ²À
³´ c µ S ~ S ~ is the number alive just
before the death at time ! . A similar analysis at time ! shows that
À
~ : ²! ³ ~ : ²! ³´ c µ ~ ²À³´ c µ S ~ S ~ is the number alive just
before the 2 deaths at time ! . Therefore there are 10 survivors just after the 2 deaths at time ! ,
and 6 of them are still alive at the next death point ! . This implies that 4 observations were
censored between ! and ! .
25. The number of passengers compensated in one year : has a compound distribution.
The frequency is 5 , the number of delayed flights in one year, which is Poisson with mean 24 (2
per month for 12 months). The severity @ is the number of passengers on a delayed flight.
We have ,´5 µ ~ = ´5 µ ~ Á ,´@ µ ~ and = ´@ µ ~ ~ .
Since the frequency has a Poisson distribution, it follows that = ´:µ ~ ,´5 µ h ,´@ µ À
From the given information we have ,´@ µ ~ = ´@ µ b ²,´@ µ³ ~ b ~ À
Then, = ´:µ ~ ²³²³ ~ Á .
Alternatively, = ´:µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ ~ Á À
Each passenger on a delayed flight receives \$100 in compensation. The total compensation paid
in one year is : , and the variance is = ´:µ ~ ² ³²Á ³ À
The standard deviation of annual compensation is jÁ ~ Á À Answer: B
26. Insurance pays
@ ~ d ² c :³b ~ d F
c :
:  :  ~ d ² c ,´: w µ³ .
: ~ ²: w ³ b ²: c ³b
S ~ ,´:µ ~ ,´: w µ b ,´²: c ³b µ ~ ,´: w µ b À
S ,´: w µ ~ À S ,´@ µ ~ ² c À³ ~ . Answer: C
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27. The data points are & ~ Á & ~ Á & ~ Á & ~ Á & ~ .
The empirical distribution is ²³ ~ , ²³ ~ , ²³ ~ , ²³ ~ , ²³ ~ The kernel smoothed estimate of - ²³ is
V ²³ ~ ²& ³ h 2& ²³
~
~ h 2 ²³ b h 2 ²³ b h 2 ²³ b h 2 ²³ b h 2 ²³ ,
where 2& ²%³ is the cdf of the normal distribution with mean & and variance 1.
%c&
2& ²%³ ~ )² ³ ~ )²% c &³ .
2 ²³ ~ )² c ³ ~ )²³ ~ À Á
2 ²³ ~ )² c ³ ~ )² c ³ ~ À Á
2 ²³ ~ )² c ³ ~ )² c ³ ~ À Á
2 ²³ ~ )² c ³ ~ )² c ³ ~ Á
2 ²³ ~ )² c ³ ~ )² c ³ ~ .
V ²³ ~ h 2 ²³ b h 2 ²³ b h 2 ²³ b h 2 ²³ b h 2 ²³
Then, ~ h ²À³ b h ²À³ b h ²À³ b h ²³ b h ²³ ~ À
28. When we have truncated and censored losses for an exponential distribution, the mle of the
total insurance payment
ground up exponential mean is V ~ number of losses (non-censored) . In this case, the insurance
payments are 100, 200 and 400. The mle of is bb
~ .
29. The Kolmogorov-Smirnov test statistic is the maximum deviation of the fitted model
distribution function from the empirical distribution function. If the model is appropriate, as the
sample size increases, the empirical distribution function should approach the true model
distribution function (which is the fitted model if it is appropriate), so the maximum deviation
will go to 0.
30. For a compound Poisson distribution of total claims cost, the standard for full credibility
= ´@ µ
based on number of claims needed is < b ²,´@ µ³ = ,
&
b
where @ is the severity random variable, and ~ ² ³ , with & being the percentile of
the standard normal distribution. The coefficient of variation of @ is
is the square of the coefficient of variation of @ .
&
j= ´@ µ
,´@ µ
= ´@ µ
= ´@ µ
, so that ²,´@ µ³
&
From the initial information given, we have ~ ² ³ < b ²,´@ µ³ = ~ ² ³ < b = .
&
After changing the assumptions, the standard for full credibility will be ² ³ < b = . The
&
ratio of new standard to old is
² ³ <b =
&
² ³ <b =
~ ²b
³²b ³ ~ .
Therefore the new standard for credibility is ² ³²³ ~ expected claims.
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31. Given years without a claim, we find the conditional density function of the probability
7 ²(O%³h ²%³
parameter %. We use the usual Bayes approach ²%O(³ ~ 7 ´(O%³ ²%³ % .
With the unconditional density of % being ²%³ ~ (uniform distribution), this becomes
7 ²no claims in first yearsO%)h ²%³
²%Ono claims in first years) ~ 7 ²no claims in first yearsO%)h ²%³ %
²c%³ h
~ ~ ² b ³² c %³ .
%
²c%³
Then, the conditional expectation of % given no claims in the first years is
%² b ³² c %³ % . With the change of variable & ~ c %, this integral becomes
² b ³ & ² c &³ & ~ ² b ³´ b
c b
µ ~ b
.
32. We first simulate the numbers needing repair in each week. The probability function for the
binomial distribution with ~ 4 trials and probability ~ .5 of a success on a trial is
4
4
( ) ~ 4 5 (1 c )c ~ 4 5 h (.5) (.5)c ~ 4 5(.0625).
Thus, (0) ~ .0625, (1) ~ .25, (2) ~ .375, (3) ~ .25 and (4) ~ .0625.
The cumulative distribution function for this binomial distribution is then
- ²0) = ²0) = .0625, - ²1) = - ²0) b ²1) = .3125, - ²2) = - ²1) + ²2) = .6875,
- ²3) ~ - ²2) b ²3) ~ .9375 and - ²4) ~ - ²3) b ²4) ~ 1.
Using the inversion method of simulation for a discrete random variable, given the uniform [0Á 1]
random number < , the simulated value is the k satisfying the relationship - ²c1)  <  - ² ).
Thus, the three given uniform numbers .3, .6 and .7 result in simulated numbers of machines
needing repairs of 1 in the first week, 2 in the second week and 3 in the third week. If < is a
random uniform [0Á 1] number, then 10< will be a random uniform on the interval [0Á 10]. Thus,
the simulation of the repair times for the machines results in times of 3 for the 1 machine in the
first week, 1 and 7 for the 2 machines in the second week, and 6, 5 and 8 for the three machines
in the third week. The total repair time for the 3 weeks is 3+1+7+6+5+8 = 30 (hours). Note that
if the inversion method is applied to = , the uniform variable on [0Á 10], the cdf of = is
#
- ²#) = 10
. Then if < is uniform on [0Á 1], solving < = - ²#) results in # = 10< as indicated
above.
33. = ´Aµ ~ = ´,´A  @ µµ b ,´= ´A  @ µµ .
,´A  @ µ ~ h @ (a sum of @ independent normal random variables each with mean 2), and
= ´A  @ µ ~ h @ (a sum of @ independent normal random variables each with variance 2).
Thus, = ´Aµ ~ = ´ h @ µ b ,´ h @ µ ~ h = ´@ µ b h ,´@ µ .
,´@ µ ~ ,´,´@  ?µµ , and if ? ~ (prob. .4) then ,´@ µ ~ À
, and if ? ~ (prob. .6) then
,´@ µ ~ À . Thus, ,´@ µ ~ ²À
³²À³ b ²À³²À
³ ~ À .
= ´@ µ ~ = ´,´@  ?µµ b ,´= ´@  ?µµ . Since ,´@ µ ~ À
with prob. .4 (? ~ ) and
,´@ µ ~ À with prob. .6 (? ~ ), it follows that
= ´,´@  ?µµ = ²À
³ ²À³ b ²À³ ²À
³ c ²À³ ~ À
.
Also, if ? ~ , the variance of @ is ²À³ b ²À³ c ²À
³ ~ À
, and if ? ~ , the variance of
@ is ²À³ b ²À³ c ²À³ ~ À
.
Thus, ,´= ´@  ?µµ ~ ²À
³²À³ b ²À
³²À
³ ~ À
, and so = ´@ µ ~ À
b À
~ À
.
Then, the variance of A is ²À
³ b ²À³ ~ À . Note that = ´@ µ can also be found
from ,´@ µ c ²,´@ µ³ . Again, ,´@ µ ~ À , as above, and now,
,´@ µ ~ ,´,´@  ?µµ ~ ²³²À³ b ²À³²À
³ ~ À, so that = ´@ µ ~ À
.
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34. The empirical cdf is - ²³ ~ Á - ²³ ~ and - ²³ ~ .
For parameter value , the K-S statistic is the maximum of
O O Á O c O Á O c O Á O c O Á O c O Á O c O .
Since there is an observed value of 4, must be at least 4.
With ~ , we see that the K-S statistic is the maximum of
O O Á O c O Á O c O Á O c O Á O c O Á O c O .
This maximum is .
If  , then O c O ~ c  , so that to minimize the K-S statistic, we want  À
Therefore   , and the six quantities are
Á c Á c Á c Á c Á c (the absolute value signs can be removed
because   .
To find the K-S statistic for a given , we must find the maximum of the six quantities above.
Suppose that we pick a value of for which c ~ c (the 2nd set equal to the third).
Solving for results in ~ . For that value of we know that the K-S statistic is at least (but might be larger when the other quantities are calculated). Any other value of will definitely
result in a larger K-S statistic than .
For each "crossover" value of (a value of that is found by setting two of the 6 quantities
equal), we find the K-S statistic, The minimum possible K-S statistic occurs at the minimum of
these.
The "crossover" values of are 4 , 4.5 , 4.8 , 5 and 6 .
The K-S statistic in each of these cases are Á Á Á Á .
The minimum of these is and it occurs when ~ À .
35. # is the group number. Hypothetical mean for group 1 is ,´?O# ~ µ ~ ²À³²³ ~ (.2 is expected frequency, and ²À³²³ b ²À³²³ b ²À³²³ ~ is the expected severity
for group 1) and for group 2 is ,´?O# ~ µ ~ ²À³²³ ~ .
Since the groups are of equal sizes, 7 ´# ~ µ ~ 7 ´# ~ µ ~ À .
Then, the expected hypothetical mean is ,´,´?O#µµ ~ ²À³²³ b ²À³²
³ ~ .
The variance of the hypothetical means is
~ = ´,´?O#µµ ~ ²À³² b ³ c ²³ ~ (also equal to ² c ³ ²À³²À³) .
The process variance for group 1 is
= ´?O# ~ µ ~ ,´? O# ~ µ c ²³
~ ´²À³²À³² ³ b ²À³²À³² ³ b ²À³²À³² ³µ c ²³ ~ , and
for group 2 it is = ´?O# ~ µ ~ ,´? O# ~ µ c ²
³
~ ´²À³²À³² ³ b ²À³²À³² ³ b ²À³²À³² ³µ c ²
³ ~ .
Then, # ~ ,´= ´?O#µµ ~ ²À³´
b µ ~ . The credibility for 1 driver for 1 year is
A~
~ À .
b
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ACTEX EXAM C/4 - PRACTICE EXAM 10
1. You are given the following:
- losses in 1999 follow the density function
²%³ ~ %c Á %  - inflation of 10% impacts all claims uniformly from 1999 to 2000
Determine the probability that losses in 2000 exceed 2.2.
A) Less than .05
B) At least .05, but less than .10
C) At least .10, but less than .15
D) At least .15, but less than .20
E) At least .20
2. For a population of individuals, you are given:
(i) Each individual has a constant hazard rate for time until death.
(ii) The hazard rates are uniformly distributed over the interval ²Á ³À
Calculate the probability that an individual drawn at random from this population dies within one
year.
A) 0.37
B) 0.43
C) 0.50
D) 0.57
E) 0.63
3. Losses for a line of insurance follow a Pareto distribution with ~ and ~ .
An insurer sells policies that pay 100% of each loss up to \$5,000. The next year the insurer
changes the policy terms so that it will pay 80% of each loss after applying a \$100 deductible.
The \$5,000 limit continues to apply to the original loss amount. That is, the insurer will pay 80%
of the loss amount between \$100 and \$5,000. Inflation will be 4%.
Calculate the decrease in the insurer's expected payment per loss.
A) Less than 23%
B) At least 23%, but less than 24%
C) At least 24%, but less than 25%
D) At least 25%, but less than 26%
E) At least 26%
4. A an aggregate claim amount random variable : can only assume non-negative integer values.
:
For integer  , a reinsurer offers an "all-or-nothing" reinsurance that pays 1 ~ F
: :
.
Which of the following is a correct expression for ,´1b µ c ,´1 µ ?
A) c h 7 ´: ~ µ
B) h 7 ´: ~ µ
C) c ² b ³ h 7 ´: ~ b µ
D) ² b ³ h 7 ´: ~ b µ
E) 7 ´: ~ µ
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5. A dental insurance plan will pay for one check-up per year. The plan will pay \$75 per year
toward the cost of the annual check-up. The probability that a randomly selected plan member
will visit a dentist for a check-up in a given year is .8 . A plan member who goes for a check-up
has a 25% chance of requiring additional dental work, and, if needed, the additional dental work
will have a mean cost of \$150 with a standard deviation of \$10. All plan members have a 5%
chance per year (independent of whether or not they went for a check-up) of requiring
emergency dental work, which, when needed, has a mean cost of \$400 and a standard deviation
of \$200. The insurer charges a premium of \$125 per year for each plan member. Using the
normal approximation, pick the smallest of the following numbers that would result in a
probability of at least .95 that the total premium collected will be greater than the total claims
for the year (note that 7 ´A  À
µ ~ À for a standard normal random variable A ).
A) 150
B) 175
C) 200
D) 225
E) 250
6. You are given the following data set for the times of death or right-censoring (+) for 25
individuals:
2, 3, 3, 3b , 4, 4, 4, 4, 4b , 5b , 6, 6, 7, 7, 7, 7b , 7b , 8, 9, 10, 12b , 13, 13, 14, 16
The underlying time until death random variable is denoted ; .
Calculate the Product-Limit estimate of the conditional probability 7 ´;  O;  µ
A) Less than .40
B) At least .40 but less than .42
C) At least .42 but less than .44
D) At least .44 but less than .46
E) At least .46
7. You are given the following random sample of 12 data points from a population
distribution ? :
7 , 12 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53
Determine the triangle kernel density estimator V ²³.
A) B) C) D) E) 8.. A portfolio of insurance policies consists of two policy types. A policy of Type 1 has a claim
number distribution that is Poisson with mean 1 and a policy of Type 2 has a claim number
distribution that is Poisson with mean 1.5. The proportion of Type 1 policies in the portfolio is .
From a random sample of observed claim numbers the sample mean is found to be 1.1 . Find by the method of moments.
A) .2
B) .4
B) .6
D) .8
E) 1.0
9. The random variable ? has the density function
²%³ ~ ²% b ³cc Á  %  B Á  Determine the limit of V as the sample mean goes to 0, where V is the maximum likelihood
estimator of based on a random sample of size .
A) 0
B) 1/2
C) 1
D) 2
E) B
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10. Data set 1 has points, and data set 2 has points. Each data set is used separately to
estimate the parameter by maximum likelihood estimation for the distribution with cdf
- ²%³ ~ c c%° . The resulting maximum likelihood estimates are V and V . The two data
sets are then pooled, and an estimate of is obtained using all data points. Which of the
following is the new estimate?
V V
A) b
2
V bV
B) b
C)
V b V
V
V
E) b b
D) b
V
V
11. The likelihood ratio test is being applied to models A and B, where model A has one
parameter, and model B has more than one parameter. After parameter estimation is done, the
loglikelihood values are M( ~ c and M) ~ c . Find the maximum number of
parameters for model B for which the test will suggest that model B is preferable to model A at a
5% level of significance.
A) 2
B) 3
C) 4
D) 5
E) 6
12. The distribution of ? is assumed to be exponential with parameter , so that
²%O³ ~ c%° À The prior distribution of # is assumed to be uniform on the interval ´Á µ
Find the mean of the marginal distribution of ? .
A) 13
B) 14
C) 15
D) 16
E) 17
13. A portfolio of risks consists of classes, each with the same number of policyholders.
For each risk class , the number of claims in one year follows a Bernoulli distribution with
probability of a claim occurring equal to . For risk class , if a claim occurs, the claim amount
has an exponential distribution with mean . A randomly chosen risk is found to have had a
claim of amount %  during the year. Find the probability that the risk is from class 1.
A)
c%°
c%°
' c%°
B) c%°
' C)
' D)
c%°
' c%°
E) '
14. Losses for the year for a risk are uniformly distributed on ²Á ³ , where is uniformly
distributed on the interval ²Á ³. The first year loss for a risk is 3. Find the Buhlmann credibility
premium for the second year's loss for the same risk in terms of 3.
A) À3 b À
B) À3 b À
C) À3 b À
D) À3 b À
E) À
3 b À
15. > is a geometric random variable with parameter ~ . > is the form of the geometric
distribution defined for > ~ Á Á Á ÀÀÀ Use the inverse transform method with uniform random
number 0.70 to simulate a value of > .
A) 0
B) 1
C) 2
D) 3
E) 4
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16. The following random sample of size 5 is taken from the distribution of ? :
1 , 3 , 4 , 7 , 10
Bootstrap approximation of the mean square error of estimators is to be based on the following 6
resamplings of size 5 from the empirical distribution:
Resample 1 : 1 , 1 , 4 , 7 , 7
Resample 2 : 3 , 4 , 4 , 7 , 10
Resample 3 : 1 , 4 , 4 , 10 , 10
Resample 4 : 3 , 3 , 3 , 4 , 10
Resample 5 : 4 , 4 , 7 , 7 , 10
Resample 6 : 1 , 7 , 7 , 10 , 10
The median of ? is estimated by the third order statistic of a sample.
Find the bootstrap approximation to the estimator of the median using the 6 resamplings.
A) Less than 2.00
B) At least 2.00 but less than 2.25
C) At least 2.25 but less than 2.50
D) At least 2.50 but less than 2.75
E) At least 2.75
17. A health plan implements an incentive to physicians to control hospitalization under which
the physicians will be paid a bonus ) equal to times the amount by which total hospital claims
are under 400 (   ). The effect the incentive plan will have on underlying hospital claims
is modeled by assuming that the new total hospital claims will follow a two parameter Pareto
distribution with ~ and ~ . You are also given ,´)µ ~ .
Calculate .
A) 0.44
B) 0.48
C) 0.52
D) 0.56
E) 0.60
18. ? has a uniform distribution on the interval ´Á j\$µ , and \$ has a uniform distribution on the
interval ´Á µ . Determine the variance of the unconditional distribution of ? .
A) B) C) D) E) 19. 5 has a Poisson distribution with mean 2. The random variable 4 is defined as follows: for
 , 7 ´4 ~ µ ~ 7 ´5 ~ µ , and 7 ´4 ~ µ ~ c 7 ´4  µ.
Find the coefficient of variation of 4 .
A) B) j
C) D) j
E) 2
20. Job offers for a college graduate arrive according to a Poisson process with mean 2 per
month. A job offer is acceptable if the wages are at least 28,000. Wages offered are mutually
independent and follow a lognormal distribution with ~ 10.12 and ~ 0.12.
Calculate the probability that it will take a college graduate more than 3 months to receive an
acceptable job offer.
A) 0.27
B) 0.39
C) 0.45
D) 0.58
E) 0.61
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21. For a compound loss model, you are given:
• The claim count follows a Poisson distribution with ~ À .
• Individual losses are distributed as follows:
%
- ²%³
À
À
À
À
À
À
À
Á À
Á À
Calculate the probability of paying at least one claim after implementing a \$500 deductible.
A) Less than .005
B) At least .005, but less than .010
C) At least .010, but less than .015
D) At least .015, but less than .020
E) At least .020
22. You are given the following random sample of 12 data points from a population
distribution ? :
7 , 12 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53
Find the smoothed empirical estimate of the 40-th percentile of ? .
A) 26.0
B) 26.1
C) 26.2
D) 26.3
E) 26.4
23. You are given the following data set for the times of death or right-censoring (+) for 25
individuals:
2, 3, 3, 3b , 4, 4, 4, 4, 4b , 5b , 6, 6, 7, 7, 7, 7b , 7b , 8, 9, 10, 12b , 13, 13, 14, 16
Find the ratio of the width of the 90% linear confidence interval over the width of the 90% logtransformation confidence interval for /²³.
A) Less than .900
B) At least .900 but less than .925
C) At least .925 but less than .950
D) At least .950 but less than .975
E) At least .975
24. The random variable ? has the density function
²%³ ~ À c%° b À c%° ,  %  B Á   À
A random sample taken of the random variable ? has mean 1 and variance .
Determine the values of for which the method of moments estimate of and exist.
A)  only
B)  only
C)  only
D)  only
E) The correct answer is not given by A, B, C, or D
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25. 5 is a discrete random variable from the ²Á Á ³ class of distributions. The following
information is known about the distribution:
• 7 ²5 ~ ³ ~ À
• 7 ²5 ~ ³ ~ À
• 7 ²5 ~ ³ ~ À
• ,´5 µ ~ À
Based on this information, which of the following are true statements?
I. 7 ²5 ~ ³ ~ À
II. 5 is from a binomial distribution
II. 5 is from a Negative Binomial distribution.
A) I only
B) II only
C) III only
D) All but I
E) All but II
26. The product-limit estimate is used to estimate survival probabilities. Analysis of the data
shows that there is unclear information regarding one of the subjects in the study. It is not clear if
the subject is to be recorded as a death at the first death point or as being right-censored at the
first death point.
• the product limit estimate of :²& ³ (survival probability to the first death point) increases by a
ratio of 1.012987 if the subject is regarded as a right-censoring rather than as a death at the first
death point &
• the Nelson-Aalen estimate of /²& ³ increases by a ratio of 1.070825 if the subject is regarded
as a death rather than as a right-censoring at the first death point
• the Nelson-Aalen estimate increases by .011765 if the subject is regarded as a death rather than
as a right-censoring at the first death point
Determine the product-limit estimate of :²& ³, the survival probability to the second death point,
assuming that the uncertain subject is a right-censored.
A) 0.80
B) 0.82
C) 0.84
D) 0.86
E) 0.88
27. An actuary obtains two independent, unbiased estimates @ and @ for a certain parameter.
The variance of @ is four times that of @ . A new unbiased estimator of the form
h @ b h @ is to be constructed. What is the value of that minimizes the variance of the
new estimate?
A) Less than 0.18
B) At least 0.18, but less than 0.23
C) At least 0.23, but less than 0.28
D) At least 0.28, but less than 0.33
E) 0.33 or more
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PRACTICE EXAM 10
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28. During a one-year period, the number of accidents per day was distributed as follows:
Number of Accidents Days
You use a chi-square test to measure the fit of a Poisson distribution with mean 0.60. The
minimum expected number of observations in any group should be 5. The maximum possible
number of groups should be used. Determine the chi-square statistic.
A) 1
B) 3
C) 10
D) 13
E) 32
29. You are given the following:
- The number of claims per period follows a Poisson distribution.
- Claim sizes follow a lognormal distribution with parameters (unknown) and ~ - The number of claims and claim sizes are independent.
- 6,600 expected claims are needed for full credibility of aggregate claims per period.
- The full credibility standard has been selected so that actual aggregate claim costs per period
will be within 10% of expected aggregate claim costs per period 7 % of the time.
Using the methods of limited fluctuation credibility to determine the value of 7 .
A) Less than 75
B) At least 75 but less than 77
C) At least 77 but less than 79
D) At least 79 but less than 81
E) At least 81
30. Annual aggregate claims for a particular policy are modeled as a compound Poisson distribution with
Poisson parameter for the frequency (number of claims per year), and a severity (individual claim size)
@ that is either 1 or 2 with 7 ²@ ~ ³ ~ 7 ²@ ~ ³ ~ À . An insurer has a large portfolio of policies,
and each policy has its own value of . For a randomly chosen policy from the portfolio, the distribution
of is exponential with a mean of 1.
The claim sizes and the numbers of claims are independent of one another given .
A policy is chosen at random from the portfolio, and : denotes the aggregate claim for that policy for one
year.
Find the posterior density of if the observed aggregate loss for the year is 1.
A) c B) 4c C) 4c D) c E) c
31. A discrete random variable X has probability function ²%³ ~ %b
for % ~ Á Á Á Á .
Successive values of ? , labeled % Á % Á À À À , are to be simulated by the inverse transform
method until
%  20 . Find if the successive uniform values used in order in the
!~
simulation are "! : .38 , .68 , .50 , .92 , .11 , .45 , .03 , .18 , .75 , .24 , .29 , .86 , .02 , .64 , .40
A) 3
B) 6
C) 9
D) 12
E) 15
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 10
32. You are given:
(i) The number of claims incurred in a month by any insured has a Poisson distribution with
mean .
(ii) The claim frequencies of different insureds are independent.
(iii) The prior distribution is gamma with probability density function:
²³
c
²³ ~
(iv)
Month
Number of Insureds
Number of Claims
1
100
6
2
150
8
3
200
11
4
300
?
Determine the Buhlmann-Straub credibility estimate of the number of claims in Month 4.
A) 16.7
B) 16.9
C) 17.3
D) 17.6
E) 18.0
33. You are given the following table of data for three policyholders over a three year period.
Policy Year S
1
2
3
Number of Claims
Average Claim Size
40
200
50
220
40
250
2
Number of Claims
Average Claim Size
100
200
120
200
120
150
3
Number of Claims
Average Claim Size
50
200
60
250
Policyholder
¨
1
Apply the nonparametric empirical Bayes credibility method to find the credibility premium per
claim in the 4-th year for Policyholder 2.
A) Less than 180
B) At least 180 but less than 185
C) At least 185 but less than 190
D) At least 190 but less than 195
E) At least 195
34. An insurance policy on the loss ? has an ordinary deductible of 40. The policy also has the
following adjustments. If the loss is between 40 and 60, the insurance policy pays the amount of
the loss above 40. If the loss is between 60 and 80, the insurance pays 20 plus 75% of the loss
above 60. If the loss is above 80, the insurance pays 35. If the distribution of ? is uniform on
the interval ²Á ³ , find the expected cost per loss .
A) Less than 10
B) At least 10 but less than 12
C) At least 12 but less than 14
D) At least 14 but less than 16
E) At least 16.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 10
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35. You are given the following information about a random sample:
(i) The sample size equals five.
(ii) The sample is from a Weibull distribution with ~ 2.
(iii) Two of the sample observations are known to exceed 50, and the remaining three
observations are 20, 30 and 45.
Calculate the maximum likelihood estimate of .
A) Less than 40
B) At least 40, but less than 45
C) At least 45, but less than 50
D) At least 50, but less than 55
E) At least 55
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PRACTICE EXAM 10
ACTEX EXAM C/4 - PRACTICE EXAM 10 SOLUTIONS
1. Let ? be the post inflation loss random variable, and ? the pre-inflation loss random
B
variable. Then ? ~ À? ,and 7 ´?  Àµ ~ 7 ´?  µ ~ %c % ~ ~ À .
2. This question involves a continuous mixing of distributions. Given the parameter , the
survival distribution of an individual who has constant hazard rate with value is exponential
with mean . The distribution of is uniformly distributed on the interval ²Á ³ . The
probability of a randomly chosen individual dying within one year is
7 ´;  µ ~ 7 ´;  Oµ h ²³ , where ²³ is the density function of the hazard rate .
Since the force is uniformly distributed on ²Á ³, we have ²³ ~ À . Since ; has constant
force given , 7 ´;  Oµ ~ c c .
Then, 7 ´;  µ ~ ² c c ³²À³ ~ c ²À³² c c ³ ~ À
3. This year the insurer's expected payment per loss is
c
,´? w µ ~ c
´ c ² b
µ ~ ³
c ´ c ² b ³µ ~ À .
Next year the insurer pays 80% of ,´À? w µ c ,´À? w µ .
,´À? w µ ~ À,´? w ³µ ~ À , and
À µ ~ ²À³² c ³´ c ² ,´À? w µ ~ À,´? w
À µ
~
²À³² c ³´ c ²
À b
À b
³µ ~ À .
The insurer's expected payment per loss next year is ²À³´
À c Àµ ~ À .
À
The decrease is À c À ~ À, which is À
~ À , a 23.1% decrease.
4. Under ordinary stop-loss reinsurance with deductible , the reinsurer pays
:
:
. Therefore, 1 ~ ²: c ³b b F
.
²: c ³b ~ F
: c : 
:
Since : is integer valued, we have ,´²: c c ³b µ ~ ,´²: c ³b µ c ´ c -: ²³µ .
We also have ,´1 µ ~ ,´²: c ³b µ b h ´ c -: ²³µ . Therefore,
,´1b µ c ,´1 µ ~ ,´²: c c ³b µ b ² b ³ h ´ c -: ² b ³µ
c ²,´²: c ³b µ b h ´ c -: ²³µ³
~ ,´²: c c ³b µ c ,´²: c ³b µ b c ² b ³ h -: ² b ³ b h -: ²³
~ c ´ c -: ²³µ b c ² b ³ h ´-: ²³ b : ² b ³µ b h -: ²³
~ c ² b ³: ² b ³ .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 10
PE-195
5. All members have an expected emergency dental cost of ²À³²³ ~ . For those that go
for a check-up there is a definite cost of 75 plus an additional expected cost of
²À³²³ ~ À for a total expected cost of À for those who go for a check-up. This
results in a total expected cost per member (per year) of b ²À³²À³ ~ . (We can
regard the claim variable per member as @ ~ ? b ? , where ? is the part of the claim due to
going for a check-up, and ? is the part of the claim due to emergency dental treatment.) Then,
,´? µ ~ ²À³²³ ~ and ,´? µ ~ ,´,´?  0µµ
where 0 is the indicator, with 0 ~ that the individual has not gone for a checkup and 0 ~ indicating that the individual has gone for a check-up. Now, let ) be the claim random variable
given that the individual has gone for a check-up. Then ,´?  0 ~ µ ~ ,´)µ ~ ,´,´)  1 µµ,
where 1 is the indicator random variable for which 1 ~ indicates no further treatment is
required. and 1 ~ indicates further treatment is required. Thus,
,´?  0 ~ µ ~ ²³²À³ b ²³²À³ ~ À , and
,´? µ ~ ,´?  0 ~ µ h 7 ´0 ~ µ b ,´?  0 ~ µ h 7 ´0 ~ µ
= ²³²À³ b ²À³²À³ ~ .)
Since ? and ? are independent, = ´? b ? µ ~ = ´? µ b = ´? µ . = ´? µ is found
in the usual way to be = ´? µ ~ ²À³²À³² ³ b ²À³² ³ ~ Á . We can find
= ´? µ from = ´? µ ~ = ´,´?  0µµ b ,´= ´?  0µµ .
,´?  0µ is equal to 0 if 0 ~ and is equal to À if 0 ~ .
Since 7 ´0 ~ µ ~ À , it follows that
= ´,´?  0µµ ~ h ²À³ b ²À ³ h ²À³ c ´²À³²À³µ ~ Á .
Also, = ´?  0 ~ µ ~ , and if 0 ~ then
= ´?  0 ~ µ ~ = ´)µ ~ = ´,´)  1 µµ b ,´= ´)  1 µµ .
Now, ,´)  1 ~ µ ~ ( prob. .75) and ,´)  1 ~ µ ~ b ~ (prob. .25), thus,
= ´,´)  1 µµ ~ ² ³²À³ b ² ³²À³ c ´²³²À³ b ²³²À³µ ~ Á À . We also
have = ´)  1 ~ µ ~ and = ´)  1 ~ µ ~ , so that ,´= ´)  1 µµ ~ .
Then we have = ´?  0 ~ µ ~ Á À b ~ Á À , and
,´= ´?  0µµ ~ h ²À³ b ²Á À³²À³ ~ Á .
It then follows that = ´? µ ~ Á b Á ~ Á ,
and then = ´? b ? µ ~ Á b Á ~ Á .
With members the aggregate claim random variable : ~ '@ , has mean and variance
Á and the probability that premium collected () will exceed total claims is
:c,´:µ
7 ´  :µ ~ 7 ´ c
jÁ  j= ´:µ µ ~ À .
:c,´:µ
Using the normal approximation, j
has a distribution which is approximately
= ´:µ
standard normal, and thus, c
~ À
. Solving for results in  À
.
j
Á
Note that it is possible to find = ´? ] in an alternative way.
(i) ? is 0 with prob. .2 (if the individual does not go for a check-up - call this event ; ~ ),
note that ,´?  ; ~ µ ~ and = ´?  ; ~ µ ~ ;
(ii) ? is 75 with prob. .6 ( if he goes for a check-up and requires no additional treatment prob. ²À³²À³ ~ À
- call this event ; ~ ), note that ,´?  ; ~ µ ~ and
= ´?  ; ~ µ ~ ; and
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 10
5. continued
(iii) ? is 75 + A (additional treatment cost) with prob. .2 (call this event ; ~ ³,
note that ,´?  ; ~ µ ~ b ~ and = ´?  ; ~ µ ~ .
Then = ´? µ ~ = ´,´?  ; µµ b ,´= ´?  ; µµ .
First, ,´= ´?  ; µµ ~ ²³²À³ b ²³²À
³ b ²³²À³ ~ , and second
= ´,´?  ; µµ ~ ² ³²À³ b ² ³²À
³ b ² ³²À³ c ´²³²À³ b ²³²À
³ b ²³²À³µ
~ Á . Then, = ´? µ ~ Á .
6.
&
: ²& ³
c ~ À
²À
³² c ³ ~ À
²À³² c ³ ~ À
²À³² c ³ ~ À
²À³² c ³ ~ À
²À³² c ³ ~ À
: ²³
The estimate of 7 ´;  O;  µ is : ²³ ~ À
À ~ À so the estimate of
7 ´;  O;  µ is c À ~ À
.
7. For the point % ~ there are four & values within the band from to ; these are
&
~ Á & ~ (repeated twice) Á & ~ and & ~ . We have & ²³ ~ for all
other & -values. Since &
~   ~ &
b , we have ²³ ~ bc
~ (since
% ~ is at the right end of the interval centered at &
~ ) ,
since &7 ~   ~ & b , we have ²³ ~ bc
~ ,
since & c ~   ~ & , we have ²³ ~ bc
~ , and
since & c ~   ~ & , we have ²³ ~ bc
~ (since % ~ is at the
left end of the interval centered at & ~ ) À
Then V ²³ ~ ²& ³ h & ²³
~
~ ²&
³ h &
²³ b ²& ³ h & ²³ b ²& ³ h & ²³ b ²& ³ h & ²³
~ ² ³²³ b ² ³² ³ b ² ³² ³ b ² ³²³ ~ .
8. The claim number distribution of a randomly chosen policy is a mixture of two Poisson
distributions. The mean claim number for a randomly chosen policy is
²³²³ b ²À³² c ³ ~ À c À . According to the method of moments, this is set equal to the
first empirical moment (the sample mean). Then À c À ~ À , from which we get V ~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 10
PE-197
9. The likelihood function based on sample % Á ÀÀÀÁ % is
~
~
3²O% Á ÀÀÀÁ % ³ ~ ²% Â ³ ~ ²% b ³cc , and the log of the likelihood is
3 ~ c ² b ³²% b ³ S 3 ~ c ²% b ³ ~ S V~ ²% b³
which will go to B as the sample mean goes to 0.
This is true because ²% b ³  % if %  . Answer: E
10. The cdf is for the exponential distribution with mean . The mle of for the exponential
'&
'%
distribution is the sample mean. Therefore, V ~ and V ~ .
V
V
'% b'&
The estimate from the combined data sets is V ~ b ~ b
b
11. Model B with parameters is preferable to model A with 1 parameter if
²M) c M( ³  À ² c ³ (95-th percentile of chi-square with c degrees of freedom). From
the given information, ²M) c M( ³ ~ ² c b ³ ~ À
From the chi-square table, we see that À ~ À ²³   À
²³ ~ À .
If c ~ , then model B is preferable, but if c ~ (or more) then model B is not
preferable. The maximum value of for which model B is preferable is ~ . Answer: C
12. # has a uniform distribution on the interval ´Á µ, and the conditional distribution of ?O#
has an exponential distribution with mean #. Using the double expectation rule, we have
,´?µ ~ ,´ ,´?O#µ µ ~ ,´#µ ~ b
13. 7 ´class 1O%µ ~
~
c%° h
c%° ' h
~
²%Oclass 1)h7 ²class 1³
²%³
c%°
c%°
' ²%Oclass 1)h7 ²class 1³
~ ' ²%Oclass ³7 ²class ³
14. Prior distribution is ²³ ~ Á   .
The hypothetical mean is ²³ ~ ,´?Oµ ~ , since ? is uniformly distributed on ²Á ³.
Process variance is #²³ ~ = ´?Oµ ~ .
Expected hypothetical mean is ~ ,´?µ ~ ,´ ,´?Oµ µ ~ , ´ µ ~ À
~ À À
Expected process variance is # ~ ,´ = ´?Oµ µ ~ , ´ µ ~ h ~ .
Variance of the hypothetical mean is
~ = ´ ,´?Oµ µ ~ = ´ µ ~ = ´µ ~ h ~ À
There is one observation, so A ~ b
~ À
À
# ~
b °
°
A3 b ² c A³ ~ À
3 b À²À³ ~ À
3 b À
.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 10
15. The probability function of the geometric distribution with ~ is
² ³
~ 7 ´> ~ µ ~ ²b ³b ~
~ ²À³²À³ .
²b ³b
To apply the inverse transform method we need the distribution function of > .
> ~
ÀÀÀ
À
À
À
À
À
- ²³
À
À
À
À
À
Since < ~ À and - ²³ ~ À
 À  - ²³ ~ À the simulated value of > is 3.
16. The median of the empirical distribution is ~ .
Resample
1,1,4,7,7
3 , 4 , 4 , 7 , 10 1 , 4 , 4 , , 3 , 3 , 3 , 4 , 10 4 , 4 , 7 , 7 , 10 1 , 7 , 7 , 10 , 10
V
²V c ³
4
² c ³ ~ ² c ³ ~ ² c ³ ~ ² c ³ ~ ² c ³ ~ ² c ³ ~ The bootstrap estimate of MSE(V ³ is
bbbbb
~ À .
17. Suppose that ? denotes total claims. Then the bonus is ) ~ F
² c ?³ ?  .
 ! ~ ?
This can be written in the form ) ~ ´ c ²? w ³µ ,
and the expected bonus is ,´)µ ~ ~ ´ c ,²? w ³µ À
For the Pareto distribution with ~ and ~ (from the Tables for Exam C) we have
