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**BEGINNING OF EXAMINATION**
1.
You are given:
(i)
A random sample of five observations from a population is:
0.2
(ii)
0.7
1.1
1.3
You use the Kolmogorov-Smirnov test for testing the null hypothesis, H 0 , that the
probability density function for the population is:
f ( x) =
(iii)
0.9
4
(1 + x )
5
, x>0
Critical values for the Kolmogorov-Smirnov test are:
Level of Significance
Critical Value
0.10
1.22
0.05
1.36
n
n
0.025 0.01
1.48 1.63
n
n
Determine the result of the test.
(A)
Do not reject H 0 at the 0.10 significance level.
(B)
Reject H 0 at the 0.10 significance level, but not at the 0.05 significance level.
(C)
Reject H 0 at the 0.05 significance level, but not at the 0.025 significance level.
(D)
Reject H 0 at the 0.025 significance level, but not at the 0.01 significance level.
(E)
Reject H 0 at the 0.01 significance level.
Exam C: Spring 2005
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2.
You are given:
(i)
The number of claims follows a negative binomial distribution with parameters r and
β = 3.
(ii)
Claim severity has the following distribution:
Claim Size
(iii)
Probability
1
0.4
10
0.4
100
0.2
The number of claims is independent of the severity of claims.
Determine the expected number of claims needed for aggregate losses to be within 10% of
expected aggregate losses with 95% probability.
(A)
Less than 1200
(B)
At least 1200, but less than 1600
(C)
At least 1600, but less than 2000
(D)
At least 2000, but less than 2400
(E)
At least 2400
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3.
You are given:
(i)
A mortality study covers n lives.
(ii)
None were censored and no two deaths occurred at the same time.
(iii)
tk = time of the k th death
(iv)
39
A Nelson-Aalen estimate of the cumulative hazard rate function is Hˆ (t2 ) =
.
380
Determine the Kaplan-Meier product-limit estimate of the survival function at time t9 .
(A)
Less than 0.56
(B)
At least 0.56, but less than 0.58
(C)
At least 0.58, but less than 0.60
(D)
At least 0.60, but less than 0.62
(E)
At least 0.62
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4.
Three observed values of the random variable X are:
1
1
4
You estimate the third central moment of X using the estimator:
g ( X1, X 2 , X 3 ) =
3
1
Xi − X )
(
∑
3
Determine the bootstrap estimate of the mean-squared error of g.
(A)
Less than 3.0
(B)
At least 3.0, but less than 3.5
(C)
At least 3.5, but less than 4.0
(D)
At least 4.0, but less than 4.5
(E)
At least 4.5
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You are given the following p-p plot:
1.0
0.8
F(x)
5.
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
Fn(x)
The plot is based on the sample:
1
2
3
15
30
50
51
99
100
Determine the fitted model underlying the p-p plot.
(A)
F(x) = 1 – x − 0.25 , x ≥ 1
(B)
F(x) = x / (1 + x), x ≥ 0
(C)
Uniform on [1, 100]
(D)
Exponential with mean 10
(E)
Normal with mean 40 and standard deviation 40
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6.
You are given:
(i)
Claims are conditionally independent and identically Poisson distributed with
mean Θ .
(ii)
The prior distribution function of Θ is:
⎛ 1 ⎞
F (θ ) = 1 − ⎜
⎟
⎝ 1+θ ⎠
2.6
,
θ >0
Five claims are observed.
Determine the Bühlmann credibility factor.
(A)
Less than 0.6
(B)
At least 0.6, but less than 0.7
(C)
At least 0.7, but less than 0.8
(D)
At least 0.8, but less than 0.9
(E)
At least 0.9
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7.
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
50
−
75
125
(1,000 , 5,000)
150
150
300
(5,000 , 10,000)
100
200
300
At 5,000
40
80
120
At 10,000
10
20
30
400
525
925
(300 , 500]
50
(500 , 1,000]
Total
50
Using the Kaplan-Meier approximation for large data sets, with α = 1 and β = 0 , estimate
F(5000).
(A)
0.25
(B)
0.32
(C)
0.40
(D)
0.51
(E)
0.55
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8.
You are given the following knots:
(0,0)
(2,0)
and derivative values:
f ′ ( 0 ) = −2
f ′( 2) = 2 .
Determine the value of the squared norm measure of curvature for the cubic spline that
satisfies these conditions.
(A)
16/15
(B)
8/3
(C)
4
(D)
8
(E)
16
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9-10.
Use the following information for questions 9 and 10.
The time to an accident follows an exponential distribution. A random sample of size two
has a mean time of 6.
Let Y denote the mean of a new sample of size two.
9.
Determine the maximum likelihood estimate of Pr (Y > 10 ) .
(A)
0.04
(B)
0.07
(C)
0.11
(D)
0.15
(E)
0.19
Exam C: Spring 2005
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9-10.
(Repeated for convenience) Use the following information for questions 9 and 10.
The time to an accident follows an exponential distribution. A random sample of size two
has a mean time of 6.
Let Y denote the mean of a new sample of size two.
10.
Use the delta method to approximate the variance of the maximum likelihood estimator of
FY (10) .
