CREDIBILITY - PROBLEM SET 3
Bayesian Credibility - Discrete Prior
1. Two bowls each contain 10 similarly shaped balls. Bowl 1 contains 5 red and 5 white balls
(equally likely to be chosen). Bowl 2 contains 2 red and 8 white balls (equally likely to be
chosen). A bowl is chosen at random with each bowl having the chance of being chosen. A ball
is chosen from that bowl. After the ball is chosen it is returned to its bowl and another ball is
chosen at random from the same bowl. Suppose that first ball chosen is red. Find the probability
that second ball chosen is red.
A) "$
B) "(
C) "*
D) #$
E) #*
(!
(!
(!
(!
(!
2. A portfolio of insurance policies consists of two types of policies. Policies of type 1 each have
a Poisson claim number per month with mean 2 per period and policies of type 2 each have a
Poisson claim number with mean 4 per period. #$ of the policies are of type 1 and "$ are of type 2.
A policy is chosen at random from the portfolio and the number of claims generated by that
policy in the following is the random variable \ . Suppose that a policy is chosen at random and
the number of claims is observed to be 1 for that month. The same policy is observed the
following month and the number of claims is \# (assumed to be independent of the first month's
claims for that policy). Find T Ò\# œ "l\" œ "Ó.
A) .15
B) .20
C) .25
D) .30
E) .35
Problems 3 and 4 refer to the distribution of \ , which is a Poisson random variable with
parameter A, where the prior distribution of A is a discrete uniform distribution on the integers
"ß #ß 3 . A single observation of \ is made.
3. Find the joint pf 0\ßA ÐBß -Ñ .
- B
A) / Bx-
- B
B) $/ Bx -
- B
C) /Ð$BÑx
- B
D) / $Bx-
- BÎ$
E) / Bx-
4. Find the mean of the posterior distribution given that \ œ ".
A) 1.5
B) 1.6
C) 1.7
D) 1.8
E) 1.9
5. (Example CR-15) A portfolio of risks is divided into three classes. The characteristics of the
annual claim distributions for the three risk classes is as follows:
Class I
Class II
Class III
Annual Claim
Poisson
Poisson
Poisson
Number Distribution mean 1
mean 2
mean 5
50% of the risks are in Class I, 30% are in Class II, and 20% are in Class III.
(a) A risk is chosen at random from the portfolio and is observed to have 2 claims in the year.
Find the probability that the risk will have 2 claims next year, and find the expected number of
claims for the risk next year.
(b) A risk is chosen at random from the portfolio and is observed to have 2 claims in the first year
and 2 claims in the second year. Find the probability that the risk will have 2 claims in the third
year, and find the expected number of claims for the risk in the third year.
Problems 6 and 7 are based on the following situation. A risk class is made up of three equally
sized groups of individuals. Groups are classified as Type A, Type B and Type C. Any
individual of any type has probability of .5 of having no claim in the coming year and has a
probability of .5 of having exactly 1 claim in the coming year. Each claim is for amount 1 or 2
when a claim occurs. Suppose that the claim distributions given that a claim occurs, for the three
types of individuals are
#Î$ B œ "
T Ðclaim of amount BlType A and a claim occursÑ œ š
ß
"Î$ B œ #
"Î# B œ "
T Ðclaim of amount BlType B and a claim occursÑ œ š
ß
"Î# B œ #
&Î' B œ "
T Ðclaim of amount BlType C and a claim occursÑ œ š
Þ
"Î' B œ #
An insured is chosen at random from the risk class and is found to have a claim of amount 2.
6. Find the probability that the insured is Type A.
A) "'
B) "$
C) "#
D) #$
E) &'
7. Find the Bayesian premium.
A) #&
B) $%
C) #*
D) $"
$'
$'
$'
E) ""
"#
8. You are given the following:
- Four shooters are available to shoot at a target some distance away that has the following
design:
R
S
T
U
V
W
X
Y
Z
- Shooter A hits areas R,S,U,V, each with probability 1/4 .
- Shooter B hits areas S,T,V,W each with probability 1/4 .
- Shooter C hits areas U,V,X,Y each with probability 1/4 .
- Shooter D hits areas V,W,Y,Z each with probability 1/4 .
Two distinct shooters are randomly selected and each fires one shot. Determine the probability
that both shots land in the same Area.
9. You are given the following:
- A portfolio consists of 75 liability risks and 25 property risks.
- The risks have identical claim count distributions.
- Loss sizes for liability risks follow a Pareto distribution with parameters ) œ 300 and α œ 4.
- Loss sizes for property risks follow a Pareto distribution ) œ 1,000 and α œ $.
(a) Determine the variance of the claim size distribution for this portfolio for a single claim.
(b) A risk is randomly selected from the portfolio and a claim of size 5 is observed. Determine
the limit of the posterior probability that this risk is a liability risk as 5 goes to zero.
10. You are given the following random sample of 8 data points from a population distribution
\:
1,2,2,2,2,3,4,8
It is assumed that \ has an exponential distribution with parameter ) , and the prior distribution of
) is discrete with T Ò@ œ "Ó œ Þ#&ß T Ò@ œ #Ó œ Þ&ß T Ò@ œ $Ó œ Þ#&
Find the mean of the posterior distribution.
A) 2.5
B) 2.6
C) 2.7
D) 2.8
E) 2.9
11. (Example CR-16) A portfolio of risks is divided into two classes. The characteristics of the
loss amount distributions for the two risk classes is as follows:
Class I
Class II
Loss Amount
Exponential
Pareto
Distribution
mean 1000
) œ "!!!ß α œ $
The portfolio is evenly divided between Class I and Class II risks.
(a) A risk is chosen at random from the portfolio and is observed to have a loss of 2000. Find the
expected value of the next loss from the same risk.
(b) A risk is chosen at random from the portfolio and is observed to have a first loss of 2000 and a
second loss of 1000. Find the probability that the risk was chosen from Class I. Find the expected
value of the third loss from the same risk.
12. A car manufacturer is testing the ability of safety devices to limit damages in car accidents.
You are given:
(i) A test car has either front air bags or side air bags (not both), each type being equally likely.
(ii) The test car will be driven into either a wall or a lake, with each accident type equally likely.
(iii) The manufacturer randomly selects 1, 2, 3 or 4 crash test dummies to put into a car with
front air bags.
(iv) The manufacturer randomly selects 2 or 4 crash test dummies to put into a car with side air
bags.
(v) Each crash test dummy in a wall-impact accident suffers damage randomly equal to either 0.5
or 1, with damage to each dummy being independent of damage to the others.
(vi) Each crash test dummy in a lake-impact accident suffers damage randomly equal to either 1
or 2, with damage to each dummy being independent of damage to the others.
One test car is selected at random, and a test accident produces total damage of 1. Determine the
expected value of the total damage for the next test accident, given that the kind of safety device
(front or side air bags) and accident type (wall or lake) remain the same.
(A) 2.44
(B) 2.46
(C) 2.52
(D) 2.63
(E) 3.09
13. Suppose that the distribution of \ given R œ 8 is binomial with parameters 8 and ; (; is
non-random), and the prior distribution of R is Poisson with parameter -. Find the posterior
distribution of R based on one observation of \ .
14. You are given:
(i) Two classes of policyholders have the following severity distributions:
Claim Amount
250
2,500
60,000
Probability of Claim
Amount for Class 1
0.5
0.3
0.2
Probability of Claim
Amount for Class 2
0.7
0.2
0.1
(ii) Class 1 has twice as many claims as Class 2.
A claim of 250 is observed. Determine the Bayesian estimate of the expected value of a second
claims from the same policyholder.
(A) Less than 10,200
(B) At least 10,200, but less than 10,400
(C) At least 10,400, but less than 10,600
(D) At least 10,600, but less than 10,800
(E) At least 10,800
15. (CAS) The Allerton Insurance Company insures 3 indistinguishable populations. The claims
frequency of each insured follows a Poisson process. Given:
Population
Expected
Probability
Claim
(class)
time between
of being in
cost
claims
class
I
12 months
1/3
1,000
II
15 months
1/3
1,000
III
18 months
1/3
1,000
Calculate the expected loss in year 2 for an insured that had no claims in year 1.
A) Less than 810
B) At least 810, but less than 910
C) At least 910, but less than 1,010
D) At least 1,010, but less than 1,110
E) At least 1,110
16. A portfolio of insurance policies consists of two types of policies. The annual aggregate loss
distribution for each type of policy is is a compound Poisson distribution. Policies of Type I have
a Poisson parameter of 1 and policies of Type 2 have a Poisson parameter of 2. For both policy
types, the claim size (severity) distribution is uniformly distributed on the integers 1, 2 and 3.
Half of the policies are of Type I and half areof Type II.
A policy is chosen at random and an aggregate annual claim of 2 is observed.
(a) Find the posterior distribution of the Poisson parameter for the two policy types.
(b) Find the mean of the aggregate claim next year for the same policy (given aggregate claim
this year was 2).
(c) Find the variance of the aggregate claim next year for the same policy (given aggregate claim
this year was 2) in two ways:
(i) Z +<ÒW# lW" œ #Ó œ IÒW## lW" œ #Ó  ÐIÒW# lW" œ "ÓÑ#
(ii) Z +<ÒW# lW" œ #Ó œ Z +<Ò IÒW# l-Ó lW" œ #Ó  IÒ Z +<ÒW#l-Ó lW" œ #Ó (this requires
using the posterior distribution of - found in part (a))
17. (SOA) You are given the following information about two classes of risks:
(i) Risks in Class A have a Poisson claim count distribution with a mean of 1.0 per year.
(ii) Risks in Class B have a Poisson claim count distribution with a mean of 3.0 per year.
(iii) Risks in Class A have an exponential severity distribution with a mean of 1.0.
(iv) Risks in Class B have an exponential severity distribution with a mean of 3.0.
(v) Each class has the same number of risks.
(vi) Within each class, severities and claim counts are independent.
A risk is randomly selected and observed to have two claims during one year. The observed claim
amounts were 1.0 and 3.0. Calculate the posterior expected value of the aggregate loss for this
risk during the next year.
(A) Less than 2.0
(B) At least 2.0, but less than 4.0
(C) At least 4.0, but less than 6.0
(D) At least 6.0, but less than 8.0
(E) At least 8.0
18. (SOA) You are given the following for a dental insurer:
(i) Claim counts for individual insureds follow a Poisson distribution.
(ii) Half of the insureds are expected to have 2.0 claims per year.
(iii) The other half of the insureds are expected to have 4.0 claims per year.
A randomly selected insured has made 4 claims in each of the first two policy years.
Determine the Bayesian estimate of this insured’s claim count in the next (third) policy year.
