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Transcript
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
12 April 2005 (am)
Subject CT3
Probability and Mathematical Statistics
Core Technical
Time allowed: Three hours
INSTRUCTIONS TO THE CANDIDATE
1.
Enter all the candidate and examination details as requested on the front of your answer
booklet.
2.
You must not start writing your answers in the booklet until instructed to do so by the
supervisor.
3.
Mark allocations are shown in brackets.
4.
Attempt all 13 questions, beginning your answer to each question on a separate sheet.
5.
Candidates should show calculations where this is appropriate.
Graph paper is required for this paper.
AT THE END OF THE EXAMINATION
Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this
question paper.
In addition to this paper you should have available the 2002 edition of the
Formulae and Tables and your own electronic calculator.
CT3 A2005
Faculty of Actuaries
Institute of Actuaries
1
Calculate the sample mean and standard deviation of the following claim amounts (£):
534
671
581
620
401
340
980
845
550
690
[3]
2
Suppose A, B and C are events with P ( A) = 12 , P ( B ) = 12 , P(C ) = 13 , P ( A
P( A
C ) = 16 , P ( B
(a)
(b)
Determine whether or not the events A and B are independent.
Calculate the probability P( A B C ).
C) =
1
6
and P ( A
B
B ) = 34 ,
1 .
C ) = 12
[4]
3
Claim sizes in a certain insurance situation are modelled by a distribution with
moment generating function M(t) given by
M(t) = (1 10t) 2.
Show that E[X 2] = 600 and find the value of E[X 3].
4
Consider a random sample of size 16 taken from a normal distribution with mean
= 25 and variance 2 = 4. Let the sample mean be denoted X .
State the distribution of X and hence calculate the probability that X assumes a
value greater than 26.
5
[3]
[3]
Consider a random sample of size 21 taken from a normal distribution with mean
= 25 and variance 2 = 4. Let the sample variance be denoted S2.
State the distribution of the statistic 5S2 and hence find the variance of the statistic S2.
[3]
6
In a survey conducted by a mail order company a random sample of 200 customers
yielded 172 who indicated that they were highly satisfied with the delivery time of
their orders.
Calculate an approximate 95% confidence interval for the proportion of the
company s customers who are highly satisfied with delivery times.
7
[3]
For a group of policies the total number of claims arising in a year is modelled as a
Poisson variable with mean 10. Each claim amount, in units of £100, is independently
modelled as a gamma variable with parameters = 4 and = 1/5.
Calculate the mean and standard deviation of the total claim amount.
CT3 A2005
2
[5]
8
9
The distribution of claim size under a certain class of policy is modelled as a normal
random variable, and previous years records indicate that the standard deviation is
£120.
(i)
Calculate the width of a 95% confidence interval for the mean claim size if a
sample of size 100 is available.
[2]
(ii)
Determine the minimum sample size required to ensure that a 95% confidence
interval for the mean claim size is of width at most £10 .
[2]
(iii)
Comment briefly on the comparison of the confidence intervals in (i) and (ii)
with respect to widths and sample sizes used.
[1]
[Total 5]
Let X1, , Xn denote a large random sample from a distribution with unknown
population mean and known standard deviation 3. The null hypothesis H0: = 1 is
to be tested against the alternative hypothesis H1: > 1, using a test based on the
sample mean with a critical region of the form X
k , for a constant k.
It is required that the probability of rejecting H0 when = 0.8 should be
approximately 0.05, and the probability of not rejecting H0 when = 1.2 should be
approximately 0.1.
(i)
Show that the test requires
k 0.8
3/ n
where
(ii)
0.95 and
k 1.2
3/ n
0.10
is the standard normal distribution function.
[4]
The values for the sample size n and the critical value k which satisfy the
requirements of part (i) are n = 482 and k = 1.025 (you are not asked to verify
these values).
Calculate the approximate level of significance of the test, and comment on
the value.
[3]
[Total 7]
CT3 A2005
3
PLEASE TURN OVER
10
A model used for claim amounts (X, in units of £10,000) in certain circumstances has
the following probability density function, f(x), and cumulative distribution function,
F(x):
f ( x) =
5(105 )
(10 x)
, x
6
0 ; F ( x) = 1
10
10 x
5
.
You are given the information that the distribution of X has mean 2.5 units (£25,000)
and standard deviation 3.23 units (£32,300).
(i)
Describe briefly the nature of a model for claim sizes for which the standard
deviation can be greater than the mean.
[2]
(ii)
(a)
Show that we can obtain a simulated observation of X by calculating
x = 10 (1 r )
0.2
1
where r is an observation of a random variable which is uniformly
distributed on (0,1).
(b)
Explain why we can just as well use the formula
x = 10 r
0.2
1
to obtain a simulated observation of X.
(c)
Calculate the missing values for the simulated claim amounts in the
table below (which has been obtained using the method in (ii)(b)
above):
r
Claim (£)
0.7423
0.0291
0.2770
0.5895
0.1131
0.9897
0.6875
0.8525
0.0016
0.5154
6,141
10,2872
29,272
11,148
54,635
207
7,782
3,243
?
?
[5]
[Total 7]
CT3 A2005
4
11
Twenty insects were used in an experiment to examine the effect on their activity
level, y, of 3 standard preparations of a chemical. The insects were randomly
assigned, 4 to receive each of the preparations and 8 to remain untreated as controls.
Their activity levels were metered from vibrations in a test chamber and were as
follows:
Activity levels (y)
Totals
Control
Preparation A
Preparation B
Preparation C
43
73
84
46
40
55
63
91
For these data
y = 1, 203 ,
65
61
51
84
51
65
72
71
33
39
54
62
387
254
270
292
y 2 = 77, 249 .
(i)
Conduct an analysis of variance test to establish whether the data indicate
significant differences amongst the results for the four treatments.
[7]
(ii)
(a)
Complete the following table of residuals for the data and analysis in
part (i) above:
Control
Preparation A
Preparation B
Preparation C
(iii)
?
9.5
16.5
?
?
8.5
?
?
16.6
2.5
16.5
?
2.6
1.5
4.5
2
15.4
9.4
5.6 13.6
(b)
Make a rough plot of the residuals against the treatment means.
(c)
State the assumptions underlying the analysis of variance test
conducted in part (i).
(d)
Comment on how well the data conform to these assumptions in the
light of the residual plot.
[8]
It is suggested that any differences can be explained in terms of a difference
between controls on the one hand and treated groups on the other.
Comment on any evidence for this and state how you would formally test for
this effect (but do not carry out the test).
[4]
[Total 19]
CT3 A2005
5
PLEASE TURN OVER
12
(i)
A random variable Y has a Poisson distribution with parameter but there is a
restriction that zero counts cannot occur. The distribution of Y in this case is
referred to as the zero-truncated Poisson distribution.
(a)
Show that the probability function of Y is given by
y
p( y ) =
(ii)
e
y !(1 e )
( y = 1, 2, ).
(b)
Show that E[Y ] = /(1 e ).
[4]
(a)
Let y1, , yn denote a random sample from the zero-truncated Poisson
distribution.
Show that the maximum likelihood estimate of
by the solution to the following equation:
y
e
may be determined
= 0,
1 e
and deduce that the maximum likelihood estimate is the same as the
method of moments estimate.
(b)
(iii)
Obtain an expression for the Cramer-Rao lower bound (CRlb) for the
variance of an unbiased estimator of .
[9]
The following table gives the numbers of occupants in 2,423 cars observed on
a road junction during a certain time period on a weekday morning.
Number of occupants
Frequency of cars
1
1,486
2
694
3
195
4
37
5
10
6
1
The above data were modelled by a zero-truncated Poisson distribution as
given in (i).
The maximum likelihood estimate of is = 0.8925 and the Cramer-Rao
lower bound on variance at = 0.8925 is 5.711574 10 4 (you do not need to
verify these results.)
CT3 A2005
(a)
Obtain the expected frequencies for the fitted model, and use a 2
goodness-of-fit test to show that the model is appropriate for the data.
(b)
Calculate an approximate 95% confidence interval for and hence
calculate a 95% confidence interval for the mean of the zero-truncated
Poisson distribution.
[9]
[Total 22]
6
13
As part of an investigation into health service funding a working party was concerned
with the issue of whether mortality rates could be used to predict sickness rates. Data
on standardised mortality rates and standardised sickness rates were collected for a
sample of 10 regions and are shown in the table below:
Region
Mortality rate m (per 10,000)
Sickness rate s (per 1,000)
1
2
3
4
5
6
7
8
9
10
125.2
119.3
125.3
111.7
117.3
100.7
108.8
102.0
104.7
121.1
206.8
213.8
197.2
200.6
189.1
183.6
181.2
168.2
165.2
228.5
Data summaries:
m = 1136.1,
(i)
m2 = 129,853.03, s = 1934.2,
s2 = 377,700.62, ms = 221,022.58
Calculate the correlation coefficient between the mortality rates and the
sickness rates and determine the probability-value for testing whether the
underlying correlation coefficient is zero against the alternative that it is
positive.
[4]
(ii)
Noting the issue under investigation, draw an appropriate scatterplot for these
data and comment on the relationship between the two rates.
[3]
(iii)
Determine the fitted linear regression of sickness rate on mortality rate and test
whether the underlying slope coefficient can be considered to be as large
as 2.0.
[5]
(iv)
For a region with mortality rate 115.0, estimate the expected sickness rate and
calculate 95% confidence limits for this expected rate.
[4]
[Total 16]
END OF PAPER
CT3 A2005
7
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
April 2005
Subject CT3
Probability and Mathematical Statistics
Core Technical
EXAMINERS REPORT
Introduction
The attached subject report has been written by the Principal Examiner with
the aim of helping candidates. The questions and comments are based around
Core Reading as the interpretation of the syllabus to which the examiners are
working. They have however given credit for any alternative approach or
interpretation which they consider to be reasonable.
M Flaherty
Chairman of the Board of Examiners
15 June 2005
Faculty of Actuaries
Institute of Actuaries
Subject CT3 (Probability and Mathematical Statistics Core Technical)
1
2
April 2005
Examiners Report
x = 6212 , x2 = 4186784
x
6212
= £621.20
10
s
1
62122
4186784
9
10
(a)
P( A
B)
327889.6
9
P( A) P( B ) P( A
P ( A) P( B) ( 12 )( 12 )
(b)
P( A
B)
P( A
B
1
2
1
3
1
2
3
4
1
2
1
4
so the events A and B are independent as
P( A) P ( B) P(C ) P( A
1 1
3 4
1
6
B
C)
P( A
1
6
1
12
5 .]
6
[OR P ( A
3
4
B)
P( A) P( B).
C)
1
2
1
4
= £190.87
1
6
1 1
6 12
B) P( A
C ) P( B C )
P( A B
C)
5
6
B ) P (C ) P ( A
C ) P( B
C ) P( A
B
C)
[OR Use a Venn diagram]
3
M(t) = (1 10t) 2
M (t) = ( 2)( 10)(1 10t) 3 = 20(1 10t) 3
M (t) = ( 60)( 10)(1 10t) 4 = 600(1 10t) 4
M (t) = ( 2400)( 10)(1 10t) 5 = 24000(1 10t)
5
Putting t = 0
Putting t = 0
E[X2] = 600
E[X3] = 24000
[OR use the power series expansion M(t) = 1 + 20t + 600t2/2! + 24000t3/3! +
[OR use the result on E[X r] for a gamma(2,0.1) variable in the Yellow Book]
4
X ~ N 25, 0.25
P X
Page 2
26
P Z
26 25
0.5
P Z
2
1 0.97725 0.02275
]
Subject CT3 (Probability and Mathematical Statistics Core Technical)
5
n 1 S2
5S 2 ~
2
x
n
p
172
200
7
2
20 =
40, so V[S2] = 40/25 = 1.6
0.86
95% CI is p 1.96
0.86
Examiners Report
2
20
V[5S2] = variance of
6
April 2005
p (1 p )
n
1.96(0.0245)
0.86
0.048 or (0.812, 0.908)
Let S = X1 + X2 + . . . + XN be the total claim amount.
Note that E[N] = V[N] = 10, E[X] = 4/(1/5) = 20, V[X] = 4/(1/5)2 = 100
E[S] = E[N]E[X] = 10(20) = 200
mean of total claim amount = £20,000.
V[S] = E[N]V[X] + V[N]{E[X]}2
=10(100) + 10(20)2 = 5000
s.d.[S] = 70.71
8
(i)
s.d. of the total claim amount = £7,071
Width of 95% confidence interval: 1.96
120
120
[or 2 1.96
n
n
3.92
120
]
n
120
23.52
100
120
[or 2 1.96
£47.04]
100
1.96
(ii)
For the width of a 95% confidence interval to be at most
120
1.96
10
n
n
1.96 120
10
23.52 , n
10 we require
553.19
i.e., take the sample size as 554.
Page 3
Subject CT3 (Probability and Mathematical Statistics Core Technical)
9
(iii)
The confidence interval in (ii) in narrower
much larger sample size.
(i)
X approx
0.05
N
,
32
n
April 2005
Examiners Report
to achieve this we require a
for large n by the central limit theorem.
P(reject H0| = 0.8) = P( X > k| = 0.8)
k 0.8
3/ n
=1
k 0.8
3/ n
0.95
and
0.1
P(do not reject H0| = 1.2)
= P(X
1.2)
k 1.2
3/ n
=
(ii)
k|
Significance level
1
= P(reject H0 when H0 is true) = P ( X
1.025 1
3 / 482
1
k|
1)
(0.18) 1 0.57 0.43.
The significance level of the test is very high (43%).
10
(i)
X takes positive values only so to have such a relatively high standard
deviation the distribution must be positively skewed with sizeable probability
associated with high values (i.e. the model embraces high claim sizes; the
density has a long or heavy tail).
(ii)
(a)
Solving r = F(x)
(b)
R ~ U(0,1) 1 R ~ U(0, 1) so (1 r) is also a random number from
(0, 1), so we can use 1 r in place of r , giving the formula
x 10 r
(c)
Page 4
0.2
r = 0.0016
r = 0.5154
(1 + x/10) = (1 r)
1
claim = 262390
claim = 14175
0.2
x = 10[(1
r)
0.2
1]
Subject CT3 (Probability and Mathematical Statistics Core Technical)
(i)
SST = 77249 12032/20 = 4888.55
SSB = 3872/8 + 2542/4 + 2702/4 + 2922/4
SSR = 2857.875
Examiners Report
12032/20 = 2030.675
H0: no treatment effects (i.e. population means are equal) v H1: not H0
Analysis of Variance
Source
DF
SS
Factor
3
2031
Error
16
2858
Total
19
4889
MS
677
179
F
3.79
F3,16(0.05) = 3.239, F3,16(0.01) = 5.292
P-value is lower than 0.05 (but higher than 0.01), so we can reject H0 at least
at the 5% level of testing (Note: actually P-value is 0.032). The data do
indicate significant differences amongst the treatment means.
(ii)
(a)
Residual = observed value treatment mean
Treatment means are: Control 48.375, A 63.5, B 67.5, C 73.0
Missing values are:
Control
Preparation A
Preparation B
Preparation C
5.4
9.5
16.5
27
8.4
8.5
4.5
18
16.6
2.5
16.5
11
2.6
1.5
4.5
2
15.4
9.4
5.6
13.6
(b)
20
10
resids
11
April 2005
0
-10
-20
-30
means
treatment
50
Control
60
70
A
B
C
Page 5
Subject CT3 (Probability and Mathematical Statistics Core Technical)
(iii)
April 2005
Examiners Report
(c)
Observations Yij (jth value for treatment i) are independent and
normally distributed with variance 2 which is constant across
treatments.
(d)
The assumptions seem reasonable
with the exception of the constant
variance assumption, which is questionable
the data for preparation
A appear to be less variable than the data for the other treatments.
The control mean is lower than all three treatment means
(48.4 v 63.5, 67.5, 73.0) so there is prima facie evidence to support the
suggestion.
One could perform a two-sample t-test of control mean = treatment mean
by combining the data for the 3 preparations (and using samples of sizes 8
and 12).
12
(i)
(a)
The probability function for the zero-truncated Poisson distribution is
given by
P (Y
y |Y
P(Y
0)
y and Y
P(Y 0)
y
e
y !(1 P(Y
y
0))
e
( y 1, 2, ).
y !(1 e )
(b)
Expectation of Y:
y
E[Y ]
y
y 1
e
y !(1 e )
z
(1 e )
(1 e
Page 6
)
z 0
[1]
e
z!
(z
.
(1 e
)
y 1)
0)
Subject CT3 (Probability and Mathematical Statistics Core Technical)
(ii)
(a)
The log likelihood function for
April 2005
Examiners Report
is:
n
log L( )
yi log
n
n log(1 e )
constant
i 1
d log L( )
d
ny
n n
e
1 e
the ML estimate is determined by the solution of the equation
d log L( )
d
0
e
y
0
1 e
As this equation may be rewritten as
y
and E[Y ]
1 e
1 e
the ML estimate is the same as the method of moments estimate.
(b)
d2
d
2
ny
log L( )
n
2
and since E[Y ] E[Y ]
e
(1 e ) 2
(1 e )
, the Cramer-Rao lower bound is
given by,
1
CRlb
E
or
d
d
2
log L( )
(1 e ) 2
n(1 e
1
2
e )
n
1
(1 e )
e
(1 e ) 2
.
Page 7
Subject CT3 (Probability and Mathematical Statistics Core Technical)
(iii)
(a)
April 2005
Examiners Report
The expected frequencies for the fitted zero-truncated Poisson model
are given by
y
n
e
( y 1, 2, ) where
0.8925 and n
2423
y !(1 e )
y
ei
fi
1
1500.48
1486
2
669.59
694
3
199.20
195
4
44.45
37
( fi ei ) 2
(1486 1500.48) 2
=
ei
1500.48
(on 4 df).
2
5
7.93
10
6
1.35
1
Total
2423.00
2423
(1 1.35) 2
= 2.99
1.35
The Yellow Book gives that the probability value is greater than 50%,
therefore there is no evidence to reject the null hypothesis, i.e. the
model seems appropriate for the data.
[OR
(b)
2
= 2.68 on 3 df if
5 combined rather than 6.]
As approx. ~ N ( , CRlb) for large n, a 95% confidence interval for
is given by
1.96 CRlb
0.8925 1.96 5.711574 10
at
4
, since CRlb
5.711574
= 0.8925,
= 0.8925 1.96(0.0238989) = 0.8925
= (0.84566, 0.93934)
0.04684
Then the 95% confidence interval for the mean of Y ,
by
0.84566
1 e
Page 8
10-4
0.84566
,
0.93934
1 e
0.93934
= (1.48, 1.54).
1
, is given
1 e
Subject CT3 (Probability and Mathematical Statistics Core Technical)
13
(i)
April 2005
Smm = 129853.03
(1136.1)2/10 = 780.709
Sss = 377700.62
(1934.2)2/10 = 3587.656
Sms = 221022.58
(1136.1)(1934.2)/10 = 1278.118
r
1278.118
(780.709)(3587.656)
H0:
= 0 v. H1:
t
r 8
1 r2
3.35
Examiners Report
0.764
>0
Prob-value = P(t8 > 3.35) = 0.005 from tables.
[OR use Fisher s transformation]
(ii)
Given the issue of whether mortality can be used to predict sickness, we
require a plot of sickness against mortality:
There seems to be an increasing linear relationship such that mortality could
be used to predict sickness.
Page 9
Subject CT3 (Probability and Mathematical Statistics Core Technical)
1278.118
1.6371 and
780.709
(iii)
2
1
[1934.2
10
April 2005
Examiners Report
(1136.1)] 7.426
1
(1278.118) 2
{3587.656
} 186.902
8
780.709
2
Var[ ]
Test H0:
t
0.2394
780.709
= 2 v. H1:
1.6371 2
0.2394
<2
0.74 on 8 df
Prob-value large; no evidence to reject H0: = 2
So we can accept that the slope is as large as 2.
(iv)
For a region with m = 115:
estimated expected s = 7.426 + 1.6371(115) = 195.69
with variance =
2
1
{
10
(115 113.61) 2
} 19.1528
780.709
95% confidence limits are:
195.69
t8(s.e.)
195.69
2.306(4.376)
195.69
10.09 or (185.60, 205.78)
END OF EXAMINERS REPORT
Page 10
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
13 September 2005 (am)
Subject CT3
Probability and Mathematical Statistics
Core Technical
Time allowed: Three hours
INSTRUCTIONS TO THE CANDIDATE
1.
Enter all the candidate and examination details as requested on the front of your answer
booklet.
2.
You must not start writing your answers in the booklet until instructed to do so by the
supervisor.
3.
Mark allocations are shown in brackets.
4.
Attempt all 14 questions, beginning your answer to each question on a separate sheet.
5.
Candidates should show calculations where this is appropriate.
Graph paper is required for this paper.
AT THE END OF THE EXAMINATION
Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this
question paper.
In addition to this paper you should have available the 2002 edition of the
Formulae and Tables and your own electronic calculator.
CT3 S2005
Faculty of Actuaries
Institute of Actuaries
1
The table below gives the number of thunderstorms reported in a particular summer
month by 100 meteorological stations.
Number of thunderstorms:
Number of stations:
0
22
1
37
2
20
3
13
4
6
5
2
(a)
Calculate the sample mean number of thunderstorms.
(b)
Calculate the sample median number of thunderstorms.
(c)
Comment briefly on the comparison of the mean and the median.
[3]
2
In an opinion poll, each individual in a random sample of 275 individuals from a large
population is asked which political party he/she supports. If 45% of the population
support party A, calculate (approximately) the probability that at least 116 of the
sample support A.
[3]
3
Claim amounts on a certain type of policy are modelled as following a gamma
distribution with parameters = 120 and = 1.2.
Calculate an approximate value for the probability that an individual claim amount
exceeds 120, giving a reason for the approach you use.
[3]
4
Calculate a 99% confidence interval for the percentage of claims for household
accidental damage which are fully settled within six months of being submitted, given
that in a random sample of 100 submitted claims of this type, exactly 83 were fully
settled within six months of being submitted.
[3]
5
A random sample of 500 claim amounts resulted in a mean of £237 and a standard
deviation of £137.
Calculate an approximate 95% confidence interval for the true underlying mean claim
amount for such claims, explaining why the normal distribution can be used.
[3]
6
A random sample of 16 observations was selected from each of four populations.
Given that the between treatments mean square is 280 and the total sum of squares is
SST = 1,500, construct an analysis of variance table and test the null hypothesis that
the means of the four populations are equal.
[3]
CT3 S2005
2
7
A sample of 20 claim amounts (£) on a group of household policies gave the
following data summaries:
x = 3,256 and
8
x2 = 866,600.
(a)
Calculate the sample mean and standard deviation for these claim amounts.
(b)
Comment on the skewness of the distribution of these claim amounts, giving
reasons for your answer.
[4]
A simple procedure for incorporating a no claims discount into an annual insurance
policy is as follows:
a premium of £400 is payable for the first year;
if no claim is made in the first year, the premium for the second year is £400k,
where k is a constant such that 0 < k < 1;
if no claim is made in the first and second years, the premium for the third
year is £400k2;
if no subsequent claims are made in future years, the premium remains as
£400k2;
if a claim is made, the premium the following year reverts to £400 and the
procedure starts again as above.
(i)
Show that the probability distribution of the premium for the fourth year, that
is, for the year following the third year, is given by
£400
with probability
£400k with probability
£400k2 with probability
p
p(1 p)
(1 p)2
where p is the probability of a claim being made in any year.
[3]
(ii)
Obtain an expression for the expected premium for the fourth year under this
procedure.
[1]
(iii)
If it is desired that this expected premium should equal £300, determine the
required value of k for the case where p = 0.1.
[3]
[Total 7]
CT3 S2005
3
PLEASE TURN OVER
9
Random samples are taken from two independent normally distributed populations
(with means 1 and 2 respectively) with the following results.
Sample from population 1:
sample size n1 = 11, sample mean x1 124 , sample variance s12
59
Sample from population 2:
sample size n2 = 15, sample mean x2 105 , sample variance s22
42
Calculate a 95% confidence interval for 1
2, the difference between the
population means (you may assume that the population variances are equal).
[5]
10
Let X denote a random variable with a continuous uniform (0, 1000) distribution.
Define a random variable Y as the minimum of X and 800.
(i)
Show that the conditional distribution of X given X < 800 is a continuous
uniform (0, 800) distribution.
[2]
(ii)
Verify (giving clear reasons) that the expectation of the random variable Y is
480.
[3]
(iii)
Suppose that Y1, , Yn are independent and identically distributed, each with
the same distribution as Y.
In the case that n is large, determine the approximate distribution of
1 n
Y =
Yi , stating its expectation. (You are not required to derive or state
n i1
the variance of Y .)
[1]
(iv)
CT3 S2005
Comment on the comparison of the conditional expectation of X given X < 800
with the expectation of Y.
[2]
[Total 8]
4
11
Consider the following simple model for the number of claims, N, which occur in a
year on a policy:
0
0.55
n
P(N = n)
1
0.25
2
0.15
3
0.05
(a)
Explain how you would simulate an observation of N using a number r, an
observation of a random variable which is uniformly distributed on (0, 1).
(b)
Illustrate your method described in (i) by simulating three observations of N
using the following random numbers between 0 and 1:
0.6221, 0.1472, 0.9862.
[4]
12
A certain type of insurance policy has a claim rate of per year and the cover ceases
and the policy expires after the first claim. Accordingly the duration of a policy is
modelled by an exponential distribution with density function e x : 0 x
.
A company has data on (m + n) policies which have expired and which may be
assumed to be independent. Of these, m policies had duration less than 5 years and n
policies had duration greater than or equal to 5 years.
(i)
An investigator makes note of the actual durations, x1, , xn, of the latter
group of n policies, but ignores the former group without even noting the
value of m.
(a)
Explain why the xi s come from a truncated exponential distribution
with density function
f ( x) = k . e
, 5
x
e5 .
and show that k
(b)
x
Write down the likelihood for the data from the point of view of this
investigator and hence show that the maximum likelihood estimate
(MLE) of is given by
n
n
.
xi 5n
i 1
(c)
The data yield the values: n = 10 and xi = 71. Calculate this
investigator s MLE of .
[8]
CT3 S2005
5
PLEASE TURN OVER
(ii)
A second investigator ignores the actual policy durations and simply notes the
values of m and n.
(a)
Write down the likelihood for this information and hence show that the
resulting MLE of is given by
=
(b)
1
m n
log
.
5
n
The same data as in part (i) yield the values: m = 120 and n = 10.
Calculate this investigator s MLE of .
[5]
(iii)
The two investigators decide to pool their data, and so have the information
that there are m policies with duration less than 5 years, and n policies with
actual durations x1, ... , xn.
(a)
Explain why the likelihood for this joint information is given by
L( ) = (1 e
5
)m .
n
e
xi
i 1
and determine an equation, the solution of which will lead to the MLE
of .
(b)
Given that this leads to an MLE of
comparison of the three MLE s.
equal to 0.508, comment on the
[5]
[Total 18]
13
A survey, carried out at a major flower and gardening show, was concerned with the
association between the intention to return to the show next year and the purchase of
goods at this year s show. There were 220 people interviewed and of these 101 had
made a purchase; 69 of these people said they intended to return next year. Of the
119 who had not made a purchase, 68 said they intended to return next year.
(i)
Suppose one of the 220 people surveyed is selected at random.
Calculate the probabilities that the selected person:
(a)
(b)
(c)
(ii)
CT3 S2005
intends to return next year, given that he/she has made a purchase
intends to return next year, given that he/she has not made a purchase
has made a purchase, given that he/she intends to return next year
[3]
By testing the difference between the proportions of purchasers and nonpurchasers who intend to return next year, examine whether there is sufficient
evidence to justify concluding that the intention to return depends on whether
or not a purchase was made.
[7]
6
14
(iii)
Present the data as a contingency table and perform a 2 test of association
between the attributes intention to return and purchasing status .
[5]
(iv)
Discuss briefly the connection between the comparison of proportions carried
out in part (ii) and the test of association performed in part (iii).
[2]
[Total 17]
The data given in the following table are the numbers of deaths from AIDS in
Australia for 12 consecutive quarters starting from the second quarter of 1983.
1
1
Quarter (i):
Number of deaths (ni):
(i)
2
2
3
3
4
1
5
4
6 7 8 9
9 18 23 31
10
20
11
25
12
37
(a)
Draw a scatterplot of the data.
(b)
Comment on the nature of the relationship between the number of
deaths and the quarter in this early phase of the epidemic.
[4]
(ii)
A statistician has suggested that a model of the form
E Ni = i 2
might be appropriate for these data, where is a parameter to be estimated
from the above data. She has proposed two methods for estimating , and
these are given in parts (a) and (b) below.
(a)
Show that the least squares estimate of , obtained by minimising
q=
12
(n
i 1 i
=
(b)
i 2 ) 2 , is given by
12 2
i ni
i 1
12 4
i
i 1
.
Show that an alternative (weighted) least squares estimate of ,
obtained by minimising q* =
(c)
=
12
n
i 1 i
12 2
i
i 1
Noting that
12 4
i
i 1
12
i 1
i2
ni
i2
2
is given by
.
= 60, 710 and
12 2
i
i 1
= 650 , calculate
and
for the above data.
[8]
CT3 S2005
7
PLEASE TURN OVER
(iii)
To assess whether the single parameter model which was used in part (ii) is
appropriate for the data, a two parameter model is now considered. The model
is of the form
E Ni = i
for i = 1,
(a)
, 12.
To estimate the parameters and , a simple linear regression model
E Yi =
xi
is used, where xi = log(i) and Yi = log(Ni) for i = 1, , 12. Relate the
parameters and to the regression parameters and .
(b)
The least squares estimates of and are 0.6112 and 1.6008 with
standard errors 0.4586 and 0.2525 respectively (you are not asked to
verify these results).
Using the value for the estimate of , conduct a formal statistical test to
assess whether the form of the model suggested in (ii) is adequate.
[7]
[Total 19]
END OF PAPER
CT3 S2005
8
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
September 2005
Subject CT3
Probability and Mathematical Statistics
Core Technical
EXAMINERS REPORT
Faculty of Actuaries
Institute of Actuaries
Subject CT3 (Probability and Mathematical Statistics Core Technical)
1
2
September 2005
Examiners Report
150
1.5
100
(a)
x
(b)
Median is (101/2)th observation i.e. the mean of the 50th and 51st observations
so median = 1
(c)
Mean is higher than the median as the data are positively skewed (skewed to
the right).
Let X denote the number in the sample who support party A.
X ~ Binomial(275, 0.45)
E[X] = 275 0.45 = 123.75
V[X] = 275 0.45 0.55 = 68.0625
The normal approximation to the binomial gives, using a continuity correction,
P( X
3
116)
P( X
115.5) 1
Gamma(120, 1.2) has mean
115.5 123.75
68.0625
1
( 1)
120
120
100 and variance
1.2
1.22
(1) 0.841
83.333
X N (100,9.1292 ) by the Central Limit theorem (since the gamma variable is the
sum of 120 independent gamma(1,1.2) variables)
P X
4
120
P Z
120 100
9.129
2.191
1 0.98578 0.0142
Sample proportion = 0.83
99% CI for the population proportion is 0.83
i.e. 0.83 0.0968 i.e. (0.733, 0.927)
99% CI for percentage is thus (73.3% , 92.7%)
Page 2
[2.5758
(0.83 0.17/100)1/2]
Subject CT3 (Probability and Mathematical Statistics Core Technical)
5
September 2005
Examiners Report
n = 500 is very large, so the Central Limit Theorem justifies normality.
95% CI is x 1.96
237 1.96
s
n
137
500
237 12.0 or (£225.0, £249.0)
6
Source of variation
Between groups
Residual
Total
d.f.
3
60
63
SS
840
660
1500
MSS
280
11
F = 280/11 which equals 25.45.
Therefore there is overwhelming evidence to suggest that the four population
means are not equal (F3,60(5%) = 2.758, F3,60(1%) = 4.126).
7
(a)
x
s
8
2
1
(3256) 162.8
20
1
32562
866600
19
20
17711.7
s 133.1
(b)
Distribution must have strong positive skewness
since the s.d. is large relative to the mean and the amounts must be positive.
(i)
Using C for a claim, N for no claim, then
P(premium = 400) = P(C in year 3, regardless of the first 2 years) = p
P(premium = 400k) = P(CN in years 2/3, regardless of the first year) = p(1 p)
P(premium = 400k2) = P(NN in years 2/3, regardless of the first year) = (1 p)2
[These probabilities may be derived in other ways, such as via a tree diagram]
Page 3
Subject CT3 (Probability and Mathematical Statistics Core Technical)
(ii)
E(premium) = 400.p + 400k.p(1 p) + 400k2.(1
= 400{p + kp(1 p) + k2(1 p)2}
(iii)
For E(premium) = 300 when p = 0.1
then 0.1 + 0.09k + 0.81k2 = 0.75
0.81k2 + 0.09k
0.092 4(0.81)(0.65)
1.62
0.09
k
k
9
September 2005
Examiners Report
p)2
0.09 1.454
1.62
0.65 = 0
for 0 k 1
0.84
Pooled estimate of common population variance =
10 59
14 42
10 14
49.0833
t24 (0.025) = 2.064
95% CI for
1
124 105
10
2
is given by
2.064
49.0833
1 1
11 15
1/ 2
i.e. 19 5.74 i.e. 13.3 , 24.7
X ~ U(0,1000) , Y = min(X,800)
(i)
P( X
x| X
800)
P( X
x and X 800)
P ( X 800)
P( X x)
for 0 < x < 800
800 /1000
x /1000
800 /1000
x
for 0
800
x 800
so the conditional distribution is U(0,800)
[other reasonable arguments were given credit, e.g. the conditional
distribution is simply a scaled version of the original uniform distribution on a
restricted range .]
(ii)
E[Y] = E[X|X < 800] P(X < 800) + 800 P(X
400
= 480
Page 4
800
1000
800
200
1000
800)
Subject CT3 (Probability and Mathematical Statistics Core Technical)
September 2005
Examiners Report
(iii)
Y is approximately normal with expectation 480 by Central Limit Theorem
(iv)
E[X | X < 800] = 400 whereas E[Y] = 480.
The higher value for E[Y] results from 20% of the Y values being 800 (and
80% being between 0 and 800).