(A)
0.08
(B)
0.12
(C)
0.16
(D)
0.19
(E)
0.22
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11.
You are given:
(i)
The number of claims in a year for a selected risk follows a Poisson distribution with
mean λ .
(ii)
The severity of claims for the selected risk follows an exponential distribution with
mean θ .
(iii)
The number of claims is independent of the severity of claims.
(iv)
The prior distribution of λ is exponential with mean 1.
(v)
The prior distribution of θ is Poisson with mean 1.
(vi)
A priori, λ and θ are independent.
Using Bühlmann’s credibility for aggregate losses, determine k.
(A)
1
(B)
4/3
(C)
2
(D)
3
(E)
4
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12.
A company insures 100 people age 65. The annual probability of death for each person is
0.03. The deaths are independent.
Use the inversion method to simulate the number of deaths in a year. Do this three times
using:
u1 = 0.20
u2 = 0.03
u3 = 0.09
Calculate the average of the simulated values.
(A)
1
(B)
1
(C)
5
(D)
7
(E)
3
3
3
3
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13.
You are given claim count data for which the sample mean is roughly equal to the sample
variance. Thus you would like to use a claim count model that has its mean equal to its
variance. An obvious choice is the Poisson distribution.
Determine which of the following models may also be appropriate.
(A)
A mixture of two binomial distributions with different means
(B)
A mixture of two Poisson distributions with different means
(C)
A mixture of two negative binomial distributions with different means
(D)
None of (A), (B) or (C)
(E)
All of (A), (B) and (C)
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14.
You are given:
(i)
Annual claim frequencies follow a Poisson distribution with mean λ.
(ii)
The prior distribution of λ has probability density function:
1
6
π ( λ ) = (0.4) e − λ / 6 + (0.6)
1 − λ /12
e
,
12
λ >0
Ten claims are observed for an insured in Year 1.
Determine the Bayesian expected number of claims for the insured in Year 2.
(A)
9.6
(B)
9.7
(C)
9.8
(D)
9.9
(E)
10.0
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15.
Twelve policyholders were monitored from the starting date of the policy to the time of first
claim. The observed data are as follows:
Time of First Claim
Number of Claims
1
2
2
1
3
2
4
2
5
1
6
2
7
2
Using the Nelson-Aalen estimator, calculate the 95% linear confidence interval for the
cumulative hazard rate function H(4.5).
(A)
(0.189, 1.361)
(B)
(0.206, 1.545)
(C)
(0.248, 1.402)
(D)
(0.283, 1.266)
(E)
(0.314, 1.437)
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16.
For the random variable X, you are given:
(i)
E[ X ] = θ ,
(ii)
Var ( X ) =
(iii)
θˆ =
(iv)
MSE θˆ (θ ) = 2 ⎡⎣ bias θˆ (θ ) ⎤⎦
θ >0
θ2
25
k
X,
k +1
k >0
2
Determine k.
(A)
0.2
(B)
0.5
(C)
2
(D)
5
(E)
25
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17.
You are given:
(i)
The annual number of claims on a given policy has a geometric distribution with
parameter β.
(ii)
The prior distribution of β has the Pareto density function
π (β ) =
α
( β + 1)(α +1)
,
0< β <∞ ,
where α is a known constant greater than 2.
A randomly selected policy had x claims in Year 1.
Determine the Bühlmann credibility estimate of the number of claims for the selected policy
in Year 2.
(A)
(B)
(C)
(D)
(E)
1
α −1
(α − 1) x
α
+
1
α (α − 1)
x
x +1
α
x +1
α −1
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18.
You are given:
(i)
The Cox proportional hazards model is used for the following data:
Gender
(Z1)
Age
(Z2)
0
0
1
1
20
30
30
40
Time to First
Accident
(X)
3
>6
7
>8
(ii)
The baseline hazard rate function is a constant equal to 1/θ.
(iii)
The regression coefficients for gender and age are β1 and β2, respectively.
Determine the loglikelihood function when θ = 18, β1 = 0.1, and β2 = 0.01.
(A)
Less than −10
(B)
At least −10, but less than −7.5
(C)
At least −7.5, but less than −5
(D)
At least −5, but less than −2.5
(E)
At least −2.5
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19.
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 the chi-square goodness-of-fit test
becomes larger with increased sample size.
(E)
None of (A), (B), (C) or (D) is true.
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20.
For a particular policy, the conditional probability of the annual number of claims given
Θ = θ , and the probability distribution of Θ are as follows:
Number of claims
Probability
0
1
2
2θ
θ
1 − 3θ
θ
0.05
0.30
Probability
0.80
0.20
Two claims are observed in Year 1.
Calculate the Bühlmann credibility estimate of the number of claims in Year 2.
(A)
Less than 1.68
(B)
At least 1.68, but less than 1.70
(C)
At least 1.70, but less than 1.72
(D)
At least 1.72, but less than 1.74
(E)
At least 1.74
Exam C: Spring 2005
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21.
You are given:
(i)
The annual number of claims for a policyholder follows a Poisson distribution with
mean Λ .
(ii)
The prior distribution of Λ is gamma with probability density function:
(2λ )5 e−2λ
f (λ ) =
,
24λ
λ >0
An insured is selected at random and observed to have x1 = 5 claims during Year 1 and
x2 = 3 claims during Year 2.