,²? w ³ ~ c
h ´ c ² b
³c µ ~ À .
Then ~ ´ c Àµ S ~ À .
18. The uniform distribution on ´Á j\$µ has pdf ?O\$ ²%O³ ~ j
for  %  j ,
or equivalently,  % . Also, the pdf of \$ is \$ ²³ ~ .
Then ? ²%³ ~ ?O\$ ²%O³ h \$ ²³ ~ % j h 1 d ~ ² c %³ for  %  .
The mean of % is ,´?µ ~ % h ² c %³ % ~ .
j\$
Alternatively, since ,´?O\$µ ~ , we have
j\$
j
j
,´?µ ~ ,´ ,´?O\$µ µ ~ ,´ µ ~ h \$ ²³ ~ ~ .
The 2nd moment of ? is ,´? µ ~ % h ² c %³ % ~ .
The variance of ? is = ´?µ ~ ,´? µ c ²,´?µ³ ~ c ~ À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 10
PE-199
18. continued
Alternatively, since the variance of a uniform distribution is the interval length squared over 12,
we have = ´?O\$µ ~ , so that
j\$
\$
= ´?µ ~ ,´ = ´?O\$µ µ b = ´ ,´?O\$µ µ ~ ,´ µ b = ´ µ .
\$
,´ µ ~ h ,´\$µ ~ h ~ .
j\$
j\$ j\$ = ´ µ ~ ,´² ³ µ c ²,´ µ³ ~ ,´ \$ µ c ² ³ ~ h c ~ .
Then = ´?µ ~ b ~ À
B
B
B
19. ,´4 µ ~ h 7 ´4 ~ µ ~ h 7 ´4 ~ µ ~ h h 7 ´5 ~ µ
~
~
B
h h 7 ´5 ~ µ ~
~
~
,´5 µ
~
~ . Since = ´5 µ ~ ,´5 µ ~ , it follows that
,´5 µ ~ = ´5 µ b ²,´5 µ³ ~ . Then,
B
B
B
,´4 µ ~ h 7 ´4 ~ µ ~ h 7 ´4 ~ µ ~ h h 7 ´5 ~ µ
~
B
~
~
h h 7 ´5 ~ µ ~
~
,´5 µ
~
~ . Therefore,
= ´4 µ ~ ,´4 µ c ²,´4 µ³ ~ c ~ .
The coefficient of variation of 4 is
j= ´4 µ
,´4 µ
j
~ ~ j .
20. There are (at least) two approaches to solving this problem.
(1) Each job offer can be classified as either Type 1 (wage offer is over 28,000, so accept) or
Type 2 (wage offer is under 28,000, so reject). Any particular offer has probability
~ 7 ´?  Á µ ~ c )²³ ~ À of being accepted (as in solution (1)) . The number
of acceptable offers within 3 months, 5 , has a Poisson distribution with mean À . The
probability that no acceptable offers arrive within 3 months is 7 ´5 ~ µ ~ cÀ ~ À
.
(2) The number of job offers 5 that will arrive in a 3-month period has a Poisson distribution
with a mean of 6. We are asked to find
7 ´no acceptable job offer arrives within three monthsµ ~ 7 ´(µ .
B
B
7 ´(µ ~ 7 ´(O5 ~ µ h 7 ´5 ~ µ ~ 7 ´(O5 ~ µ h c
h [ .
~
~
A job offer is unacceptable if the wage ? is less than 28,000 . Since ? has a lognormal
distribution with ~ À , and ~ À, the probability that a particular job offer is
unacceptable is 7 ´?  Á µ ~ 7 ´ ?  Á µ
cÀ
cÀ
~ 7 ´ ? À
 ÀÀ
µ ~ )²³ ~ À .
Since the wages on different offers are mutually independent,
7 ´(O5 ~ µ ~ ²7 ´?  Á µ³ ~ ²À³ .
B
B
´
h²À³µ
Then, 7 ´(µ ~ ²À³ h c
h [ ~ c
h ~ c
h h²À³ ~ À
.
[
~
~
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 10
21. the claim count 5 is Poisson with ~ .À before the deductible is applied.
When the deductible of 500 is applied, the probability of a loss being above the deductible is
7 ´?  µ ~ c - ²³ ~ À . The number of claims 5 Z that are above the deductible also
has a Poisson distribution, but with mean Z ~ À ~ À .
The probability of paying at least one claim after the deductible is implemented is
7 ´5 Z  µ ~ c 7 ´5 Z ~ µ ~ c cÀ ~ À .
We are using the following property of the Poisson distribution. Suppose 5 has a Poisson
distribution with mean and is a count of the number of events occurring in a unit time period.
Suppose that the events can be classified as Type 1 and Type 2 events, and each time an event
occurs the probability of being Type 1 is , and Type 2 is c . If we define 5 Z to be the number
of Type 1 events occurring in the unit time period, then 5 Z also has a Poisson distribution, with
mean Z ~ . In this question, claims can be described as Type 1 (over 500) and Type 2
(  500) . Then ~ 7 ²?  ³ ~ À , and the number of Type 1 claims is Poisson with mean
~ ²À³²À³ ~ À .
22. With ~ , ~ À
 À  À
~ , which is the same proportionally as
 À²³ ~ À  . Therefore, the smoothed empirical estimate of the 40-th percentile is
linearly interpolated .2 of the way from % ~ to %
~ . This will be VÀ ~ À .
V
23. The Nelson-Aalen estimate of /²³ is /²³
~ b b b b ~ À
À
V
The Aalen estimate of the variance of /²³
is
V
=V
´/²³µ
~ b b b b ~ À .
V
The linear interval is /²³
f À
jÀ ,
j
which has width d À
À ~ À
.
For the log-transformation interval, we have < ~ %>
V
À
k= V
´/²³µ
?
V
/²³
~ À
À
V
/²³
The lower limit of the log-transformation interval is < ~ À
À
~ À , and the upper limit
V
is /²³ h < ~ À , so the log-transformed interval has width À c À ~ À
.
Then À
À
~ À
.
24. ? is a mixture of two exponentials with means and and mixing weights of .5 each.
The mean of ? is ²À³² b ³ . According to the method of moments, we set the first
theoretical moment (the mean) equal to the first empirical moment (sample mean) which is given
as 1. Therefore ²À³² b ³ ~ , which is the same as b ~ .
The second moment of an exponential with mean is , and therefore the second moment of
? is ²À³² b ³ ~ b . The empirical variance is , so that the second empirical
moment is b ~ b (2nd moment ~ variance b 1st moment ). Following the method of
moments we set b ~ b . Substituting ~ c , this equation becomes
b ² c ³ ~ b . The method of moments estimates will exist if this quadratic equation
has real roots. This is equivalent to c b c ~ having real roots and distinct
positive solutions. This is equivalent to ² c ³ c ²³²³² c ³  , and
c ² c ³  j² c ³ c ²³²³² c ³ , or equivalently,  , and  . Answer: E
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 10
PE-201
25. The ²Á Á ³ class satisfies ~ b , so that À
À
~ b and
c
À
À
~ b . The resulting equations in and are
b ~ and b ~ À
. Solving for and results in ~ À and ~ À .
À
Then, ~ b ~ À
, so that ²À
³² ³ ~ À .
Therefore, Statement I is false.
There are three²Á Á ³ distributions. The Poisson has ~ , the binomial has  and the negative binomial had  . Since  in this example, the distribution must be
Negative Binomial, so that Statement II is false and Statement III is true. Answer: C
26. Let
denote the number of deaths at time & , not including the uncertain subject.
We are given
b
c c
b
~ À ²³ Á
b b b ~ À
²³ and
b c ² b ³ ~ À
²³. From the third equation we get ~ À
.
Applying this to the first equation, we get
c c
b
c ~ c² bÀ
³
~ À and it follows
that ~ À. Applying
~ À
and ~ À to the second equation results
in ~ À . The product-limit estimate of :²& ³ is
² c ³² c ³ ~ ² c À³² c À³ ~ À .
27. Suppose that the parameter being estimated is . Then ,´@ µ ~ ,´@ µ ~ .
We are also told that = ´@ µ ~ = ´@ µ.
In order for @ b @ to be an unbiased estimator, we must have
,´ @ b @ µ ~ ,´@ µ b ,´@ µ ~ ² b ³ ~ , which means that we must
have b ~ .
= [ @ b @ ] ~ = ´@ µ b = ´@ µ ~ ² b ³= ´@ µ
~ ´ b ² c ³ µ= ´@ µ ~ ² c b ³= ´@ µ .
This variance is minimized if c b is minimized. The minimum of
² ³ ~ c b occurs where Z ² ³ ~ c ~ , so that the minimum variance
occurs when ~ À .
28. With a mean of .6, the Poisson probability of having accidents on a particular day is
²À
³
7 ´5 ~ µ ~ cÀ
h [ . The expected number of days in each of the groups (365 day year) is:
Expected Number of Days
Number of Accidents
h 7 ´5 ~ µ ~ À
À
À
À
À
À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 10
28. continued
We are told that the minimum expected number of observations for any group should be 5.
Therefore, we combine groups 3, 4 and 5 in the group 5  , which has probability
7 ´5  µ ~ c 7 ´5 ~ Á or µ ~ c À ~ À .
²6 c, ³
Then the chi-square statistic is , where 6 is the actual number of observations in
,
group and , is the expected number for group . This is
²cÀ³
À
b
²cÀ³
À
b
²c
À
³
À
b
²cÀ³
À
~ À
.
29. When the claim number distribution is Poisson, the standard for full credibility for
= ´@ µ
,´@ µ
aggregate claims per period based on number of claims is h ´ b ²,´@ µ³ µ ~ h ´ ²,´@ µ³ µ ,
where @ is the claim amount random variable.
For the lognormal, ,´@ µ ~ %² b ³ and ,´@ µ ~ %² b ³ .
%²b ³
Therefore, ~ h ´%²b ³µ ~ ~ S ~ À .
&
Since ~ ² À ³ , we have & ~ À . But & is the b7
percentile of the standard normal
distribution. From the normal table, we have
7 ´>  Àµ ~ À
~ b7
>  5 ²Á ³ . Therefore, 7 ~ À .
, where
²Á³
7 ²:~O³h²³
30. ²O: ~ ³ ~ 7 ²:~³ ~
. 7 ²: ~ O³ ~ 7 ²5 ~ q @ ~ O³ ~ c h ,
7 ²:~³
and ²Á ³ ~ 7 ²: ~ O³ h ²³ ~ c h c ~ c .
B
B
Then 7 ²: ~ ³ ~ ²Á ³ ~ c ~ , and
²O: ~ ³ ~
c
8
~ 4 c
.
8
31. The probability function and distribution function are
%
²%³ À
À À À
À
- ²%³ À
À À À
À
The uniform number .38 simulates an %-value of 2, since
- ²³ ~ À  À  À ~ - ²³ . Similarly, % ~ (total 6), % ~ (total 9),
% ~ (total 13), % ~ (total 14), %
~ (total 17), % ~ (total 17),
% ~ (total 18), %& ~ (total 22). Therefore ~ simulated values are needed.
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32. This is a combination of a gamma prior distribution and Poisson model distribution. For this
combination, the Buhlmann credibility premium is the same as the Bayesian credibility premium
estimate (called exact credibility). The Buhlmann-Straub estimate in this case is the same as the
Buhlmann estimate (which is the same as the Bayesian estimate). The first three months of data
are combined so that we have a total of ~ insureds with a total number of claims of
% ~ b b ~ . Using the Bayesian method, the predictive mean for the next policy of
~
the same type is ² b '% ³² b ³ (actually, the predictive distribution for the number of
claims on the next policy of the same type is a negative binomial distribution with parameters
~ b '% and ~ b ).
In this problem where ~ and ~ ~ À are the gamma parameters from the prior
À
distribution. Therefore, ~ b ~ and ~ ²³²À³b ~ À .
The predictive expectation is ~ À
, This is the expected number of claims for a single
insured in Month 4. For 300 insureds in Month 4 we would expect ²³²À
³ ~ À .
If this had not involved the specific combination of the gamma prior and Poisson model
distributions we would have had to use the Buhlmann method.
Note that we are implicitly assuming that all 450 insureds have the same (unknown) , and we
wish to find the expected number of claims from 300 more policies with the same . That is the
standard Bayesian approach.
²³²³b²³²³b²³²³
c
33. ? ~
~ À Á ~ Á ~ bb
²³²³b²³²³b²³²³
c
? ~
~ À Á ~ bb
²³²³b²
³²³
c
? ~
~ À Á ~ b
²³²³b²³²³b²³²³b²³²³b²³²³b²³²³b²³²³b²
³²³
c
V~?~
bb
~ , ~ .
V# ~ c h ´² c À³ b ² c À³ b ² c À³ µ ~ Á À
Á
V# ~ c h ´² c À³ b ² c À³ b ² c À³ µ ~ Á À Á
V# ~ c h ´² c À³ b ² c À³ µ ~ Á À
#
V b#
V b#
V
~ Á À .
V# ~ bb
~
V
c
~
c
c
h ´ ²? c ? ³ c V#² c ³ µ
~
~ c ²b b ³
d ´ ´²À c ³ b ²À c ³ b ²À c ³ µ c Á À²³µ
~ À .
The credibility premium for policyholder 2 is
V c b ² c A³
V A?
V~²
Á
À ³ h ²À³ b ² c
Á
À ³²³ ~ À .
b
À
b
À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 10
?  ? c  ?  34. @3 ~ H
b À²? c ³  ?  ?  ,´@3 µ ~ ²% c ³²À³ % b ´ b À²% c ³µ²À³ % b ²À³ %
~ b À b ~ À . Answer: D
35. For the Weibull distribution with given, if the data consists of uncensored observations
% Á ÀÀÀÁ % and right-censored values " Á ÀÀÀÁ " , then the mle of is
V ~ ´ % bÄb% b" bÄb" µ ° . Note that in this formulation, the denominator is , the number
of uncensored observations. This approach does not apply if there is truncation involved. The
details of this approach reviewed in the notes section of this study guide.
For this problem, we are given ~ uncensored observations, 20,30,45, and 2 censored
observations, both 50. We are also given ~ . The mle of is
V ~ ´ b b b b µ ° ~ À .
We can solve the problem from basic principles by setting up the likelihood function.
²%°³ c²%°³
²%°³ c²%°³
The pdf is ²%³ ~
~
% %
c²%°³
c²%°³
the cdf is - ²%³ ~ c ~c
3 ~ ²³ h ²³ h ²³ h ´ c - ²³µ
, and
. The likelihood function is
² ³ c²°³
² ³ c²°³
² ³ c²°³
~ h h h ´c²°³ µ
h c² b b bh ³°
² h
³
~
. The loglikelihood function is
hh
² b b bh ³
3 ~ b ² h h ³ c c
c ² h h ³ .
² b b bh ³
Then 3 ~ c b
.
Setting 3 ~ and solving for results in ~ À . Note that from the pdf we get
%
%
²%³ ~ b % c c ² ³ , and ´ c - ²%³µ ~ c ² ³ À Since we are trying to
maximize with respect to , we can ignore the additive factors in 3 that do not involve .
%
%
We maximize '´ c c ² ³ µ b '´ c ² ³ µ , where the first sum is over the known
observations, and the second sum is over the censored observations that are  .
² b b ³
c ² ³ .
² b b bh ³
This reduces to setting c b
~ , as before. Answer: D
We maximize ² c ³ c
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 11
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ACTEX EXAM C/4 - PRACTICE EXAM 11
1. Suppose that the continuous random variable ? has the density function
!²b³
²%³ ~ !²³!²³ %c ² c %³c for  %  , where  and  .
If ~ and ~ , what is the expected value of ² c ?³c ?
A) B) C) D) E) 2. ? has an exponential distribution with mean 1 and ? has an exponential distribution with
mean 2. In a portfolio of risks, half of them have loss variable ? and the other half have loss
variable ? . A policy limit of 1 is imposed on all losses. We define A ~ ? w , and
A ~ ? w . We define @ the following two random variables:
(i) @ is the mixture of ? and ? with mixing weights .5 for each component, and
(ii) > is the mixture of A and A , with mixing weights .5 for each component.
Find ,´@ w µ*,´> ].
A) .25
B) .5
C) 1
D) 2
E) 4
3. Which of the following statements are true about the sums of discrete, independent random
variables?
1. The sum of two Poisson random variables is always a Poisson random variable.
2. The sum of two negative binomial random variables with parameters ²Á ³ and ²Z Á Z ³ is a
negative binomial random variable if ~ Z .
3. The sum of two binomial random variables with parameters ²Á ³ and ²Z Á Z ³ is a binomial
random variable if ~ Z .
A) None if 1, 2, or 3 is true.
B) 1 and 2 only
C) 1 and 3 only
D) 2 and 3 only
E) 1, 2, and 3
Use the following information for questions 4 and 5. You are the producer of a television quiz
show that gives cash prizes. The number of prizes, 5 , and prize amounts, ? , have the following
distributions:
7 ²5 ~ ³
% 7 ²? ~ %³
À
À
À
À
À
4. Your budget for prizes equals the expected prizes plus the standard deviation of prizes.
A) 384
B) 394
C) 494
D) 588
E) 596
5. You buy stop-loss insurance for prizes with a deductible of 200. The cost of insurance
includes a 175% relative security load (the relative security load is the percentage of expected
payment that is added). Calculate the cost of the insurance.
A) 204
B) 227
C) 245
D) 273
E) 357
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6. A farmer develops a model for his seeding season for a particular crop. The number of days in
which crops can be seeded during the season has a Poisson distribution with a mean of 20. On a
day suitable for seeding, the number of acres than can be seeded is either 1 or 2, each with
probability .5. Apply the normal approximation to the total number of acres that can be seeded
during the season to find the probability that the farmer will seed at least 40 acres in the season.
A) .07
B) .08
C) .09
D) .10
E) .11
Questions 7 and 8 are based on the following data set for the times of death or right-censoring (+)
for 25 individuals:
2, 3, 3, 3b , 4, 4, 4, 4, 4b , 5b , 6, 6, 7, 7, 7, 7b , 7b , 8, 9, 10, 12b , 13, 13, 14, 16
7. Find Greenwood's approximation to the variance of : ²³ .
A) À B) À
C) À
D) À E) À
8. * ~ lower limit of the 95% linear confidence interval for :²³
+ ~ lower limit of the 95% log-transformation confidence interval for :²³
Find * c + (nearest .005).
A) .025
B) .005
C) c .005
D) c .025
E) c .050
9. A box contains 2 coins: one has probability 12 of heads, while the other has unknown
probability , 0   1, of heads. A coin is selected at random and flipped. It is then replaced
in the box, and this entire procedure is repeated another time. If 2 heads are observed, what is
the value of the maximum likelihood estimator of ?
A) 14
B) 12
C) 34
D) 1
E) Does not exist
10. You are given the following random sample of 12 data points from a population
distribution ? :
7 , 12 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53
Determine the method of percentile matching estimate of 7 ´?  µ for the lognormal
distribution using the smoothed empirical estimates of 20-th and 80-th percentiles.
A) .25
B) .27
C) .29
D) .31
E) .33
Use the following information for Questions 11 and 12.
The random variable ? has density function ²%³ ~ c%° for %  .
A random sample of three observations of ? yields the values % Á % and % .
11. Find the maximum likelihood estimator of .
A)
D)
% b % b %
% b% b%
B)
% b% b%
%´ c
µ
E) ²% % % ³°
C)
% b % b % SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 11
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V, the maximum
12. If % Á % and % are 10, 20 and 30, respectively, estimate the variance of likelihood estimator of .
A) Less than 50
B) At least 50, but less than 150
C) At least 150, but less than 250
D) At least 250, but less than 350
E) At least 350
13. For an individual risk in a population, the number of claims per month follows a Poisson
distribution with mean #. For the population, # is distributed according to the exponential
distribution, with pdf ²³ ~ c . A randomly selected risk in the population is found to
have had 1 claim in the final 6 months of 2000 and 1 claim for all 2001. Find the BuhlmannStraub credibility premium for this individual for the first three months of 2002.
A) B) C) D) E) 14. A random sample of 1000 observations from a loss distribution has been grouped into five
intervals as follows:
Number of
Observations
Interval
´ Á À³
´À Á À³
´À Á À³
´À Á À³
´À Á B³
A Pareto distribution is fit to the data using the minimum chi-square estimator. The estimated
À
parameter values are ~ À and ~ (- ²%³ ~ c ² %b
³ ).
Find the value of the chi-square goodness-of-fit statistic for testing the fit of the model to the
data.
A) Less than 9.0
B) At least 9.0, but less than 9.2
C) At least 9.2, but less than 9.4
D) At least 9.4, but less than 9.6
E) At least 9.6
15. All policyholders of a particular insurer have a binomial distribution for claim number in one
year. The policyholders are divided into two classes, A and B. Class A is divided equally into
two groups. Policyholders in Group A1 have Bernoulli claim number distribution with mean .1
and policyholders in Group A2 have Bernoulli claim number distribution with mean .2. Class B
policyholders all have an annual claim number distribution which is binomial with ~ and
~ À . Eighty percent of all policyholders are in Class A. A policyholder is selected at random
from the portfolio of all policyholders, and is found to have 0 claims for the year. Three actuaries
wish to determine the expected number of claims next year from this policyholder, using the
following methodologies. Actuary ? ignores first year claim number. Actuary @ calculates the
expected claim number for another policyholder drawn from the same Class as the first
policyholder. Actuary A assumes the same policyholder in the second year as in the first year.
Find the correct ranking of the expectations found by the three actuaries.
A) ?  @  A
B) @  A  ?
C) A  ?  @
D) ?  A  @
E) A  @  ?
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 11
16. A portfolio of insureds consists of two types of insureds. Losses from the two types are:
Type 1 insured loss: exponential with a mean of 1 ,
Type 2 insured loss: exponential with a mean of 2 ,
In the portfolio, of the insureds are of Type 1 and of the insureds are of Type 2.
Two insureds are chosen at random and one loss is observed from each insured.
The first insured is observed to have a loss of 1 and the second insured is observed to have a
loss of 2. Find the probability that the two insured are of the same Type. It is assumed that the
losses of the two insureds are independent of one another.
A) Less than .400
B) At least .40 but less than .425
C) At least .425 but less than .450
D) At least .450 but less than .475
E) At least .475.
17. The number of losses arising from b individual insureds over a single period of
observation is distributed as follows:
Number of Insureds
Number of Losses
0
1
3
2
1
3 or more
0
The number of losses for each insured follows a Poisson distribution, but the mean of each such
distribution may be different for individual insureds. You estimate the variance of the
hypothetical means using Empirical Bayes semiparametric estimation.
Suppose that ~ and suppose that a particular insured is observed for two periods of
observation and is found to have 0 losses in the first period and 2 losses in the second period.
Find the credibility premium for the number of losses for that individual in the next period of
observation.
A) .3
B) .4
C) .5
D) .6
D) .7
18. Two actuaries are simulating the number of automobile claims for a book of business. For
the population that are studying:
i) The claim frequency for each individual driver has a Poisson distribution.
ii) The means of the Poisson distributions are distributed as a random variable, \$.
iii) \$ has a gamma distribution.
In the first actuary's simulation, a driver is selected and one year's experience is generated. This
process of selecting a driver and simulating one year is repeated 5 times. In the second actuary's
simulation, a driver is selected and 5 years of experience are generated for that driver.
Which of the following is/are true?
I. The ratio of the number of claims the first actuary simulates to the number of claims the
second actuary simulates should tend towards 1 as 5 tends to infinity.
II. The ratio of the number of claims the first actuary simulates to the number of claims the
second actuary simulates will equal 1, provided that the same uniform random numbers are used.
III. When the variances of the two sequences of claim counts are compared the first actuary's
sequence will have a smaller variance because more random numbers are used in computing it.
A) I only
B) I and II only
C) I and III only
D) II and II only
E) None of I, II, or III is true
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 11
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19. You are given the following random sample of ten claims:
46
121
493
738
775
1078
1452
2054
2199
3207
Determine the smoothed empirical estimate of the 90! percentile.
A) Less than 2150
B) At least 2150, but less than 2500
C) At least 2500, but less than 2850
D) At least 2850, but less than 3200
E) At least 3200
20. On his walk to work, Lucky Tom finds coins on the ground at a Poisson rate. The Poisson
rate, expressed in coins per minute, is constant during any one day, but varies from day to day
according to a gamma distribution with mean 2 and variance 4. Calculate the probability that
Lucky Tom finds exactly one coin during the sixth minute of today's walk.
A) 0.22
B) 0.24
C) 0.26
() 0.28
E) 0.30
21. For a certain insurance, individual losses in 2000 were uniformly distributed over ² Á ³.
A deductible of 100 is applied to each loss. In 2001, individual losses have increased 5%, and are
still uniformly distributed. A deductible of 100 is still applied to each loss. Determine the
percentage increase in the standard deviation of amount paid.
A) Less than 5.5%
B) At least 5.5% but less than 6.0%
C) At least 6.0% but less than 6.5%
D) At least 6.5% but less than 7.0%
E) At least 7.0%
22. For a discrete probability distribution, you are given the recursion relation
²³ ~ h ²c³,
~ Á Á Ã À
Find the variance of the associated zero-truncated distribution.
A) Less than 1.00
B) At least 1.00 but less than 1.25
C) At least 1.25 but less than 1.50
D) At least 1.50 but less than 1.75
E) At least 1.75
23. Lucky Tom finds coins on his way to work at a Poisson rate of 0.5 coins per minute.
The denominations are randomly distributed:
(i)
60% of the coins are worth 1;
(ii) 20% of the coins are worth 5;
(iii) 20% of the coins are worth 10.
Calculate the variance of the value of the coins Tom finds during his one-hour walk to work.
A) 566
B) 594
C) 632
D) 670
E) 768
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 11
24. You are given
(i) Losses are uniformly distributed on ²Á ³ with  .
(ii) There is a deductible of 20, and a maximum covered loss of 170.
(iii) A sample of payments is: 14, 33, 72, 94, 120, 135, 150, 150
Estimate by matching the average sample cost per payment to the expected cost per payment .
A) Less than 200
B) At least 200 but less than 210
C) At least 210 but less than 220
D) At least 220 but less than 230
E) At least 230
25. The following random sample of size 2 is taken from a Pareto distribution:
18.7 , 51.5
An iterative maximum likelihood estimation method is applied to estimate and .
A starting estimate of is ~ . is estimated using maximum likelihood estimation.
The value of found is denoted and it used to find the mle of , denoted .
The value of is used to find the next mle of denoted .
Find .
A) Less than 3.00
B) At least 3.00 but less than 3.25
C) At least 3.25 but less than 3.50
D) At least 3.50 but less than 3.75
E) At least 3.75
26. You are given the following:
- the parameters of a loss distribution are and V , have information matrix
- the maximum likelihood estimates of these parameters , V and c 0 ² Á ³ ~ >
c ?
Determine the approximate variance of V.
A) .12
B) .4
C) 1.5
D) .6
E) .75
27. The aggregate loss in one week, : , follows a compound negative binomial distribution, and
the severity distribution is exponential. Limited fluctuation credibility is being applied to : so
that the full credulity standard is to be within 5% of expected aggregate losses 95% of the time.
It is found that the expected number of claims needed for full credibility is 5,412 .
Suppose that the frequency distribution is modified (but still negative binomial) so that mean and
variance of the frequency both increase by 20%. Find the full credibility standard for the number
of claims needed for the new compound negative binomial distribution (severity is the same
exponential distribution as before).
A) Less than 5400
B) At least 5400 but less than 5500
C) At least 5500 but less than 5600
D) At least 5600 but less than 5700
E) At least 5700
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 11
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28. For a coin chosen at random from a large collection of coins, the probability of tossing a head
with that randomly chosen coin is , where has pdf ²³ ~ c defined on the interval
  (it is assumed that  ) . Suppose that a coin is chosen at random from the collection of
coins. Suppose that the coin is tossed twice and there is one head and one tail observed.
Find the posterior density of in terms of .
A) ² b ³ ² c ³ B) ² b ³² b ³ ² c ³ C) ² b ³ ² c ³
D) ² b ³ c ² c ³ E) ² b ³² b ³
29. An individual insured has a frequency distribution per year that follows a Poisson
distribution with mean . The prior distribution for is a mixture of two exponential
distributions with means of 1 and 3, and with mixing weights both equal to .5 . An individual is
observed to have 0 claims in a year. Find the Buhlmann credibility estimated number of claims
for the same individual for the following year.
A) B) C) D) E) 30. A scientist perform experiments, each with a 60% success rate. Let ? represent the number
of trials until the first success. Use the inverse transform method to simulate the random variable,
?, and the following ransom numbers (where low numbers correspond to a high number of
trials): 0.15 , 0.62 , 0.37, 0.78 . Generate the total number of trials until three successes result.
A) 3
B) 4
C) 5
D) 6
E) 7
31. For a portfolio of independent risks, you are given:
(i) The risks are divided into two classes, Class A and Class B.
(ii) Equal numbers of risks are in Class A and Class B.
(iii) For each risk, the probability of having exactly 1 claim during the year is 20% and the
probability of having 0 claims is 80%.
(iv) All claims for Class A are of size 2.
(v) All claims for Class B are of size , an unknown but fixed quantity.
One risk is chosen at random, and the total loss for one year for that risk is observed. You wish to
estimate the expected loss for that same risk in the following year.
Determine the limit of the Buhlmann credibility factor as goes to infinity.
A) 0
B) 1/9
C) 4/5
D) 8/9
E) 1
32. You are given a random sample of two values from a distribution function - :
1
3
2
?1 b ?2 .
2
You estimate = (? ) using the estimator 1
2 ~1(? c ? ) , where ? ~
2
Determine the bootstrap approximation to the mean square error.
A) 0.0
B) 0.5
C) 1.0
D) 2.0
E) 2.5
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 11
33. A coin is chosen at random from a collection of coins. The probability of tossing a head is .
Suppose that is a random variable with pdf ²³ ~ ² c ³ for   .
The coin is tossed 3 times. Find the probability of tossing 3 heads.
A) B) C) D) E) 34. The times of death in the data set being analyzed are
! ¢ 10 , 20 , 30 , 40 , 50 , 60 , 70 , 80 , 90 , 100 .
Use the triangle kernel with bandwidth 20 to find the kernel density estimator of the distribution
V ²!³ of time until death at the point ! ~ .
function A) .40
B) .42
C) .44
D) .46
E) .48
35. A portfolio of insurance policies consists of two types of policies. The annual aggregate loss
distribution for each type of policy is a compound Poisson distribution. Policies of Type I have a
Poisson parameter of 1 and policies of Type 2 have a Poisson parameter of 2. For both policy
types, the claim size (severity) distribution is uniformly distributed on the integers 1, 2 and 3.
Half of the policies are of Type I and half are of Type II. A policy is chosen at random and an
aggregate annual claim of 2 is observed.
Find the probability that the policy is of Type 1.
A) Less than .49
B) At least .49, but less than .51
C) At least .51, but less than .53
D) At least .53, but less than .55
E) At least .55
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 11
PE-213
ACTEX EXAM C/4 - PRACTICE EXAM 11 SOLUTIONS
1. ²%³ is the density function of a beta distribution with parameters and .
!²³!²³
Thus, since ²%³ % ~ it follows that %c ² c %³c % ~ !²b³ .
!²b³
Then, ,´² c ?³c µ ~ ² c %³c ²%³ % ~ !²³!²³ %c ² c %³cc %
!²b³
!²³!²c³
!²b³
!²c³
~ !²³!²³ h !²bc³ ~ !²³ h !²bc³ .
Then using the relationship !²³ ~ ² c ³[ for any integer  , or using the relationship
!²!³ ~ ²! c ³ h !²! c ³ for !  , the expression
reduces to
²bc³²bc³²bc³²bc³
²c³²c³²c³²c³
becomes 42.
!²b³
!²³
!²c³
h !²bc³
. With ~ and ~ , this
2. The cdf of @ is the mixture of the cdf's of ? and ? ,
[email protected] ²%³ ~ ²À³² c c% ³ b ²À³² c c%° ³ ~ c ²À³²c% b c%° ³ .
Then ,´@ w µ ~ ´ c [email protected] ²%³µ % ~ ²À³²c% b c%° ³ %
~ ²À³´² c c ³ b ² c c° ³µ ~ À c ²À³²c b c° ³ ~ À .
Since [email protected] ²%³ ~ ²À³´-? ²%³ b -? ²%³µ , it follows that
c [email protected] ²%³ ~ c ²À³´-? ²%³ b -? ²%³µ ~ ²À³´² c -? ²%³³ b ² c -? ²%³³µ .
Then ,´@ w µ ~ ´ c [email protected] ²%³µ % ~ ²À³´² c -? ²%³³ b ² c -? ²%³³µ %
~ ²À³´ ´ c -? ²%³µ % b ´ c -? ²%³µ %µ ~ ²À³²,´? w µ b ,´? w µ³
~ ²À³²,´A µ b ,´A µ³ ~ ,´> µ . This does not depend on the ? 's having exponential
distributions.
3. 1. True. This can be verified by looking at the moment generating function of the sum of two
Poisson random variables and seeing that it is also Poisson.
2. False. The sum is negative binomial if ~ Z (in that case the sum is negative binomial with
parameters b Z and . This can be thought of in another way. If is an integer, then the
negative binomial is the number of failures until the -th success, where the probability of success
on any given trial is b . The sum of the two negative binomial random variables with ~ Z is
the number of failures until the -th success, plus the number of failures after that until the Z -th
success; this total is the number of failures until the b Z -th success).
3. True. The sum is binomial with parameters b Z and . This can be thought as follows.
The binomial random variable with parameters and is the number of successes in trials of
an experiment, where is the probability of success in any given trial. If the success probability
is the same for two binomial random variable, ( ~ Z ), then the sum of the two independent
binomial random variables is the number of successes in the first trials, plus the number of
successes in the next Z trials, which is the number of successes in b Z trials. Answer: C
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 11
4. This problem involves a compound distribution. The frequency (number of prizes) is 5 and
the severity (prize amount) is ? . The aggregate prize amount is
: ~ ? b ? b Ä b ?5 , with mean ,´:µ ~ ,´5 µ h ,´?µ ~ ²À³²³ ~ and
variance = ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ À
In this case, = ´5 µ ~ ,´5 µ c ²,´5 µ³ ~ À
c ²À³ ~ À
, and
= ´?µ ~ ,´? µ c ²,´?µ³ ~ Á c ²³ ~ Á .
Then, = ´:µ ~ ²À³²Á ³ b ²À
³²³ ~ Á À
The budget is ,´:µ b Àj= ´:µ ~ b ÀjÁ ~ .
5. This problem involves stop-loss insurance. The aggregate amount of prizes, : , is integervalued, with multiples of 100. The cost of the stop-loss insurance is ² b À³,´²: c ³b µ,
where ,´²: c ³b µ ~ ,´:µ c ,´: w µ .
From Problem 29, we have ,´:µ ~ . We can find probabilities for : by looking at
combinations of 5 and : .
7 ´: ~ µ ~ 7 ´5 ~ µ h 7 ´? ~ µ b 7 ´5 ~ µ h 7 ´? ~ µ h 7 ´? ~ µ
~ ²À³²À³ b ²À³²À³²À³ ~ À
,
7 ´: ~ µ ~ 7 ´5 ~ µ h 7 ´? ~ µ b 7 ´5 ~ µ h 7 ´? ~ µ h 7 ´? ~ µ
b 7 ´5 ~ µ h 7 ´? ~ µ h 7 ´? ~ µ ~ ²À³²À³ b ²À³²À³²À³ ~ À
.
Then, ,´: w µ ~ 7 ²: ~ ³ b 7 ²:  ³³
~ ²À
³ b ² c À
c À
³ ~ À ,
and ,´²: c ³b µ ~ c À ~ À .
Therefore, the cost of the insurance is ² b À³²À³ ~ À. Answer: D
6. : ~ number of acres seeded has a compound Poisson distribution with mean
,´:µ ~ ,´?µ ~ ²³²À³ ~ , and variance
= ´:µ ~ ,´? µ ~ ²³² ³ ~ . : is an integer, so
:c,´:µ
= ´:µ
7 ´:  µ ~ 7 ´:  Àµ ~ 7 ´ j
 Àc
µ ~ 7 ´A  Àµ ~ c )²À³ ~ À
.
j
7. ~ Á ~ Á ~ Á ~ Á ~ Á ~ Á ~ Á ~ Á ~ ~ Á ~ Á ~ Á ~ Á ~ Á ~ Á ~ Á ~ À
² c ³ ~ ²³²³ b ²³²³ b ²³²³ b ²³²³ b ²³²³
!  b ²³²³
b ²³²
³
b ²
³²³
~ À . Greenwood's approximation is
=V
´: ²³µ ~ ´: ²³µ ²À³ ~ ²À
³ ²À³ ~ À . Answer: C
8.
² c ³
!  ~ À (from Problem 4 above), : ²³ ~ À
À
j= V
´: ²³µ
²À
³hjÀ
< ~ %> : ²³h´: ²³µ ? ~ %< ²À
³h´: ²³µ = ~ À
À
* ~ : ²³ c ²À
³ h j= V
´: ²³µ ~ À
c ²À
³j À ~ À.
°<
°À
V
+ ~ ´:²³µ
~ ²À
³
~ À À * c + ~ c À