11
(a)
For 0 r < 0.55
0.55 r < 0.8
0.8 r < 0.95
0.95 r 1
n=0
n=1
n=2
n=3
[OR any equivalent allocation which reflects the probabilities of the 4 values
of N.]
12
(b)
0.6221
0.1472
0.9862
(i)
(a)
n=1
n=0
n=3
The xi s are known to be such that xi
is a scaled form of e
x
5 , therefore have density which
for 5 < x < .
The scaling constant k is such that
k. e
x
dx 1
5
k[ e
x
]5
1
k .e
5
1
k
e5
[Note: this can be argued in other ways; e.g. by referring to a
conditional density and dividing by P(X > 5)]
Page 5
Subject CT3 (Probability and Mathematical Statistics Core Technical)
n
(b)
xi
e5 e
L( )
n 5n
e
September 2005
Examiners Report
xi
e
i 1
log L( )
n log
d
log L( )
d
5n
n
xi
5n
xi
n
equate to zero for MLE
n
xi 5n
i 1
[OR It could be noted that X 5 ~ exp( ) and that the MLE is
therefore the reciprocal of the mean of the data ( xi 5) giving the
required answer]
(ii)
(c)
n = 10, xi = 71
(a)
L( ) (1 e
log L( )
5
10
71 50
) m (e
m log(1 e
d
5me
log L( )
d
1 e
5
5
5
n
m n
)n
) n log(e
5
)
5
5
equate to zero for MLE
e
0.476
5n
e
5
1 e
5
n
m
1
m n
log(
)
5
n
[OR Reason via the MLE for a binomial p = P(X > 5) such that
n
p
and p e 5 ]
m n
(b)
Page 6
m = 120, n = 10
0.513
Subject CT3 (Probability and Mathematical Statistics Core Technical)
(iii)
(a)
(1 e
5
n
e
September 2005
Examiners Report
)m is the likelihood of observing m policies with duration < 5
xi
is the likelihood of observing the actual durations x1,
, xn
i 1
and independence leads to the product of these
log L( )
m log(1 e
d
5me
log L( )
d
1 e
5
5
) n log
n
5
xi
xi
equate to zero and the solution gives the MLE.
(b)
All three are re-assuringly close.
The pooled estimate is between the first two (as expected, but it is
closer to 0.513).
13
return
no return
purchase
69
32
101
no purchase
68
137
51
83
119
220
(i)
(a)
(b)
(c)
69/101 (= 0.6832)
68/119 (= 0.5714)
69/137 (= 0.5036)
(ii)
H0: population proportions of those who intend to return are equal
v H1: not H0
Proportion of purchasers
2
1
69 /101 ; proportion of non-purchasers
68 /119
Under H0, estimate of common proportion who intend to return = 137/220
Observed value of D
1
2
0.1117
137 83 1
1
Estimated standard error of D =
220 220 101 119
1/ 2
0.06558
Page 7
Subject CT3 (Probability and Mathematical Statistics Core Technical)
P-value = 2
=2
September 2005
Examiners Report
P(D > 0.1117) = 2 P(Z > 0.1117/0.06558) = 2
(1 0.95543) = 0.08914 (i.e. 8.9%)
P(Z > 1.70)
There is not sufficient evidence (using a two-sided test) to justify rejecting H0
i.e. there is not sufficient evidence to justify concluding that the intention to
return depends on whether or not a purchase was made.
(iii)
H0: no association between attributes v H1: not H0
Expected frequencies under H0 in brackets:
return
no return
Test statistic = 6.12
P-value = P
2
1
1
62.9
2.90
purchase
69 (62.9)
32 (38.1)
101
no purchase
68 (74.1)
51 (44.9)
119
1
1
1
74.1 38.1 44.9
137
83
220
2.90
1 0.9114 0.0886 (i.e. 8.9%)
There is not sufficient evidence to justify rejecting H0 i.e. there is not
sufficient evidence to justify concluding that the intention to return is
associated with purchasing status.
(iv)
The two approaches complement each other:
the P-values are the same
the conclusions are the same.
[Note: there is a formal connection: the
the z value in (ii) (1.70).]
Page 8
2
1 value
in (iii) (2.90) is the square of
Subject CT3 (Probability and Mathematical Statistics Core Technical)
(i)
(a)
Examiners Report
Points are shown on scatterplot.
40
Number of deaths
14
September 2005
30
20
10
0
0
5
10
Quarter
(b)
The mean number of deaths increases with an increasing rate with
quarter.
The variance also appears to increase with quarter.
(ii)
(a)
q
i 2 )2
(ni
dq
d
2 i 2 (ni
dq
d
2 i 2 (ni
0
i 2 ni
i4
i 2 ni
i4
(
d 2q
d
2
i2 )
i2 ) 0
0
.
2 i4
0
minimum.)
Page 9
Subject CT3 (Probability and Mathematical Statistics Core Technical)
(b)
q*
(ni
dq*
d
2
i 2 )2
ni
i
i2
dq*
d
i
0
ni
i
September 2005
2
i
i
i2
ni
0
ni
i2
(
d 2 q*
d
i2
2
2
i 2 ni
(c)
i
4
ni
i
(iii)
(a)
E Ni
15694
60710
174
650
2
minimum.)
0
0.259
0.268
i
Taking logs gives
log E Ni
Thus
[OR
(b)
log( i ) = log
= log and
log(i ) log
= .
e and
.]
= 1.6008 s.e.( ) = 0.2525
H0 :
t
= 2 v H1 :
2
s.e.( )
2
1.6008 2
0.2525
1.58
Compare with a t-distribution with 10 d.f.
Page 10
xi
Examiners Report
Subject CT3 (Probability and Mathematical Statistics Core Technical)
September 2005
Examiners Report
As the 5% critical value of a two-tailed test is 2.228, do not reject the
null hypothesis.
Therefore, the model used in (ii) with
=
= 2 seems appropriate.
END OF EXAMINERS REPORT
Page 11
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
30 March 2006 (am)
Subject CT3
Probability and Mathematical Statistics
Core Technical
Time allowed: Three hours
INSTRUCTIONS TO THE CANDIDATE
1.
Enter all the candidate and examination details as requested on the front of your answer
booklet.
2.
You must not start writing your answers in the booklet until instructed to do so by the
supervisor.
3.
Mark allocations are shown in brackets.
4.
Attempt all 12 questions, beginning your answer to each question on a separate sheet.
5.
Candidates should show calculations where this is appropriate.
Graph paper is required for this paper.
AT THE END OF THE EXAMINATION
Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this
question paper.
In addition to this paper you should have available the 2002 edition of the
Formulae and Tables and your own electronic calculator.
CT3 A2006
Faculty of Actuaries
Institute of Actuaries
1
The stem and leaf plot below gives the surrender values (to the nearest 1,000) of 40
endowment policies issued in France and recently purchased by a dealer in such
policies in Paris. The stem unit is 10,000 and the leaf unit is 1,000.
5
5
6
6
7
7
8
8
9
9
3
6
02
5779
122344
556677899
1123444
567778
024
6
Determine the median surrender value for this batch of policies.
2
[2]
In a certain large population 45% of people have blood group A. A random sample of
300 individuals is chosen from this population.
Calculate an approximate value for the probability that more than 115 of the sample
have blood group A.
[3]
3
A random sample of size 10 is taken from a normal distribution with mean
variance 2 = 1.
= 20 and
Find the probability that the sample variance exceeds 1, that is find P(S2 > 1).
4
[3]
In a one-way analysis of variance, in which samples of 10 claim amounts (£) from
each of three different policy types are being compared, the following means were
calculated:
y1
276.7 ,
y2
254.6 ,
y3
296.3
with residual sum of squares SSR given by
3 10
SS R
( yij
yi ) 2 15,508.6
i 1 j 1
Calculate estimates for each of the parameters in the usual mathematical model, that
is, calculate , 1, 2 , 3 , and 2 .
[4]
CT3 A2006 2
5
6
A large portfolio of policies is such that a proportion p (0 < p < 1) incurred claims
during the last calendar year. An investigator examines a randomly selected group of
25 policies from the portfolio.
(i)
Use a Poisson approximation to the binomial distribution to calculate an
approximate value for the probability that there are at most 4 policies with
claims in the two cases where (a) p = 0.1 and (b) p = 0.2.
[3]
(ii)
Comment briefly on the above approximations, given that the exact values of
the probabilities in part (i), using the binomial distribution, are 0.9020 and
0.4207 respectively.
[2]
[Total 5]
One variable of interest, T, in the description of a physical process can be modelled as
T = XY where X and Y are random variables such that X ~ N(200, 100) and Y depends
on X in such a way that Y|X = x ~ N(x, 1).
Simulate two observations of T, using the following pairs of random numbers
(observations of a uniform (0, 1) random variable), explaining your method and
calculations clearly:
Random numbers
0.5714 , 0.8238
0.3192 , 0.6844
[6]
7
Let (X1, X2 , , Xn) be a random sample from a uniform distribution on the interval
( , ), where is an unknown positive number.
A particular sample of size 5 gives values 0.87, 0.43, 0.12, 0.92, and 0.58.
(i)
Draw a rough graph of the likelihood function L( ) against
(ii)
State the value of the maximum likelihood estimate of .
CT3 A2006 3
for this sample.
[3]
[2]
[Total 5]
PLEASE TURN OVER
8
The events that lead to potential claims on a policy arise as a Poisson process at a rate
of 0.8 per year. However the policy is limited such that only the first three claims in
any one year are paid.
(i)
Determine the probabilities of 0, 1, 2 and 3 claims being paid in a particular
year.
[2]
(ii)
The amounts (in units of £100) for the claims paid follow a gamma
distribution with parameters = 2 and = 1.
Calculate the expectation of the sum of the amounts for the claims paid in a
particular year.
[3]
(iii)
9
Calculate the expectation of the sum of the amounts for the claims paid in a
particular year, given that there is at least one claim paid in the year.
[2]
[Total 7]
The total claim amount on a portfolio, S, is modelled as having a compound
distribution
S = X1 + X2 +
+ XN
where N is the number of claims and has a Poisson distribution with mean , Xi is the
amount of the ith claim, and the Xi s are independent and identically distributed and
independent of N. Let MX(t) denote the moment generating function of Xi.
(i)
Show, using a conditional expectation argument, that the cumulant generating
function of S, CS(t), is given by
CS(t) =
MX(t) 1}.
Note: You may quote the moment generating function of a Poisson random
variable from the book of Formulae and Tables.
[4]
(ii)
Calculate the variance of S in the case where
variance 10.
CT3 A2006 4
= 20 and X has mean 20 and
[2]
[Total 6]
10
A marketing consultant was commissioned to conduct a questionnaire survey of the
clients of a financial company. The total number of respondents was 650, of whom
220 had investments above a specified threshold.
(i)
Each respondent who had investments above the threshold was asked about
the percentage of these investments that was held in the form of a certain type
of trust. The respondents answered by ticking appropriate boxes and the
results led to the following frequency distribution.
percentage
frequency
(ii)
< 10
22
10 25
76
25 50
73
> 50
49
(a)
Present these data graphically using a carefully drawn histogram.
(b)
Calculate the mean percentage for the full set of 220 such respondents,
assuming that the frequencies in each category are uniformly spread
over the corresponding range.
[5]
Calculate a 95% confidence interval for the percentage of such investors who
would have investments above the threshold.
[4]
The same respondents with investments referred to in part (i) were also asked to
specify their satisfaction with the current return received from their full portfolio of
investment. This was in the form of a four-point qualitative scale: very satisfied, quite
satisfied, a little disappointed, very disappointed. The following two-way table of
frequencies was obtained.
<10
percentage in type of trust
10 25
25 50
>50
1
8
10
3
very satisfied
quite satisfied
a little disappointed
very disappointed
6
29
37
4
7
36
28
2
6
27
15
1
In order to investigate whether there is any relationship between the percentage in
such trusts and satisfaction with current return, a 2 test is to be performed.
(iii)
Calculate the expected frequencies for the above table under an appropriate
hypothesis (which should be stated) and comment on why it would be
inappropriate to carry out a 2 test directly with these data.
[3]
(iv)
Combining the very satisfied and quite satisfied categories together and
the a little disappointed and very disappointed categories together results
in the following reduced two-way table.
<10
satisfied
disappointed
9
13
percentage in type of trust
10 25
25 50
>50
35
41
43
30
33
16
Perform the required 2 test at the 5% level using this reduced table and
comment on your conclusion.
[7]
[Total 19]
CT3 A2006 5
PLEASE TURN OVER
11
An actuary has been advised to use the following positively-skewed claim size
distribution as a model for a particular type of claim, with claim sizes measured in
units of £100,
f ( x; )
x2
2
3
exp
x
: 0
x
,
with moments given by E[X] = 3 , E[X2] = 12
0
2
and E[X3] = 60 3.
(i)
Determine the variance of this distribution and calculate the coefficient of
skewness.
[4]
(ii)
Let X1, X2,
, Xn be a random sample of n claim sizes for such claims.
Show that the maximum likelihood estimator (MLE) of
is given by
and show that it is unbiased for .
(iii)
A sample of n = 50 claim sizes yields xi
X
3
[5]
313.6 and xi2
2, 675.68 .
(a)
Calculate the MLE
.
(b)
Calculate the sample variance and comment briefly on its comparison
with the variance of the distribution evaluated at .
(c)
Given that the sample coefficient of skewness is 1.149, comment
briefly on its comparison with the coefficient of skewness of the
distribution.
[4]
(iv)
12
(a)
Write down a large-sample approximate 95% confidence interval for
the mean of the distribution in terms of the sample mean x and the
sample variance s2. Hence obtain an approximate 95% confidence
interval for and evaluate this for the data in part (iii) above.
(b)
Evaluate the variance of the distribution at both the lower and upper
limits of this confidence interval and comment briefly with reference to
your answer in part (iii)(b) above.
[5]
[Total 18]
In an experiment to compare the effects of vaccines of differing strengths intended to
give protection to children against a particular condition, twelve batches of vaccine
were tested in twelve equal-sized groups of children. The percentages of children
who subsequently remained healthy after exposure to the condition, named the PRH
values, were recorded. The strength of each batch of vaccine was measured by an
independent test and recorded as the SV value.
CT3 A2006 6
The recorded values are:
Batch:
1
PRH (y): 16
SV (x):
0.9
x 38.4 ;
(i)
2
3
4
68 23 35
1.6 2.3 2.7
y
528 ;
5
6
7
8
9
42 41 46 48 52
3.0 3.3 3.7 3.8 4.1
x 2 137.16 ;
y2
10 11 12
50 54 53
4.2 4.3 4.5
25, 428 ;
xy 1, 778.4
Draw a rough plot of the data to show the relationship between the SV and
PRH values.
[2]
It is evident that one of the observations is out of line and so may have an undue
effect on any regression analysis. You are asked to investigate this as follows.
(ii)
(a)
Calculate the total, regression, and error sums of squares for a leastsquares linear regression analysis for predicting PRH values from SV
values using all 12 data observations.
(b)
Determine the coefficient of determination R2.
(c)
Determine the equation of the fitted regression line.
(d)
Examine whether or not there is evidence, at the 5% level of testing, to
enable one to conclude that the slope of the underlying regression
equation is non-zero.
[11]
The details of the regression analysis after removing the data for batch 2 are given in
the box below.
Regression equation: y = 3.76 + 11.4 x
Coef
Stdev
t-ratio
p-val
Intercept
x
0.255
0.000
3.757
11.377
Analysis of Variance
Source
df
SS
Regression 1
Error
9
Total
10
(iii)
1486.9
80.8
1567.6
3.092
0.8838
MS
1486.9
8.98
1.22
12.9
F
p-val
165.69
0.000
(a)
Comment on the main differences in the results of the regression
analysis resulting from removing the data for batch 2.
(b)
Calculate a 95% confidence interval for the expected (mean) PRH
value for a batch of vaccine with SV value 3.5.
[9]
[Total 22]
END OF PAPER
CT3 A2006 7
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
April 2006
Subject CT3
Probability and Mathematical Statistics
Core Technical
EXAMINERS REPORT
Introduction
The attached subject report has been written by the Principal Examiner with the aim of
helping candidates. The questions and comments are based around Core Reading as the
interpretation of the syllabus to which the examiners are working. They have however given
credit for any alternative approach or interpretation which they consider to be reasonable.
M Flaherty
Chairman of the Board of Examiners
June 2006
Faculty of Actuaries
Institute of Actuaries
Subject CT3 (Probability and Mathematical Statistics Core Technical)
April 2006
Examiners Report
Comments
Comments on answers presented by candidates are given below. Note that in some cases
variations on the solutions given are possible
the examiners gave credit for all sensible
comments and correct solutions.
Question 6
This question was the worst one on the paper as regards quality of answers. The
question linked the concept of a conditional distribution with the simulation of
observations of normal random variables (Core Reading Unit 6, section 1.3 and Unit 4,
section 5.2). There were few good answers. Many candidates simply did not submit
answers, suggesting that they were not familiar with the basic approach to the
simulation of observations despite the fact that there were short questions on this
topic in both immediately previous papers, for which solutions are readily available.
Question 7
The likelihood function in this question is not of standard form and expressing and
graphing it correctly requires a good understanding of the likelihood concept. Many
candidates did not think clearly about the range of values of for which the likelihood is
positive and for which it is zero and so got the wrong graph.
Question 8
Many candidates ignored the fact that only the first three claims in any one year are
paid . Suppose Y denotes the numbers of claims which arise, then
Y ~ Poisson(0.8). Suppose X denotes the number of claims which are paid. Many
candidates worked with the set of probabilities P(X = i) = P(Y = i), i = 0,1,2,3.
But these four probabilities do not sum to 1 and so do not provide a proper probability
distribution for X. What is required is the set of four probabilities
P(X = i) = P(Y = i), i = 0,1,2 together with P(X = 3) = P(Y 3).
Question 9
The wording of the question made it clear that candidates could assume the mgf of a
Poisson random variable and, armed with this information, should use a conditional
expectation argument . Full marks were not awarded to candidates who jumped into
the middle of the argument by assuming the mgf of a compound Poisson random
variable.
Question 10
Candidates should be aware that when constructing a histogram with unequal group
widths one must ensure that the areas (and not the heights) of the rectangles are
proportional to the frequencies.
In part (ii), many candidates calculated a confidence interval for a different proportion
to the one asked for.
Page 2
Subject CT3 (Probability and Mathematical Statistics Core Technical)
April 2006
Examiners Report
Question 11
Many candidates were unsure of the definition of the coefficient of skewness (Core
Reading Unit 3, section 3.4).
Question 12
In part (ii)(a), many candidates calculated S yy , S xx , and S xy but did not go on to
calculate the regression and error sums of squares (SSREG, SSRES) as asked for.
In part (iii)(a) many candidates failed to make any of the most pertinent possible
comments (but credit was given for relevant comments other than those given here).
Page 3
Subject CT3 (Probability and Mathematical Statistics Core Technical)
April 2006
Examiners Report
1
n = 40, so the median is the 20.5th observation, which is ½(7.7+7.8) = 7.75.
This represents 77,500.
2
If X is the number in the sample with group A, then X has a binomial (300, 0.45)
distribution, so
E[X] = 300
0.45 = 135 and Var[X] = 300
0.45
0.55 = 74.25.
Then, using the continuity correction,
P(X > 115) = P(X > 115.5) 1
3
n 1 S2
2
~
P S2 1
1 0.5627
4
1
2
3
2
5
Page 4
(i)
so here 9S 2 ~
2
n 1
P
2
9
0.437
115.5 135
=1
74.25
( 2.26) =
2
9
9
(tables p165)
1
(276.7 254.6 296.3) 275.87
3
276.7 275.87 0.83
254.6 275.87
21.27
296.3 275.87 20.43
SS R
27
15508.6
27
574.4
Let X = number of policies with claims
So X ~ binomial(25, p).
Poisson approximation is X Poisson(25p).
(a)
using Poisson(2.5)
P(X 4) = 0.89118 from tables [or evaluation]
(b)
using Poisson(5)
P(X 4) = 0.44049 again from tables
(2.26) = 0.99.
Subject CT3 (Probability and Mathematical Statistics Core Technical)
(ii)
April 2006
Examiners Report
in (a) error is 0.8912 0.9020 = 0.0108
in (b) error is 0.4405 0.4207 = 0.0198
The approximation is valid for small p , and, as p is smaller in (a), this gives
a better approximation as noted with the smaller error.
[Note: candidates may also comment on the fact that the sample size 25 is not
large and so we would not expect the Poisson approximations to be very
good anyway. In fact the key to the approximations is the small p and here the
given approximations are quite good]
7
Solving P(Z < z) = 0.5714 z = 0.180
Solving P(Z < z) = 0.8238 z = 0.930
t = 201.80 202.73 = 40911
Solving P(Z < z) = 0.3192 z = 0.470
Solving P(Z < z) = 0.6844 z = 0.480
t = 195.3 195.78 = 38236
(i)
L( )
n
1
2
c
1
x = 200 + 10(0.180) = 201.80
y = 201.80 + 0.930 = 202.73
x = 200 + 10( 0.470) = 195.3
y = 195.3 + 0.480 = 195.78
n
for
< xi < , i = 1, 2,
, n and L( ) = 0 otherwise
So, as increases from zero, L( ) is zero until it reaches the largest
observation in absolute value i.e. max |xi|, i = 1, 2, , n. For the data given,
this value is 0.92.
It is then a decreasing function . Hence the graph is as below:
L(theta)
6
0
0
(ii)
0.92
The maximum value of L( ) is attained at the largest absolute value of the
data. The ML estimate of is 0.92.
Page 5
Subject CT3 (Probability and Mathematical Statistics Core Technical)
8
(i)
April 2006
Examiners Report
By subtraction using entries in tables for Poisson(0.8), the probabilities for the
Poisson distribution for 0, 1, 2 and 3 are: [or by evaluation]
0.44933, 0.35946, 0.14379 and (1 0.95258) = 0.04742
(ii)
Let N = number of claims paid and let X1,
S = Xi is the sum of the amounts.
, Xn be the claim amounts then
E[S] = E[N]E[X]
Here E[N] = 1(0.35946) + 2(0.14379) + 3(0.04742) = 0.7893
and E[X] = 2/1 = 2 from gamma(2,1)
So E[S] = (0.7893)(2) = 1.5786 = £157.86
(iii)
Given that N > 0, divide the probabilities in part (i) by (1 0.44933) =
0.55067 to give the probabilities for 1, 2 and 3 claims paid as:
0.6528, 0.2611 and 0.0861
E[N] = 1(0.6528) + 2(0.2611) + 3(0.0861) = 1.4333
So E[S] = (1.4333)(2) = 2.8666 = £286.66
9
(i)
MS(t) = E[etS] = E[E[etS|N]]
Now E[etS|N = n] = E[exp(tX1 +
+ tXn)] = E[exp(tXi)] = {MX(t)}n
MS(t) = E[{MX(t)}N] = E[exp{NlogMX(t)}] = MN{logMX(t)}
= exp[ MX(t) 1}] since N ~ Poisson( )
CS(t) = logMS(t) = MX(t) 1}
(ii)
V[S] = CS (0) =
MX (0)} = E[X2] = 20(10 + 202) = 8200
OR V[S] = E[N]V[X] + V[N]{E[X]}2 = 20 10 + 20 202 = 8200
Page 6
Subject CT3 (Probability and Mathematical Statistics Core Technical)
10
(i)
(a)
April 2006
Examiners Report
The key feature of the histogram is that the areas of the four rectangles
should be proportional to the frequencies.
See histogram below.
(b)
Mean is calculated from the following frequency distribution:
5
22
x
f
f = 220,
(ii)
17.5
76
fx = 7852.5
Estimated proportion is p
220
650
37.5
73
7852.5
220
x
0.338
75
49
35.7%
(or 33.8%)
95% confidence interval for underlying proportion is
p 1.96
p (1 p )
650
0.338 1.96
0.338(0.662)
650
as a percentage: 33.8%
0.338 0.036
3.6% or (30.2%, 37.4%)
Page 7
Subject CT3 (Probability and Mathematical Statistics Core Technical)
(iii)
Examiners Report
Under the null hypothesis of no association between percentage in type of trust
and satisfaction with current return, expected frequencies are
2.0
10.0
9.0
1.0
22
6.9
34.5
31.1
3.5
76
6.6
33.2
29.9
3.3
73
six are less than 5 which would invalidate a
(iv)
April 2006
4.5
22.3
20.0
2.2
49
2
20
100
90
10
220
test
expected frequencies (e) are
12.000
10.000
41.455
34.545
39.818
33.182
26.727
22.273
6.455
+6.455
+3.182
3.182
+6.273
6.273
1.005
1.206
0.254
0.305
1.472
1.767
table of residuals (o e) is
3.000
+3.000
table of contributions to
2
is
0.750
0.900
giving
2
= 7.659 on 3 d.f.
2
3 (5%)
7.815
must accept the null hypothesis that there is no
relationship between percentage in type of trust and satisfaction with current
return.
However this decision to accept is marginal at the 5% level and there is some
evidence, but not strong, to suggest that satisfaction improves as the
percentage increases.
Page 8
Subject CT3 (Probability and Mathematical Statistics Core Technical)
11
(i)
2
= E[X 2 ]
[E(X)]2 = 12
2
= E[X 3] 3 E[X2] + 2
3
= 60
(3 )2 = 3
April 2006
Examiners Report
2
3
3(3 )(12 2) + 2(3 )3
= (60
108 + 54)
3
3
=6
coefficient of skewness =
3
3
3
6
1.155
2 3
( 3
)
[OR: note that X ~ gamma(3,1/ ) and use formulae in tables
2
so var = 3 2 and coef. of skew. =
]
3
n
(ii)
L( )
i 1
xi2
2
exp(
3
xi
xi2
)
2
n 3n
log L( ) log( xi2 ) n log 2 3n log
xi
xi
3n
log L( )
xi
exp
2
equate to zero:
xi
3n
xi
3n
2
2
this clearly maximises L( )
Xi
3n
So MLE is
(iii)
E
1
E X
3
(a)
x
(b)
s2
2
2
log L( ) ]
X
3
1
E X
3
313.6
50
[or consider
6.272
1
3
3
unbiased
6.272
3
2.091
1
313.62
(2675.68
) 14.465
49
50
3
2
and 3
2
13.117
s2 is a bit larger but still quite close
Page 9
Subject CT3 (Probability and Mathematical Statistics Core Technical)
(iv)
April 2006
Examiners Report
(c)
sample coefficient 1.149 is very close to the distribution value 1.155
(a)
s2
is x 1.96
n
approximate 95% CI for
as
= 3 , divide by 3 for an approximate 95% CI for
1
s2
x 1.96
3
n
for data:
1
14.465
6.272 1.96
3
50
2.091 0.351
(b)
2
=3
2
or
(1.740, 2.442)
= 9.083 at lower limit of 1.740
= 17.890 at upper limit of 2.442
s2 = 14.465 is well within these values
confirming that s 2 is quite close to 3 2 .
12
(i)
70
60
PRH
50
40
30
20
1
2
3
4
SV
(ii)
Page 10
(a)
SSTOT = Syy = 25428
5282/12 = 2196
Sxx = 137.16 38.42/12 = 14.28 , Sxy = 1778.4
(38.4 528)/12 = 88.8
SSREG = 88.82/14.28 = 552.20, SSRES = 2196
552.20 = 1643.80
Subject CT3 (Probability and Mathematical Statistics Core Technical)
(b)
R2 = 552.2/2196 = 0.251 (25.1%)
(c)
y = a + bx:
April 2006
b 88.8 /14.28 6.2185
a 528 /12 88.8 /14.28 (38.4 /12)
Examiners Report
24.101
Fitted line is y = 24.101 + 6.2185x
(d)
s.e. b
1643.8 /10
14.28
1/ 2
3.3928
Observed t = (6.2185 0)/3.3928 = 1.833 < t10(0.025) = 2.228
so we do not have evidence at the 5% level of testing to justify
rejecting b = 0 and concluding that the underlying slope is non-zero.
(iii)
(a)
Large change in slope (and intercept) of fitted line.
The total and error sums of squares are much reduced.
The fit of the linear regression model is much improved (R2 is much
increased from 25.1% to 94.9%).
We have overwhelming evidence to justify concluding that the slope is
non-zero.
(b)
Fitted PRH value at SV = 3.5 is 3.757 + (11.377 3.5) = 43.577
n 11,
x 38.4 1.6 36.8,
x 2 137.16 1.62 134.6
Sxx = 11.4873
s.e. of estimation =
1
11
3.5 36.8 /11
11.4873
1/ 2
2
8.98
0.9138
t9(0.025) = 2.262
95% CI for expected PRH is 43.577
i.e. 43.577
2.067
(2.262
0.9138)
or (41.51, 45.64)
END OF EXAMINERS REPORT
Page 11
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
7 September 2006 (am)
Subject CT3
Probability and Mathematical Statistics
Core Technical
Time allowed: Three hours
INSTRUCTIONS TO THE CANDIDATE
1.
Enter all the candidate and examination details as requested on the front of your answer
booklet.
2.
You must not start writing your answers in the booklet until instructed to do so by the
supervisor.
3.
Mark allocations are shown in brackets.
4.
Attempt all 12 questions, beginning your answer to each question on a separate sheet.
5.
Candidates should show calculations where this is appropriate.
Graph paper is required for this paper.
AT THE END OF THE EXAMINATION
Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this
question paper.
In addition to this paper you should have available the 2002 edition of the
Formulae and Tables and your own electronic calculator.
CT3 S2006
Faculty of Actuaries
Institute of Actuaries
1
A bag contains 8 black and 6 white balls. Two balls are drawn out at random, one
after the other and without replacement.
Calculate the probabilities that:
2
(i)
The second ball drawn out is black.
(ii)
The first ball drawn out was white, given that the second ball drawn out is
black.
[2]
[Total 3]
Let A and B denote independent events.
Show that A and B , the complement of event B, are also independent events.
3
[1]
[3]
Consider 12 independent insurance policies, numbered 1, 2, 3, , 12, for each of
which a maximum of 1 claim can occur. For each policy, the probability of a claim
occurring is 0.1.
Find the probability that no claims arise on the group of policies numbered 1, 2, 3, 4,
5 and 6, and exactly 1 claim arises in total on the group of policies numbered 7, 8, 9,
10, 11, and 12.
[3]
4
In a large portfolio 65% of the policies have been in force for more than five years.
An investigation considers a random sample of 500 policies from the portfolio.
Calculate an approximate value for the probability that fewer than 300 of the policies
in the sample have been in force for more than five years.
[3]
5
In a random sample of 200 policies from a company s private motor business, there
are 68 female policyholders and 132 male policyholders.
Let the proportion of policyholders who are female in the corresponding population of
all policyholders be denoted .
Test the hypotheses
H0:
0.4 v H1:
< 0.4
stating clearly the approximate probability value of the observed statistic and your
conclusion.
[4]
CT3 S2006
2
6
7
It is assumed that claims arising on an industrial policy can be modelled as a Poisson
process at a rate of 0.5 per year.
(i)
Determine the probability that no claims arise in a single year.
[1]
(ii)
Determine the probability that, in three consecutive years, there is one or more
claims in one of the years and no claims in each of the other two years.
[2]
(iii)
Suppose a claim has just occurred. Determine the probability that more than
two years will elapse before the next claim occurs.
[2]
[Total 5]
A commuter catches a bus each morning for 100 days. The buses arrive at the stop
according to a Poisson process, at an average rate of one per 15 minutes, so if Xi is the
waiting time on day i, then Xi has an exponential distribution with parameter
1
15
so
E[Xi] = 15, Var[Xi] = 152 = 225.
(i)
Calculate (approximately) the probability that the total time the commuter
spends waiting for buses over the 100 days exceeds 27 hours.
[3]
(ii)
At the end of the 100 days the bus frequency is increased, so that buses arrive
at one per 10 minutes on average (still behaving as a Poisson process). The
commuter then catches a bus each day for a further 99 days. Calculate
(approximately) the probability that the total time spent waiting over the
whole 199 days exceeds 40 hours.
[3]
[Total 6]
CT3 S2006
3
PLEASE TURN OVER
8
Let X1 denote the mean of a random sample of size n from a normal population with
mean
2
1,
and variance
and let X 2 denote the mean of a random sample also of
size n from a normal population with the same mean
two samples are independent.
but with variance
2
2.
The
Define W as the weighted average of the sample means
W
X1 (1
)X2
(i)
Show that W is an unbiased estimator of .
[1]
(ii)
Obtain an expression for the mean square error of W.
[2]
(iii)
Show that the value of
given by
2
2
2
1
2
2
for which W has minimum mean square error is
,
and verify that the optimum corresponds to a minimum.
9
[3]
(iv)
Consider the special case when the variances of the two random samples are
equal to a common value 2 . State (do not derive) the maximum likelihood
estimator of calculated from the combined samples, and compare it with the
estimator obtained in (iii).
[2]
[Total 8]
(i)
Show that for continuous random variables X and Y:
E[Y] = E[E(Y|X)].
(ii)
Suppose that a random variable X has a standard normal distribution, and the
conditional distribution of a Poisson random variable Y, given the value of
X = x, has expectation g(x) = x2 + 1.
Determine E[Y] and Var[Y].
CT3 S2006
[3]
4
[5]
[Total 8]
10
Let (X1, X2,
, Xn) be a random sample of a gamma(4.5, ) random variable, with
sample mean X .
(i)
(ii)
2
9n .
(a)
Using moment generating functions, show that 2 nX ~
(b)
Construct a 95% confidence interval for , based on X and the result
in (i)(a) above.
(c)
Evaluate the interval in (i)(b) above in the case for which a random
sample of 10 observations gave a value
xi 21.47
[9]
(a)
Show that the maximum likelihood estimator of
is given by
4.5
.
X
(b)
Show that the asymptotic standard error of
4.5n
(c)
1/ 2
is estimated by
.
Construct a 95% confidence interval for based on the asymptotic
distribution of , and evaluate this interval in the case for which a
random sample of 100 observations gave a value
xi 225.3.
[9]
[Total 18]
CT3 S2006
5
PLEASE TURN OVER
11
A survey of financial institutions which offered tax-efficient savings accounts was
conducted. These accounts had limits on the amounts of money that could be
invested each year. The study was interested in comparing the maturity values
achieved by investing the maximum possible amount each year over a certain time
period.