Determine E ( Λ x1 = 5, x2 = 3) .
(A)
3.00
(B)
3.25
(C)
3.50
(D)
3.75
(E)
4.00
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22.
You are given the kernel:
⎧2
2
⎪π 1 − ( x − y) ,
⎪
ky (x) = ⎨
⎪ 0,
⎪
⎩
y −1 ≤ x ≤ y +1
otherwise
You are also given the following random sample:
1
3
3
5
Determine which of the following graphs shows the shape of the kernel density estimator.
(A)
(B)
(C)
(D)
(E)
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23.
You are fitting a curve to a set of n + 1 data points.
Which of the following statements is true?
(A)
The collocation polynomial is recommended for extrapolation.
(B)
Collocation polynomials pass through all data points, whereas cubic splines do not.
(C)
The curvature-adjusted cubic spline requires that m0 = mn = 0 where m j = f ′′( x j ).
(D)
For a cubic runout spline, the second derivative is a linear function throughout the
intervals [ x0 , x 2 ] and [ x n − 2 , x n ] .
(E)
If f ( x) is the natural cubic spline passing through the n + 1 data points, and h( x) is
any function with continuous first and second derivatives that passes through the
same points, then
xn
∫x
0
Exam C: Spring 2005
xn
[ f ′′( x)]2 dx > ∫ [h′′( x)]2 dx .
x0
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24.
The following claim data were generated from a Pareto distribution:
130 20 350 218 1822
Using the method of moments to estimate the parameters of a Pareto distribution, calculate
the limited expected value at 500.
(A)
Less than 250
(B)
At least 250, but less than 280
(C)
At least 280, but less than 310
(D)
At least 310, but less than 340
(E)
At least 340
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25.
You are given:
Group
Total Claims
Year 1
1
Number in Group
Average
Total Claims
Number in Group
Average
2
16,000
100
160
Total Claims
Number in Group
Average
Year 2
Year 3
Total
10,000
15,000
25,000
50
200
60
250
110
227.27
18,000
90
200
34,000
190
178.95
59,000
300
196.67
You are also given â = 651.03.
Use the nonparametric empirical Bayes method to estimate the credibility factor for
Group 1.
(A)
0.48
(B)
0.50
(C)
0.52
(D)
0.54
(E)
0.56
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26.
You are given the following information regarding claim sizes for 100 claims:
Claim Size
0 - 1,000
1,000 - 3,000
3,000 - 5,000
5,000 - 10,000
10,000 - 25,000
25,000 - 50,000
50,000 - 100,000
over 100,000
Number of Claims
16
22
25
18
10
5
3
1
Use the ogive to estimate the probability that a randomly chosen claim is between 2,000 and
6,000.
(A)
0.36
(B)
0.40
(C)
0.45
(D)
0.47
(E)
0.50
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27.
You are given the following 20 bodily injury losses (before the deductible is applied):
Loss
750
200
300
>10,000
400
Number of
Losses
3
Deductible
Policy Limit
200
3
4
6
4
0
0
0
300
∞
10,000
20,000
10,000
∞
Past experience indicates that these losses follow a Pareto distribution with parameters α
and θ = 10,000 .
Determine the maximum likelihood estimate of α .
(A)
Less than 2.0
(B)
At least 2.0, but less than 3.0
(C)
At least 3.0, but less than 4.0
(D)
At least 4.0, but less than 5.0
(E)
At least 5.0
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28.
You are given:
(i)
During a 2-year period, 100 policies had the following claims experience:
Total Claims in
Years 1 and 2
0
Number of Policies
1
30
2
15
3
4
4
1
50
(ii)
The number of claims per year follows a Poisson distribution.
(iii)
Each policyholder was insured for the entire 2-year period.
A randomly selected policyholder had one claim over the 2-year period.
Using semiparametric empirical Bayes estimation, determine the Bühlmann estimate for the
number of claims in Year 3 for the same policyholder.
(A)
0.380
(B)
0.387
(C)
0.393
(D)
0.403
(E)
0.443
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29.
For a study of losses on two classes of policies, you are given:
(i)
The Cox proportional hazards model was used with baseline hazard rate function:
h0 ( x) =
2x
θ
, 0< x<∞
(ii)
A single covariate Z was used with Z = 0 for policies in Class A and Z = 1 for policies
in Class B.
(iii)
Two policies in Class A had losses of 1 and 3 while two policies in Class B had losses
of 2 and 4.
Determine the maximum likelihood estimate of the coefficient β.
(A)
− 0.7
(B)
0.5
(C)
0.7
(D)
1.0
(E)
1.6
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30.
A natural cubic spline is fit to h ( x ) = x5 at the knots x0 = −2, x1 = 0 and x2 = 2 .
Determine f ′′ ( −0.5 ) .
(A)
–1.0
(B)
–0.5
(C)
0.0
(D)
0.5
(E)
1.0
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31.
Personal auto property damage claims in a certain region are known to follow the Weibull
distribution:
F ( x) = 1− e
(θ )
− x
0.2
,
x>0
300
540
A sample of four claims is:
130
240
The values of two additional claims are known to exceed 1000.
