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CREDIBILITY - PROBLEM SET 3Bayesian 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 bechosen). A bowl is chosen at random with each bowl having the chance of being chosen. A ballis chosen from that bowl. After the ball is chosen it is returned to its bowl and another ball ischosen at random from the same bowl. Suppose that first ball chosen is red. Find the probabilitythat 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 havea Poisson claim number per month with mean 2 per period and policies of type 2 each have aPoisson 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 thatpolicy 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 thefollowing 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 , where the prior distribution of is a discrete uniform distribution on the integersA A"ß #ß \3 . A single observation of is made.
3. Find the joint pf .0 ÐBß Ñ\ßA -
A) B) C) D) E) / $/ / / /Bx Bx Ð$BÑx $Bx Bx
B B B B BÎ$- - - - -- - - - -
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 theannual claim distributions for the three risk classes is as follows: Class I Class II Class IIIAnnual Claim Poisson Poisson PoissonNumber Distribution mean 1 mean 2 mean 550% 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 ofclaims 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 yearand 2 claims in the second year. Find the probability that the risk will have 2 claims in the thirdyear, 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 equallysized groups of individuals. Groups are classified as Type A, Type B and Type C. Anyindividual of any type has probability of .5 of having no claim in the coming year and has aprobability of .5 of having exactly 1 claim in the coming year. Each claim is for amount 1 or 2when a claim occurs. Suppose that the claim distributions given that a claim occurs, for the threetypes of individuals are
TÐ Bl Ñ œ ß#Î$ B œ ""Î$ B œ #
claim of amount Type A and a claim occurs šTÐ Bl Ñ œ ß
"Î# B œ ""Î# B œ #
claim of amount Type B and a claim occurs šTÐ Bl Ñ œ Þ
&Î' B œ ""Î' B œ #
claim of amount Type C and a claim occurs šAn 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 probabilitythat 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 300 and 4.) αœ œ - Loss sizes for property risks follow a Pareto distribution 1,000 and .) αœ 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 is observed. Determine5the limit of the posterior probability that this risk is a liability risk as goes to zero.5
10. You are given the following random sample of 8 data points from a population distribution\: 1 , 2 , 2 , 2 , 2 , 3 , 4 , 8It 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 theloss amount distributions for the two risk classes is as follows: Class I Class II Loss Amount Exponential ParetoDistribution mean 1000 ) αœ "!!!ß 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 theexpected 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 asecond loss of 1000. Find the probability that the risk was chosen from Class I. Find the expectedvalue 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 withfront air bags.(iv) The manufacturer randomly selects 2 or 4 crash test dummies to put into a car with side airbags.(v) Each crash test dummy in a wall-impact accident suffers damage randomly equal to either 0.5or 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 1or 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 theexpected 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 is binomial with parameters and ( is\ R œ 8 8 ; ;non-random), and the prior distribution of is Poisson with parameter . Find the posteriorR -distribution of based on one observation of .R \
14. You are given:(i) Two classes of policyholders have the following severity distributions:
Claim Amount Probability of Claim Probability of ClaimAmount for Class 1 Amount for Class 2
250 0.5 0.72,500 0.3 0.2
60,000 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 secondclaims 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 claimsfrequency 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 1,000Calculate 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 910C) At least 910, but less than 1,010 D) At least 1,010, but less than 1,110E) At least 1,110
16. A portfolio of insurance policies consists of two types of policies. The annual aggregate lossdistribution for each type of policy is is a compound Poisson distribution. Policies of Type I havea Poisson parameter of 1 and policies of Type 2 have a Poisson parameter of 2. For both policytypes, 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 claimthis year was 2).(c) Find the variance of the aggregate claim next year for the same policy (given aggregate claimthis year was 2) in two ways: (i) Z +<ÒW lW œ #Ó œ IÒW lW œ #Ó ÐIÒW lW œ "ÓÑ# " " # "#
# #
(ii) (this requiresZ +<ÒW lW œ #Ó œ Z +<Ò IÒW l Ó lW œ #Ó IÒ Z +<ÒW l Ó lW œ #Ó# " # " # "- - 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 claimamounts were 1.0 and 3.0. Calculate the posterior expected value of the aggregate loss for thisrisk 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 Paretodistributionwith parameters 10 and or 1, 2 or 3, with each value being equally likely. You then) αœ œobserve one claim of 20 for a randomly selected risk. Determine the posterior probability thatthe 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 1/31 4/9 1235 2/32 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 2/31 4/9 328 1/32 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 thatinsured for the year.
DieA 1/4 3/4B 3/4 1/4
T<ÐG6+37= œ !Ñ T<ÐG6+37= œ "Ñ
Two spinners, and , are used to determine claim cost. Spinner has two areas marked 12\ ] ]and . Spinner has only one area marked 12.- ]
Spinner1/2 1/2
T<ÐG9=> œ "#Ñ T<ÐG9=> œ -Ñ\] " !
To determine the losses for the year, a die is randomly selected from A and B and rolled. If aclaim 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 uponthe 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. The annual aggregate lossA portfolio of insurance policies consists of two types of policies. distribution for each type of policy is a compound Poisson distribution. Policies of Type I have aPoisson parameter of 1 and policies of Type 2 have a Poisson parameter of 2. For both policytypes, 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 anaggregate annual claim of 2 is observed. Find the Bayesian premium for the same policy for nextyear.
23. A portfolio of insurance policies consists of three types of policies. The distribution of thenumber of losses in one year for each type of policy is summarized as follows:Policy Type Type I Type II Type IIIAnnual Number Poisson Poisson Poissonof Losses With Mean 1 With Mean 2 With Mean 4Half 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 each of the following two ways:Z +<Ò\Ó (i) Z +<Ò\Ó œ IÒ\ Ó ÐIÒ\ÓÑ# #
(ii) , where is the randomZ +<Ò\Ó œ Z +<Ò IÒ\lX Ó Ó IÒ Z +<Ò\lX Ó Ó X œ ÖMß MMß MMM× 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 two ways:Z +<Ò\ l\ œ "Ó# "
(i) , andIÒ\ l\ œ "Ó ÐIÒ\ l\ œ "ÓÑ# ## " # "
(ii) Type Type .IÒ Z +<Ò\ l Ó l\ œ "Ó Z +<Ò IÒ\ l Ó l\ œ "Ó# " # "
Again assume conditional independence of 's given policy type.\
CREDIBILITY - PROBLEM SET 3 SOLUTIONS
1. 2nd red 1st redT Ò l Ó œT Ò ∩ Ó
T Ò Ó2nd red 1st red
1st red
œTÒ ∩ l Ó†T Ò ÓT Ò ∩ l Ó†T Ò Ó
(Î#!2nd red 1st red bowl 1 bowl 1 2nd red 1st red bowl 2 bowl 2
œ œÐ ÑÐ ÑÐ ÑÐ ÑÐ ÑÐ Ñ
(Î#! (!#*
& & " # # ""! "! # "! "! # .