The following values (in units of £1,000 and rounded to 2 decimal places) are the
maturity values for such investments for 8 high street banks and 12 other banks (i.e.
those without high street branches).
High street banks (x1): 11.91, 11.87, 11.83, 11.66, 11.53, 11.49, 11.49, 11.42
( x1 = 93.20, x12 = 1,086.0470)
Other banks (x2): 12.23, 12.17, 12.16, 11.90, 11.82, 11.77, 11.74, 11.70, 11.64, 11.60,
11.55, 11.50
( x2 = 141.78, x22 = 1,675.8224)
(i)
Draw a diagram in which the maturity values for high street and other banks
may be compared.
[2]
(ii)
Calculate a 95% confidence interval for the difference between the means of
the maturity values for high street and other banks, and comment on any
implications suggested by the interval.
[6]
(iii)
(a)
Show that a test of the equality of variances of maturity values for high
street banks and other banks is not significant at the 5% level.
(b)
Comment briefly on the validity of the assumptions required for the
interval in (ii).
[4]
The following values (in units of £1,000 and rounded to 2 decimal places) are the
maturity values for the maximum possible investment for a random sample of 12
building societies (a different kind of financial institution).
Building societies (x3): 12.40, 12.19, 12.06, 12.01, 12.00, 11.97, 11.94, 11.92, 11.88,
11.86, 11.81, 11.79
( x3 = 143.83, x32 = 1,724.2449)
(iv)
Add further points to your diagram in part (i) such that the maturity values for
all three types of financial institution may be compared.
[1]
(v)
Use one-way analysis of variance to compare the maturity values for the three
different types of financial institution, and comment briefly on the validity of
the assumptions required for analysis of variance.
[6]
(vi)
Interpret the results of the statistical analyses conducted in (ii) and (v).
[2]
[Total 21]
CT3 S2006
6
12
A development engineer examined the relationship between the speed a vehicle is
travelling (in miles per hour, mph), and the stopping distance (in metres, m) for a new
braking system fitted to the vehicle. The following data were obtained in a series of
independent tests conducted on a particular type of vehicle under identical road
conditions.
Speed of vehicle (x):
Stopping distance (y):
x = 280
(i)
y = 241
10
5
20
10
x 2 = 14,000
30
23
40
34
50
40
y 2 = 11,951
60
54
70
75
xy = 12,790
Construct a scatterplot of the data, and comment on whether a linear
regression is appropriate to model the relationship between the stopping
distance and speed.
[4]
(ii)
Calculate the equation of the least-squares fitted regression line.
[5]
(iii)
Calculate a 95% confidence interval for the slope of the underlying regression
line, and use this confidence interval to test the hypothesis that the slope of the
underlying regression line is equal to 1.
[5]
(iv)
Use the fitted line obtained in part (ii) to calculate estimates of the stopping
distance for a vehicle travelling at 50 mph and for a vehicle travelling at 100
mph.
Comment briefly on the reliability of these estimates.
END OF PAPER
CT3 S2006
7
[4]
[Total 18]
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
September 2006
Subject CT3 — Probability and Mathematical Statistics
Core Technical
EXAMINERS’ REPORT
Introduction
The attached subject report has been written by the Principal Examiner with the aim of
helping candidates. The questions and comments are based around Core Reading as the
interpretation of the syllabus to which the examiners are working. They have however given
credit for any alternative approach or interpretation which they consider to be reasonable.
M A Stocker
Chairman of the Board of Examiners
November 2006
© Faculty of Actuaries
© Institute of Actuaries
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2006 — Examiners’ Report
Comments
Comments on answers presented by candidates are given below. Note that in some cases
variations on the solutions given are possible — the examiners gave credit for all
sensible comments and correct solutions.
The most common problems noted by the examiners are summarised below.
Some candidates were unsure of basic concepts in probability (such as the independence
of two events) and gave poor answers to Questions 2 and 3.
Question 5
Many candidates used πˆ = 0.34 (wrongly) rather than π = 0.4 (correctly) in the
expression for the standard error of the estimate (the sample proportion) under H0.
However, it makes little difference numerically, and the examiners were generous on this
point when marking.
Question 7
was poorly attempted, with many candidates failing to realise that the distribution of the
total waiting time can be approximated by a normal distribution, by virtue of the central
limit theorem.
Question 8
Some candidates did not know the result on the variance of the mean of a random
sample of size n, namely Var ⎡⎣ X ⎤⎦ =
σ2
n
.
Question 9
Some candidates displayed a lack of familiarity with the use of conditional expectations,
and in particular with the application of the result
Var[Y] = Var[E(Y|X)] + E[Var(Y|X)]
Question 10
Some candidates did not know that the asymptotic standard error of a maximum
likelihood estimator is found from evaluating 1/ I , where
⎡ d 2A ⎤
I = E ⎢ − 2 ⎥ and A ( λ ) is the log-likelihood.
⎣ dλ ⎦
Page 2
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2006 — Examiners’ Report
Question 11
s12
s22
(which is the reciprocal
In the part on equality of variances (part (iii)(a)) some candidates who worked with
(= 0.607) did not know how to find the lower 2.5% point of F7,11
of the upper 2.5% point of F11,7 , and is approximately
1/4.71 = 0.212).
Page 3
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2006 — Examiners’ Report
1
(i)
P(second ball drawn is B) = P(first ball drawn is B) = 8/14 = 0.571
OR P(1st B and 2nd B) + P(1st W and 2nd B)
= (8/14)×(7/13) + (6/14)×(8/13) = 8/14
(ii)
2
P(1st W | 2nd B) = P(1st W and 2nd B)/P(2nd B) = (6/14)×(8/13)/(8/14)
= 6/13 = 0.462
Since A and B are independent, P ( A ) = P ( A | B ) = P ( A | B )
(
Noting that ( B ) = B , it follows immediately that P ( A ) = P ( A | B ) = P A | ( B )
)
and so A and B are independent.
[OR
Since A and B are independent P (A ∩ B) = P(A)P(B).
Thus,
P(A ∩ B ) = P(A) − P(A ∩ B) = P(A) − P(A)P(B) = P(A){1 − P(B)} = P(A)P( B )
∴ A and B are independent.
3
P(no claims on 6 policies) = 0.5314 (from tables p186 — or using 0.96)
P(1 claim on 6 policies) = 0.8857 − 0.5314 = 0.3543 (or using 6(0.1)(0.95))
So required probability = 0.5314 × 0.3543 = 0.188.
4
Let X be the number in force for more than five years
then X ~ binomial(500,0.65)
Using a normal approximation, X ≈ N(325, 10.6652)
P(X < 300) becomes P(X < 299.5) using continuity correction
P( Z <
299.5 − 325
) where Z ~ N(0,1)
10.665
= P ( Z < −2.39) = 1 − 0.99158 = 0.0084
Page 4
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2006 — Examiners’ Report
5
Under H0: sample proportion P is approximately normally distributed with mean 0.4
and standard error (0.4×0.6/200)1/2 = 0.03464
∴ P-value of observed proportion (68/200 = 0.34)
0.34 − 0.4 ⎞
⎛
= P⎜Z <
⎟ = P ( Z < −1.732 ) = 0.042
0.03464 ⎠
⎝
We reject H0 at the 5% level of testing and conclude that the proportion of
policyholders who are female is less than 0.4.
[OR This is actually better - working with the number of female policyholders
(observed = 68), the P-value is
⎛
⎞
68.5 − 80
P⎜Z <
= −1.660 ⎟ = 0.048
⎜
⎟
200(0.4)(0.6)
⎝
⎠
]
Note: We can word the conclusion: we reject H0 at levels of testing down to 4.2% (or
4.8%) and conclude …
6
(i)
P(no claims) = P(X = 0) where X ~ Poisson(0.5)
= 0.60653 from tables [or evaluation]
(ii)
Let Y = number of years with a claim
then Y ~ binomial(3,0.3935) [or just directly as below]
P(Y = 1) = 3(0.3935)(0.6065)2 = 0.434
(iii)
Let T = time until next claim
then T ~ exp(0.5)
P(T > 2) = e–0.5(2) [or by integration]
= e–1 = 0.368
[OR: answer = {P(no claim)}2 = 0.606532 = 0.368]
[OR: claim rate for period of 2 years = 1, so P(no claim in 2 years)
= e-1 = 0.368]
7
(i)
As stated in the question, if Xi is the waiting time on day i, then Xi has an
exponential distribution with parameter
1
15
so E(Xi) = 15, Var(Xi) = 152 = 225.
If X is the total waiting time over the 100 days, X = ∑ i =1 X i ,
100
Page 5
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2006 — Examiners’ Report
so E[X] = 1500 and Var[X] = 22500 and by the CLT
X has approximately an N(1500, 22500) distribution,
⎛ 1620 − 1500 ⎞
so P(X > 1620) ≈ 1 − Φ ⎜
⎟ = 1 − Φ(0.8) = 0.2119.
150
⎝
⎠
(ii)
If Yj is the waiting time on day j of the extra 99 days, then E(Yj) = 10 and
Var(Yj) = 100 so that if Y =
∑ j =1Y j
99
is the total waiting time over the 99 days,
then Y is approximately N(990,9900) by CLT.
If Z = X + Y (so that Z is the total waiting time over the whole 199 days), then
since X and Y are independent, Z is approximately N(1500+990, 22500+9900),
i.e. N(2490, 32400).
⎛ 2400 − 2490 ⎞
Hence P(Z > 2400) ≈ 1 − Φ ⎜
⎟ = 1 − Φ(−0.5) = Φ(0.5) = 0.6915.
180
⎝
⎠
8
(i)
E (W ) = E (αX1 + (1 − α) X 2 )
= αE ( X1 ) + (1 − α ) E ( X 2 ) = αμ + (1 − α )μ = μ
Therefore W is unbiased.
(ii)
MSE(W) = var(W) + {bias(W)}2
W is unbiased
∴ MSE(W) = var(W)
= var(αX1 + (1 − α ) X 2 )
= α 2 var( X1 ) + (1 − α) 2 var( X 2 ) (independent samples)
= α2
(iii)
σ12
σ2
+ (1 − α) 2 2
n
n
σ2
σ2
dMSE
= 2α 1 − 2(1 − α) 2
dα
n
n
dMSE
= 0 ⇒ (σ12 + σ22 )α = σ22
dα
Page 6
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2006 — Examiners’ Report
σ22
∴ α=
σ12 + σ 22
d 2 MSE
d α2
(iv)
=2
σ12
σ2
+ 2 2 > 0 ∴ minimum
n
n
The maximum likelihood estimator of μ in the special case with
σ12 = σ22 = σ2 is
μˆ =
=
nX + nX 2
sum of observations
= 1
number of observations
2n
1
1
X1 + X 2
2
2
This is the same as W since
α=
9
(i)
σ22
σ12
+ σ 22
=
σ2
2
σ +σ
2
=
1
1
1
⇒ W = X1 + X 2 .
2
2
2
E[E(Y|X)] = ∫ E[Y|X = x] f(x)dx
= ∫ ∫ yf(y|x)dy f(x)dx
= ∫ ∫ y f(y|x) f(x)dydx
but f(y|x) f(x) = f(x,y), the joint pdf of X and Y, so
E[E(Y|X)] = ∫ ∫ y f(x,y)dxdy = E[Y]
(ii)
E[Y]
= E[E(Y|X)] = E[X2 + 1]
= V[X] + {E[X]}2 + 1 = 1 + 0 + 1 = 2
Var[Y]
= Var[E(Y|X)] + E[Var(Y|X)] = Var[X2 + 1] + E[X2 + 1]
= Var[X2] + E[X2] + 1
but Z = X2 is χ12 so has variance 2 and expectation 1
Thus Var[Y] = 2 + 1 + 1 = 4
Page 7
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2006 — Examiners’ Report
10
(i)
(a)
Mgf of Xi is (1 – t/λ)−4.5 so mgf of
n
∏ (1 − t / λ )
−4.5
= (1 − t / λ )
n
∑ Xi
is
i =1
−4.5n
i =1
n
Hence mgf of 2λ ∑ X i = 2λnX is (1 − 2λt / λ )
−4.5n
= (1 − 2t )
−4.5 n
i =1
This is the mgf of a χ2 variable — with 9n degrees of freedom.
(b)
b ⎞
⎛ a
<λ<
P ( a < 2λnX < b ) = 0.95 ⇒ P ⎜
⎟ = 0.95
2nX ⎠
⎝ 2nX
where a and b are such that
(
)
(
)
P χ92n < a = 0.025 and P χ92n > b = 0.025.
b
⎛ a
so a 95% CI for λ is given by ⎜
,
⎝ 2nX 2nX
(c)
⎞
⎟.
⎠
9n = 90, and from tables of χ2 with 90df we have a = 65.65, b = 118.1
118.1 ⎞
⎛ 65.65
CI is ⎜
,
⎟ = (1.53 , 2.75).
⎝ 2 × 21.47 2 × 21.47 ⎠
(ii)
(a)
(
L ( λ ) ∝ λ 4.5n exp −λ ∑ xi
)
so
A ( λ ) = ( 4.5n ) log λ − λ ∑ xi + constant
⇒
dA
= 4.5n / λ − ∑ xi
dλ
d 2A
Setting
= 4.5n / λ 2 so s.e.(λˆ ) ≅
(b)
−
(c)
95% CI is λˆ ± 1.96 × s.e. λˆ
dλ
2
{
λˆ
( 4.5n )1/ 2
( )}
In the case n = 100, Σx = 225.3,
Page 8
4.5n 4.5
dA
=
= 0 ⇒ λˆ =
dλ
∑ Xi X
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2006 — Examiners’ Report
4.5 / 2.253
λˆ = 4.5 / 2.253 = 1.9973 and s.e.(λˆ ) ≅
= 0.0942
( 450 )1/ 2
so CI is 1.9973 ±(1.96 × 0.0942) i.e. (1.81 , 2.18).
11
(i)
Maturity values for high street banks and other banks
High street
Other banks
11.4
(ii)
11.5
11.6
11.7
11.8
11.9
12.0
12.1
12.2
x1: maturity value for high street bank
x2: maturity value for other bank
x1 =
93.20
= 11.650
8
x2 =
141.78
= 11.815
12
1⎛
93.202 ⎞
s12 = ⎜ 1086.0470 −
⎟ = 0.038143
7 ⎜⎝
8 ⎟⎠
s22 =
1⎛
141.782 ⎞
1675.8224
−
⎜
⎟ = 0.062882
11 ⎜⎝
12 ⎟⎠
Pooled estimate of σ:
s 2p
(n1 − 1) s12 + (n2 − 1) s22 7(0.038143) + 11(0.062882)
=
=
= 0.053261
n1 + n2 − 2
18
∴ s p = 0.2308
95% confidence interval for μ1 − μ2 is
11.650 − 11.815 ± t18 (2½%) s p
= −0.165 ± (2.101)(0.2308)
1 1
+
8 12
1 1
+
8 12
Page 9
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2006 — Examiners’ Report
i.e. −0.165 ± 0.2213
i.e. (−0.386, 0.056)
i.e. the confidence interval for the difference between the means for high street
banks and other banks (μ1 − μ2) is −£386 to £56.
As zero is within the confidence interval, there is insufficient evidence, at 5%
level, to reject the null hypothesis that the mean maturity values do not differ
for the accounts offered by high street banks and other banks.
(iii)
(a)
S 22
S12
~ F11,7
under the assumption that the variances are equal for high street and
other banks,
i.e. H0: σ12 = σ22
s22
s12
=
0.062882
= 1.65
0.038143
We cannot reject the null hypothesis at the 5% level as the two-sided
critical value of a 5% level test is approximately 4.71 (by interpolation
using 2½% one-sided F table in Yellow Book).
[OR probability value is p > 0.20 as a two-sided 20% level test has a
critical value of approximately 2.69.]
(b)
(iv)
The plot in (i) indicates that the assumption of a normal distribution for
maturity values is reasonable (but small samples) for both high street
and other banks. The assumption of equal variance also seems valid as
the test in (iii)(a) is not significant (and the plot above supports this).
Adding points for building societies to previous plot in (i).
Maturity values
High street
Other banks
Building soc
11.4
Page 10
11.9
12.4
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2006 — Examiners’ Report
(v)
Σx = 93.20 + 141.78 + 143.83 = 378.81
Σx2 = 1086.0470 + 1675.8224 + 1724.2449 = 4486.114
SST = 4486.114 −
378.812
= 1.832
32
93.202 141.782 143.832 378.812
+
+
−
= 0.551
8
12
12
32
∴ SSR = SST − SSB = 1.832 − 0.551 = 1.281
SSB =
Analysis of variance table
Source of variation
Financial institution types
Residual
Total
F=
df
2
29
31
SS
0.551
1.281
1.832
MSS
0.276
0.044
0.276
= 6.27 on (2, 29) degrees of freedom
0.044
F2,29 (5%) = 3.328 and F2,29 (1%) = 5.42
Reject H0: μ1 = μ2 = μ3 (population means are equal) at 1% level.
Strong evidence of differences between the 3 financial institutions.
The plot shows nothing strong enough to invalidate the assumptions of
normality and equal variances, even though the variability for the building
societies is a bit smaller than for the banks.
(vi)
Part (ii) indicates that there are no differences between the mean maturity
values of the two types of bank, but (v) indicates that there are differences
between the mean maturity values of the 3 types of financial institution.
Therefore, in conclusion, it seems that the mean maturity value for building
societies is not equal to the mean maturity values of the banks. Also, the plot
in (iv) suggests that the maturity value for building societies is higher than the
mean maturity values for the banks.
Page 11
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2006 — Examiners’ Report
12
(i)
Plot of stopping distance against speed
80
Stopping distance
70
60
50
40
30
20
10
0
10
20
30
40
50
60
70
Speed
There is a suggestion of a curve but linear regression might still be reasonable.
(ii)
n=7
Sxx = Σx2 − (Σx)2/n = 14000 − (280)2/7 = 2800
Syy = Σy2 − (Σy)2/n = 11951 − (241)2/7 = 3653.714
Sxy = Σxy − (Σx)( Σy)/n = 12790 − (280)(241)/7 = 3150
Model: E[Y] = α + βx
Slope: βˆ =
S xy
S xx
=
3150
= 1.125
2800
Intercept: αˆ = y − βˆ x = 241/7 − (1.125)(280/7) = −10.571
The equation of the least-squares fitted regression line is:
Distance = −10.571 + 1.125 Speed
Page 12
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2006 — Examiners’ Report
(iii)
σˆ 2 =
S2
1 ⎛
⎜ S yy − xy
n−2⎜
S xx
⎝
s.e.( β̂ ) =
σˆ 2
=
S xx
⎞ 1⎛
(3150) 2 ⎞
⎟ = ⎜ 3653.714 −
⎟ = 21.99
⎟ 5 ⎜⎝
2800 ⎟⎠
⎠
21.99
= 0.0886
2800
95% confidence interval for slope:
βˆ ± tn− 2 (0.025) s.e.( β̂ )
(df = n − 2 = 5)
= 1.125 ± (2.571)(0.0886) = 1.125 ± 0.228 or (0.897, 1.353)
β = 1 is within this 95% confidence interval, therefore we would not reject the
null hypothesis β = 1 at the 5% significance level.
(iv)
When x = 50: y = −10.571 + 1.125(50) = 45.7 m
When x = 100: y = −10.571 + 1.125(100) = 101.9 m
The stopping distance of 45.7 m when the speed is 50 mph can be regarded as
a reliable estimate as x = 50 is well within the range of the x data values.
However, the stopping distance for a speed of 100 mph may be unreliable as
x = 100 is outside the range of the data and involves extrapolation.
END OF EXAMINERS’ REPORT
Page 13
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
23 April 2007 (am)
Subject CT3 — Probability and Mathematical Statistics
Core Technical
Time allowed: Three hours
INSTRUCTIONS TO THE CANDIDATE
1.
Enter all the candidate and examination details as requested on the front of your answer
booklet.
2.
You must not start writing your answers in the booklet until instructed to do so by the
supervisor.
3.
Mark allocations are shown in brackets.
4.
Attempt all 13 questions, beginning your answer to each question on a separate sheet.
5.
Candidates should show calculations where this is appropriate.
Graph paper is required for this paper.
AT THE END OF THE EXAMINATION
Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this
question paper.
In addition to this paper you should have available the 2002 edition of the
Formulae and Tables and your own electronic calculator.
CT3 A2007
© Faculty of Actuaries
© Institute of Actuaries
1
Consider the following two random samples of ten observations which come from the
distributions of random variables which assume non-negative integer values only.
Sample 1: 7 4 6 11 5 9 8 3 5 5
sample mean = 6.3, sample variance = 6.01
Sample 2: 8 3 5 11 2 4 6 12 3 9
sample mean = 6.3, sample variance = 12.46
One sample comes from a Poisson distribution, the other does not.
State, with brief reasons, which sample you think is likely to be which.
2
[2]
A random sample of 200 policy surrender values (in units of £1,000) yields a mean of
43.6 and a standard deviation of 82.2.
Determine a 99% confidence interval for the true underlying mean surrender value for
such policies.
[3]
3
It is assumed that claims on a certain type of policy arise as a Poisson process with
claim rate λ per year.
For a group of 150 independent policies of this type, the total number of claims during
the last calendar year was recorded as 123.
Determine an approximate 95% confidence interval for the true underlying annual
claim rate for such a policy.
[4]
4
The sample correlation coefficient for the set of data consisting of the three pairs of
values
(−1,−2) , (0,0) , (1,1)
is 0.982. After the x and y values have been transformed by particular linear functions,
the data become:
(2,2) , (6,−4) , (10,−7).
State (or calculate) the correlation coefficient for the transformed data.
CT3 A2007—2
[2]
5
The number of claims arising in one year from a group of policies follows a Poisson
distribution with mean 12. The claim sizes independently follow an exponential
distribution with mean £80 and they are independent of the number of claims.
The current financial year has six months remaining.
Calculate the mean and the standard deviation of the total claim amount which arises
during this remaining six months.
[4]
6
Consider the discrete random variable X with probability function
f ( x) =
(i)
4
5 x +1
,
x = 0, 1, 2, ...
Show that the moment generating function of the distribution of X is given by
M X (t ) = 4(5 − et ) −1 ,
for et < 5.
(ii)
7
[3]
Determine E[X] using the moment generating function given in part (i).
[3]
[Total 6]
A charity issues a large number of certificates each costing £10 and each being
repayable one year after issue. Of these certificates, 1% are randomly selected to
receive a prize of £10 such that they are repaid as £20. The remaining 99% are repaid
at their face value of £10.
(i)
Show that the mean and standard deviation of the sum repaid for a single
purchased certificate are £10.1 and £0.995 respectively.
[2]
Consider a person who purchases 200 of these certificates.
(ii)
Calculate approximately the probability that this person is repaid more than
£2,040 by using the Central Limit Theorem applied to the total sum repaid.
[3]
(iii)
An alternative approach to approximating the probability in (ii) above is based
on the number of prize certificates the person is found to hold. This number
will follow a binomial distribution.
Use a Poisson approximation to this binomial distribution to approximate the
probability that this person is repaid more than £2,040.
[3]
(iv)
Comment briefly on the comparison of the two approximations above given
that the exact probability using the binomial distribution is 0.0517.
[1]
[Total 9]
CT3 A2007—3
PLEASE TURN OVER
8
A random sample of size n is taken from a distribution with probability density
function
α
f ( x) =
(1 + x)α+1
,
0< x<∞
where α is a parameter such that α > 0.
(i)
Show by evaluating the appropriate integral that, in the case α > 1, the mean
1
.
of this distribution is given by
α −1
[Hint: when integrating, write x = (1 + x) - 1 and exploit the fact that the
integral of a density function is unity over its full range.]
(ii)
9
10
Determine the method of moments estimator of α.
[3]
[2]
[Total 5]
Consider three random variables X, Y, and Z with the same variance σ2 = 4. Suppose
that X is independent of both Y and Z, but Y and Z are correlated, with correlation
coefficient ρYZ = 0.5.
(i)
Calculate the covariance between X and U, where U = Y+Z.
[1]
(ii)
Calculate the covariance between Z and V, where V = 3X – 2Y.
[2]
(iii)
Calculate the variance of W, where W = 3X – 2Y + Z.
[2]
[Total 5]
A random sample of insurance policies of a certain type was examined for each of
four insurance companies and the sums insured (yij, for companies i = 1, 2, 3, 4)
under each policy are given in the table below (in units of £100):
Company
1
2
3
4
For these data,
CT3 A2007—4
Total
58.2
56.3
50.1
52.9
57.2
54.5
54.2
49.9
58.4
57.0
55.5
50.0
55.8
55.3
51.7
54.9
284.5
223.1
159.8
204.5
∑ i ∑ j yij = 871.9 and ∑ i ∑ j yij2 = 47, 633.53
Consider the ANOVA model Yij = μ + τi + eij , i = 1,..., 4, j = 1,..., ni , where Yij is the jth
sum insured for company i, ni is the number of responses for company i,
eij ~ N (0, σ2 ) are independent errors, and
11
∑ i=1 ni τi = 0 .
4
(i)
Calculate estimates of the parameters μ and τi , i = 1, 2, 3, 4 .
(ii)
Test the hypothesis that there are no differences in the means of the sums
insured under such policies by the four companies.
[5]
[Total 8]
[3]
The number of claims, X, which arise in a year on each policy of a particular class is
to be modelled as a Poisson random variable with mean λ. Let X = (X1, X2, …, Xn) be
1 n
a random sample of size n from the distribution of X, and let X = ∑ X i .
n i =1
Suppose that it is required to estimate λ, the mean number of claims on a policy.
(i)
Show that λ̂ , the maximum likelihood estimator of λ, is given by λˆ = X . [3]
(ii)
Derive the Cramer-Rao lower bound (CRlb) for the variance of unbiased
estimators of λ.
(iii)
[4]
(a)
Show that λ̂ is unbiased for λ and that it attains the CRlb.
(b)
Explain clearly why, in the case that n is large, the distribution of λ̂ can
be approximated by
⎛ λ⎞
λˆ ~ N ⎜ λ, ⎟ .
⎝ n⎠
[3]
(iv)
(a)
Show that, in the case n = 100, an approximate 95% confidence
interval for λ is given by
x ± 0.196 x .
(b)
Evaluate the confidence interval in (iv)(a) based on a sample with the
following composition:
observation
frequency
0
11
1
28
2
19
3
28
4
9
5
2
6
2
7
1
[6]
[Total 16]
CT3 A2007—5
PLEASE TURN OVER
12
An insurance company is investigating past data for two household claims assessors,
A and B, used by the company. In particular claims resulting from similar types of
water damage were extracted. The following table shows the assessors’ initial
estimates of the cost (in units of £100) of meeting each claim.
A:
B:
4.6
5.7
6.6
3.4
2.8
4.7
5.8
3.6
2.1
6.5
5.2
3.3
5.9
3.8
3.4
2.4
7.8
7.0
3.5
4.0
1.6
4.4
8.6
2.7
for the A data: nA = 13, Σx = 60.6 and Σx2 = 340.92
for the B data: nB = 11, Σx = 48.8 and Σx2 = 236.80
(i)
Draw a suitable diagram to represent these data so that the initial estimates of
the two assessors can be compared.
[2]
(ii)
You are required to perform an appropriate test to compare the means of the
assessors’ initial estimates for this type of water damage.
(a)
State your hypotheses clearly.
(b)
Use your diagram in part (i) to comment briefly on the validity of your
test.
(c)
Calculate your test statistic and specify the resulting P-value.
(d)
State your conclusion clearly.
[10]
(iii)
You are required to perform an appropriate test to compare the variances of
the assessors’ initial estimates.
(a)
State your hypotheses clearly.
(b)
Use your diagram in part (i) to comment briefly on the validity of your
test.
(c)
Calculate your test statistic and specify the resulting P-value.
(d)
State your conclusion clearly and hence comment further on the
validity of your test in part (ii).
[7]
(iv)
Use your answers in parts (i) to (iii) to comment on the overall comparison of
the two assessors as regards their initial estimates for this type of water
damage.
[1]
[Total 20]
CT3 A2007—6
13
In a study of the relation between the amount of information available and use of
buses in eight comparable test cities, bus route maps were given to residents of the
cities at the beginning of the test period. The increase in average daily bus use during
the test period was recorded. The numbers of maps and the increase in bus use are
given in the table below (both in thousands).
Number of maps (x)
80 220 140 120 180 100 200 160
Increase in bus use (y) 0.60 6.70 5.30 4.00 6.55 2.15 6.60 5.75
For these data:
∑ x = 1, 200 , ∑ x2 = 196,800 , ∑ y = 37.65 , ∑ y 2 = 213.4875 , ∑ xy = 6,378
(i)
Construct a scatterplot of the data and comment on the relationship between
the increase in bus use and the number of maps distributed.
[4]
(ii)
The equation of the fitted linear regression is given by y = −1.816 + 0.04348 x .
Perform an appropriate statistical test to assess the hypothesis that the slope in
this fitted model suggests no relationship between the increase in bus use and
the number of maps distributed. Any assumptions made should be clearly
stated.
[6]
(iii)
The fitted responses and the residuals from the linear regression model fitted
in part (ii) are given below:
Fitted values ( yˆ )
Residuals (eˆ)
1.66 7.75 4.27 3.40 6.01 2.53 6.88 5.14
-1.06 -1.05 1.03 0.60 0.54 -0.38 -0.28 0.61
Plot the residuals against the values of the fitted responses and comment on
the adequacy of the model.
[4]
(iv)
A new city is added to the study, and 250,000 maps are distributed to its
citizens.
Calculate the prediction of the increase in bus use in this city according to the
model fitted in part (ii) and comment on the validity of this prediction.
[2]
[Total 16]
END OF PAPER
CT3 A2007—7
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
April 2007
Subject CT3 — Probability and Mathematical Statistics
Core Technical
EXAMINERS’ REPORT
Introduction
The attached subject report has been written by the Principal Examiner with the aim of
helping candidates. The questions and comments are based around Core Reading as the
interpretation of the syllabus to which the examiners are working. They have however given
credit for any alternative approach or interpretation which they consider to be reasonable.
M A Stocker
Chairman of the Board of Examiners
June 2007
Comments
Comments are given in the solutions that follow. Note that in some cases variations on the
solutions given are possible — the examiners gave credit for all sensible comments and
correct solutions.
The paper was well-answered overall and there are no particular topics that stand out as being
poorly attempted. Similarly there were no particular misunderstandings evident widely, and
no particular errors were made repeatedly and are worthy of comment.
In Question 13(iii) candidates were asked to plot a given set of residuals and comment. There
was in fact a negative sign missing from the first residual (quoted as 1.6). The error was
noted before marking commenced. No candidate was disadvantaged — all answers using
1.06 or -1.06 were accepted as being equally valid. There was no evidence in the scripts of
any problem for candidates. The examiners wish to apologise for the minor error.
© Faculty of Actuaries
© Institute of Actuaries
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2007 — Examiners’ Report
1
A Poisson random variable has mean = variance and this will be reflected in the
sample mean and variance for a random sample.
Sample 2 has a very much higher variance than mean, whereas sample 1 has mean
and variance approximately the same, so sample 1 is likely to be the one which comes
from a Poisson distribution.
2
Approximate large sample confidence interval for the mean is given by
s
n
x ± zα / 2
for 99% CI, zα / 2 = 2.5758
leading to 43.6 ± 2.5758
or
3
(28.6, 58.6)
or
82.2
200
⇒ 43.6 ± 15.0
(£28,600, £58,600)
The mean number of claims per policy is X =
123
= 0.82
150
Using the normal approximation to the Poisson distribution
approximate 95% confidence interval for λ is X ± 1.96
→ 0.82 ± 1.96
0.82
150
X
n
→ 0.82 ± 1.96(0.0739)
→ 0.82 ± 0.145 or (0.675, 0.965)
4
Answer = −0.982
(The relationship is now a negative one; the only change is the sign. An answer of
+0.982 gets 1 mark.)
Page 2
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2007 — Examiners’ Report
5
Let N = number of claims in the six months
Let X = a single claim size
Let S = sum of claim sizes for the six months
Then N ~ Poisson(6)
E ( S ) = E ( N ) E ( X ) = (6)(80) = £480
V ( S ) = E ( N )V ( X ) + V ( N )[ E ( X )]2 = (6)(802 ) + (6)(802 ) = 76800
∴ sd ( S ) = £277
6
(i)
M X (t ) = E[etx ]
∞
4⎛1⎞
4 ∞
= ∑e
=
∑
⎜ ⎟
5⎝5⎠
5 x =0
x =0
x
tx
⎛ et
⎜⎜
⎝5
x
⎞
⎟⎟ ,
⎠
and for et < 5 ,
M X (t ) =
(ii)
4 1
5 1 − et
(
(
= 4 5 − et
)
−1
.
5
M '(t ) = 4et 5 − et
)
−2
Mean is given by E ( X ) = M '(0)
(
∴ E[ X ] = 4e0 5 − e0
)
−2
=
1
.
4
[OR, by expansion as a power series.]
Page 3
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2007 — Examiners’ Report
7
(i)
Let X = the sum repaid for a single certificate.
E ( X ) = 10(0.99) + 20(0.01) = 10.1
E ( X 2 ) = 102 (0.99) + 202 (0.01) = 103
∴V ( X ) = 103 − 10.12 = 0.99 ∴ sd ( X ) = 0.9950
(ii)
Let S = the sum repaid for 200 certificates.
∴ E ( S ) = 200(10.1) = 2020, V ( S ) = 200(0.99) = 198 ∴ sd ( X ) = 14.07
P ( S > 2040) = P( Z >
2040 − 2020
= 1.42)
14.07
= 1 − 0.9222 = 0.0778
(iii)
N ~ binomial(200, 0.01) ≈ Poisson(2)
P ( S > 2040) = P( N > 4)
= 1 − P( N ≤ 4) = 1 − 0.94735 = 0.0527
(iv)
Clearly the Poisson approximation to the binomial is better than the Central
Limit Theorem approximation.
OR:
Since S is discrete and increases in steps of 10, one can argue for the use of a
continuity correction in (ii) above:
2045 − 2020 ⎞
⎛
P ( S > 2040 ) = P ⎜ Z ≥
⎟ = P ( Z > 1.78 )
14.07
⎝
⎠
= 1 − 0.96246 = 0.0375
(Either approach is acceptable for the marks.)