Determine the maximum likelihood estimate of θ.
(A)
Less than 300
(B)
At least 300, but less than 1200
(C)
At least 1200, but less than 2100
(D)
At least 2100, but less than 3000
(E)
At least 3000
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32.
For five types of risks, you are given:
(i)
The expected number of claims in a year for these risks ranges from 1.0 to 4.0.
(ii)
The number of claims follows a Poisson distribution for each risk.
During Year 1, n claims are observed for a randomly selected risk.
For the same risk, both Bayes and Bühlmann credibility estimates of the number of claims in
Year 2 are calculated for n = 0,1,2, ... ,9.
Which graph represents these estimates?
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(A)
(B)
(C)
(D)
(E)
Exam C: Spring 2005
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33.
You test the hypothesis that a given set of data comes from a known distribution with
distribution function F(x). The following data were collected:
Interval
x<2
0.035
Number of
Observations
5
2≤x<5
0.130
42
5≤x<7
0.630
137
7≤x<8
0.830
66
8≤x
Total
1.000
50
F ( xi )
300
where xi is the upper endpoint of each interval.
You test the hypothesis using the chi-square goodness-of-fit test.
Determine the result of the test.
(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.
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34.
Unlimited claim severities for a warranty product follow the lognormal distribution with
parameters µ = 5.6 and σ = 0.75 .
You use simulation to generate severities.
The following are six uniform (0, 1) random numbers:
0.6179
0.4602
0.9452
0.0808
0.7881
0.4207
Using these numbers and the inversion method, calculate the average payment per claim for a
contract with a policy limit of 400.
(A)
Less than 300
(B)
At least 300, but less than 320
(C)
At least 320, but less than 340
(D)
At least 340, but less than 360
(E)
At least 360
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35.
You are given:
(i)
The annual number of claims on a given policy has the geometric distribution with
parameter β.
(ii)
One-third of the policies have β = 2, and the remaining two-thirds have β = 5.
A randomly selected policy had two claims in Year 1.
Calculate the Bayesian expected number of claims for the selected policy in Year 2.
(A)
3.4
(B)
3.6
(C)
3.8
(D)
4.0
(E)
4.2
**END OF EXAMINATION**
Exam C: Spring 2005
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STOP
Exam C, Spring 2005
PRELIMINARY ANSWER KEY
Question #
Answer
Question #
Answer
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
D
E
A
E
A
E
E
D
D
A
B
B
A
D
A
D
D
C
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
E
B
B
D
D
C
B
B
C
C
A
C
E
A
C
A
C
Exam C: Spring 2005
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Exam C Solutions
Spring 2005
Question #1
Key: D
1
(1 + x) 4
F(x)
compare to:
0.518
0, 0.2
0.880
0.2, 0.4
0.923
0.4, 0.6
0.949
0.6, 0.8
0.964
0.8, 1.0
The CDF is F ( x) = 1 −
Observation (x)
0.2
0.7
0.9
1.1
1.3
Maximum difference
0.518
0.680
0.523
0.349
0.164
The K-S statistic is the maximum from the last column, 0.680.
Critical values are: 0.546, 0.608, 0.662, and 0.729 for the given levels of significance. The test
statistic is between 0.662 (2.5%) and 0.729 (1.0%) and therefore the test is rejected at 0.025 and
not at 0.01.
Question #2
Key: E
For claim severity,
µ S = 1(0.4) + 10(0.4) + 100(0.2) = 24.4,
σ S2 = 12 (0.4) + 102 (0.4) + 1002 (0.2) − 24.42 = 1, 445.04.
For claim frequency,
µ F = r β = 3r , σ F2 = r β (1 + β ) = 12r.
For aggregate losses,
µ = µ S µ F = 24.4(3r ) = 73.2r ,
σ 2 = µ S2σ F2 + σ S2 µ F = 24.42 (12r ) + 1, 445.04(3r ) = 11, 479.44r.
For the given probability and tolerance, λ0 = (1.96 / 0.1) 2 = 384.16.
The number of observations needed is
λ0σ 2 / µ 2 = 384.16(11, 479.44r ) /(73.2r ) 2 = 823.02 / r.
The average observation produces 3r claims and so the required number of claims is
(823.02 / r )(3r ) = 2, 469.
Question #3
Key: A
1
1
2n − 1
39
Hˆ (t2 ) = +
=
=
=> 39n 2 − 799n + 380 = 0 => n = 20, n = 0.487 .
n n − 1 n(n − 1) 380
Discard the non-integer solution to have n = 20.
The Kaplan-Meier Product-Limit Estimate is:
19 18 11 11
Sˆ (t9 ) =
... =
= 0.55.
20 19 12 20
Question #4
Key: E
There are 27 possible bootstrap samples, which produce four different results. The results, their
probabilities, and the values of g are:
Bootstrap
Sample
1, 1, 1
1, 1, 4
1, 4, 4
4, 4, 4
Prob
8/27
12/27
6/27
1/27
g
0
2
-2
0
The third central moment of the original sample is 2. Then,
12
6
1
44
2
2
2
2⎤
⎡8
.