À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 11
PE-215
9. Given the likelihood function 3²³ for a parameter based on an observed random sample or
random experiment, the maximum likelihood estimate for is that value of that maximizes
3²³. We first determine the likelihood function for the parameter . Let coin 1 be the one with
probability of heads, and let coin 2 be the one with probability of heads.
Let ( denote the event that the coin chosen at random and then flipped turns up as a head. Let
) and ) denote the events that a coin chosen at random is coin 1 () ) or coin 2 () ),
respectively. Then, since ) and ) are mutually exclusive and exhaustive events (i.e., exactly
one of ) and ) must occur), it follows that
7 ´(µ ~ 7 ´( q ²) r ) ³µ ~ 7 ´²( q ) ³ r ²( q ) ³µ
~ 7 ´( q ) µ b [( q ) µ ~ 7 ´(  ) µ h 7 ´) µ b 7 ´(  ) ] h 7 ) µ
~ 12 h 12 b h ~ ²b).
The likelihood function for based on the occurrence of event ( on two independent trials of the
procedure outlines above is [ ²b³µ . Since 0   , this is maximized at the point ~ .
10. The sample has ~ data points, so the smoothed empirical percentiles for the sample are
%
ÀÀÀ
...
Smoothed Perc. À À À
À
À
À
Since À d ~ À
, the smoothed empirical estimate of the 20-th percentile is 60% of the way
from the 2nd to the 3rd data point, which is 13.8 . Similarly, since À d ~ À, the smoothed
empirical estimate of the 80-th percentile is 40% of the way from the 10-th to the 11-th data
% c point, which is 35. The cdf of the lognormal with parameters and is - ²%³ ~ )² ³ .
The 20-th and 80-th percentiles of the standard normal distribution are 'À ~ c À and
'À
~ À .
À c c The percentile matching equations are
~ c À and
~ À À
Solving these two equations results in V ~ À Á ~ À .
c À
Then, 7 ´?  µ ~ c - ²³ ~ c )² À
³ ~ c )²À
³ ~ c À ~ À .
11. The mle of the mean of an exponential random variable is the sample mean.
c²% b% b% ³°
The likelihood function is 3²³ ~ ² c% ° ³² c% ° ³² c% ° ³ ~
.
% b% b%
The loglikelihood is M²³ ~ 3²³ ~ c
c , and the maximum likelihood
c
% b% b%
V ~ % b% b% ~ ?
equation is
M²³ ~ S
c ~S
.
This is just the mle for the exponential distribution, which is the sample mean.
= ´?µ
c
c
V~?
Vµ ~ = ´?µ
12. Since , it follows that = ´
~
À For the exponential
distribution with mean , the variance is the square of the mean, = ´?µ ~ . In this case,
c
V~?
V is
~ , and the estimate of is ~ bb
~ , so the estimated variance of ~ .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 11
13. Let * represent the number of claims in a month for a randomly chosen individual. Then
²³ ~ ,´* O# ~ µ ~ (mean of the Poisson distribution with parameter ),
~ ,´*µ ~ ,´²#³µ ~ ,´#µ ~ À (mean of the exponential),
#²³ ~ = ´*O# ~ µ ~ (variance of the Poisson distribution with parameter ),
# ~ ,´#²#³µ ~ ,´#µ ~ À, and ~ = ´²#³µ ~ = ´#µ ~ À (variance of the exponential is
the square of the mean).
Let *Á Á *Á Á ÀÀÀÁ *Á
denote the number of claims in each of the final 6 months of 2000 for the
individual, and let *Á Á *Á Á ÀÀÀÁ * Á denote the number of claims in each of the 12 months of
2001. We have ~ 18 monthly claim numbers and we can apply the Buhlmann method to * .
À
~ # ~ À
~ Á A ~ b
~ b
~ ,
c *Á b*Á bÄb*Á
b*Á b*Á bÄÀb* Á
and * ~
~ .
Then the Buhlmann(-Straub) credibility premium is
c
A* b ² c A³ ~ ² ³² ³ b ² ³²À³ ~ per month.
The credibility premium for the first three months of 2001 is ² ³ ~ À Answer: A
²6 c, ³
À 6 ~ Á 6 ~ Á 6 ~ Á 6 ~ Á 6 ~ ,
~
14. ~ , ~ - ²³ ~ ´ c ² ³À µ ~ À Á
, ~ ´- ²À³ c - ²³µ ~ ´² ³À c ² À ³À µ ~ À Á
, ~ ´- ²³ c - ²À³µ ~ ´² À ³À c ² ³À µ ~ À Á
, ~ ´- ²³ c - ²³µ ~ ´² ³À c ² ³À µ ~ À Á
, ~ ´- ²B³ c - ²³µ ~ ´² ³À µ ~ À À Á
²cÀ³
²cÀ³
²cÀ³
b
b
À
À
À
²cÀ³
²cÀ³
b
b
À
À
~
15. 5 ~ number of claims in one year.
For Actuary ? , ,´5 µ ~ ,´5 O(µ h 7 ²(³ b ,´5 O)µ h 7 ´)µ À
,´5 O(µ ~ ,´5 O(µ h 7 ²(O(³ b ,´5 O(µ h 7 ²(O(³ ~ ²À³²À³ b ²À³²À³ ~ À À
Also, ,´5 O)µ ~ ²³²À³ ~ À . We are given 7 ²(³ ~ À Á 7 ²)³ ~ À , so that
,´5 µ ~ ²À³²À³ b ²À³²À³ ~ À
.
For Actuary @ , we wish to find ,´5 O5 ~ µ , where 5 is the number of claims from someone
drawn randomly from the same class as for 5 . We condition over the class:
,´5 O5 ~ µ ~ ,´5 O(µ h 7 ²(O5 ~ ³ b ,´5 O)µ h 7 ²)O5 ~ ³ .
Since Class A has two types of policyholder, we condition over the two types to get
,´5 O(µ ~ ,´5 O(µ h 7 ²(O(³ b ,´5 O(µ h 7 ²(O(³ ~ ²À³²À³ b ²À³²À³ ~ À À
Class B has one type of policyholder, ,´5 O)µ ~ ²³²À³ ~ À .
We must now find 7 ²(O5 ~ ³ and 7 ²)O5 ~ ³ . Since the Class is selected at random and
two-thirds are in Class A, 7 ²(³ ~ À and 7 ²)³ ~ À , and since Class A is divided equally into
the two types, 7 ²(³ ~ 7 ²(³ ~ À (and 7 ²(³ ~ 7 ²(³ b 7 ²(³ ~ À).
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 11
PE-217
15. continued
We find 7 ²(O5 ~ ³ in the usual Bayesian way:
7 ´²5 ~³q(µ
7 ²5 ~³
7 ²5 ~|(³h7 ²(³
~ 7 ²5 ~|(³h7 ²(³b7
²5 ~|)³h7 ²)³ .
7 ²5 ~ |(³ ~ 7 ²5 ~ O( ³ h 7 ²( O(³ b 7 ²5 ~ O( ³ h 7 ²( O(³
~ ²À³²À³ b ²À³²À³ ~ À Á and 7 ²5 ~ |)³ ~ ²À³ ~ À À
7 ²(O5 ~ ³ ~
²À³²À³
Then, 7 ²(O5 ~ ³ ~ ²À³²À³b²À³²À³ ~ À
Á 7 ´)O5 ~ µ ~ À .
Finally, ,´5 O5 ~ µ ~ ,´5 O(µ h 7 ²(O5 ~ ³ b ,´5 O)µ h 7 ²)O5 ~ ³
~ ²À³²À
³ b ²À³²À³ ~ À
For Actuary A , we wish to find ,´5 O5 ~ µ , where 5 is the number of claims from the same
policyholder. We condition over policyholder type.
,´5 O5 ~ µ ~ ,´5 O(µ h 7 ²(O5 ~ ³ b ,´5 O(µ h 7 ²(O5 ~ ³
b ,´5 O)µ h 7 ²)O5 ~ ³ .
As above, ,´5 O(µ ~ À Á ,´5 O(µ ~ À Á ,´5 O)µ ~ À , 7 ´(µ ~ 7 ´(µ ~ À Á
7 ²5 ~ O(³ ~ À , 7 ²5 ~ O(³ ~ À and
7 ²5 ~ ³ ~ ²À³²À³ b ²À³²À³ ~ À .
Then, 7 ²(O5 ~ ³ ~
²À³²À³
7 ´²5 ~³q(µ
7 ²5 ~³
7 ²5 ~|(³h7 ²(³
~ 7 ²5 ~|(³h7²(³b7 ²5 ~|)³h7 ²)³
~ À ~ À
, and similarly 7 ²(O5 ~ ³ ~ À .
As above, 7 ²)O5 ~ ³ ~ À .
Then ,´5 O5 ~ µ ~ ²À³²À
³ b ²À³²À³ b ²À³²À³ ~ À .
Then A  @  ? .
16. We will define ? and ? to be the loss amounts from the first and second insured,
respectively.
We are given that ? ~ and ? ~ .
We will define \$ to be the ¸Á ¹ random variable that identifies Type, and \$ and \$ are the types of
insured 1 and insured 2.
We wish to find 7 ²\$ ~ \$ O? ~ Á ? ~ ³ . Using the definition of conditional probability, this is
7 ´²\$ ~\$ ³q²? ~Á? ~³µ
7 ²? ~Á? ~³
À
Since \$ must be 1 or 2, the numerator can be formulated as
7 ´²\$ ~ \$ ³ q ²? ~ Á ? ~ ³µ
~ 7 ´²\$ ~ \$ ~ ³ q ²? ~ Á ? ~ ³µ b 7 ´²\$ ~ \$ ~ ³ q ²? ~ Á ? ~ ³µ
~ 7 ²? ~ Á ? ~ O\$ ~ \$ ~ ³ h 7 ²\$ ~ \$ ~ ³
b 7 ²? ~ Á ? ~ O\$ ~ \$ ~ ³ h 7 ²\$ ~ \$ ~ ³
Since both policies are chosen at random, 7 ²\$ ~ \$ ~ ³ ~ and 7 ²\$ ~ \$ ~ ³ ~
Also, since the insureds are independent of one another,
7 ²? ~ Á ? ~ O\$ ~ \$ ~ ³
~ 7 ²? ~ O\$ ~ \$ ~ ³ h 7 ²? ~ O\$ ~ \$ ~ ³ ~ c h c ~ c
(the notation 7 really means density of ? , not probability, in this situation).
.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 11
16. continued
Similarly, 7 ²? ~ Á ? ~ O\$ ~ \$ ~ ³
~ 7 ²? ~ O\$ ~ \$ ~ ³ h 7 ²? ~ O\$ ~ \$ ~ ³ ~ c° h c° ~ c° .
Then, 7 ´²\$ ~ \$ ³ q ²? ~ Á ? ~ ³µ ~ c h b c° h ~ À
.
Also, 7 ²? ~ Á ? ~ ³ ~ 7 ²? ~ ³ h 7 ²? ~ ³ because of independence of ? and ? .
Each of the ? 's is a mixture of two exponentials with pdf ²c% b c%° ³ , so
7 ²? ~ ³ ~ ²c b c° ³ , and 7 ²? ~ ³ ~ ²c b c° ³ , and
7 ²? ~ Á ? ~ ³ ~ ²c b c° ³ h ²c b c° ³ ~ À .
Finally, 7 ²\$ ~ \$ O? ~ Á ? ~ ³ ~ À
À ~ À .
c
17. V ~ V# ~ ? ~ ~ À Á
=V
´?µ ~ ´² c ³ b ² c ³ b ² c ³ µ ~ À Á
V
~ À c À ~ À À A ~
V
À ~ À
b À
The credibility premium is ²À³² b
³ b ²À³²À³ ~ À .
18. I. The combination of a Poisson claim count with mean \$ and a gamma distribution for \$
results in a negative binomial distribution being simulated by actuary 1. The average number of
claims simulated by actuary 1 in 5 trials is 5 h ,´\$µ . The second actuary selects a driver with
Poisson parameter and the average number of claims in 5 years for that driver will be 5 .
The ratio
5 ,´\$µ
5
tends to 1 only if the second actuary's driver's is equal to ,´\$µ. False
II. This is false for the same reason as I. False
III. For actuary 1, the variance of the sequence generated is the variance of a negative binomial
distribution. For actuary 2, the variance of the sequence generated is the variance of the Poisson
distribution with parameter (the for the driver chosen by actuary 2). Either variance could be
larger than the other depending on the value of for actuary 2's driver. False
19. With ~ ordered numbers, we find the empirical estimate of the 90-th percentile as
b
follows. First identify the integer value of such that b  À  b . For this example, we
b
find so that  À  , from which it follows that  À  b , and therefore,
~ À We then perform a linear interpolation between the 9-th (-th) and 10-th ( b -st)
numbers in the series. The 9-th and 10-th numbers (in order) are 2199 and 3207, and we
interpolate to the "9.9-th number". Since 9.9 is .9 of the way from 9 to 10, the smoothed
empirical estimate is ² c À³²³ b ²À³²³ ~ À . Answer: D
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 11
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20. 5 is the number of coins found per minute today. We are told that the distribution of 5
given is Poisson with mean , and the distribution of is gamma with mean ~ and
variance ~ . It follows that ~ and ~ À The unconditional distribution of 5 is
negative binomial with parameters ~ ~ and ~ ~ .
The probability 7 ´5 ~ µ ~ ²b ³ ~ .
21. In 2000, with no policy limit and with a deductible of 100, the expected amount paid per loss
is ,´²? c ³b µ . The density function of ? is ? ²%³ ~ À for  %  (0
otherwise) and the distribution function -? ²%³ ~ À% for  %  .
The expectation can be formulated in each of the following ways:
(i) ²% c ³²À³ % ~ (ii) ,´?µ c ,´? w µ ~ c ´ %²À³ % b ² c -? ²³³µ
~ c ´ b ² c À³µ ~ .
(iii) ´ c -? ²%³µ % ~ ´ c À%µ % ~ .
The variance of amount paid is ,´²? c ³b µ c ²,´²? c ³b µ³ .
,´²? c ³b µ ~ ²% c ³ ²À³ % ~ Á , so that the standard deviation of
amount paid per loss in 2000 is jÁ c ²³ ~ .0 .
In 2001, the loss @ is uniform on ² Á ³ . The variance of amount paid per loss is
,´²@ c ³b µ c ²,´²@ c ³b µ³
~ ²% c ³ ² ³ % c ´
²% c ³ ² ³ %µ ~ Á c ²À
³ ~ Á ,
and the standard deviation is 295. The percentage increase in standard deviation from 2000 to
2001 is À
À c ~ À .
22. This question can be solved by referring to the ²Á Á ³ class of distributions. A discrete non
negative integer-valued random variable with probability function ²³ ~ Á ~ Á Á Á Á ÀÀÀ
is a member of the ²Á Á ³ class is there are constants and such that for all
~ Á Á Á ÀÀÀ, the probability function satisfies the relationship ~ b .
The Poisson with parameter has
c
~
²c ³°[
²c c ³°²c³[
~
c
, so that ~ Á ~ À
Therefore, the distribution in this problem is Poisson with ~ .
The probability the ? ~ is c . The zero-truncated distribution has probabilities
7 ²? A ~ ³ ~ c
c d 7 ²? ~ ³ for ~ Á Á ÀÀÀ .
Then ,´? A µ ~ c
,´²? A ³ µ ~ c
c d ,´?µ ~ cc and
c d ,´? µ ~ cc
(note that ,´? µ ~ = ´?µ b ²,´?µ³ ~ b ~ ³.
c
c
= ´? A µ ~ c
c c ² cc ³ ~ ²cc ³ ~ À .
23. The overall rate at which coins are found is 30 per hour (.5 per minute). The rate at which
coins worth 1 are found is ²³²À
³ ~ per hour,
for coins worth 5 the rate is ²³²À³ ~ per hour, and for coins worth 10 the rate is
²³²À³ ~ per hour. Since the overall way in which coins are found is a Poisson process, and
since the denominations are randomly distributed (and therefore independent) it is also true that
for each denomination of coin amount, coins are found according to a Poisson process.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 11
23. continued
Suppose that 5 is the number of coins worth 1 found in one hour, 5 is the number of coins
worth found in one hour, 5 is the number of coins worth 10 found in one hour.
Then = ´5 µ ~ Á = ´5 µ ~ and = ´5 µ ~ , and 5 Á 5 and 5 are mutually
independent. The total value of all coins found in the hour is 5 b 5 b 5 , and the
variance is = ´5 b 5 b 5 µ ~ = ´5 µ b = ´5 µ b = ´5 µ
~ b ²
³ b ²
³ ~ .
24. The average sample cost per payment is bbbbbbb
~ .
For a uniform distribution on the interval ²Á ³, the expected cost per loss, with deductible 20
and maximum covered loss 170 is
,´? w µ c ,´? w µ ~ ´ c - ²%³µ % ~ ´ c % µ % ~ c c À
The expected cost per payment is
,´?wµc,´?wµ
7 ²?³
~
We set this equal to 96 and solve for , which results in
c Á
c
c Á
c ~
c Á
À
c ~ , so that ~ À .
25. The log of the density of the Pareto distribution is
²%³ ~ b c ² b ³ ²% b ³ and
C
C
b
²%³ ~ c %b
,
C
C
²%³ ~ b c ²% b ³
.
Given ~ , with 2 sample values, to find the mle of , we solve
C
C 3 ~ b c '²% b ³ ~ b c ² À b À³ ~ .
Solving for results in ~ À .
Now, with ~ À , to find the mle of , we solve
²À³
À
À
3 ~ c ' %b
~ c Àb
c Àb ~ .
b
À²Àb³²Àb³cÀ²Àb³cÀ ²Àb ³
This expression becomes
~.
²Àb³²Àb³
C
C
The numerator is À b À c .
Setting this equal to 0 results in two solutions for , 113.5 and c 28.9 .
We ignore the negative root and use ~ À as the new estimate for .
The next estimate of is found from
b À c ² À b À³ ~ .
Solving for results in ~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 11
PE-221
26. The variance/covariance matrix of the maximum likelihood estimates is the inverse of the
c
information matrix. The inverse of >
is c
.
?
>
c
?
À À
The inverse of the given matrix is ²³²
³c²c³²c³
> ? ~ > À À ? .
V . Therefore, the variance of This is the variance/covariance matrix of V,
V is .12 .
27. We will denote the frequency by 5 and the severity will be @ .
The full credibility standard for the expected number of claims needed is
À h
= ´5 µh²,´@ µ³ b,´5 µh= ´@ µ
,´5 µh²,´@ µ³
~ À .
Since the severity is exponential, we have ,´@ µ ~ and = ´@ µ ~ ~ ²,´@ µ³ .
The full credibility standard for expected number of claims needed becomes
À h
= ´5 µb,´5 µ
,´5 µ
. Since both the mean and variance of 5 are increasing by 20%,
this full credibility standard is unchanged at 5412.
28. 7 ²? ~ ³ ~ ²Á ³ ~ 7 ²? ~ O³ ²³ ~ ² c ³ h c ~ ² c b ³ ~ h ´ b
c b
µ ~ ²b³²
b³ .
Alternatively, ² c ³ h c ~ h ² c ³ ~ h )² b Á ³
~ h
!²b³h!²³
!²b³
h!²b³h
~ ²b³²b³h!²b³ ~ ²b³²
b³ .
7 ²? ~ ³ ~ ²Á ³ ~ 7 ²? ~ O³ ²³ ~ h c ~ b ~ b
.
We wish to find ²O? ~ ³ .
This can be formulated as
²?~O³h²³
7 ²?~³
~
²c³hc
²b³²b³
~ ² b ³² b ³ ² c ³ .
Alternatively, the prior is a beta distribution with ~ and ~ , and since the model distribution of
? is binomial with ~ and ~ , and since we have observed % ~ ,
the posterior distribution of is also beta with
Z ~ b % ~ b and Z ~ b c % ~ b c ~ , with pdf
!²Z bZ ³
!²b³
²O% ~ ³ ~ !²Z ³h!²Z ³ h c ² c ³ c ~ !²b³h!²³ h ² c ³
~ ² b ³² b ³ ² c ³ . Answer: B
Z
Z
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 11
29. ,´?Oµ ~ Á = ´?Oµ ~ . Since is a mixture, ,´µ ~ ²À³²³ b ²À³²³ ~ and ,´ µ ~ ²À³² d ³ b ²À³² d ³ ~ (each component of the mixture has an
exponential distribution, and the second moment of an exponential is two times the square of the
mean). Then ~ ,´²³µ ~ ,´µ ~ Á # ~ ,´#²³µ ~ ,´µ ~ Á
~ = ´²³µ ~ = ´µ ~ ,´ µ c ²,´µ³ ~ .
For a single observation of ? , ~ , the Buhlmann credibility factor is A ~ b ~ .
With ? ~ , the Buhlmann credibility estimate for the next year is
c
A? b ² c A³ ~ b ² ³²³ ~ .
30. To say that low numbers correspond to a low number of trials is the standard form of the
inverse transform method: given a uniform random number we find the integer such that
-? ² c ³   -? ²³ , and the simulated value of ? is . In this problem we are asked to
apply the inverse transform method in the form where low numbers correspond to a high number
of trials, so that given a uniform random number , we find the integer such that
-? ² c ³  c  -? ²³ .
The random variable ? is the number of trials until the first success. This is a version of the
geometric distribution with probability function ²%³ and distribution function - ²%³ as follows:
?~
ÀÀÀ
²³
À
²À³²À
³ ~ À
²À³ ²À
³ ~ À
²À³ ²À
³ ~ À ÀÀÀ
- ²³
À
À
À
À
Each uniform number simulates the number of trials until the next success. We must simulate
? three times to get ? (the simulated number of trials until the first success), ? (the additional
number of trials until the second success) and ? (the additional number of trials until the third
success). Then the total number of trials until the third success is ? b ? b ? .
The first random number is ~ À, so that c ~ À . We see that
- ²³ ~ À  À  À
~ - ²³ so that the simulated value of ? is 3. This is ? , the
simulated number of trials until the first success. The second random number is ~ À
, with
c ~ À . We see that À  À
~ - ²³ , so that the simulated value of ? is 1.
~ À , so that c ~ À
. We see that - ²³ ~ À
 À
 À ~ - ²³, so that the
simulated value of ? is 2. Then ? b ? b ? ~ b b ~ . Answer: D
31. ? is the total loss for one year on the randomly chosen claim.
# is either A or B, denoting the two classes. 7 ´# ~ (µ ~ 7 ´# ~ )µ ~ .
²³ ~ ,´?Oµ S ²(³ ~ ,´?O(µ ~ ²À³ ~ À Á ²)³ ~ ,´?O)µ ~ À Á
#²³µ ~ = ´?Oµ S #²(³ ~ = ´?O(µ ~ ²À³²À³ ~ À
Á
#²)³ ~ = ´?O)µ ~ ²À³²À³ ~ À
. The credibility factor is A ~ b # ,
where # ~ ,´#²#³µ ~ h #²(³ b h #²)³ ~ À b À , and
~ = ´,´?O#µµ ~ = ´²#³µ ~ ´ h ²À³ b h ²À³ µ c ´ h ²À³ b h ²À³µ
~ À c À b À . Credibility is based on ~ year.
ÀbÀ Then A ~ b # ~
. As SB , ÀcÀbÀ S ,
ÀbÀ b ÀcÀbÀ
and A S b ~ .
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32. Suppose that a random sample from a distribution is given: ? Á ? Á ÀÀÀÁ ? , and suppose that
the sample is used to estimate some parameter of the distribution. If the estimator is V , then the
bootstrap approximation to the mean square error of this estimator is , <²V c ³ =.
In this expression is the value in the empirical distribution of the parameter being estimated, and
the expected value is taken within the empirical distribution. In this case, the parameter being
b
estimated is the distribution variance. The mean of the empirical distribution is ~ , and
the variance of the empirical distribution is ´² c ³ b ² c ³ ³ ~ .
To find the expectation, we must consider all samples of size 2 from the empirical distribution
(since the empirical distribution was based on a sample of size 2 itself).
The mean square error of the estimator is , <²V c ³ =.
From the empirical distribution on the set ¸Á ¹, there are four possible samples, which are
2
²? Á ? ³ ¢ ²Á ³ Á ²Á ³ Á ²Á ³ and ²Á ³ . For each sample, we must calculate 1 (? c ? )2
2 ~1
2
c
for that sample. For instance, for the sample ²Á ³, we have ? ~ , and (? c ? )2 ~ .
~1
The bootstrap approximation to the mean square error of the estimator is the average of the 4
2
values of (? c ? )2 that we get. This is summarized in the following table.
~1
Sample ²? Á ? ³
c ²? c ?³
~
²Á ³
²Á ³
²Á ³
²Á ³
Since each of the four possible samples is equally likely to occur, the bootstrap estimate of
´² c ³ b ² c ³ b ² c ³ b ² c ³ µ ~ .
33. 7 ´? ~ µ ~ 7 ´? ~ Oµ h ²³ ~ h ² c ³ ~ ´ c µ ~ .
V ²%³ ~ ²& ³ h 2& ²%³ À
34. The kernel density estimator of - ²%³ is ~
There are 10 distinct data points, so that ²&³ ~ À for each & . The estimate is at ! ~ with
%&c
bandwidth ~ . From 2& ²%³ ~
H
²b%c&³
²b&c%³
c
&c %&
we have
& %&b
%&b
2& ²³ ~ for any &  b ~ (for & ~ Á and , in this case), and 2& ²³ ~ for any &  c ~ (for & ~ Á and ). .
V ²³ ~ ²À³´2 ²³ b 2 ²³ b 2 ²³ b 2 ²³ b 2 ²³ b 2
²³ b 2 ²³µ .
2 ²³ ~ c
²bc³
²³
~ À , since & ~  % ~  b ~ & b .
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PRACTICE EXAM 11
34. continued
²bc³
~ À ,
²³
²bc
³
~ À , since
²³
²bc³
~ À .
²³
2 ²³ ~ c
2
²³ ~
& c ~ c  % ~  ~ & , and
2 ²³ ~
V ²³ ~ ²À³´ b b b À b À b À b Àµ ~ À
-
35. The prior parameter has distribution ~ F
prob. .
2 prob. The model distribution : has a compound distribution with Poisson frequency with mean , and the
stated severity distribution.
7 ² ~ O: ~ ³ ~
7 ²: ~q~³
7 ²: ~³
7 ²: ~ O ~ ³ ~ 7 ²1 claim for amount 2O ~ ³ b 7 ²2 claims for amount 1 eachO ~ ³
c
c
~ c h b h h ~ .
c
c
7 ²: ~ q ~ ³ ~ 7 ²: ~ O ~ ³ h 7 ² ~ ³ ~ h ~ À
7 ²: ~ O ~ ³ ~ 7 ²1 claim for amount 2O ~ ³ b 7 ²2 claims for amount 1 eachO ~ ³
c c
~ c h h b h h h ~ .
c
c
7 ²: ~ q ~ ³ ~ 7 ²: ~ O ~ ³ h 7 ² ~ ³ ~ h ~ À
c
c
7 ²: ~ ³ ~ 7 ²: ~ q ~ ³ b 7 ²: ~ q ~ ³ ~ b .
7 ² ~ O: ~ ³ ~
7 ²: ~q~³
7 ²: ~³
c
c
c
~ ² ³,² b ³ ~ À . Answer: D
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 12
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ACTEX EXAM C/4 - PRACTICE EXAM 12
1. A customer service operator accepts calls continuously throughout the work day. The length of
each call is exponentially distributed with an average of 3 minutes. Calculate the probability that
at least one call will be completed in the next 2 minutes.
A) Less than 0.50
B) At least 0.50, but less than 0.55
C) At least 0.55, but less than 0.60
D) At least 0.60, but less than 0.65
E) At least 0.65
2. An insurance agent gets a bonus based on the underlying losses, 3, from his book of business.
3 follows a Pareto distribution with parameters ~ and ~ Á . His bonus, ) , is
calculated as ²
Á c 3³° if this quantity is positive and - otherwise. Calculate his
expected bonus.
A) Less than 100,000
B) At least 100,000, but less than 120,000
C) At least 120,000, but less than 140,000
D) At least 140,000, but less than 160,000
E) At least 160,000
3. An insurer has a portfolio of 1000 independent policies. Each policy has a probability of .01
of making a claim, and the claim size is uniformly distributed between 5,000 and 25,000. The
insurer purchases reinsurance to cover any amount above 15,000 for individual claims. The
reinsurer charges a relative security loading of to ensure (using the normal approximation) a
95% probability that reinsurance premium collected will exceed reinsurance claim. The ceding
insurer also charges a relative security loading (before reinsurance is taken into account) of .
What is the ceding insurer's expected gain on the portfolio of policies? (7 ´A  À
µ ~ À
for the standard normal random variable A .) The relative security loading is the proportion that
the premium is above the expected loss.
A) 100,000
B) 102,000
C) 104,000
D) 106,000
E) 108,000
4. A compound Poisson aggregate claim distribution has a discrete claim amount distribution
whose values are only positive integers. It so happens : has a geometric distribution with
parameter ~ , so that 7 ´: ~ µ ~ ²À³²À³ . What is 7 ´?  ,´?µµ ?
A) .62
B) .67
C) .72
D) .77
E) .82
5. A mortality study of 50 impaired lives has no censored data. It is found that in the first four
years, there is a death at time 1, 2 deaths at time 2.5 and a death at time 4.
The cumulative hazard function at time 4 is estimated two ways:
A: based on the Product-Limit estimate of :²³
B: Nelson-Aalen estimate.
estimate A
Find the ratio estimate B À
A) À
B) À
C) À
D) À
E) À
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PRACTICE EXAM 12
6. A random sample of size 4 has the following values 1 , 2 , 2 , 3 .
Kernel smoothing is applied to the data using a Pareto kernel. For sample point &, the kernel
density function is Pareto with ~ and ~ &. Find the kernel smoothed estimate of :²³.
A) Less than .200
B) At least .2 but less than .225
C) At least .225 but less than .250
D) At least .250 but less than .275
E) At least .275
7. As a result of a hard disk crash, the individual claim amounts were lost for a set of 1000
claims. The only surviving claim information shows that 800 claims are in the interval ²Á µ
and 200 claims are in the interval ²Á B³ . There is some surviving analysis of the data set.
The surviving analysis shows that an exponential distribution with mean was fitted to the data
using the method of moments, resulting in an estimate of of 760. The surviving analysis also
shows that an exponential distribution with mean Z was fitted to cost per payment after a
deductible of 1000 was applied to the claim amounts; the method of moments was used in this
estimation also, and resulted in an estimate of Z of 1000. Determine the average claim amount
for the 800 claims that are in the interval ²Á µ .
A) 400
B) 425
C) 450
D) 475
E) 500
8. You are given:
(i) Losses follow an exponential distribution with mean .
(ii) A random sample of 25 losses is distributed as follows:
Loss Range
Frequency
´ Á µ
´² Á µ
² Á B ³
Calculate the maximum likelihood estimate of .
A) Less than 1400
B) At least 1400, but less than 1450
C) At least 1450, but less than 1500
D) At least 1500, but less than 1550
E) At least 1550
9. The inflation rate for the coming year is assumed to have a distribution that is uniform on the
interval ²ÀÁ À³ . Employee bonuses are paid at the end of the year. If there will be no
inflation in the coming year, the per employee bonus to be paid at the end of the coming year is
assumed to have a uniform distribution on the interval ´Á µ . The inflation factor for the
coming year will be applied to determine the actual bonus per employee at year end. It is
assumed that the baseline distribution of bonus (in the absence of inflation) is independent of the
inflation rate for the coming year. Find the standard deviation of the after inflation bonus to be
paid at the end of the coming year.
A) 290
B) 293
C) 296
D) 299
E) 302
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10. The prior distribution of the parameter is a gamma distribution with parameters ~ and
~ . The model distribution of ? given is Poisson with a mean of . A sample value of ? is
observed, say ? , based on a random value of from the prior distribution. Another sample
value is then obtained, say ? , based on the same (unknown) value of . Find the probability that
? is at least 1 given that ? is at least 1.
A) .90
B) .92
C) .94
D) .96
11. A portfolio of insurance policies consists of two types of policies. The annual aggregate loss
distribution for each type of policy is a compound Poisson distribution. Policies of Type I have a
Poisson parameter of 1 and policies of Type 2 have a Poisson parameter of 2. For both policy
types, the claim size (severity) distribution is uniformly distributed on the integers 1, 2 and 3.
Half of the policies are of Type I and half are of Type II. A policy is chosen at random and an
aggregate annual claim of 2 is observed. Find the Bayesian premium for the same policy for next
year.
A) Less than 2.0
B) At least 2.0 but less than 2.2
C) At least 2.2 but less than 2.4
D) At least 2.4 but less than 2.6
E) At least 2.6
12. Type A risks have each year's losses uniformly distributed on the interval ²Á ³ .
Type B risks have each year's losses uniformly distributed on the interval ²Á ³ .
A risk is selected at random, with each type being equally likely. The first year's losses equal 3.
Find the Buhlmann credibility premium for the second year's losses in terms of 3.
B) À3 b À
A) À3 b À
C) À3 b À
D) À3 b À
E) À3 b À
13. An insurer has data on losses for four policyholders for 7 years. The loss from the -th
policyholder for year is ? . You are given:
c c
c ²? c ?
²?
³ ~ À
c ?³ ~ À
~ ~
~
Using nonparametric empirical Bayes estimation, calculate the Bühlmann credibility factor for an
individual policyholder.
A) Less than 0.74
B) At least 0.74, but less than 0.77
C) At least 0.77, but less than 0.80
D) At least 0.80, but less than 0.83
E) At least 0.83
14. An actuary is simulating annual aggregate loss for a product liability policy, where claims
occur according to a binomial distribution with parameters ~ and ~ À, and severity is
given by an exponential distribution with parameter ~ Á . The number of claims is
simulated using the inverse transform method (where small random numbers correspond to small
claim numbers) and a random value of 0.58 from the uniform distribution on ´Á µ. The claim
severities are simulated using the inverse transform method (where small random numbers
correspond to small claim sizes) using the following values from the uniform distribution on
´Á µ: 0.35 , 0.70 , 0.61 , 0.20 . Calculate the simulated annual aggregate loss for the product
liability policy.
A) Less than 250,000
B) At least 250,000 but less than 500,000
C) At least 500,000 but less than 750,000
D) At least 750,000 but less than 1,000,000
E) At least 1,000,000
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15. ? has an exponential distribution with a mean of 2. Find *; ,À , the Conditional Tail
Expectation at confidence level .8 for ? (rounded to nearest integer).
A) 2
B) 3
C) 4
D) 5
E) 6
16. You are given the following information about a study of individual claims:
(i) 20! percentile = 18.25
(ii) 80! percentile = 35.80
Parameters and of a lognormal distribution are estimated using percentile matching.
Determine the probability that a claim is greater than 30 using the fitted lognormal distribution.
A) 0.34
B) 0.36
C) 0.38
D) 0.40
E) 0.42
17. In analyzing the behavior of the Loss Elimination Ratio for a particular loss distribution, it is
found that at deductible level , the derivative of the LER with respect to is
3,9²³ ~ h ² b ³ , for all  .
Find the mean excess loss for this loss distribution if there is deductible of 1000.
A) 500
B) 1000
C) 1500
D) 2000
E) 2500
18. The following information is available for a collective risk model:
• ? is a random variable representing the size of each loss.
• ? follows a Gamma distribution with ~ and ~ • 5 is a random variable representing the number of claims.
• : is a random variable representing aggregate losses.
• : ~ ? b Ä b ?5
Calculate the mode of : when 5 ~ .
A) Less than 950
B) At least 950, but less than 1050
C) At least 1050, but less than 1150
D) At least 1150, but less than 1250
E) At least 1250
19. The Loss Elimination Ratio for a Pareto loss distribution ? with ~ and ordinary
deductible 100 is .578125. Find the mean of ? .
A) 50
B) 100
C) 150
D) 200
E) 250
20. Aggregate claims per period follow a compound Poisson distribution with claim severity ?
uniformly distributed on the interval [0,100] and with expected number of claims per period
being 10. The insurer is considering putting a deductible of amount on each individual loss,
and determining a modified frequency distribution which counts only the number of losses per
period that are above the deductible. What value of would result in an expected modified
frequency of 5 losses above the deductible per period?
A) j
B) C) D) E) c j
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21. For a stop-loss insurance on a three person group:
(i) Loss amounts are independent.
(ii) The distribution of loss amount for each person is:
Loss Amount Probability
0
0.4
1
0.3
2
0.2
3
0.1
(iii) The stop-loss insurance has a deductible of 1.5 for the group.
Calculate the expected stop-loss payment.
A) Less than 1.5
B) At least 1.5 but less than 1.6
C) At least 1.6 but less than 1.7
D) At least 1.7 but less than 1.8
E) At least 1.8
22. You are given the following claim data for automobile policies:
250 305 345 370 410 470 490 540 550 570 1070
Calculate the smoothed empirical estimate of the 45-th percentile.
A) 408
B) 421
C) 434
D) 440
E) 446
23. In a mortality study for the year of age ²%Á % b µ , 100 individuals are under observation at
age %. 10 deaths occur before age % b À, and 15 individuals are right-censored at age % b À .
3 more deaths occur between ages % b À and % b .
Determine the Product-Limit estimate of % , the one year mortality probability for someone
whose age is %.
A) .106
B) .116
C) .126
D) .136
E) .146
24. ? is a mixture of 3 exponential random variables, with means 1, 2 and 3.
A random sample of observed values of ? has sample mean 2. and sample median 1..
A combination of the method of moments and the method percentile matching is applied to
estimate the mixing weights. Find the variance of the estimated distribution.
A) Less than 6.8
B) At least 6.8 but less than 7.0
C) At least 7.0 but less than 7.2
D) At least 7.2 but less than 7.4
E) At least 7.4
25. You are given the following random sample of claim frequencies that occurred during a 100
day period: ~ Á ~ Á ~ Á ~ Á ~ .
is the number of days that had claims.
If the number of claims per day follows a negative binomial distribution with ~ , find the
estimated variance of the distribution using maximum likelihood estimation.
A) 3.0
B) 2.8
C) 2.6
D) 2.4
E) 2.2
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PRACTICE EXAM 12
26. ? is uniformly distributed on the interval ² Á ³ where  . ? Á ÀÀÀÁ ? is a random
sample from the distribution of ? . Find the bias in the maximum likelihood estimator of .
A) c B) c b
C) c b
D) c b
E) c b
27. You are given:
(i) A sample of claim payments is:
44
79
105
1150 1197
(ii) Claim sizes are assumed to follow an exponential distribution.
(iii) The mean of the exponential distribution is estimated using the method of moments.
Calculate the value of the Kolmogorov-Smirnov test statistic.
A) Less than .15
B) At least .15 but less than .20
C) At least .20 but less than .25
D) At least .25 but less than .30
E) At least .30
28. An insurer has two separate classes of policies. The characteristics of the loss per insured in
each of the two classes during a one year period are as follows:
Class I: Expected claim per insured is 100. To be within 5% of expected loss 90% of the time,
the standard for number of insureds needed for full credibility is 1082.4 .
Class II: Expected claim per insured is 200. To be within 5% of expected loss 90% of the time,
the standard for number of insureds needed for full credibility is 1082.4 .
Class I has twice the number of insureds as Class II. The two classes of insureds are combined
and regarded as a single class with the appropriate adjusted loss per insured during a one year
period. Find the full credibility standard for the minimum number of insureds required in the
combined portfolio, where the full credibility is to be within 5% of expected loss 90% of the time.
A) 1082.4
B) 1190.7
C) 1244.8
D) 1298.9
E) 1353.0
29. The number of claims in one exposure period follows a Bernoulli distribution with mean .
The prior density function of is assumed to be ²³ ~ Á   .
The claims experience is observed for one exposure period and no claims are observed.
Determine the posterior density function of .
Hint: ~ and ~ ² c ³ .
A) Á   D) Á   Á