Page 4
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2007 — Examiners’ Report
∞
8
(i)
Mean =
α
∫ x (1 + x)α+1 dx
0
∞
=
∫ (1 + x)
0
α
(1 + x)
∞
dx − ∫ 1
α+1
0
α
(1 + x)α+1
dx
∞
=
α
(α − 1)
dx − 1
∫
α − 1 (1 + x)(α−1)+1
0
=
(ii)
α
1
−1 =
α −1
α −1
Equate population mean to sample mean:
Solve to get α = 1 +
9
(i)
1
=x
α −1
1
1
, so MME = 1 +
x
X
Cov(X,Y+Z) = Cov(X,Y) + Cov(X,Z) = 0
[Note: The simple statement of the answer “0” is acceptable for the single
mark available.]
(ii)
Cov(Z, 3X - 2Y) = 3Cov(Z,X) – 2Cov(Z,Y) = 0 – 2ρYZ σ2
= -2 × 0.5 × 4 = -4
(iii)
V[3X – 2Y + Z] = (9 + 4 + 1)σ2 – 12Cov(X,Y) + 6Cov(X,Z) – 4Cov(Y,Z)
= 14(4) – 4(0.5)(4) = 48
Page 5
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2007 — Examiners’ Report
(i)
10
(ii)
μˆ = Y.. =
τˆ1 = Y1. − Y.. =
284.5 871.9
−
= 2.406
5
16
τˆ 2 = Y2. − Y.. =
223.1 871.9
−
= 1.281
4
16
τˆ 3 = Y3. − Y.. =
159.8 871.9
−
= −1.227
3
16
τˆ 4 = Y4. − Y.. =
204.5 871.9
−
= −3.369
4
16
SST = ∑∑ yij2 −
i
SSB =
871.9
= 54.494
16
j
Y..2
= 120.430
n
∑ ni (Yi. − Y.. )2 = (5 × 2.4062 ) + (4 ×1.2812 ) + (3 ×1.2272 ) + (4 × 3.3692 ) = 85.425
i
⎛Y2 ⎞ Y2
[OR : SS B = ∑ ⎜ i. ⎟ − .. = 85.428]
⎜
⎟ n
i ⎝ ni ⎠
SS R = SST − SS B = 35.002
The ANOVA table is:
Source
DF
SS
MS
F
Company (between treatments) 3
85.428 28.476 9.763
Residual
12 35.002 2.917
Total
15 120.430
At the 5% significance level, F0.05,3,12 = 3.490 (or F0.01,3,12 = 5.953 )
Since F = 9.763 > 3.490, there is evidence against the null hypothesis, and we
conclude that there are differences in the mean sums insured by the
companies.
Page 6
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2007 — Examiners’ Report
11
(i)
L( x) =
( λ ) = log L ( λ ) = −nλ + ( ∑ xi ) log λ + constant
⇒
⇒
(ii)
e − nλ λ ∑ xi
∏ xi !
d
∑ xi = 0 ⇒ λˆ =
= −n +
λ
dλ
d2
d λ2
=−
∑ Xi = X
n
∑ xi
λ2
⎡ d2 ⎤ 1
nλ n
⇒ − E ⎢ 2 ⎥ = 2 E ⎡⎣ ∑ X i ⎤⎦ = 2 =
λ
λ
⎢⎣ d λ ⎥⎦ λ
⇒ CRlb =
(iii)
(a)
λ
.
n
E ⎡⎣λˆ ⎤⎦ = E ⎡⎣ X ⎤⎦ = E [ X ] = λ
V ⎡⎣λˆ ⎤⎦ = V ⎡⎣ X ⎤⎦ =
(iv)
V [X ]
n
=
λ
,which is CRlb.
n
(b)
The theory of asymptotic distributions of MLEs (and in this case the
⎛ λ⎞
CLT) gives λˆ ~ N approximately, for large n so λˆ ~ N ⎜ λ, ⎟ ,
⎝ n⎠
approximately.
(a)
Large sample approximate 95% CI for λ is given by
(
( ))
λˆ ± 1.96 × s.e. λˆ
i.e. x ± (1.96 × s.e. ( x ) )
()
λ
s.e. λˆ =
which we can estimate by using x to estimate λ, giving
n
()
the estimated standard error e.s.e. λˆ =
x
n
With n = 100, we get the 95% CI as
⎛
x ⎞
x ± ⎜⎜ 1.96 ×
⎟ i.e. x ± 0.196 x .
100 ⎟⎠
⎝
Page 7
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2007 — Examiners’ Report
(b)
x = 215 /100 = 2.15
CI is 2.15 ± 0.196(2.15)1/2 i.e. 2.15 ± 0.287 i.e. (1.86, 2.44)
12
(i)
dotplots on same scale are most suitable
[alternatively boxplots are acceptable]
(ii)
(a)
Let μA = mean initial estimate for this type of water damage for
assessor A and μB = mean initial estimate for this type of water damage
for assessor B.
H0 : μA = μB v H1 : μA ≠ μB
(b)
dotplots show that normality assumption is reasonably valid
dotplots perhaps cast doubt on equal variances assumption
(c)
test statistic is t =
From data: x A =
60.6
1
60.62
) = 4.8692
= 4.662 , s 2A = (340.92 −
13
12
13
xB =
48.8
1
48.82
) = 2.0305
= 4.436 , sB2 = (236.80 −
11
10
11
s 2p =
12(4.8692) + 10(2.0305)
= 3.5789 ∴ s p = 1.8918
22
and
observed t =
Page 8
x A − xB
∼ tnA + nB −2 under H0
1
1
sp
+
n A nB
4.662 − 4.436
0.226
=
= 0.29 on 22 df
0.775
1 1
1.8918
+
13 11
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2007 — Examiners’ Report
Clearly P-value is very large, or noting that t22(40%) = 0.2564, then
P-value is just a bit less than 0.8.
(iii)
(d)
So there is no evidence at all of any difference between assessors A
and B as regards their mean initial estimates for this type of water
damage.
(a)
H0 : σ2A = σ2B v H1 : σ2A ≠ σ2B
(b)
as in (i) dotplots show that normality assumption is reasonably valid
(c)
test statistic is F =
observed F =
s 2A
sB2
∼ Fn A −1,nB −1 under H0
4.8692
= 2.40 on 12,10 df
2.0305
F12,10(10%) = 2.284 and F12,10(5%) = 2.913
Thus P-value is between 0.10 and 0.20.
(d)
So there is no real evidence of any difference between assessors A and
B as regards the variances of their initial estimates for this type of
water damage.
This validates the possibly doubtful assumption required in part (ii).
(iv)
Overall there is no real evidence to distinguish any differences in the initial
estimates for this type of water damage for the two assessors A and B.
Page 9
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2007 — Examiners’ Report
(i)
The scatterplot is shown below.
4
1
2
3
trips
5
6
13
80
100
120
140
160
180
200
220
maps
The plot suggests that there is a positive relationship between the increase in
bus use and the number of maps distributed. The increase seems to be
reasonably linear up to around 180000 maps, after which point it seems to
level off (overall, relationship seems curved, possibly quadratic).
(ii)
Sxx = Σx2 − (Σx)2/n = 196800 − (1200)2/8 = 16800
Syy = Σy2 − (Σy)2/n = 213.4875 − (37.65)2/8 = 36.29719
Sxy = Σxy − (Σx)( Σy)/n = 6378 − (1200)(37.65)/8 = 730.5
2⎞
S xy2 ⎞ 1 ⎛
1 ⎛⎜
⎟ = ⎜ 36.29719 − (730.5) ⎟ = 0.75558
σ̂ =
S yy −
n − 2 ⎜⎝
S xx ⎟⎠ 6 ⎜⎝
16800 ⎟⎠
2
s.e.( β̂ ) =
0.75558
σˆ 2
=
= 0.006706
16800
S xx
To test H 0 : β = 0 v H1 : β ≠ 0 , the test statistic is
0.04348
βˆ − 0
=
= 6.484 ,
s.e.(βˆ ) 0.006706
and under the assumption that the errors of the regression are i.i.d. N (0, σ2 )
random variables, it has a t distribution with n - 2 = 6 df.
Page 10
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2007 — Examiners’ Report
From statistical tables we find t6,0.025 = 2.447 (or, t6,0.005 = 3.707 ).
Therefore, there is strong evidence against H 0 . We conclude that a straight
line representation of the relationship between the increase in bus use and the
number of maps distributed would have a non-zero slope.
The plot is shown below.
0.0
-1.0
-0.5
Residuals
0.5
1.0
(iii)
2
3
4
5
6
7
Fitted
Negative residuals are associated with the fitted values at the two ends of the
data set, suggesting that the model is inadequate. Pattern suggests that a
quadratic model might be appropriate.
(iv)
Predicted value is yˆ = −1.816 + 0.04348 × 250 = 9.054
This uses extrapolation on the fitted regression line. The prediction is probably
not valid, especially as the linear model does not seem adequate.
END OF EXAMINERS’ REPORT
Page 11
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
4 October 2007 (am)
Subject CT3 — Probability and Mathematical Statistics
Core Technical
Time allowed: Three hours
INSTRUCTIONS TO THE CANDIDATE
1.
Enter all the candidate and examination details as requested on the front of your answer
booklet.
2.
You must not start writing your answers in the booklet until instructed to do so by the
supervisor.
3.
Mark allocations are shown in brackets.
4.
Attempt all 13 questions, beginning your answer to each question on a separate sheet.
5.
Candidates should show calculations where this is appropriate.
Graph paper is required for this paper.
AT THE END OF THE EXAMINATION
Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this
question paper.
In addition to this paper you should have available the 2002 edition of the
Formulae and Tables and your own electronic calculator.
CT3 S2007
© Faculty of Actuaries
© Institute of Actuaries
1
Data collected on claim amounts (£) for two postcode regions give the following
values for n (the number of claims), x (the mean claim amount) and s (the sample
standard deviation) of the claim amounts.
Region 1
25
120.2
58.1
n
x
s
Region 2
18
142.7
62.2
Calculate, to one decimal place, the mean and sample standard deviation of the claim
amounts for both regions combined.
[4]
2
The random variable X has probability density function
f ( x) =
2
x3
,
x >1
and cumulative distribution function
⎧0,
⎪
F ( x) = ⎨
1
⎪1 − 2 ,
⎩ x
x <1
x ≥1
.
Use the following uniform(0,1) random numbers
0.5719, 0.8612, 0.3028
to simulate three observations of X, explaining your method and calculations clearly.
[4]
3
It is known that 24% of the customers in a bank holding a current account also have
another type of account with the bank.
Calculate an approximate value for the probability that fewer than 50 customers in a
random sample of 250 customers with a current account also have another type of
account.
[3]
4
In a random sample of 200 policies from a company’s private motor business, there
are 68 female policyholders and 132 male policyholders.
Calculate an approximate 99% confidence interval for the proportion of policyholders
who are female in the corresponding population of all policyholders.
[3]
CT3 S2007—2
5
For a particular insurance company a sample of eight claim amounts (in units of
8
£1,000) on household contents is taken. The data give
∑ xi = 56.7 and
i =1
8
∑ xi2 = 403.95 .
The claim amounts are assumed to follow a normal distribution.
i =1
6
(i)
Calculate a 90% confidence interval for the true mean claim amount.
[3]
(ii)
Use the confidence interval calculated in (i) above to comment on an expert’s
assessment that the average claim amount for the company is £6,500.
[1]
[Total 4]
In an investigation into the relationship between two normally distributed variables, X
and Y, based on a sample of 15 points, it is desired to perform the following test
concerning the true underlying correlation coefficient ρ
H0 : ρ = 0 v H1 : ρ > 0.
Use the t distribution to determine an upper critical value for the sample correlation
coefficient r for this test at the 1% level.
[4]
7
Let N denote the number of claims which arise in a portfolio of business and let Xi be
the amount of the ith claim. Let each of the Xi’s be independently modelled as a
normal variable with mean £10,000 and standard deviation £2,000 and let N be
independently modelled as a Poisson variable with parameter 20.
Calculate the mean and standard deviation of the total claim amount S = X1+…+XN.
[3]
8
Claim sizes in a certain insurance situation are modelled by a normal distribution with
mean μ = £30,000 and standard deviation σ = £4,000 The insurer defines a claim to
be a large claim if the claim size exceeds £35,000.
(i)
Calculate the probabilities that the size of a claim exceeds
(a)
(b)
£35,000, and
£36,000
[2]
(ii)
(iii)
Calculate the probability that the size of a large claim (as defined by the
insurer) exceeds £36,000.
[2]
Calculate the probability that a random sample of 5 claims includes 2 which
exceed £35,000 and 3 which are less than £35,000.
[2]
[Total 6]
CT3 S2007—3
PLEASE TURN OVER
9
10
For a certain class of policies issued by a large insurance company it is believed that
the probability of each policy giving rise to any claims is 0.5, independently of all
other policies. A random sample of 250 such policies is selected.
(i)
Determine approximately the probability that at least 139 of the policies in the
sample will each give rise to any claims.
[4]
(ii)
Suppose we do observe that 139 policies in our sample give rise to at least one
claim. Use your answer to part (i) to determine whether this suggests at the
1% level of significance that the probability of any claims arising from a
policy of this certain class is greater than initially believed.
[3]
[Total 7]
A chi-square test of association for the frequency data in the following 2 × 3 table
Factor B
B1
B2
Factor A
A1
A2
A3
40
30
50
80
30
70
produces a chi-square statistic with value 4.861 and associated P-value 0.089.
Consider a chi-square test of association for the data in the following 2 × 3 table, in
which all frequencies are twice the corresponding frequencies in the first table:
Factor B
B1
B2
Factor A
A1
A2
A3
80
60
100
160
60
140
(i)
State, or calculate, the value of the chi-square test statistic for the second table.
[2]
(ii)
Find the P-value associated with the test statistic in (i).
(iii)
Comment on the results.
CT3 S2007—4
[1]
[2]
[Total 5]
11
Suppose that the random variable X follows an exponential distribution with
probability density function
f ( x) = λe −λx , 0 < x < ∞
Define a new random variable Y =
(i)
(a)
(λ > 0) .
1
X3.
Show that the cumulative density function of Y is given by
⎧⎪1 − exp(−λy 3 ),
FY ( y ) = ⎨
⎪⎩0,
y≥0
y<0
and hence, or otherwise, find the probability density function of Y.
(ii)
(b)
Explain how you would simulate a value of Y given a value u from the
uniform U(0,1) distribution.
[7]
(a)
Find an expression for the maximum likelihood estimator of the
parameter λ, using a sample y1, y2, …, yn, from the distribution of Y.
(b)
Eight observed values of the random variable Y are given below:
0.72 1.15 1.26 1.03 1.69 1.30 1.42 1.15
Calculate the maximum likelihood estimate of λ using these values. [6]
(iii)
(a)
The hazard function of a continuous random variable T is defined
f (t )
as h(t ) =
, where f(t) denotes the probability density function and
S (t )
S(t) denotes the survival function defined as S (t ) = P (T > t ) .
Derive the hazard functions of the random variables X and Y defined
above.
(b)
If a random variable T represents the lifetime of an individual, then the
hazard function h(t), as defined in part (iii)(a), gives the instantaneous
mortality rate (that is, the force of mortality) at time t for that
individual.
State (with reasons) which of the two random variables (X and Y) you
would use to model the lifetime of pensioners for a period of time
longer than one year, basing your answer on the form of the
corresponding hazard functions derived in part (iii)(a).
[5]
[Total 18]
CT3 S2007—5
PLEASE TURN OVER
12
A series of n geomagnetic readings are taken from a meter, but the readings are
judged to be approximate and unreliable. The chief scientist involved does know
however that the true values are all positive and she suggests that an appropriate
model for the readings is that they are independent observations of a random variable
which is uniformly distributed on (0, θ), where θ > 1.
(i)
(ii)
Suppose that the chief scientist knows only that the number, M, of the readings
which are less than 1 is m, with the remaining n − m being greater than 1 and
that she adopts the model as suggested above.
1
.
θ
(a)
Show that the probability that a single reading is less than 1 is
(b)
n
Demonstrate that the maximum likelihood estimate of θ is θˆ = .
m
(c)
Demonstrate that the Cramer-Rao lower bound (CRlb) for estimating
θ2 ( θ − 1)
and hence state the large sample distribution of θ̂ .
θ is
n
[10]
Suppose that exactly 45 readings in a random sample of 100 readings are less
than 1.
(a)
Calculate an estimate of the standard error of θ̂ and hence calculate an
approximate two-sided 95% confidence interval for θ.
(b)
Use the large sample distribution of θ̂ to test the hypotheses
H0: θ = 3 v H1: θ < 3.
[9]
[Total 19]
CT3 S2007—6
13
In a laboratory experiment a response variable (yield, y) is thought to be affected by a
quantitative factor (percentage of catalyst, x). The experiment involved making four
observations of y at each of four values of x, being 12%, 14%, 16% and 18%, and
resulted in the following observed response data.
12%
46
51
47
42
14%
56
57
63
60
16%
56
63
60
64
18%
47
51
54
55
These data are analysed by two statisticians, A and B, who use an analysis of variance
approach and a linear regression approach, respectively.
(i)
Statistician A’s approach:
You are given the following data summaries:
sub-totals Σy = 186, 236, 243 and 207 at x = 12, 14, 16 and 18,
respectively, and overall totals Σy = 872 and Σy2 = 48,196.
(a)
Apply a one-way analysis of variance to these data and obtain the
resulting F-value for the usual test.
(b)
Show that the P-value for the test is substantially less than 0.01, by
referring to tables of percentage points for the F distribution.
(c)
The result of part (b) above shows that there is very strong evidence of
an effect on y due to the quantitative factor x. Suggest a suitable
diagram that statistician A could now use to describe the effect of x on
y. Draw this diagram and hence comment on the effect of x on y.
(d)
The graph below shows the residuals plotted against the values of x:
Comment briefly on any implications of this graph.
CT3 S2007—7
[10]
PLEASE TURN OVER
(ii)
Statistician B’s approach:
You are given the following data summaries:
Σx = 240 Σy = 872 Σx2 = 3,680 Σy2 = 48,196 Σxy = 13,150.
(a)
Perform a linear regression analysis on these data to show that the
fitted line is given by y = 41.4 + 0.875x.
(b)
Perform the hypothesis test on the slope coefficient
H0 : β = 0 v H1 : β ≠ 0
showing that the P-value is greater than 0.20.
Comment on what this implies about the relationship between x and y.
(c)
The graph below shows the residuals plotted against the values of x:
Comment briefly on what this graph implies about the effect of x on y.
(d)
Suggest an additional analysis statistician B could now use to describe
the effect of x on y.
[10]
[Total 20]
END OF PAPER
CT3 S2007—8
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
September 2007
Subject CT3 — Probability and Mathematical Statistics
Core Technical
EXAMINERS’ REPORT
Introduction
The attached subject report has been written by the Principal Examiner with the aim of
helping candidates. The questions and comments are based around Core Reading as the
interpretation of the syllabus to which the examiners are working. They have however given
credit for any alternative approach or interpretation which they consider to be reasonable.
M A Stocker
Chairman of the Board of Examiners
December 2007
Comments
The paper was answered quite well overall and there are no particular topics that stand out as
being poorly attempted. Similarly there were no particular misunderstandings widely evident,
and no particular errors were made so repeatedly as to be worthy of comment.
© Faculty of Actuaries
© Institute of Actuaries
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2007 — Examiners’ Report
1
x=
Σx 25(120.2) + 18(142.7) 5573.6
=
=
= 129.6
n
25 + 18
43
Using the fact that Σx 2 = (n − 1) s 2 + nx 2 , then for the combined set
Σx 2 = [24(58.1) 2 + 25(120.2) 2 ] + [17(62.2) 2 + 18(142.7) 2 ] = 874525.14
∴ s2 =
2
874525.14 − (5573.6) 2 / 43
= 3621.02 ∴ s = 60.2
42
Method: set uniform(0,1) random number r = F(x) = 1 – 1/x2
⇒ simulated observation x = [1/(1 − r)]1/2
Here we get x = 1.528 , 2.684, 1.198.
Note: We can do away with the step of subtracting r from 1 and use x = (1/r)1/2.
This gives x = 1.322, 1.078, 1.817.
3
Let N be the number who have another type of account.
∴ N ~ binomial(250, 0.24) ≈ N (60, 45.6) = N (60, 6.7532 )
P ( N < 50) → P( N < 49.5) with continuity correction
= P( Z <
4
49.5 − 60
= −1.55) = 1 − 0.93943 = 0.061
6.753
Sample proportion P = 68/200 = 0.34
99% CI is 0.34 ± 2.576
Page 2
0.34 × 0.66
i.e. 0.34 ± 0.086 i.e. (0.254, 0.426)
200
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2007 — Examiners’ Report
5
(i)
56.7
= 7.0875.
8
1⎛
56.7 2 ⎞
s 2 = ⎜ 403.95 −
⎟ = 0.298 ⇒ s = 0.546 .
7 ⎜⎝
8 ⎟⎠
x=
90% CI for the true mean is given by:
x ± t7,0.05
s
0.546
= 7.0875 ± 1.895
= 7.0875 ± 0.3658
n
8
i.e. the 90% CI is (6.722, 7.453), or (£6722, £7453).
(ii)
6
The value £6500 is not included in the CI above, and therefore we conclude
that the data are not consistent with the expert’s assessment at the 10%
significance level.
Use the result
r n−2
1− r2
~ tn−2 under H0.
From tables t0.01,13 = 2.650
so critical value is solution of
r
1− r2
13 = 2.65
2.652
13
= 0.592
Solving gives r =
2.652
1+
13
7
From the Yellow Book:
Mean:
E(S) = E(N)E(X) = (20)(10000) = £200,000
Variance: var(S) = E(N) var(X) + var(N)[E(X)]2
= 20(20002) + 20(100002) = 2080 × 106
∴ Standard deviation = £45607
[OR, using compound Poisson results (in Yellow Book)
E(S) = λm1 and var(S) = λm2 where λ = E(N) and mr = E(X r)]
Page 3
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2007 — Examiners’ Report
8
X ~ N with mean μ = 30 and σ = 4 (working in units of £1000)
(i)
(a)
(b)
(ii)
35 − 30 ⎞
⎛
P ( X > 35 ) = P ⎜ Z >
⎟ = P ( Z > 1.25 ) = 1 − 0.89435 = 0.10565
4 ⎠
⎝
36 − 30 ⎞
⎛
P ( X > 36 ) = P ⎜ Z >
⎟ = P ( Z > 1.5 ) = 1 − 0.93319 = 0.06681
4 ⎠
⎝
P(X > 36 | X > 35) = P(X > 36 and X > 35) / P(X > 35)
= P(X > 36) / P(X > 35)
= P(Z > 1.5)/P(Z > 1.25) = 0.06681/0.10565 = 0.632
(iii)
9
(i)
⎛5⎞
2
3
⎜ ⎟ × 0.1056 × 0.8944 = 0.0798
2
⎝ ⎠
If X is the random variable denoting the number of policies giving a claim,
then X ~ binomial(250,0.5).
Using the normal approximation (CLT), X ≈ N (125, 62.5) .
Using the appropriate continuity correction we have:
P ( X ≥ 139) = P( X > 138.5)
138.5 − 125 ⎞
⎛
= P⎜Z >
⎟ = 1 − Φ (1.7076 ) = 0.044 .
62.5 ⎠
⎝
(ii)
This is a one-sided test of H 0 : p = 0.5 v H1 : p > 0.5 .
P-value of the test is 0.044 from part (i).
The evidence against the hypothesis that p = 0.5 (and in favour of
p > 0.5) is not strong enough to justify rejecting it at the 1% level of testing
— we cannot conclude “p > 0.5”.
Page 4
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2007 — Examiners’ Report
10
(i)
Chi-square statistic is doubled and has value 9.722
OR work it out
(ii)
(
)
P-value is given by P χ 22 > 9.722 = 0.0077
Note: answer = 0.008 is acceptable for the mark
11
(iii)
Comment: With the first table we do not have strong enough evidence to
justify rejecting the hypothesis of no association. In the second table, we have
the same proportions in the columns, but based on more data, and now we do
have strong enough evidence (P-value < 1%) to justify rejecting the
hypothesis of no association.
(i)
(a)
Y=X
1
3
⇒ X = Y 3 , and range of Y is (0, ∞).
The cdf is given by
FY ( y ) = P (Y ≤ y ) = P( X ≤ y 3 ) = FX ( y 3 )
⎪⎧1 − exp(−λy 3 ), y ≥ 0
∴ FY ( y ) = ⎨
y<0
⎪⎩0,
(using formulae or by integration).
Then, the pdf of Y can be derived as
fY ( y ) =
(
)
d
FY ( y ) = 3λy 2 exp −λy 3 .
dy
[OR, directly as
fY ( y ) = f X ( x)
3
dx
= λe −λy 3 y 2
dy
(
)
⇒ fY ( y ) = 3λy 2 exp −λy 3 ,
OR, from formulae, identifying the cdf as that of a Weibull distribution
with c = λ, γ = 3. ]
Page 5
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2007 — Examiners’ Report
(b)
First simulate X ~ exp(λ) as
1
u = 1 − e−λx ⇒ x = − log(1 − u ) ,
λ
then set y = x
1
3.
[OR, use cdf of Y directly, i.e. u = 1 − e
−λy 3
⎧ 1
⎫
⇒ y = ⎨− log(1 − u ) ⎬
⎩ λ
⎭
(ii)
n
(a)
n
{
(
L(λ) = ∏ f ( yi ;λ ) = ∏ 3λyi 2 exp −λyi 3
i =1
i =1
)} = 3 λ ∏ y
n n
i
(λ) = log L(λ ) = n log(λ ) − λ ∑ yi 3 + constant
i
n
′(λ ) = − ∑ yi 3
λ i
′(λ ) = 0 ⇒ λˆ =
n
∑ yi3
i
[Check that ′′(λ ) = −
(b)
n
λ2
< 0. ]
For the given data we have
∑ yi3 = 16.3952
i
∴λˆ =
n
∑ yi
3
=
8
= 0.488 .
16.3952
i
(iii)
(a)
For X ~ exp(λ) we have
h( x ) =
f ( x)
λe −λx
=
=λ
S ( x) 1 − 1 − e −λx
(
)
For Y (using pdf and cdf derived above):
(
(
) = 3λy
))
3λy 2 exp −λy 3
f ( y)
h( y ) =
=
S ( y ) 1 − 1 − exp −λy 3
(
Page 6
i
2
.
2
⎛
⎞
exp ⎜ −λ ∑ yi 3 ⎟
⎜
⎟
i
⎝
⎠
1
3
]
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2007 — Examiners’ Report
(b)
12
(i)
X has a constant hazard rate h( x) = λ , and therefore should only be
used when the force of mortality can be assumed constant, e.g. over a
one-year period of time in mortality studies. For longer periods of
lifetime the r.v. Y is more suitable, as it gives an increasing hazard
function with time.
Let X be a reading and M be the number of readings which are less than 1
(a)
Since X ~ U(0, θ) , P(X < 1) = length of [0,1]/ length of [0, θ] = 1/θ
(b)
⎛1⎞ ⎛ 1⎞
L ( θ ) ∝ ⎜ ⎟ ⎜1 − ⎟
⎝θ⎠ ⎝ θ⎠
m
⇒
⇒
n−m
⇒
θ −1 ⎞
⎟
⎝ θ ⎠
( θ ) = −m log θ + ( n − m ) log ⎛⎜
( θ ) = ( n − m ) log ( θ − 1) − n log θ
∂
n−m n
=
−
set to zero ⇒ θˆ = n / m
∂θ θ − 1 θ
OR Since M ~ bi(n,1/θ) , MLE of 1/θ is the sample proportion of
readings which are <1, namely m/n, so
(1/ θ ) = m / n
(c)
⇒ 1/ θˆ = m / n ⇒ θˆ = n / m
( n − m) − n
∂
n−m n
∂2
=
−
⇒− 2 =
∂θ θ − 1 θ
∂θ
( θ − 1)2 θ2
⎡ n−M ⎤ n n−n/θ n
⎡ ∂2 ⎤
n
⎥−
=
− 2= 2
∴ E ⎢− 2 ⎥ = E ⎢
2
2
2
⎢⎣ ( θ − 1) ⎥⎦ θ
( θ − 1) θ θ ( θ − 1)
⎣⎢ ∂θ ⎦⎥
∴ CRlb =
(ii)
(a)
θ2 ( θ − 1)
n
⎛ θ2 ( θ − 1) ⎞
Large sample distribution of θ̂ is θˆ ∼ N ⎜ θ,
⎟
⎜
⎟
n
⎝
⎠
ˆ
n = 100, m = 45, θ = 100 / 45 = 2.222
Estimate of standard error of
θ̂ = [(100/45)2(100/45 – 1)/100]1/2 = 0.2457
⇒ approximate 95% CI for θ is given by 2.222 ± 1.96 × 0.2457
i.e. 2.222 ± 0.482 i.e. (1.74, 2.70).
Page 7
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2007 — Examiners’ Report
(b)
Under H0: θˆ ∼ N ( 3, 0.18 )
2.222 − 3 ⎞
⎛
P -value = P θˆ < 2.222 = P ⎜ Z <
⎟ = P ( Z < −1.834 )
0.4243 ⎠
⎝
(
)
= 0.033 (by interpolation in the table)
We can reject H0 (at levels of testing down to 3.3%) and conclude that
θ < 3.
[OR note that P(Z < −1.834) is less than 0.05, so “reject H0 at 5% level”]
13
(i)
(a)
SST = 48196 – 8722/16 = 48196 – 47524 = 672
SSB = (1862 + 2362 + 2432 + 2072)/4 – 47524 = 523.5
SSR = 672 – 523.5 = 148.5
F=
523.5 / 3 174.5
=
= 14.10
148.5 /12 12.375
[or construct an ANOVA table]
(b)
F3,12(1%) = 5.953 from tables
since 14.10 >> 5.953, P-value << 0.01
(c)
Statistician A could plot either the individual y values or the four
means of y against x to see what “shape” the effect might take.
Here is a plot of the individual y values:
Page 8
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2007 — Examiners’ Report
The shape of the effect seems to be curved, initially increasing, then
decreasing.
(d)
(ii)
(a)
The implications are simply that there is nothing to invalidate the
assumptions required for the analysis.
2402
= 3680 −
= 80
16
8722
= 48196 −
= 672
16
S xx
S yy
[or could state it is the same as SST from (i)]
S xy = 13150 −
(240)(872)
= 70
16
70
βˆ =
= 0.875 as required
80
αˆ =
(b)
1
(872 − 0.875(240)) = 41.375 as required.
16
σˆ 2 =
1
702
(672 −
) = 43.625
14
80
43.625
= 0.7385
s.e. (βˆ ) =
80
t=
0.875 − 0
= 1.185 on 14 d.f.
0.7385
P-value = 2 × P(t14 >1.185)
As P(t14 >1.345) = 0.10 from tables, P-value > 2(0.10), i.e. > 0.20
This implies that there is no evidence against H0, and hence that there
is no linear relationship between x and y.
[not that there is no “relationship”]
(c)
Residual plot suggests that there may be a curved, rather than linear,
relationship between x and y.
(d)
Statistician B could try a quadratic regression (or some other curved
form) of y on x.
END OF EXAMINERS’ REPORT
Page 9
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
10 April 2008 (am)
Subject CT3 — Probability and Mathematical Statistics
Core Technical
Time allowed: Three hours
INSTRUCTIONS TO THE CANDIDATE
1.
Enter all the candidate and examination details as requested on the front of your answer
booklet.
2.
You must not start writing your answers in the booklet until instructed to do so by the
supervisor.
3.
Mark allocations are shown in brackets.
4.
Attempt all 13 questions, beginning your answer to each question on a separate sheet.
5.
Candidates should show calculations where this is appropriate.
Graph paper is required for this paper.
AT THE END OF THE EXAMINATION
Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this
question paper.
In addition to this paper you should have available the 2002 edition of the Formulae
and Tables and your own electronic calculator from the approved list.
CT3 A2008
© Faculty of Actuaries
© Institute of Actuaries
1
The number of claims which arose during the calendar year 2005 on each of a group
of 80 private motor policies was recorded and resulted in the following frequency
distribution:
Number of claims x
Number of policies f
0
64
1
12
2
3
3
0
4
1
For these data Σfx = 22, Σfx2 = 40
Calculate the sample mean and standard deviation of the number of claims per policy.
[3]
2
Data on a sample of 29 claim amounts give a sample mean of £461.5 and a sample
standard deviation of £618.8.
One claim amount of £3,657.50 is identified as an outlier and after investigation is
found to be in error. Calculate the revised sample mean and standard deviation if this
erroneous amount is removed.
[4]
3
The following sample contains claim amounts (£) on a particular class of insurance
policies:
1,717
1,614
4
1,595
1,524
1,764
4,320
1,464
1,626
1,854
1,440
1,560
1,602
1,698
(i)
Determine the mean and the median of the claim amounts.
[2]
(ii)
State, with reasons, which of the two measures considered above you would
prefer to use to estimate the central point of the claim amounts.
[1]
[Total 3]
Consider two events A and B, such that P ( A) = 0.3 and P ( A ∩ B ) = 0.1 .
Find the minimum and maximum possible values of the conditional probability
P( A | B) .
CT3 A2008—2
[4]
5
An insurance company covers claims from four different non-life portfolios, denoted
as G1, G2, G3 and G4. The number of policies included in each portfolio is given
below:
G1
G2
G3
G4
Portfolio
No. of policies 4,000 7,000 13,000 6,000
It is estimated that the percentages of policies that will result in a claim in the
following year in each of the portfolios are 8%, 5%, 2% and 4% respectively.
Suppose a policy is chosen at random from the group of 30,000 policies comprising
the four portfolios after one year and it is found that a claim did arise on this policy
during the year. Calculate the probability that the selected policy comes from
portfolio G3.
[3]
6
Consider two random variables X and Y with joint probability density function (pdf)
f ( x, y ) =
4
(1 − xy ) , 0 < x < 1, 0 < y < 1 .