Note also thatT Ò l Ó2nd red 1st redœ TÒ l Ó † T Ò l Ó T Ò l Ó † T Ò l Ó 2nd red bowl 1 bowl 1 1st red 2nd red bowl 2 bowl 2 1st redœ Ð ÑÐ Ñ Ð ÑÐ Ñ œ& & # # #*
"! ( "! ( (! . Answer: E
2. T Ò\ œ "l\ œ "Ó œ# T Ò\ œ"∩\ œ"Ó
T Ò\ œ"Ó# "
"
œ Type 1 Type 1 Type Type T Ò\ œ"∩\ œ"l Ó†T Ò ÓT Ò\ œ"∩\ œ"l #Ó†T Ò #Ó
T Ò\ œ"Ó# " # "
"
œ œ Þ#%(Ð/ † Ñ †Ð ÑÐ/ † Ñ †Ð Ñ
Ð/ † цРÑÐ/ † цРÑ
# # % ## # % "" "
"x $ "x $
# %# # % "" "
"x $ "x $
.
Alternatively, T Ò\ œ "l\ œ "Ó# "
œ T Ò\ œ "l Ó † T Ò l\ œ "Ó T Ò\ œ "l #Ó † T Ò l\ œ "Ó# " # "Type 1 Type 1 Type Type 2œ Ð/ † ÑÐÞ))"Ñ Ð/ † ÑÐÞ"!*Ñ œ Þ#%'# %# %
"x "x
" "
. Answer: C
3. Answer: D0 ÐBß Ñ œ 0 ÐBl Ñ † Ð Ñ œ † Þ\ß \lA A- - 1 - / "Bx $
B--
4. The marginal pf for is \ 0 ÐBÑ œ 0 ÐBß Ñ œ ß\œ"
$
\ß-
A - / / # / $$Bx $Bx $Bx
" # B $ B
and the posterior distribution of has pfA
1 -Al\Ð lBÑ œ œ0 ÐBß Ñ
0 ÐBÑ
\ß
\
/ B
$Bx/ / # / $" # B $ B
$Bx $Bx $Bx
A--- , and
1 1A Al\ l\Ð"l"Ñ œ œ Þ%'( ß Ð l"Ñ œ œ Þ$%% ß/ /"
$ $/ / # / $ / / # / $" # $ " # $
$ $ $ $ $ $ 2
22
1 AAl\Ð$l"Ñ œ œ Þ"*! Þ IÐ l\ œ "Ñ œ ÐÞ%'(Ñ #ÐÞ$%%Ñ $ÐÞ"*!Ñ œ "Þ(# Þ/ $$
$/ / # / $" # $
$ $ $
Answer: C
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 ∩ Ð œ "ÑÓ T ÒÐ\ œ #Ñ ∩ Ð œ #ÑÓ T ÒÐ\ œ #Ñ ∩ Ð œ &ÑÓ" " " "A A Aœ TÒ\ œ #l œ "Ó † T Ò œ "Ó T Ò\ œ #l œ #Ó † T Ò œ #Ó" "A A A A
TÒ\ œ #l œ &Ó † T Ò œ &Ó" A A
œ / †" / †# / †&#x #x #x
" # # # & #
† ÐÞ&Ñ † ÐÞ$Ñ † ÐÞ#Ñ œ Þ"*!! .T ÒÐ\ œ #Ñ ∩ Ð\ œ #ÑÓ œ T ÒÐ\ œ #Ñ ∩ Ð\ œ #Ñl œ "Ó † T Ò œ "Ó# " # " A A TÒÐ\ œ #Ñ ∩ Ð\ œ #Ñl œ #Ó † T Ò œ #Ó T ÒÐ\ œ #Ñ ∩ Ð\ œ #Ñl œ &Ó † T Ò œ &Ó# " # "A A A A
œ ˆ ‰ ˆ ‰ ˆ ‰/ †" / †# / †&#x #x #x
" # # # & ## # #† ÐÞ&Ñ † ÐÞ$Ñ † ÐÞ#Ñ œ Þ!%!$ .
T Ò\ œ #l\ œ #Ó œ œ œ Þ#"# "T ÒÐ\ œ#Ñ∩Ð\ œ#ÑÓ
T Ò\ œ#Ó Þ"*!!Þ!%!$# "
" .
IÒ\ l\ œ #Ó œ IÒ\ l œ "l † T Ò œ "l\ œ #Ó IÒ\ l œ #l † T Ò œ #l\ œ #Ó# " # " # "A A A A .IÒ\ l œ &l † T Ò œ &l\ œ #Ó# "A AIn Example CR-14 we found
T Ò œ "l\ œ #Ó œ œ œ Þ%)%AT Ò\œ#l œ"Ó†T Ò œ"Ó
T Ò\œ#Ó
†ÐÞ&ÑA A / †"" #
#x/ †" / †# / †&" # # # & #
#x #x #x†ÐÞ&Ñ †ÐÞ$Ñ †ÐÞ#Ñ .
In a similar way, we get , andT Ò œ #l\ œ #Ó œ œ Þ%#(A/ †## #
#x/ †" / †# / †&" # # # & #
#x #x #x
†ÐÞ$Ñ
†ÐÞ&Ñ †ÐÞ$Ñ †ÐÞ#Ñ
T Ò œ &l\ œ #Ó œ œ Þ!)*A/ †&& #
#x/ †" / †# / †&" # # # & #
#x #x #x
†ÐÞ#Ñ
†ÐÞ&Ñ †ÐÞ$Ñ †ÐÞ#Ñ .
Then, since and , we haveIÒ\ l œ "l œ " ß IÒ\ l œ #l œ # IÒ\ l œ &l œ &# # #A A AIÒ\ l\ œ #Ó œ Ð"ÑÐÞ%)%Ñ Ð#ÑÐÞ%#(Ñ Ð&ÑÐÞ!)*Ñ œ "Þ()$# " .
(b) We are to find and .T Ò\ œ #l\ œ #ß\ œ #Ó 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!%!$# "
We find in a similar way.T ÒÐ\ œ #Ñ ∩ Ð\ œ #Ñ ∩ Ð\ œ #ÑÓ$ # "
T ÒÐ\ œ #Ñ ∩ Ð\ œ #Ñ ∩ Ð\ œ #ÑÓ$ # "
œ ˆ ‰ ˆ ‰ ˆ ‰/ †" / †# / †&#x #x #x
" # # # & #$ $ $† ÐÞ&Ñ † ÐÞ$Ñ † ÐÞ#Ñ œ Þ!!*") .
Then T Ò\ œ #l\ œ #ß\ œ #Ó œ œ Þ##)$ " #Þ!!*")Þ!%!$ .