MSE = ⎢ ( 0 − 2 ) + ( 2 − 2 ) + ( −2 − 2 ) + ( 0 − 2 ) ⎥ =
27
27
27
⎣ 27
⎦ 9
Question #5
Key: A
Pick one of the points, say the fifth one. The vertical coordinate is F(30) from the model and
should be slightly less than 0.6. Inserting 30 into the five answers produces 0.573, 0.096, 0.293,
0.950, and something less than 0.5. Only the model in answer A is close.
Question #6
Key: C, E
The distribution of Θ is Pareto with parameters 1 and 2.6. Then,
1
2
= 0.625, a = VHM = Var (Θ) =
− 0.6252 = 1.6927,
v = EPPV = E (Θ) =
2.6 − 1
1.6(0.6)
5
k = v / a = 0.625 /1.6927 = 0.3692, Z =
= 0.9312.
5 + 0.3692
Question #7
Key: E
At 300, there are 400 policies available, of which 350 survive to 500. At 500 the risk set
increases to 875, of which 750 survive to 1,000. Of the 750 at 1,000, 450 survive to 5,000. The
350 750 450
= 0.45. The distribution function is 1 – 0.45 =
probability of surviving to 5,000 is
400 875 750
0.55.
Alternatively, the formulas in Loss Models could be applied as follows:
j
Interval (c j , c j +1 ]
0
1
2
3
(300, 500]
(500, 1,000]
(1,000, 5,000]
(5,000, 10,000]
qj
Fˆ (c j )
400
0 50
0 400
0.125
525
0 125 350 875 0.142857
0 120 300 750 750
0.4
0 30 300 330 330 0.90909
0
0.125
0.25
0.55
dj
uj
xj
Pj
rj
where
dj = number of observations with a lower truncation point of cj,
uj = number of observations censored from above at cj+1,
xj = number of uncensored observations in the interval (c j , c j +1 ] .
Then, Pj +1 = Pj + d j − u j − x j , rj = Pj + d j (using α = 1 and β = 0 ), and q j = x j / rj .
j −1
Finally, Fˆ (c j ) = 1 − ∏ (1 − qi ) .
i =0
Question #8
Key: D
Let the function be f ( x) = a + bx + cx 2 + dx 3 . The four conditions imply the following:
f (0) = 0
⇒ a = 0,
f (2) = 0
⇒ 2b + 4c + 8d = 0,
f '(0) = −2 ⇒ b = −2,
f '(2) = 2 ⇒ b + 4c + 12d = 2.
The solution is c = 1 and d = 0 and thus the function is f ( x) = x 2 − 2 x . The curvature measure is
∫
2
0
2
[ f "( x)]2 dx = ∫ 22 dx = 8.
0
Question #9
Key: D
For an exponential distribution, the maximum likelihood estimate of θ is the sample mean, 6.
Let Y = X 1 + X 2 where each X has an exponential distribution with mean 6. The sample mean is
Y/2 and Y has a gamma distribution with parameters 2 and 6. Then
−x/6
∞ xe
Pr(Y / 2 > 10) = Pr(Y > 20) = ∫
dx
20
36
∞
20e −20 / 6
xe − x / 6
=−
− e− x / 6 =
+ e −20 / 6 = 0.1546.
6
6
20
Question #10
Key: A
From Question 9, F (10) = 1 −
g '(θ ) = −
20
θ2
e
−20 / θ
10e −10 / θ
(1 + 20θ ) + e
−1
θ
−20 / θ
− e −10 /θ = 1 − e −10 /θ (1 + 10θ −1 ) = g (θ ).
20
θ2
=−
400e −20 / θ
θ3
.
At the maximum likelihood estimate of 6, g '(6) = −0.066063.
The maximum likelihood estimator is the sample mean. Its variance is the variance of one
observation divided by the sample size. For the exponential distribution the variance is the
square of the mean, so the estimated variance of the sample mean is 36/2 = 18. The answer is
(−0.066063) 2 (18) = 0.079.
Question #11
Key B
µ (λ ,θ ) = E ( S | λ ,θ ) = λθ ,
v(λ , θ ) = Var ( S | λ , θ ) = λ 2θ 2 ,
v = EVPV = E (λ 2θ 2 ) = 1(2)(1 + 1) = 4,
a = VHM = Var (λθ ) = E (λ 2 ) E (θ 2 ) − [ E (λ ) E (θ )]2 = 2(2) − 1 = 3,
k = v / a = 4 / 3.
Question #12
Key: B
The distribution is binomial with m = 100 and q = 0.03. The first three probabilities are:
p0 = 0.97100 = 0.04755, p1 = 100(0.97)99 (0.03) = 0.14707,
100(99)
(0.97)98 (0.03) 2 = 0.22515.
2
Values between 0 and 0.04755 simulate a 0, between 0.04755 and 0.19462 simulate a 1, and
between 0.19462 and 0.41977 simulate a 2. The three simulated values are 2, 0, and 1. The
mean is 1.
p2 =
Question #13
Key: A
A mixture of two Poissons or negative binomials will always have a variance greater than its
mean. A mixture of two binomials could have a variance equal to its mean, because a single
binomial has a variance less than its mean.
Question #14
Key: D
The posterior distribution can be found from
e − λ λ 10 ⎛ 0.4 − λ / 6 0.6 − λ /12 ⎞
−7 λ / 6
10
+
+ 0.6e−13λ /12 ) .