3
The marginal pdf of X is given by
f ( x) =
2
(2 − x) , 0 < x < 1
3
with a corresponding marginal pdf for Y by symmetry (you are not asked to verify
these marginal densities).
(i)
Show that the conditional pdf of Y given X = x is given by
f ( y | x) = 2
(ii)
(1 − xy )
, 0 < y < 1.
(2 − x)
[2]
(a)
Determine the conditional expectation E (Y | X = x) as a function of x
and hence determine E (Y ) .
(b)
Verify your answer in part (a) by determining E (Y ) directly from the
marginal pdf of Y.
[5]
[Total 7]
CT3 A2008—3
PLEASE TURN OVER
7
8
The claim amount X in units of £1,000 for a certain type of industrial policy is
modelled as a gamma variable with parameters α = 3 and λ = ¼.
1
X ~ χ62 .
2
(i)
Use moment generating functions to show that
(ii)
Hence use tables to find the probability that a claim amount exceeds £20,000.
[2]
[Total 5]
[3]
A woodcutter has to cut 100 fence posts of a standard length and he has a metal bar of
the required length to act as the standard. The woodcutter decides to vary his
procedure from post to post − he cuts the first post using the metal standard, then uses
this post as his standard for the cut of the next post. He continues in a similar manner,
each time using the most recently cut post as the standard for the next cut.
Each time the woodcutter cuts a post there is an error in the length cut relative to the
standard being employed for that cut − you should assume that the errors are
independent observations of a random variable with mean 0 and standard deviation
3mm.
Calculate, approximately, the probability that the length of the final post differs from
the length of the original metal standard by more than 15mm.
[5]
9
10
A researcher wishes to investigate whether a coin is balanced or not, that is if
P (heads) = 0.5 . She throws the coin four times and decides to accept the hypothesis
H 0 : P(heads) = 0.5 in a test against the alternative H1 : P(heads ) ≠ 0.5 , if the number
of times that the coin lands “heads” is 1, 2, or 3.
(i)
Calculate the probability of the type I error of this test.
[3]
(ii)
Calculate the probability of the type II error of this test, if the true probability
that the coin lands “heads” is 0.7.
[3]
[Total 6]
Pressure readings are taken regularly from a meter. It transpires that, in a random
sample of 100 such readings, 45 are less than 1, 35 are between 1 and 2, and 20 are
between 2 and 3.
Perform a χ2 goodness of fit test of the model that states that the readings are
independent observations of a random variable that is uniformly distributed on (0, 3).
[5]
CT3 A2008—4
11
In an investigation about the duration of insurance policies of a certain type, a sample
of n policies is studied. All n policies have been initiated at the same time, which is
also the time of the start of the investigation. For each policy, the time T (in months)
until the policy expires can be modelled as an exponential random variable with
parameter λ, independently of the times for all other policies.
(i)
Suppose that the investigation is terminated as soon as k policies have expired,
where k is a known (predetermined) constant. The observed policy expiry
times are denoted by t1, t2, …, tk with 0 < k ≤ n and t1< t2 < … < tk.
(a)
Show that the probability that any randomly selected policy is still in
force at the time of the termination of the investigation is e −λtk .
(b)
Show that the likelihood function of the parameter λ, using information
from all n policies, is given by
k
L (λ ) = λ k e
−λ ∑ ti
i =1
e − ( n − k ) λtk .
Hence find the maximum likelihood estimate (MLE) of λ.
(c)
Consider an investigation on 20 policies which is terminated when five
policies have expired, giving the following observed expiry times (in
months):
1.03 6.67 12.70 12.88 21.54
Calculate the MLE of λ based on this sample.
[9]
(ii)
Suppose instead that the investigation is terminated after a fixed length of time
t0. The number of policies that have expired by time t0 is considered to be a
random variable, denoted by K.
(a)
Explain clearly why the distribution of K is binomial and determine its
parameters.
(b)
Hence find the MLE of λ in this case.
(c)
Consider an investigation on 20 policies that is terminated after 24
months. By the time of termination five policies have expired.
Use this information to calculate the MLE of λ in this case.
[9]
[Total 18]
CT3 A2008—5
PLEASE TURN OVER
12
The members of the computer games clubs of three neighbouring schools decide to
take part in a light-hearted competition. Each club selects five of its members at
random under a procedure agreed and supervised by the clubs. Each selected student
then plays a particular game at the end of which the score he/she has attained is
displayed and recorded – the standard set by the games designers is such that
reasonably competent players should score about 100.
The results are as follows:
School 1
School 2
School 3
(i)
105
103
137
134
81
115
96
91
105
147
100
123
116
110
149
An analysis of variance is conducted on these results and gives the following
ANOVA table:
Source of variation
Between schools
Residual
Total
(a)
d.f.
SS
MSS
2
12
14
2,298
3,468
5,766
1,149
289
Test the hypothesis that there are no school effects against a general
alternative.
You should quote a narrow range of values within which the
probability-value of the data lies, and state your conclusion clearly.
(b)
CT3 A2008—6
Calculate a 95% confidence interval for the underlying mean score for
club members in School 1, using the information available from all
three schools.
[7]
(ii)
The members of the computer games club of a nearby fourth school hear about
the competition and ask to be included in the overall comparison. Scores for a
random sample of five of the club members at this school (School 4) are
obtained and are:
112 140 88 103 123.
The scores obtained by all twenty students are shown in the display below:
(a)
Carry out an analysis of variance on the results for all four schools
together − you should construct the ANOVA table and test the
hypothesis that there are no school effects against a general alternative.
You should quote an approximate value for the probability-value of the
data, and state your conclusion clearly.
(b)
Comment briefly on the comparison of the results of the analysis
involving Schools 1−3 only conducted in part (i)(a) and the results here
of the analysis involving all four schools.
(c)
Calculate a 95% confidence interval for the difference in the
underlying mean scores for club members in Schools 1 and 2, using the
information available from all four schools.
[13]
[Total 20]
CT3 A2008—7
PLEASE TURN OVER
In a medical experiment concerning 12 patients with a certain type of ear condition,
the following measurements were made for blood flow (y) and auricular pressure (x):
13
8.5
3
x:
y:
9.8
12
10.8
10
11.5
14
11.2
8
9.6
7
10.1
9
13.5
13
14.2
17
11.8
10
8.7
5
6.8
5
Σx = 126.5 Σx 2 = 1,381.85 Σy = 113 Σy 2 = 1, 251 Σxy = 1, 272.2
(i)
Construct a scatterplot of blood flow against auricular pressure and comment
briefly on any relationship between them.
[3]
(ii)
Calculate the equation of the least-squares fitted regression line of blood flow
on auricular pressure.
[4]
(iii)
(a)
Use a suitable pivotal quantity with a t distribution to show how to
derive the usual 95% confidence interval for the slope coefficient of
the underlying regression line, and calculate the interval.
(b)
Use your calculated confidence interval to comment on the hypothesis
that the true underlying slope coefficient is equal to 1.5.
[5]
(a)
Use a suitable pivotal quantity with a χ2 distribution to derive a 95%
confidence interval for the underlying error variance σ2, and calculate
the interval.
(b)
Hence calculate a 95% confidence interval for the error standard
deviation σ.
[5]
[Total 17]
(iv)
END OF PAPER
CT3 A2008—8
Faculty of Actuaries
Institute of Actuaries
Subject CT3 — Probability and Mathematical Statistics
Core Technical
EXAMINERS’ REPORT
April 2008
Introduction
The attached subject report has been written by the Principal Examiner with the aim of
helping candidates. The questions and comments are based around Core Reading as the
interpretation of the syllabus to which the examiners are working. They have however given
credit for any alternative approach or interpretation which they consider to be reasonable.
M A Stocker
Chairman of the Board of Examiners
June 2008
© Faculty of Actuaries
© Institute of Actuaries
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2008 — Examiners’ Report
1
n = 80, Σfx = 22, Σfx2 = 40
x = 22 / 80 = 0.275
s2 =
2
1 ⎛
222 ⎞ 33.95
= 0.42975 ⇒ s = 0.656
⎜⎜ 40 −
⎟=
79 ⎝
80 ⎟⎠
79
Σx = nx = 29(461.5) = 13383.5
Σx 2 = (n − 1) s 2 + nx 2 = 28(618.8) 2 + 29(461.5) 2 = 16898062
removing the outlier of 3657.5 gives
Σx = 13383.5 − 3657.5 = 9726
Σx 2 = 16898062 − 3657.52 = 3520756
∴x =
9726
= £347.4
28
1
97262
s = [3520756 −
] = 5272.6 ∴ s = £72.6
27
28
2
3
(i)
x=
23778
= 1829.08.
13
Median = 7th ordered observation = 1614.
(ii)
4
The median should be preferred, as it is not sensitive to the extreme observed
claim of £4320.
P ( A | B) =
P ( A ∩ B)
P ( B)
Maximum value of P(B) is 0.8 in which case P ( A | B ) =
0.1
= 0.125
0.8
Minimum value of P(B) is 0.1, in the case B ⊂ A. Then P ( A | B ) = 1
Page 2
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2008 — Examiners’ Report
5
Define the following events:
C: Policy results in a claim;
Bi: Policy comes from portfolio i, i = 1, 2, 3, 4.
Then the required probability is P(B3|C), and using Bayes’ theorem:
P ( B3 | C ) =
P(C | B3 ) P( B3 )
P(C | B3 ) P( B3 )
,
=
P(C )
∑ P(C | Bi ) P( Bi )
i
which gives
P ( B3 | C ) =
0.02 ×
0.08 ×
13
30
4
7
13
6
+ 0.05 × + 0.02 × + 0.04 ×
30
30
30
30
=
0.26 / 30 0.26
=
= 0.222 .
1.17 / 30 1.17
[OR It is possible to argue straight to
260/(320 + 350 + 260 + 240) = 260/1170 = 0.222
which is correct and gets full marks.]
6
(i)
f ( y | x) =
f ( x, y )
f ( x)
4
(1 − xy )
(1 − xy )
3
, 0 < y <1
=
=2
2
(2 − x)
(2 − x)
3
(ii)
(a)
E (Y | X = x) =
1
2
y (1 − xy )dy
(2 − x) ∫ 0
2
y2
y3 1
2
1 x
(3 − 2 x)
=
[ − x ]0 =
( − )=
(2 − x) 2
3
(2 − x) 2 3 3(2 − x)
E (Y ) = ∫
=
1 (3 − 2 x )
2
(2 − x)dx
0 3(2 − x ) 3
2 1
2
4
2 1
−
=
−
=
(3
2
x
)
dx
[3
x
x
]
0
9 ∫0
9
9
Page 3
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2008 — Examiners’ Report
(b)
7
(i)
E (Y ) =
2 1
2 2 y3 1 2 2 4
y
(2
−
y
)
dy
=
[ y − ]0 = ⋅ =
3 ∫0
3
3
3 3 9
M X (t ) = E (etX ) = (1 − 4t ) −3 from yellow book
Let Y =
1
X.
2
∴ M Y (t ) = E (etY ) = E (etX / 2 ) = M X (t / 2) = (1 − 2t ) −6 / 2
which is the m.g.f. of a gamma(3,1/2) or χ62 variable
(ii)
P ( X > 20) = P(Y > 10)
= 1 – 0.8753 = 0.1247
8
Let L be the length of the metal bar and Zi be the error that arises at the ith cut.
Length of 1st post cut = L + Z1
Length of 2nd post cut = L + Z1 + Z2
Length of 100th post cut = L + Z1 + Z2 + … + Z100
Error in length of last post cut is E = Z1 + Z2 + … + Z100
E ~ N(0,900) approximately, by CLT
P(|E| <15) ≈ P(|Z| < 15/30) = P(|Z| < 0.5) = 2 × 0.1915 = 0.383
So P(error exceeds 15mm) ≈ 1 – 0.383 = 0.617
9
(i)
Probability of type I error is
α = P(reject H 0 H 0 is true) = P ( X = 0 or X = 4 H 0 is true) ,
which gives
α = P{ X = 0 P( Heads) = 0.5)} + P{ X = 4 P( Heads ) = 0.5)}
4
4
⎛1⎞ ⎛1⎞
= ⎜ ⎟ + ⎜ ⎟ = 0.125.
⎝2⎠ ⎝2⎠
Page 4
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2008 — Examiners’ Report
(ii)
Probability of type II error of the test at P(Heads) = 0.7 is
β = P(accept H 0 H1 is true) = 1 − P{ X = 0 or X = 4 P( Heads) = 0.7}
(
)
⇒ β = 1 − 0.34 + 0.7 4 = 0.7518.
[OR using P{ X = 1 or X = 2 or X = 3 | P (heads ) = 0.7}]
10
range
observed frequency
expected frequency
0–1
45
100/3
1–2
35
100/3
2–3
20
100/3
χ2 = [(45 – 100/3)2 + (35 – 100/3)2 + (20 – 100/3)2]/(100/3) = 9.50 on 2df
P-value = P (χ 22 > 9.50) < 0.01
Reject model (at the 1% level of testing) as not providing a good fit to the data.
OR 5% point of χ22 is 5.991, so we reject model at 5%
OR 1% point of χ22 is 9.210, so we reject model at 1%
11
(i)
(a)
The required probability is
P (T > tk ) = 1 − P(T ≤ tk ) = 1 − FT (tk )
= 1 − (1 − e −λtk ) = e −λtk (using formulae or by integration).
(b)
The likelihood function is given by:
k
L(λ) = ∏ f (ti )
i =1
k
(
= ∏ λe−λti
i =1
n
∏
j = k +1
P(T > tk )
) ∏ (e ) = λ e
n
−λtk
k
k
−λ ∑ ti
i =1
e − ( n − k ) λ tk
j = k +1
Page 5
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2008 — Examiners’ Report
For the MLE:
k
l (λ ) = log L(λ ) = k log(λ ) − λ ∑ ti − (n − k )λtk
i =1
l ′(λ ) =
k k
− ∑ ti − (n − k )tk
λ i =1
l ′(λ ) = 0 ⇒ λˆ =
k
.
k
∑ ti + (n − k )tk
i =1
[And l ′′(λ ) = −
(c)
k
<0]
λ2
For the observed data,
k
n = 20, k = 5, tk = 21.54,
∑ ti = 54.82 .
i =1
λˆ =
k
=
k
∑ ti + (n − k )tk
5
= 0.0132 .
54.82 + 15 × 21.54
i =1
(ii)
(a)
We have n policies with independent durations, and each will have
expired by the time of termination with probability
p = P(T ≤ t0 ) = 1 − e−λt0 ,
or will have not expired with probability 1 - p.
(
Therefore, K ~ bin n, 1 − e −λt0
(b)
(
L(λ) ∝ 1 − e −λt0
)
) (e )
k
−λt0 n − k
(
)
l (λ ) = log L(λ ) = k log 1 − e−λt0 − (n − k )λt0
l ′(λ ) =
kt0 e−λt0
1 − e−λt0
− (n − k )t0
l ′(λ ) = 0 ⇒ e−λt0 =
Page 6
n−k
1
⎛ k⎞
⇒ λˆ = − log ⎜1 − ⎟
n
t0
⎝ n⎠
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2008 — Examiners’ Report
[OR, observed proportion (k/n) is the MLE of corresponding
proportion/probability {1 − exp(−λt0)}; solving for λ leads to same
estimate as above.]
(c)
Now t0 = 24 and all other involved quantities are as before.
1
1
5 ⎞
⎛ k⎞
⎛
λˆ = − log ⎜ 1 − ⎟ = − log ⎜1 − ⎟ = 0.0120 .
t0
24
⎝ n⎠
⎝ 20 ⎠
12
(i)
(a)
F = 1149/289 = 3.98 on (2, 12) degrees of freedom
From Yellow Tables pages 172/3, P-value of the data is between 0.05
and 0.025.
We can reject H0 (the “no schools effects” hypothesis) at the 5% level
of testing but not at the 1% level. We have some evidence against the
“no schools effects” hypothesis − and conclude that there are school
effects (i.e. differences among the underlying means).
(b)
School 1 mean = 598/5 = 119.6
t12(0.025) = 2.179
95% CI for school 1 mean is 119.6 ± 2.179×(289/5)1/2
i.e. 119.6 ± 16.6 or
(ii)
(a)
(103.0, 136.2)
y1• = 598, y2• = 485, y3• = 629, y4• = 566
y•• = 2278, Σy2 = 266,788
SST = 266788 – 22782/20 = 7323.8
SSB = (5982 + 4852 + 6292 + 5662)/5 − 22782/20 = 2301
⇒ SSR = 7324 − 2301 = 5023
Source of variation
d.f.
Between schools
Residual
Total
3
16
19
SS
MSS
2301
5023
7324
767
314
F = 767/314 = 2.44 on (3, 16) degrees of freedom
Page 7
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2008 — Examiners’ Report
From Yellow Tables pages 172/3, P-value of the data is just more than
0.1 (>10%)
We do not have sufficiently strong evidence against the “no schools
effects” hypothesis, which can stand.
(b)
With only three schools involved, the results from one of them (School
2) are sufficiently different from those of the other two to allow us to
detect a difference among underlying means. However, the results for
the fourth school range across the results for the original three schools
− with all four schools in the comparison, the “between schools” sum
of squares is no longer so high relative to the residual and we fail to
detect differences.
(c)
t16(0.025) = 2.120
0.5
⎧
⎛ 1 1 ⎞⎫
95% CI is (119.6 − 97 ) ± 2.120 ⎨314 ⎜ + ⎟ ⎬
⎝ 5 5 ⎠⎭
⎩
i.e. 22.6 ± 23.76 or (−1.2, 46.4)
13
(i)
A clearly labelled scatterplot:
There seems to be a positive linear relationship between blood flow and
auricular pressure.
Page 8
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2008 — Examiners’ Report
(ii)
n = 12
S xx = 1381.85 −
S yy = 1251 −
1132
= 186.9167
12
S xy = 1272.2 −
βˆ =
S xy
S xx
=
126.52
= 48.3292
12
(126.5)(113)
= 80.9917
12
80.9917
= 1.676
48.3292
1
αˆ = y − βˆ x = (113 − 1.6758*126.5) = −8.249
12
Fitted line is y = -8.249 + 1.676x
(iii)
(a)
βˆ − β
σˆ 2
S xx
~ tn −2 where σˆ 2 =
P[−tn −2 (2.5%) <
βˆ − β
σˆ 2
S xx
2
S xy
1
( S yy −
)
n−2
S xx
< tn−2 (2.5%)] = 0.95
Rearrangement results in the 95% confidence interval for β
2
ˆβ ± t (2.5%) σˆ
n−2
S xx
Here: σˆ 2 =
1
(80.9917) 2
(186.9167 −
) = 5.1188
10
48.3292
95% CI is 1.676 ± 2.228(0.3254) ⇒ 1.676 ± 0.725 ⇒ (0.95, 2.40)
(b)
As 1.5 lies comfortably inside this confidence interval, then there is no
evidence at all against the hypothesis that β = 1.5.
Page 9
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2008 — Examiners’ Report
(iv)
(a)
(n − 2)σˆ 2
σ
2
~ χ 2n−2
P[χ n2− 2 (97.5%) <
(n − 2)σˆ 2
σ
2
< χ 2n− 2 (2.5%)] = 0.95
Rearrangement results in the 95% confidence interval for σ2
(n − 2)σˆ 2
χ n2 − 2 (2.5%)
2
<σ <
Here 95% CI is
(n − 2)σˆ 2
χ n2 −2 (97.5%)
10(5.1188)
10(5.1188)
< σ2 <
20.48
3.247
⇒ (2.50,15.76)
(b)
95% CI for σ is ( 2.50, 15.76) ⇒ (1.58,3.97)
END OF EXAMINERS’ REPORT
Page 10
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
18 September 2008 (am)
Subject CT3 — Probability and Mathematical Statistics
Core Technical
Time allowed: Three hours
INSTRUCTIONS TO THE CANDIDATE
1.
Enter all the candidate and examination details as requested on the front of your answer
booklet.
2.
You must not start writing your answers in the booklet until instructed to do so by the
supervisor.
3.
Mark allocations are shown in brackets.
4.
Attempt all 12 questions, beginning your answer to each question on a separate sheet.
5.
Candidates should show calculations where this is appropriate.
Graph paper is required for this paper.
AT THE END OF THE EXAMINATION
Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this
question paper.
In addition to this paper you should have available the 2002 edition of the Formulae
and Tables and your own electronic calculator from the approved list.
CT3 S2008
© Faculty of Actuaries
© Institute of Actuaries
1
The mean of a sample of 30 claim amounts arising from a certain kind of insurance
policy is £5,200. Six of these claim amounts have mean £8,000 while ten others have
mean £3,100.
Calculate the mean of the remaining claim amounts in this sample.
2
[3]
Five years ago a financial institution issued a specialised type of investment bond and
investors had the option to cash in after 1, 2, 3, 4 or 5 years. The following table
gives a frequency distribution showing the numbers of those investors who cashed in
at each stage.
duration (length of time held before being cashed in)
1 year
2 years
3 years
4 years
5 years
130
151
97
64
98
Calculate the sample mean and standard deviation of the duration of these bonds
before being cashed in.
3
(i)
[4]
Let Y be the sum of two independent random variables X1 and X2, that is,
Y = X1 + X 2 .
Show that the moment generating function (mgf) of Y is the product of the
mgfs of X1 and X2.
[2]
(ii)
Let X1 and X2 be independent gamma random variables with parameters
(α1 , λ) and (α 2 , λ ) , respectively .
Use mgfs to show that Y = X1 + X 2 is also a gamma random variable and
specify its parameters.
[2]
[Total 4]
4
A random sample of 15 observations is taken from a normally distributed population
of values. The sample mean is 94.2 and the sample variance is 24.86.
Calculate a 99% confidence interval for the population mean.
5
[3]
The number of claims, X, arising on each policy in a certain portfolio depends on
another random variable Y. X is considered to follow a Poisson distribution with
mean Y. The variable Y itself is assumed to have a gamma distribution with
parameters (a,b).
Find expressions for the unconditional moments E[X] and E[X 2] using appropriate
conditional moments.
[4]
CT3 S2007—2
6
Suppose that the time T, measured in days, until the next claim arises under a
portfolio of non-life insurance policies, follows an exponential distribution with
mean 2.
(i)
Find the probability that no claim is made in the next one day period.
(ii)
The median of a random variable is defined as the value for which the
cumulative distribution function of the variable is equal to 0.5.
Find the median time until the next claim arises.
(iii)
[2]
[2]
Now let T1, T2, …, T30 be the times (in days) until the next claim arises under
each one of 30 similar portfolios of non-life insurance policies, and assume
that each Ti, i = 1,…,30, follows an exponential distribution with mean 2,
independently of all others.
Calculate, approximately, the probability that the total of all 30 times which
elapse until a claim arises on each of the portfolios exceeds 45 days.
[4]
[Total 8]
7
Let N be the number of claims arising on a group of policies in a period of one week
and suppose that N follows a Poisson distribution with mean 60.
Let X1, X2, . . , XN be the corresponding claim amounts and suppose that,
independently of N, these are independent and identically distributed with mean £500
and standard deviation £400.
N
Let S = ∑ X i be the total claim amount for the period of one week.
i =1
(i)
Determine the mean and the standard deviation of S.
(ii)
Explain why the distribution of S can be taken as approximately normal, and
hence calculate, approximately, the probability that S is greater than £40,000.
[3]
[Total 5]
CT3 S2008—3
[2]
PLEASE TURN OVER
8
(i)
Use the following uniform(0,1) random numbers
0.9236 , 0.2578
and a suitable table of probabilities to simulate two observations of the random
variable X, where X ~ N(200,100).
[3]
(ii)
Use the following uniform(0,1) random numbers
0.3287 , 0.9142
to simulate two observations of the random variable Y, where Y has an
exponential distribution with mean 100.
[3]
[Total 6]
9
A random sample of four insurance policies of a certain type was examined for each
of three insurance companies and the sums insured were recorded. An analysis of
variance was then conducted to test the hypothesis that there are no differences in the
means of the sums insured under such policies by the three companies.
The total sum of squares was found to be SST = 420.05 and the between-companies
sum of squares was found to be SSB = 337.32.
Perform the analysis of variance to test the above hypothesis and state your
conclusion.
[4]
(ii)
State clearly any assumptions that you made in performing the analysis in (i).
[2]
(iii)
The plot of the residuals of this analysis of variance against the associated
fitted values, is given below.
0
-4
-2
Residuals
2
4
(i)
18
20
22
24
26
28
30
32
Fitted
Comment briefly on the validity of the test performed in (i), basing your
answer on the above plot.
[2]
[Total 8]
CT3 S2007—4
10
When a new claim comes into an office it is screened at a first stage and has a
probability θ of being cleared for progress, otherwise it is rejected. If it clears the first
stage, it is then independently screened at a second stage and has the same probability
θ of being cleared for progress, otherwise it is rejected.
(i)
Explain clearly why the probability of a claim being rejected at the first stage
is 1 - θ, of being rejected at the second stage is θ (1 - θ) and of progressing
after the two stages is θ2.
[3]
(ii)
For a sample of n independent claims which came into the office x1 were
rejected at the first stage, x2 were rejected at the second stage and x3
progressed after the two stages (x1 + x2 + x3 = n).
(a)
Write down the likelihood L(θ) for this sample and hence show that the
derivative of the log-likelihood is given by
x + 2 x3 x1 + x2
∂
.
log L(θ) = 2
−
1− θ
∂θ
θ
(b)
Show that the maximum likelihood estimator (MLE) is given by
θˆ =
x2 + 2 x3
.
x1 + 2 x2 + 2 x3
[7]
(iii)
(iv)
∂
2
(a)
log L(θ) of the log-likelihood in
∂θ2
part (ii) above and hence show that the Cramer-Rao lower bound
θ(1 − θ)
(CRlb) is given by
.
n(1 + θ)
(b)
Use the asymptotic distribution for the MLE θ̂ with the CRlb
evaluated at θ̂ to obtain an approximate large-sample 95% confidence
interval for θ expressing it simply in terms of θ̂ and n.
[7]
Determine the second derivative
For a sample of 1,000 independent claims, 110 were rejected at the first stage,
96 were rejected at the second stage and 794 progressed after the two stages.
Calculate the MLE θ̂ together with an approximate 95% confidence interval
for θ.
[3]
[Total 20]
CT3 S2008—5
PLEASE TURN OVER
11
Male
4
6
10
1
1
A study was conducted to investigate lengths of stay, in days, of short-term stay
patients in a particular hospital. Independent random samples of 40 male patients and
35 female patients were selected and the lengths of stay of these patients are given in
the following tables:
8
6
7
7
8
2
8
5
2
1
6
7
7
8
3
9
3
7
11
9
6
5
6
5
3
4
6
9
6
1
Male: Σx = 215 Σx2 = 1,481
10
1
2
2
3
Female
2
7
8
6
3
5
7
2
3
6
5
4
4
4
5
1
1
6
4
9
9
5
5
11
6
1
4
1
6
2
3
4
8
8
3
Female: Σx = 168 Σx2 = 1,026
The male observations are assumed to be normally distributed with mean μ1 and
standard deviation σ1 , and independently the female observations are assumed to be
normally distributed with mean μ2 and standard deviation σ 2 .
(i)
Suppose that it is known that σ1 = 3.0 days and σ2 = 2.5 days.
(a)
Construct a 95% confidence interval for the difference between the
mean length of stay for males and the mean length of stay for females,
that is for μ1 − μ2.
(b)
Comment briefly on any implications of this confidence interval.
[6]
(ii)
Suppose now that σ1 and σ 2 are unknown.
(a)
Perform a two-sample t-test to investigate whether there is a difference
between the mean length of stay for males and the mean length of stay
for females, assuming that σ1 and σ2 are equal.
(b)
Show that the variances in the male and female samples are not
significantly different at the 5% level, and comment briefly with
reference to the validity of the test conducted in (ii)(a).
(c)
Suppose you are not prepared to assume more than you feel is
absolutely necessary – in particular you do not want to assume that
σ1 and σ2 are equal, nor that the observations necessarily come from
normal populations.
Perform an alternative (large-sample) test to that conducted in part
(ii)(a), “to investigate whether there is a difference between the mean
length of stay for males and the mean length of stay for females”, and
compare the results of the test with the results of the test obtained in
part (ii)(a).
[11]
[Total 17]
CT3 S2007—6
12
Consider a situation in which the data consist of two responses at each of five values
of an explanatory variable (x = 1, 2, 3, 4, 5), so we have a data set with ten responses
(y), as in the following table:
1
12
x
y
1
19
2
18
2
35
3
19
3
44
4
32
4
53
5
44
5
65
For these data Σx = 30, Σy = 341, Σx2 = 110, Σy2 = 14,345 , Σxy = 1,211
(i)
(ii)
You are asked to carry out a linear regression analysis using these data.
(a)
Draw a plot of the data to show the relationship between the response
and explanatory values.
(b)
Calculate the total, regression, and residual sums of squares for a leastsquares linear regression analysis of y on x, and hence calculate the
value of R2, the coefficient of determination.
(c)
Determine the equation of the fitted regression line.
(d)
Calculate a 95% confidence interval for the slope of the underlying
regression line.
[12]
A colleague suggests that it will be simpler and will produce the same results
if we use the following reduced data, in which the two responses at each x
value are replaced by their mean:
x
y
1
2
3
4
5
15.5 26.5 31.5 42.5 54.5
The details of the regression analysis for these data are given in the box below.
Regression equation: y = 5.90 + 9.40 x
Coef
Stdev
t-ratio
p-val
Intercept
x
5.900
9.400
2.233
0.673
2.64
13.96
0.078
0.001
s = 2.129
R-sq = 98.5%
Analysis of Variance
Source
df
SS
Regression 1 883.60
Error
3
13.60
Total
4 897.20
MS
F
p-val
883.60 194.91 0.001
4.53
Discuss the similarities and the differences between the two approaches and their
results, in particular addressing the claim by the colleague that the two analyses will
produce “the same results”.
[6]
[Total 18]
END OF PAPER
CT3 S2008—7
Faculty of Actuaries
Institute of Actuaries
Subject CT3 — Probability and Mathematical Statistics
Core Technical
EXAMINERS’ REPORT
September 2008
Introduction
The attached subject report has been written by the Principal Examiner with the aim of
helping candidates. The questions and comments are based around Core Reading as the
interpretation of the syllabus to which the examiners are working. They have however given
credit for any alternative approach or interpretation which they consider to be reasonable.
R D Muckart
Chairman of the Board of Examiners
November 2008
Comments
The paper was answered well overall and there are no particular topics that stand out as being
poorly attempted. Similarly there were no particular misunderstandings widely evident, and
no particular errors were made so repeatedly as to be worthy of comment.
© Faculty of Actuaries
© Institute of Actuaries
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2008 — Examiners’ Report
1
n = 30, x = 5200
n1 = 6, x1 = 8000
n2 = 10, x2 = 3100
n3 = 14
x=
∑ x = n1x1 + n2 x2 + n3 x3
n1 + n2 + n3
n
⇒ x3 =
2
nx − n1x1 − n2 x2 30 × 5200 − 6 × 8000 − 10 × 3100 77000
=
=
= 5500
n3
14
14
data: Σf = 540, Σfx = 1469, Σfx2 = 5081
mean =
1469
= 2.72 years
540
variance =
3
(i)
1
14692
(5081 −
) = 2.0126
539
540
∴ s.d. = 1.42 years
M Y (t ) = E (etY ) = E (et ( X1 + X 2 ) )
= E (etX1 ) E (etX 2 ) = M X1 (t ) M X 2 (t )
(ii)
t
M X i (t ) = (1 − ) −αi
λ
t
∴ M Y (t ) = (1 − ) −(α1 +α 2 )
λ
so that Y is a gamma r.v. with parameters (α1 + α 2 , λ) .
4
t14(0.005) = 2.977
99% CI is 94.2 ± 2.977
Page 2
24.86
15
i.e. 94.2 ± 3.83 i.e. (90.37,98.03)
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September2008 — Examiners’ Report
5
E ( X ) = EY { E X ( X Y )} = E (Y ) =
a
.
b
E ( X 2 ) = var( X ) + E 2 ( X )
with
var( X ) = EY {varX ( X Y )} + varY { E X ( X Y )}
= E (Y ) + var(Y ) =
a a
+
b b2
giving
a a a2
E( X ) = + 2 + 2 .
b b
b
2
{
{var
}
[OR E ( X 2 ) = EY E X ( X 2 Y )
= EY
X
}
( X Y ) + E X2 ( X Y ) = E (Y ) + E (Y 2 )
= E (Y ) + var(Y ) + E 2 (Y )
=
6
a a a2
.]
+ +
b b2 b2
(i)
T ~ Exp(0.5) and therefore P (T > 1) = e −0.5×1 = 0.6065 .
(ii)
The median, M, is such that
M
∫
0
M
f (t )dt = 0.5 ⇒ ∫ 0.5e −0.5t dt = 0.5
0
which gives
1 − e−0.5 M = 0.5 ⇒ M = −2 log(0.5) , or M = 2 log(2) = 1.386 .
(Note: the cdf is available from the Yellow Book, p11.)
30
(iii)
From CLT, Y = ∑ Ti ~ N (30 × 2, 30 × 4), i.e. N (60,120) , approximately.
i =1
Then,
45 − 60 ⎞
⎛
P (Y > 45) = P ⎜ Z >
⎟ = P( Z > −1.3693) = P( Z < 1.3693) = 0.915.
120 ⎠
⎝
Page 3
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2008 — Examiners’ Report
2
, from which we can then use the
[OR Y ~ gamma(30,1/2), that is Y ~ χ60
normal approximation as above, or get P(Y > 45) = 0.922 (approximately) by
2
(Yellow Book p168).]
interpolating in tables of percentage points of χ60
7
(i)
E ( S ) = E ( N ) E ( X ) = (60)(500) = £30, 000
V ( S ) = E ( N )V ( X ) + V ( N )[ E ( X )]2
= (60)(4002 ) + (60)(5002 ) = 24, 600, 000 ∴ sd ( S ) = £4,960
(ii)
As S is the sum of a large number of i.i.d. variables, then the central limit
theorem gives an approximate normal distribution for S.
P ( S > 40000) = P ( Z >
40000 − 30000
= 2.016)
4960
= 1 − 0.9781 = 0.0219
[Note: 2.02 leading to 0.0217 is also acceptable.]