IÒ\ l\ œ #ß\ œ #Ó œ IÒ\ l œ "l † T Ò œ "l\ œ #ß\ œ #Ó$ " # $ " #A AIÒ\ l œ #l † T Ò œ #l\ œ #ß\ œ #Ó IÒ\ l œ &l † T Ò œ &l\ œ #ß\ œ #Ó# " # # " #A A A A
T Ò œ "l\ œ #ß\ œ #Ó œ œ Þ%#! ßA " #
ˆ ‰ˆ ‰ ˆ ‰ ˆ ‰
/ †"" #
#x/ †" / †# / †&" # # # & #
#x #x #x
#
# # #
†ÐÞ&Ñ
†ÐÞ&Ñ †ÐÞ$Ñ †ÐÞ#Ñ
T Ò œ #l\ œ #ß\ œ #Ó œ Þ&%& ß T Ò œ &l\ œ #ß\ œ #Ó œ Þ!$&A A" # " # .Then .IÒ\ l\ œ #ß\ œ #Ó œ Ð"ÑÐÞ%#!Ñ Ð#ÑÐÞ&%&Ñ Ð&ÑÐÞ!$&Ñ œ "Þ')&$ " #
6. Type AT Ò l\ œ #Ó œ œTÒ\œ#l Ó†T ÒEÓ T Ò\œ#l Ó†T ÒEÓ
T Ò\œ#Ó T Ò\œ#lEÓ†T ÒEÓT Ò\œ#lFÓ†T ÒFÓT Ò\œ#lGÓ†T ÒGÓA A
œ œÒÐÞ&ÑÐ ÑӆРÑ
ÒÐÞ&ÑÐ ÑӆРÑÒÐÞ&ÑÐ ÑӆРÑÒÐÞ&ÑÐ ÑӆРÑ"$
" "$ $
" " " " " "$ $ # $ ' $
. Answer: B
7. Type TypeIÒ\ l\ œ #Ó œ IÒ\ l Ó † T Ò l\ œ #Ó# " # "Type
œ IÒ\ l Ó † T Ò l\ œ #Ó IÒ\ l Ó † T Ò l\ œ #Ó IÒ\ l Ó † T Ò l\ œ #Ó# " # " # "A A B B C C .As in Problem 6, we find B and C .T Ò l\ œ #Ó œ T Ò l\ œ #Ó œ" "
# '
Also, A , and similarly,IÒ\ l Ó œ ÐÞ&ÑÐ!Ñ ÐÞ&ÑÒ" † # † Ó œ## " #$ $ $
IÒ\ l Ó œ IÒ\ l Ó œ Þ# #B and C $ (% "#
Then, . Answer: AIÒ\ l\ œ #Ó œ Ð ÑÐ Ñ Ð ÑÐ Ñ Ð ÑÐ Ñ œ# "# " $ " ( " #&$ $ % # "# ' $'
8. No matter which shooter is chosen first, two of the other shooters have 2 target Areas incommon (call this event ), and the last has 1 target Area in common (call this event ).#G "G
Therefore ThenT Ò#GÓ œ ß T Ò"GÓ œ Þ# "$ $
T Ò Óboth shots land in same Areaœ TÒ l#GÓ † T Ò#GÓ T Ò l"GÓ † T Ò"GÓboth shots land in same Area both shots land in same Area
If the two shooters have two Areas in common, then there is probability that both hit#Ð Ñ œ" ""' )
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 mixof the corresponding Pareto moments, with weights .75 and .25.IÒ] Ó œ ÐÞ(&Ñ ÐÞ#&Ñ œ (& "#& œ #!!
Ð$!!Ñ Ð#Ñ Ð%"Ñ Ð"!!!Ñ Ð#Ñ Ð$"ÑÐ%Ñ Ð$Ñ
> > > >> >
"
.
IÒ] Ó œ ÐÞ(&Ñ ÐÞ#&Ñ œ ##ß &!! #&!ß !!! œ #(#ß &!!# Ð$!!Ñ Ð$Ñ Ð%#Ñ Ð"!!!Ñ Ð$Ñ Ð$#ÑÐ%Ñ Ð$Ñ
# #> > > >> > .
Z +<Ò] Ó œ #(#ß &!! Ð#!!Ñ œ #$#ß &!!# .
(b) 0ÐPl5Ñ œ œ Þ0Ð5lPÑ0ÐPÑ
0Ð5lPÑ0ÐPÑ0Ð5lT Ñ0ÐT Ñ
†ÐÞ(&Ñ
†ÐÞ(&Ñ †ÐÞ#&Ñ
%Ð$!!Ñ%
Ð5$!!Ñ&
%Ð$!!Ñ $Ð"!!!Ñ% $
Ð5$!!Ñ& Ð5"!!!Ñ%
As the limit is .5p! œ Þ*$!%
$!!% $
$!! "!!!
ÐÞ(&Ñ
ÐÞ(&Ñ ÐÞ#&Ñ
10. The model distribution is .0Ð l Ñ œ / œ / œ /B ) #3œ"
)B Î B Î #%Î" " "
) ) )3 3) D ) )
) )
The joint distribution of and is\ @
0 Ð ß Ñ œ 0ÐBl œ Ñ † T Ò œ Ó œ
Þ#& ‚ / œ "
Þ& ‚ # / œ #
Þ#& ‚ $ / œ $\ß
B
) B Î#
) B Î$@
D
D
D
B ) @ ) @ )
@
@
@
3
3
3
.
The marginal distribution of is\0 Ð Ñ œ 0 Ð ß "Ñ 0 Ð ß #Ñ 0 Ð ß $Ñ\ \ß \ß \ßB B B B@ @ @
œ ÐÞ#&Ñ/ ÐÞ&Ñ# / ÐÞ#&Ñ$ / B ) B Î# ) B Î$D D D3 3 3 .The posterior distribution of is@
1@l\Ð"l Ñ œBÐÞ#&Ñ/
ÐÞ#&Ñ/ ÐÞ&Ñ# / ÐÞ#&Ñ$ /
B3
B B Î# B Î$3 3 3) )
D
D D D ,
1@l\Ð#l Ñ œBÐÞ&Ñ# /
ÐÞ#&Ñ/ ÐÞ&Ñ# / ÐÞ#&Ñ$ /
) B Î#3
B B Î# B Î$3 3 3) )
D
D D D ,
10. continued
1@l\Ð$l Ñ œ ÞBÐÞ#&Ñ$ /
ÐÞ#&Ñ/ ÐÞ&Ñ# / ÐÞ#&Ñ$ /
) B Î$3
B B Î# B Î$3 3 3) )
D
D D D
For the given vector of values, this becomes\1 1 1@ @ @l\ l\ l\Ð"l Ñ œ Þ!!!$)!( ß Ð#l Ñ œ Þ%)%! ß Ð$l Ñ œ Þ&"&' ÞB B B
The posterior mean is .ÐÞ!!!$)!(ÑÐ"Ñ ÐÞ%)%!ÑÐ#Ñ ÐÞ&"&'ÑÐ$Ñ œ #Þ&"&Answer: A
11.(a) Class ] Class IÒ\ l\ œ #!!!Ó œ IÒ\ l M † T Ò Ml\ œ #!!!Ó# " # "
] Class IÒ\ lG6+== MM † T Ò MMl\ œ #!!!Ó# "
IÒ\ l M œ "!!! IÒ\ l MM œ œ &!!# #Class ] , and Class ] .)α"
In Example CR-15 it was found that Class , andT Ò Ml\ œ #!!!Ó œ Þ()&"
T Ò MMl\ œ #!!!Ó œ Þ#"&Class . Thus,"
IÒ\ l\ œ #!!!Ó œ Ð"!!!ÑÐÞ()&Ñ Ð&!!ÑÐÞ#"&Ñ œ )*#Þ&!# "
(b) We wish to find Class , . This probability isT Ò Ml\ œ #!!! \ œ "!!!Ó" #0 ÒB œ#!!! B œ"!!!ß œMÓ
0 ÒB œ#!!! B œ"!!!Ó" #
" #
,,
@ . The numerator is
0ÒB œ #!!! B œ "!!!l œ MÓ † T Ò œ MÓ œ Ð ÑÐ ÑÐÞ&Ñ œ" #, .@ @ / / Þ&/"!!! "!!! "!!!