π (λ |10) ∝
e
e
⎜
⎟ ∝ λ ( 0.8e
10! ⎝ 6
12
⎠
The required constant is found from
∫
∞
0
λ 10 ( 0.8e −7 λ / 6 + 0.6e−13λ /12 ) d λ = 0.8(10!)(6 / 7)11 + 0.6(10!)(12 /13)11 = 0.395536(10!).
The posterior mean is
∞
1
E (λ |10) =
λ 11 ( 0.8e −7 λ / 6 + 0.6e−13λ /12 ) d λ
∫
0
0.395536(10!)
=
0.8(11!)(6 / 7)12 + 0.6(11!)(12 /13)12
= 9.88.
0.395536(10!)
Question #15
Key: A
2 1 2 2
2
1
2 2
Hˆ (4.5) = + + + = 0.77460. The variance estimate is
+ 2 + 2 + 2 = 0.089397.
2
12 10 9 7
12 10 9 7
The confidence interval is 0.77460 ± 1.96 0.089397 = 0.77460 ± 0.58603. The interval is from
0.1886 to 1.3606.
Question #16
Key: D
k
θ
bias = E (θˆ) − θ =
,
θ −θ = −
k +1
k +1
k 2θ 2
⎛ kθ ⎞
Var (θˆ) = Var ⎜
,
=
⎟
2
⎝ k + 1 ⎠ 25(k + 1)
MSE = Var (θˆ) + bias 2 =
k 2θ 2
θ2
,
+
25(k + 1) 2 (k + 1) 2
2θ 2
.
(k + 1) 2
Setting the last two equal and canceling the common terms gives
k2
+ 1 = 2 for k = 5.
25
MSE = 2bias 2 =
Question #17
Key: D
For the geometric distribution µ ( β ) = β and v( β ) = β (1 + β ) . The prior density is Pareto with
parameters α and 1. Then,
1
µ = E (β ) =
,
α −1
α
1
2
v = EVPV = E[ β (1 + β )] =
+
=
,
α − 1 (α − 1)(α − 2) (α − 1)(α − 2)
α
2
1
a = VHM = Var ( β ) =
−
=
,
2
(α − 1)(α − 2) (α − 1)
(α − 1) 2 (α − 2)
1
1
k = v / a = α − 1, Z =
= .
1+ k α
The estimate is
1
x +1
⎛ 1⎞ 1
x + ⎜1 − ⎟
=
.
α
α
⎝ α ⎠ α −1
Question #18
Key: C
The hazard rate function is h( x) = h =
1
θ
e β1z1 + β2 z2 , a constant. For the four observations, those
1 .1(0)+.01(20)
1
= 0.06786, e.1(0) +.01(30) = 0.07449,
e
18
18
constants are
1 .1(1) +.01(30)
1
= 0.08288, e.1(1) +.01(40) = 0.09160.
e
18
18
Each observation is from an exponential distribution with density function he − hx and survival
function e − hx . The contributions to the loglikelihood function are ln(h) − hx and − hx ,
respectively. The answer is,
ln(0.06786) − 0.06786(3) − 0.07499(6) + ln(0.08288) − 0.08288(7) − 0.09160(8) = −7.147.
Question #19
Key: E
A is false. Using sample data gives a better than expected fit and therefore a test statistic that
favors the null hypothesis, thus increasing the Type II error probability. The K-S test works only
on individual data and so B is false. The A-D test emphasizes the tails, thus C is false. D is false
because the critical value depends on the degrees of freedom which in turn depends on the
number of cells, not the sample size.
Question #20
Key: B
E (θ ) = 0.05(0.8) + 0.3(0.2) = 0.1,
E (θ 2 ) = 0.052 (0.8) + 0.32 (0.2) = 0.02,
µ (θ ) = 0(2θ ) + 1(θ ) + 2(1 − 3θ ) = 2 − 5θ ,
v(θ ) = 02 (2θ ) + 12 (θ ) + 22 (1 − 3θ ) − (2 − 5θ ) 2 = 9θ − 25θ 2 ,
µ = E (2 − 5θ ) = 2 − 5(0.1) = 1.5,
v = EVPV = E (9θ − 25θ 2 ) = 9(0.1) − 25(0.02) = 0.4,
a = VHM = Var (2 − 5θ ) = 25Var (θ ) = 25(0.02 − 0.12 ) = 0.25,
1
5
= ,
k = v / a = 0.4 / 0.25 = 1.6, Z =
1 + 1.6 13
5
8
P = 2 + 1.5 = 1.6923.
13
13
Question #21
Key: B
e − λ λ e − λ λ 3 25 λ 5e−2 λ
∝ λ 12 e −4 λ . This is a gamma distribution with parameters 13
5! 3!
24λ
and 0.25. The expected value is 13(0.25) = 3.25.
f (λ | 5,3) ∝
Alternatively, if the Poisson-gamma relationships are known, begin with the prior parameters
α = 5 and β = 2 where β = 1/ θ if the parameterization from Loss Models is considered. Then
the posterior parameters are α ' = 5 + 5 + 3 = 13 where the second 5 and the 3 are the observations
and β ' = 2 + 2 = 4 where the second 2 is the number of observations. The posterior mean is then
13/4 = 3.25.