(A) 3.2
(B) 3.4
(C) 3.6
(D) 3.8
(E) 4.0
19. (SOA) Prior to observing any claims, you believed that claim sizes followed a Pareto
distribution
with parameters ) œ 10 and or α œ 1, 2 or 3, with each value being equally likely. You then
observe one claim of 20 for a randomly selected risk. Determine the posterior probability that
the next claim for this risk will be greater than 30.
(A) 0.06
(B) 0.11
(C) 0.15
(D) 0.19
(E) 0.25
20. (SOA) The claim count and claim size distributions for risks of type A are:
Number of Claims Probabilities
Claim Size Probabilities
0
4/9
500
1/3
1
4/9
1235
2/3
2
1/9
The claim count and claim size distributions for risks of type B are:
Number of Claims Probabilities
Claim Size Probabilities
0
1/9
250
2/3
1
4/9
328
1/3
2
4/9
Risks are equally likely to be type A or type B.
Claim counts and claim sizes are independent within each risk type.
The variance of the total losses is 296,962.
A randomly selected risk is observed to have total annual losses of 500.
Determine the Bayesian premium for the next year for this same risk.
(A) 493
(B) 500
(C) 510
(D) 513
(E) 514
21. Two eight-sided dice, A and B, are used to determine the number of claims for an insured.
The faces of each die are marked with either 0 or 1, representing the number of claims for that
insured for the year.
Die T <ÐG6+37= œ !Ñ T <ÐG6+37= œ "Ñ
A
1/4
3/4
B
3/4
1/4
Two spinners, \ and ] , are used to determine claim cost. Spinner ] has two areas marked 12
and - . Spinner ] has only one area marked 12.
Spinner T <ÐG9=> œ "#Ñ T <ÐG9=> œ -Ñ
\
1/2
1/2
]
"
!
To determine the losses for the year, a die is randomly selected from A and B and rolled. If a
claim occurs, a spinner is randomly selected from \ and ] and spun. For subsequent years, the
same die and spinner are used to determine losses. Losses for the first year are 12. Based upon
the results of the first year, you determine that the expected losses for the second year are 10.
Calculate - .
(A) 4
(B) 8
(C) 12
(D) 24
(E) 36
22. A portfolio of insurance policies consists of two types of policies. The annual aggregate loss
distribution for each type of policy is a compound Poisson distribution. Policies of Type I have a
Poisson parameter of 1 and policies of Type 2 have a Poisson parameter of 2. For both policy
types, the claim size (severity) distribution is uniformly distributed on the integers 1, 2 and 3.
Half of the policies are of Type I and half are of Type II. A policy is chosen at random and an
aggregate annual claim of 2 is observed. Find the Bayesian premium for the same policy for next
year.
23. A portfolio of insurance policies consists of three types of policies. The distribution of the
number of losses in one year for each type of policy is summarized as follows:
Policy Type
Type I
Type II
Type III
Annual Number Poisson
Poisson
Poisson
of Losses
With Mean 1
With Mean 2
With Mean 4
Half of the policies are of Type I, one-quarter of the policies are of Type II and one-quarter are Type III.
A policy is chosen at random, and the number of losses in one year is \ .
(a) Find IÒ\Ó .
(b) Find Z +<Ò\Ó each of the following two ways:
(i) Z +<Ò\Ó œ IÒ\ # Ó  ÐIÒ\ÓÑ#
(ii) Z +<Ò\Ó œ Z +<Ò IÒ\lX Ó Ó  IÒ Z +<Ò\lX Ó Ó , where X œ ÖMß MMß MMM× is the random
variable describing the Type of policy chosen.
(c) An observed value of \ is equal to ". Find the posterior probabilities that the policy type is
I, II or III.
(d) Again, an observed value of \ is equal to ". Find the Bayesian premium (the expected
number of claims next year for the same policy). Assume conditional independence of \ 's given policy
type.
(e) Find Z +<Ò\# l\" œ "Ó two ways:
(i) IÒ\## l\" œ "Ó  ÐIÒ\# l\" œ "ÓÑ# , and
(ii) IÒ Z +<Ò\# lTypeÓ l\" œ "Ó  Z +<Ò IÒ\# lTypeÓ l\" œ "Ó .
Again assume conditional independence of \ 's given policy type.
CREDIBILITY - PROBLEM SET 3 SOLUTIONS
T Ò2nd red∩1st redÓ
T Ò1st redÓ
T Ò2nd red∩1st redlbowl 1Ó†T Òbowl 1ÓT Ò2nd red∩1st redlbowl 2Ó†T Òbowl 2Ó
(Î#!
&
&
#
#
Ð "!
ÑÐ "!
ÑÐ "# ÑÐ "!
ÑÐ "!
ÑÐ "# Ñ
#*
œ (! .
(Î#!
1. T Ò2nd redl1st redÓ œ
œ
œ
Note also that
T Ò2nd redl1st redÓ
œ T Ò2nd redlbowl 1Ó † T Òbowl 1l1st redÓ  T Ò2nd redlbowl 2Ó † T Òbowl 2l1st redÓ
&
#
œ Ð "!
ÑÐ &( Ñ  Ð "!
ÑÐ #( Ñ œ #*
Answer: E
(! .
T Ò\# œ"∩\" œ"Ó
T Ò\" œ"Ó
T Ò\# œ"∩\" œ"lType 1Ó†T ÒType 1ÓT Ò\# œ"∩\" œ"lType #Ó†T ÒType #Ó
T Ò\" œ"Ó
2. T Ò\# œ "l\ œ "Ó œ
œ
"
œ
"
Ð/# † #"x Ñ# †Ð $# ÑÐ/% † %"x Ñ# †Ð $" Ñ
"
"
Ð/# † #"x цР$# ÑÐ/% † %"x цР$" Ñ
œ Þ#%( .
Alternatively, T Ò\# œ "l\" œ "Ó
œ T Ò\# œ "lType 1Ó † T ÒType 1l\" œ "Ó  T Ò\# œ "lType #Ó † T ÒType 2l\" œ "Ó
"
"
œ Ð/# † #"x ÑÐÞ))"Ñ  Ð/% † %"x ÑÐÞ"!*Ñ œ Þ#%' .
Answer: C
- B
3. 0\ßA ÐBß -Ñ œ 0\lA ÐBl-Ñ † 1Ð-Ñ œ / Bx- † $" Þ Answer: D
$
4. The marginal pf for \ is 0\ ÐBÑ œ
"
# B
$ B
0\ßA ÐBß -Ñ œ /$Bx  / $Bx#  / $Bx$ ß
-œ"
and the posterior distribution of A has pf
1Al\ Ð-lBÑ œ
1Al\ Ð"l"Ñ œ
1Al\ Ð$l"Ñ œ
Answer: C
0\ßA ÐBß-Ñ
0\ ÐBÑ
œ
/ - - B
$Bx
/" /# #B /$ $B
$Bx  $Bx  $Bx
/"
$
"
#
/
/ # /$ $

$
$  $
/$ $
$
/" /# # /$ $

$
$  $
, and
œ Þ%'( ß 1Al\ Ð2l"Ñ œ
/ 2 2
$
"
/
/# # /$ $

$
$  $
œ Þ$%% ß
œ Þ"*! Þ IÐAl\ œ "Ñ œ ÐÞ%'(Ñ  #ÐÞ$%%Ñ  $ÐÞ"*!Ñ œ "Þ(# Þ
5. (a) T Ò\# œ #l\" œ #Ó œ
T ÒÐ\# œ#Ñ∩Ð\" œ#ÑÓ
T Ò\" œ#Ó
T Ò\" œ #Ó œ T ÒÐ\" œ #Ñ ∩ ÐA œ "ÑÓ  T ÒÐ\" œ #Ñ ∩ ÐA œ #ÑÓ  T ÒÐ\" œ #Ñ ∩ ÐA œ &ÑÓ
œ T Ò\" œ #lA œ "Ó † T ÒA œ "Ó  T Ò\" œ #lA œ #Ó † T ÒA œ #Ó
 T Ò\" œ #lA œ &Ó † T ÒA œ &Ó
"
#
#
#
&
#
œ / #x†" † ÐÞ&Ñ  / #x†# † ÐÞ$Ñ  / #x†& † ÐÞ#Ñ œ Þ"*!! .
T ÒÐ\# œ #Ñ ∩ Ð\" œ #ÑÓ œ T ÒÐ\# œ #Ñ ∩ Ð\" œ #ÑlA œ "Ó † T ÒA œ "Ó
 T ÒÐ\# œ #Ñ ∩ Ð\" œ #ÑlA œ #Ó † T ÒA œ #Ó  T ÒÐ\# œ #Ñ ∩ Ð\" œ #ÑlA œ &Ó † T ÒA œ &Ó
" # #
# # #
& # #
œ ˆ / †" ‰ † ÐÞ&Ñ  ˆ / †# ‰ † ÐÞ$Ñ  ˆ / †& ‰ † ÐÞ#Ñ œ Þ!%!$ .
#x
T Ò\# œ #l\" œ #Ó œ
#x
T ÒÐ\# œ#Ñ∩Ð\" œ#ÑÓ
T Ò\" œ#Ó
#x
Þ!%!$
œ Þ"*!!
œ Þ#" .
IÒ\# l\" œ #Ó œ IÒ\# lA œ "l † T ÒA œ "l\" œ #Ó  IÒ\#lA œ #l † T ÒA œ #l\" œ #Ó
 IÒ\# lA œ &l † T ÒA œ &l\" œ #Ó .
In Example CR-14 we found
T ÒA œ "l\ œ #Ó œ
T Ò\œ#lAœ"Ó†T ÒAœ"Ó
T Ò\œ#Ó
œ
/" †"#
#x †ÐÞ&Ñ
/" †"#
/# †##
/& †&#
#x †ÐÞ&Ñ #x †ÐÞ$Ñ #x †ÐÞ#Ñ
#
#
/ †#
#x †ÐÞ$Ñ
/" †"#
/# †##
/& †&#
†ÐÞ&Ñ
#x
#x †ÐÞ$Ñ #x †ÐÞ#Ñ
œ Þ%)% .
In a similar way, we get T ÒA œ #l\ œ #Ó œ
T ÒA œ &l\ œ #Ó œ
/& †&#
#x †ÐÞ#Ñ
"
#
/ †"
/# †##
/& †&#
#x †ÐÞ&Ñ #x †ÐÞ$Ñ #x †ÐÞ#Ñ
œ Þ%#( , and
œ Þ!)* .