²c³
E) ²c³ Á  B)
C)
²c³
Á   
30. Aggregate claims for an exposure period for an individual in a risk class with parameter
# ~ has a binomial distribution with parameters ~ and ~ . The parameter # has pdf
²³ ~ ² c ³ for   . A portfolio of insurance policies has 10 individuals in each
risk class in 1997, 12 individuals in each risk class in 1998 and 15 individuals in each risk class in
1999. An individual from a randomly chosen risk class is observed for the three years, and it is
found that aggregate claims for the three consecutive years are 18 (1997), 20 (1998) , 27 (1999).
Using the Buhlmann-Straub model find the credibility premium in 2000 for aggregate claims for
this risk class, if there are 20 individuals in the risk class in 2000.
A) 33
B) 34
C) 35
D) 36
E) 37
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PRACTICE EXAM 12
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31. Using the Inverse Transform Method, a Binomial (10 , 0.20) random variable is generated,
with 0.65 from U(0,1) as the initial random number. Determine the simulated result.
A) 0
B) 1
C) 2
D) 3
E) 4
32. An actuary uses the following algorithm, where < is a random number generated from the
uniform distribution on ´Á µ, to simulate a random variable ? :
(1) If <  À, set ? ~ , then stop.
(2) If <  À
, set ? ~ , then stop.
(1) If <  À, set ? ~ , then stop.
(2) Otherwise, set ? ~ , then stop.
What are the probabilities for ? ~ Á Á Á respectively?
A) 0.40 , 0.25 , 0.15 , 0.20
B) 0.25 , 0.40 , 0.20 , 0.15
C) 0.15 , 0.25 , 0.20 , 0.40
D) 0.15 , 0.20 , 0.40 , 0.25
E) 0.20 , 0.25 , 0.25 , 0.40
33. The number of claims occurring in one week, 5 , has a distribution with the following
probability function: 7 ²5 ~ ³ ~ c b for ~ Á Á Á ÀÀÀÀÀ .
The parameter is a random variable with pdf ²³ ~ ² c b ³ for   .
Find the coefficient of variation of 5 .
j
A) B) C)
D) j
E) j
j
j
34. Phil and Sylvia are competitors in the light bulb business. Sylvia advertises that her light
bulbs burn twice as long as Phil’s. You were able to test 20 of Phil’s bulbs and 10 of Sylvia’s.
You assumed that the distribution of the lifetime (in hours) of a light bulb is exponential, and
separately estimated Phil’s parameter as V7 ~ 1000 and Sylvia’s parameter as V: ~ 1500 using
maximum likelihood estimation. Determine i , the maximum likelihood estimate of 7 restricted
by Sylvia’s claim that : ~ 27 .
A) Less than 900
B) At least 900, but less than 950
C) At least 950, but less than 1000
D) At least 1000, but less than 1050
E) At least 1050
35. You are given:
(i) The parameter A has an inverse gamma distribution with probability density function:
() ~ 500c4 c10/ ,  0
(ii) The size of a claim has an exponential distribution with probability density function:
(%|\$ ~ ) ~ c1 c%/ , %  0,  0
For a single insured, two claims were observed that totaled 50.
Determine the expected value of the next claim from the same insured.
A) 5
B) 12
C) 15
D) 20
E) 25
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 12
ACTEX EXAM C/4 - PRACTICE EXAM 12 SOLUTIONS
1. Since calls are accepted continuously, we assume that the operator starts a call as soon as the
previous one is completed. From the lack of memory property of the exponential distribution, it is
irrelevant how long the current call has taken, the remaining time until the call is completed is
exponential with a mean of 3 minutes. 7 ²;  ³ ~ -; ²³ ~ c c° ~ À . Answer: A
2. The bonus payment is ) ~ ²
Á c 3³ if 3  Á and 0 otherwise.
This is the same as ) ~ ´
Á c ²3 w Á ³µ , and the expected bonus is
,´3µ ~ ´
Á c ,²3 w Á ³µ .
For the given Pareto random variable 3,
Á
Á
,²3 w Á ³ ~ c h ´ c ² Áb
Á ³c µ ~ Á .
The expected bonus is ´
Á c Á µ ~ Á .
3. On an individual claim, say ? , to the ceding insurer, the amount paid by the reinsurer has
distribution of ? Z , where 7 ´? Z ~ µ ~ À and ²%³ ~ À for  %  Á . On an
individual claim to the ceding insurer, the expected claim on the reinsurer is ,´? Z µ ~ ~ ,
with variance ~ Á Á . Since the probability of an individual policy making a claim
is .01, the per policy claim random variable on the reinsurer has a mean of ²À³ ~ and a
variance of ²À³²À³ b ~ Á . With 1000 independent policies, the aggregate
claim on the reinsurer has a mean of Á and a variance of Á Á . In order to ensure
a 95% probability that premium will exceed claims (using the normal approximation), the
reinsurer must charge a premium of Á b À
h j
Á Á ~ Á . This results
Á
in a relative security loading of ~ Á c ~ À .
The ceding insurer has expected claims, before reinsurance, of
h ²À³²Á ³ ~ Á and so charges a premium of Á h ²À³ ~ Á .
After paying the reinsurance premium, the ceding insurer's expected gain is
Á c Á c ²Á c Á ³ ~ Á .
4. ²³ ~ À ~ c ¦ ~ À. Since ,´:µ ~ ~ ~ h ,´?µ, we have ,´?µ ~ À.
B
Using the recursive formula, : ²%³ ~ % &? ²&³ h : ²% c &³ , % ~ Á Á Á ÀÀÀ
&~
²À³²À³ ~ À ~ : ²³ ~ h ? ²³ h : ²³ ~ À h ? ²³ ¦ ? ²³ ~ À ,
²À³²À³ ~ : ²³ ~ h ´? ²³: ²³ b ? ²³: ²³µ ¦ ? ²³ ~ À
,
²À³²À³ ~ : ²³ ~ h ´? ²³: ²³ b ? ²³: ²³ b ? ²³: ²³µ ¦ ? ²³ ~ À
.
7 ´?  Àµ ~ ? ²³ b ? ²³ b ? ²³ ~ À
.
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5. Estimate A is c : ²³ , where : ²³ is the Product-Limit estimate
: ²³ ~ ~ À (the Product-Limit estimate is ² c ³² c ³² c ³ ~ ,
when there is no censoring, the Product-Limit estimate at time ! will be the fraction of those still
alive at time !). Estimate A is c À ~ À .
V
V
The Nelson-Aalen estimate is /²!³
~ , so that /²³
~ b b ~ À
! !
The ratio is
À
À
~ À
.
6. The Pareto survival function is :²%³ ~ ² %b
³ .
With the given points, there is empirical probability of at & ~ and , and empirical
probability of
at & ~ . With ~ and ~ &, the kernel survival functions are:
for & ~ , : ²%³ ~ ² %b
³ , for & ~ , : ²%³ ~ ² %b
³ , and for & ~ , : ²%³ ~ ² %b
³ .
The kernel smoothed estimate of :²%³ is
V
:²%³
~ : ²%³ b : ²%³b : ²%³ ~ ² %b
³ b ² %b
³ b ² %b
³ .
V
Then :²³
~ ² b
³ b ² b
³ b ² b
³ ~ À . Answer: C
7. Denote by ; the total of the 800 claims in ²Á µ , and denote by ; the total of the 200
claims in the interval ²Á B³ . Then the total for all 1000 claims is ; b ; , and
b;
the average (sample mean) is ;
. According to the method of moments for a one parameter
distribution, the sample mean is set equal to the distribution mean. We are given that is the
method of moments estimate of the exponential distribution parameter. The exponential
b;
distribution parameter is , the mean. Thus, ;
~ .
After applying the deductible of 1000 to the 200 claims in the interval ²Á B³ , the total
amount paid (after deductible) is ; c Á , and the average cost per payment made is
; cÁ
; cÁ
; we are given that this is equal to 1000, so that ~ . It follows
that ; ~ Á . Then, from ; b ; ~ Á , we get ; ~ Á . The average
Á
amount per claim for the 800 claims in the interval ²Á µ is ~ . Answer: C
8. Since the data is in interval grouped form, the likelihood function is
3 ~ ´- ²³ c - ²³µ h ´- ²³ c - ²³µ h ´- ²B³ c - ²³µ
~ ´ c c° µ h ´c° c c° µ h ´c° µ .
If we let & ~ c° , then 3 ~ ² c &³ ²& c & ³ ²& ³ ,
which can be factored into 3 ~ ² c &³ ² b &³ & .
Since   B , the range for & is  &  .
3 ~ ² c &³ b ² b &³ b & , and
3
c
& ~ c& b b& b & .
Setting this equal to 0 results in
c²b&³&b²c&³&b²c&³²b&³
²c&³²b&³&
~.
Setting the numerator equal to 0 results in the quadratic equation c & c & b ~ .
The solutions are & ~ À and c À. We reject the negative root for & since  &  .
Then from & ~ ~° ~ À , we get ~ .
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PRACTICE EXAM 12
9. The bonus is ) ~ ² b 9³ h ? , where 9 is the coming year's inflation rate, and ? is the
bonus in the absence of inflation. = ´)µ ~ ,´) µ c ²,´)µ³ .
Because of independence,
,´)µ ~ ,´² b 9³?µ ~ ² b ,´9µ³ h ,´?µ ~ ²À³²³ ~ À , and
,´) µ ~ ,´² b 9³ ? µ ~ ,´² b 9³ µ,´? µ
~ ²À
³²Á Á ³ ~ Á Á À
(,´² b 9³ µ ~ À ² b ³ ² À
³ ~ À
Á ,´? µ ~ % ² ³ % ~ Á Á ).
Then, = ´)µ ~ Á Á c ²À³ ~ Á , and the standard deviation is
j= ´)µ ~ jÁ ~ À .
10. The unconditional distribution of ? is negative binomial with ~ and ~ .
7 ´²? ³q²? ³µ
.
7 ´? µ
c 7 ´? ~ µ ~ c ² b ³
7 ´?  O?  µ ~
~ c ² b
³ ~ c ²³ ~ .
B
7 ´²?  ³ q ²?  ³µ ~ 7 ´²?  ³ q ²?  ³Oµ h ²³ B
~ 7 ´?  µ h 7 ´?  Oµ h ²³ 7 ´?  µ ~
c c°
c°
B
B
~ ² c c ³² c c ³ !²³ ~ ² c c ³² c c ³ !
²³ B
~ h ²c° c c² ³ b c² ³ ³ ~ h ´ ² ³ c ² ³ b ² ³ µ ~ c b ~ À
7 ´?  O?  µ ~
°
°
~ À .
11. We first find the posterior probabilities for the parameter :
7 ² ~ O: ~ ³ ~
7 ²: ~q~³
7 ²: ~³
7 ²: ~ O ~ ³ ~ 7 ²1 claim for amount 2O ~ ³ b 7 ²2 claims for amount 1 eachO ~ ³
~ c h
b
c
h
h
c
~ .
c
c
7 ²: ~ q ~ ³ ~ 7 ²: ~ O ~ ³ h 7 ² ~ ³ ~ h ~ À
7 ²: ~ O ~ ³ ~ 7 ²1 claim for amount 2O ~ ³ b 7 ²2 claims for amount 1 eachO ~ ³
c c
b h h h ~ .
c
7 ²: ~ q ~ ³ ~ 7 ²: ~ O ~ ³ h 7 ² ~ ³ ~ h
~ c h h
c
~ À
c
c
7 ²: ~ ³ ~ 7 ²: ~ q ~ ³ b 7 ²: ~ q ~ ³ ~ b .
c
c
c
7 ²: ~q~³
7 ² ~ O: ~ ³ ~ 7²: ~³ ~ ² ³,² b ³ ~ À
and 7 ² ~ O: ~ ³ ~ c 7 ² ~ O: ~ ³ ~ À
.
,´: O: ~ µ ~ ,´: O ~ µ h 7 ² ~ O: ~ ³ b ,´: O ~ µ h 7 ² ~ O: ~ ³
~ ²³²³²À³ b ²³²³²À
³ ~ À
.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 12
PE-235
12. Prior distribution is 7 ²(³ ~ 7 ²)³ ~ À
Hypothetical means are ²(³ ~ ,´?O(µ ~ À Á ²)³ ~ ,´?O)µ ~ .
Process variances are #²(³ ~ = ´?O(µ ~ Á #²)³ ~ ~ .
~ ,´?µ ~ expected hypothetical mean ~ ²À³² ³ b ²³² ³ ~ .
# ~ expected process variance ~ ² ³² ³ b ² ³² ³ ~ .
~ variance of hypothetical mean ~ ² c À³ ² ³² ³ ~ .
A ~ b
# ~
° ~ À .
b °
A3 b ² c A³ ~ À3 b À
² ³ ~ À3 b À
.
13. Under the nonparametric empirical Bayes method applied to the Buhlmann credibility model,
c ²? c ?
the estimated expected process variance is V# ~
³ ,
²c³
~ ~
where is the number of policyholders and is the number of exposures per policyholder,
c
c
and the estimated variance of the hypothetical means: V
~ c
²? c ? ³ c V# .
V ~
The estimated credibility factor for each policyholder is A
~
b VV#
.
In this problem we have ~ policyholders and ~ exposure periods (years) per
policyholder. From the given values we get V# ~ ²³²c³
h ²À
³ ~ À and
V
~ c
h ²À³ c À
V
~ À . Then, A ~
À
b À
~ À .
14. The distribution of the binomial with ~ and ~ À is
5
À
À
À
À
À
À
À
À
À
A uniform value of .58 is between .3125 and .6875 so the inverse transform method results in a
simulated number of claims of 2 .
The cdf of the exponential distribution is - ²%³ ~ c c%°Á . Since small random numbers
correspond to small claim amounts, for a uniform random number ", we solve for % from the
relationship " ~ - ²%³ to get the simulated value of %.
The simulated value of the first claim % is the solution of À ~ c c% °Á , so that
% ~ Á . The simulated value of the second claim is % is the solution of
À ~ c c% °Á , so that % ~ Á .
The simulated aggregate loss is Á b Á ~ Á .
15. *; ,À ~ ,´?O?  8À µ . By the lack of memory property of the exponential distribution,
,´?O?  8À µ ~ ,´?µ ~ . Answer: A
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 12
%c
16. If ? has a lognormal distribution with parameters and , then - ²%³ ~ )² ³ .
The method of percentile matching sets - ²À³ ~ À , and - ²À³ ~ À .
À c c
Therefore, )²
³ ~ À , and )² À
³ ~ À . From the standard normal table, we
get that
we get
Àc
Àc
~ c À , and
~ À . Dividing the second equation by the first
Àc
Àc ~ c , from which we get ~ À and then ~ À . We are asked to
find 7 ´?  µ ~ c - ²³ . This is
À
7 ´?  µ ~ c - ²³ ~ c )² c
³ ~ c )²À³ ~ c À
~ À .
À
,´?wµ
,´?µ
17.
~
%h% ²%³ %bh´c-? ²³µ
,´?µ
.
Using the product rule for differentiation we get
c-? ²³
,´?µ À
,´?µc,´?wµ
 µ ~
c- ²³
3,9²³ ~ ,´?µ
h ´ h ? ²³ b c -? ²³ c h ? ²³µ ~
The mean excess loss with a deductible of is ,´? c O?
If ~ , then 3,9²³c
c-? ²³
,´?µ
.
~ ,´?µ
À From the given derivative, substituting in
~ we get ,´?µ
~ , so that ,´?µ ~ . Then, substituting ~ into the
given derivative formula, we have
~
~
c-? ²³
c-? ²³
~
Á from which we get -? ²³ ~ À .
,´?µ
c²³
c²³
?
?
We also have 3,9²³ ~ ,´?µ
~ ~ h ² b
³ , so that
-? ²³ ~ c ² b ³ . This is a Pareto distribution with ~ and ~ . Once the
h ² b
³ ~
identification of the Pareto distribution is made, the limited expected value ,´? w µ can be
found from the Exam C Table of Distributions, from the relationship
c
,´? w µ ~ c
h ´ c ² b
µ , so that
³
c
,´? w µ ~ µ ~ .
c h ´ c ² b ³
Alternatively, if it had not been noticed that the distribution was Pareto, once -? ²³ is known, it
is possible to find ,´? w µ from ,´? w µ ~ ´ c -? ²%³µ % .
The mean excess loss for this loss distribution with a deductible of 1000 is
,´? c O?  µ ~
,´?µc,´?wµ
c-? ²³
~ c
18. If 5 ~ , : ~ ? b ? b ? b ? b ? , where each ? has a Gamma distribution with
~ and ~ . Since the ? 's are independent, : has a Gamma distribution with ~ and ~ . This is true because, just as the sum of independent exponential distributions
with common mean has a Gamma distribution with ~ and , the sum of independent
Gamma random variables with common is a Gamma in which the 's are added.
c °
The pdf of : is ² ³ ~ h[ . The mode occurs where ² ³ is maximized.
We can find this be differentiating ² ³, setting Z ² ³ equal to and solving for :
cÀ
Z ² ³ ~ h[ h ´ cÀ c À cÀ µ ~ h[ h ´ c À µ ~ .
Solving for results in the mode of ~ .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 12
PE-237
c
19. Since ? is Pareto, ,´?µ ~ c
and ,´? w µ ~ c
h ´ c ² b
µ.
³
The loss elimination ration for a deductible of 100 is
,´?wµ
c
~ c ² b
~ À . Then, ² b
³
³ ~ À ,
,´?µ
and b
~ À , so that ~ . Finally, ,´?µ ~ c ~ À Answer:
B
20. With deductible , the probability of a loss being above the deductible is
c
i
~ c À . If 5 is Poisson with parameter , then 5 is Poisson with
parameter ~ ² c À³ ~ S ~ .
21. We use the formulation ,´²: c À³b µ ~ ,´:µ c ,´: w Àµ ,
7 ²: ~ ³
where : w À ~ H 7 ²: ~ ³ . .
À 7 ²:  ³
,´:µ ~ ,´? µ b ,´? µ b ,´? µ ~ h ,´?µ ~ h ´²³²À³ b ²³²À³ b ²³²À³µ ~ À
7 ´: ~ µ ~ 7 ´²? ~ ³ q ²? ~ ³ q ²? ~ ³µ ,
and by independence of the ? 's this becomes
7 ´? ~ µ h 7 ´? ~ µ h 7 ´? ~ µ ~ ²À³ ~ À
.
7 ´: ~ µ ~ 7 ´²? ~ ³ q ²? ~ ³ q ²? ~ ³µ b 7 ´²? ~ ³ q ²? ~ ³ q ²? ~ ³µ
b 7 ´²? ~ ³ q ²? ~ ³ q ²? ~ ³µ ~ d ²À³ d ²À³ ~ À .
Then ,´: w Àµ ~ d 7 ²: ~ ³ b À d 7 ²:  ³
~ À b À d ² c À
c À³ ~ À
Then, ,´²: c À³b µ ~ ,´:µ c ,´: w Àµ ~ c À ~ À
.
22. Since there are 11 data points, we assign percentiles of , , . . . , , to
the successive data points. We must find the interval to b
which contains .45 .
We see that .45 lies in the interval À
~  À  ~ À , between the 5-th and
6-th data values. The 5-th and 6-th data values are 360 and 420 . The smoothed empirical
estimate is found by linear interpolation between 360 and 420 . The calculation can be slightly
simplified interpolating between ² ³  ²À³  ² ³ , i.e.,  À  .
This means that the smoothed empirical estimate of the 45-th percentile is the interpolated value
that is 40% of the way between the 5-th and 6-th sample values. This will be
b ²À³² c ³ ~ .
23. The Product-Limit estimate of survival for the year is ´ c µ´ c µ ~ À
.
The exposure is 100 for the first half-year (the censorings at time .5 are assumed to occur after the
deaths), and the exposure in the second half-year is c c ~ .
V % ~ c À
~ À
À
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PRACTICE EXAM 12
24. ²%³ ~ c% b h c%° b ² c c ³ h c%° .
- ²%³ ~ ² c c% ³ b ² c c%° ³ b ² c c ³² c c%° ³ À
,´?µ ~ b b ² c c ³ ~ c c ~ À
S b ~ À
.
- ²À³ ~ ² c cÀ ³ b ² c cÀ° ³ b ² c c ³² c cÀ° ³ ~ À
S À b À
~ À
Solving for and results in ~ À
Á ~ À
Á c c ~ À
À
The second moment of ? is
,´? µ ~ h b h b ² c c ³ h ~ À ,
and the variance of ? is À c ²À³ ~ À .
V ~c
25. Since is known, the mle equation for is % , so that ~ À .
V ² b V ³ ~ À . Answer: B
The variance of the estimated distribution is 26. The pdf of ? is ²%³ ~ for  %  , so the likelihood function is
3²³ ~ ²% ³ ~ . 3²³ will be maximized when is minimized.
~
?
?
For each % it must be true that   À The smallest possible value of that is consistent
with these inequalities for all % sample values is
V ~ 4 %² ? Á ? Á À À À Á ? ³ ~ h 4 %²? Á À À À Á ? ³ . This is the mle.
The bias in the estimator V is ,´ V µ c .
To find ,´ V µ we first find the cdf of V, -V ²!³ ~ 7 ´ V  !µ .
This is 7 ´ V  !µ ~ 7 ´ h 4 %²? Á À À À Á ? ³  !µ ~ 7 ´4 %²? Á À À À Á ? ³  !µ
~ 7 ´²?  !³ q ²?  !³ q Ä q ²?  !³µ ~ ´7 ²?  !³µ
(since the ? 's are mutually independent).
For !  , we have !  , so that 7 ²?  !³ ~ ,
and for !  , we have !  , so that 7 ²?  !³ ~ .
For  !  we have  !  , so that 7 ²?  !³ ~ !c
(? has a uniform distribution from to ).
Therefore, for  !  we have -V ²!³ ~ 7 ´ V  !µ ~ ´7 ²?  !³µ ~ ´ !c
µ .
Since  , the ? 's are all  , and V  . Therefore,
B
°
B
,´ V µ ~ ´ c -V ²!³µ ! ~ ´ c µ ! b ° ² c ´ !c
µ ³ ! b ² c ³ !
²!c³
~ ! c ° ´ !c
µ ! ~ c < ²b³
Then the bias in the estimator V is ,´ V µ c ~
b !~
c
~ ² b
= ~ c ²b³
b ³ .
!~°
b
² b ³ c ~ c b
.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 12
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27. According to the method of moments, the exponential parameter of the estimated
distributions is equal to the sample mean, V ~ .
The Kolmogorov-Smirnov statistic is found by first calculating the empirical cdf for ~ observations. The model cdf is - i ²% ³ ~ c c%° .
%
- ²%c
- ²%b
- i ²% ³
Maximum absolute difference
³
³
À
À
À
À
À
À
À
À
À
À
À
À
À
À
The maximum absolute difference is the K-S statistic. This is . .
À
À
À
À
À
0
28. Class I: À ~ ² À
À ³ h ²³ S 0 ~ Á ~ ,´?0 µ c ²³
S ,´?0 µ ~ Á .
00
Class II: À ~ ² À
À ³ h ²³ S 00 ~ Á ~ ,´?00 µ c ²³
S ,´?00
µ ~ Á .
For the combined portfolio of policies, the loss per insured is ? , is a mixture of ?0 and ?00 ,
with and weighting applied, respectively. Then ,´?µ ~ ² ³²³ b ² ³²³ ~ and
,´? µ ~ ² ³²Á ³ b ² ³²Á ³ ~ Á , so that
= ´?µ ~ Á c ² ³ ~
Á
.
= ´?µ
Á°
À
The full credibility standard for ? is ² À
À ³ h ²,´?µ³ ~ ² À ³ h ²°³ ~ .
?O# ²%O³²³
7 ´?~Oµ ²³
29. ²O? ~ ³ ~ #O? ²O%³ ~ ²%O³h²³ ~ ?O#
7 ´?~Oµ ²³ ²c³h ~ ~
h
²c³
²c³h c,´µ
~
²c³h c ²c³
~ ²c³ Á   .
30. Let > denote the claims for an individual in a single year. Then
²³ ~ ,´> O# ~ µ ~ , and #²³ ~ = ´> O# ~ µ ~ ² c ³
(binomial mean and variance with 3 trials).
Then, using the distribution of #, we have
~ ,´?µ ~ ,´²#³µ ~ ,´#µ ~ h ² c ³ ~ ;
# ~ ,´#²#³µ ~ ,´#² c #³µ ~ ² c ³ h ² c ³ ~ ; and
~ = ´²#³µ ~ = ´#µ ~ = ´#µ ~ ´ ,´# µ c ²,´#µ³ µ
~ ´ h ² c ³ c ² ³ µ ~ .
With 10 policyholders per risk class in 1997, the average aggregate claim per policyholder in the
randomly selected risk class is ? ~
>Á b>Á bÄb>Á
~ , ~ .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 12
30. continued
c
Similarly, ? ~ , ~ , and ? ~ , ~ are the observed average aggregate
claim amounts per individual for 1998 and 1999.
Applying the Buhlmann-Straub model, ~ # ~ , and the credibility factor is
A ~ b
~ bb ~ À
. Also,
bbb c
? ~ h ? ~ ² ³² ³ b ² ³² ³ b ² ³² ³ ~ ~ À , so that the credibility
~
premium per individual in the risk class is
c
A? b ² c A³ ~ ²À
³²À³ b ²À³²À³ ~ À À With 20 individuals in the risk class
in 2000, the credibility premium is ²À³ ~ À
À
c
Note that we have a total of 37 exposures ( b b ³, with an ? of bb
~ ~ À,
and we have applied the basic Buhlmann method, without really needing to refer to
Buhlmann-Straub.
31. To apply the Inverse Transform Method we must formulate the cumulative distribution
function. First we find the probability function for this binomial distribution.
The probability function is ~ 7 ²? ~ ³ ~ 2 3 ² c ³c ~ 2 3²À³ ²À³c .
?~
- ²³
À
À
À
À
À
À
À
À
ÀÀÀ
À
À
The simulated value of ? is , where - ² c ³  <  - ²³ .
We see that À ~ - ²³  À
 À
~ - ²³ . Therefore, the simulated value of ? is 2.
32. The probability for ? ~ is .4, the probability for ? ~ is À
c À ~ À,
the probability for ? ~ is À c À
~ À , and the probability for ? ~ is c À ~ À.
33. The conditional distribution of 5 given is geometric, with ~ c , so the conditional
mean and variance of 5 are ,´5 Oµ ~ ~ c and = ´5 Oµ ~ ² b ³ ~ ²c³ .
Then, ,´5 µ ~ ,´ ,´5 Oµ µ ~ , ´ c µ ~ c h ² c b ³ ~ c h ² c ³ ~ ² c ³ ~ , and
b
b
,´5 µ ~ ,´ ,´5 Oµ µ ~ , ´ ²c³ µ ~ ²c³ h ² c ³ ~ ² b ³ ~ À
= ´5 µ ~ c ~ .
The coefficient of variation of 5 is
j= ´5 µ
,´5 µ
~
j
~ j .
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34. The likelihood function for Phil's mle is 37 ²7 ³ ~ c% ° ~ c'% ° ,
~ and the likelihood function for Sylvia's mle is 3: ²: ³ ~ c& °: ~ c'& °S .
S
~ :
The mle of the exponential distribution mean is the sample mean of the data points.
Therefore, '% ~ V7 ~ Á , and '& ~ V: ~ Á .
The combined likelihood function, using the restriction : ~ 7 is
3 ~ c'% ° h ( ³ c'& °7 ~ cÁ° h ( ³ cÁ°7
7
~ h cÁ° .
7
Á
7 , and
Á
. Setting 3 ~ and solving for 7 results in
7 3 ~ c 7 b 7
7
Then, 3 ~ c c 7 c
V7 ~ À .
Note that once the function we wish to maximize is in the form c° , the maximum occurs
at ~ . Thus, when we have 3 ~ h cÁ° , we see that the maximum will occur at
Á
~ ~ À . Answer: B
35. We wish to find ,´? O? b ? ~ µ , where ? is the size of the 3rd claim and ? , ?
are the first two claim sizes. Since the claim size distribution is described as a conditional
distribution given \$, we can find the expectation by conditioning over \$.
B
,´? O? b ? ~ µ ~ ,´? O\$ ~ µ h ²O? b ? ~ ³ .
Since ? has an exponential distribution with mean , we have ,´? O\$ ~ µ ~ .
B
Therefore, ,´? O? b ? ~ µ ~ h ²O? b ? ~ ³ , which is the mean of the
posterior distribution, which has pdf ²O? b ? ~ ³ .
We see that the prior distribution has density () ~ 500c4 c10/ , which is the pdf of an 29
inverse gamma distribution with ~ and ~ .
In this study guide's notes on Bayesian estimation, the combination of the inverse gamma prior
and exponential model distribution was one of the various prior/model combinations considered.
It was seen there that this combination results in a posterior distribution that is also inverse
gamma. If there are sample values available % Á % Á ÀÀÀÁ % , then the posterior distribution has
Z ~ b and Z ~ b '% . This can be seen from the joint distribution of % Á % Á ÀÀÀÁ % and
which has pdf ²% Á % Á ÀÀÀÁ % Á ³ ~ ²% O³ h ²% O³Ä ²% O³ h ²³
c²b'% ³°
~ c% ° h c% ° Ä c% ° h b!²³ c° which is proportional to bb .
The posterior density is proportional to this, and therefore must have an inverse gamma
distribution with Z ~ b and Z ~ b '% .
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35. continued
In this case we have an inverse gamma prior with ~ and ~ , and we are given ~ sample values of % with % b % ~ . Therefore, the posterior distribution is inverse gamma
with Z ~ b ~ and Z ~ b ~ . As noted above, the predictive expectation for
the next claim is equal to the posterior mean, which is the mean of the inverse gamma, which is
Z
Z c ~ c ~ . Notice that we only needed the sum of the %-values, not the individual
values.
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ACTEX EXAM C/4 - PRACTICE EXAM 13
1. Two independent random variables, ? and ? , follow the negative binomial distribution with
parameters ² Á ³ and ² Á ³, respectively. Under which of the following circumstances will
? b ? always be negative binomial?
1. ~ 2. ~ 3. The coefficients of variation ? and ? are equal.
A) 1 only
B) 2 only
C) 3 only
D) 1 and 3 only
E) 2 and 3 only
2. A frailty model has a base age-at-death distribution that follows an exponential distribution
with mean 100, and associated hazard rate function ²%³. The conditional hazard rate for the ageat-death random variable ? for an individual with parameter is ?O ²%O³ ~ ²%³. For a
new-born individual in the frailty model group, the value of is uniformly distributed between .8
and 1.5. Find the probability that a randomly selected new-born from the frailty group will
survive to at least age 80.
A) Less than .1
B) At least .1, but less than .3
C) At least .3, but less than .5
D) At least .5, but less than .7
E) At least .7
Questions 3 and 4 are based on a continuous loss random variable ? is uniformly distributed on
the interval ²Á ³.
3. If a loss is over 25 but less than 75, an insurance policy pays 60% of the loss amount that is
over 25. If a loss is over 75, the insurance pays the full loss amount over 25. Find the expected
cost per payment for this insurance.
A) Less than 25
B) At least 25, but less than 28
C) At least 28, but less than 31
D) At least 31, but less than 34
E) At least 34
4. If ?  , a risk manager is paid a bonus equal to 75% of the difference between ? and 60.
Find variance of the bonus received by the risk manager.
A) 10.5
B) 11.5
C) 12.5
D) 13.5
E) 14.5
5. The claims department of an insurance company receives envelopes with claims for insurance
coverage at a Poisson rate of ~ envelopes per week. For any period of time, the number of
envelopes and the numbers of claims in the envelopes are independent. The numbers of claims in
the envelopes have the following distribution:
Number of Claims Probability
1
0.20
2
0.25
3
0.40
4
0.15
!
Using the normal approximation, calculate the 90 percentile of the number of claims received in
13 weeks.
A) Less than 1700
B) At least 1700, but less than 1720
C) At least 1720, but less than 1740
D) At least 1740, but less than 1760
E) At least 1760
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6. The 95% log-transformed confidence interval for /²! ³ is ²ÀÁ À³ . Find the upper limit
of the 90% linear confidence interval for /²! ³
A) Less than 3.00
B) At least 3.00 but less than 3.04
C) At least 3.04 but less than 3.08
D) At least 3.08 but less than 3.12
E) At least 3.12
7. A mortality study of 100 individuals with impaired health has resulted in the following
observations (time measured in years). There is no censoring of any data points.
Time
.1
.3
.4
.6
.7
.8
.9
1
Number of Deaths
2
1
3
2
4
4
5
3
Determine the kernel-smoothed estimate of the density function of time at death at time .7 using a
%&c
bandwidth of ~ À years and the uniform kernel & ²%³ ~ H & c  %  & b .
%&b
A) .1
B) .2
C) .3
D) .4
E) .5
8. You are given:
(i) Losses follow a Pareto distribution with ~ .
(ii) A random sample of losses is distributed as follows:
Loss Range
Number of Losses
² Á µ
² Á µ
² Á µ
² Á µ
² Á µ
² Á µ
Total
Estimate by matching at the 80-th percentile. Determine the mean of the estimated distribution.
A) Less than 250
B) At least 250 but less than 260
C) At least 260 but less than 270
D) At least 270 but less than 280
E) At least 280
9. You observe the following five ground-up claims from a data set that is truncated from below
at 100:
125
150
165
175
250
You fit a ground-up two Parameter Pareto distribution with ~ using maximum likelihood
estimation. Determine the mean of the fitted distribution.
A) Less than 25
(B) At least 25 but less than 50
C) At least 50 but less than 75
D) At least 75 but less than 100
E) At least 100
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10. An urn has a large number of balls numbered from 1 to 10. The proportion of balls with the
number 1 on them is not known. You wish to draw a sample (with replacement) of balls from the
urn to estimate the proportion of balls in the urn that have the number 1. You want to choose
enough balls so that the standard deviation of the estimated proportion is no more than .01 . Find
the minimum number of balls that must be drawn in general to satisfy this requirement.
A) 1500
B) 2000
C) 2500
D) 3000
E) 3500
11. You are given the following:
(i) the claims frequency rate for a group of insureds is believed to be an exponential
distribution with unknown parameter; - ²%³ ~ c c% , %  (ii) ten random observations of ? yield the following sample in ascending order:
.001 , .003 , .053 , .062 , .127 , .131 , .377 , .382 , .462 . .481
(iii) summary statistics for the sample data are:
% ~ À Á
~
~
~
% ~ À Á ²% ³ ~ c À
(iv) is the maximum likelihood estimator for Use the normal distribution to determine a 95% confidence interval for based upon the sample
data.
A) (.20 , .22)
B) (.27, 9.36)
C) (1.83 , 7.79)
D) (2.50 , 7.12)
E) (3.20 , 6.33)
12. > is a random variable with mean ,´> µ and variance = ´> µ. In a partial credibility
analysis of > , the manual premium used is 4 ~ . A sample of 350 observations of > is available
and the sum of the observed values is 300,000. Partial credibility is applied to determine a credibility
premium based on the "5% closeness" and "90% probability" criteria.
If the credibility standard used is the one based on the expected number of observations of > needed,
then the partial readability premium is 884.40. If the credibility standard used is the one based on the
expected sum of the observed values of > needed, then the partial credibility premium is 887.19. Using
this information, determine the mean of > .
A) Less than 750
B) At least 750 but less than 850
D) At least 850 but less than 950
D) At least 950 but less than 1050
E) At least 1050
13. The prior distribution of the parameter has pdf ²³ ~ for  .
The model distribution has a uniform distribution on the interval ´ Á µ .
Find the mean of the posterior distribution.
A) %
B) %
C) %
D) %
E) %
14. The severity distribution of ? has pdf ²%O³ ~ %c% for %  , where # has pdf
²³ ~ ec for  . Find the Bayesian premium ,´?b O % À À À % µ.
'%
A) b
B)
²b'% ³
b
C)
²b'% ³
b
'%
D) b
b'%
E) b
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PRACTICE EXAM 13
15. ? is a Poisson random variable with parameter \$, where the prior distribution of \$ is a
discrete uniform distribution on the integers Á Á 3 . A single observation of ? is made. Find the
Buhlmann factor A .
A) B) C) D) E) 16. The per year claim amount random variable has an exponential distribution with an unknown
mean that varies by individual. The experience of 200 individuals for one year is summarized in
the following information: % ~ Á
~
~ ²% c c
%³ ~ À
.
~
Determine the credibility factor A of one year's experience for a single individual using
semiparametric empirical Bayes estimation.
A) .03
B) .04
C) .05
D) .06
E) .07
17. You are given the following random sample of three values from the distribution function - :
c
c
? b? b?
You are to estimate = ²?³ using the estimator ²? c ?³ , where ? ~ .
~
Find the bootstrap approximation to the MSE of the estimator.
A) .1
B) .2
C) .3
D) .4
E) .5
18. Maximum likelihood estimation is applied to a data set of 20 observations to estimate in the
distribution with pdf ²%³ ~ h c%° Á %  . The sample mean of the data set is c
% ~.
The probability 7 ´  ?  µ is estimated using the mle of . Find the estimated variance of
that estimated probability using the delta method.
A) .0005
() .0010
C) .0015
() .0020
E) .0025
19. You are given the following information about two classes of business, where ? is the loss
for an individual insured:
Class 1 Class 2
Number of insureds
25
50
, (? )
380
23
, (? 2 )
365,000
---You are also given that an analysis has resulted in a Buhlmann value of 2.65.
Calculate the process variance for Class 2.
A) 2,280
B) 2,810
C) 7,280
D) 28,320
E) 75,050
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20. A population has 30% who are smokers with a constant hazard rate for time until death of 0.2
and 70% who are non-smokers with a constant hazard rate for time until death 0.1. Calculate the
75-th percentile of the distribution of the future lifetime of an individual selected at random from
this population.
A) 10.7
B) 11.0
C) 11.2
D) 11.6
E) 11.8
21. The loss random variable ? follows a two-parameter Pareto distribution with ~ and
~ . Calculate mean excess loss based on an ordinary deductible of 20..
A) 5
B) 15
C) 25
D) 35
E) 45
22. Dental Insurance Company sells a policy that covers two types of dental procedures: root
canals and fillings. There is a limit of one root canal per year and a separate limit of two fillings
per year. The number of root canals a person needs in a year follows a Poisson distribution with
~ , and the number of fillings a person needs in a year follows a Poisson distribution with
~ . The company is considering replacing the single limits with a combined limit of 3 claims
per year, regardless of the type of claim. Determine the change in the expected number of claims
per year if the combined limit is adopted.
A) No change
B) More than 0.00 but less than 0.20 more claims
C) At least 0.20 but less than 0.25 more claims
D) At least 0.25 but less than 0.30 more claims
E) At least 0.30 more claims
23. A compound Poisson claim distribution : has Poisson parameter ~ and severity
prob. .4
distribution ? ~ H 2 prob. .2 .
3 prob. .4
Stop loss insurance with a deductible of 2 is applied to : . Find ,´²: c ³b µ .
A) Àc
B) b Àc C) b Àc
D) c Àc E) c Àc
24. You are given the following data in grouped form for the amount of each of 100 separate
claims.
Interval
Number of Claims
² Á µ
² Á µ
² Á µ
² Á µ
² Á µ
² Á µ
Find the empirical estimate of the standard deviation of the loss .
A) 723
B) 763
C) 803
D) 843
E) 883
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25. The following data is from the first five years of a cohort life study.
The study began with 50 individuals. The year by year numbers of deaths and censorings are as
follows:
Year
# deaths
# censorings
1
3
2
2
5
4
3
6
3
4
6
5
5
8
4
It is assumed that censorings occur at mid-year and deaths for a particular year occur just before
year-end. Find the estimated variance of the cumulative hazard rate at the end of the 4-th year.
A) .02
B) .03
C) .04
D) .05
E) .06
26. You are given the following random sample of observations:
3 , 4 , 6 , 6 , 7 , 7 , 8 , 8 , 9 , 12
The data is a sample of 10 observations from a compound Poisson distribution for which the
severity distribution is exponential. Apply the method of moments to estimate the Poisson
parameter (mean) and the exponential parameter (mean) . What is the estimate of ?
A) 9
B) 11
C) 13
D) 15
E) 17
27. The following claim payment information is available for a collection of policies for which
the ground up loss distribution is exponential with mean (all policies have the same ground up
loss distribution).
Claim payments for 10 policies with no deductible, and no policy limit:
22 , 28 , 31 , 38 , 38 , 45 , 49 , 50 , 55 , 73
Claim payments for 5 policies with no deductible, but with policy limit 50:
18 , 29 , 37 , 42 , 50 (limit payment)
Claim payments for 5 policies with deductible of 10 and with policy limit 40
(maximum covered loss is 50): 21 , 24 , 33 , 40 (limit payment) , 40 (limit payment)
c(°
The likelihood function is of the form )