#!!!Î"!!! "!!!Î"!!! $
#
The denominator is0ÒB œ #!!! B œ "!!!l œ MÓ † T Ò œ MÓ 0ÒB œ #!!! B œ "!!!l œ MMÓ † T Ò œ MMÓ" # " #, ,@ @ @ @
Ð ÑÐ ÑÐÞ&Ñ † † ÐÞ&Ñ œ Þ/ / $†"!!! $†"!!! Þ&/"!!! "!!! "!!!Ð#!!!"!!!Ñ Ð"!!!"!!!Ñ Ð$!!!‚#!!!Ñ
ÐÞ&ÑÐ*ÑÐ"!!! Ñ#!!!Î"!!! "!!!Î"!!! $ $ $
% % %#
'
Then Class , , andT Ò Ml\ œ #!!! \ œ "!!!Ó œ œ Þ)()" #
Þ&/$
"!!!#
Þ&/$
"!!!#ÐÞ&ÑÐ*ÑÐ"!!! Ñ'
Ð$!!!‚#!!!Ñ%
We are also asked to find , . This predictive expectation isIÒ\ l\ œ #!!! \ œ "!!!Ó$ " #
IÒ\ l Ml † T Ò Ml Ó$ Class Class \ œ #!!! \ œ "!!!" #, IÒ\ l MMl † T Ò MMl Ó$ Class Class \ œ #!!! \ œ "!!!" #, œ Ð"!!!ÑÐÞ)()Ñ Ð&!!ÑÐÞ"##Ñ œ *$* .
12. We wish to find . We condition over the combinations ofIÒ\ l\ œ "Ó# "
Front/Side air bags , Wall/Lake accident , so thatIÒ\ 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\ lJPÓ † T ÒJPl\ œ "Ó Þ# " # "
We immediately get that since with side air bags there will be 2 or 4T ÒWPl\ œ "Ó œ !"
dummies and with a lake crash the minimum damage per dummy is 1, so that the minimumoverall damage with is 2. Therefore,WPIÒ\ l\ œ "Ó œ IÒ\ lW[ Ó † T ÒW[ l\ œ "Ó# " # "
IÒ\ lJ[ Ó † T ÒJ[ l\ œ "Ó IÒ\ lJPÓ † T ÒJPl\ œ "Ó Þ# " # "
We find (2 or 4 dummies equally likely, averageIÒ\ lW[ Ó œ Ð ÑÐ# %ÑÐ Ñ œ #Þ#&#" Þ&"# #
damage of for each dummy) ,Þ&"#
12. continuedIÒ\ lJ[ Ó œ Ð ÑÐ" # $ %ÑÐ Ñ œ "Þ)(& ß#
" Þ&"% # and
IÒ\ lJPÓ œ Ð ÑÐ" # $ %ÑÐ Ñ œ $Þ(&#" "#% # .
We then calculate ,T ÒW[ l\ œ "Ó œ"T Ò\ œ"lW[ Ó†T ÒW[ Ó
T Ò\ œ"Ó"
"
T ÒJ[ l\ œ "Ó œ"T Ò\ œ"lJ[ Ó†T ÒJ[ Ó
T Ò\ œ"Ó"
" ,
and , whereT ÒJPl\ œ "Ó œ"T Ò\ œ"lJPÓ†T ÒJPÓ
T Ò\ œ"Ó"
"
T ÒW[ Ó œ T ÒWPÓ œ T ÒJ[ Ó œ T ÒJPÓ œ Þ#& , andT Ò\ œ "lW[ Ó œ ÐÞ&ÑÐÞ&Ñ œ Þ"#&"
# since there must be 2 dummies (prob. .5) and each sustainsdamage of .5 (prob. .5 each),T Ò\ œ "lJ[ Ó œ ÐÞ#&ÑÐÞ&Ñ ÐÞ#&ÑÐÞ&Ñ œ Þ")(&"
# (either 1 dummy and damage of 1, or twodummies and damage of .5 each), andT Ò\ œ "lJPÓ œ ÐÞ#&ÑÐÞ&Ñ œ Þ"#&" .
Then T Ò\ œ "Ó œ T Ò\ œ "lW[ Ó † T ÒW[ Ó T Ò\ œ "lJ[ Ó † T ÒJ[ Ó" " "
. TÒ\ œ "lJPÓ † T ÒJPÓ œ ÐÞ"#&ÑÐÞ#&Ñ ÐÞ")(&ÑÐÞ#&Ñ ÐÞ"#&ÑÐÞ#&Ñ œ Þ"!*$(&"
Then T ÒW[ l\ œ "Ó œ œ Þ#)&( ß T ÒJ[ l\ œ "Ó œ œ Þ%#)' ß" "ÐÞ"#&ÑÐÞ#&Ñ ÐÞ")(&ÑÐÞ#&ÑÞ"!*$(& Þ"!*$(&
and .T ÒJPÒ\ œ "Ó œ œ Þ#)&("ÐÞ"#&ÑÐÞ#&ÑÞ"!*$(&
We can summarize these conditional probability calculations as follows.
WP ß Þ#& W[ ß Þ#& JP ß Þ#& J[ ß Þ#& Ì Ì Ì Ì
T Ò\ œ "lWPÓ T Ò\ œ "lW[ Ó T Ò\ œ "lJPÓ T Ò\ œ "lJ[ Ó" " " " œ ! œ ÐÞ&ÑÐÞ&Ñ œ Þ"#& œ ÐÞ#&ÑÐÞ&Ñ œ Þ"#& œ ÐÞ#&ÑÐÞ& Þ& Ñ œ Þ")(& # #
Ì Ì Ì Ì T Ò\ œ " ∩ WPÓ T Ò\ œ " ∩ W[ Ó T Ò\ œ " ∩ JPÓ T Ò\ œ " ∩ J[Ó" " " " œ Ð!ÑÐÞ#&Ñ œ ÐÞ"#&ÑÐÞ#&Ñ œ ÐÞ"#&ÑÐÞ&Ñ œ ÐÞ")(&ÑÐÞ#&Ñ
ÌTÒ\ œ "Ó œ Ð!ÑÐÞ#&Ñ ÐÞ"#&ÑÐÞ#&Ñ ÐÞ"#&ÑÐÞ&Ñ ÐÞ")(&ÑÐÞ#&Ñ œ ÐÞ#&ÑÐÞ%$(&Ñ"
Ì Ì Ì Ì T ÒWPl\ œ "Ó T ÒW[ l\ œ "Ó T ÒJPl\ œ "Ó T ÒJ[ l\ œ "Ó" " " "
œ œ œ œ!