Then, since IÒ\# lA œ "l œ " ß IÒ\# lA œ #l œ # and IÒ\# lA œ &l œ & , we have
IÒ\# l\" œ #Ó œ Ð"ÑÐÞ%)%Ñ  Ð#ÑÐÞ%#(Ñ  Ð&ÑÐÞ!)*Ñ œ "Þ()$ .
(b) We are to find T Ò\$ œ #l\" œ #ß \# œ #Ó and IÒ\$ l\" œ #ß \# œ #Ó .
T Ò\$ œ #l\" œ #ß \# œ #Ó œ
T ÒÐ\$ œ#Ñ∩Ð\# œ#Ñ∩Ð\" œ#ÑÓ
T ÒÐ\# œ#Ñ∩Ð\" œ#ÑÓ
.
In part (a) we found T ÒÐ\# œ #Ñ ∩ Ð\" œ #ÑÓ œ Þ!%!$ .
We find T ÒÐ\$ œ #Ñ ∩ Ð\# œ #Ñ ∩ Ð\" œ #ÑÓ in a similar way.
T ÒÐ\$ œ #Ñ ∩ Ð\# œ #Ñ ∩ Ð\" œ #ÑÓ
" # $
# # $
& # $
œ ˆ / †" ‰ † ÐÞ&Ñ  ˆ / †# ‰ † ÐÞ$Ñ  ˆ / †& ‰ † ÐÞ#Ñ œ Þ!!*") .
#x
#x
#x
Then T Ò\$ œ #l\" œ #ß \# œ #Ó œ Þ!!*")
Þ!%!$ œ Þ##) .
IÒ\$ l\" œ #ß \# œ #Ó œ IÒ\$ lA œ "l † T ÒA œ "l\" œ #ß \# œ #Ó
 IÒ\# lA œ #l † T ÒA œ #l\" œ #ß \# œ #Ó  IÒ\# lA œ &l † T ÒA œ &l\" œ #ß \# œ #Ó
T ÒA œ "l\" œ #ß \# œ #Ó œ
" #
ˆ / †" ‰# †ÐÞ&Ñ
#x
"
#
# #
& #
ˆ / †" ‰# †ÐÞ&Ñˆ / †# ‰# †ÐÞ$Ñˆ / †& ‰# †ÐÞ#Ñ
#x
#x
#x
œ Þ%#! ß
T ÒA œ #l\" œ #ß \# œ #Ó œ Þ&%& ß T ÒA œ &l\" œ #ß \# œ #Ó œ Þ!$& .
Then IÒ\$ l\" œ #ß \# œ #Ó œ Ð"ÑÐÞ%#!Ñ  Ð#ÑÐÞ&%&Ñ  Ð&ÑÐÞ!$&Ñ œ "Þ')& .
T Ò\œ#lAÓ†T ÒEÓ
T Ò\œ#lAÓ†T ÒEÓ
œ T Ò\œ#lEÓ†T ÒEÓT Ò\œ#lFÓ†T ÒFÓT Ò\œ#lGÓ†T ÒGÓ
T Ò\œ#Ó
ÒÐÞ&ÑÐ "$ ÑӆР"$ Ñ
œ "$ .
Answer: B
"
"
ÒÐÞ&ÑÐ $ ÑӆР$ ÑÒÐÞ&ÑÐ "# ÑӆР"$ ÑÒÐÞ&ÑÐ "' ÑӆР"$ Ñ
6. T ÒType Al\ œ #Ó œ
œ
7. IÒ\# l\" œ #Ó œ
IÒ\# lTypeÓ † T ÒTypel\" œ #Ó
Type
œ IÒ\# lAÓ † T ÒAl\" œ #Ó  IÒ\# lBÓ † T ÒBl\" œ #Ó  IÒ\#lCÓ † T ÒCl\" œ #Ó .
As in Problem 6, we find T ÒBl\ œ #Ó œ "# and T ÒCl\ œ #Ó œ "' .
Also, IÒ\# lAÓ œ ÐÞ&ÑÐ!Ñ  ÐÞ&ÑÒ" † #$  # † "$ Ó œ #$ , and similarly,
(
IÒ\# lBÓ œ $% and IÒ\# lCÓ œ "#
Þ
(
Then, IÒ\# l\" œ #Ó œ Ð #$ ÑÐ "$ Ñ  Ð $% ÑÐ "# Ñ  Ð "#
ÑÐ "' Ñ œ #&
$' .
Answer: A
8. No matter which shooter is chosen first, two of the other shooters have 2 target Areas in
common (call this event #G ), and the last has 1 target Area in common (call this event "G ).
Therefore T Ò#GÓ œ #$ ß T Ò"GÓ œ "$ Þ Then
T Òboth shots land in same AreaÓ
œ T Òboth shots land in same Areal#GÓ † T Ò#GÓ  T Òboth shots land in same Areal"GÓ † T Ò"GÓ
"
If the two shooters have two Areas in common, then there is #Ð "'
Ñ œ ") probability that both hit
"
the same Area, and if the two shooters have one Area in common, then there is "'
probability
" #
"
"
&
that both hit the same area. The probability in question is then Ð ) ÑÐ $ Ñ  Ð "' ÑÐ $ Ñ œ %)
.
9. (a) The unconditional claim model is a mixture of two Paretos. The moments will be the mix
of the corresponding Pareto moments, with weights .75 and .25.
Ð$!!Ñ >Ð#Ñ>Ð%"Ñ
Ð"!!!Ñ" >Ð#Ñ>Ð$"Ñ
 ÐÞ#&Ñ
œ (&  "#& œ #!! .
>Ð%Ñ
>Ð$Ñ
#
#
Ð$!!Ñ >Ð$Ñ>Ð%#Ñ
Ð"!!!Ñ >Ð$Ñ>Ð$#Ñ
ÐÞ(&Ñ
 ÐÞ#&Ñ
œ ##ß &!!  #&!ß !!!
>Ð%Ñ
>Ð$Ñ
IÒ] Ó œ ÐÞ(&Ñ
IÒ] # Ó œ
#
Z +<Ò] Ó œ #(#ß &!!  Ð#!!Ñ œ #$#ß &!! .
(b) 0 ÐPl5Ñ œ
0 Ð5lPÑ0 ÐPÑ
0 Ð5lPÑ0 ÐPÑ0 Ð5lT Ñ0 ÐT Ñ
As 5p! the limit is
œ
%
$!! ÐÞ(&Ñ
%
$
$!! ÐÞ(&Ñ "!!! ÐÞ#&Ñ
%Ð$!!Ñ%
†ÐÞ(&Ñ
Ð5$!!Ñ&
%Ð$!!Ñ%
$Ð"!!!Ñ$
†ÐÞ(&Ñ Ð5"!!!Ñ
% †ÐÞ#&Ñ
Ð5$!!Ñ&
Þ
œ Þ*$! .
10. The model distribution is 0 ÐBl)Ñ œ # ") /B3 Î) œ )") /DB3 Î) œ )") /#%Î) .
3œ"
)
The joint distribution of \ and @ is
0\ß@ ÐBß )Ñ œ 0 ÐBl@ œ )Ñ † T Ò@ œ )Ó œ
Þ#& ‚ /DB3
Þ& ‚ #) /DB3 Î#
Þ#& ‚ $) /DB3 Î$
The marginal distribution of \ is
0\ ÐBÑ œ 0\ß@ ÐBß "Ñ  0\ß@ ÐBß #Ñ  0\ß@ ÐBß $Ñ
œ ÐÞ#&Ñ/DB3  ÐÞ&Ñ#) /DB3 Î#  ÐÞ#&Ñ$) /DB3 Î$ .
The posterior distribution of @ is
ÐÞ#&Ñ/DB3
1@l\ Ð"lBÑ œ ÐÞ#&Ñ/DB3 ÐÞ&Ñ#) /DB3Î# ÐÞ#&Ñ$) /DB3Î$ ,
ÐÞ&Ñ#) /DB3 Î#
1@l\ Ð#lBÑ œ ÐÞ#&Ñ/DB3 ÐÞ&Ñ#) /DB3Î# ÐÞ#&Ñ$) /DB3Î$ ,
@œ"
@œ# .
@œ$
œ #(#ß &!! .
10. continued
ÐÞ#&Ñ$) /DB3 Î$
1@l\ Ð$lBÑ œ ÐÞ#&Ñ/DB3 ÐÞ&Ñ#) /DB3Î# ÐÞ#&Ñ$) /DB3Î$ Þ
For the given vector of \ values, this becomes
1@l\ Ð"lBÑ œ Þ!!!$)!( ß 1@l\ Ð#lBÑ œ Þ%)%! ß 1@l\ Ð$lBÑ œ Þ&"&' Þ
The posterior mean is ÐÞ!!!$)!(ÑÐ"Ñ  ÐÞ%)%!ÑÐ#Ñ  ÐÞ&"&'ÑÐ$Ñ œ #Þ&"& .
Answer: A
11.(a) IÒ\# l\" œ #!!!Ó œ IÒ\# lClass M ] † T ÒClass Ml\" œ #!!!Ó
 IÒ\# lG6+== MM ] † T ÒClass MMl\" œ #!!!Ó
)
IÒ\# lClass M ] œ "!!! , and IÒ\# lClass MM ] œ α"
œ &!! .
In Example CR-15 it was found that T ÒClass Ml\" œ #!!!Ó œ Þ()&, and
T ÒClass MMl\" œ #!!!Ó œ Þ#"& . Thus,
IÒ\# l\" œ #!!!Ó œ Ð"!!!ÑÐÞ()&Ñ  Ð&!!ÑÐÞ#"&Ñ œ )*#Þ&!
(b) We wish to find T ÒClass Ml\" œ #!!!, \# œ "!!!Ó . This probability is
0 ÒB" œ#!!!,B# œ"!!!ß @œMÓ
0 ÒB" œ#!!!, B# œ"!!!Ó
. The numerator is
#!!!Î"!!!
"!!!Î"!!!
$
Þ&/
0 ÒB" œ #!!!,B# œ "!!!l@ œ MÓ † T Ò@ œ MÓ œ Ð / "!!! ÑÐ / "!!! ÑÐÞ&Ñ œ "!!!
# .
The denominator is
0 ÒB" œ #!!!,B# œ "!!!l@ œ MÓ † T Ò@ œ MÓ  0 ÒB" œ #!!!,B# œ "!!!l@ œ MMÓ † T Ò@ œ MMÓ
#!!!Î"!!!
Ð / "!!!
"!!!Î"!!!
ÑÐ / "!!!