8
(i)
From Yellow Book Table
P(Z < 1.43) = 0.9236 giving x value (10*1.43) + 200 = 214.3
P(Z < −0.65) = 0.2578 giving x value (10*(−0.65)) + 200 = 193.5
(ii)
Setting r = P(Y < y) = 1 – exp(−y/100) ⇒ y = −100*log(1 − r)
r = 0.3287 ⇒ y = −100log(0.6713) = 39.85
r = 0.9142 ⇒ y = −100log(0.0858) = 245.6
Note: We can do away with the step of subtracting r from 1 and use.
y = −100*log(r). This gives y = 111.3, 8.971.
9
(i)
SSR = SST – SSB = 420.05 – 337.32 = 82.73.
The degrees of freedom are 3 – 1 = 2 for the treatment (company) SS, and
12 – 1 – 2 = 9 for the residual SS.
Page 4
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September2008 — Examiners’ Report
These give F =
337.32 2
= 18.348.
82.73 9
From tables, F0.01,2,9 = 8.022, and therefore we have strong evidence against
the hypothesis that the means of the insured sums are equal for the 3
companies.
10
(ii)
To perform the ANOVA we assume that the data follow normal distributions
and that their variance is constant.
(iii)
The variance of the residuals seems to depend on the company from which the
data come. This violates the assumption of constant variance in the response
variable, and therefore the analysis may not be valid.
(i)
P(rejected at 1st) = 1 – P(cleared at 1st) = 1 – θ
P(rejected at 2nd) = P(cleared at 1st)P(rejected at 2nd | cleared at 1st)
= θ (1 – θ)
P(progressing after two) = P(cleared at 1st) P(cleared at 2nd) = θ2
(ii)
(a)
L(θ) = [(1 − θ)]x1 [θ(1 − θ)]x2 [θ2 ]x3
= θ x2 + 2 x3 (1 − θ) x1 + x2
∴ log L(θ) = ( x2 + 2 x3 ) log θ + ( x1 + x2 ) log(1 − θ)
∴
(b)
x + 2 x3 x1 + x2
∂
log L(θ) = 2
−
1− θ
∂θ
θ
equate to zero for MLE
∴θ( x1 + x2 ) = (1 − θ)( x2 + 2 x3 )
∴θ( x1 + 2 x2 + 2 x3 ) = x2 + 2 x3
x2 + 2 x3
x1 + 2 x2 + 2 x3
∴θˆ =
(iii)
(a)
∂2
x2 + 2 x3
∂θ
θ2
log L(θ) = −
2
E{
−
x1 + x2
(1 − θ) 2
∂2
nθ(1 − θ) + 2nθ2
∂θ
θ2
log L(θ)} = −
2
−
n(1 − θ) + nθ(1 − θ)
(1 − θ) 2
Page 5
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2008 — Examiners’ Report
n
n
n(1 + θ)
1
1
= − (1 + θ) −
(1 + θ) = − n(1 + θ)( +
)=−
θ
θ 1 − θ)
θ(1 − θ)
(1 − θ)
1
CRlb =
− E[
∂
log L(θ)]
∂θ2
2
=
θ(1 − θ)
n(1 + θ)
θˆ ≈ N (θ, CRlb) for large n
(b)
using CRlb =
θˆ (1 − θˆ )
θˆ (1 − θˆ )
, then θˆ ≈ N (θ,
)
n(1 + θˆ )
n(1 + θˆ )
θˆ (1 − θˆ )
95% CI is θˆ ± 1.96
n(1 + θˆ )
(iv)
θˆ =
96 + 2(794)
1684
=
= 0.8910
110 + 2(96) + 2(794) 1890
CRlb ≈
0.8910(1 − 0.8910)
= 0.0000514 ∴ CRlb = 0.00717
1000(1 + 0.8910)
∴95% CI is 0.8910 ± 1.96(0.00717)
⇒ 0.891 ± 0.014 or (0.877, 0.905)
11
(i)
(a)
Males: n1 = 40
x1 = 215/40 = 5.375
Females: n2 = 35 x2 = 168/35 = 4.8
95% CI:
x1 − x2 ± z0.025
σ12 σ22
+
n1 n2
= 5.375 – 4.8 ± 1.96
32 2.52
+
40 35
= 0.575 ± (1.96)(0.6353)
= 0.575 ± 1.245 or (–0.67, 1.82)
(b)
Page 6
As this CI includes the value 0 we would not eliminate the possibility
that the males and females have the same expected length of stay.
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September2008 — Examiners’ Report
(ii)
(a)
s12 = (1481 – 2152/40)/39 = 8.34295
s22 = (1026 – 1682/35)/34 = 6.45882
s 2p =
(n1 − 1) s12 + (n2 − 1) s22
(39)(8.34295) + (34)(6.45882)
=
= 7.46541
40 + 35 − 2
n1 + n2 − 2
Two sample t-test
x1 − x2
t=
⎛1 1 ⎞
s 2p ⎜ + ⎟
⎝ n1 n2 ⎠
=
5.375 − 4.8
1 ⎞
⎛ 1
7.46541⎜ + ⎟
⎝ 40 35 ⎠
= 0.909
t73(0.025) = 1.996 (by interpolation of 2.000 and 1.980 for 60 df and
120 df)
[OR just quote the N(0,1) value 1.96 in place of the t73 value.]
Therefore there is no evidence to reject the null hypothesis that the
means for males and females do not differ at the 5% significance level,
and we conclude that the mean lengths of stay are the same.
(b)
s12
s22
=
8.34295
= 1.29
6.45882
Comparing this to an F39,34 distribution, which has a 5% critical point
between 2.075 and 1.717 (two-sided test), there is no evidence that the
population variances differ.
The assumption of common variance was made when conducting the
test in (ii)(a), and this seems valid given the result of the test in (ii)(b).
(c)
z=
=
x1 − x2
s12 s22
+
n1 n2
5.375 − 4.8
8.34295 6.45882
+
40
35
= 0.917
Page 7
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2008 — Examiners’ Report
Compare with N(0,1), e.g. 1.96 for 5% level test. Therefore we reach
exactly the same conclusion (as in (ii)(a) but without making the
assumptions of equal variances and normal distributions – we have
large samples and can rely on CLT).
12
(i)
(a)
(b)
SSTOT = Syy = 14345 – 3412/10 = 2716.9
Σx = 30, Σx2 = 110 so Sxx = 110 – 302/10 = 20
Sxy = 1211 − 30*341/10 = 188
∴SSREG = 1882/20 = 1767.2
SSRES = 2716.9 − 1767.2 = 949.7
R2 = 1767.2/2716.9 = 0.650 (65.0%)
(c)
y = a + bx:
bˆ = 188 / 20 = 9.4
aˆ = 341/10 − 9.4 × (30 /10) = 5.9
Fitted line is y = 5.9 + 9.4x
(d)
()
t8(0.025) = 2.306
Page 8
1/ 2
⎛ 949.7 / 8 ⎞
s.e. bˆ = ⎜
⎟
⎝ 20 ⎠
= 2.4363
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September2008 — Examiners’ Report
95% confidence interval for b is given by 9.4 ± 2.306 × 2.4363
i.e. 9.4 ± 5.62 i.e. (3.78, 15.02)
(ii)
When we replace the pair of responses by their mean:
the equation of the fitted line remains the same
but otherwise the analyses do not produce “equivalent results”
the “fit” of the line is very much better [the goodness-of-fit measure R2 increases
to a very high value − from 65% to 98.5%]
plus, for example:
the estimate of the slope has a much lower standard error (2.436 drops to 0.6733)
the SSTOT drops hugely (from 2716.9 on 9df to 897.2 on 4df)
the residual error (SSRES) drops hugely [from 949.7 on 8df (error variance
estimate 118.7 ) to 13.60 on 3 df (error variance estimate 4.53)]
BUT we lose all information on the variation of the response for a given value of
the explanatory variable
Note: these and other relevant comments will receive credit.
END OF EXAMINERS’ REPORT
Page 9
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
30 April 2009 (am)
Subject CT3 — Probability and Mathematical Statistics
Core Technical
Time allowed: Three hours
INSTRUCTIONS TO THE CANDIDATE
1.
Enter all the candidate and examination details as requested on the front of your answer
booklet.
2.
You must not start writing your answers in the booklet until instructed to do so by the
supervisor.
3.
Mark allocations are shown in brackets.
4.
Attempt all 13 questions, beginning your answer to each question on a separate sheet.
5.
Candidates should show calculations where this is appropriate.
Graph paper is required for this paper.
AT THE END OF THE EXAMINATION
Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this
question paper.
In addition to this paper you should have available the 2002 edition of the Formulae
and Tables and your own electronic calculator from the approved list.
CT3 A2009
© Faculty of Actuaries
© Institute of Actuaries
1
A random sample of 12 claim amounts (in units of £1,000) on a general insurance
portfolio is given by:
14.9
12.4
19.4
3.1
17.6
21.5
15.3
20.1
18.8
11.4
46.2
16.2
For these data: Σx = 216.9, Σx2 = 5,052.13, sample mean x = £18,075
sample median = £16,900, sample standard deviation s = £10,143.
Calculate the sample mean, median, and standard deviation of the sample (of size 10)
which remains after we remove the claim amounts 3.1 and 46.2 from the original
sample (you should show intermediate working and/or give justifications for your
answers).
[6]
2
Consider three events A, B, and C for which A and C are independent, and B and C are
mutually exclusive. You are given the probabilities P(A) = 0.3, P(B) = 0.5, P(C) = 0.2
and P(A∩B) = 0.1.
Find the probability that none of A, B, or C occurs.
3
[3]
The random variable X has probability density function
f ( x) = k (1 − x)(1 + x),
0 < x <1,
where k is a positive constant.
4
5
(i)
Show that k = 1.5.
(ii)
Calculate the probability P(X > 0.25).
[2]
[2]
[Total 4]
Let the random variable Y denote the size (in units of £1,000) of the loss per claim
sustained in a particular line of insurance. Suppose that Y follows a chi-square
distribution with 2 degrees of freedom. Two such claims are randomly chosen and
their corresponding losses are assumed to be independent of each other.
(i)
Determine the mean and the variance of the total loss from the two claims. [2]
(ii)
Find the value of k such that there is a probability of 0.95 that the total loss
from the two claims exceeds k.
[2]
[Total 4]
The human resources department of a large insurance company currently estimates
that 82% of new employees recruited by their call centres will still be employed by
the company after one year. A recent extension to the call centre business led to 280
new employees being recruited.
Calculate an approximate value for the probability that at least 240 of these new
employees will still be employed by the company after one year.
CT3 A2009—2
[3]
6
7
The variables X1, X2, …, X40 give the size (in units of £100) of each of 40 claims in a
random sample of claims arising from damage to cars by vandals. The size of each
claim is assumed to follow a gamma distribution with parameters α = 4 and λ = 0.5
1 40
and each is independent of all others. Let X =
∑ X i be the random variable
40 i =1
giving the mean size of such a sample.
(i)
State the approximate sampling distribution of X and determine its parameters.
[2]
(ii)
Determine approximately the median of X .
[1]
[Total 3]
A survey is undertaken to investigate the frequency of motor accidents at a certain
intersection.
It is assumed that, independently for each week, the number of accidents follows a
Poisson distribution with mean λ.
(i)
8
In a single week of observation two accidents occur. Determine a 95%
confidence interval for λ, using tables of “Probabilities for the Poisson
distribution”.
[3]
(ii)
In an observation period of 30 weeks an average of 2.4 accidents is recorded.
Determine a 95% confidence interval for λ, using a normal approximation. [3]
(iii)
Comment on your answers in parts (i) and (ii) above.
[1]
[Total 7]
A random sample of 25 recent claim amounts in a general insurance context is taken
from a population that you may assume is normally distributed. In units of £1,000,
the sample mean is x = 9.416 and the sample standard deviation is s = 2.105.
Calculate a 95% one-sided upper confidence limit (that is, the upper limit k of a
confidence interval of the form (0,k)) for the standard deviation of the claim amounts
in the population.
[5]
CT3 A2009—3
PLEASE TURN OVER
9
An analysis of variance investigation with samples of size eight for each of four
treatments results in the following ANOVA table.
Source of variation
d.f.
SS
MSS
Between treatments
Residual
Total
3
28
31
6716
3362
10078
2239
120
(i)
Calculate the observed F statistic, specify an interval in which the resulting
P-value lies, and state your conclusion clearly.
[3]
(ii)
The four treatment means are:
y1 = 85.0, y2 = 66.5, y3 = 59.0, and y4 = 95.5 .
(a)
Calculate the least significant difference between pairs of means using
a 5% level.
(b)
List the means in order, illustrate the non-significant pairs using
suitable underlining, and comment briefly.
[3]
[Total 6]
10
For a group of policies the probability distribution of the total number of claims, N,
arising during a period of one year is given by
P(N = 0) = 0.70, P(N = 1) = 0.15, P(N = 2) = 0.10, P(N = 3) = 0.05.
Each claim amount, X (in units of £1,000), follows a gamma distribution with
parameters α = 2 and λ = 0.1 independently of each other claim amount and of the
number of claims.
Calculate the expected value and the standard deviation of the total of the claim
amounts for a period of one year.
CT3 A2009—4
[5]
11
The number of claims, X, which arise in a year on each policy of a particular class is
to be modelled as a Poisson random variable with mean λ. Let X = (X1, X2, …, Xn) be
a random sample from the distribution of X, and let X =
1 n
∑ Xi .
n i =1
n
(i)
(a)
Use moment generating functions to show that
∑ X i has a Poisson
i =1
distribution with mean nλ.
(b)
State, with a brief reason, whether or not the variable 2X1 + 5 has a
Poisson distribution.
(c)
State, with a brief reason, whether or not X has a Poisson distribution
in the case that n = 2.
(d)
State the approximate distribution of X in the case that n is large.
[8]
An actuary is interested in the level of claims being experienced and wants in
particular to test the hypotheses
H0: λ = 1 v H1: λ > 1 .
He decides to use a random sample of size n = 100 and the best (most powerful)
available test. You may assume that this test rejects H0 for x > k , for some
constant k.
(ii)
(a)
Show that the value of k for the test with level of significance 0.01 is
k = 1.2326.
(b)
Calculate the power of the test in part (ii)(a) in the case λ = 1.2 and
then in the case λ = 1.5.
(c)
Comment briefly on the values of the power of the test obtained in part
(ii)(b).
[9]
[Total 17]
CT3 A2009—5
PLEASE TURN OVER
12
In a genetic plant breeding experiment a total of 1,500 plants were categorised into
one of four classes (labelled A, B, C and D) with the following results:
class:
frequency:
A
1071
B
62
C
68
D
299
A genetic model specifies that the probability that an individual plant belongs to each
class is given by:
class:
probability:
A
B
C
D
1
(2 + θ)
4
1
(1 − θ)
4
1
(1 − θ)
4
1
θ
4
where θ is an unknown parameter such that 0 < θ < 1.
(i)
(a)
Write down the likelihood for these data and determine the loglikelihood.
(b)
Show that the maximum likelihood estimate (MLE) of θ is a solution
of the quadratic equation
750θ2 − 256θ − 299 = 0
and hence that the MLE is given by θˆ = 0.825 .
(ii)
(iii)
[7]
(a)
Determine the second derivative of the log-likelihood and use this,
evaluated at θ̂ = 0.825, to obtain an approximation for the Cramer-Rao
lower bound for this situation.
(b)
Hence calculate an approximate 95% confidence interval for θ, using
the asymptotic distribution of the MLE.
[5]
An extension of the genetic model suggests that the value of θ should be equal
to 0.775.
(a)
Carry out an appropriate χ2 test to investigate the extent to which the
current data support the extended model with this value of θ (you
should calculate and comment on the P-value).
(b)
Comment briefly on how this relates to your approximate confidence
interval in part (ii)(b).
[6]
[Total 18]
CT3 A2009—6
13
The following table gives the scores (out of 100) that 10 students obtained on a
midterm test (x) and the final examination (y) in a course in statistics.
65
44
Midterm x
Final y
62
49
50
54
82
59
80
66
68
67
88
71
67
81
90
89
92
98
For these data you are given: Sxx = 1,760.4, Syy = 2,737.6, Sxy = 1,529.8
(i)
(a)
Draw a scatterplot of the data and comment briefly on the relationship
between the score in the final examination and that in the midterm test.
(b)
The equation of the line of best fit is given by y = 3.146 + 0.869x.
Perform a suitable test involving the slope parameter β, to test the null
hypothesis H0: β = 0 against H1: β > 0.
(c)
Calculate a 95% confidence interval for the mean final examination
score for a midterm score of 75.
(d)
Consider now that we require a 95% confidence interval for an
individual predicted final examination score for a midterm score of 75.
State (giving reasons) whether this interval will be narrower or wider
than the one calculated in part (i)(c) above. (You are not asked to
calculate the interval.)
[13]
The lecturer of this course decides to assess the linear relationship between the score
in the final examination and that in the midterm test, by using the sample correlation
coefficient r.
The hypothesis H0: ρ = 0 (where ρ denotes the population correlation coefficient) can
be tested against H1: ρ > 0, by using the result that under H0 the sampling distribution
of the statistic
(ii)
r n−2
1− r2
is the tn-2 distribution (where n is the size of the sample).
(a)
Show algebraically, that is without referring to the specific data given
here, that in general the above statistic and the statistic involving β that
you used in (i)(b) produce equivalent tests.
(b)
Calculate the value of r for the given data and hence verify numerically
the result of part (ii)(a) above.
[6]
[Total 19]
END OF PAPER
CT3 A2009—7
Faculty of Actuaries
Institute of Actuaries
Subject CT3 — Probability and Mathematical Statistics
Core Technical
EXAMINERS’ REPORT
April 2009
Introduction
The attached subject report has been written by the Principal Examiner with the aim of
helping candidates. The questions and comments are based around Core Reading as the
interpretation of the syllabus to which the examiners are working. They have however given
credit for any alternative approach or interpretation which they consider to be reasonable.
R D Muckart
Chairman of the Board of Examiners
June 2009
Comments
The paper was answered quite well overall. Some questions were answered less well (or by
noticeably fewer candidates) than others – they were:
Question 7(i) – confidence intervals based on small samples
Question 9(ii) – differences between pairs of treatment means in an ANOVA context
Question 11(i) – deciding whether or not certain variables derived from Poisson variables are
themselves Poisson variables
Question 12 – writing down the correct likelihood function, in which the stated probabilities
are raised to powers given by the observed frequencies of occurrence (not multiplied by the
frequencies)
There were no other misunderstandings widely evident, and no particular errors were made so
repeatedly as to be worthy of comment.
© Faculty of Actuaries
© Institute of Actuaries
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2009 — Examiners’ Report
1
Revised mean = (216.9 – 3.1 – 46.2)/10 = 16.76 i.e. £16,760
Revised median = original median = £16,900
Revised Σx2 = 5052.13 – 3.12 – 46.22 = 2908.08
1/ 2
⎧⎪ 1 ⎛
167.62 ⎞ ⎫⎪
so revised standard deviation = ⎨ ⎜ 2908.08 −
⎟⎬
⎜
10 ⎟⎠ ⎪⎭
⎪⎩ 9 ⎝
2
= 3.31837 i.e. £3,318.37
P(A∪B∪C) = P(A) + P(B) + P(C) – P(A∩B) – P(A∩C) – P(B∩C) + P(A∩B∩C)
= 0.3 + 0.5 + 0.2 – 0.1 – (0.3 × 0.2) = 0.84
(OR via a Venn diagram)
So P(none occur) = 1 – 0.84 = 0.16
1
3
(i)
∫
0
(ii)
1
1
⎡
x3 ⎤
⎛ 1⎞
f ( x)dx = 1 ⇒ ∫ k (1 − x 2 )dx = 1 ⇒ k ⎢ x − ⎥ = 1 ⇒ k ⎜1 − ⎟ = 1 ⇒ k = 1.5
3 ⎥⎦
⎝ 3⎠
⎣⎢
0
0
P(X > 0.25) =
1
∫ 0.25 f ( x)dx
1
⎡
x3 ⎤
= 1.5 × 0.422 = 0.633.
= ∫ 1.5(1 − x )dx = 1.5 ⎢ x − ⎥
0.25
3 ⎥⎦
⎢⎣
0.25
1
4
(i)
2
E[Yi ] = 2, Var[Yi ] = 4
Therefore E[Y1 + Y2] = 4 and (since Y1, Y2 independent) Var[Y1 + Y2] = 8.
So, for total loss, mean = £4,000 and variance = 8 ×106 (£2).
(OR from Y1 + Y2 ~ χ 24 )
(ii)
For the total loss we have Y1 + Y2 ~ χ 42 , so we want a constant k
(
)
2
such that P χ 4 > k = 0.95.
(
)
From tables of the χ 24 distribution P χ 24 > 0.7107 = 0.95.
∴The total losses will exceed £710.7 with probability 0.95.
Page 2
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2009 — Examiners’ Report
5
Let X be the number still in employment after one year.
∴ X ~ bin(280, 0.82) ≈ N (229.6, 6.432 )
P ( X ≥ 240) = P( X > 239.5) applying a continuity correction
= P( Z >
6
(i)
239.5 − 229.6
) = P ( Z > 1.54) = 1 − 0.93822 = 0.062
6.43
E[Xi] = 4/0.5 = 8 and Var[Xi] = 4/0.52 = 16 (or by noting X i ~ χ82 ).
Var[ X i ] ⎞
∑ Xi ⎛
X = i =1 ≈ N E[ X ],
, i.e. N(8, 0.4)
40
Using the CLT:
⎜
⎝
40
i
40
⎟
⎠
approximately.
[Note: The exact distribution of X is Gamma(160,20)]
(ii)
The symmetry of the distribution gives: median ⎡⎣ X ⎤⎦ = mean ⎡⎣ X ⎤⎦ = 8,
i.e. £800.
7
(i)
For a single observation x from Poisson(λ) a 95% confidence interval for λ is
(λ1,λ2) where
∞
∑ p(r; λ1) = 0.025
x
and
r=x
∑ p(r; λ 2 ) = 0.025
r =0
So for x = 2
λ1 is s.t.
∞
∑ p(r; λ1) = 0.025
r =2
1
i.e.
∑ p(r; λ1) = 0.975
r =0
From tables λ1 is between 0.20 and 0.30 being about 0.24.
λ2 is s.t.
2
∑ p(r; λ 2 ) = 0.025
r =0
From tables λ2 is between 7.00 and 7.25 being about 7.23.
95% confidence interval for λ is (0.24, 7.23).
Page 3
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2009 — Examiners’ Report
(ii)
For a sample of n with observed mean x
X −λ
≈ N (0,1)
ˆλ
where λˆ = X
[OR: could use
ΣX − nλ
≈ N (0,1) ]
ˆ
nλ
n
giving an approximate 95% confidence interval as X ± 1.96
X
n
So for n = 30, x =2.4
95% CI is 2.4 ± 1.96
(iii)
2.4
⇒ 2.4 ± 0.55 ⇒ (1.85, 2.95)
30
The CI in (ii) is much narrower due to having more data.
(It is also centred higher due to the larger estimate.)
8
Let σ2 be the population variance.
24 S 2
σ2
⎛ 24S 2
⎞
~ χ 224 ⇒ P ⎜ 2 > 13.85 ⎟ = 0.95
⎜ σ
⎟
⎝
⎠
⎛ 2 24S 2 ⎞
2
2
⇒ P⎜σ <
⎟⎟ = 0.95 ⇒ k = 24 × 2.105 / 13.85 = 7.678
⎜
13.85 ⎠
⎝
⇒ k = 2.771 so upper confidence limit for σ is 2.771 i.e. £2771
(OR: CI is (0, 2.771) i.e. (0, £2771)).
9
(i)
F=
2239
= 18.66 on 3,28 d.f.
120
from tables F3,28(1%) = 4.568
∴ P-value < 0.01 or 0 < P-v < 0.01
So there is overwhelming evidence of a difference between the underlying
treatment means.
Page 4
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2009 — Examiners’ Report
(ii)
(a)
1 1
1 1
LSD = t28 (2.5%) σˆ 2 ( + ) = 2.048 120( + ) = 11.2
8 8
8 8
(b)
means in order
y3. < y2. < y1. < y4.
underlined thus:
Treatments 2 & 3 are separate from treatments 1 & 4 which have
significantly higher means.
10
E [ N ] = 0.7(0) + 0.15(1) + 0.1(2) + 0.05(3) = 0.5
E ⎡ N 2 ⎤ = 0.15(1) + 0.1(4) + 0.05(9) = 1.00 ∴Var [ N ] = 1.00 − 0.52 = 0.75
⎣ ⎦
E[X ] =
2
2
= 20, Var [ X ] = 2 = 200
0.1
0.1
Let S = total of the claim amounts.
E [ S ] = E [ N ] E [ X ] = (0.5)(20) = 10 , i.e. £10,000.
V [ S ] = E [ N ]Var [ X ] + Var [ N ][ E ( X )]2 = (0.5)(200) + (0.75)(20) 2 = 400
∴ sd ( S ) = 20 , i.e. £20,000.
n
11
(i)
(a)
Let S = ∑ X i
i =1
{(
)}
M X ( t ) = exp λ et − 1
{(
)}
n
{ (
)}
n
⇒ M S ( t ) = {M X ( t )} = ⎡ exp λ et − 1 ⎤ = exp nλ et − 1
⎥⎦
⎣⎢
⇒ S ~ Poisson(nλ)
(b)
No
One reason is that E[2X1 + 5] = 2λ + 5, which is not equal
to V[2X1 + 5] = 4λ
[Note: another obvious reason is that 2X1 + 5 can only takes values 5,
7, 9, … , not 0, 1, 2, 3,… ]
Page 5
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2009 — Examiners’ Report
(c)
No
One reason is that E ⎣⎡ X ⎦⎤ = λ , which is not equal to V ⎡⎣ X ⎤⎦ = λ / 2
[Note: another obvious reason is that X can take values 0.5, 1.5, 2.5,
… , which a Poisson variable cannot.]
(d)
(ii)
(a)
(b)
⎛ λ⎞
X ≈ N ⎜ λ, ⎟
⎝ n⎠
1 ⎞
⎛
Under H0 , X ~ N ⎜1 ,
⎟ approximately
⎝ 100 ⎠
k −1
k is such that P ( X > k | H 0 ) = 0.01 so
= 2.3263
0.1
⇒ k = 1.2326
Power(λ) = P(reject H0|λ)
Power(λ = 1.2) = P ( X > 1.2326 ) where X ~ N (1.2, 0.012)
= P(Z > 0.298) = 0.383
Power(λ = 1.5) = P ( X > 1.2326 ) where X ~ N (1.5, 0.015)
= P(Z > −2.183) = 0.985
(c)
12
(i)
(a)
Power of test increases as the value of λ increases further away from
λ = 1.
1
1
1
1
L(θ) = [ (2 + θ)]1071[ (1 − θ)]62 [ (1 − θ)]68[ θ]299 (× constant)
4
4
4
4
∝ (2 + θ)1071 (1 − θ)130 θ299
log L(θ) = const + 1071log(2 + θ) + 130 log(1 − θ) + 299 log θ
(b)
d
1071 130 299
log L(θ) =
−
+
dθ
2 + θ 1− θ
θ
=
1071θ(1 − θ) − 130θ(2 + θ) + 299(2 + θ)(1 − θ)
(2 + θ)(1 − θ)θ
numerator = 1071θ − 1071θ2 − 260θ − 130θ2 + 598 − 299θ − 299θ2
equate to zero: 750θ2 − 256θ − 299 = 0
Page 6
1
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2009 — Examiners’ Report
256 ± 2562 − 4(750)(−299)
ˆ
∴θ =
= 0.17067 ± 0.65406
2(750)
So MLE θˆ = 0.82473 (or 0.825 to 3dp) as other root is negative.
(ii)
(a)
d2
dθ
2
log L(θ) = −
1071
(2 + θ)
2
−
130
(1 − θ)
2
−
299
θ2
d2
at θˆ = 0.825 ,
log L(θ) = −134.20 − 4244.90 − 439.30 = −4818.4
d θ2
CRlb =
(b)
1
⎡ d2
⎤
− E ⎢ 2 log L(θ) ⎥
⎢⎣ d θ
⎥⎦
≈
1
= 0.0002075
4818.4
θˆ ≈ N (θ, CRlb) for large samples
and so an approximate 95% CI for θ is θˆ ± 1.96 CRlb
Here: 0.825 ± 1.96 0.0002075 ⇒ 0.825 ± 0.028 or (0.797, 0.853)
(iii)
(a)
With θ = 0.775 the four probabilities are
0.69375, 0.05625, 0.05625, 0.19375 respectively
and the corresponding expected frequencies are
1040.625, 84.375, 84.375, 290.625.
∴χ 2 = ∑
( o − e) 2
= 0.887 + 5.934 + 3.178 + 0.241 = 10.24 on 3 df
e
P-value = P (χ32 > 10.24) = 1 − 0.983 = 0.017
These data do not support the model with the value θ = 0.775 in that
the probability of observing these data when θ = 0.775 is only 0.017.
[OR: could say “do not support at 5% level, but do support at 1%
level”]
(b)
This is consistent with the fact that θ = 0.775 is well outside the
approximate 95% CI.
Page 7
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2009 — Examiners’ Report
(i)
(a)
The plot is given below.
70
60
30
40
50
Final exam score
80
90
100
13
50
60
70
80
90
Midterm score
There seems to be a positive relationship between final and midterm
score. However it is not clear if this relationship is linear.
[Following comment also valid: the relationship looks linear but with
substantial scatter.]
(b)
2
S xy
1 ⎛
⎜ S yy −
σˆ =
n−2⎜
S xx
⎝
2
s.e. (βˆ ) =
⎞ 1⎛
(1529.8)2 ⎞
⎟ = ⎜ 2737.6 −
⎟ = 176.0241
⎟ 8 ⎜⎝
1760.4 ⎟⎠
⎠
σˆ 2
176.0241
=
= 0.3162
1760.4
S xx
To test H 0 : β = 0 v H1 : β > 0 , the test statistic is
0.869
βˆ − 0
=
= 2.748 ,
s.e.(βˆ ) 0.3162
and under the assumption that the errors of the regression are
i.i.d. N (0, σ2 ) random variables, it has a t distribution
with n - 2 = 8 df.
Page 8
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2009 — Examiners’ Report
From statistical tables we find t8,0.05 = 1.860 , and t8,0.01 = 2.896 .
We reject the hypothesis H0: β = 0 in favour of H1: β > 0 at the 5%
level (but not at the 1% level).
(c)
The mean response, yˆ new for xˆnew = 75 is
yˆ new = 3.146 + 0.869 × 75 = 68.321 .
Its standard error is calculated as
1 ( xnew − x ) 2
+
n
S xx
s.e.(yˆ new ) = σˆ
= 13.26741
1 (75 − 74.4) 2
+
= 13.26741× 0.31655
10
1760.4
= 4.1998
The 95% CI is given by yˆ new ± t0.025,8 × s.e.( yˆ new ) ,
i.e. 68.321 ± 2.306 × 4.1998 = 68.321 ± 9.6847
or (58.636, 78.006)
(ii)
(d)
The CI for the individual predicted score will be wider than the CI for
the mean score in (i)(c), because the variance for the individual
predicted value is larger.
(a)
Both statistics follow a tn-2 distribution under the null hypothesis.
In addition
βˆ − 0
σˆ 2
S xx
(b)
r=
Then
S xy
S xx
=
2
⎛
S xy
1
⎜ S yy −
(n − 2) S xx ⎜
S xx
⎝
S xy
S xx S yy
1− r
2
S yy 1 −
S xy
S xx
2
S xy
=
r n−2
1− r2
S xx S yy
1529.8
= 0.6969.
1760.4 × 2737.6
=
r n−2
⎞
⎟
⎟
⎠
=
(n − 2) S xx
=
0.6969 × 8
1 − 0.6969
2
= 2.748 , same as in (i)(b).
END OF EXAMINERS’ REPORT
Page 9
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
9 October 2009 (am)
Subject CT3 — Probability and Mathematical Statistics
Core Technical
Time allowed: Three hours
INSTRUCTIONS TO THE CANDIDATE
1.
Enter all the candidate and examination details as requested on the front of your answer
booklet.
2.
You must not start writing your answers in the booklet until instructed to do so by the
supervisor.
3.
Mark allocations are shown in brackets.
4.
Attempt all 12 questions, beginning your answer to each question on a separate sheet.
5.
Candidates should show calculations where this is appropriate.
Graph paper is not required for this paper.
AT THE END OF THE EXAMINATION
Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this
question paper.
In addition to this paper you should have available the 2002 edition of the Formulae
and Tables and your own electronic calculator from the approved list.
CT3 S2009
© Faculty of Actuaries
© Institute of Actuaries
1
In a sample of 100 households in a specific city, the following distribution of number
of people per household was observed:
Number of people x
Number of households fx
1
7
2
f2
3
20
4
f4
5
18
6
10
7
5
The mean number of people per household was found to be 4.0. However, the
frequencies for two and four members per household (f2 and f4 respectively) are
missing.
2
3
(i)
Calculate the missing frequencies f2 and f4.
[2]
(ii)
Find the median of these data, and hence comment on the symmetry of the
data.
[2]
[Total 4]
Two tickets are selected at random, one after the other and without replacement, from
a group of six tickets, numbered 1, 2, 3, 4, 5, and 6.
(i)
Calculate the probability that the numbers on the selected tickets add up to 8.
[2]
(ii)
Calculate the probability that the numbers on the selected tickets differ by
3 or more.
[2]
[Total 4]
Let X be a random variable with moment generating function MX(t) and cumulant
generating function CX(t), and let Y = aX + b, where a and b are constants. Let Y have
moment generating function MY(t) and cumulant generating function CY(t).
(i)
Show that CY(t) = bt + CX(at).
(ii)
Find the coefficient of skewness of Y in the case that MX(t) = (1 – t)–2 and
Y = 3X + 2 (you may use the fact that CY′″(0) = E[(Y − μY)3]).
[5]
[Total 7]
CT3 S2009—2
[2]
4
Let the random variables (X,Y) have the joint probability density function
f X ,Y ( x, y ) = exp{−( x + y )}, x > 0, y > 0.