ÐÞ#&ÑÐÞ%$(&Ñ ÐÞ#&ÑÐÞ%$(&Ñ ÐÞ#&ÑÐÞ%$(&Ñ ÐÞ#&ÑÐÞ%$(&ÑÐÞ"#&ÑÐÞ#&Ñ ÐÞ"#&ÑÐÞ#&Ñ 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 ÐBl8Ñ œ ; Ð" ;Ñ œ\lR8B
B 8Bˆ ‰ 8x; Ð";ÑBxÐ8BÑx
B 8B
and the prior distribution of is Poisson with , so that the joint distribution of R Ð8Ñ œ \1 /8x
8--
and is .R 0 ÐBß 8Ñ œ † œ †\ßR8x; Ð";Ñ / Ð; Ñ ÒÐ";Ñ Ó
BxÐ8BÑx 8x Bx Ð8BÑx/B 8B B 8B 8- -- - -
The marginal distribution of is (the summation\ 0 ÐBÑ œ †\8œB
∞/ Ð; Ñ ÒÐ";Ñ ÓBx Ð8BÑx
B 8B- - -
starts at because as a binomial distribution with parameter , it must be the case that8 œ B 8! Ÿ B Ÿ 8 5 œ 8 B). Applying the change of variable to the summation results in
0 ÐBÑ œ † œ † / œ\5œ!
∞Ð";Ñ/ Ð; Ñ ÒÐ";Ñ Ó / Ð; Ñ / Ð; Ñ
Bx 5x Bx Bx
B 5 B ; B- - -- - - -- .
The marginal distribution of is Poisson with parameter .\ ;-
The posterior distribution of has probability functionR
1Rl\Ð8lBÑ œ œ † Î œ0 ÐBß8Ñ0 ÐBÑ Bx Ð8BÑx Bx Ð8BÑx
/ Ð; Ñ ÒÐ";Ñ Ó / Ð; Ñ / ÒÐ";Ñ Ó\ßR
\
B 8B ; B Ð";Ñ 8B’ “- - -- - - -
for (this is a translated Poisson; a Poisson with parameter translated by ).8 B Ð" ;Ñ B-
14. Class 1 Class 1IÒ\ l\ œ #&!Ó œ IÒ\ l Ó † T Ò l\ œ #&!Ó# " # "
Class 2 Class 2 .IÒ\ l Ó † T Ò l\ œ #&!Ó# "
IÒ\ l Ó œ Ð#&!ÑÐÞ&Ñ Ð#ß &!!ÑÐÞ$Ñ Ð'!ß !!!ÑÐÞ#Ñ œ "#ß )(&# Class 1 ,IÒ\ l Ó œ Ð#&!ÑÐÞ(Ñ Ð#ß &!!ÑÐÞ#Ñ Ð'!ß !!!ÑÐÞ"Ñ œ 'ß '(&# Class 2 .Given Class 1 Class 2T Ò Ó œ T Ò Ó œ
# "$ $
Given Class 1 Class 2T Ò\ œ #&!l Ó œ Þ& T Ò\ œ #&!l Ó œ Þ(" "
Ì Ì Class 1 Class 2T ÒÐ\ œ #&!Ñ ∩ Ó T ÒÐ\ œ #&!Ñ ∩ Ó" "
Class 1 Class 1 Class 2 Class 2œ TÒ\ œ #&!l Ó † T Ò Ó œ T Ò\ œ #&!l Ó † T Ò Ó" "
œ ÐÞ&ÑÐ Ñ œ œ ÐÞ(ÑÐ Ñ œ# " "$ $ $ (
$! Ì Class 1 Class 2T Ò\ œ #&!Ó œ T ÒÐ\ œ #&!Ñ ∩ Ó T ÒÐ\ œ #&!Ñ ∩ Ó" " "
œ œ" ( "($ $! $!
Ì
Class 1 ,T Ò l\ œ #&!Ó œ œ œ"T ÒÐ\ œ#&!Ñ∩ Ó "Î$
T Ò\ œ#&!Ó "(Î$! "("!"
"
Class 1
Class 2 .T Ò l\ œ #&!Ó œ " œ""! ("( "(
Then . Answer: BIÒ\ l\ œ #&!Ó œ Ð"#ß )(&ÑÐ Ñ Ð'ß '(&ÑÐ Ñ œ "!ß $### ""! ("( "(
15. For each Class the claims follow a Poisson process. For a particular Class, if the expectedamount of time between claims is (in years), then the expected number of claims per year is .>!
">!
Class I has an expected amount of time of 1 year between claims, so that the expected number ofclaims per year for Class 1 is . Class II has an expected amount of time of 1.25 years (15-" œ "months) between claims, so that the expected number of claims per year for Class 1 is-# œ œ Þ)"
"Þ#& . Class III has an expected amount of time of 1.5 years (15 months) betweenclaims, so that the expected number of claims per year for Class 1 is .-$ œ œ" #
"Þ& $
Let us denote the number of claims in year 1 by and the number of claims in year 2 is denotedR"
by . We wish to find , and then, since each claim cost is 1,000 in all classes,R IÒR lR œ !Ó# # "
the expected loss in year 2 is ."!!!IÒR lR œ !Ó# "
We find by conditioning over the Class type.IÒR lR œ !Ó# "
IÒR lR œ !Ó œ IÒR l Ó † T Ò lR œ !Ó# " # "Class I Class I Class II Class II Class III Class III .IÒR l Ó † T Ò lR œ !Ó IÒR l Ó † T Ò lR œ !Ó# " # "
The conditional expectations are the expected number of claims in a year for a given class:IÒR l Ó œ " IÒR l Ó œ Þ) IÒR l Ó œ# # #Class I , Class II and Class III .#
$The conditional probabilities can be found from the following probability table:
Class I , I Class II , II Class III , IIIT Ò Ó œ T Ò Ó œ T Ò Ó œ" " "$ $ $
I II IIIR œ ! T ÒR œ !l Ó œ T ÒR œ !l Ó œ T ÒR œ !l Ó œ" " " "/ †" / †Þ)!x !x !x
/ †Ð Ñ" ! Þ) ! #$ #
$!
œ Þ$'()) œ Þ%%*$$ œ Þ&"$%#
I II IIIT ÒR œ ! ∩ Ó T ÒR œ ! ∩ Ó T ÒR œ ! ∩ Ó" " "
I I II II III III œ TÒR œ !l Ó † T Ò Ó œ T ÒR œ !l Ó † T Ò Ó œ T ÒR œ !l Ó † T Ò Ó" " "
œ ÐÞ$'())Ñ œ ÐÞ%%*$$Ñ œ ÐÞ&"%$#ÑÐ Ñ Ð Ñ Ð Ñ" " "$ $ $
Then, I II III , andT ÒR œ !Ó œ T ÒR œ ! ∩ Ó T ÒR œ ! ∩ Ó T ÒR œ ! ∩ Ó œ Þ%%$)%" " " "
T Ò lR œ !Ó œ œ œ Þ#('$Class I ."T Ò ∩R œ!Ó
T ÒR œ!Ó Þ%%$)%
ÐÞ$'())ÑÐ ÑClass I "
"
"$
T Ò lR œ !Ó œ œ œ Þ$$(&Class II .#T Ò ∩R œ!Ó
T ÒR œ!Ó Þ%%$)%
ÐÞ%%*$$ÑÐ ÑClass II "
"
"$
T Ò lR œ !Ó œ œ œ Þ$)'$Class III .#T Ò ∩R œ!Ó
T ÒR œ!Ó Þ%%$)%
ÐÞ&"%$#ÑÐ ÑClass III "
"
"$
Then , and the expected lossIÒR lR œ !Ó œ Ð"ÑÐÞ#('$Ñ ÐÞ)ÑÐÞ$$(&Ñ ÐÞ$)'$Ñ œ Þ)!%# " Ð Ñ#$for year 2 is . Answer: A"!!!ÐÞ)!%Ñ œ )!%
16. The prior parameter has distribution .prob.
2 prob.- - œ
"œ
"#"#
The model distribution has a compound distribution with Poisson frequency with mean , andW -the stated severity distribution.(a) TÐ œ "lW œ #Ñ œ- "
T ÐW œ#∩ œ"ÑT ÐW œ#Ñ"
"
-
TÐW œ #l œ "Ñ œ TÐ l œ "Ñ TÐ l œ "Ñ" - - -1 claim for amount 2 2 claims for amount 1 eachœ / † † † œ" " / " " (/
$ # $ $ ")
" "
.