ÐÞ&ÑÐ*ÑÐ"!!!' Ñ
$†"!!!$
$†"!!!$
Þ&/$
ÑÐÞ&Ñ  Ð#!!!"!!!Ñ
% † Ð"!!!"!!!Ñ% † ÐÞ&Ñ œ "!!!#  Ð$!!!‚#!!!Ñ% Þ
Then T ÒClass Ml\" œ #!!!, \# œ "!!!Ó œ
Þ&/$
"!!!#
Þ&/$ ÐÞ&ÑÐ*ÑÐ"!!!' Ñ

"!!!#
Ð$!!!‚#!!!Ñ%
œ Þ)() , and
We are also asked to find IÒ\$ l\" œ #!!!, \# œ "!!!Ó . This predictive expectation is
IÒ\$ lClass Ml † T ÒClass Ml\" œ #!!!, \# œ "!!!Ó
 IÒ\$ lClass MMl † T ÒClass MMl\" œ #!!!, \# œ "!!!Ó
œ Ð"!!!ÑÐÞ)()Ñ  Ð&!!ÑÐÞ"##Ñ œ *$* .
12. We wish to find IÒ\# l\" œ "Ó . We condition over the combinations of
Front/Side air bags , Wall/Lake accident , so that
IÒ\# l\" œ "Ó œ IÒ\# lW[ Ó † T ÒW[ l\" œ "Ó  IÒ\#lWPÓ † T ÒWPl\" œ "Ó
 IÒ\# lJ [ Ó † T ÒJ [ l\" œ "Ó  IÒ\# lJ PÓ † T ÒJ Pl\" œ "Ó Þ
We immediately get that T ÒWPl\" œ "Ó œ ! since with side air bags there will be 2 or 4
dummies and with a lake crash the minimum damage per dummy is 1, so that the minimum
overall damage with WP is 2. Therefore,
IÒ\# l\" œ "Ó œ IÒ\# lW[ Ó † T ÒW[ l\" œ "Ó
 IÒ\# lJ [ Ó † T ÒJ [ l\" œ "Ó  IÒ\# lJ PÓ † T ÒJ Pl\" œ "Ó Þ
"
Þ&"
We find IÒ\# lW[ Ó œ Ð # ÑÐ#  %ÑÐ # Ñ œ #Þ#& (2 or 4 dummies equally likely, average
Þ&"
damage of # for each dummy) ,
12. continued
"
Þ&"
IÒ\# lJ [ Ó œ Ð % ÑÐ"  #  $  %ÑÐ # Ñ œ "Þ)(& ß and
"
"#
IÒ\# lJ PÓ œ Ð % ÑÐ"  #  $  %ÑÐ # Ñ œ $Þ(& .
T Ò\" œ"lW[ Ó†T ÒW[ Ó
We then calculate T ÒW[ l\" œ "Ó œ
,
T Ò\" œ"Ó
T Ò\" œ"lJ [ Ó†T ÒJ [ Ó
,
T Ò\" œ"Ó
T Ò\" œ"lJ PÓ†T ÒJ PÓ
and T ÒJ Pl\" œ "Ó œ
, where
T Ò\" œ"Ó
T ÒJ [ l\" œ "Ó œ
T ÒW[ Ó œ T ÒWPÓ œ T ÒJ [ Ó œ T ÒJ PÓ œ Þ#& , and
T Ò\" œ "lW[ Ó œ ÐÞ&ÑÐÞ&Ñ# œ Þ"#& since there must be 2 dummies (prob. .5) and each sustains
damage of .5 (prob. .5 each),
T Ò\" œ "lJ [ Ó œ ÐÞ#&ÑÐÞ&Ñ  ÐÞ#&ÑÐÞ&Ñ# œ Þ")(& (either 1 dummy and damage of 1, or two
dummies and damage of .5 each), and
T Ò\" œ "lJ PÓ œ ÐÞ#&ÑÐÞ&Ñ œ Þ"#& .
Then T Ò\" œ "Ó œ T Ò\" œ "lW[ Ó † T ÒW[ Ó  T Ò\" œ "lJ [ Ó † T ÒJ [ Ó
 T Ò\" œ "lJ PÓ † T ÒJ PÓ œ ÐÞ"#&ÑÐÞ#&Ñ  ÐÞ")(&ÑÐÞ#&Ñ  ÐÞ"#&ÑÐÞ#&Ñ œ Þ"!*$(& .
ÐÞ"#&ÑÐÞ#&Ñ
ÐÞ")(&ÑÐÞ#&Ñ
Þ"!*$(& œ Þ#)&( ß T ÒJ [ l\" œ "Ó œ Þ"!*$(& œ Þ%#)' ß
ÐÞ"#&ÑÐÞ#&Ñ
and T ÒJ PÒ\" œ "Ó œ Þ"!*$(& œ Þ#)&( .
Then T ÒW[ l\" œ "Ó œ
We can summarize these conditional probability calculations as follows.
WP ß Þ#&
W[ ß Þ#&
J P ß Þ#&
J [ ß Þ#&
Ì
Ì
Ì
Ì
T Ò\" œ "lWPÓ
T Ò\" œ "lW[ Ó
T Ò\" œ "lJ PÓ
T Ò\" œ "lJ [ Ó
œ!
œ ÐÞ&ÑÐÞ&Ñ# œ Þ"#&
œ ÐÞ#&ÑÐÞ&Ñ œ Þ"#&
œ ÐÞ#&ÑÐÞ&  Þ&# Ñ œ Þ")(&
Ì
Ì
Ì
Ì
T Ò\" œ " ∩ WPÓ
T Ò\" œ " ∩ W[ Ó
T Ò\" œ " ∩ J PÓ
T Ò\" œ " ∩ J [ Ó
œ Ð!ÑÐÞ#&Ñ
œ ÐÞ"#&ÑÐÞ#&Ñ
œ ÐÞ"#&ÑÐÞ&Ñ
œ ÐÞ")(&ÑÐÞ#&Ñ
Ì
T Ò\" œ "Ó œ Ð!ÑÐÞ#&Ñ  ÐÞ"#&ÑÐÞ#&Ñ  ÐÞ"#&ÑÐÞ&Ñ  ÐÞ")(&ÑÐÞ#&Ñ œ ÐÞ#&ÑÐÞ%$(&Ñ
Ì
Ì
Ì
Ì
T ÒWPl\" œ "Ó
T ÒW[ l\" œ "Ó
T ÒJ Pl\" œ "Ó
T ÒJ [ l\" œ "Ó
!
ÐÞ"#&ÑÐÞ#&Ñ
ÐÞ"#&ÑÐÞ#&Ñ
ÐÞ")(&ÑÐÞ#&Ñ
œ ÐÞ#&ÑÐÞ%$(&Ñ
œ ÐÞ#&ÑÐÞ%$(&Ñ
œ ÐÞ#&ÑÐÞ%$(&Ñ
œ ÐÞ#&ÑÐÞ%$(&Ñ
#
#
$
œ!
œ (
œ (
œ (
Finally, IÒ\# l\" œ "Ó œ Ð#Þ#&ÑÐÞ#)&(Ñ  Ð"Þ)(&ÑÐÞ%#)'Ñ  Ð$Þ(&ÑÐÞ#)&(Ñ œ #Þ&# .
Answer: C
13. The model distribution is 0\lR ÐBl8Ñ œ ˆ 8B ‰; B Ð"  ;Ñ8B œ
8x; B Ð";Ñ8B
BxÐ8BÑx
,
- 8
and the prior distribution of R is Poisson with 1Ð8Ñ œ / 8x- , so that the joint distribution of \
- 8
8x; B Ð";Ñ8B
/- Ð; -ÑB ÒÐ";Ñ-Ó8B
and R is 0\ßR ÐBß 8Ñ œ BxÐ8BÑx † / 8x- œ
† Ð8BÑx .
Bx
The marginal distribution of \ is 0\ ÐBÑ œ
/- Ð; -ÑB
Bx
∞
†
8œB
ÒÐ";Ñ-Ó8B
Ð8BÑx
(the summation
starts at 8 œ B because as a binomial distribution with parameter 8, it must be the case that
! Ÿ B Ÿ 8). Applying the change of variable 5 œ 8  B to the summation results in
0\ ÐBÑ œ
/- Ð; -ÑB
Bx
∞
†
5œ!
ÒÐ";Ñ-Ó5
5x
œ
/- Ð; -ÑB
Bx
† /Ð";Ñ- œ
/; - Ð; -ÑB
Bx
.
The marginal distribution of \ is Poisson with parameter ;-.
The posterior distribution of R has probability function
1R l\ Ð8lBÑ œ
0\ßR ÐBß8Ñ
0\ ÐBÑ
œ’
/- Ð; -ÑB
Bx
†
ÒÐ";Ñ-Ó8B
/; - Ð; -ÑB
Î
“
Ð8BÑx
Bx
œ
/Ð";Ñ- ÒÐ";Ñ-Ó8B
Ð8BÑx
for 8 B (this is a translated Poisson; a Poisson with parameter Ð"  ;Ñ- translated by B).
14. IÒ\# l\" œ #&!Ó œ IÒ\# lClass 1Ó † T ÒClass 1l\" œ #&!Ó
 IÒ\# lClass 2Ó † T ÒClass 2l\" œ #&!Ó .
IÒ\# lClass 1Ó œ Ð#&!ÑÐÞ&Ñ  Ð#ß &!!ÑÐÞ$Ñ  Ð'!ß !!!ÑÐÞ#Ñ œ "#ß )(& ,
IÒ\# lClass 2Ó œ Ð#&!ÑÐÞ(Ñ  Ð#ß &!!ÑÐÞ#Ñ  Ð'!ß !!!ÑÐÞ"Ñ œ 'ß '(& .
#
"
Given
T ÒClass 1Ó œ $
T ÒClass 2Ó œ $
Given
T Ò\" œ #&!lClass 1Ó œ Þ&
T Ò\" œ #&!lClass 2Ó œ Þ(
Ì
Ì
T ÒÐ\" œ #&!Ñ ∩ Class 1Ó
T ÒÐ\" œ #&!Ñ ∩ Class 2Ó
œ T Ò\" œ #&!lClass 1Ó † T ÒClass 1Ó
œ T Ò\" œ #&!lClass 2Ó † T ÒClass 2Ó
#
"
"
(
œ ÐÞ&ÑÐ $ Ñ œ $
œ ÐÞ(ÑÐ $ Ñ œ $!
Ì
T Ò\" œ #&!Ó œ T ÒÐ\" œ #&!Ñ ∩ Class 1Ó  T ÒÐ\" œ #&!Ñ ∩ Class 2Ó
"
(
"(
œ $  $! œ $!
Ì
T ÒÐ\" œ#&!Ñ∩Class 1Ó
"Î$
"!
œ "(Î$! œ "( ,
T Ò\" œ#&!Ó
"!
(
T ÒClass 2l\" œ #&!Ó œ "  "( œ "( .