(i)
Derive the marginal probability density functions of X and Y, and hence
determine (giving reasons) whether or not the two variables are independent.
[3]
(ii)
Derive the joint cumulative distribution function FX ,Y ( x, y ).
[2]
[Total 5]
5
Let X be a random variable with probability density function given by
f ( x) = 2 xθ−2 ,
0 < x < θ.
Find an unbiased estimator of θ , based on a single observation of X.
6
[4]
A random sample of size n is taken from an exponential distribution with parameter
λ, that is, with probability density function
f ( x) = λe −λx ,
(i)
0< x<∞.
Determine the maximum likelihood estimator (MLE) of λ.
[3]
Claim sizes for certain policies are modelled using an exponential distribution with
parameter λ. A random sample of such claims results in the value of the MLE of λ as
λˆ = 0.00124 .
A large claim is defined as one greater than £4,000 and the claims manager is
particularly interested in p, the probability that a claim is a large claim.
(ii)
Determine p̂ , the MLE of p, explaining why it is the MLE.
CT3 S2009—3
[3]
[Total 6]
PLEASE TURN OVER
7
A scientific investigation involves a linear regression with the usual assumptions that
the response variable y follows a normal distribution with mean α + βx and variance
σ2. Twenty data points were recorded, corresponding to four observations of y at
x = 1, three observations of y at x = 2, six observations of y at x = 3, and seven
observations of y at x = 4. The resulting means of these sets of y observations are
given in the table below.
x
no. of y's
mean of y's
8
1
4
18.6
3
6
23.2
4
7
27.1
(i)
Determine the fitted regression line of y on x.
(ii)
Suppose that you have been asked to provide a 95% confidence interval for
the slope coefficient.
[5]
(a)
Comment briefly on any problems you might encounter in the
computation of the required confidence interval.
(b)
Indicate briefly any further information that you would need in order
to overcome these problems.
[3]
[Total 8]
The table below shows a bivariate probability distribution for two discrete random
variables X and Y:
Y=1
Y=2
Find the value of E[X|Y = 2].
9
2
3
21.7
X=0
0.15
0.05
X=1
0.20
0.15
X=2
0.25
0.20
[3]
In a group of motor insurance policies issued by a company, 80% of claims are made
on comprehensive policies and 20% are made on third-party-only policies.
(i)
Calculate the average amount paid out on a claim, given that the average
amount paid out by the company on a comprehensive policy claim is £1,650,
and the average amount paid out on a third-party-only policy claim is £625.
[1]
(ii)
Calculate the total expected amount paid out in claims by the company in one
year, given that the total number of policies is 150,000 and, on average, the
claim rate is 0.15 claims per policy per year.
[2]
[Total 3]
CT3 S2009—4
10
Consider a population in which a proportion θ of members have some specified
characteristic. Let P denote the corresponding proportion of members in a random
sample of size n from the population.
(i)
Explain clearly why the mean and standard error of P are given by
E [ P ] = θ,
s.e.[ P ] =
θ (1 − θ )
n
.
[3]
An insurance company uses a questionnaire to monitor the satisfaction of its
customers.
In one part customers are asked to answer “yes” or “no” to a particular question.
Suppose that a random sample of 200 responses is examined.
(ii)
Calculate the approximate probability that at least 150 “yes” answers are
found in the sample, on the assumption that the true (population) proportion of
“yes” answers is 0.7.
[4]
Suppose the true (population) proportion of “yes” answers (θ) is unknown, and for a
random sample of 200 responses, the number of “yes” answers is found to be 146.
(iii)
(a)
Calculate an upper (one-sided) 95% confidence interval of the form
(0, L) for θ.
(b)
Calculate a lower (one-sided) 95% confidence interval of the form
(L, 1) for θ.
(c)
A test of the hypotheses:
H0: θ = 0.7 v H1: θ > 0.7
results in a P-value of 0.198.
Comment on how this result relates to the confidence interval in part
(iii)(b).
[9]
[Total 16]
CT3 S2009—5
PLEASE TURN OVER
11
Three insurance company colleagues had just completed an investigation which
involved the application of a two-sample t-test to compare two independent samples,
each of size 11. They were concerned about the validity of the equal variance
assumption required for this test. Their data were as follows.
A:
B:
21
19
22
18
28
38
27
33
20
24
23
39
26
22
32
20
25
28
21
26
30
30
Σx A = 275, Σx 2A = 7, 033, ΣxB = 297, ΣxB2 = 8,559
(i)
One of the colleagues suggested a graphical approach for the comparison of
the variances.
Draw a suitable diagram to represent these data so that the variability of the
samples can be compared, and comment briefly on that comparison.
[3]
(ii)
(iii)
Another colleague suggested using an F-test for the comparison of variances.
(a)
Perform this F-test at the 5% level to compare the variances and
express your conclusion clearly.
(b)
In addition obtain an approximate value of the P-value for this test by
linearly interpolating between suitable entries in the tables.
[6]
The third colleague suggested another procedure using a two-sample t-test in
the following way:
“For each sample calculate the absolute values of the
deviations of the observations from the mean of that sample;
then apply a two-sample t-test to the two sets of absolute
deviations.”
(a)
Discuss the possible reasoning behind this suggested procedure by
considering the potential values of such absolute deviations when the
assumption of equal variances is valid and when it is not valid.
(b)
(1)
Calculate the required sets of absolute deviations for the given
data.
(2)
Perform the suggested two-sample t-test at the 5% level stating
your conclusion clearly.
(3)
Obtain, in addition, an approximate P-value for this test by
linearly interpolating between suitable entries in the tables.
[11]
(iv)
Comment briefly on the conclusions that may have been reached by the three
colleagues.
[2]
[Total 22]
CT3 S2009—6
12
A bank has a free telephone number for its customer services department. Often the
call volume is heavy and customers are placed on hold until a staff member is
available to answer. The bank hopes that a caller remains on hold until the call is
answered, so as not to upset or lose an existing or potential customer.
A survey was conducted to analyse whether callers would remain on hold longer (on
average), if they heard a recorded message containing: (A) an advertisement about the
bank’s products; (B) “easy listening” music; or (C) classical music. The bank
randomly selected a sample of five unanswered calls under each recorded message,
and the length of time (in minutes) that the caller remained on hold before hanging up
is given in the table below.
Recorded message
Time
Total
5 1 11 2 8
A: advertisement
B: easy listening music 0 1 4 6 3
C: classical music
13 9 8 15 7
27
14
52
For these data Σy = 93, Σy2 = 865
Let μ A , μ B , μC denote the mean telephone holding times under recorded message A,
B and C respectively.
(i)
(a)
Perform an analysis of variance to test the hypothesis that the nature of
the recorded message has no effect on the length of time that callers
remain on hold. You should construct an appropriate ANOVA table
and state your conclusion clearly.
(b)
Calculate a 95% confidence interval for μ A − μC , using information
available from all three samples.
[11]
An equivalent approach for analysing the effects of the recorded messages on holding
time is the following:
consider the regression model E[Yi ] = a + b1xi1 + b2 xi 2 , i = 1, 2, …, 15, where
Yi is the telephone holding time and xi1, xi 2 are indicator variables such that
xi1 = 1 if the message for caller i contains an advertisement (and 0 otherwise),
and xi 2 = 1 if the message contains easy listening music (and 0 otherwise).
The results from fitting this model are given below:
Coef.
Intercept
x1
x2
s = 3.406
CT3 S2009—7
10.400
−5.000
−7.600
Std. Error t-value
1.523
2.154
2.154
p-value
6.828 1.8 * 10-5
−2.321
0.039
−3.528
0.004
R-sq = 51.7%
PLEASE TURN OVER
(ii)
Using the fitted model:
(a)
Calculate the predicted value for the telephone holding time when the
message contains classical music.
(b)
Test the hypothesis H0: b1 = 0 against H1: b1 ≠ 0 at the 5% level of
significance.
(c)
Derive an expression relating b1 with μ A and μC , and hence verify
your result from the test in (ii)(b) using the confidence interval
obtained in (i)(b).
[7]
[Total 18]
END OF PAPER
CT3 S2009—8
Faculty of Actuaries
Institute of Actuaries
Subject CT3 — Probability and Mathematical Statistics
Core Technical
September 2009 Examinations
EXAMINERS REPORT
Introduction
The attached subject report has been written by the Principal Examiner with the aim of
helping candidates. The questions and comments are based around Core Reading as the
interpretation of the syllabus to which the examiners are working. They have however given
credit for any alternative approach or interpretation which they consider to be reasonable.
R D Muckart
Chairman of the Board of Examiners
December 2009
Comments for individual questions are given with the solutions that follow.
Faculty of Actuaries
Institute of Actuaries
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2009 — Examiners’ Report
Comments
The paper was answered quite well overall and there are no topics that stand out as being
particularly poorly attempted. Similarly there were no particular misunderstandings widely
evident, and no particular errors were made so repeatedly as to be worthy of comment.
1
(i)
We have 60
These give f 2
f 4 100 and
f2
40
4.
1
f 4 and 2 f 2 4 f 4 148
from which we obtain f 2
(ii)
7 2 f 2 60 4 f 4 90 60 35
100
6 and f 4
34 .
1
Median is equal to the midpoint between the 50th and 51st ordered
observations, i.e. median = 4.
1
We have mean = median, suggesting that the distribution of these data is
roughly symmetric.
1
2
(i)
With sample space {(i,j), i = 1, …, 6, j = 1, …, 6, j
i}
(that is, i is the number on the first ticket selected, j that on the second
selected) there are 30 equally likely outcomes.
Favourable outcomes are (2,6), (3,5), (5,3), (6,2)
so probability = 4/30 = 2/15 = 0.133
(ii)
2
Favourable outcomes are
(1,4), (4,1), (1,5), (5,1), (1,6), (6,1), (2,5), (5,2), (2,6), (6,2), (3,6), (6,3)
so probability = 12/30 = 0.4
2
OR: Use a sample space of size 15: {(i,j)} where i is smaller number selected,
j is larger.
Then event (i) has 2 favourable outcomes and event (ii) has 6.
3
(i)
MY(t) = E[etY] = E[et(aX+b)] = etbE[eatX] = ebtMX(at)
CY(t) = log MY(t) = bt + log MX(at) = bt + CX(at)
(ii)
CY(t) = 2t + log(1 – 3t)–2 = 2t – 2log(1 – 3t)
1
CY (t) = 2 + 6(1 – 3t)-1 , CY (t) = 18(1 – 3t)–2 , CY (t) = 108(1 – 3t)–3
2
E[(Y
Page 2
2
μY)2] = CY (0) = 18, E[(Y
μY)3] = CY (0) = 108
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2009 — Marking Schedule
coefficient of skewness of Y = 108/183/2 = 21/2 = 1.414
2
OR note that coefficient of skewness of Y = coefficient of skewness of X and
just work with X (some candidates may recognise X ~ Gamma(2,1) and
comment on the formula for coefficient of skewness (2/ ) given in the
Yellow Book).
4
(i)
f X ( x)
f ( x, y)dy
0
fY ( y)
e
x y
dy e
x
e
x y
dx e
y
e
y
e
x
0
0
f ( x, y)dx
0
0
x y
Since f X ,Y ( x, y ) e
0
e
x
.
1
e
y
[OR by symmetry].
1
f X ( x) fY ( y ) , X and Y are independent.
1
xy
(ii)
FX ,Y ( x, y)
e
u v
1
dvdu
00
y
x
u
FX ,Y ( x, y)
e v dv
e du
0
e
u x
0
0
e
v y
0
1 e
x
1 e
y
1
5
E[ X ]
0
xf ( x)dx
2
3
Consider Z
2
x3
3
X
2
0
2x2
2
1
dx
2
.
3
0
1
3
E[ X ]
2
E[Z ]
Z is an unbiased estimator of
.
.
2
6
(i)
L( )
Π e
log L( )
xi
n log
n
xi
e
xi
1
d
log L( )
d
and
Equate to zero for the MLE
ˆ
n
Xi
n
xi
1
1
X
(second derivative is clearly negative, so maximum)
1
Page 3
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2009 — Examiners’ Report
(ii)
p = P(X > 4000) = exp(–4000λ) for the exponential distribution
1
pˆ
exp( 4000 ˆ ) using the invariance property of MLE’s
1
pˆ
exp( 4000(0.00124)) 0.0070
1
7
(i)
We need to calculate the basics sums ∑x, ∑x2, ∑y, ∑xy
n = 20
∑x = 4(1) + 3(2) + 6(3) + 7(4) = 56
∑x2 = 4(12) + 3(22) + 6(32) + 7(42) = 182
∑y = 4(18.6) + 3(21.7) + 6(23.2) + 7(27.1) = 468.4
∑xy = 1(4)(18.6) + 2(3)(21.7) + 3(6)(23.2) + 4(7)(27.1) = 1381.0
2
S xy
69.48
25.2
ˆ
ˆ
y
1
(56)(468.4) 69.48 and S xx
20
1381.0
ˆx
182
1
(56)2
20
25.2
1
1
2.757
1
[468.4 (2.757)56] 15.7
20
1
yˆ 15.7 2.757 x
(ii)
(a)
95% CI for β is ˆ t0.025,18
ˆ2
S xx
So we need to calculate
ˆ2
1
n 2
( yi
ˆ
ˆ x )2
i
or
1
n 2
( S yy
2
S xy
S xx
).
1
Problem as we do not have the individual yi values, only means of sets
of them.
(b)
We would need these individual yi values (or the s.d. or
set).
y 2 for each
2
8
X|Y = 2 takes values 0, 1, 2 with probabilities 1/8, 3/8, 4/8
Page 4
(being in the ratios 1:3:4)
2
So E[X|Y = 2] = 1(3/8) + 2(4/8) = 11/8 = 1.375
1
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2009 — Marking Schedule
9
(i)
E(amount) = 0.8(1650) + 0.2(625) = 1,320 + 125 = £1,445
1
(ii)
E(number of claims) = 150000(0.15) = 22,500
1
E(total claim amount) = 22500(1445) = £32,512,500
10
(i)
(ii)
(iii)
Number of sample members with the characteristic X ~ bi(n,) with mean
n and variance n (1 – ). P = X/n.
1
E[P] = n / n =
1
s.e.[P] = {V[P]}1/2 = {V[X]/n2}1/2 = {n (1 – )/n2}1/2 = { (1 – )/n}1/2
1
X ~ bi(200, 0.7) with mean 140 and variance 42
2
P Z
149.5 140
42
P( X
150)
P( Z
1.466) 0.071
(a)
Sample proportion P ~ N( (1 – )/200)
2
Observed P = 146/200 = 0.73
Estimated standard error(P) = (0.73×0.27/200)1/2 = 0.03139
Pr
P
e.s.e. P
Pr
1.645
1
0.95
P 1.645 e.s.e. P
0.95
2
Upper 95% CI is given by (0, 0.73 + 1.645×0.03139)
i.e. (0, 0.782)
(b)
1
By analogy with (i),
Lower 95% CI is given by (0.73 – 1.645×0.03139, 1)
(c)
i.e. (0.678, 1)
2
The P–value indicates that the null hypothesis “ = 0.7” can stand
and we do not have to conclude that > 0.7.
1
The CI in (iii)(b) includes values down to 0.678, so all such values,
including 0.7, are consistent with the data when considering how
low a value of is reasonable.
1
The two results complement each other.
1
11
(i)
Dotplots on same scale are most suitable
[alternatively boxplots or histograms are acceptable]
Page 5
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2009 — Examiners’ Report
2
The spread of the B data appears to be greater than that of the A data and so
casts some doubt on the equal variance assumption.
1
(ii)
(a)
s 2A
1
2752
7033
10
11
15.8
sB2
1
297 2
8559
10
11
54.0
1
3.418 on 10, 10 df
1
F
sB2
s 2A
54.0
15.8
For a two-sided test at the 5% level, critical value is
F10,10(2.5%) = 3.717
So we accept H 0 :
(b)
2
A
1
2
B
at the 5% level.
1
F10,10(2.5%) = 3.717 and F10,10(5%) = 2.978
So P-value is between 0.05 and 0.10
1
By interpolation: P-value is
0.05
(iii)
(a)
3.717 3.418
(0.10 0.05) 0.05 (0.405)(0.05) 0.070
3.717 2.978
If the samples have equal variances, then the absolute deviations will
be similar in size for both samples; if one sample has a larger variance
than the other, then the deviations will be more extreme such that the
absolute deviations will be larger for that sample.
A two-sided two-sample t-test applied to these absolute deviations
will test for a difference in the means of these absolute deviations
and hence for a difference in the variances in the original samples.
(b)
(1)
xA
275
11
2
25
So the deviations for sample A , i.e. d A
4 3 3 2 5 2 1 7 0 4 5
Page 6
1
| xA xA | , are
1
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2009 — Marking Schedule
xB
297
11
27
So the deviations for sample B , i.e. d B
| xB
xB | are
8 9 11 6 3 12 5 7 1 1 3
(2)
1
Calculations:
dA
36,
d A2 158 and
dB
66,
d B2
540
dA
36
11
2
3.273 and sdA
1
362
158
10
11
4.018
dB
66
11
2
6.000 and sdB
1
662
540
10
11
14.400
1
2
sdp
10(4.018) 10(14.400)
20
obs. t =
3.273 6.000
1 1
3.035
11 11
3.035
1
2.107 on 20 df
1
9.209
2.727
1.294
sdp
For the two-sided test at the 5% level, critical value is
t20(2.5%) = 2.086
So we just reject H 0 :
at the 5% level.
(3)
dA
dB
and hence H 0 :
2
A
1
2
B
1
t20(2.5%) = 2.086 and t20(1%) = 2.528
So P-value is between 0.02 and 0.05
1
By interpolation: P-value is
0.02
(iv)
2.528 2.107
(0.03) 0.02 (0.952)(0.03) 0.049
2.528 2.086
1
Tests in (ii) and (iii) give different results at the 5% level, but in fact have
quite similar P-values.
Graphical approach in (i) casts doubt on H0.
So all three are fairly consistent.
2
12
(i)
(a)
2
yA = 27, yB = 14, yC = 52 , y = 93, y = 865
SST = 865 – 932/15 = 288.4
Page 7
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2009 — Examiners’ Report
2
2
2
2
SSB = (27 + 14 + 52 )/5
SSR = 288.4
93 /15 = 149.2
2
149.2 = 139.2
Source of variation
d.f.
SS
MSS
Between
2
149.2
74.6
Residual
12
139.2
11.6
Total
14
288.4
2
(b)
F = 74.6/11.6 = 6.431 on (2,12) degrees of freedom.
1
From yellow tables, F2,12(0.05) = 3.885 and F2,12(0.01) = 6.927.
1
We can reject the hypothesis of “no message effect” at the 5%
significance level, but not at the 1% level. We have some evidence
against the “no message effect” hypothesis and conclude that there
is a message effect.
1
t12(0.025) = 2.179
1
1
95% CI is (5.4 10.4) 2.179 11.6
5
i.e. 5 4.694
(ii)
(a)
or ( 9.694, 0.306)
10.4
2
1
2
The P-value is 0.039. We have evidence to reject the hypothesis that
b1 = 0 at the 5% level of significance.
(c)
0.5
We have yˆi 10.4 5.0 xi1 7.6 xi 2 and for classical music
message we need xi1 xi 2 0 .
This gives yˆi
(b)
1
5
For xi1 1, xi 2
and xi1
b1
xi 2
A
0 we have
0 gives
C
A
1
a b1
a
C
1
1
The 95% CI for A C in (i)(b) can be used for testing H0:
0 and equivalently H0: b1 = 0. The interval does not include
A
C
the value 0, and thus we reject H0 at the 5% level.
2
END OF EXAMINERS’ REPORT
Page 8
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
22 April 2010 (am)
Subject CT3 — Probability and Mathematical Statistics
Core Technical
Time allowed: Three hours
INSTRUCTIONS TO THE CANDIDATE
1.
Enter all the candidate and examination details as requested on the front of your answer
booklet.
2.
You must not start writing your answers in the booklet until instructed to do so by the
supervisor.
3.
Mark allocations are shown in brackets.
4.
Attempt all 12 questions, beginning your answer to each question on a separate sheet.
5.
Candidates should show calculations where this is appropriate.
Graph paper is NOT required for this paper.
AT THE END OF THE EXAMINATION
Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this
question paper.
In addition to this paper you should have available the 2002 edition of the Formulae
and Tables and your own electronic calculator from the approved list.
CT3 A2010
© Faculty of Actuaries
© Institute of Actuaries
1
The mean height of the women in a large population is 1.671m while the mean height
of the men in the population is 1.758m. The mean height of all the members of the
population is 1.712m.
Calculate the percentage of the population who are women.
2
[2]
Consider a group of 10 life insurance policies, seven of which are on male lives and
three of which are on female lives. Three of the 10 policies are chosen at random
(one after the other, without replacement).
Find the probability that the three selected policies are all on male lives.
3
[2]
Let X1 , X 2 , …, X n be a random sample of size n from a population with mean μ and
variance σ2.
Let the sample mean be X and the sample variance be S 2 =
1
{ΣX i2 − nX 2 } .
n −1
σ2
.
You may assume that E ⎣⎡ X ⎦⎤ = μ and V ⎡⎣ X ⎤⎦ =
n
Show that E ⎡ S 2 ⎤ = σ2 .
⎣ ⎦
4
[3]
It is assumed that the numbers of claims arising in one year from motor insurance
policies for young male drivers and young female drivers are distributed as Poisson
random variables with parameters λm and λf respectively.
Independent random samples of 120 policies for young male drivers and 80 policies
for young female drivers were examined and yielded the following mean number of
claims per policy in the last calendar year: xm = 0.24 and x f = 0.15 .
Calculate an approximate 95% confidence interval for λm − λf , the difference between
the respective Poisson parameters.
[3]
5
A computer routine selects one of the integers 1, 2, 3, 4, 5 at random and replicates
the process a total of 100 times. Let S denote the sum of the 100 numbers selected.
Calculate the approximate probability that S assumes a value between 280 and 320
inclusive.
[5]
CT3 A2010—2
6
Let X1 , X 2 ,…, X n be a random sample of claim amounts which are modelled using a
gamma distribution with known parameter α = 4 and unknown parameter λ.
n
(i)
(a)
Specify the distribution of
∑ Xi .
i =1
(b)
Justify the fact that 2nλX has a χ 2k distribution, where X is the mean
of the sample, by using a suitable relationship between the gamma and
the χ 2 distribution, and specify the degrees of freedom k.
[3]
A random sample of five such claim amounts yields a mean of x = 17.5 .
(ii)
7
Use the pivotal method with the χ 2 result from part (i)(b) to obtain a 95%
confidence interval for λ.
[3]
[Total 6]
An employment survey is carried out in order to determine the percentage, p, of
unemployed people in a certain population in a way such that the estimation has a
margin of error less than 0.5% with probability at least 0.95. In a similar study
conducted a year ago it was found that the percentage of unemployed people in the
population was 6%.
Calculate the sample size, n, that is required to achieve this margin of error, by
constructing an appropriate confidence interval (or otherwise).
8
[6]
For a sample of 100 insurance policies the following frequency distribution gives the
number of policies, f, which resulted in x claims during the last year:
x:
f:
0
76
1
22
2
1
3
1
(i)
Calculate the sample mean, standard deviation and coefficient of skewness for
these data on the number of claims per policy.
[4]
A Poisson model has been suggested as appropriate for the number of claims per
policy.
(ii)
(a)
State the value of the estimated parameter when a Poisson distribution
is fitted to these data using the method of maximum likelihood.
(b)
Verify that the coefficient of skewness for the fitted model is 1.92, and
hence comment on the shape of the frequency distribution relative to
that of the corresponding fitted Poisson distribution.
[3]
[Total 7]
CT3 A2010—3
PLEASE TURN OVER
9
The number of claims, N, arising over a period of five years for a particular policy is
assumed to follow a “Type 2” negative binomial distribution (as in the book of
k (1 − p)
Formulae and Tables page 9) with mean E [ N ] =
and variance
p
k (1 − p )
.
V [N ] =
p2
Each claim amount, X (in units of £1,000), is assumed to follow an exponential
distribution with parameter λ independently of each other claim amount and of the
number of claims.
Let S be the total of the claim amounts for the period of five years, in the case
k = 2, p = 0.8 and λ = 2.
(i)
Calculate the mean and the standard deviation of S based on the above
assumptions.
[4]
Now assume that:
N follows a Poisson distribution with parameter μ = 0.5, that is, with the same
mean as N above;
X follows a gamma distribution with parameters α = 2 and λ = 4, that is, with
the same mean as X above.
(ii)
(iii)
10
Calculate the mean and the standard deviation of S based on these
assumptions.
Compare the two sets of answers in (i) and (ii) above.
[3]
[2]
[Total 9]
The size of claims (in units of £1,000) arising from a portfolio of house contents
insurance policies can be modelled using a random variable X with probability density
function (pdf) given by:
ac a
f X ( x) = a +1 , x ≥ c
x
where a > 0 and c > 0 are the parameters of the distribution.
ac
, for a > 1 .
a −1
(i)
Show that the expected value of X is E [ X ] =
(ii)
Verify that the cumulative distribution function of X is given by
[2]
a
⎛c⎞
FX ( x) = 1 − ⎜ ⎟ ,
⎝x⎠
CT3 A2010—4
x≥c
(and = 0 for x < c).
[2]
Suppose that for the distribution of claim sizes X it is known that c = 2.5, but a is
unknown and needs to be estimated given a random sample x1, x2, …, xn.
(iii)
(iv)
Show that the maximum likelihood estimate (MLE) of a is given by:
n
aˆ = n
.
⎛ xi ⎞
∑ log ⎜⎝ 2.5 ⎟⎠
i =1
Derive the asymptotic variance of the MLE â , and hence determine its
approximate asymptotic distribution.
[3]
[4]
Consider a sample of 30 observations from this distribution, for which:
30
∑ log( xi ) = 32.9 .
i =1
(v)
Calculate the MLE â in this case, together with an approximate 95%
confidence interval for a.
[5]
In the current year, claim sizes are assumed to follow the distribution of X with a = 6,
c = 2.5. Inflation for the following year is expected to be 5%.
(vi)
Calculate the probability that the size of a claim arising from this portfolio in
the following year will exceed £4,000.
[3]
[Total 19]
CT3 A2010—5
PLEASE TURN OVER
11
Consider the following three independent random samples from a normally
distributed population with unknown mean μ:
Sample 1:
19.9 20.4 20.3 22.3 16.7 18.7 20.5 19.0 20.1 16.4 21.5 21.4 17.8 22.5 15.2
For these data: n = 15,
∑ xi = 292.7,
∑ xi2 = 5, 778.69
Sample 2:
20.8 25.9 22.1 21.7 16.0 12.1 27.6 16.1 16.8 17.1 21.3 18.6 24.9 14.8 22.2
For these data: n = 15,
∑ xi = 298.0,
sample mean = 19.867,
∑ xi2 = 6,192.32
sample variance = 19.432
Sample 3:
20.6 18.5 21.5 16.9 21.5 21.2 20.9 22.4 14.5 22.0 20.2 17.0 20.3 23.0 19.3
18.9 20.6 20.9 15.3 21.5 16.8 18.5 21.6 16.8 20.4
For these data: n = 25, ∑ xi = 491.1,
sample mean = 19.644,
∑ xi2 = 9, 773.77
sample variance = 5.275
Consider t-tests of the hypotheses H0: μ = 18 v H1: μ > 18.
(i)
(ii)
(a)
Calculate the sample mean and variance for Sample 1.
(b)
Carry out a t-test of the stated hypotheses using the Sample 1 data
(stating the approximate P–value) and show that H0 can be rejected at
the 1% level of testing.
[6]
(a)
Carry out a t-test of the stated hypotheses using the Sample 2 data
(stating the approximate P–value and the conclusion clearly).
(b)
Discuss the comparison of the results with those based on Sample 1
(include reasons for any difference or similarity in the test
conclusions).
[6]
(iii)
(a)
Carry out a t-test of the stated hypotheses using the Sample 3 data
(stating the approximate P–value and your conclusion clearly).
(b)
Discuss the comparison of the results with those based on Sample 1
(include reasons for any difference or similarity in the test
conclusions).
[6]
[Total 18]
CT3 A2010—6
12
As part of a project in a modelling module, a statistics student is required to submit a
report on the sums insured on home contents insurance policies based on samples of
such policies covering risks in five medium-sized towns in each of England, Wales,
and Scotland. Data are provided on the average sum insured (Y, in units of £1,000)
for each of the 15 towns and are as follows:
England
y 11.9 11.1 9.5 9.2
For these data:
∑y
13.9
5.9
Wales
9.1 8.0 5.7
8.1
9.3
Scotland
9.1 7.7 8.2
10.4
= 55.6 (England), 36.8 (Wales), 44.7 (Scotland)
overall
∑ y = 137.1, ∑ y 2
= 1,316.63
The student decides to use an analysis of variance approach.
(i)
Suggest brief comments the student should make on the basis of the plot
below:
Individual and mean value plot of England, Wales, Scotland
14
Average sum insured
13
12
11
10
9
8
7
6
5
England
Wales
Scotland
[2]
(ii)
(a)
(b)
Carry out the analysis of variance on the average sums insured.
Comment on your conclusions.
[6]
The lecturer of the module decides to provide further information. It has been
suggested that the value of a UK index of the town’s prosperity (X) might also be a
useful explanatory variable (in addition to the country in which the town is situated).
The data on the index are as follows (for the towns in the same order as in the first
table):
x
23
27
England
14 19
For these data: overall
CT3 A2010—7
29
15
27
Wales
24 18
22
22
16
Scotland
20 25
28
∑ x = 329 , ∑ x2 = 7,543 , ∑ xy = 3, 091.7
PLEASE TURN OVER
A graph of average sum insured against index (with country identified) is given
below:
12
14
Average sum insured v index
*
* England
+ Scotland
% Wales
*
+
10
y
*
*
+
+
*
8
%
6
+
%
10
%
+
%
%
15
20
25
30
x
The student decides to add the results of a regression approach to her report, using
“index” as an explanatory variable, so she fits the regression model
Y = a + bX + e
using the least squares criterion.
Part of the output from fitting the model using a statistics package on a computer is as
follows:
Coefficients: Estimate
(Intercept) 3.46166
x
0.25889
Residual standard error:
R-Squared: 0.3449
(iii)
Std. Error t value Pr(>|t|)
2.21919
1.560 0.1428
0.09896
2.616 0.0213*
1.789 on 13 degrees of freedom
Verify (by performing your own calculations) the following results for the
fitted model as given in the output above:
(a)
the fitted regression line is y = 3.462 + 0.2589x
(b)
the percentage of the variation in the response (y) explained by the
model (x) is 34.5%
(c)
the standard error of the slope estimate is 0.09896
[8]
(iv)
Comment briefly on the usefulness of “index” as a predictor for the average
sum insured.
[2]
(v)
Suggest another model which you think might be more successful in
explaining the variability in the values of the average sum insured and provide
a better predictor.
[2]
[Total 20]
END OF PAPER
CT3 A2010—8
Faculty of Actuaries
Institute of Actuaries
EXAMINERS’ REPORT
April 2010 Examinations
Subject CT3 — Probability and Mathematical Statistics
Core Technical
Introduction
The attached subject report has been written by the Principal Examiner with the aim of
helping candidates. The questions and comments are based around Core Reading as the
interpretation of the syllabus to which the examiners are working. They have however given
credit for any alternative approach or interpretation which they consider to be reasonable.
R D Muckart
Chairman of the Board of Examiners
July 2010
© Faculty of Actuaries
© Institute of Actuaries
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2010 — Examiners’ Report
The paper was answered well overall and there is only one question that stands out as being
very poorly attempted (and poorly answered by those who did attempt it), namely question 7.
The question is about the precision of estimation - it involves applying the standard error of
estimation of a sample proportion to find the sample size required to ensure a stated margin
of error. The standard error of estimation of a sample proportion is an important and widelyused quantity.
1
Let p be the proportion of women.
Then, using a weighted average, 1.671p + 1.758(1− p) = 1.712
⇒ 0.087p = 0.046 ⇒ p = 0.529 so percentage is 52.9%
2
P(all 3 on male lives) =
7 6 5
× ×
10 9 8
=
7
24
= 0.292
⎛ 7 ⎞ ⎛10 ⎞
[OR ⎜ ⎟ / ⎜ ⎟ = 35 /120 = 7 / 24 ]
⎝ 3⎠ ⎝ 3 ⎠
3
4
1
E ⎡S 2 ⎤ =
{ΣE ( X i2 ) − nE ( X 2 )}
⎣ ⎦
n −1
σ2
1
=
{Σ(σ2 + μ 2 ) − n( + μ 2 )}
n −1
n
1
{n(σ2 + μ 2 ) − σ2 − nμ 2 }
=
n −1
1
{(n − 1)σ2 } = σ2
=
n −1
Approximate 95% CI for λ m − λ f is ( xm − x f ) ± 1.96
⇒ (0.24 − 0.15) ± 1.96
xm x f
+
120 80
0.24 0.15
+
120
80
⇒ 0.09 ± 1.96(0.062) ⇒ 0.09 ± 0.122 or ⇒ (−0.032, 0.212)
Page 2
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2010 — Examiners’ Report
5
S = ΣXi where Xi has a uniform distribution on 1, 2, 3, 4, 5, with mean 3 and variance
(25 – 1)/12 = 2 (result known, or calculated via E[X2] = 11, or from book of formulae,
p10, with a = 1, b = 5, h = 1).
So S ~ N(300, 200) approximately
320.5 − 300 ⎞
⎛ 279.5 − 300
P (280 ≤ S ≤ 320) = P ⎜
<Z<
⎟
200
200 ⎠
⎝
= P ( −1.450 < Z < 1.450 ) = 0.853
6
(i)
(a)
ΣX i ~ gamma(4n, λ)
(b)
If Y ~ gamma(α, λ) and 2α is an integer, then 2λY ~ χ 22α (from book
of formulae, p12)
So 2λnX ~ χ 2 with df 8n.