16. continuedTÐW œ # ∩ œ "Ñ œ TÐW œ #l œ "Ñ † T Ð œ "Ñ œ † œ Þ" "- - - (/ " (/
") # $'
" "
T ÐW œ #l œ #Ñ œ TÐ l œ #Ñ TÐ l œ #Ñ" - - -1 claim for amount 2 2 claims for amount 1 eachœ / † # † † † œ# " / †# " " )/
$ # $ $ *
# # #
.
TÐW œ # ∩ œ #Ñ œ TÐW œ #l œ #Ñ † T Ð œ #Ñ œ † œ Þ" "- - - )/ " %/* # *
# #
T ÐW œ #Ñ œ TÐW œ # ∩ œ "Ñ TÐW œ # ∩ œ #Ñ œ " " "- - (/ %/$' *
" #
.
TÐ œ "lW œ #Ñ œ œ Ñ œ Þ&%$#- "%/*
T ÐW œ#∩ œ"ÑT ÐW œ#Ñ $' $'
(/ (/"
"
" "- Ð Ñ Ð‚ #
and .TÐ œ #lW œ #Ñ œ " TÐ 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) (from part (b) ).IÒW lW œ #Ó œ #Þ*"$'# "
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 has a compound Poisson distribution with , the mean is 2,W l œ " œ " IÒW l œ "Ó œ# #- - -and the variance is , where is the severity distribution.Z +<ÒW l œ "Ó œ † IÒ\ Ó œ IÒ\ Ó \#
# #- -
From the distribution of we have , and then\ IÒ\ Ó œ# "%$
.IÒW l œ "Ó œ Ð"Ñ Ð#Ñ œ## #- Ð Ñ"% #'
$ $Similarly,IÒW l œ #Ó œ Z +<ÒW l œ #Ó ÐIÒW l œ #ÓÑ œ Ð#Ñ Ð%Ñ œ#
# # ## #- - - Ð Ñ"% ('
$ $ .
From part (a), we know the posterior distribution of given , so that- W œ #"
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 ÐÞ&%$#Ñ ÐÞ%&')Ñ œ "'Þ#(*'Ð Ñ Ð Ñ#' ('$ $ .
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 , we first note that .Z +<Ò IÒW l Ó lW œ #Ó 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),TÐ œ "lW œ #Ñ œ Þ&%$# T Ð œ #lW œ #Ñ œ Þ%&')- -" " and .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 , we note that .IÒ Z +<ÒW l Ó lW œ #Ó 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 thatIÒ 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 riskR 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 riskWe 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 IÒWl Ó † T Ò lR œ # ß \ œ " ß \ œ $Óselected risk is from Class A Class A " #
selected risk is from Class B Class B .IÒWl Ó † T Ò lR œ # ß \ œ " ß \ œ $Ó" #
If the selected risk is from Class A, then has a compound Poisson distribution with PoissonWparameter (frequency) 1.0 and has a claim amount distribution (severity) with a mean of 1.0 (weare also told that the claim amount has an exponential distribution). It follows thatIÒWl Ó œ œ "Class A (Poisson parameter)(expected claim amount) .In a similar way, we get Class B .IÒWl Ó œ Ð$Þ!ÑÐ$Þ!Ñ œ *
Once the conditioning relationship is set up, most of the work in this problem is in finding theconditional probabilities Class AT Ò lR œ # ß \ œ " ß \ œ $Ó" #
and Class B .T Ò lR œ # ß \ œ " ß \ œ $Ó" #
This requires the use of rules for conditional probability.T Ò lR œ # ß \ œ " ß \ œ $Ó œClass A and" #
T Ò ∩ÐRœ# ß\ œ" ß\ œ$ÑÓT ÒRœ# ß\ œ" ß\ œ$Ó
Class A " #
" #
T Ò ∩ ÐR œ # ß \ œ " ß \ œ $ÑÓ œ T ÒR œ # ß \ œ " ß \ œ $l Ó † T Ò ÓClass A A A ." # " #
If it is known that the risk is from Class A then we know the distributions of and , so we canR \calculate A . We are told that a risk is selected randomly. ThisT ÒR œ # ß \ œ " ß \ œ $l Ó" #
means that risk Classes A and B are equally likely to be chosen, so that A . ThenT Ò Ó œ "#
T Ò ∩ ÐR œ # ß \ œ " ß \ œ $ÑÓ œ T ÒR œ # ß \ œ " ß \ œ $l Ó † T Ò ÓClass A A A" # " #
From independence of and the 's, it follows thatR \TÒR œ # ß \ œ " ß \ œ $l Ó œ T ÒR œ #l Ó † T Ò l Ó" # A A Two Claim amounts are 1, 3 A .
Since claim amount has a continuous (exponential distribution), we use the density\0Ð\ œ "lEÑ T Ò\ œ "l Ó \" " # instead of probability A (and the same goes for ).From the table of distributions made available with Exam C, we know the probability function forthe Poisson distribution with mean is , and the density function for the- T ÒR œ 5Ó œ / †- -5
5x
exponential distribution with mean is .) 0ÐBÑ œ /")BÎ)
Then AT ÒR œ # ß \ œ " ß \ œ $l Ó" #
œ T ÒR œ #l Ó † T Ò l Ó œ Ð/ † Ñ ‚ #Ð" † / ÑÐ" † / Ñ œ # †A Two Claim amounts are 1, 3 A ." " $" /#x #
# &
The factor of 2 arises from the two combinations of claims of amounts 1 and 3 (first claim isamount 1 and second claims is amount 3, and first claim is amount 3 and second claim isamount 1).
It follows that Class AT Ò ∩ ÐR œ # ß \ œ " ß \ œ $ÑÓ œ # œ # †" # Ð ÑÐ Ñ/ " /# # %
& &
(this is the numerator of the probability Class A that we are tryingT Ò lR œ # ß \ œ " ß \ œ $Ó" #
to find).