"!
(
Then IÒ\# l\" œ #&!Ó œ Ð"#ß )(&ÑÐ "( Ñ  Ð'ß '(&ÑÐ "( Ñ œ "!ß $## .
Answer: B
T ÒClass 1l\" œ #&!Ó œ
15. For each Class the claims follow a Poisson process. For a particular Class, if the expected
amount of time between claims is >! (in years), then the expected number of claims per year is >" .
!
Class I has an expected amount of time of 1 year between claims, so that the expected number of
claims per year for Class 1 is -" œ " . Class II has an expected amount of time of 1.25 years (15
months) between claims, so that the expected number of claims per year for Class 1 is
"
-# œ "Þ#&
œ Þ) . Class III has an expected amount of time of 1.5 years (15 months) between
"
claims, so that the expected number of claims per year for Class 1 is -$ œ "Þ&
œ $# .
Let us denote the number of claims in year 1 by R" and the number of claims in year 2 is denoted
by R# . We wish to find IÒR# lR" œ !Ó , and then, since each claim cost is 1,000 in all classes,
the expected loss in year 2 is "!!!IÒR# lR" œ !Ó .
We find IÒR# lR" œ !Ó by conditioning over the Class type.
IÒR# lR" œ !Ó œ IÒR# lClass IÓ † T ÒClass IlR" œ !Ó
 IÒR# lClass IIÓ † T ÒClass IIlR" œ !Ó  IÒR# lClass IIIÓ † T ÒClass IIIlR" œ !Ó .
The conditional expectations are the expected number of claims in a year for a given class:
IÒR# lClass IÓ œ " , IÒR# lClass IIÓ œ Þ) and IÒR# lClass IIIÓ œ #$ .
The conditional probabilities can be found from the following probability table:
Class I , T Ò I Ó œ "$
Class II , T Ò II Ó œ "$
" !
R" œ ! T ÒR" œ !lIÓ œ / !x†"
œ Þ$'())
Þ)
!
/ $ †Ð $ Ñ!
T ÒR" œ !lIIÓ œ / !x†Þ) T ÒR" œ !lIIIÓ œ
!x
œ Þ%%*$$
œ Þ&"$%#
T ÒR" œ ! ∩ IÓ
œ T ÒR" œ !lIÓ † T Ò I Ó
œ ÐÞ$'())ÑÐ "$ Ñ
Class III , T Ò III Ó œ "$
#
#
T ÒR" œ ! ∩ IIÓ
T ÒR" œ ! ∩ IIIÓ
œ T ÒR" œ !lIIÓ † T Ò II Ó
œ T ÒR" œ !lIIIÓ † T Ò III Ó
œ ÐÞ%%*$$ÑÐ "$ Ñ
œ ÐÞ&"%$#ÑÐ "$ Ñ
Then, T ÒR" œ !Ó œ T ÒR" œ ! ∩ IÓ  T ÒR" œ ! ∩ IIÓ  T ÒR" œ ! ∩ IIIÓ œ Þ%%$)% , and
ÐÞ$'())ÑÐ "$ Ñ
T ÒClass I ∩ R" œ!Ó
œ
œ Þ#('$ .
T ÒR" œ!Ó
Þ%%$)%
"
ÐÞ%%*$$ÑÐ Ñ
T ÒClass II ∩ R" œ!Ó
IIlR# œ !Ó œ
œ Þ%%$)% $ œ Þ$$(& .
T ÒR" œ!Ó
ÐÞ&"%$#ÑÐ " Ñ
T ÒClass III ∩ R" œ!Ó
IIIlR# œ !Ó œ
œ Þ%%$)% $ œ Þ$)'$
T ÒR" œ!Ó
T ÒClass IlR" œ !Ó œ
T ÒClass
T ÒClass
.
Then IÒR# lR" œ !Ó œ Ð"ÑÐÞ#('$Ñ  ÐÞ)ÑÐÞ$$(&Ñ  Ð #$ ÑÐÞ$)'$Ñ œ Þ)!% , and the expected loss
for year 2 is "!!!ÐÞ)!%Ñ œ )!% .
Answer: A
16. The prior parameter - has distribution - œ œ
"
prob. "#
.
2 prob. "#
The model distribution W has a compound distribution with Poisson frequency with mean -, and
the stated severity distribution.
(a) T Ð- œ "lW" œ #Ñ œ
T ÐW" œ#∩-œ"Ñ
T ÐW" œ#Ñ
T ÐW" œ #l- œ "Ñ œ T Ð1 claim for amount 2l- œ "Ñ  T Ð2 claims for amount 1 eachl- œ "Ñ
"
"
œ /" † "$  /# † "$ † "$ œ (/") .
16. continued
"
"
T ÐW" œ # ∩ - œ "Ñ œ T ÐW" œ #l- œ "Ñ † T Ð- œ "Ñ œ (/") † "# œ (/$' Þ
T ÐW" œ #l- œ #Ñ œ T Ð1 claim for amount 2l- œ #Ñ  T Ð2 claims for amount 1 eachl- œ #Ñ
# #
#
œ /# † # † "$  / #†# † "$ † "$ œ )/* .
#
#
T ÐW" œ # ∩ - œ #Ñ œ T ÐW" œ #l- œ #Ñ † T Ð- œ #Ñ œ )/* † "# œ %/* Þ
"
#
T ÐW" œ #Ñ œ T ÐW" œ # ∩ - œ "Ñ  T ÐW" œ # ∩ - œ #Ñ œ (/$'  %/* .
T Ð- œ "lW" œ #Ñ œ
T ÐW" œ#∩-œ"Ñ
T ÐW" œ#Ñ
œ Ð (/$' Ñ‚Ð (/$' 
"
"
%/#
* Ñ
œ Þ&%$#
and T Ð- œ #lW" œ #Ñ œ "  T Ð- œ "lW" œ #Ñ œ Þ%&') .
(b) IÒW# lW" œ #Ó œ IÒW# l- œ "Ó † T Ð- œ "lW" œ #Ñ  IÒW#l- œ #Ó † T Ð- œ #lW" œ #Ñ
œ Ð"ÑÐ#ÑÐÞ&%$#Ñ  Ð#ÑÐ#ÑÐÞ%&')Ñ œ #Þ*"$' .
(c)(i) IÒW# lW" œ #Ó œ #Þ*"$' (from part (b) ).
IÒW## lW" œ #Ó œ IÒW## l- œ "Ó † T Ð- œ "lW" œ #Ñ  IÒW##l- œ #Ó † T Ð- œ #lW" œ #Ñ Þ
IÒW## l- œ "Ó œ Z +<ÒW# l- œ "Ó  ÐIÒW# l- œ "ÓÑ# .
Since W# l- œ " has a compound Poisson distribution with - œ ", the mean is IÒW# l- œ "Ó œ 2,
and the variance is Z +<ÒW# l- œ "Ó œ - † IÒ\ # Ó œ IÒ\ # Ó , where \ is the severity distribution.
From the distribution of \ we have IÒ\ # Ó œ "%
$ , and then
#'
#
IÒW## l- œ "Ó œ Ð"ÑÐ "%
$ Ñ  Ð#Ñ œ $ .
Similarly,
('
#
IÒW## l- œ #Ó œ Z +<ÒW# l- œ #Ó  ÐIÒW# l- œ #ÓÑ# œ Ð#ÑÐ "%
$ Ñ  Ð%Ñ œ $ .
From part (a), we know the posterior distribution of - given W" œ #, so that
IÒW## lW" œ #Ó œ IÒW## l- œ "Ó † T Ð- œ "lW" œ #Ñ  IÒW##l- œ #Ó † T Ð- œ #lW" œ #Ñ
('
œ Ð #'
$ ÑÐÞ&%$#Ñ  Ð $ ÑÐÞ%&')Ñ œ "'Þ#(*' .
Then Z +<ÒW# lW" œ #Ó œ IÒW## lW" œ #Ó  ÐIÒW# lW" œ #Ñ# œ "'Þ#(*'  Ð#Þ*"$'Ñ# œ (Þ(* .
(ii) To find Z +<Ò IÒW# l-Ó lW" œ #Ó , we first note that IÒW# l-Ó œ - † IÒ\Ó œ #- .
Then, Z +<Ò IÒW# l-Ó lW" œ #Ó œ Z +<Ò#-lW" œ #Ó œ %Z +<Ò-lW" œ #Ó .
We found the posterior in part (a),
T Ð- œ "lW" œ #Ñ œ Þ&%$# and T Ð- œ #lW" œ #Ñ œ Þ%&') .
Then, Z +<Ò-lW" œ #Ó œ Ð"# ÑÐÞ&%$#Ñ  Ð## ÑÐÞ%&')Ñ  ÒÐ"ÑÐÞ&%$#Ñ  Ð#ÑÐÞ%&')ÑÓ# œ Þ#%)" ,
and Z +<ÒW# lW" œ #Ó œ %ÐÞ#%)"Ñ œ Þ**#% .
To find IÒ Z +<ÒW# l-Ó lW" œ #Ó, we note that Z +<ÒW# l-Ó œ - † IÒ\ # Ó œ "%$- .
Then, IÒ Z +<ÒW# l-Ó lW" œ #Ó œ IÒ "%$- lW" œ #Ó œ "%
$ † IÒ-lW" œ #Ó Þ
From the posterior distribution of - found in part (a), we have
IÒ-lW" œ #Ó œ Ð"ÑÐÞ&%$#Ñ  Ð#ÑÐÞ%&')Ñ œ "Þ%&') , so that
IÒ Z +<ÒW# l-Ó lW" œ #Ó œ "%
$ † Ð"Þ%&')Ñ œ 'Þ(*)% .
Then,
Z +<ÒW# lW" œ #Ó œ Z +<Ò IÒW# l-Ó lW" œ #Ó  IÒ Z +<ÒW#l-Ó lW" œ #Ó œ Þ**#%  'Þ(*)% œ (Þ(*
17. Suppose we use the following notation:
W œ aggregate claims in second year for the selected risk
R œ number of claims in first year for the selected risk
\" œ amount of first claim in first year for the selected risk
\# œ amount of second claim in first year for the selected risk
We wish to find IÒWlR œ # ß \" œ " ß \# œ $Ó .
This can be found by conditioning over the risk class.
IÒWlR œ # ß \" œ " ß \# œ $Ó
œ IÒWlselected risk is from Class AÓ † T ÒClass AlR œ # ß \" œ " ß \# œ $Ó
 IÒWlselected risk is from Class BÓ † T ÒClass BlR œ # ß \" œ " ß \# œ $Ó .