A) 780
B) 783
C) 786
D) 789
. Determine ( b ) .
E) 792
28. A sample of ten losses has the following statistics:
? c ~ À
? À ~ À
~
~
~
~
~
~
? c ~ À
? cÀ ~ À
? ~ Á ? ~ Á Á You assume that the losses come from a Weibull distribution with ~ À
Determine the maximum likelihood estimate of the Weibull parameter À
A) Less than 500
B) At least 500, but less than 1500
C) At least 1500, but less than 2500
D) At least 2500, but less than 3500
E) At least 3500
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29. You are given the following random sample of size 7: 1 , 2 , 4 , 5 , 7 , 9 , 14
A uniform distribution on the interval ´Á µ is the estimated distribution.
Which of the following is the -plot for the data and estimated distribution?
A)
B)
C)
D)
E) None of A, B, C or D is similar to the -plot for this data and estimated distribution.
30. A portfolio of risks is divided into three classes. The characteristics of the annual claim
distributions for the three risk classes is as follows:
Class I
Class II
Class III
Annual Claim
Poisson
Poisson
Poisson
Number Distribution mean 1
mean 2
mean 5
50% of the risks are in Class I, 30% are in Class II, and 20% are in Class III.
A risk is chosen at random from the portfolio and is observed to have 2 claims in the first year
and 2 claims in the second year. Find the expected number of claims for the risk in the third year.
A) Less than 1.5
B) At least 1.5 but less than 1.6
C) At least 1.6 but less than 1.7
D) At least 1.7 but less than 1.8
E) At least 1.8
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 13
31. You are given:
(i) The number of claims made by an individual insured in a year has a Poisson distribution with
mean .
(ii) The prior distribution for is gamma with parameters ~ and ~ À .
Three claims are observed in Year 1, and no claims are observed in Year 2.
Determine the estimated variance of the posterior distribution of the number of claims in Year 3.
A) 1.85
B) 1.86
C) 1.90
D) 1.91
E) 1.93
32. You are making credibility estimates for regional rating factors. You observe that the
Buhlmann-Straub nonparametric empirical Bayes method can be applied, with rating factor
playing the role of pure premium. ? denotes the rating factor for region and year , where
~ Á Á and ~ Á Á Á . Corresponding to each rating factor is the number of reported
claims, , measuring exposure. You are given:
_
_
_
_
~ ? ~ ? V
# ~ ²? c? ³ ²? c?³
~
~
~
À
À
À
À
À
À
Determine the credibility estimate of the rating factor for region 1.
A) 1.31
B) 1.33
C) 1.35
D) 1.37
E) 1.39
À
À
À
33. The inverse transform method is used to simulate the number of failures before a success in a
series of independent trials each with success probability ~ À, low values of uniform
random numbers correspond to small numbers of failures. What is the simulated number of
failures that corresponds to a uniform random number of 0.78 .
A) 0
B) 1
C) 2
D) 3
E) 4
34. Bob is an overworked underwriter. Applications arrive at his desk at a Poisson rate of 60 per
day. Each application has a 1/3 chance of being a “bad” risk and a 2/3 chance of being a “good”
risk. Since Bob is overworked, each time he gets an application he flips a fair coin. If it comes up
heads, he accepts the application without looking at it. If the coin comes up tails, he accepts the
application if and only if it is a “good” risk. The expected profit on a “good” risk is 300 with
variance 10,000. The expected profit on a “bad” risk is –100 with variance 90,000. Calculate the
variance of the profit on the applications he accepts today.
A) 4,000,000
B) 4,500,000
C) 5,000,000
D) 5,500,000
E) 6,000,000
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35. A sample of size is used to estimate the parameters in two possible models for the data.
The maximized log-likelihood for the 3-parameter generalized Pareto model is M( , and the
maximized log-likelihood for the exponential model is M) . You are given that according to the
Schwarz Bayesian Criterion, model A is preferred to model B. You are also given that according
to the likelihood ratio test, in which the null hypothesis is that model B is acceptable, and the
alternative hypothesis is that model A is preferable to model B, the null hypothesis is rejected at
the 5% level of significance but not at the 1% level of significance. Find the maximum value of that is compatible with these results.
A) 99
B) 101
C) 103
D) 105
E) 107
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ACTEX EXAM C/4 - PRACTICE EXAM 13 SOLUTIONS
2. The base hazard rate function is ²%³ ~ À.
The conditional survival probability to age 80 for an individual with parameter is
:²O³ ~ c ²%³ % ~ ´:²³µ ~ ²c À % ³ ~ cÀ .
The pdf of the parameter is \$ ²³ ~ À
(uniform distribution on ²ÀÁ À³).
The survival probability to age 80 for a randomly chosen individual is
²cÀ ³À c²cÀ ³À
À
À
:²³ ~ À :²O³ \$ ²³ ~ À cÀ h À
~
²À³²À³
3. À
²% c ³²À³ % b ²% c ³²À³ % ~ À b À
~ À is expected cost
À
per loss. Expected cost per payment is 7À
´?µ ~ À ~ À .
c ? ?  ~ À´
c ²? w ³µ
?  ,´Bonusµ ~ À´
c ,´? w µµ ~ À´
c ´ %²À³ % b 7 ´?  µµµ
~ À´
c ´ b µµ ~ À or
,´Bonusµ ~ À d ²
c %³²À³ % ~ À Answer: D
4. Bonus ~ À d F
5. The number of claims received in one week has a compound Poisson distribution with
,´5 µ ~ , and @ ~ Á Á Á with the given probabilities.
The expected amount of claims received in one week is
,´5 µ h ,´@ µ ~ ²³´²³²À³ b ²³²À³ b ²³²À³ b ²³²À³µ ~ .
The variance of the amount of claims received in one week is
,´5 µ h ,´@ µ ~ ²³´² ³²À³ b ² ³²À³ b ² ³²À³ b ² ³²À³µ ~ .
The expected number of > claims in 13 weeks is ²³²³ ~ Á , and
the variance of the number of claims in 13 weeks is ²³²
³ ~ Á .
The 90-th percentile is , where 7 ´>  µ ~ 7 ´ >jc
 c
µ ~ )² c
³ ~ À .
j
j
Using the normal approximation
c
j
~ À S ~ . Answer: B
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PRACTICE EXAM 13
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6. The log-transformed interval for /²! ³ has lower and upper limits of
V ³ ~ À . Therefore,
< h /²!
V ³
/²!
<
V ³
/²!
<
~ À and
V ³ ~ ´/²!
V ³µ ~ ²À³²À³ ~ À ,
d < h /²!
V
V ³ ~ jÀ ~ À . Then, since /²! ³ ~ À ~ À , it follows that
and /²!
<
<
< ~ À
~ %>
V ³µ
À
k= V
´/²!
?
V ³
/²!
~ %>
V ³µ
À
k= V
´/²!
?.
À
V ³µ
À
k= V
´/²!
V ³µ ~ À .
Then, À
~
, and it follows that À
k= V
´/²!
À
The 90% linear confidence interval for /²! ³ is
V ³ f À
k= ´/²!
V ³µ ~ À f À
h À
k= V
V ³µ ~ À f À
/²!
´/²!
~ ²ÀÁ À
³ .
À
7. The kernel density estimator of the density function is
V ²%³ ~ ²& ³ h & ²%³ , and the uniform kernel with bandwidth is & ²%³ ~ for
~
& c  %  & b (and 0 otherwise).
In this case, the kernel function will be & ²À³ ~ for & c À  À  & b À ,
or equivalently, for À  &  À . This captures the death times at
& ~ .6, .7, .8 and .9 (& ²À³ ~ at all other death times).
Since there are 100 individuals, the empirical density function values at those points are
²À
³ ~ Á ²À³ ~ Á ²À³ ~ Á ²À³ ~ .
V
Then, ²À³ ~ ²³´À b À b À b Àµ ~ À .
8. The 80-th percentile of the empirical distribution is 400. We find this by first identifying the
empirical percentiles at the interval endpoints. These are - ²³ ~ À Á - ²³ ~ À Á
- ²³ ~ À , etc. Since we are matching at the 80-th percentile, we use 400 as the 80-th
empirical percentile (if we were asked for the 75-th percentile for example, then it would be
interpolated between the 52-nd empirical percentile which is 200 and the 80-th which is 400).
The distribution function of ? is c ² %b
³ . We use 400 as the 80-th percentile, so that
c ² b
³ ~ À S ~ À . The mean is
c
~ À
9. For the two parameter Pareto we have a shortcut for finding the MLE of when is given.
% b
If there is a deductible of and ground up loss values % are given, then we let \$ ~ ² b
³.
'\$
and the MLE if is where is the number of sample values. In this case there are 5 \$
b
values: \$ ~ ² b
b ³ Á À À À Á \$ ~ ² b ³ and the MLE of is '\$ ~ À .
The mean of the fitted distribution is c
~ Àc
~ À
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PRACTICE EXAM 13
10. After draws, the estimate of the proportion of balls that have number 1 is
c
? ~ number of 1's drawn ~ @ . @ represents the number of 1's drawn in draws. @ has a
binomial distribution with parameters and , where is the unknown probability
of drawing a 1
_
on any given draw (this is the proportion of 1's in the urn). The variance of ? is
= ´@ µ
²c³
²c³
c
c
= ´? µ ~ = ´ @ µ ~ ~
~ , and the standard deviation of ? is
k ²c³
. Since is unknown, the value of ² c ³ can be as large as (the maximum of
c for   occurs at ~ , and that maximum is 4 )À In order to ensure that the
°
c
standard deviation of ? is no more than .01, we must have k  À , or equivalently,
 .
c
11. The mle of the exponential mean is ? . In this case the distribution has mean which is
c
V ~ ~ À .
estimated using ? ~ ÀbÀbÄbÀ
~ À ~ S À
The (asymptotic) variance of the mle estimate is c
,<
C
3²³=
C ~
~
In this case, 3²³ , the likelihood function, is 3²³ ~ ²% Â ³ ~ c% ,
and 3²³ ~ c '% . Then CC 3²³ ~ c , and
, < CC 3²³= ~ ,´ c µ ~ c . The asymptotic variance of the mle estimate is ,
V
which is estimated to be ~ 2.314 . The 95% normal confidence interval for is then
À f À
h jÀ , which is the interval (1.83 , 7.79) .
= ´> µ
12. The full credibility standard based on the number of observations needed is À ²,´> µ³
= ´> µ
and based on the sum of the observed values it is À ,´> µ .
Á
c
From the given information, the sample mean of the observed values is > ~ ~ À À
c
The credibility premium based on partial credibility using is A> b ² c A³4 ,
c
where > ~ À Á 4 ~ and A is the partial credibility factor.
Using the credibility standard based on the expected number of observations needed,
A~
= ´> µ , so that ÀA b ² c A³ ~ À , from which we get
l À ²,´>
µ³
A ~ À , and therefore
= ´> µ
À ²,´>
µ³
= ´> µ
~ À
, so that ²,´> µ³ ~ À .
Using the credibility standard based on the expected sum of the observed values needed,
Á
A ~l
´> µ , so that ÀA b ² c A³ ~ À , from which we get
À =,´>
µ
Á
´> µ ~ À
À =,´>
µ
= ´> µ°,´> µ
= ´> µ°²,´> µ³ ~ À ~ A ~ À , and therefore
, so that
Then, ,´> µ ~
.
= ´> µ
,´> µ
~ .
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13. ²%O³ ~ for  %  .
²%Á ³ ~ ²%O³ h ²³ ~ h ~ (on the appropriately defined region for ? and ) .
%
If  %  , then   % , and ? ²%³ ~ ~ c % ~ %%c
.
% %
If %  , then
  % , and ? ²%³ ~ %° ~ c ~ .
%
Since %  , ? ²%³ ~
and ²O%³ ~
%
%
,´O%µ ~ %° h %
²%Á³
? ²%³
~
%
°
°%
~
%
%
%
on the region %   % .
14. The Bayesian premium can be found in two ways:
(i) ,´?b O? ~ %µ ~ %b ?b O? ²%b O%³ %b , or
(ii) ,´?b O? ~ %µ ~ ,´?b Oµ h #O? ²O%³ .
In the general -sample case, approach (i) can become quite complicated due to the complicated
nature of the pdf of the predictive distribution, ?b O? ²%b O%³.
Approach (ii) is often more straightforward.
B
B
,´?Oµ ~ % h %c% % ~ % c% % ~ h ~ .
Also, ? Á# ²% Á ÀÀÀÁ % Á ³ ~ ? O# ²% Á ÀÀÀÁ % O³ h ²³ ~ < ? O# ²% O ³=² ³
~ ´ ² % c%
~
c
³µ h e
~
b
~
h ² % ³ h c²b'% ³
~
c
~ b h ² % ³ h c²b%³
~
B
and ? ²%³ ~ ? Á# ²% Á ÀÀÀÁ % Á ³ ~ b h ² % ³ h c²b'% ³ ~
B b
h ² % ³ h c
c²b%³
~
~ ² % ³ h
~
²b³[
~
c
B b
h c²b%³
~ ² % ³ h ²b%³
c b À We have used the rule
~
B ! h c! ! ~ [
if is an integer  and  .
b
Then,
#O? ²O%³ ~
? Á# ²%Á³
? ²%³
~
c
b h² % ³hc²b% ³
~
²b³[
² % ³h ²b%
c³b
~
c b hb hc²b%c³
²b%³
²b³[
.
~
Then
~
c b hb hc²b%c³
²b%³
B
,´?b O? ~ %µ ~ h
²b³[
c b
²b%³
²b³[
B
c
h h c²b%³ ~
c b
²b%³
²b³[
²³[
c
²b%³
h ²b%³
c b ~ b
(in this integral, the following general integration form was used B
if is an integer  and  , then & c& & ~ [
b , in this case ~ and
c
~ b % ). Answer: B
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PRACTICE EXAM 13
15. ²\$³ ~ ,´?O\$µ ~ \$ , #²\$³ ~ = ´?O\$µ ~ \$ Á
~ ,´²\$³µ ~ ,´\$µ ~ ² ³² b b ³ ~ .
~ = ´²\$³µ ~ = ´\$µ ~ ² ³² b b ³ c ~ .
# ~ ,´#²\$³µ ~ ,´\$µ ~ .
~ (one observation of ? ), so A ~ b
~ .
# ~
°
16. The following comments review semiparametric estimation. The model for the portfolio may
have a parametric distribution for ? given # ~ , but an unspecified non-parametric distribution
for #. In this case, we may be able to use the relationships linking ²#³ ~ ,´?O#µ and
#²#³ ~ = ´?O#µ and the fact that = ´?µ ~ # b in order to get estimates for Á # and to
use in the credibility premium formulation. In this problem, ? given # is exponential, so that
²#³ ~ ,´?O#µ ~ # and #²#³ ~ = ´?O#µ ~ # . Then, ,´?µ ~ ,´,´?O#µµ ~ ,´#µ is
c
estimated by ? ~ ~ À , so our estimate of ,´#µ is 2.2.
Also, = ´?µ ~ ,´= ´?O#µµ b = ´,´?O#µµ
~ ,´# µ b = ´#µ ~ ,´# µ b ,´# µ c ²,´#µ³ ~ ,´# µ c ²,´#µ³ .
= ´?µ is estimated by ~ À
, and this becomes our estimate for ,´# µ c ²,´#µ³ .
The estimate for ,´# µ then comes from À
~ ,´# µ c ²,´#µ³ ~ ,´# µ c ²À³ ,
so that # ~ ,´# µ is estimated to be V# ~ À
bÀ
~ À.
Finally, we estimate ~ = ´?µ c # using V
~ c V# ~ À
c À ~ À .
V ~ V# ~ À ~ À
.
The credibility factor for a single individual is A
b V
b
À
17. The variance of the empirical distribution is
~ = ²? ³ ~ ´² c ³ b ² c ³ b ² c ³ µ ~ .
There are ~ possible bootstrap samples (of size 3) from the original sample.
These 27 samples can be described as follows:
c
- all 1's, 8 samples , ? ~ Á V ~ Á ²V c ³ ~ ² c ³ ~ Â
c
V
- two 1's and one 3, 12 samples , ? ~ Á ~ ´² c ³ b ² c ³ b ² c ³ µ ~ ²V c ³ ~ ² c ³ ~ Â
c
- one 1 and two 3's, 6 samples , ? ~ Á V ~ ´² c ³ b ² c
²V c ³ ~ ² c ³ ~ Â
c
- all 3's, 1 sample , ? ~ Á V ~ Á ²V c ³ ~ ² c ³ ~ ³
b ² c ³ µ ~ Á
Â .
The bootstrap estimate to the MSE of the estimator is
² ³² ³ b ² ³² ³ b ² ³² ³ b ² ³² ³ ~ ~ À .
An example of the count of the number of samples given above is the number of samples with
one 1 and two 3's: ²Á Á ³ Á ²Á Á ³ Á ²Á Á ³ Á ²Á Á ³ Á ²Á Á ³ Á ²Á Á ³ .
The reason that there are two samples of the form ²Á Á ³ is that there are two sample points
equal to 1. If we label them 1a and 1b, then ²Á Á ³ and ²Á Á ) should be regarded as
separate samples. The same applies to the other samples.
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18. 7 ´  ?  µ ~ - ²³ c - ²³ ~ ´ c c° µ c ´ c c° µ ~ c° c c° ~ ²³ .
The pdf is for the exponential distribution with mean , and the mle of is V ~ c
% ~.
Using the delta method, the variance of ²V³ is approximately equal to
´Z ²³µ h = ´Vµ ~ ´ h c° c h c° µ h = ´Vµ .
Since V ~ '? , which is the sample mean, the variance of the mle is the variance of the
5
sample mean, which is
= ´?µ
5
. For the exponential distribution, the variance is the square of
the mean, so that = ´?µ ~ , and therefore, = ´Vµ ~ 5
À
In this example, we know that the mle of is 1. The estimated variance of the estimate of the
V
V
probability 7 ´  ?  µ is ´ h c° c h c° µ h = ´Vµ .
V
V
V
The estimate of = ´Vµ is ~
, and the estimated variance of the probability is
c°
c° ´ h c h µ h ² ³ ~ À .
19. This is an exercise in Buhlmann's credibility model. We can put this situation in the
Buhlmann context, with parameter random variable #, and conditional distribution of loss
amount ? given #. The loss random variable ? depends on the class of business. There are two
classes of business. Class of business is represented by the parameter #. Since there are 75
insureds of which 25 are Class 1 and 50 are class 2, the probability distribution for the class
parameter # is 7 ´# ~ µ ~ and 7 ´# ~ µ ~ (when a claim occurs, the chance is
~ that the policy was from class 1, or in other words, that # ~ ). The Buhlmann value
is ~ # , where # is the expected process variance, and is the variance of the hypothetical
mean. The process variance for Class 1 is
#²³ ~ = ´?O# ~ µ ~ ,´? O# ~ µ c ²,´? O# ~ µ³ ~ Á c ²³ ~ Á ,
and for Class 2 it is #²³ ~ = ´?O# ~ µ .
In general for the Buhlmann credibility approach, #²³ ~ = ´?O# ~ µ and
# ~ ,´#²#³µ ~ ,´ = ´?O# ~ µ µ , ²³ ~ ,´?O# ~ µ and
~ = ´²#³µ ~ = ´ ,´?O# ~ µ µ .
We are given ,´?O# ~ µ ~ Á ,´?O# ~ µ ~ , and ,´? O# ~ µ ~ Á and
~ À
. Therefore, ²³ ~ ,´?O# ~ µ ~ Á ²³ ~ ,´?O# ~ µ ~ .
# ~ , prob. ²#³ is a discrete 2-point random variable, ²#³ ~ F
,
# ~ , prob. and is the variance of ²#³, ~ = ´²#³µ . A simple rule for the variance of a discrete 2point random variable is the following: if A is the discrete two-point random variable with
outcomes ¸Á ¹ with probabilities Á c , then = ´Aµ ~ ² c ³ ² c ³ .
Therefore, ~ = ´²#³µ ~ ² c ³ ² ³² ³ ~ Á .
Note that we can also find = ´²#³µ in the conventional way as the 2nd moment minus the
square of the first moment:
= ´²#³µ ~ ² ³² ³ b ² ³² ³ c ´²³² ³ b ²³² ³µ ~ Á ~ # S # ~ ~ ²À
³²Á ³ ~ Á ~ ,´= ´?O# ~ µµ À
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PRACTICE EXAM 13
19. continued
Since = ´?O# ~ µ ~ ,´? O# ~ µ c ²,´? O# ~ µ³
~ Á c ²³ ~ Á , we get
Á ~ ~ = ´?O# ~ µ h 7 ´# ~ µ b = ´?O# ~ µ h 7 ´# ~ µ
~ ²Á ³² ³ b = ´?O# ~ µ² ³ , from which we get = ´?O# ~ µ ~ .
This is the process variance for Class 2.
20. The future lifetime is ; ²%³, where ; ²%³ is a mixture of two distributions. The two mixing
distributions are ; ²%³, the future lifetime of a smoker, and ; ²%³, the future lifetime of a nonsmoker. The mixing weights are .3 for smokers and .7 for non-smokers. The 75-th percentile of
; ²%³ is , where 7 ´; ²%³  µ ~ À . For a mixture distribution,
7 ´; ²%³  µ ~ ²À³ h 7 ´; ²%³  µ b ²À³ h 7 ´; ²%³  µ .
Since ; ²%³ and ; ²%³ have constant forces of mortality, their survival probabilities have an
exponential form, c! .
7 ´; ²%³  µ ~ c cÀ , and 7 ´; ²%³  µ ~ c cÀ .
Then, 7 ´; ²%³  µ ~ ²À³² c cÀ ³ b ²À³² c cÀ ³ ~ À .
This becomes a quadratic equation with @ ~ cÀ : À@ b À@ c À ~ .
The solutions are @ ~ À Á c À
. Since @ ~ cÀ , we ignore the negative root.
Then cÀ ~ À S ~ À
.
Note that once we have formulated the quadratic equation, we can substitute in each of the five
21. ,´? c O?  µ is the mean excess loss with a deductible of 20.
,´²?c³ µ
,´?µc,´?wµ
This can be written as 7 ´; ?µb ~ 7 ´; ?µ .
Using the Pareto distribution, we have
,´?µ ~ c
~ , and ,´? w µ ~ ² c
³´ c ² b
³c µ ~ À .
Also, 7 ´?  µ ~ c -? ²³ ~ ² b
³ ~ À
.
The mean excess loss is cÀ
À
~ .
The quick solution to this problem uses that fact that for the Pareto distribution with parameters and , based on an ordinary deductible of , the cost per payment random variable is also Pareto
with parameters and b , so the expected cost per payment (which is the mean excess loss) is
b
b
c . In this problem, that becomes
22. Under the current limits, with 5 denoting the number of root canals a person needs in a
year, the number of root canal claims is either 0 or 1, with
7 ² root canal claims in one year³ ~ 7 ²5 ~ ³ ~ c and
7 ² root canal claim in one year³ ~ 7 ´5  µ ~ c 7 ´5 ~ µ ~ c c ~ À
.
The expected number of root canal claims in one year is c c (this is not the same as the
expected number of root canals a person needs, which is Poisson with a mean of 1).
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22. continued
Under the current limits, with 5 denoting the number of fillings a person needs in a year, the
number of filling claims is either 0 or 1 or 2, with
7 ² filling claims in one year³ ~ 7 ²5 ~ ³ ~ c and
c
7 ² filling claim in one year³ ~ 7 ´5 ~ µ ~ [h ~ c ,
and P²2 filling claims in one year³ ~ 7 ´5  µ ~ c 7 ´5 ~ Á µ ~ c c c c
The expected number of filling claims in one year is
c b ² c c c c ³ ~ c c ~ À
.
Under the new limit, the number of root canals and fillings combined needed in one year is 5 ,
which is Poisson with a mean of 3. The number of claims in one year is 0,1,2 or 3, with
probabilities 7 ² claims in one year³ ~ 7 ²5 ~ ³ ~ c Á
7 ² claim in one year³ ~ 7 ²5 ~ ³ ~ c h Á
c
7 ²2 claims in one year³ ~ 7 ²5 ~ ³ ~ [h Á and
7 ²2 claims in one year³ ~ 7 ²5  ³ ~ c 7 ´5 ~ Á Á µ ~ c Àc .
The expected number of claims in 1 year is
c
c h b ² [h ³ b ² c Àc ³ ~ c Àc ~ À .
Under the original limit scheme, the expected number of claims per year is
À
b À
~ À. The new limit scheme has an expected number of claims per year
that is .237 larger than the original limit scheme.
23. ,´²: c ³b µ ~ ,´:µ c ´ c - ²³µ c ´ c - ²³µ ~ c ² c ²³³ c ² c ²³ c ²³³
²³ ~ 7 ´5 ~ µ ~ c Á ²³ ~ 7 ´5 ~ µ h 7 ´? ~ µ ~ ²c h ³²À³ ~ Àc À
,´²: c ³b µ ~ c ² c c ³ c ² c c c Àc ³ ~ b Àc .
Alternatively, we can find ,´²: c ³b µ ~ ,´:µ c ,´: w µ ~ c ´ ²³ b ² c - ²³³µ
~ c ´Àc b ² c c c Àc ³µ ~ b Àc . Answer: C
24. The empirical estimate of the mean is 4 h
~
bc
5
, which is
b
b
³ b ² b
³ b ² b
³
´² ³ b ²
b ² b
³ b ² b
³µ ~ À .
² c ³
c
The empirical estimate of the second moment is 4 h ²b³²
5 , which is
cc ³
~
c
c
c
c
´² ²c³ ³ b ² ²c³ ³ b ² ²c³ ³ b ² ²c³ ³
c
c
b ² ²c³
³ b ² ²c³
³µ ~ Á Á .
The empirical variance is Á Á c ²
À³ ~ Á .
The empirical estimate of the standard deviation is jÁ ~ . Answer: B
25. ~ Á ~ Á ~ Á ~ Á ~ Á
V
=V
´/²³µ
~ b b b ~ À .
~ Á ~ Á ~ .
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PRACTICE EXAM 13
26. Let : denote the compound Poisson random variable, and let 5 denote the Poisson
frequency random variable with ,´5 µ ~ , and let ? denote the exponential severity random
variable with ,´?µ ~ . Then ,´:µ ~ ,´5 µ h ,´?µ ~ and
= ´:µ ~ ,´5 µ h ,´? µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³
~ ² ³ ~ h b h ~ .
The first moment of the empirical distribution based on the given sample is
bb
b
bbbbbb
~ À .
The variance of the empirical distribution is ´² c ³ b ² c ³ b Ä b ² c ³ µ ~ À .
According to the method of moments, we have the moment equations
,´:µ ~ ,´5 µ h ,´?µ ~ ~ À and = ´:µ ~ ,´5 µ h ,´? µ ~ ~ À .
Dividing the second equation by the first results in ~ ~ À
À , from which we get the
À
V~
estimate of , V ~ À . Then Answer: D
V ~ À .
27. For a claim payment of amount % that is not a limit payment from a policy with no
c%°
deductible, the likelihood factor is the pdf . For a limit payment of 50 from a policy with
no deductible but with limit 50, the likelihood factor is c° . For a claim payment of amount %
that is not a limit payment from a policy with deductible 10, the likelihood factor is the
c²%b³° °
c%°
conditional pdf
~ . For a limit payment of 40 from a policy with deductible 10
c°
and policy limit 40 (maximum covered loss 50), the likelihood factor is the conditional
c²b³°
probability c° ~ c° . The overall likelihood function will be
c°
c°
c°
c°
c°
c°
d dÄd d dÄd d
c°
c°
c
°
d d Ä d d c° d c° ~ .
( ~ Á ) ~ , so that ( b ) ~ .
28. The quick way of dealing with maximum likelihood estimation for the Weibull distribution
with given and data that includes known % 's and right-censored (limit) payments is as follows
(this is reviewed in the mle section of the notes in this study guide). For each data value calculate
% ( or " ) for a limit payment. Find the sample mean of those values. The mle of is that
sample mean raised to the power . For this problem, we have ~ À , and there are ~ data values, so we find '%À
. From the given information, this is ²À³ ~ À .
Then ~ ²À³° ~ ²À³ ~ .
We now solve the problem from basic principles by first setting up the likelihood function.
The density function of the Weibull distribution with parameters and is
²%°³ c²%°³
²%³ ~
. The likelihood function for the 10 observed losses is
%
²% ° ³ c²% °³
% ³ c h c h'% .
3 ~ ²% ³ ~ ~
h
²
%
~
~
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28. continued
With ~ À the likelihood function becomes
3~
²À³
h ²% ³cÀ h 3~
²À³
c
h ²% ³cÀ h c À h'%À
À
À
, and then using %À ~ À , we get
~
.
The natural log of the likelihood is 3 ~ À c c À ²% ³ c À
, and the
À
C
maximum likelihood estimate is found by setting C
3 ~ c b À
À ~ .
The result is an estimate of V ~ Á . Answer: C
29. The -plot uses the following smoothed empirical distribution function values, and model
probabilities:
%
Empirical
Dist. Fn. (left tail)
Model
Dist. Fn. (left tail)
Graph D is an accurate representation of the -plot.
30. Posterior probabilities are found first.
7 ´\$ ~ O? ~ µ ~
7 ´?~O\$~µh7 ´\$~µ
7 ´?~µ
~
c h
[ h²À³
c h
c h
c h
h²À³b
[
[ h²À³b [ h²À³
c h
[ h²À³
c
h
c h
c h
[ h²À³b [ h²À³b [ h²À³
~ À .
In a similar way, we get 7 ´\$ ~ O? ~ µ ~
7 ´\$ ~ O? ~ µ ~
c h
[ h²À³
c h
c h
c h
[ h²À³b [ h²À³b [ h²À³
~ À , and
~ À .
Then the Bayesian premium is found:
,´? O? ~ Á ? ~ µ ~ ,´? O\$ ~ O h 7 ´\$ ~ O? ~ Á ? ~ µ
b ,´? O\$ ~ O h 7 ´\$ ~ O? ~ Á ? ~ µ b ,´? O\$ ~ O h 7 ´\$ ~ O? ~ Á ? ~ µ
7 ´\$ ~ O? ~ Á ? ~ µ ~
c h 3 h²À³
[
c h c h c 2
3 h²À³b2
3 h²À³b2 h 3 h²À³
[
[
[
2
~ À Á
7 ´\$ ~ O? ~ Á ? ~ µ ~ À Á 7 ´\$ ~ O? ~ Á ? ~ µ ~ À .
Then ,´? O? ~ Á ? ~ µ ~ ²³²À³ b ²³²À³ b ²³²À³ ~ À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 13
31. The model distribution ? is Poisson with mean , where has a gamma distribution with
~ and ~ À . The posterior distribution of the number of claims in year 3 is negative
binomial with ~ b '% and ~ b . We are given ~ with % ~ and % ~ .
Therefore, the posterior distribution of ? is negative binomial with ~ b b ~ and
À
~ ²À³b
À The variance of ? is ² b ³ ~ À .
V ~ ~ V# ,
32. The estimated credibility factor for group is A
V
b V
b
c
V
V
and the estimated rating factor is A ? b ² c A ³
V.
We are given ~ Á ~ Á ~ Á ~ and ~ ~ ~ Á
so that ~ b b ~ À
c
c
c
c
? b ? From the given information we can find ~ À Á and
V ~ ? ~ ? b
b b
c
h ²? c ? ³ ~ ²À
b À b À³ ~ À Á and
V# ~ ² c³
~
~
V
c
~
~ ~
c
c
h ´ ²? c ? ³ c V#² c ³ µ
~
h ´²À b À b À³ c ²À³²³µ ~ À
.
c ² b b ³
V ~ V# ~
V
Then A
À ~ À Á A ~
À ~ À and
b À
b À
b V
V ~
A
À ~ À . Then the estimated rating factor for region 1 is
b À
~
c
V ?
V ³
A
V ~ ²À³²À
³ b ²À³²À³ ~ À À Answer: C
b ² c A
33. The distribution of the number of failures before the first success is geometric with
distribution
No. Failures À
²À³²À³
²À³ ²À³
²À³ ²À³
!
À
À
À
À
The uniform random number is " ~ À. We see that À  À  À , so the simulated
number of failures is .
34. The total profit : on a the applications accepted in a given day can be modeled as a
compound Poisson random variable with mean frequency rate ~ , and with severity
distribution @ . @ represents the profit on an arriving application. @ can be modeled as a mixture
of three component distributions. Good risks will always be accepted no matter what the result of
the coin flip. Since good risks arrive of the time, it follows that of the accepted risks are
"good". Bad risks arrive of the time, but half of the arriving bad risks will be rejected since the
coin flip is "tails" half of the time, and in that case the bad risk is rejected.
Therefore, d ~ of the accepted risks are "bad". The other of the accepted risks are 0
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 13
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34. continued
The accepted risks are a mixture of
(i) "good" risks (say @ ) with mixing weight , (ii) "bad" risks (say @ ), with mixing weight ,
and (iii) risks of amount 0 (say @ ) with mixing weight .
Since : has a compound Poisson distribution, the variance of : can be formulated as
= ´:µ ~ h ,´@ µ ~ h ,´@ µ .
Since @ is a mixture distribution, ,´@ µ ~ ² ³,´@ µ b ² ³,´@ µ b ² ³,´@ µ .
,´@ µ ~ = ´@ µ b ²,´@ µ³ ~ Á ,
and ,´@ µ ~ = ´@ µ b ²,´@ µ³ ~ Á , and ,´@ µ ~ À
Then ,´@ µ ~ ² ³²Á ³ b ² ³²Á ³ b ² ³²³ ~ Á .
Finally, = ´:µ ~ ²
³²Á ³ ~ Á Á .
35. The Schwarz Bayesian Criterion compares M( c ln²³ and M) c ln²³ ,
and since model A is preferable to model B this means that M( c ln²³ c ´M) c ln²³µ  ,
which can be rewritten as M( c M)  ²³ .
The likelihood ratio test has test statistic ²M( c M) ³ . Since model A has 3 parameters and
model B has 1 parameter, the number of degrees of freedom in the chi-square statistic is
c ~ . The critical value for a test with significance level 5% is À ²³ ~ À
and the critical value for a test with significance level 1% is À ²³ ~ À .
Since the null hypothesis is not rejected at the 1% level, it must be true that ²M( c M) ³  À,
so that M( c M)  À
. From the Schwartz Bayesian Criterion, we had ²³  M( c M) ,
and therefore, ²³  À
. It then follows that  À
~ À .
The maximum (integer) value for is .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 13
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 14
PE-265
ACTEX EXAM C/4 - PRACTICE EXAM 14
1. You are given:
(i)
Annual claim counts follow a Poisson distribution with mean -.
(ii)
The parameter - has a prior distribution with probability density function:
0 Ð-Ñ œ "\$ /-Î\$ ß -  !
Two claims were observed during the first year.
Determine the variance of the posterior distribution of -.
A) 9/16
() 27/16
C) 9/4
D) 16/3
E) 27/4
2. \ is a mixture of two exponential distributions.
Distribution 1 has a mean of 1 and a mixing weight of .25 and distribution 2 has a mean of 2 and
a mixing weight of .75. \ is simulated using the inverse transformation method with a uniform
Ð!ß "Ñ value of .7. Find the simulated value of \ .
A) Less than 2.0
B) At least 2.0 but less than 2.1
C) At least 2.1 but less than 2.2
D) At least 2.2 but less than 2.3
E) At least 2.3
3. The parameter - has prior distribution 1Ð-Ñ œ "# /-  "# Ð "# /-Î# Ñ
(mixture of two exponentials). The model distribution \ has a conditional distribution given - that is
Poisson with mean -. Find the Buhlmann credibility premium if there is a single observation of 2.
A) \$"
B) \$\$
C) \$'
D) \$*
E) %#
"(
"(
"(
"(
"(
4. A random sample of eight times until failure has the following Nelson-Aalen
estimates for the cumulative hazard function:
s
LÐ>Ñ
>
>\$
!
\$Ÿ>&
!Þ"#&
&Ÿ>'
!Þ%"!(
'Ÿ>)
!Þ'"!(
)Ÿ>*
!Þ)'!(
* Ÿ >  "!
"Þ"*%!
"! Ÿ >
#Þ"*%!
There is no censoring or truncation of the data.
Find the empirical estimate of the variance of the time until failure.
A) Less than 5.0
B) At least 5.0, but less than 5.2
C) At least 5.2, but less than 5.4
D) At least 5.4, but less than 5.6
E) At least 5.6
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 14
5. An insurance company writes a book of business that contains several classes of
policyholders. You are given:
(i) The average claim frequency for a policyholder over the entire book is 0.425.
(ii) The variance of the hypothetical means is 0.370.
(iii) The expected value of the process variance is 1.793.
One class of policyholders is selected at random from the book. Nine policyholders are selected at
random from this class and are observed to have produced a total of seven claims. Five additional
policyholders are selected at random from the same class. Determine the Buhlmann credibility
estimate for the total number of claims for these five policyholders.
A) 2.5
B) 2.8
C) 3.0
D) 3.3
E) 3.9
6. Suppose that \" ß \# ß á are independent random variables, each with probability of successes
: and probability of failure "  :, where !  : Ÿ ". Let R be the number of observations
needed to obtain the first success. What is a maximum likelihood estimator of : ?
A) R "#
B) R ""
C) R"
D) R ""
E) R "#
7. A survival study of 100 individuals with no censoring results in the following Nelson-Aalen
estimates for the cumulative hazard rates at successive death points. It is known that there were 3
deaths at the third death point, >\$ .
Death Point >3
>"
>#
>\$
s 3Ñ
LÐ>
.03
.103495
Determine the number of deaths at the second death point, ># .
A) 0
B) 1
C) 2
D) 3
E) 4
8. A particular type of individual health insurance policy models the annual loss per policy as an
exponential distribution with a mean that varies with individual insured. A sample of 1000
randomly selected policies results in the following data regarding annual loss amounts in interval
grouped form.
Interval
Number of Losses
Ò!ß "!!Ó
&!!
Ð"!!ß #!!Ó
#&!
Ð#!!ß &!!Ó
"&!
Ð&!!ß "!!!Ó
'!
Ð"!!!ß #!!!Ó
%!
It is assumed that the loss amounts are uniformly distributed within each interval.
Apply semiparametric empirical Bayes credibility to estimate the loss in the 3rd year for a
particular individual who had annual policy losses of 150 in the first year and 0 in the second
year.
A) 150
B) 152
C) 154
D) 156
E) 158
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 14
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9. You are simulating the gain/loss from insurance where:
(i) Claim occurrence follows a Poisson distribution with - œ #Î\$ per year.
(ii) Each claim amount is 1, 2 or 3, with T Ð"Ñ œ !Þ#& ß :Ð#Ñ œ !Þ#& and :Ð\$Ñ œ !Þ&! .
(iii) Claim occurrences and amounts are independent.
(iv) The annual premium equals expected annual claims plus 1.8 times the standard deviation of
annual claims.
(v) 3 œ !Þ
You use 0.75, 0.60, 0.40, and 0.20 from the unit interval to simulate time between claims, where
small numbers correspond to longer times. You use 0.30, 0.60, 0.20, and 0.70 from the unit
interval to simulate claim size, where small numbers correspond to smaller claims.
Calculate the gain or loss for the first two years from this simulation.
A) loss of 5
B) loss of 4
C) 0
D) gain of 4
E) gain of 5
10. The Allerton Insurance Company insures 3 indistinguishable populations. The claims
frequency of each insured follows a Poisson process. Given:
Population
Expected
Probability
Claim
(class)
time between
of being in
cost
claims
class
I
12 months
1/3
1,000
II
15 months
1/3
1,000
III
18 months
1/3
1,000
Calculate the expected loss in year 2 for an insured that had no claims in year 1.
A) Less than 810
B) At least 810, but less than 910
C) At least 910, but less than 1,010
D) At least 1,010, but less than 1,110
E) At least 1,110
11. The following table that was obtained by fitting both a Poisson distribution and a binomial
distribution to a data set of 100,000 integer-valued observations.
s œ Þ""&#(
7
s œ % ß s; œ Þ!#))"(&
Fitted Poisson
Fitted Binomial
85
expected
;#
expected
;#
5
!
)*,!!!
)*,""#Þ'
Þ"%##
))ß *'"Þ)
Þ!"'%
"
"!,%)(
"!ß #(#Þ!
%Þ&!!
"!ß &&)Þ*
Þ%)*(
#
&!!
&*#Þ!
"%Þ\$!&!
%(!Þ!
"Þ*"*&
#\$Þ%
%Þ'##
*Þ\$
"Þ\$()(
\$ "\$
Totals "!!ß !!!
"!!ß !!!
#\$Þ&(
"!!ß !!!
\$Þ)!
Degrees of freedom
%""œ#
%#"œ"
:  Þ!!"
Þ!&  :  Þ"
:-value
Which of the following pair of statements has both statements true regarding the chi-square
goodness-of-fit test of the null hypothesis that the model is a good fit to the data?
Poisson
Binomial
A)
Accept at 1% level
Reject at 1% level
B)
Accept at 5% level
Reject at 5% level
C)
Reject at 10% level
Reject at 10% level
D)
Reject at 5% level
Reject at 5% level
E)
Reject at 1% level
Reject at 1% level
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 14
12. Non-parametric empirical Bayes estimation has been applied to a data set with three
policyholders, with four exposure periods for each policyholder, and with one exposure unit for