Question #22
Key: D
Because the kernel extends one unit each direction there is no overlap. The result will be three
replications of the kernel. The middle one will be twice has high due to having two observations
of 3 while only one observation at 1 and 5. Only graphs A and D fit this description. The kernel
function is smooth, which rules out graph A.
Question #23
Key: D
A is false, the collocation polynomial is too wiggly. Both collocation polynomials and cubic
splines pass through all points, so B is false. C is false because the curvature adjusted spline
sets the values, but not necessarily to zero. The inequality in E should be the other direction.
Answer D is true.
Question #24
Key: C
The two moment equations are
θ
2θ 2
508 =
, 701, 401.6 =
. Dividing the square of the first equation into the
α −1
(α − 1)(α − 2)
701, 401.6
2(α − 1)
second equation gives
. The solution is α = 4.785761. From
= 2.7179366 =
2
α −1
508
the first equation, θ = 1,923.167. The requested LEV is
3.785761
⎤
1,923.167 ⎡ ⎛ 1,923.167 ⎞
E ( X ∧ 500) =
⎢1 − ⎜
⎥ = 296.21.
⎟
3.785761 ⎣⎢ ⎝ 1,923.167 + 500 ⎠
⎦⎥
Question #25
Key: B
2
2
2
2
n = 50(200 − 227.27) + 60(250 − 227.27) + 100(160 − 178.95) + 90(200 − 178.95)
vˆ = EVPV
1+1
= 71,985.647,
kˆ = 71,985.647 / 651.03 = 110.57,
Zˆ =
110
= 0.499.
110 + 110.57
Question #26
Key: B
F100 (1, 000) = 0.16, F100 (3, 000) = 0.38, F100 (5, 000) = 0.63, F100 (10, 000) = 0.81,
F100 (2, 000) = 0.5(0.16) + 0.5(0.38) = 0.27,
F100 (6, 000) = 0.8(0.63) + 0.2(0.81) = 0.666.
Pr(2, 000 < X < 6, 000) = 0.666 − 0.27 = 0.396.
Question #27
Key: C
3
⎡ f (750) ⎤
⎡ f (400) ⎤
L=⎢
f (200)3 f (300) 4 [1 − F (10, 000)]6 ⎢
⎥
⎥
⎣1 − F (200) ⎦
⎣1 − F (300) ⎦
3
3
4
6
4
⎡ α10, 200α ⎤ ⎡ α10, 000α ⎤ ⎡ α10, 000α ⎤ ⎡ 10, 000α ⎤ ⎡ α 10,300α ⎤
=⎢
α +1 ⎥ ⎢
α +1 ⎥ ⎢
α +1 ⎥ ⎢
α ⎥ ⎢
α +1 ⎥
⎣ 10, 750 ⎦ ⎣ 10, 200 ⎦ ⎣ 10,300 ⎦ ⎣ 20, 000 ⎦ ⎣ 10, 400 ⎦
= α 1410, 200−310, 00013α10,300−410, 750−3α −3 20, 000−6α10, 400−4α − 4
4
∝ α 1410, 00013α10, 750−3α 20, 000−6α10, 400−4α ,
ln L = 14 ln α + 13α ln(10, 000) − 3α ln(10, 750) − 6α ln(20, 000) − 4α ln(10, 400)
= 14 ln α − 4.5327α .
The derivative is 14 / α − 4.5327 and setting it equal to zero gives αˆ = 3.089.
Question #28
Key: C
30 + 30 + 12 + 4
= 0.76.
100
50(0 − 0.76) 2 + 30(1 − 0.76) 2 + 15(2 − 0.76) 2 + 4(3 − 0.76) 2 + 1(4 − 0.76) 2
aˆ = VHM =
− 0.76
99
= 0.090909,
vˆ = EVPV = x =
0.76
1
= 8.36, Zˆ =
= 0.10684,
0.090909
1 + 8.36
P = 0.10684(1) + 0.89316(0.76) = 0.78564.
The above analysis was based on the distribution of total claims for two years. Thus 0.78564 is
the expected number of claims for the next two years. For the next one year the expected
number is 0.78564/2 = 0.39282.
kˆ =
Question #29
Key: A
For members of Class A, h( x) = 2 x / θ , S A ( x) = e − x
2
/θ
, f A ( x) = (2 x / θ )e− x
2
/θ
and for members of
Class B, h( x) = 2 xe β / θ = 2 xγ / θ , S B ( x) = e − x γ /θ , f B ( x) = (2 xγ / θ )e− x γ / θ where γ = e β . The
likelihood function is
L(γ ,θ ) = f A (1) f A (3) f B (2) f B (4)
2
2
= 2θ −1e −1/θ 6θ −1e−9 / θ 4γθ −1e−4γ /θ 8γθ −1e−16γ / θ
∝ θ −4γ 2 e − (10+ 20γ ) /θ .
The logarithm and its derivatives are
l (γ ,θ ) = −4 ln θ + 2 ln γ − (10 + 20γ )θ −1
∂l / ∂γ = 2γ −1 + 20θ −1
∂l / ∂θ = −4θ −1 + (10 + 20γ )θ −2 .