If the selected risk is from Class A, then W has a compound Poisson distribution with Poisson
parameter (frequency) 1.0 and has a claim amount distribution (severity) with a mean of 1.0 (we
are also told that the claim amount has an exponential distribution). It follows that
IÒWlClass AÓ œ (Poisson parameter)(expected claim amount) œ " .
In a similar way, we get IÒWlClass BÓ œ Ð$Þ!ÑÐ$Þ!Ñ œ * .
Once the conditioning relationship is set up, most of the work in this problem is in finding the
conditional probabilities T ÒClass AlR œ # ß \" œ " ß \# œ $Ó
and T ÒClass BlR œ # ß \" œ " ß \# œ $Ó .
This requires the use of rules for conditional probability.
T ÒClass AlR œ # ß \" œ " ß \# œ $Ó œ
T ÒClass A∩ÐR œ# ß \" œ" ß \# œ$ÑÓ
T ÒR œ# ß \" œ" ß \# œ$Ó
and
T ÒClass A ∩ ÐR œ # ß \" œ " ß \# œ $ÑÓ œ T ÒR œ # ß \" œ " ß \# œ $lAÓ † T ÒAÓ .
If it is known that the risk is from Class A then we know the distributions of R and \ , so we can
calculate T ÒR œ # ß \" œ " ß \# œ $lAÓ . We are told that a risk is selected randomly. This
means that risk Classes A and B are equally likely to be chosen, so that T ÒAÓ œ "# . Then
T ÒClass A ∩ ÐR œ # ß \" œ " ß \# œ $ÑÓ œ T ÒR œ # ß \" œ " ß \# œ $lAÓ † T ÒAÓ
From independence of R and the \ 's, it follows that
T ÒR œ # ß \" œ " ß \# œ $lAÓ œ T ÒR œ #lAÓ † T ÒTwo Claim amounts are 1, 3lAÓ .
Since claim amount \ has a continuous (exponential distribution), we use the density
0 Ð\" œ "lEÑ instead of probability T Ò\" œ "lAÓ (and the same goes for \# ).
From the table of distributions made available with Exam C, we know the probability function for
5
the Poisson distribution with mean - is T ÒR œ 5Ó œ /- † -5x , and the density function for the
exponential distribution with mean ) is 0 ÐBÑ œ ") /BÎ) .
Then T ÒR œ # ß \" œ " ß \# œ $lAÓ
#
&
œ T ÒR œ #lAÓ † T ÒTwo Claim amounts are 1, 3lAÓ œ Ð/" † "#x Ñ ‚ #Ð" † /" ÑÐ" † /$ Ñ œ # † /# .
The factor of 2 arises from the two combinations of claims of amounts 1 and 3 (first claim is
amount 1 and second claims is amount 3, and first claim is amount 3 and second claim is
amount 1).
&
&
It follows that T ÒClass A ∩ ÐR œ # ß \" œ " ß \# œ $ÑÓ œ #Ð /# ÑÐ "# Ñ œ # † /%
(this is the numerator of the probability T ÒClass AlR œ # ß \" œ " ß \# œ $Ó that we are trying
to find).
17 continued
In order to find the denominator T ÒR œ # ß \" œ " ß \# œ $Ó we use the fact that the selected
risk must be either from Class A or from Class B, so that
T ÒR œ # ß \" œ " ß \# œ $Ó
œ T ÒClass A ∩ ÐR œ # ß \" œ " ß \# œ $ÑÓ  T ÒClass B ∩ ÐR œ # ß \" œ " ß \# œ $ÑÓ
We find T ÒClass B ∩ ÐR œ # ß \" œ " ß \# œ $ÑÓ in the same way we found
T ÒClass A ∩ ÐR œ # ß \" œ " ß \# œ $ÑÓ .
T ÒClass B ∩ ÐR œ # ß \" œ " ß \# œ $ÑÓ œ T ÒR œ # ß \" œ " ß \# œ $lBÓ † T ÒBÓ .
Using the Poisson frequency (mean 3) and exponential severity (mean 3) from risk Class B, we
$#
/"$Î$
get T ÒR œ # ß \" œ " ß \# œ $lBÓ œ Ð/$ † #x Ñ ‚ #Ð "$ † /"Î$ ÑÐ "$ † /$Î$ Ñ œ # † #
/"$Î$
/"$Î$
and then T ÒClass B ∩ ÐR œ # ß \" œ " ß \# œ $ÑÓ œ #Ð # ÑÐ "# Ñ œ # † % .
/&
We then have T ÒR œ # ß \" œ " ß \# œ $Ó œ # † Ò % 
,
/"$Î$
% ÓÞ
Now we can find
T ÒClass AlR œ # ß \" œ " ß \# œ $Ó œ
T ÒClass A∩ÐR œ# ß \" œ" ß \# œ$ÑÓ
T ÒR œ# ß \" œ" ß \# œ$Ó
&
œ
#† / %
&
#†Ò / %  /
"$Î$
Ó
%
œ Þ$$*# .
We can use the same reasoning to find T ÒClass BlR œ # ß \" œ " ß \# œ $Ó , which will turn
out to be
"$Î$
%
/& /"$Î$
#†Ò %  % Ó
#† /
œ Þ''!) .
Alternatively, and more efficiently, once T ÒClass AlR œ # ß \" œ " ß \# œ $Ó is known,
T ÒClass BlR œ # ß \" œ " ß \# œ $Ó is equal to its complement
T ÒClass BlR œ # ß \" œ " ß \# œ $Ó œ "  T ÒClass AlR œ # ß \" œ " ß \# œ $Ó .
Finally, the posterior expected value we are looking for is
IÒWlR œ # ß \" œ " ß \# œ $Ó
œ IÒWlselected risk is from Class AÓ † T ÒClass AlR œ # ß \" œ " ß \# œ $Ó
 IÒWlselected risk is from Class BÓ † T ÒClass BlR œ # ß \" œ " ß \# œ $Ó
œ Ð"ÑÐÞ$$*#Ñ  Ð*ÑÐÞ''!)Ñ œ 'Þ#* .
The main work in this problem was in finding T ÒClass AlR œ # ß \" œ " ß \# œ $Ó .
The following "table" summarizes the steps outlined to find that probability.
"
T ÒEÓ œ # , given
"
T ÒFÓ œ # , given
Ì
#
/&
T ÒR œ # ß \" œ " ß \# œ $lAÓ œ #Ð/" † "#x ÑÐ" † /" ÑÐ" † /$ Ñ œ # † #
and
#
"
"
T ÒR œ # ß \" œ " ß \# œ $lBÓ œ #Ð/$ † $#x ÑÐ $ † /"Î$ ÑÐ $ † /$Î$ Ñ œ # †
are both know from the given distributions
Ì
/&
"
/&
T ÒClass A ∩ ÐR œ # ß \" œ " ß \# œ $ÑÓ œ #Ð # ÑÐ # Ñ œ # † %
and
/"$Î$ "
/"$Î$
T ÒClass B ∩ ÐR œ # ß \" œ " ß \# œ $ÑÓ œ #Ð # ÑÐ # Ñ œ # † % .
(this calculation uses the compound Poisson distribution)
/"$Î$
#
17 continued
Ì
T ÒR œ # ß \" œ " ß \# œ $Ó
œ T ÒClass A ∩ ÐR œ # ß \" œ " ß \# œ $ÑÓ  T ÒClass B ∩ ÐR œ # ß \" œ " ß \# œ $ÑÓ
&
/"$Î$
œ # † /%  # † %
Ì
T ÒClass AlR œ # ß \" œ " ß \# œ $Ó œ
&
œ
#† / %
&
#†Ò / %  /
"$Î$
Ó
%
T ÒClass A∩ÐR œ# ß \" œ" ß \# œ$ÑÓ
T ÒR œ# ß \" œ" ß \# œ$Ó
œ Þ$$*# and
T ÒClass BlR œ # ß \" œ " ß \# œ $Ó œ "  T ÒClass AlR œ # ß \" œ " ß \# œ $Ó œ Þ''!) .
Answer: D
18. This is a standard Bayesian estimation question. We are given that the conditional
distribution of R (claim count for an individual) given A is Poisson with mean A; and we are
given that A is a two point random variable that can take on the values 2 and 4 with
T ÒA œ #Ó œ Þ& and T ÒA œ %Ó œ Þ& (this means that any randomly chosen individual has an
equal chance of A being 2 or 4). We are given that R" œ R# œ % for a randomly chosen
individual (R" and R# being the claim count for years 1 and 2, respectively).
The Bayesian estimate of the insured's claim count in year 3 is the expected number of claims for
that insured in year 3 given the information about years 1 and 2, so we are asked to find
IÒR$ lR" œ R# œ %Ó. The typical way to find this expectation is to condition over A; we can
write the expectation as
IÒR$ lA œ #Ó † T ÒA œ #lR" œ R# œ %Ó  IÒR$ lA œ %Ó † T ÒA œ %lR" œ R# œ %Ó .
(Note that if the distribution of A had been continuous, say uniform on the interval Ð#ß %Ñ,
%
then we could write IÒR$ lR" œ R# œ %Ó as '# IÒR$ lA œ -Ó † 0 Ð-lR" œ R# œ %Ñ . - ,
and we would have to find the conditional density of A given R" œ R# œ %; this conditional
distribution of A would not necessarily be uniform).
Since R given A has a Poisson distribution, we have IÒR$ lA œ #Ó œ # and IÒR$ lA œ %Ó œ %.
We must find the conditional probabilities T ÒA œ #lR" œ R# œ %Ó and
T ÒA œ %lR" œ R# œ %Ó . We now use standard methods of conditional and joint probability (or
density in the continuous case). We use the relationship
T ÒElFÓ œ
T ÒE∩FÓ
T ÒFÓ
T ÒFlEÓ†T ÒEÓ
T ÒFlEÓ†T ÒEÓ
œ T ÒF∩EÓT ÒF∩EÓ
 œ

 ,
T ÒFlEÓ†T ÒEÓT ÒFlEÓ†T ÒEÓ
with events E and F defined as follows:
E: A œ # , F : R" œ R# œ % , and Ew (complement of E): A œ % . Then
T ÒÐAœ#Ñ∩ÐR" œR# œ%ÑÓ
T ÒR" œR# œ%Ó
T ÒR" œR# œ%|Aœ#Ó†T ÒAœ#Ó
T ÒR" œR# œ%|Aœ#Ó†T ÒAœ#ÓT ÒR" œR# œ%|Aœ%Ó†T ÒAœ%Ó
T ÒA œ #lR" œ R# œ %Ó œ
œ
.