each exposure period. You are also given that \ " œ % ß \ # œ ' ß \ \$ œ ( , and
s is found based on this
estimated expected process variance is # . The credibility factor ^
information. It is decided that a fourth policyholder will be added to the data set and the
credibility factor will be recalculated. The fourth policyholder has four exposure periods with

s
one unit for each exposure period, and \ % œ \$ and [email protected]% œ " . Find the percentage change in ^
from original to recalculated value.
A)  ##%
B)  ""%
C) !%
D) ""%
E) ##%
13. A spliced distribution is defined to have the following density function.
+ † 0" ÐBÑ !  B  "!!
.
0 ÐBÑ œ œ
, † 0# ÐBÑ "!! Ÿ B  #!!
0" ÐBÑ is the density function of a uniform random variable on the interval Ð!ß "!!Ñ , + œ Þ%, and
0# ÐBÑ is the density function of the uniform distribution on the interval Ò"!!ß #!!Ñ. Find the mean
of the spliced distribution.
A) 100
B) 110
C) 120
D) 130
E) 140
14. The loss random variable \ has an exponential distribution and an ordinary deductible is
applied to all losses. You are given the following:
• The variance of the cost per loss random variable is 20,480.
• The variance of the cost per payment random variable (excess loss random variable) is 25,600.
Find the average cost per loss.
A) Less than 80
B) At least 80 but less than 83
C) At least 83 but less than 86
D) At least 86 but less than 89
E) At least 89
15. W is the mixture of two compound Poisson distributions, W" and W# , with mixing weights
" prob. "#
of "# each. W" has Poisson frequency R" with -" œ " and severity ]" œ œ
.
# probÞ "#
W# has Poisson frequency R# with -# œ " and severity ]# œ œ
Find Z +<ÒWÓ .
A) "\$"
B) "*(
%)
(#
5
C) \$*
"%%
D) ""
%
"
#
prob. "\$
.
probÞ #\$
E) \$*(
"%%
16. Let \" ß \# ß \\$ ß \% and \& be a random sample of size 8 œ & from the uniform distribution
on the interval Ð!ß )Ñ where ) is unknown. Let the estimator for ) be 5 † \7+B , where 5 is some
constant and \7+B is the largest observation. In order for 5 † \7+B to be an unbiased estimator,
the constant 5 must equal:
A) %&
B) &'
G ) " D) '&
E) (&
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 14
PE-269
17. Given the distribution 0 ÐBÑ œ )B)" ß !  B  ", and the random sample
.7 , .14 , .8 , .9 , .65 ; determine the method of moments estimator for ).
A) Less than 1.0
B) At least 1.0, but less than 1.3
C) At least 1.3, but less than 1.6
D) At least 1.6, but less that 1.9
E) At least 1.9
18. An estimate is to be made of the parameter ) in the distribution with pdf
0 ÐBÑ œ #B
)# for !  B  ) . Given a single sample value B, the estimate of ) is
s) œ #B . Find the mean square error of this estimator, Q WI) Ðs)Ñ .
A) !
#
B) )*
#
C) )'
#
D) )\$
E) )#
19. The method of moments is applied to estimate the mixing weights and Poisson parameters for
a mixture of two Poisson distributions. One of the Poisson mixture components has a mean of 5.
The estimates of the first two moments of the mixed distribution are:
• estimated first moment is 3
• estimated second moment is 14
What is the mixing weight for the Poisson mixture component with mean 5?
A) "%
B) "\$
C) "#
D) #\$
E) \$%
20. An aggregate claims random variable W has a compound distribution for which the frequency
"
prob. ;
has a geometric distribution, and the severity distribution is \ œ œ
.
2 prob. "  ;
The mean of W is 2.55 and the stop loss premium with a deductible of 1 is 1.95 .
Find the stop loss premium with a deductible of 2.
A) Less than 1.0
B) At least 1.0 but less than 1.1
C) At least 1.1 but less than 1.2
D) At least 1.2 but less than 1.3
E) At least 1.3
21. A school board is administering a standardized math test to students in a large number of
schools. The board estimated that the grade of a student is normally distributed with mean ) and
standard deviation 10, where ) varies from one school to another, but is constant for all students
within a school. The board also estimates that ) is normally distributed with a mean of 70 and a
standard deviation of 5. A particular school has 30 students taking the test and it is found that the
average grade for those 30 students is 65. The school board uses Bayesian analysis to determine
the posterior distribution of ) for students in that school.
Find the mean of that posterior distribution.
A) 65.0
B) 65.2
C) 65.4
D) 65.6
E) 65.8
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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PRACTICE EXAM 14
22. A model suggests that the mutually exclusive and exhaustive events E! ß E" ß E# and E\$ have
\$
probabilities given by T ÒE5 Ó œ Š 5 ‹ :5 Ð":Ñ\$5 , 5 œ !ß "ß #ß \$. The hypothesis L! À : œ #" is
to be tested using the chi-square goodness-of-fit statistic. The experiment is repeated 80
independent times and the respective frequencies of E! ß E" ß E# and E\$ are 15, 25, 35 and 5.
What are the values of the chi-square statistic and the degrees of freedom respectively?
A) (1.33, 3)
B) (6.67, 2)
C) (6.67, 3)
D) (25, 2)
E) (5, 3)
23. When survival data is truncated on the left at time P, so that events prior to time P are
unobservable, the Product-Limit estimator at time >  P is an estimate of which of the following?
A) WÐ>Ñ
WÐ>Ñ
B) WÐPÑ
C) WÐPÑ  WÐ>Ñ
WÐ>Ñ
D) "WÐPÑ
E)
WÐPÑWÐ>Ñ
"WÐ>Ñ
24. A company has a machine that occasionally breaks down. An insurer offers a warranty for
this machine. The number of breakdowns and their costs are independent. The number of
breakdowns each year is given by the following distribution:
Probability
# of breakdowns
0
50%
1
20%
2
20%
3
10%
The cost of each breakdown is given by the following distribution:
Probability
Cost
1,000
50%
2,000
10%
3,000
10%
5,000
30%
To reduce costs, the insurer imposes a deductible of 1,000 for each breakdown. Compute the
standard deviation of the insurer's losses for this year.
A) 1,359
B) 2,280
C) 2,919
D) 3,092
E) 3,434
25. R is the distribution of the number of claims occurring per week. R has a Poisson
distribution with an unknown mean. The standard for full credibility for R is based on the sample
mean of R being within 5% of the true mean of R with probability 90%.
With 400 observed claims in 20 weeks, the credibility premium based on partial credibility
is T . With 500 observed claims in 30 weeks, the credibility premium based on partial credibility is
T  "Þ*" . Find the credibility premium based on partial credibility if there are 550 observed claims in
35 weeks. Assume that the same manual premium is used in all cases.
A) Less than 15.1
B) At least 15.1, but less than 15.2
C) At least 15.2, but less than 15.3
D) At least 15.3, but less than 15.4
E) At least 15.4
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 14
PE-271
26. An insurer covers 60 independent risks. Each risk has a 4% probability of a loss in a year.
Calculate how often 5 or more risks would be expected to have losses in the same year.
A) Once every 3 years
B) Once every 7 years
C) Once every 11 years
D) Once every 14 years
E) Once every 17 years
27. The following data have been collected for a large insured:
Number of Average Claim
Year
Claims
Size
1
100
10,000
2
200
12,500
Inflation increases the size of all claims by 10% per year.
A Pareto distribution with parameters or α œ 3 and ) is used to model the claim size distribution.
Estimate ) for Year 3 using the method of moments.
A) 22,500
B) 23,333
C) 24,000
D) 25,850
E) 26,400
28. An insurer has data for 8 independent policies. Each policy has a ground up loss distribution
that is exponential. The policy information is as follows:
Policy
Payment
Deductible
Limit
Mean of Ground up Loss
1
10
None
None
)
2
10
None
10
)
3
10
5
None
)
4
10
5
10
)
5
10
None
None
#)
6
10
None
10
#)
7
10
5
None
\$)
8
10
5
10
\$)
Find the mle of ).
A) Less than 14.0
B) At least 14.0 but less than 14.5
C) At least 14.5 but less than 15.0
D) At least 15.0 but less than 15.5
E) At least 15.5
29. You are given the following grouped data on 100 losses based on underlying loss random
variable \ :
Interval
Number of Losses
Ð!ß "!!Ó
+
Ð"!!ß #!!Ó
,
Ð#!!ß &!!Ó
Ð&!!ß "!!!Ó
.
+ ß , ß - and . are non-negative integers.
You are also given IÒ\ • "!!Ó œ )! ß IÒ\ • #!!Ó œ "#! and IÒ\ • \$!!Ó œ "\$(Þ& .
Find J"!! Ð&!!Ñ the empirical estimate of the distribution function value J\ Ð&!!Ñ.
A) .91
B) .93
C) .95
D) .97
E) .99
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-272
PRACTICE EXAM 14
30. A driver is selected at random, If the driver is a "good" driver, he is from a Poisson
population with a mean of 1 claim per year. If the driver is a "bad" driver, he is from a Poisson
population with a mean of 5 claims per year. There is equal probability that the driver is either a
"good" driver or a "bad" driver. If the driver had 3 claims last year, calculate the probability that
the driver is a "good" driver".
A) Less than 0.325
B) At least 0.325, but less than 0.375
C) At least 0.375, but less than 0.425
D) At least 0.425, but less than 0.475
E) At least 0.475
31. \" has a normal distribution with a known mean of . and a known variance of 5 # .
For 5 œ #ß \$ß ÞÞÞ , we define the conditional distribution of \5 given \5" to be normal with a
#
mean of \5" and a variance of \5"
. Which of the following is the variance of \"! ?
#
#
#
A) &""5  &"".
B) &""5  &"#.#
C) &"#5 #  &"".#
#
#
#
D) &"#5  &"#.
E) &"#5
32. You are given the following:
- Claim sizes follow a gamma distribution with parameters α and ) œ Þ&.
- The prior distribution of α is assumed to be uniform on the interval (0, 4).
Determine the value of Buhlmann’s 5 for estimating the expected value of a claim.
A) .5
B) 1.0
C) 1.5
D) 2.0
E) 2.5
33. You are given the following random observations: 0.1
0.2
0.5
1.0
1.3
You test whether the sample comes from a distribution with probability density function:
0 ÐBÑ œ
#
,
Ð"BÑ\$
Calculate the Kolmogorov-Smirnov statistic.
A) 0.01
B) 0.06
C) 0.12
B!
D) 0.17
E) 0.19
34. Claim sizes this year are described by a 2-parameter Pareto distribution with parameters
) œ \$ß !!! and α œ %. What is the expected claim size per loss next year after 20% inflation
and the introduction of a \$200 deductible?
A) Less than 980
B) At least 980, but less than 1000
C) At least 1000, but less than 1020
D) At least 1020, but less than 1040
E) At least 1040
35. The random variable \ has pdf 0 ÐBÑ œ Ð)  "ÑB) for !  B  " . From a sample of
values, the maximum likelihood estimate of ) is found, say s), and its variance is estimated, say
s Òs)Ó. Which of the following is the correct expression for the estimated variance of T Ð\  " Ñ?
Z
#
s
s
s
s Òs)Ó
A) Ð " Ñ )" † 68Ð " Ñ
B) ÒÐ " Ñ )" † 68Ð " ÑÓ#
C) ÒÐ " Ñ )" † 68Ð " ÑÓ † Z
#
#
s Òs)Ó
D) ÒÐ "# Ñ )" † 68Ð "# ÑÓ# † Z
s
#
#
#
s Òs)Ó
E) ÒÐ "# Ñ )" † 68Ð "# ÑÓ\$ † Z
s
#
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 14
PE-273
ACTEX EXAM C/4 - PRACTICE EXAM 14 SOLUTIONS
1. This problem involves Bayesian estimation of a posterior distribution.
The prior distribution of - is exponential with ) œ \$ (mean \$) and the model distribution (of \
given -) is Poisson with mean -. This combination of prior and model distributions is a special
case of the more general situation in which the prior distribution of - is a gamma distribution
with parameters α and ) , and the model distribution (of \ given -) is Poisson with mean -. In
this general case, if sample data values \" ß \# ß ÞÞÞß \5 are obtained, then the posterior distribution
of - given \" ß \# ß ÞÞÞß \5 is also a gamma distribution, with parameters
αw œ α 
5
\3 , and )w œ 5 ))" .
3œ"
The exponential distribution is a special case of the gamma distribution in which α œ " .
In this problem, - has a gamma distribution with α œ " and ) œ \$ . \ , the annual claim count
distribution, has a Poisson distribution with mean -. The data available is the claim count during
the first year, a single observation of \ . Therefore, 5 œ " sample value, and the value of \ is
given to be 2. The posterior distribution of - is also a gamma distribution, with αw œ "  # œ \$
\$
and )w œ Ð"ÑÐ\$Ñ"
œ Þ(& . The first moment of this gamma distribution is (from the distribution
table) )'αw œ #Þ#& , and the second moment is Ð)w Ñ# Ðαw  "Ñαw œ 'Þ(& . The variance is the
second moment minus the square of the first 'Þ(&  Ð#Þ#&Ñ# œ #(
"' .
2. J ÐBÑ œ ÐÞ#&ÑÐ"  /B Ñ  ÐÞ(&ÑÐ"  /BÎ# Ñ .
We must solve for B from the equation ÐÞ#&ÑÐ"  /B Ñ  ÐÞ(&ÑÐ"  /BÎ# Ñ œ Þ( .
The equation can be written as Þ#&/B  Þ(&/BÎ#  Þ\$ œ ! .
Substituting C œ /BÎ# results in the quadratic equation C#  \$C  "Þ# œ ! .
\$„È\$# %Ð"Þ#Ñ
Solving for C we get C œ
œ Þ\$&(% ß  \$Þ\$&(% .
#
BÎ#
We ignore the negative root, since C œ /
must be  ! .
Solving for B, we get B œ  # 68ÐÞ\$&(%Ñ œ #Þ!' .
3. Hypothetical mean œ IÒ\l-Ó œ - and process variance œ Z +<Ò\l-Ó œ - .
. œ Expected hypothetical mean œ IÒ-Ó œ "# Ð"Ñ  "# Ð#Ñ œ \$# .
+ œ Variance of hypothetical mean œ Z +<Ò-Ó œ IÒ-# Ó  ÐIÒ-ÓÑ# Þ
IÒ-# Ó œ "# Ð# ‚ "# Ñ  "# Ð# ‚ ## Ñ œ & p + œ &  Ð \$# Ñ# œ ""
% .
@ œ Expected process variance œ IÒ-Ó œ \$# .
"
""
\$Î# œ "( .
" ""Î%
'
\$
Ð ""
"( ÑÐ#Ñ  Ð "( ÑÐ # Ñ
There is 8 œ " observed value, so ^ œ
œ \$"
"( .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-274
PRACTICE EXAM 14
4. The sample values can be reconstructed from the Nelson-Aalen estimate.
s
The first failure time is time 3, and LÐ\$Ñ
œ =)" œ Þ"#& , from which we get =" œ ".
Therefore, there is one failure at time 3.
s
Then, LÐ&Ñ
œ ")  =(# œ Þ%"!( , from which we get =# œ #.
There are two failures at time 5.
=
s
LÐ'Ñ
œ ")  #(  &\$ œ Þ'"!( , so =\$ œ "; there is one failure at time 6.
s
LÐ)Ñ
œ "  #  "  =% œ Þ)'!( , so =% œ "; there is one failure at time 8.
)
(
&
%
=
s
LÐ*Ñ
œ ")  #(  "&  "%  \$& œ "Þ"*%! , so =& œ "; there is one failure at time 9.
=
s
LÐ"!Ñ
œ "  #  "  "  "  ' œ #Þ"*%! , so =' œ #; there are two failures at time 10.
)
(
&
%
\$
#
The random sample values are \$ ß & ß & ß ' ß ) ß * ß "! ß "! .
# Ó œ " Ò%%!  )Ð(# ÑÓ œ ' .
The empirical estimate of the variance is ") ÒDB#3  )B
)
8
5. The Buhlmann credibility factor is ^ œ 8 @ , and the Buhlmann credibility estimate is
+