17 continuedIn order to find the denominator we use the fact that the selectedT ÒR œ # ß \ œ " ß \ œ $Ó" #
risk must be either from Class A or from Class B, so thatT ÒR œ # ß \ œ " ß \ œ $Ó" #
œ T Ò ∩ ÐR œ # ß \ œ " ß \ œ $ÑÓ T Ò ∩ ÐR œ # ß \ œ " ß \ œ $ÑÓClass A Class B" # " #
We find Class B in the same way we foundT Ò ∩ ÐR œ # ß \ œ " ß \ œ $ÑÓ" #
T Ò ∩ ÐR œ # ß \ œ " ß \ œ $ÑÓClass A ." #
T Ò ∩ ÐR œ # ß \ œ " ß \ œ $ÑÓ œ T ÒR œ # ß \ œ " ß \ œ $l Ó † T Ò ÓClass B B B ." # " #
Using the Poisson frequency (mean 3) and exponential severity (mean 3) from risk Class B, we
get B ,T ÒR œ # ß \ œ " ß \ œ $l Ó œ Ð/ † Ñ ‚ #Ð † / ÑÐ † / Ñ œ # †" #$ "Î$ $Î$$ /
#x #
# "$Î$" "$ $
and then Class B .T Ò ∩ ÐR œ # ß \ œ " ß \ œ $ÑÓ œ #Ð ÑÐ Ñ œ # †" #/ /
# %
"$Î$ "$Î$"#
We then have T ÒR œ # ß \ œ " ß \ œ $Ó œ # † Ò Ó Þ" #/ /% %
& "$Î$
Now we can findT Ò lR œ # ß \ œ " ß \ œ $Ó œClass A " #
T Ò ∩ÐRœ# ß\ œ" ß\ œ$ÑÓT ÒRœ# ß\ œ" ß\ œ$Ó
Class A " #
" #
œ œ Þ$$*##†
#†Ò Ó
/&
%
/ /&
% %
"$Î$ .
We can use the same reasoning to find Class B , which will turnT Ò lR œ # ß \ œ " ß \ œ $Ó" #
out to be .#†
#†Ò Ó
/"$Î$
%
/ /&
% %
"$Î$ œ Þ''!)
Alternatively, and , once Class A is known,more efficiently T Ò lR œ # ß \ œ " ß \ œ $Ó" #
T Ò lR œ # ß \ œ " ß \ œ $ÓClass B is equal to its complement" #
T Ò lR œ # ß \ œ " ß \ œ $Ó œ " T Ò lR œ # ß \ œ " ß \ œ $ÓClass B Class A ." # " #
Finally, the posterior expected value we are looking for isIÒWlR œ # ß \ œ " ß \ œ $Ó" #
œ IÒWl Ó † T Ò lR œ # ß \ œ " ß \ œ $Óselected risk is from Class A Class A " #
selected risk is from Class B Class BIÒWl Ó † T Ò lR œ # ß \ œ " ß \ œ $Ó" #
. œ Ð"ÑÐÞ$$*#Ñ Ð*ÑÐÞ''!)Ñ œ 'Þ#*The main work in this problem was in finding Class A .T Ò lR œ # ß \ œ " ß \ œ $Ó" #
The following "table" summarizes the steps outlined to find that probability.
, given , givenT ÒEÓ œ T ÒFÓ œ" "# #
Ì
TÒR œ # ß \ œ " ß \ œ $l Ó œ #Ð/ † ÑÐ" † / ÑÐ" † / Ñ œ # †" #" " $A "
#x
# /#
&
and
T ÒR œ # ß \ œ " ß \ œ $l Ó œ #Ð/ † ÑÐ † / ÑÐ † / Ñ œ # †" #$ "Î$ $Î$B $
#x
# " " /$ $ #
"$Î$
are both know from the given distributions Ì
TÒ ∩ ÐR œ # ß \ œ " ß \ œ $ÑÓ œ #Ð ÑÐ Ñ œ # †Class A and" #/ " /# # %
& &
T Ò ∩ ÐR œ # ß \ œ " ß \ œ $ÑÓ œ #Ð ÑÐ Ñ œ # †Class B ." #/ " /
# # %
"$Î$ "$Î$
(this calculation uses the compound Poisson distribution)
17 continued ÌTÒR œ # ß \ œ " ß \ œ $Ó" #
œ T Ò ∩ ÐR œ # ß \ œ " ß \ œ $ÑÓ T Ò ∩ ÐR œ # ß \ œ " ß \ œ $ÑÓClass A Class B" # " #
œ # † # †/%
& /%
"$Î$
Ì
TÒ lR œ # ß \ œ " ß \ œ $Ó œClass A " #T Ò ∩ÐRœ# ß\ œ" ß\ œ$ÑÓ
T ÒRœ# ß\ œ" ß\ œ$ÓClass A " #
" #
œ œ Þ$$*##†
#†Ò Ó
/&
%
/ /&
% %
"$Î$ and
T Ò lR œ # ß \ œ " ß \ œ $Ó œ " T Ò lR œ # ß \ œ " ß \ œ $Ó œ Þ''!)Class B Class A ." # " #
Answer: D
18. This is a standard Bayesian estimation question. We are given that the conditionaldistribution of (claim count for an individual) given is Poisson with mean ; and we areR A Agiven that is a two point random variable that can take on the values 2 and 4 withAT Ò œ #Ó œ Þ& T Ò œ %Ó œ Þ&A A and (this means that any randomly chosen individual has anequal chance of being 2 or 4). We are given that for a randomly chosenA R œ R œ %" #
individual ( and being the claim count for years 1 and 2, respectively).R R" #
The Bayesian estimate of the insured's claim count in year 3 is the expected number of claims forthat insured in year 3 given the information about years 1 and 2, so we are asked to findIÒR lR œ R œ %Ó$ " # . The typical way to find this expectation is to condition over ; we canAwrite the expectation asIÒR l œ #Ó † T Ò œ #lR œ R œ %Ó IÒR l œ %Ó † T Ò œ %lR œ R œ %Ó$ " # $ " #A A A A .(Note that if the distribution of had been continuous, say uniform on the interval ,A Ð#ß %Ñ
then we could write as ,IÒR lR œ R œ %Ó IÒR l œ Ó † 0Ð lR œ R œ %Ñ .$ " # $ " ##
%' A - - -
and we would have to find the conditional density of given ; this conditionalA R œ R œ %" #
distribution of would not necessarily be uniform).A
Since given has a Poisson distribution, we have and .R IÒR l œ #Ó œ # IÒR l œ %Ó œ %A A A$ $
We must find the conditional probabilities andT Ò œ #lR œ R œ %ÓA " #
T Ò œ %lR œ R œ %ÓA " # . We now use standard methods of conditional and joint probability (ordensity in the continuous case). We use the relationshipT ÒElFÓ œ œ œ
TÒE∩FÓ T ÒFlEÓ†T ÒEÓ T ÒFlEÓ†T ÒEÓT ÒFÓ T ÒF∩EÓT ÒF∩EÓ T ÒFlEÓ†T ÒEÓT ÒFlEÓ†T ÒEÓ
,with events and defined as follows:E FE œ # F R œ R œ % E E œ %: , : , and (complement of ): . ThenA A" #
w
T Ò œ #lR œ R œ %Ó œA " #T ÒÐ œ#Ñ∩ÐR œR œ%ÑÓ
T ÒR œR œ%ÓA " #
" #
œT ÒR œR œ% œ#Ó†T Ò œ#Ó
T ÒR œR œ% œ#Ó†T Ò œ#ÓT ÒR œR œ% œ%Ó†T Ò œ%Ó" #
" # " #
|| |
A AA A A A .
We are given that , and the distribution of given isT Ò œ #Ó œ T Ò œ %Ó œ Þ& R œA A A -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 areindependent of one another also.