Setting each partial derivative equal to zero and solving the first equation for θ gives θ = 10γ .
Substituting in the second equation yields
4 10 + 20γ
−
+
= 0, 400γ 2 = 100γ + 200γ 2 , γ = 0.5. Then, β = ln(γ ) = ln(0.5) = −0.69.
2
10γ
100γ
Question #30
Key: C
Let the two spline functions be
f 0 ( x) = a + b( x + 2) + c( x + 2) 2 + d ( x + 2)3 ,
f1 ( x) = e + fx + gx 2 + hx 3 .
The eight spline conditions produce the following eight equations.
f 0 (−2) = −32 ⇒ a = −32, f1 (0) = 0 ⇒ e = 0,
f 0'' (−2) = 0 ⇒ c = 0,
f 0 (0) = 0 ⇒ −32 + 2b + 4c + 8d = 0,
f1 (2) = 32 ⇒ 2 f + 4 g + 8h = 32,
f 0' (0) = f1' (0) ⇒ b + 4c + 12d = f ,
f 0'' (0) = f1'' (0) ⇒ 2c + 12d = 2 g , f1'' (2) = 0 ⇒ 2 g + 12h = 0.
Eliminating one variable at a time leads to b = 16, d = 0, f = 16, g = 0, and h = 0. Then,
f 0 ( x) = −32 + 16( x + 2) = 16 x and the second derivative is zero.
The above work could be avoided by noting that the three knots are (-2, -32), (0,0), and (2, 32).
These points lie on a straight line and so the spline will be that straight line.
Question #31
Key: E
The density function is f ( x) =
0.2 x −0.8
θ
0.2
e − ( x /θ ) . The likelihood function is
0.2
L(θ ) = f (130) f (240) f (300) f (540)[1 − F (1000)]2
=
0.2(130) −0.8
∝θ
θ
0.2
e − (130 / θ )
0.2
0.2(240) −0.8
θ
0.2
e − (240 / θ )
−0.8 −θ −0.2 (1300.2 + 2400.2 + 3000.2 + 5400.2 +10000.2 +10000.2 )
e
l (θ ) = −0.8ln(θ ) − θ
−0.2
(130 + 240
0.2
0.2
+ 300
0.2
0.2(300) −0.8
θ
0.2
e − (300 / θ )
0.2
0.2(540) −0.8
θ
0.2
e− (540 /θ ) e− (1000 /θ ) e− (1000 /θ )
0.2
0.2
0.2
,
0.2
+ 5400.2 + 10000.2 + 10000.2 )
= −0.8ln(θ ) − 20.2505θ −0.2 ,
l '(θ ) = −0.8θ −1 + 0.2(20.2505)θ −1.2 = 0,
θ −0.2 = 0.197526, θˆ = 3,325.67.
Question #32
Key: A
Buhlmann estimates are on a straight line, which eliminates E. Bayes estimates are never outside
the range of the prior distribution. Because graphs B and D include values outside the range 1 to
4, they cannot be correct. The Buhlmann estimates are the linear least squares approximation to
the Bayes estimates. In graph C the Bayes estimates are consistently higher and so the
Buhlmann estimates are not the best approximation. This leaves A as the only feasible choice.
Question #33
Key: C
The expected counts are 300(0.035) = 10.5, 300(0.095) = 28.5, 300(0.5) = 150, 300(0.2) = 60,
and 300(0.17) = 51 for the five groups. The test statistic is
(5 − 10.5) 2 (42 − 28.5) 2 (137 − 150) 2 (66 − 60) 2 (50 − 51) 2
+
+
+
+
= 11.02.
10.5
28.5
150
60
51
There are 5 – 1 = 4 degrees of freedom. From the table, the critical value for a 5% test is 9.488
and for a 2.5% test is 11.143. The hypothesis is rejected at 5%, but not at 2.5%.
Question #34
Key: A
To simulate a lognormal variate, first convert the uniform random number to a standard normal.
This can be done be using the table of normal distribution values. For the six values given, the
corresponding standard normal values are 0.3, -0.1, 1.6, -1.4, 0.8, and -0.2. Next, multiply each
number by the standard deviation of 0.75 and add the mean of 5.6. This produces random
observations from the normal 5.6, 0.752 distribution. These values are 5.825, 5,525, 6.8, 4.55,
6.2, and 5.45. To create lognormal observations, exponentiate these values. The results are 339,
251, 898, 95, 493, and 233. After imposing the policy limit of 400, the mean is (339 + 251 + 400
+ 95 + 400 + 233)/6 = 286.
Question #35
Key: C
β2
and the expected value is β.
(1 + β )3
Pr( X 1 = 2 | β = 2) Pr( β = 2)
Pr( β = 2 | X 1 = 2) =
Pr( X 1 = 2 | β = 2) Pr( β = 2) + Pr( X 1 = 2 | β = 5) Pr( β = 5)
For the geometric distribution, Pr( X 1 = 2 | β ) =
4 1
27 3
=
= 0.39024.
4 1 25 2
+
27 3 216 3
The expected value is then 0.39024(2) + 0.60976(5) = 3.83.