^\  Ð"  ^Ñ. . We are given . œ Þ%#& ß + œ Þ\$(! and @ œ "Þ(*\$ . For the 8 œ *

(
policyholders, the sample mean for number of claims is \ œ * . The credibility factor is
^œ
*
œ Þ'&! . The Buhlmann credibility estimate for number of claims for another
* "Þ(*\$
Þ\$(!
policyholder of the same class is ÐÞ'&ÑÐ (* Ñ  ÐÞ\$&ÑÐÞ%#&Ñ œ Þ'&%\$ . The credibility estimate for
5 policyholders of the same class is Ð&ÑÐÞ'&%\$Ñ œ \$Þ#( . Answer: D
6. The distribution of the random variable R is given by T ÒR œ 5Ó œ Ð":Ñ5" † : Ð5  "
failures followed by a success). The likelihood function (i.e., probability) that R trials are
required to obtain the first success is therefore Ð":ÑR " † : . The maximum likelihood estimator
for : is the value of : which maximizes Ð":ÑR " † : (given R ). Thus, if
.
R "
† : œ Ð":ÑR "  ÐR "Ñ † Ð":ÑR # † : œ !, then : œ R" . The second
.: Ð":Ñ
derivative test may be used to show that this value of : maximizes Ð":ÑR " † :. Answer: C
7. The Nelson-Aalen estimate of the cumulative hazard rate function to time >" is
="
"!! œ Þ!\$ p =" œ \$ . Then, the Nelson-Aalen estimate of the cumulative hazard rate function to
\$
"!!
=
\$
 *(#  *(= œ Þ"!\$%*& Þ By trial and error (substitution)
#
we find that =# œ % .
time >\$ is
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 14
PE-275
8. \ is the random variable for annual loss. We are given that the conditional distribution of
\ given ) is exponential with a mean of ), where ) has an unspecified distribution.
Therefore, the hypothetical mean is LQ œ IÒ\l)Ó œ )
and the process variance is T Z œ Z +<Ò\l)Ó œ )# .
The expected hypothetical mean is . œ ILQ œ IÒ IÒ\l)Ó Ó œ IÒ)Ó œ IÒ\Ó
(using the double expectation rule IÒ IÒ\l)Ó Ó œ ) ).
The expected process variance is @ œ IT Z œ IÒ Z +<Ò\l)Ó Ó œ IÒ)# Ó .
The variance of the hypothetical mean is
+ œ Z LQ œ Z +<Ò IÒ\l)Ó Ó œ Z +<Ò)Ó œ IÒ)# Ó  ÐIÒ)ÓÑ# .
From this we see that @  + œ ÐIÒ)ÓÑ# œ .# .
In general, Z +<Ò\Ó œ IÒ Z +<Ò\l)Ó Ó  Z +<Ò IÒ\l)Ó Ó œ @  + .
From the data set we can use empirical estimation to estimate IÒ\Ó:
&!!
!"!!
#&!
"!!#!!
"&!
.
Ñ  Ð "!!!
ÑÐ #!!&!!
Ñ
s œ Ð "!!! ÑÐ # Ñ  Ð "!!! ÑÐ
#
#
'!
%!
 Ð "!!!
ÑÐ &!!"!!!
Ñ  Ð "!!!
ÑÐ "!!!#!!!
Ñ œ ##! .
#
#
From the data set we can use empirical estimation to estimate IÒ\ # Ó:
\$
\$
\$
\$
\$
\$
&!!
"!! !
#&!
#!! "!!
"&!
&!! #!!
Ð "!!!
ÑÐ \$Ð"!!!Ñ
Ñ  Ð "!!!
ÑÐ \$Ð#!!"!!Ñ
Ñ  Ð "!!!
ÑÐ \$Ð&!!#!!Ñ
Ñ
\$
\$
\$
\$
'!
"!!! &!!
%!
#!!! "!!!
 Ð "!!!
ÑÐ \$Ð"!!!&!!Ñ
Ñ  Ð "!!!
ÑÐ \$Ð#!!!"!!!Ñ
Ñ œ "&&ß \$\$\$ .
The empirical estimate of the variance of \ is "&&ß \$\$\$  Ð##!Ñ# œ "!'ß *\$\$ .
In the semiparametric empirical Bayes credibility model, we use the empirical estimate of IÒ\Ó for .,
so that .
s œ ##!. We also know that Z +<Ò\Ó œ @  +, so using the empirical estimate of Z +<Ò\Ó gives
"!'ß *\$\$ œ [email protected]  s
+ . But we also know that, for this model, @  + œ .# , so using our sample estimate of .,
we have [email protected]  s
+ œ Ð##!Ñ# . We can then solve the two equations
+ œ "!'ß *\$\$ and [email protected]  s
+ œ %)ß %!! to get [email protected] œ ((ß ''( and s
+ œ #*ß #'( .
[email protected]  s
We can now find the estimated loss in the 3rd year for a policy that had losses of
s   Ð"  ^Ñ
s .
]" œ "&! in the first year and ]# œ ! in the second year. The estimate is ^]
s,
#
sœ
where ^
s œ ##!. The credibility premium is
[email protected] œ Þ%#*) and .
# s+
ÐÞ%#*)ÑÐ(&Ñ  ÐÞ&(!#ÑÐ##!Ñ œ "&) .
9. Since claims follow a Poisson process with rate - œ #\$ , the time between successive claims
has an exponential distribution with mean -" œ \$# Þ We wish to simulate exponential time between
claims. Since we are told that small numbers correspond to longer times, we use the simulation
algorithm ? œ "  J Ð>Ñ , where ? is the uniform number and > is the simulated claim time. For
the exponential distribution with mean 1.5, the cdf is J Ð>Ñ œ "  />Î"Þ& . The first simulated
claim time is found from Þ(& œ />Î"Þ& so > œ Þ%\$ is the time of the first claim. The time
between the first claim and second claim is the solution of Þ' œ />Î"Þ& so > œ Þ(( is the time
between the first and second claim, which implies that the second claim occurs at time 1.2. The
time between the second and third claim is the solution of Þ% œ />Î"Þ& so > œ "Þ\$( , so the third
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-276
PRACTICE EXAM 14
9. continued
claim occurs after the end of the second year. There are two claims simulated in the first two
years. We now simulate claim amounts. The claim amount distribution is
\
"
#
\$
:
Þ#&
Þ#&
Þ&
J
Þ#&
Þ&!
"Þ!!
We are told that small numbers correspond to small claim amounts, so we apply the usual inverse
transformation approach to simulate \ (using J ÐBÑ). The first uniform number for simulating
claim amounts is .3. Since Þ#&  Þ\$ Ÿ Þ& , the simulated claim amount is \ œ # . The second
uniform number for simulating claim amounts is .6. Since Þ&  Þ' Ÿ "Þ! , the simulated claim
amount is \ œ \$. Aggregate claims for the two years is 5.
The expected aggregate claim per year is
IÒR Ó † IÒ\Ó œ Ð #\$ ÑÒÐ"ÑÐÞ#&Ñ  Ð#ÑÐÞ#&Ñ  Ð\$ÑÐÞ&ÑÓ œ "Þ&
The variance of the compound Poisson annual aggregate claim distribution is
IÒR Ó † IÒ\ # Ó œ Ð #\$ ÑÒÐ"# ÑÐÞ#&Ñ  Ð## ÑÐÞ#&Ñ  Ð\$# ÑÐÞ&ÑÓ œ \$Þ)\$\$.
The annual premium is "Þ&  "Þ)È\$Þ)\$\$ œ &Þ!# .
The gain for the two years is premium minus claims, which is #Ð&Þ!#Ñ  & œ &Þ!% . Answer: E
10. For each Class the claims follow a Poisson process. For a particular Class, if the expected
amount of time between claims is >! (in years), then the expected number of claims per year is >" .
!
Class I has an expected amount of time of 1 year between claims, so that the expected number of
claims per year for Class 1 is -" œ " . Class II has an expected amount of time of 1.25 years (15
months) between claims, so that the expected number of claims per year for Class 1 is
"
-# œ "Þ#&
œ Þ) . Class III has an expected amount of time of 1.5 years (15 months) between
"
claims, so that the expected number of claims per year for Class 1 is -\$ œ "Þ&
œ \$# .
Let us denote the number of claims in year 1 by R" and the number of claims in year 2 is denoted
by R# . We wish to find IÒR# lR" œ !Ó , and then, since each claim cost is 1,000 in all classes,
the expected loss in year 2 is "!!!IÒR# lR" œ !Ó .
We find IÒR# lR" œ !Ó by conditioning over the Class type.
IÒR# lR" œ !Ó œ IÒR# lClass IÓ † T ÒClass IlR" œ !Ó
 IÒR# lClass IIÓ † T ÒClass IIlR" œ !Ó  IÒR# lClass IIIÓ † T ÒClass IIIlR" œ !Ó .
The conditional expectations are the expected number of claims in a year for a given class:
IÒR# lClass IÓ œ " , IÒR# lClass IIÓ œ Þ) and IÒR# lClass IIIÓ œ #\$ .
The conditional probabilities can be found from the following probability table:
Class I , T Ò I Ó œ "\$
Class II , T Ò II Ó œ "\$
" !
R" œ ! T ÒR" œ !lIÓ œ / !x†"
œ Þ\$'())
Þ)
!
/ \$ †Ð \$ Ñ!
T ÒR" œ !lIIÓ œ / !x†Þ) T ÒR" œ !lIIIÓ œ
!x
œ Þ%%*\$\$
œ Þ&"\$%#
T ÒR" œ ! ∩ IÓ
œ T ÒR" œ !lIÓ † T Ò I Ó
œ ÐÞ\$'())ÑÐ "\$ Ñ
Class III , T Ò III Ó œ "\$
#
#
T ÒR" œ ! ∩ IIÓ
T ÒR" œ ! ∩ IIIÓ
œ T ÒR" œ !lIIÓ † T Ò II Ó
œ T ÒR" œ !lIIIÓ † T Ò III Ó
œ ÐÞ%%*\$\$ÑÐ "\$ Ñ
œ ÐÞ&"%\$#ÑÐ "\$ Ñ
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 14
PE-277
10. continued
Then, T ÒR" œ !Ó œ T ÒR" œ ! ∩ IÓ  T ÒR" œ ! ∩ IIÓ  T ÒR" œ ! ∩ IIIÓ œ Þ%%\$)% , and
ÐÞ\$'())ÑÐ "\$ Ñ
T ÒClass I ∩ R" œ!Ó
œ
œ Þ#('\$ .
T ÒR" œ!Ó
Þ%%\$)%
"
ÐÞ%%*\$\$ÑÐ Ñ
T ÒClass II ∩ R" œ!Ó
IIlR# œ !Ó œ
œ Þ%%\$)% \$ œ Þ\$\$(& .
T ÒR" œ!Ó
ÐÞ&"%\$#ÑÐ "\$ Ñ
T ÒClass III ∩ R" œ!Ó
IIIlR# œ !Ó œ
œ
œ Þ\$)'\$
T ÒR" œ!Ó
Þ%%\$)%
T ÒClass IlR" œ !Ó œ
T ÒClass
T ÒClass
.
Then IÒR# lR" œ !Ó œ Ð"ÑÐÞ#('\$Ñ  ÐÞ)ÑÐÞ\$\$(&Ñ  Ð #\$ ÑÐÞ\$)'\$Ñ œ Þ)!% , and the expected loss
for year 2 is "!!!ÐÞ)!%Ñ œ )!% .
11. Since the :-value for the Poisson model is  .001, the Poisson model is rejected as a good
fit at any significance level above .001 (.1%). Since the :-value for the Binomial model is
between .05 and .1, the binomial model is rejected at the 10% level of significance or above, and
is accepted at any level of significance below 5%. Statement C is the only one satisfying both
requirements.
8