(ii)
P (χ 240 (97.5) < 10λX < χ 240 (2.5)) = 0.95
⎛ χ 2 (97.5) χ 240 (2.5) ⎞
,
giving the 95% CI as ⎜ 40
⎟⎟
⎜ 10 X
10
X
⎝
⎠
⎛ 24.43
59.34 ⎞
,
Data ⇒ ⎜
⎟ = (0.140, 0.339)
⎝ 10(17.5) 10(17.5) ⎠
Page 3
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2010 — Examiners’ Report
7
The 95% CI for the population percentage p is
giving | p − pˆ |≤ 1.96
pˆ ± 1.96
pˆ (1 − pˆ )
n
pˆ (1 − pˆ )
n
For the margin of error to be less than 0.5% we need to solve
0.005 = 1.96
1.962 pˆ (1 − pˆ )
pˆ (1 − pˆ )
⇒n=
.
n
0.0052
Using the percentage from the previous study as the value for p̂ , i.e. pˆ = 0.06 , we
obtain n = 8,666.6.
So we need a sample of (at least) 8667 people.
⎛ p (1 − p ) ⎞
(OR, solution can be based on pˆ ~ N ⎜ p,
⎟ and
n
⎝
⎠
P ( −0.005 < pˆ − p < 0.005 ) > 0.95 without referring to the CI.)
8
(i)
Σf = 100, Σfx = 27, Σfx 2 = 35
x=
27
= 0.27
100
s2 =
1
27 2
{35 −
} = 0.2799 ∴ s = 0.529
99
100
Third moment about mean is
m3 =
1
{76(0 − 0.27)3 + 22(1 − 0.27)3 + (2 − 0.27)3 + (3 − 0.27)3} = 0.3259
100
[OR: using Σfx3 = 57, m3 =
So coefficient of skewness is
1
{57 − 3(0.27)(35) + 2(100)(0.27)3} ]
100
0.3259
(0.2799)3/2
= 2.20
[OR: can use m2 = 0.2771 in denominator to give 2.23 ]
(ii)
Page 4
(a)
μˆ = x = 0.27
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2010 — Examiners’ Report
(b)
1
= 1.92 (from book of formulae, p7)
0.27
Coefficient of skewness is
so, the data distribution is slightly more positively skewed than the
fitted Poisson.
9
(i)
E[N] =
k (1 − p) 2(0.2)
k (1 − p ) 2(0.2)
=
= 0.5 and V [ N ] =
=
= 0.625
p
0.8
p2
0.82
E[X ] =
1 1
1
1
= = 0.5 and V [ X ] = 2 = 2 = 0.25
λ 2
2
λ
E [ S ] = E [ N ] E [ X ] = 0.5 × 0.5 = 0.25 , i.e. £250
V [ S ] = E [ N ] V [ X ] + V [ N ]{ E [ X ]} = 0.5 × 0.25 + 0.625 × 0.52 = 0.28125
2
∴ SD [ S ] = 0.530 , i.e. £530
(ii)
E [ N ] = V [ N ] = μ = 0.5
E[X ] =
α 2
α
2
= = 0.5 and V [ X ] = 2 = 2 = 0.125
λ 4
λ
4
E [ S ] = E [ N ] E [ X ] = 0.5 × 0.5 = 0.25 , i.e. £250
V [ S ] = E [ N ] V [ X ] + V [ N ]{ E [ X ]} = 0.5 × 0.125 + 0.5 × 0.52 = 0.1875
2
∴ SD [ S ] = 0.433 , i.e. £433
(iii)
As expected the means are the same,
but the standard deviation in (i) is larger than that in (ii) due to the fact that
both N and X have larger variances.
Page 5
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2010 — Examiners’ Report
10
(i)
We have:
∞
∞
c
c
∞
ac a
E [ X ] = ∫ xf X ( x)dx = ∫ x
x
dx = ac a ∫ x − a dx = −
a +1
c
ac a ⎡ −( a −1) ⎤ ∞
x
⎦c
a −1 ⎣
and for a > 1
ac a
ac
E[X ] = −
(0 − c − a +1 ) =
.
a −1
a −1
(ii)
x
x
c
t a +1
c
ac a
FX ( x) = ∫ f X (t )dt = ∫
dt
which gives
a
x
⎛c⎞
FX ( x) = −c ⎡t − a ⎤ = −c a ( x − a − c − a ) = 1 − ⎜ ⎟ ,
⎣ ⎦c
⎝x⎠
a
[OR differentiate FX ( x) to obtain f X ( x ) ]
(iii)
The likelihood function is given by:
n
n
ac a
i =1
i =1
xi a +1
L(a) = ∏ f X ( xi ) =∏
n
= a n c na ∏ xi−( a +1)
i =1
and
n
l (a ) = n log(a ) + na log(c) − (a + 1)∑ log( xi )
i =1
For the MLE:
l ′(a ) = 0 ⇒
⇒ aˆ =
n
n
+ n log(c) − ∑ log( xi ) = 0
a
i =1
n
=
n
∑ log( xi ) − n log(c)
i =1
and for c = 2.5, aˆ =
⎛x ⎞
∑ log ⎜⎝ ci ⎟⎠
i =1
n
n
⎛ x ⎞
∑ log ⎜⎝ 2.5i ⎟⎠
i =1
Page 6
n
n
,
x≥c
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2010 — Examiners’ Report
(iv)
For the asymptotic variance we use the Cramer-Rao lower bound:
l ′′(a ) = −
n
a
2
, and E ⎡⎣l ′′ ( a ) ⎤⎦ = −
n
a2
giving
{
}
V [ aˆ ] = − E ⎡⎣l ′′ ( a ) ⎤⎦
−1
a2
=
.
n
Hence, asymptotically, aˆ ~ N (a, a 2 n) .
(v)
MLE is
aˆ =
n
n
⎛x ⎞
∑ log ⎜⎝ ci ⎟⎠
i =1
=
n
n
∑ log( xi ) − n log(c)
=
30
= 5.544 .
32.9 − 30 × log(2.5)
i =1
Using the asymptotic normal distribution given above, an approximate 95% CI
is given by
aˆ ± 1.96
a2
aˆ
= aˆ ± 1.96
n
n
i.e. 5.544 ± 1.96
(vi)
5.544
, giving (3.560, 7.528).
30
Size of claim in the following year will be given by 1.05X
4 ⎞
⎛
⎛ 4 ⎞
So we want P (1.05 X > 4) = P ⎜ X >
⎟ = 1 − FX ⎜
⎟
1.05 ⎠
⎝
⎝ 1.05 ⎠
and using FX given in the question
6
⎛ 1.05 × 2.5 ⎞
P (1.05 X > 4) = ⎜
⎟ = 0.0799 .
4
⎝
⎠
Page 7
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2010 — Examiners’ Report
11
(i)
(a)
x = 19.513, s 2 =
(b)
Test statistic is
1⎛
292.7 2 ⎞
5778.69
−
⎜
⎟ = 4.7955
14 ⎜⎝
15 ⎟⎠
X −μ
2
S /n
~ tn−1
Here t = (19.513 – 18)/(4.7955/15)1/2 = 2.68
P-value = P(t14 > 2.68), which is just less than 0.01 (1%)
We reject H0 and accept “μ > 18” at the 1% level of testing.
(ii)
(a)
Here t = (19.867 – 18)/(19.432/15)1/2 = 1.64
P-value = P(t14 > 1.64), which is between 0.05 and 0.1.
P-value exceeds 5% and so we cannot reject H0, so “μ = 18” can stand.
(b)
Sample 2 does not provide enough evidence to justify rejecting H0,
despite having the same size and a similar mean to Sample 1.
The reason for the loss of significance is the much greater variation in
the data in Sample 2 – the variance is four times bigger than in
Sample 1 (19.432 v 4.7955)
– this greatly increases the standard error of estimation and reduces the
value of the t-statistic (1.64 v 2.68).
(iii)
(a)
Here t = (19.644 – 18)/(5.275/25)1/2 = 3.58
P-value = P(t24 > 3.58), which is less than 0.001 (0.1%)
We reject H0 and accept “μ > 18” at a level lower than 0.1%.
(b)
Sample 3 provides even stronger evidence against H0, despite having a
similar mean and variance to Sample 1.
The main reason for the much greater level of significance is the
increased sample size (25 v 15)
– this decreases the standard error of estimation and increases the value
of the t-statistic considerably (3.58 v 2.68).
Page 8
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2010 — Examiners’ Report
12
(i)
(ii)
•
the three sets of points are positioned at different levels (the means are
shown), so there is a prima facie case for suggesting that the underlying
means are different (i.e. there are country effects)
•
the means are in the order England (highest), Scotland, Wales (lowest)
•
the variation in the data for Scotland is perhaps lower than that for
England, but with only 5 observations for each country, we cannot be sure
that there is a real underlying difference in variance
(a)
SST = 1316.63 – 137.12/15 = 63.536, SSB = (55.62 + 36.82 + 44.72) /
5 – 137.12/15 = 35.644
∴ SSR = 63.536 – 35.644 = 27.892
Source of variation
Between countries
Residual
Total
Df
2
12
14
SS
MSS
35.644 17.82
27.892 2.324
Under H0: no country effects F = 17.82/2.324 = 7.67 on (2,12) df
P-value of F = 7.67 is less than 0.01, so we reject H0 and conclude that
there are differences among the population means of the average sum
insured
(iii)
(b)
We have strong evidence that country effects exist − the means appear
to be in the order England (highest), Scotland, Wales (lowest).
(a)
Sxx = 7543 – 3292/15 = 326.9333, Syy = 63.536 (from (i)(b) above)
Sxy = 3091.7 – 329×137.1/15 = 84.64
ˆ ( 329 /15 ) = 3.4617
βˆ = 84.64 / 326.9333 = 0.25889 , αˆ = 137.1/15 − β×
So fitted line is y = 3.462 + 0.2589x
(b)
R2 = Sxy2/(SxxSyy) = 84.642/(326.9333×63.536) = 0.34488 so 34.5%
(c)
SSRES = Syy – Sxy2/Sxx = 63.536 – 84.642/326.9333 = 41.62349
⇒ σˆ 2 = 41.62349 /13 = 3.201807
()
1/2
⇒ s.e. βˆ = ( 3.201807 / 326.9333) = 0.09896
Page 9
Subject CT3 (Probability and Mathematical Statistics Core Technical) — April 2010 — Examiners’ Report
(iv)
From the plot we see that the relationship between “index” and “average sum
insured” is weak, positive (and possibly linear) – the percentage of the
variation in “average sum insured” explained by the relationship with “index”
is only 34.5%.
So “index” is of some, but limited, use as a predictor of “average sum
insured”.
(v)
We should try a “multiple regression” model which includes “country” and
“index” in the model.
[Note: although not explicitly in the syllabus, a comment to the effect that
“Country” should be included as a qualitative variable (a “factor”) e.g. by
using a text vector (with entries “E”, “W”, “S” say) or a pair of (Bernoulli)
dummy variables, may attract a bonus for a borderline candidate.]
END OF EXAMINERS’ REPORT
Page 10
Faculty of Actuaries
Institute of Actuaries
EXAMINATION
4 October 2010 (am)
Subject CT3 — Probability and Mathematical Statistics
Core Technical
Time allowed: Three hours
INSTRUCTIONS TO THE CANDIDATE
1.
Enter all the candidate and examination details as requested on the front of your answer
booklet.
2.
You must not start writing your answers in the booklet until instructed to do so by the
supervisor.
3.
Mark allocations are shown in brackets.
4.
Attempt all 12 questions, beginning your answer to each question on a separate sheet.
5.
Candidates should show calculations where this is appropriate.
Graph paper is required for this paper.
AT THE END OF THE EXAMINATION
Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this
question paper.
In addition to this paper you should have available the 2002 edition of the Formulae
and Tables and your own electronic calculator from the approved list.
CT3 S2010
© Faculty of Actuaries
© Institute of Actuaries
1
The marks of a sample of 25 students from a large class in a recent test have sample
mean 57.2 and standard deviation 7.3. The marks are subsequently adjusted: each
mark is multiplied by 1.1 and the result is then increased by 8.
Calculate the sample mean and standard deviation of the adjusted marks.
2
[2]
In a survey, a sample of 10 policies is selected from the records of an insurance
company. The following data give, in ascending order, the time (in days) from the
start date of the policy until a claim has arisen from each of the policies in the sample.
297 301 312 317 355 379 404 419 432+ 463+
Some of the policies have not yet resulted in any claims at the time of the survey, so
the times until they each give rise to a claim are said to be censored. These values are
represented with a plus sign in the above data.
3
4
(i)
Calculate the median of this sample.
[2]
(ii)
State what you can conclude about the mean time until claims arise from the
policies in this sample.
[2]
[Total 4]
Suppose that in a group of insurance policies (which are independent as regards
occurrence of claims), 20% of the policies have incurred claims during the last year.
An auditor is examining the policies in the group one by one in random order until
two policies with claims are found.
(i)
Determine the probability that exactly five policies have to be examined until
two policies with claims are found.
[2]
(ii)
Find the expected number of policies that have to be examined until two
policies with claims are found.
[1]
[Total 3]
For a certain class of business, claim amounts are independent of one another and are
distributed about a mean of μ = £4,000 and with standard deviation σ = £500.
Calculate an approximate value for the probability that the sum of 100 such claim
amounts is less than £407,500.
[3]
5
A random sample of 200 travel insurance policies contains 29 on which the
policyholders made claims in their most recent year of cover.
Calculate a 99% confidence interval for the proportion of policyholders who make
claims in a given year of cover.
[3]
CT3 S2010—2
6
The random variable X has a Poisson distribution with mean Y, where Y itself is
considered to be a random variable. The distribution of Y is lognormal with
parameters μ and σ2.
Derive the unconditional mean E[X] and variance V[X] using appropriate conditional
moments. (You may use any standard results without proof, including results from
the book of Formulae and Tables.)
[4]
7
Let X be a discrete random variable with the following probability distribution:
X:
P(X = x):
(i)
0
0.4
1
0.3
2
0.2
3
0.1
Simulate three observations of X using the following three random numbers
from a uniform distribution on (0,1) (you should explain your method briefly
and clearly).
Random numbers: 0.4936, 0.7269, 0.1652
[3]
Let X be a random variable with cumulative distribution function:
FX ( x ) =
(ii)
8
1
1 − e −1
(1 − e ) ,
− x2
0 < x < 1 (FX(x) = 0 for x ≤ 0 and FX(x) = 1 for x ≥ 1).
Derive an expression for a simulated value of X using a random number u
from a uniform distribution on (0,1) and hence simulate an observation of X
using the random number u = 0.8149.
[3]
[Total 6]
A certain type of claim amount (in units of £1,000) is modelled as an exponential
random variable with parameter λ = 1.25. An analyst is interested in S, the total of 10
such independent claim amounts. In particular he wishes to calculate the probability
that S exceeds £10,000.
(i)
(a)
Show, using moment generating functions, that:
(1)
(2)
(b)
S has a gamma distribution, and
2.5S has a χ 220 distribution.
Use tables to calculate the required probability.
[5]
(ii)
(a)
Specify an approximate normal distribution for S by applying the
central limit theorem, and use this to calculate an approximate value
for the required probability.
(b)
Comment briefly on the use of this approximation and on the result.
[3]
[Total 8]
CT3 S2010—3
PLEASE TURN OVER
9
Let the random variable X have the Poisson distribution with probability function:
f ( x) =
(i)
e−λ λ x
,
x!
x = 0,1, 2,...
Show that P ( X = k + 1) =
λ
P ( X = k ), k = 0,1, 2,...
k +1
[2]
It is believed that the distribution of the number of claims which arise on insurance
policies of a certain class is Poisson. A random sample of 1,000 policies is taken
from all the policies in this class which have been in force throughout the past year.
The table below gives the number of claims per policy in this sample.
No. of claims, k: 0
1
2
3 4 5 6 7 8 or more
No. of policies, fk: 310 365 202 88 26 6 2 1
0
For these data the maximum likelihood estimate (MLE) of the Poisson parameter λ is
λˆ = 1.186.
10
(ii)
Calculate the frequencies expected under the Poisson model with parameter
given by the MLE above, using the recurrence formula of part (i) (or
otherwise).
[3]
(iii)
Perform an appropriate statistical test to investigate the assumption that the
numbers of claims arising from this particular class of policies follow a
Poisson distribution.
[5]
[Total 10]
In the collection of questionnaire data, randomised response sampling is a method
which is used to obtain answers to sensitive questions. For example a company is
interested in estimating the proportion, p, of its employees who falsely take days off
sick. Employees are unlikely to answer a direct question truthfully and so the
company uses the following approach.
Each employee selected in the survey is given a fair six-sided die and asked to throw
it. If it comes up as a 5 or 6, then the employee answers yes or no to the question
“have you falsely taken any days off sick during the last year?”. If it comes up as a 1,
2, 3 or 4, then the employee is instructed to toss a coin and answer yes or no to the
question “did you obtain heads?”. So an individual’s answer is either yes or no, but it
is not known which question the individual has answered.
For the purpose of the following analysis you should assume that each employee
answers the question truthfully.
(i)
Show that the probability that an individual answers yes is
1
( p + 1) .
3
Suppose that 100 employees are surveyed and that this results in 56 yes answers.
CT3 S2010—4
[2]
(ii)
(a)
Show that the likelihood function L(p) can be expressed in the form:
L( p ) ∝ ( p + 1)56 (2 − p ) 44 .
(b)
Hence show that the maximum likelihood estimate (MLE) of p is
pˆ = 0.68 .
[5]
1
56
Let θ = ( p + 1) and note that, using binomial results, the MLE of θ is θˆ =
.
3
100
(iii)
Explain why p̂ can be obtained as the solution of
verify that pˆ = 0.68 .
(iv)
1
( pˆ + 1) = θˆ , and hence
3
[2]
(a)
Determine the second derivative of the log likelihood for p and
evaluate it at pˆ = 0.68 .
(b)
State an approximate large-sample distribution for the MLE p̂ .
(c)
Hence calculate approximate 95% confidence limits for p.
[5]
Now suppose that the same numerical estimate, that is pˆ = 0.68 , had been obtained
from a sample of the same size, that is 100, without using the randomised response
method but relying on truthful answers. So the number of yes answers was 68 and
68
using binomial results.
pˆ =
100
(v)
(a)
Calculate approximate 95% confidence limits for p for this situation.
(b)
Suggest why the confidence limits in part (iv)(c) are wider than these
limits.
[3]
[Total 17]
CT3 S2010—5
PLEASE TURN OVER
11
A life insurance company issuing critical illness insurance wants to compare the delay
times from the date when a claim is made until it is settled, for different causes of
illness covered. Random samples of 12 claims associated with two types of illness (A
and B) related to heart disease have been collected. The logarithms of the delay times
are given below (where the original times were measured in days):
Cause A, yA: 4.0 5.4 4.6 3.5 4.2 4.5 4.2 4.9 5.1 5.2 5.1 5.4
Cause B, yB: 5.7 5.6 4.2 5.1 4.4 5.9 5.4 3.9 5.7 4.5 4.8 3.9
For these data:
∑ y A = 56.1, ∑ y 2A = 266.33,
∑ yB = 59.1, ∑ yB2 = 297.03
(i)
Use a suitable t-test to investigate the hypothesis that the mean delay time is
the same for claims related to the two causes of illness and state clearly your
conclusion.
[6]
(ii)
Give a possible reason why the logarithms of the original delay time
observations are used in this analysis.
(iii)
[2]
(a)
Calculate an equal-tailed 95% confidence interval for σ2A σ2B , the
ratio of the variances of the delay times for the two causes of illness.
(b)
Comment on the validity of the test in part (i) based on this confidence
interval.
[4]
The company collects a third sample of 12 claims associated with an illness (C)
related to brain disease, and the logarithms of the delay times are given below:
Cause C, yc: 5.6 6.2 6.0 5.6 7.1 5.0 4.5 6.4 4.6 6.0 5.5 5.3
For these data:
∑ yC = 67.8, ∑ yC2 = 389.28
For data in all three samples:
∑∑ y = 183.0, ∑∑ y 2 = 952.64
(iv)
Use analysis of variance to test the hypothesis that the mean delay times are
the same for all three causes of illness.
[6]
(v)
State the assumptions made for this analysis of variance.
CT3 S2010—6
[2]
Comment briefly on the validity of the test in (iv), using the plot of the
residuals of the analysis given below.
[2]
B
A
Cause
C
(vi)
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
Residuals
[Total 22]
CT3 S2010—7
PLEASE TURN OVER
12
An investigation concerning the improvement in the average performance of female
track athletes relative to male track athletes was conducted using data from various
international athletics meetings over a period of 16 years in the 1950s and 1960s. For
each year and each selected track distance the observation y was recorded as the
average of the ratios of the twenty best male times to the corresponding twenty best
female times.
The data for the 100 metres event are given below together with some summaries.
year t:
ratio y:
1
0.882
2
0.879
3
0.876
4
0.888
5
0.890
6
0.882
7
0.885
8
0.886
year t:
ratio y:
9
0.885
10
0.887
11
0.882
12
0.893
13
0.878
14
0.889
15
0.888
16
0.890
Σt = 136, Σt 2 = 1496, Σy = 14.160, Σy 2 = 12.531946, Σty = 120.518
(i)
(ii)
Draw a scatterplot of these data and comment briefly on any relationship
between ratio and year.
Verify that the equation of the least squares fitted regression line of ratio on
year is given by:
y = 0.88105 + 0.000465t.
(iii)
[3]
[4]
(a)
Calculate the standard error of the estimated slope coefficient in part
(ii).
(b)
Determine whether the null hypothesis of “no linear relationship”
would be accepted or rejected at the 5% level.
(c)
Calculate a 95% confidence interval for the underlying slope
coefficient for the linear model.
[5]
Corresponding data for the 200 metres event resulted in an estimated slope coefficient
of:
βˆ = 0.000487 with standard error 0.000220.
(iv)
(a)
Determine whether the “no linear relationship” hypothesis would be
accepted or rejected at the 5% level.
(b)
Calculate a 95% confidence interval for the underlying slope
coefficient for the linear model and comment on whether or not the
underlying slope coefficients for the two events, 100m and 200m, can
be regarded as being equal.
(c)
Discuss why the results of the tests in parts (iii)(b) and (iv)(a) seem to
contradict the conclusion in part (iv)(b).
[6]
[Total 18]
END OF PAPER
CT3 S2010—8
INSTITUTE AND FACULTY OF ACTUARIES
EXAMINERS’ REPORT
September 2010 examinations
Subject CT3 — Probability and Mathematical Statistics
Core Technical
Introduction
The attached subject report has been written by the Principal Examiner with the aim of
helping candidates. The questions and comments are based around Core Reading as the
interpretation of the syllabus to which the examiners are working. They have however given
credit for any alternative approach or interpretation which they consider to be reasonable.
T J Birse
Chairman of the Board of Examiners
December 2010
© Institute and Faculty of Actuaries
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2010 — Examiners’ Report
Comments
The paper was answered well and overall performance was satisfactory. However, some
questions were poorly attempted. A number of candidates could not answer Question 1
correctly and efficiently – the question required basic knowledge of data summaries.
Question 8 was not answered very well – answers to questions that require candidates to
“show” a particular statement, need to demonstrate intermediate steps clearly and
accurately. The same applies to Question 10, where deriving specific results regarding
maximum likelihood estimation was not performed accurately by many candidates.
1
Sample mean = (1.1 × 57.2) + 8 = 70.92
Sample standard deviation = 1.1 × 7.3 = 8.03
2
(i)
Sample median is not affected by the fact that the last two observations are
censored.
It is therefore given by the 5.5th ranked observation, i.e. (355 + 379) / 2 = 367
days.
(ii)
We know that the last two observations have minimum values 432 and 463.
Using these two values the sample mean would be equal to 3679/10 = 367.9.
So, the sample mean is at least equal to 367.9 days.
3
(i)
(ii)
4
Using the negative binomial distribution, or from first principles,
⎛ 5 − 1⎞
2
3
P(5 policies required) = ⎜
⎟ (0.2) (0.8) = 0.0819
⎝ 2 − 1⎠
2
= 10
Expected number = mean of negative binomial distribution =
0.2
Working in units of £1,000, sum of 100 claim amounts S has E[S] = 100×4 = 400 and
V[S] = 100 × 0.52 = 25, and so S ~ N(400, 52) approximately.
P(S < 407.5) = P(Z < 1.5) = 0.933
5
Sample proportion = 29/200 = 0.145
99% CI is given by 0.145 ± 2.5758
i.e. (0.081, 0.209).
Page 2
0.145 × 0.855
i.e. 0.145 ± 0.064
200
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2010 — Examiners’ Report
6
E [ X ] = E[ E ( X | Y )] = E [Y ] = eμ+σ
2
/2
V [ X ] = E[V ( X | Y )] + V [ E ( X | Y )] = E [Y ] + V [Y ] = eμ+σ
7
(i)
2
/2
2
2
+ e2μ+σ (eσ − 1)
Method
0 < u ≤ 0.4
0.4 < u ≤ 0.7
0.7 < u ≤ 0.9
0.9 < u ≤ 1
⇒x=0
⇒x=1
⇒x=2
⇒x=3
We get x = 1, 2, 0
(ii)
Setting u =
1
1− e
−1
(1 − e ) ⇒ e
− x2
(
− x2
)
(
)
= 1 − 1 − e −1 u
1/2
⇒ x = ⎡ − log ⎡1 − 1 − e −1 u ⎤ ⎤
⎢⎣
⎣
⎦ ⎥⎦
u = 0.8149 ⇒ x = 0.851
8
(i)
(a)
(1)
Let Xi be a claim amount.
t ⎞
⎛
Mgf of Xi is M X (t ) = ⎜1 −
⎟
⎝ 1.25 ⎠
−1
10
t ⎞
⎛
Mgf of S = ∑ X i is M S (t ) = [ M X (t )]10 = ⎜ 1 −
⎟
⎝ 1.25 ⎠
i =1
−10
,
which is the mgf of a gamma(10, 1.25) variable.
(2)
Mgf of 2.5S is E[et (2.5 S ) ] = E[e(2.5t ) S ] = M S (2.5t ) = (1 − 2t )−10 ,
which is the mgf of a gamma(10, ½) variable, i.e. χ220 .
(ii)
(b)
P (total > £10, 000) = P ( S > 10) = P (χ 220 > 25) = 1 − 0.7986 = 0.2014
(a)
S has mean
10
10
= 8 and variance
= 6.4 . So S ≈ N (8, 6.4)
1.25
1.252
10 − 8
P ( S > 10) ≅ P ( Z >
= 0.791) = 1 − 0.786 = 0.214
6.4
Page 3
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2010 — Examiners’ Report
(b)
9
(i)
(ii)
n is not particularly large for the use of the CLT, but the approximation
is still quite close to the true probability.
P ( X = k + 1) = e −λ
λ k +1
λk λ
λ
= e−λ
=
P( X = k ) .
k ! k +1 k +1
( k + 1)!
Using P ( X = 0) = e
−1.186
7
, P(X = 8 or more) = 1 − ∑ P ( X = i ) , and the
i =0
recurrent formula, we obtain:
K
P(X = k)
Expected, ek
(iii)
0
1
2
3
4
5
6
7
0.3054 0.3623 0.2148 0.0849 0.0252 0.0060 0.0012 0.0002
305.4 362.3 214.8
84.9
25.2
6.0
1.2
0.2
8 or more
4 × 10−5
0.0
Combining the last 4 categories to obtain expected frequencies greater than 5,
we have:
k
No. of policies, fk
Expected, ek
0
310
305.4
1
365
362.3
2
202
214.8
3
88
84.9
4
26
25.2
5 or more
9
7.4
This gives
χ =∑
2
( f k − ek )2
ek
= 0.0693 + 0.0201 + 0.7628 + 0.1132 + 0.0254 + 0.3459 = 1.3367
2
DF = 6 − 1 − 1 = 4, and from statistical tables, χ0.05,4
= 9.488 .
Therefore, we do not have evidence against the hypothesis that the number of
claims comes from a Poisson(1.186) distribution.
(Alternatively if we only combine the last 3 categories, the expected
frequencies for 5 and 6 or more policies are 6 and 1.4, with observed
frequencies 6 and 3 respectively. These give χ 2 = 2.819 on 5 DF, and with
2
χ0.05,5
= 11.071 the conclusion is the same as before.)
Page 4
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2010 — Examiners’ Report
10
(i)
P (yes) = P (5, 6) P (yes | main question) + P (1, 2,3, 4) P (yes | coin question)
=
(ii)
(a)
(b)
2
4 1 1
p + ⋅ = ( p + 1)
6
6 2 3
1
1
L( p) ∝ [ ( p + 1)]56 [1 − ( p + 1)]100−56
3
3
56
44
∝ ( p + 1) (2 − p )
log L = 56 log( p + 1) + 44 log(2 − p ) + const.
d
56
44
log L =
−
dp
p +1 2 − p
=
56(2 − p) − 44( p + 1)
68 − 100 p
=
( p + 1)(2 − p)
( p + 1)(2 − p)
Equate to zero => 68 − 100 p = 0 ∴ pˆ =
(iii)
68
= 0.68
100
1
( pˆ + 1) = θˆ
3
Due to the invariance property of MLEs
1
56
168
68
∴ ( pˆ + 1) =
∴ pˆ =
−1 =
= 0.68
3
100
100
100
(iv)
(a)
d2
dp
2
log L = −
at pˆ = 0.68 ,
(v)
56
( p + 1)
d2
dp
2
2
−
44
(2 − p ) 2
log L = −
56
1.68
2
−
−1
= 0.02218 and
−45.0938
44
1.322
(b)
CRlb =
(c)
Approximate 95% CI for p is
giving 0.68 ± 0.292
(a)
Approximate 95% CI for p is pˆ ± 1.96
= −45.0938
pˆ ≈ N ( p, 0.02218)
pˆ ± 1.96 0.02218
pˆ (1 − pˆ )
100
giving
0.68 ± 1.96
0.68(1 − 0.68)
100
⇒ 0.68 ± 1.96(0.0466) ⇒ 0.68 ± 0.091
Page 5
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2010 — Examiners’ Report
(b)
11
(i)
Less accurate estimation is the penalty paid for using the randomised
response method.
We want to test H0: μ A = μ B against H1: μ A ≠ μ B .
Data give: y A = 56.1/12 = 4.675 , yB = 59.1/12 = 4.925
s 2A = (266.33 − 56.12 /12) /11 = 0.36932 ,
sB2 = (297.03 − 59.12 /12) /11 = 0.54205
Assuming that the two samples come from normal distributions with the same
variance,
11s 2A + 11sB2
2
= 0.455685
we first compute the pooled variance as s p =
22
y − yB
which gives t = A
= −0.907 .
s p 2 /12
Critical values at 5% level are t22(0.025) = −2.074 and t22(0.975) = 2.074
so we don’t have evidence against H0 and conclude that the mean delay time is
the same for claims associated with the two causes of illness.
(ii)
(iii)
Distribution of times can be skewed to the right, and we need a log
transformation to normalise the data (for test to be valid).
(a)
CI is given by
⎛ s 2A / sB2
⎞
, ( s A2 / sB2 ) * F11,11 (0.025) ⎟
⎜⎜
⎟
⎝ F11,11 (0.025)
⎠
F11,11 (0.025) = 3.478 (using interpolation in the tables)
giving CI as (0.68134/3.478, 0.68134*3.478) = (0.196, 2.370).
(b)
(iv)
Page 6
The value “1” is included in the 95% CI, meaning that the assumption
of common variance made for the test is valid.
SST = 952.64 – 1832/36 = 22.39
SSB = (56.12 + 59.12 + 67.82 )/12 − 1832/36 = 6.155
⇒ SSR = 22.39 − 6.155 = 16.235
Source of variation
d.f.
SS
MSS
Between
Residual
Total
2
33
35
6.155
16.235
22.390
3.078
0.492
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2010 — Examiners’ Report
F = 3.078/0.492 = 6.256 on (2,33) degrees of freedom.
From tables, F2,33(0.05) is between 3.276 and 3.295, and F2,33(0.01) is
between 5.289 and 5.336.
We have strong evidence against the null hypothesis and conclude that the
three mean delay times are not equal.
12
(v)
The assumptions are that the data come from normal populations with
constant variance.
(vi)
The plot suggests that the normality assumption is reasonable and that
variance does not depend on cause. Test seems valid.
(i)
Scatterplot with suitable axes and clearly labelled:
There does not appear to be much of a relationship, perhaps a slight increasing
linear relationship but it is weak with quite a bit of scatter.
(ii)
n = 16
Stt = 1496 −
1362
= 340
16
S yy = 12.531946 −
Sty = 120.518 −
14.1602
= 0.000346
16
(136)(14.160)
= 0.158
16
Page 7
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2010 — Examiners’ Report
βˆ =
Sty
Stt
=
0.158
= 0.0004647
340
14.160
136
αˆ = y − βˆ t =
− (0.0004647)
= 0.88105
16
16
Fitted line is y = 0.88105 + 0.000465t
(iii)
(a)
s.e. ( β̂ ) =
σˆ 2 =
σˆ 2
Stt
2
Sty
1
where σˆ =
( S yy −
)
n−2
Stt
2
1
0.1582
(0.000346 −
) = 0.0000195
14
340
0.0000195
∴ s.e.(βˆ ) =
= 0.000239
340
(b)
Null hypothesis of “no linear relationship” is equivalent to H0: β = 0
We use t =
βˆ
~ t14 under H0: β = 0
s.e.(βˆ )
Observed t =
0.000465
= 1.95 and
0.000239
t0.025,14 = 2.145
So we must accept H0: no linear relationship at the 5% level.
(iv)
(c)
95% CI is 0.000465 ± 2.145 × 0.000239
giving 0.000465 ± 0.000513 or (−0.000048, 0.000978)
(a)
Observed t =
0.000487
= 2.21 – this is greater than t0.025,14 = 2.145
0.000220
So we reject H0: no linear relationship at the 5% level.
(b)
95% CI is 0.000487 ± 2.145 × 0.000220
giving 0.000487 ± 0.000472 or (0.000015, 0.000959)
The two CIs overlap substantially, so there is no evidence to suggest
that the slopes are different.
Page 8
Subject CT3 (Probability and Mathematical Statistics Core Technical) — September 2010 — Examiners’ Report
(c)
Although the tests have different conclusions at the 5% level, the 100m
observed t is only just inside the critical value of 2.145 and the 200m
one is just outside. This in fact agrees with, rather than contradicts, the
conclusion that the slopes are not different.
END OF EXAMINERS’ REPORT
Page 9