We are given that T ÒA œ #Ó œ T ÒA œ %Ó œ Þ& , and the distribution of R given A œ - is
Poisson with parameter-. It must be implicitly assumed that for a randomly chosen individual
given the value of A, the (conditional) distributions of number of claims in separate years are
independent of one another also.
18 continued
This assumption is always made in the Bayesian estimation context, but note that it is not
generally true that the unconditional distributions of numbers of claims in separate years are
independent. What is meant here is that if the value of A is known, then the conditional
distributions R" lA and R# lA are independent, but the unconditional distributions of R" and R#
are not generally independent.
Therefore
# %
# %
T ÒR" œ R# œ %lA œ #Ó œ T ÒR" œ %lA œ #Ó † T ÒR# œ %lA œ #Ó œ Ð / %x†# ÑÐ / %x†# Ñ , and
% %
% %
T ÒR" œ R# œ %lA œ %Ó œ Ð / %x†% ÑÐ / %x†% Ñ .
Then, T ÒA œ #lR" œ R# œ %Ó œ
# †#% /# †#%
%x ÑÐ %x ÑÐÞ&Ñ
% %
% %
/# †#% /# †#%
Ð %x ÑÐ %x ÑÐÞ&ÑÐ / %x†% ÑÐ / %x†% ÑÐÞ&Ñ
Ð/
"
œ "Ð/#
œ Þ"(&) .
†#% Ñ#
Since A is either 2 or 4, T ÒA œ %lR" œ R# œ %Ó is the complement of T ÒA œ #lR" œ R# œ %Ó
so that T ÒA œ %lR" œ R# œ %Ó œ "  T ÒA œ #lR" œ R# œ %Ó œ Þ)#%# .
Finally, IÒR$ lR" œ R# œ %Ó
œ IÒR$ lA œ #Ó † T ÒA œ #lR" œ R# œ %Ó  IÒR$ lA œ %Ó † T ÒA œ %lR" œ R# œ %Ó .
œ Ð#ÑÐÞ"(&)Ñ  Ð%ÑÐÞ)#)%Ñ œ $Þ'%) .
The main work in the solution of this problem was to find T ÒA œ #lR" œ R# œ %Ó and
T ÒA œ %lR" œ R# œ %Ó . The calculations needed to find these probabilities can be summarized
as follows.
T ÒA œ #Ó œ Þ&
T ÒA œ %Ó œ Þ&
and
and
#
%
%
%
T ÒR" œ R# œ %lA œ #Ó œ Ð / %x†# Ñ#
T ÒR" œ R# œ %lA œ %Ó œ Ð / %x†% Ñ#
The probabilities for A were given explicitly, and the conditional probabilities follow since R is
Poisson given A.
Ì
Ì
# %
T ÒÐR" œ R# œ %Ñ ∩ ÐA œ #ÑÓ œ T ÒR" œ R# œ %lA œ #Ó † T ÒA œ #Ó œ Ð / %x†# Ñ# Ð "# Ñ and
% %
T ÒÐR" œ R# œ %Ñ ∩ ÐA œ %ÑÓ œ T ÒR" œ R# œ %lA œ %Ó † T ÒA œ % œ ÓÐ / %x†% Ñ# Ð "# Ñ .
Ì
T ÒR" œ R# œ %Ó œ T ÒÐR" œ R# œ %Ñ ∩ ÐA œ #ÑÓ  T ÒÐR" œ R# œ %Ñ ∩ ÐA œ #ÑÓ
# %
% %
œ Ð / %x†# Ñ# Ð "# Ñ  Ð / %x†% Ñ# Ð "# Ñ .
T ÒA œ #lR" œ R# œ %Ó œ
Ì
# †#% # "
%x Ñ Ð # Ñ
% %
/# †#% # "
Ð %x Ñ Ð # ÑÐ / %x†% Ñ# Ð "# Ñ
Ð/
œ Þ"(&) and
T ÒA œ %lR" œ R# œ %Ó œ "  T ÒA œ #lR" œ R# œ %Ó œ
% †%% # "
%x Ñ Ð # Ñ
% %
/# †#% # "
Ð %x Ñ Ð # ÑÐ / %x†% Ñ# Ð "# Ñ
Ð/
Answer: C
19. We wish to find T Ò\#  $!l\" œ #!Ó .
T Ò\#  $!l\" œ #!Ó œ
0 ÒÐ\# $!Ñ∩Ð\" œ#!ÑÓ
0 Ð\" œ#!Ñ
This can be written as
.
The numerator is
0 ÒÐ\#  $!Ñ ∩ Ð\" œ #!Ñlα œ "ÓT Ðα œ "Ñ
 0 ÒÐ\#  $!Ñ ∩ Ð\" œ #!Ñlα œ #ÓT Ðα œ #Ñ
 0 ÒÐ\#  $!Ñ ∩ Ð\" œ #!Ñlα œ $ÓT Ðα œ $Ñ ,
œ Þ)#%# .
19 continued
"
œ Ð $ ÑÒ0 ÒÐ\#  $!Ñ ∩ Ð\" œ #!Ñlα œ "Ó  0 ÒÐ\#  $!Ñ ∩ Ð\" œ #!Ñlα œ #Ó
 0 ÒÐ\#  $!Ñ ∩ Ð\" œ #!Ñlα œ $ÓÑ
and the denominator is
0 Ð\" œ #!lα œ "ÓT Ðα œ "Ñ  0 Ð\" œ #!lα œ #ÑT Ðα œ "Ñ  0 Ð\" œ #!lα œ $ÑT Ðα œ $Ñ .
"
œ Ð $ ÑÒ0 Ð\" œ #!lα œ "Ñ  0 Ð\" œ #!lα œ #Ñ  0 Ð\" œ #!lα œ #ÑÓ
α
α"!
The Pareto distribution with ) œ "! has pdf 0 ÐBÑ œ ÐB"!Ñ
α" ,
"! α
and cdf J ÐBÑ œ "  Ð B"!
Ñ .
"!
#!!
$!!!
0 Ò\" œ #!lα œ "Ó œ $!
# ß 0 Ò\" œ #!lα œ #Ó œ $!$ ß 0 Ò\" œ #!l α œ $Ó œ $!% ,
"!
#!!
$!!!
so that 0 Ò\" œ #!Ó œ Ð "$ ÑÒ $!
#  $!$  $!% Ó œ Þ!!(%" is the denominator.
From conditional independence of the \ 's given α, we have
0 ÒÐ\#  $!Ñ ∩ Ð\" œ #!Ñlα œ "Ó œ 0 Ò\#  $!lα œ "Ó † 0 Ò\" œ #!lα œ "Ó ,
and the same for α œ #ß $, so that
"!
"!
0 ÒÐ\#  $!Ñ ∩ Ð\" œ #!Ñlα œ "Ó œ T Ò\#  $!lα œ "Ó † 0 Ò\" œ #!lα œ "Ó œ Ð $!"!
ÑÐ $!
# Ñ,
"!
# #!!
and 0 ÒÐ\#  $!Ñ ∩ Ð\" œ #!Ñlα œ #Ó œ Ð $!"! Ñ Ð $!$ Ñ and
"!
0 ÒÐ\#  $!Ñ ∩ Ð\" œ #!Ñlα œ 3Ó œ Ð $!"!
Ñ3 Ð $!!!
Ñ .
$!%
"!
"!
"!
"!
# #!!
3 $!!!
Then, 0 ÒÐ\#  $!Ñ ∩ Ð\" œ #!ÑÓ œ Ð "$ ÑÒÐ $!"!
ÑÐ $!
# Ñ  Ð $!"! Ñ Ð $!$ Ñ  Ð $!"! Ñ Ð $!% ÑÓ
œ Þ!!""! is the numerator, and then
T Ò\#  $!l\" œ #!Ó œ
0 ÒÐ\# $!Ñ∩Ð\" œ#!ÑÓ
0 Ð\" œ#!Ñ
Þ!!""!
œ Þ!!(%"
œ Þ"%) .
Answer: C
20. Since we are given one observation (total annual losses of \" œ 500), the Bayesian
premium is IÒ\# l\" œ &!!Ó . Since the "parameter" in this Bayesian situation is the class A or
B, the prior distribution is T ÐEÑ œ T ÐFÑ œ Þ& (risks A and B are equally likely). For each of
the two classes, total annual loss has a compound distribution.
The Bayesian premium can be formulated as
IÒ\# l\" œ &!!Ó œ IÒ\# lEÓ † T ÒEl\" œ &!!Ó  IÒ\#lFÓ † T ÒFl\" œ &!!Ó .
Within class A, IÒ\# lEÓ œ Ð! † %*  " † %*  # † "* ÑÐ&!! † "$  "#$& † #$ Ñ œ ''! .
Within class B, IÒ\# lFÓ œ Ð! † "*  " † %*  # † %* ÑÐ#&! † #$  $#) † "$ Ñ œ $') .
T ÒE∩Ð\" œ&!!ÑÓ
T Ò\ œ&!!lEÓ†T ÒEÓ
T ÒEl\" œ &!!Ó œ T Ò\ œ&!!Ó
œ T Ò\ œ&!!lEÓ†T" ÒEÓT Ò\ œ&!!lFÓ†T ÒFÓ Þ
"
"
"
%
T Ò\" œ &!!lEÓ œ T Ò1 claimÓ † T Òclaim of 500Ó œ Ð %* ÑÐ "$ Ñ œ #(
,
T Ò\" œ &!!lFÓ œ T Ò2 claimsÓ † ÐT Òclaim of 250ÓÑ# œ Ð %* ÑÐ #$ Ñ# œ "'
)" .
Then, T ÒEl\" œ &!!Ó œ
%
Ð #(
ÑÐ "# Ñ
$
%
" œ ( .
Ð #(
ÑÐ "# ÑÐ "'
ÑÐ
Ñ
)" #
In a similar way, T ÒFl\" œ &!!Ó œ %( .
Then, IÒ\# l\" œ &!!Ó œ Ð''!ÑÐ $( Ñ  Ð$')ÑÐ %( Ñ œ %*$Þ" .
Answer: A
"
21. There are 4 possible die/spinner pairs, each with prior probability of % :
"
T ÐEß \Ñ œ T ÐEß ] Ñ œ T ÐFß \Ñ œ T ÐFß ] Ñ œ % . The associated expected losses are
$
"
IÒPlEß \Ó œ Ð % ÑÐ # ÑÐ"#  -Ñ ß IÒPlEß ] Ó œ Ð $% ÑÐ"#Ñ ß
IÒPlFß \Ó œ Ð "% ÑÐ "# ÑÐ"#  -Ñ ß IÒPlFß ] Ó œ Ð "% ÑÐ"#Ñ .