12. Original data set: \ œ "< \ 3 œ \$" Ò%  '  (Ó œ "(
@ œ # (given), 8 œ % ß
\$ . s
4œ"
"
+ œ <"
s
sœ
^
<


"( #
"( #
#
""
#
Ð\ 3  \ Ñ#  [email protected] œ #" ÒÐ%  "(
\$ Ñ  Ð'  \$ Ñ  Ð(  \$ Ñ Ó  % œ ' ,
3œ"
8
8 [email protected]
œ
%
#
% ""Î'
œ Þ()' .
8


Revised data set: < œ % ß \ œ "< \ 3 œ %" Ò%  '  (  \$Ó œ #!
% œ &Þ
4œ"
Original data set: # œ [email protected] œ "<
8
"
[email protected] œ \$ Ò@
s"  [email protected]#  [email protected]\$ Ó p [email protected]"  [email protected]#  [email protected]\$ œ ' .
4œ"
"
Revised data set: [email protected] œ "% Ò@
s"  [email protected]#  [email protected]\$  [email protected]% Ó œ % Ò'  "Ó œ "Þ(& .
<


"
+ œ <"
Ð\ 3  \ Ñ#  [email protected] œ \$" ÒÐ%  &Ñ#  Ð'  &Ñ#  Ð(  &Ñ#  Ð\$  &Ñ# Ó  "Þ(&
s
% œ #Þ*! ,
3œ"
sœ
^
8
8 [email protected]
œ
%
% "Þ(&
#Þ*!
œ Þ)'* . Percentage change is "!!Ð Þ)'*
Þ()'  "Ñ œ ""% . Answer: D
13. 0 ÐBÑ œ œ
Þ!!%
!  B  "!!
.
, † 0# ÐBÑ "!! Ÿ B  #!!
#!!
'!#!! 0 ÐBÑ .B œ " p '!"!! ÐÞ!!%Ñ .B  '"!!
Þ!", .B œ " p , œ Þ' .
#!!
"!!
#!!
IÒ\Ó œ '! B 0 ÐBÑ .B œ '! Þ!!%B .B  '"!! Þ!!'B .B œ #!  *! œ ""!Þ Answer: B
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-278
PRACTICE EXAM 14
14. \ has an exponential distribution with mean ) and an ordinary deductible of . is applied.
The cost per loss random variable ]P has mean IÒ]P Ó œ IÒÐ\  .Ñ Ó œ )/.Î) ,
and has variance Z +<Ò]P Ó œ )# /.Î) † Ð#  /.Î) Ñ œ #!ß %)! .
Since \ is exponential, the cost per payment (excess loss) random variable ]T is exponential
with mean ), and variance )# œ #&ß '!! . Therefore, ) œ "'! Þ
We then get the equation #&ß '!!/.Î) Ð#  /.Î) Ñ œ #!ß %)! , so that
/.Î) Ð#  /.Î) Ñ œ Þ)! . If we let - œ /.Î) , then we get a quadratic equation in - :
-Ð#  -Ñ œ Þ) , or - #  #-  Þ) œ ! . The roots are - œ Þ&&#) ß "Þ%%(# .
Since - œ /.Î) , we ignore the root greater than 1.
Then /.Î"'! œ Þ&&#) Ðso that . œ *%Þ)%Ñ .
The average cost per loss is IÒ]P Ó œ "'!ÐÞ&&#)Ñ œ ))Þ% . Answer: D
15. Z +<ÒWÓ œ IÒZ +<ÒWl>C:/ÓÓ  Z +<ÒIÒWl>C:/ÓÓ
Z +<ÒWl>C:/ "Ó œ Ð"ÑÐ &# Ñ ß Z +<ÒWl>C:/ #Ó œ Ð"ÑÐ\$Ñ p IÒZ +<ÒWl>C:/ÓÓ œ "# Ò &#  \$Ó œ ""
%Þ
\$
&
\$
& # " "
"
IÒWl>C:/ "Ó œ Ð"ÑÐ # Ñ ß IÒWl>C:/ #Ó œ Ð"ÑÐ \$ Ñ p Z +<ÒIÒWl>C:/ÓÓ œ Ð #  \$ Ñ Ð # ÑÐ # Ñ œ "%%
"
\$*(
Z +<ÒWÓ œ ""
%  "%% œ "%% .
Alternatively,
"
#
#
#
#
IÒW # Ó  ÐIÒWÓÑ# œ IÒW # Ó  Ð "*
"# Ñ ß IÒW Ó œ # ÐIÒW" Ó  IÒW # ÓÑ
\$(*
œ "# ÒZ +<ÒW" Ó  IÒW" Ó#  Z +<ÒW# Ó  IÒW# Ó# Ó œ "# Ò &#  *%  \$  #&
* Ó œ (# Þ
"* #
\$*(
Z +<ÒWÓ œ \$(*
(#  Ð "# Ñ œ "%% Þ
16. Using the methods of order statistics, \7+B œ ]& , the fifth (i.e., largest) order statistic from
a random sample of size 5. The density function for \ is 0 ÐBÑ œ ") , !  B  ), and the
cumulative distribution function for \ is J ÐBÑ œ B) for !  B  ), and J ÐBÑ œ ! for B  !
and J ÐBÑ œ " for B  ). Using the methods of order statistics, the density function of ]& can be
&C%
found to be 1& ÐC& Ñ œ &ÒJ ÐC& ÑÓ&" † 0 ÐC& Ñ œ )&& if !  C&  ), and it is 0 otherwise. Then,
&C%
)
IÒ]& Ó œ '! C& † )&& .C& œ &') . In order for 5 † \7+B œ 5 † ]& to be an unbiased estimator for )
we must have IÒ5 † \7+B Ó œ 5 † IÒ]& Ó œ &5') œ ). Thus, 5 œ '& .
17. The distribution has one parameter so that according to the method of moments, we set the
distribution mean equal to the sample mean:
"
"
"

)
IÒ\Ó œ '! B 0 ÐBÑ .B œ '! B )B)" .B œ '! )B) .B œ )"
ß \ œ Þ'\$)
p ) œ Þ'\$) p s) œ "Þ(' .
)"
18. Q WI) Ðs)Ñ œ Z +<Ðs)Ñ  Ò,) Ðs)ÑÓ# œ Z +<Ðs)Ñ  ÒIÐs)Ñ  )Ó# .
)
#)
For the given distribution, IÒ\Ó œ '! B † #B
)# .B œ \$ ,
)
)
#) #
)
#
#
IÒ\ # Ó œ '! B# † #B
)# .B œ # , Z +<Ò\Ó œ IÒ\ Ó  ÐIÒ\ÓÑ œ #  Ð \$ Ñ œ ") .
#
#
s) œ #B p IÒs)Ó œ #IÒ\Ó œ %) , Z +<Òs)Ó œ %Z +<Ò\Ó œ % † ) œ #) .
\$
")
*
#
#
Q WI) Ðs)Ñ œ #)  Ò %)  )Ó# œ ) .
#
)
*
\$
#
#
\$
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 14
PE-279
19Þ The second moment of a Poisson distribution with mean - is -#  - .
Suppose the second mixture component has a mean of -, and suppose that the mixture component
with mean 5 has mixing weight +. Then the mixture component with mean - has mixing
weight "  +. We have the following moment equations:
&+  -Ð"  +Ñ œ \$ ß Ð#&  &Ñ+  Ð-#  -ÑÐ"  +Ñ œ "% .
From the second equation we get \$!+  Ð"  -Ñ-Ð"  +Ñ œ "% ,
and from the first equation we get -Ð"  +Ñ œ \$  &+
so that \$!+  Ð"  -ÑÐ\$  &+Ñ œ "% , from which we get - œ ""#&+
\$&+ .
But from the first equation we have - œ \$&+
"+ .
""#&+
"
Therefore, \$&+
"+ œ \$&+ resulting in an equation + which has solution + œ \$ . Answer: B
20. Let us denote the geometric distribution parameter with the usual notation " .
Then IÒR Ó œ " and IÒ\Ó œ "  #Ð"  ;Ñ œ #  ; ,
and IÒWÓ œ IÒR Ó ‚ IÒ\Ó œ " ‚ Ð#  ;Ñ œ #Þ&& .
"
IÒW • "Ó œ " ‚ T ÐW "Ñ œ "  T ÐW œ !Ñ œ "  T ÐR œ !Ñ œ "  ""
Þ
We are given that IÒÐW  "Ñ Ó œ "Þ*& , so that
"
IÒÐW  "Ñ Ó œ IÒWÓ  IÒW • "Ó œ " ‚ Ð#  ;Ñ  Ò"  ""
Ó œ "Þ*& .
"
It follows that "Þ&&  ""
œ "Þ*& from which we get " œ "Þ& ,
and then from " ‚ Ð#  ;Ñ œ #Þ&& we get ; œ Þ\$ .
The stop loss premium with a deductible of 2 is
IÒÐW  #Ñ Ó œ IÒWÓ  IÒW • #Ó œ #Þ&&  Ò" ‚ 0W Ð"Ñ  # ‚ T ÐW #ÑÓ .
"
"Þ&
0W Ð"Ñ œ T ÐW œ "Ñ œ T ÐR œ "Ñ ‚ T Ð\ œ "Ñ œ Ð"" Ñ# ‚ ; œ Ð#Þ&Ñ
# ‚ ÐÞ\$Ñ œ Þ!(# ,
and T Ð\ #Ñ œ "  0W Ð!Ñ  0W Ð"Ñ œ "  Þ%  Þ!(# œ Þ&#) .
Then IÒÐW  #Ñ Ó œ #Þ&&  ÒÞ!(#  #ÐÞ&#)ÑÓ œ "Þ%## . Answer: E
21. We use the following Bayesian relationship. If \ given ) is normal with mean ) and
variance 5 # (5 # is non-random), and if the prior distribution of @ is normal with mean . and
variance +, then the posterior distribution of @ is normal with mean

.
"
8
"
#
( 8B
5 #  + )Î ( 5 #  + ) and variance 8 " . We are given that 5 œ "!! ,
5#
+
. œ (! and + œ #& . The 8 œ \$! students have an average grade of 
B œ '&.
Ð\$!ÑÐ'&Ñ
The posterior distribution of ) has mean Ò "!!
(!
\$!
"
 #&
ÓÎÐ "!!
 #&
Ñ œ '&Þ' . Answer: D
22. Note that T ÒE! Ó œ T ÒE\$ Ó œ ") and T ÒE" Ó œ T ÒE# ] œ \$) under the hypothesis L! . Define
the random variables \3 , 3 œ 0ß "ß #ß \$ as follows: \3 œ the number of occurrences of E3 in 80
trials. Then under the assumption of L! , the random variables \! ß \" ß \# ß \\$ have a
multinomial distribution with parameters 80, ") ß \$) ß \$) ß ") . Let us define the statistic U as follows:
\$
Uœ
3œ!
Ð\3 )!:3 Ñ#
)!:3
where :3 œ T ÒE3 Ó under L! . Since there is no estimation, U has a chi-square
distribution with % œ  " œ \$ degrees of freedom (this is the chi-square goodness-of-fit test
statistic). The value of U based on the given results of the 80 trials is:
U œ
Ò"&)!Ð ") ÑÓ#
8!Ð ") Ñ

Ò2&)!Ð \$) ÑÓ#
80Ð \$) Ñ

Ò3&)!Ð \$) ÑÓ#
8!Ð \$) Ñ

Ò&)!Ð ") ÑÓ#
8!Ð ") Ñ
œ #!
\$ œ 'Þ'(.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-280
PRACTICE EXAM 14
23. Truncation of the left at time P means that only events occurring after time P can be
observed, and therefore must have survived to time P. For >  P, the Product-Limit estimator
estimates is the probability of survival to time > given that the individual had survived to time P,
WÐ>Ñ
WÐPÑ
Þ
24. After the deductible is imposed on a breakdown, the cost per breakdown \ has distribution
Cost:
0
1000
2000
4000
Probability:
50%
10%
10%
30%
The insurer's losses in a year has a compound distribution with frequency R as indicated in the
question and severity \ given above. The variance of the insurer's total losses for the year is
IÒR Ó † Z +<Ò\Ó  Z +<ÒR Ó † ÐIÒ\ÓÑ# Þ
IÒR Ó œ Ð!ÑÐÞ&Ñ  Ð"ÑÐÞ#Ñ  Ð#ÑÐÞ#Ñ  Ð\$ÑÐÞ"Ñ œ Þ* ß
IÒR # Ó œ Ð!# ÑÐÞ&Ñ  Ð"# ÑÐÞ#Ñ  Ð## ÑÐÞ#Ñ  Ð\$# ÑÐÞ"Ñ œ "Þ*ß
Z +<ÒR Ó œ "Þ*  ÐÞ*Ñ# œ "Þ!* Þ
IÒ\Ó œ Ð!ÑÐÞ&Ñ  Ð"!!!ÑÐÞ"Ñ  Ð#!!!ÑÐÞ"Ñ  Ð%!!!ÑÐÞ\$Ñ œ "ß &!! ß
IÒ\ # Ó œ Ð!# ÑÐÞ&Ñ  Ð"!!!# ÑÐÞ"Ñ  Ð#!!!# ÑÐÞ"Ñ  Ð%!!!# ÑÐÞ\$Ñ œ &ß \$!!ß !!! ß
Z +<Ò\Ó œ &ß \$!!ß !!!  Ð"ß &!!Ñ# œ \$ß !&!ß !!! Þ
Then the variance of the insurer's losses for the year is
ÐÞ*ÑÐ\$ß !&!ß !!!Ñ  Ð"Þ!*ÑÐ"ß &!!Ñ# œ &ß "*(ß &!! Þ
The standard deviation is È&ß "*(ß &!! œ #ß #)! Þ
25. Since R is Poisson, the full credibility standard for estimating the mean of R is either
Z +<ÐR Ñ
(i) "!)#Þ% † ÒIÐR ÑÓ# œ "!)#Þ% † --# œ "!)#Þ%
as the expected number of exposures of R (weeks) needed, or
Z +<ÐR Ñ
(ii) "!)#Þ% † IÐR Ñ œ "!)#Þ% † -- œ "!)#Þ% as the total expected number of claims needed.
Since we do not know the value of -, the only standard we can apply is (ii).
With 400 claims in 20 weeks, the average number of claims per week (sample mean) is

R œ %!!
#! œ #! . Using credibility standard (ii) above, the partial credibility factor is
%!!
^ œ É "!)#Þ%
œ Þ'!(* , and the partial credibility premium is

^ † R  Ð"  ^Ñ † Q œ "#Þ"'  Þ\$*#"Q œ T , where Q is the manual premium .
With 500 claims in 30 weeks, the average number of claims per week (sample mean) is

R œ &!!
\$! œ "'Þ'''( . Using credibility standard (ii) above, the partial credibility factor is
&!!
^ œ É "!)#Þ%
œ Þ'(*( , and the partial credibility premium is

^ † R  Ð"  ^Ñ † Q œ ""Þ\$\$  Þ\$#!\$Q œ T  "Þ*" , where Q is the manual premium.
From the two equations, "#Þ"'  Þ\$*#"Q œ T and ""Þ\$\$  Þ\$#!\$Q œ T  "Þ*" ,
we get Q œ "&Þ!% and T œ ")Þ!% . Then, with 550 claims in 35 weeks, we have

R œ &&!
\$& œ "&Þ("%\$ . Using credibility standard (ii) above, the partial credibility factor is
&&!
^ œ É "!)#Þ%
œ Þ("#) , and the partial credibility premium is

^ † R  Ð"  ^Ñ † Q œ ""Þ#!  ÐÞ#)(#ÑÐ"&Þ!%Ñ œ "&Þ&# .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 14
PE-281
26. We define the event I œ 5 or more losses occur in the same year,
and we define a year to be a success if event I occurs, and let U œ T ÒIÓ. We let ] be the year
in which the first success occurs. Then ] has a geometric distribution with probabilities
T Ò] œ "Ó œ U ß T Ò] œ #Ó œ Ð"  UÑU ß T Ò] œ \$Ó œ Ð"  UÑ#U ß ÞÞÞ
"
The expected value of ] is U
(in the Exam C table the geometric starts at 0; in this problem we
consider this is the version of the geometric that starts at 1, it is the geometric in the table +1).
We now find the probability U œ T Ò5 or more losses in one yearÓ
\ , the number of risks experiencing a loss in a year has a binomial distribution with 8 œ '! and
; œ Þ% (each risk either has a loss or doesn't; let us define success for a particular risk meaning
that the risk has a loss for in the year).
U œ T Ò\ &Ó œ "  T Ò\ œ !ß "ß #ß \$ß %Ó
'!
'!
'!
œ "  Òˆ ! ‰ÐÞ!%Ñ! ÐÞ*'Ñ'!  ˆ " ‰ÐÞ!%Ñ" ÐÞ*'Ñ&*  ˆ # ‰ÐÞ!%Ñ# ÐÞ*'Ñ&)
 ˆ \$ ‰ÐÞ!%Ñ\$ ÐÞ*'Ñ&(  ˆ % ‰ÐÞ!%Ñ% ÐÞ*'Ñ&' Ó
"
œ "  Þ*!)#'# œ Þ!*"(\$) . Then U
œ "!Þ* years.
'!
'!
27. The 100 claims in year 1 would have an average claim size of "!ß !!!Ð"Þ"ÑÐ"Þ"Ñ œ "#ß "!!
in year three after including the effect of inflation for two years. The 200 claims in year 2 would
have an average claim size of "#ß &!!Ð"Þ"Ñ œ "\$ß (&! in year three after including the effect of
inflation for one year.
The 300 claims can be regarded as data for year 3 with an average (after inflation is included) per
Ð"!!ÑÐ"#ß"!!ÑÐ#!!ÑÐ"\$ß&!!Ñ
œ "\$ß #!! . The Pareto distribution
\$!!
)
mean of α"
, which becomes #) since we are given that α œ \$.
claim of
with parameters α and )
has a
According to the method
of moments, when a distribution has one unknown parameter, the estimate of the parameter is
found by equating the first moment of the distribution (the mean of the distribution) to the first
empirical moment (the sample mean of the data). The moment equation is )# œ "\$ß #!! , so that
) œ #'ß %!! . Answer: E
28. The likelihood function will be
0 Ð"&Ñ
"J Ð"&Ñ
0 Ð"&Ñ
"J Ð"&Ñ
\$
(
0" Ð"!Ñ † Ò"  J# Ð"!ÑÓ † "J
† "J% Ð&Ñ † 0& Ð"!Ñ † Ò"  J' Ð"!ÑÓ † "J
† "J) Ð&Ñ †
\$ Ð&Ñ
%
( Ð&Ñ
)
For policy 3, the payment is 10 after a deductible of 5, so the ground up loss is 15. A similar
comment applies to other factors in the likelihood. Also, for policy 4, the policy limit is ?  . ,
where ? œ "& is the maximum covered loss, so that the policy limit is "&  & œ "!, but a limit
payment is triggered by a ground up loss greater than 15.
This likelihood becomes
" "&Î\$)
" "&Î)
"&Î)
"&Î\$)
/
/
PÐ)Ñ œ )" /"!Î) † /"!Î) † )/&Î) † //&Î) † #") /"!Î#) † /"!Î#) † \$)/&Î\$) † //&Î\$)
(!Î)
œ '")% //"\$Þ\$\$Î) œ '")% /&'Þ'(Î) Þ The mle of ) is s) œ &'Þ'(
% œ "%Þ"(.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-282
PRACTICE EXAM 14
29. If we find +ß ,ß and - , then J"!! Ð&!!Ñ œ +,"!! .
&!+"!!Ð"!!+Ñ
œ "!!  Þ&+ œ )! p + œ %! .
"!!
%!Ð&!Ñ"&!,#!!Ð'!,Ñ
IÒ\ • #!!Ó œ
œ "%!  Þ&, œ "#! p , œ %! .
"!!
%!Ð&!Ñ%!Ð"&!Ñ \$ Ð#&!ÑÐ#! \$- ÑÐ\$!!Ñ
IÒ\ • \$!!Ó œ
œ "%!  Þ&"!!
\$ œ "\$(Þ&
%!%!"&
Then J"!! Ð&!!Ñ œ
œ Þ*& .
"!!
IÒ\ • "!!Ó œ
p - œ "& .
30. The distribution of the number of claims of a randomly chosen driver is a mixture of the two
Poisson distributions. Since there is an equal probability of choosing a "good" or "bad" driver,
the mixing probabilities are .5 for each of those two driver types. We wish to find
T Ògood driverl3 claims last yearÓ . We use the rules for conditional probability in this classical
T ÒF∩EÓ
T ÒEÓ .
T Ò(good driver)∩(3 claims last year)Ó
yearÓ œ
T Ð3 claims last yearÑ
Bayesian question. T ÒFlEÓ œ
T Ògood driverl3 claims last
.
Using the rule T ÐG ∩ HÑ œ T ÐGl.Ñ † T ÐHÑ , the numerator can be written as
T Ò(good driver) ∩ (3 claims last year)Ó œ T Ð3 claims last year l good driverÑ † T Ðgood driverÑ
"
\$
"
œ / \$x†" † "# œ /"# œ Þ!\$!( .
To find the denominator, we use the rule T ÐIÑ œ T ÐI ∩ J Ñ  T ÐI ∩ J w Ñ , so that
T Ð3 claims last yearÑ œ T Ò(3 claims last year) ∩ (good driver)Ó
 T Ò(3 claims last year) ∩ (bad driver)Ó .
"
We already have T Ò(3 claims last year) ∩ (good driver)Ó œ /"# œ Þ!\$!( .
In a similar way we get
T Ò(3 claims last year) ∩ (bad driver)Ó œ T Ð3 claims last year lbad driverÑ † T Ðbad driverÑ
&
\$
œ / \$x†& † "# œ Þ!(!# .
Then T Ð3 claims last yearÑ œ Þ!\$!(  Þ!(!# œ Þ"!!* , and
T Ðgood driverl3 claims last yearÑ œ Þ!\$!(
Þ"!!* œ Þ\$!% .
These calculations can be summarized in the following probability table, where we have
F œ bad driver , K œ good driver , \$G œ 3 claims last year.
\$G
K , T ÐKÑ œ Þ&
F , T ÐFÑ œ Þ&
" \$
T Ð\$GlKÑ œ / \$x†" œ Þ!'"\$
& \$
T Ð\$GlFÑ œ / \$x†& œ Þ"%!%
T Ð\$G ∩ KÑ œ T Ð\$GlKÑ † T ÐKÑ
œ ÐÞ!'"\$ÑÐÞ&Ñ œ Þ!\$!(
T Ð\$G ∩ FÑ œ T Ð\$GlFÑ † T ÐFÑ
œ ÐÞ"%!%ÑÐÞ&Ñ œ Þ!(!#
T Ð\$GÑ œ T Ð\$G ∩ KÑ  T Ð\$G ∩ FÑ œ Þ!\$!(  Þ!(!# œ Þ"!!* .
T ÐKl\$GÑ œ
T ÐK∩\$GÑ
T Ð\$GÑ
Þ!\$!(
œ Þ"!!*
œ Þ\$!%
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PRACTICE EXAM 14
PE-283
#
31. Z +<Ò\5 Ó œ Z +<ÒIÒ\5 l\5" ÓÓ  IÒZ +<Ò\5 l\5" ÓÓ œ Z +<Ò\5"Ó  IÒ\5"
Ó,
and IÒ\5 Ó œ IÒIÒ\5 l\5" ÓÓ œ IÒ\5" Ó .
#
Note that IÒ\5"
Ó œ Z +<Ò\5" Ó  ÐIÒ\5" ÓÑ# , so that
Z +<Ò\5 Ó œ #Z +<Ò\5" Ó  ÐIÒ\5" ÓÑ# .
We also note that sinceIÒ\5 Ó œ IÒ\5" Ó , it follows that IÒ\5 Ó œ . for all 5 so that
Z +<Ò\5 Ó œ #Z +<Ò\5" Ó  .# .
We find these successively for 5 œ #ß \$ß ÞÞÞ to se if a pattern is established.
Z +<Ò\# Ó œ #Z +<Ò\" Ó  .# œ #5 #  .# .
Z +<Ò\\$ Ó œ #Z +<Ò\# Ó  .# œ #Ð#5 #  .# Ñ  .# œ ## 5 #  Ð##  "Ñ.# .
Z +<Ò\% Ó œ #Z +<Ò\\$ Ó  .# œ #Ð## 5 #  Ð##  "Ñ.# Ñ  .# œ #\$ 5 #  Ð#\$  "Ñ.# . . .
Z +<Ò\"! Ó œ #* 5 #  Ð#*  "Ñ.# œ &"#5 #  &"".# . Answer: C
32. The hypothetical mean given α is )α œ Þ&α , and the variance of the hypothetical mean is
#
%
Z LQ œ Z +<ÐÞ&αÑ œ Þ#&Z +<ÐαÑ œ ÐÞ#&ÑÐ "#
Ñ œ "\$ œ + .
The process variance is T Z œ Z +<Ò\lαÓ, the variance of the gamma distribution, which is
)# α œ Þ#&α , and the expected value of the process variance is
"Î#
IT Z œ IÒÞ#&αÓ œ ÐÞ#&ÑÐ#Ñ œ Þ& œ @ . Then 5 œ [email protected] œ "Î\$ œ "Þ& .Answer: C
#
"
33. The distribution function is J ÐBÑ œ '! 0 Ð>Ñ .> œ '! Ð">Ñ
\$ .> œ "  Ð">Ñ# .
The Kolmogorov-Smirnov statistic is Q +BÖlJ ÐB3 Ñ  J8 ÐB
3 Ñl ß lJ ÐB3 Ñ  J8 ÐB3 Ñl×
where the maximum is taken over all data points. J8 ÐB
Ñ
œ
J8 ÐB3" Ñ .
3
J8 is the empirical distribution function.

B3
J ÐB3 Ñ
J8 ÐB
lJ ÐB3 Ñ  J8 ÐB3Ñl
3 Ñß J8 ÐB3 Ñ lJ ÐB3 Ñ  J8 ÐB3 Ñl
Þ"
Þ"(\$'
!
ß Þ#
Þ"(\$'
Þ!#'%
Þ#
Þ\$!&'
Þ#
ß Þ%
Þ"!&'
Þ!*%%
Þ&
Þ&&&'
Þ%
ß Þ'
Þ"&&'
Þ!%%%
"Þ!
Þ(&!!
Þ'
ß Þ)
Þ"&
Þ!&
"Þ\$
Þ)""!
Þ)
ß "Þ!
Þ!""!
Þ")*
The maximum is .189 .
B
B
34. Expected claim size per loss next year will be
IÒÐ"Þ#\  #!!Ñ Ó œ IÒ"Þ#\Ó  IÒ"Þ#\ • #!!Ó œ "Þ#ÐIÒ\Ó  IÒ\ • #!!
"Þ# ÓÑ .
)
)
) α"
IÒ\Ó œ α"
œ \$!!!
Ó (from the Exam C Table),
%" œ "!!! ß IÒ\ • .Ó œ α" Ò"  Ð .) Ñ
#!!
\$!!!
\$!!!
%"
so that IÒ\ • "Þ# Ó œ %" Ò"  Ð #!! \$!!! Ñ Ó œ "%*Þ(% Þ
"Þ#
IÒÐ"Þ#\  #!!Ñ Ó œ Ð"Þ#ÑÒ"!!!  "%*Þ(%Ó œ "!#! .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
PE-284
PRACTICE EXAM 14
"Î#
35. T Ò\  "# Ó œ '! Ð)  "ÑB) .B œ Ð "# Ñ)" œ 1Ð)Ñ .
Z +<Ò1Ðs)ÑÓ œ Ò1w Ð)ÑÓ# † Z Òs)Ó . 1w Ð)Ñ œ Ð " Ñ)" † 68Ð " Ñ .
#
#