In order to find the posterior expectation IÒP# lP" œ "#Ó we use the relationship
IÒP# lP" œ "#Ó œ IÒP# lEß \Ó † T ÒEß \lP" œ "#Ó  IÒP#lEß ] Ó † T ÒEß ] lP " œ "#Ó
 IÒP# lFß \Ó † T ÒFß \lP" œ "#Ó  IÒP# lFß ] Ó † T ÒFß ] lP" œ "#Ó
$
œ Ð ) ÑÐ"#  -Ñ † T ÒEß \lP" œ "#Ó  Ð*Ñ † T ÒEß ] lP" œ "#Ó
"
 Ð ) ÑÐ"#  -Ñ † T ÒFß \lP" œ "#Ó  Ð$Ñ † T ÒFß ] lP" œ "#Ó .
To find the posterior probabilities, we first find
T ÒP" œ "#Ó œ T ÒEß \ß "#Ó  T ÒEß ] ß "#Ó  T ÒFß \ß "#Ó  T ÒFß ] ß "#Ó
œ T Ò"#lEß \Ó † T ÐEß \Ñ  T Ò"#lEß ] Ó † T ÐEß ] Ñ
 T Ò"#lFß \Ó † T ÐFß \Ñ  T Ò"#lFß ] Ó † T ÐFß ] Ñ
$ "
$
" "
"
"
$
œ ÒÐ % ÑÐ # Ñ  Ð % ÑÐ"Ñ  Ð % ÑÐ # Ñ  Ð % ÑÐ"ÑÓÐ % Ñ œ ) Þ
The posterior probabilities are then
Ð $% ÑÐ "# цР"% Ñ
T ÒP" œ"#lEß\Ó†T ÐEß\Ñ
"
œ
œ % ß
$
T ÒP" œ"#Ó
)
Ð $% ÑÐ"цР"% Ñ
T ÒP" œ"#lEß] Ó†T ÐEß] Ñ
"
T ÒEß ] lP" œ "#Ó œ
œ
œ # ß
$
T ÒP" œ"#Ó
)
Ð "% ÑÐ "# цР"% Ñ
T ÒP" œ"#lFß\Ó†T ÐFß\Ñ
"
T ÒFß \lP" œ "#Ó œ
œ
œ "# ß
$
T ÒP" œ"#Ó
)
Ð "% ÑÐ"цР"% Ñ
T ÒP" œ"#lFß] Ó†T ÐFß] Ñ
"
T ÒFß ] lP" œ "#Ó œ
œ
œ ' .
$
T ÒP" œ"#Ó
)
T ÒEß \lP" œ "#Ó œ
The posterior expectation is
$
"
"
"
"
"
IÒP# lP" œ "#Ó œ Ð ) ÑÐ"#  -Ñ † Ð % Ñ  Ð*Ñ † Ð # Ñ  Ð ) ÑÐ"#  -Ñ † Ð "# Ñ  Ð$Ñ † Ð ' Ñ .
We are given that this is equal to 10. Solving for - results in - œ $' . Answer: E
22. We first fine the posterior probabilities for the parameter -:
T Ð- œ "lW" œ #Ñ œ
T ÐW" œ#∩-œ"Ñ
T ÐW" œ#Ñ
T ÐW" œ #l- œ "Ñ œ T Ð1 claim for amount 2l- œ "Ñ  T Ð2 claims for amount 1 eachl- œ "Ñ
œ /" †
"
$

/"
#
†
"
$
†
"
$
"
œ (/") .
"
"
T ÐW" œ # ∩ - œ "Ñ œ T ÐW" œ #l- œ "Ñ † T Ð- œ "Ñ œ (/") † "# œ (/$' Þ
T ÐW" œ #l- œ #Ñ œ T Ð1 claim for amount 2l- œ #Ñ  T Ð2 claims for amount 1 eachl- œ #Ñ
# #
#
 / #†# † "$ † "$ œ )/* .
#
T ÐW" œ # ∩ - œ #Ñ œ T ÐW" œ #l- œ #Ñ † T Ð- œ #Ñ œ )/* †
œ /# † # †
"
$
"
#
#
œ %/* Þ
"
#
T ÐW" œ #Ñ œ T ÐW" œ # ∩ - œ "Ñ  T ÐW" œ # ∩ - œ #Ñ œ (/$'  %/* .
"
"
#
T ÐW œ#∩-œ"Ñ
T Ð- œ "lW" œ #Ñ œ T"ÐW œ#Ñ œ Ð (/$' Ñ‚Ð (/$'  %/* Ñ œ Þ&%$#
"
and T Ð- œ #lW" œ #Ñ œ "  T Ð- œ "lW" œ #Ñ œ Þ%&') .
The Bayesian premium is
IÒW# lW" œ #Ó œ IÒW# l- œ "Ó † T Ð- œ "lW" œ #Ñ  IÒW#l- œ #Ó † T Ð- œ #lW" œ #Ñ
œ Ð"ÑÐ#ÑÐÞ&%$#Ñ  Ð#ÑÐ#ÑÐÞ%&')Ñ œ #Þ*"$' .
23.(a) IÒ\Ó œ IÒ\lMÓ † T ÐMÑ  IÒ\lMMÓ † T ÐMMÑ  IÒ\lMMMÓ † T ÐMMMÑ
œ Ð"ÑÐ "# Ñ  Ð#ÑÐ "% Ñ  Ð%ÑÐ "% Ñ œ #
(b)(i) IÒ\ # Ó œ IÒ\ # lMÓ † T ÐMÑ  IÒ\ # lMMÓ † T ÐMMÑ  IÒ\ #lMMMÓ † T ÐMMMÑ
œ Ð"  "# ÑÐ "# Ñ  Ð#  ## ÑÐ "% Ñ  Ð%  %# ÑÐ "% Ñ œ (Þ&
Z +<Ò\Ó œ (Þ&  ## œ $Þ&
"
X œ M ß prob.
(ii) IÒ\lX Ó œ
#
%
"
#
X œ MM ß prob. "%
X œ MMM ß prob. "%
"
Z +<Ò\lX Ó œ
and
#
%
"
#
X œ MM ß prob. "%
X œ MMM ß prob. "%
X œ M ß prob.
.
Z +<Ò IÒ\lX Ó Ó œ Ð"# ‚ "#  ## ‚ "%  %# ‚ "% Ñ  Ð" ‚ "#  # ‚ "%  % ‚ "% Ñ# œ "Þ& .
IÒ Z +<Ò\lX Ó Ó œ Ð" ‚ "#  # ‚ "%  % ‚ "% Ñ œ #
Z +<Ò\Ó œ Z +<Ò IÒ\lX Ó Ó  IÒ Z +<Ò\lX Ó Ó œ "Þ&  # œ $Þ& Þ
(c) T ÐType Il\ œ "Ñ œ
Type I, "#
T Ð\œ"lType I)†T ÐType IÑ
T Ð\œ"Ñ
Type II , "%
.
Type III, "%
T Ð\ œ "lType I) œ /"
T Ð\ œ "lType II) œ #/#
T Ð\ œ "lType I) œ %/%
T Ð\ œ " ∩ Type IÑ
T Ð\ œ " ∩ Type IIÑ
T Ð\ œ " ∩ Type IIIÑ
œ
" "
#/
œ
" #
#/
œ /%
T Ð\ œ "Ñ œ "# /"  "# /#  /%
" "
/
T ÐType Il\ œ "Ñ œ " /" #" /# /% œ Þ')" ß
#
#
" #
/
T ÐType IIl\ œ "Ñ œ " /" #" /# /% œ Þ#&" ß
#
#
%
T ÐType IIIl\ œ "Ñ œ " /" /" /# /% œ Þ!') .
#
#
(d) IÒ\# l\" œ "Ó œ IÒ\# lType IÓ † T ÐType Il\" œ "Ñ  IÒ\# lType IIÓ † T ÐType IIl\" œ "Ñ
 IÒ\# lType IIIÓ † T ÐType IIIl\" œ "Ñ
œ Ð"ÑÐÞ')"Ñ  Ð#ÑÐÞ#&"Ñ  Ð%ÑÐÞ!')Ñ œ "Þ%&& .
23. (e) Z +<Ò\# l\" œ "Ó œ IÒ\## l\" œ "Ó  ÐIÒ\# l\" œ "ÓÑ# œ IÒ\##l\" œ "Ó  Ð"Þ%&&Ñ# .
IÒ\## l\" œ "Ó œ IÒ\## lType IÓ † T ÐType Il\" œ "Ñ  IÒ\## lType IIÓ † T ÐType IIl\" œ "Ñ
 IÒ\## lType IIIÓ † T ÐType IIIl\" œ "Ñ
œ Ð"  "# ÑÐÞ')"Ñ  Ð#  ## ÑÐÞ#&"Ñ  Ð%  %# ÑÐÞ!')Ñ œ %Þ##) .
Z +<Ò\# l\" œ "Ó œ %Þ##)  Ð"Þ%&&Ñ# œ #Þ"" .
Alternatively,
Z +<Ò\# l\" œ "Ó œ IÒ Z +<Ò\# lTypeÓ l\" œ "Ó  Z +<Ò IÒ\#lTypeÓ l\" œ "Ó .
Z +<Ò\# lTypeÓ œ
Z +<Ò\# lType IÓ œ "
Z +<Ò\# lType IIÓ œ #
Z +<Ò\# lType IIIÓ œ 4
prob. T ÐType Il\" œ "Ñ œ Þ')"
prob. T ÐType IIl\" œ "Ñ œ Þ#&"
prob. T ÐType IIIl\" œ "Ñ œ Þ!')
p IÒ Z +<Ò\# lTypeÓ l\" œ "Ó œ Ð"ÑÐÞ')"Ñ  Ð#ÑÐÞ#&"Ñ  Ð%ÑÐÞ!')Ñ œ "Þ%&& .
IÒ\# lType IÓ œ "
prob. T ÐType Il\" œ "Ñ œ Þ')"
IÒ\# lTypeÓ œ IÒ\# lType IIÓ œ # prob. T ÐType IIl\" œ "Ñ œ Þ#&"
IÒ\# lType IIIÓ œ 4 prob. T ÐType IIIl\" œ "Ñ œ Þ!')
Z +<Ò IÒ\# lTypeÓ l\" œ "Ó
œ ÒÐ"# ÑÐÞ')"Ñ  Ð## ÑÐÞ#&"Ñ  Ð%# ÑÐÞ!')ÑÓ  ÒÐ"ÑÐÞ')"Ñ  Ð#ÑÐÞ#&"Ñ  Ð%ÑÐÞ!')ÑÓ# œ Þ'&' .
Z +<Ò\# l\" œ "Ó œ "Þ%&&  Þ'&' œ #Þ"" .