18 continuedThis assumption is always made in the Bayesian estimation context, but note that it is notgenerally true that the unconditional distributions of numbers of claims in separate years areindependent. What is meant here is that if the value of is known, then the conditionalAdistributions and are independent, but the unconditional distributions of and R l R l R R" # " #A Aare not generally independent.ThereforeT ÒR œ R œ %l œ #Ó œ T ÒR œ %l œ #Ó † T ÒR œ %l œ #Ó œ" # " #A A A Ð ÑÐ Ñ/ †# / †#
%x %x
# % # %
, and
T ÒR œ R œ %l œ %Ó œ" # A Ð ÑÐ Ñ/ †% / †%%x %x
% % % % .
Then, T Ò œ #lR œ R œ %Ó œA " #Ð ÑÐ ÑÐÞ&Ñ
Ð ÑÐ ÑÐÞ&ÑÐ ÑÐ ÑÐÞ&Ñ
/ †# / †## % # %
%x %x/ †# / †# / †% / †%# % # % % % % %
%x %x %x %x
œ œ Þ"(&)""Ð/ †# Ñ# % # .
Since is either 2 or 4, is the complement of A A AT Ò œ %lR œ R œ %Ó T Ò œ #lR œ R œ %Ó" # " #
so that .T Ò œ %lR œ R œ %Ó œ " T Ò œ #lR œ R œ %Ó œ Þ)#%#A A" # " #
Finally, IÒR lR œ R œ %Ó$ " #
œ IÒR l œ #Ó † T Ò œ #lR œ R œ %Ó IÒR l œ %Ó † T Ò œ %lR œ R œ %Ó$ " # $ " #A A A A .œ Ð#ÑÐÞ"(&)Ñ Ð%ÑÐÞ)#)%Ñ œ $Þ'%) .
The main work in the solution of this problem was to find andT Ò œ #lR œ R œ %ÓA " #
T Ò œ %lR œ R œ %ÓA " # . The calculations needed to find these probabilities can be summarizedas follows. T Ò œ #Ó œ Þ& T Ò œ %Ó œ Þ&A A and and T ÒR œ R œ %l œ #Ó œ T ÒR œ R œ %l œ %Ó œ" # " #
#A AÐ Ñ Ð Ñ/ †# / †%%x %x
## % % %
The probabilities for were given explicitly, and the conditional probabilities follow since isA RPoisson given .A Ì Ì
TÒÐR œ R œ %Ñ ∩ Ð œ #ÑÓ œ T ÒR œ R œ %l œ #Ó † T Ò œ #Ó œ Ñ" # " ##A A A Ð Ñ Ð/ †# "
%x #
# %
and
T ÒÐR œ R œ %Ñ ∩ Ð œ %ÑÓ œ T ÒR œ R œ %l œ %Ó † T Ò œ % œ Ó" # " ##A A A Ð Ñ Ð Ñ/ †% "
%x #
% %
. ÌTÒR œ R œ %Ó œ T ÒÐR œ R œ %Ñ ∩ Ð œ #ÑÓ T ÒÐR œ R œ %Ñ ∩ Ð œ #ÑÓ" # " # " #A A
œ Ð Ñ Ð Ñ Ð Ñ Ð Ñ/ †# " / †% "%x # %x #
# % % %# # .
Ì
TÒ œ #lR œ R œ %Ó œ œ Þ"(&)A " #Ð Ñ Ð Ñ
Ð Ñ Ð ÑÐ Ñ Ð Ñ
/ †# "# %
%x ##
/ †# " / †% "# % % %
%x # %x ## #
and
T Ò œ %lR œ R œ %Ó œ " T Ò œ #lR œ R œ %Ó œ œ Þ)#%#A A" # " #Ð Ñ Ð Ñ
Ð Ñ Ð ÑÐ Ñ Ð Ñ
/ †% "% %
%x ##
/ †# " / †% "# % % %
%x # %x ## #
.
Answer: C
19. We wish to find . This can be written asT Ò\ $!l\ œ #!Ó# "
T Ò\ $!l\ œ #!Ó œ# "0 ÒÐ\ $!Ñ∩Ð\ œ#!ÑÓ
0Ð\ œ#!Ñ# "
" .
The numerator is0ÒÐ\ $!Ñ ∩ Ð\ œ #!Ñl œ "ÓT Ð œ "Ñ# " α α 0ÒÐ\ $!Ñ ∩ Ð\ œ #!Ñl œ #ÓT Ð œ #Ñ# " α α , 0ÒÐ\ $!Ñ ∩ Ð\ œ #!Ñl œ $ÓT Ð œ $Ñ# " α α
19 continuedœ Ð ÑÒ0 ÒÐ\ $!Ñ ∩ Ð\ œ #!Ñl œ "Ó 0ÒÐ\ $!Ñ ∩ Ð\ œ #!Ñl œ #Ó
"$ # " # "α α
0ÒÐ\ $!Ñ ∩ Ð\ œ #!Ñl œ $ÓÑ# " α
and the denominator is0Ð\ œ #!l œ "ÓT Ð œ "Ñ 0Ð\ œ #!l œ #ÑT Ð œ "Ñ 0Ð\ œ #!l œ $ÑT Ð œ $Ñ" " "α α α α α α .œ Ð ÑÒ0Ð\ œ #!l œ "Ñ 0Ð\ œ #!l œ #Ñ 0Ð\ œ #!l œ #ÑÓ
"$ " " "α α α
The Pareto distribution with has pdf ,) œ "! 0ÐBÑ œ α"!ÐB"!Ñ
α
α"
and cdf .JÐBÑ œ " Ð Ñ"!B"!
α
0Ò\ œ #!l œ "Ó œ ß 0Ò\ œ #!l œ #Ó œ ß 0Ò\ œ #!l œ $Ó œ" " "α α α"! #!! $!!!$! $! $!# $ % ,
so that is the denominator.0Ò\ œ #!Ó œ œ Þ!!(%"" Ð ÑÒ Ó" "! #!! $!!!$ $! $! $!# $ %
From conditional independence of the 's given , we have\ α0ÒÐ\ $!Ñ ∩ Ð\ œ #!Ñl œ "Ó œ 0Ò\ $!l œ "Ó † 0 Ò\ œ #!l œ "Ó# " # "α α α ,and the same for , so thatα œ #ß $
0ÒÐ\ $!Ñ ∩ Ð\ œ #!Ñl œ "Ó œ T Ò\ $!l œ "Ó † 0 Ò\ œ #!l œ "Ó œ# " # "α α α Ð ÑÐ Ñ"! "!$!"! $!# ,
and and0ÒÐ\ $!Ñ ∩ Ð\ œ #!Ñl œ #Ó œ# " α Ð Ñ Ð Ñ"! #!!$!"! $!
#$
0 ÒÐ\ $!Ñ ∩ Ð\ œ #!Ñl œ Ó œ# " α 3 .Ð Ñ Ð Ñ"! $!!!$!"! $!
3%
Then, 0ÒÐ\ $!Ñ ∩ Ð\ œ #!ÑÓ œ Ó# "#Ð ÑÒÐ ÑÐ Ñ Ð Ñ Ð Ñ Ð Ñ Ð Ñ" "! "! "! #!! "! $!!!
$ $!"! $! $!"! $! $!"! $!# $ %3
œ Þ!!""! is the numerator, and thenT Ò\ $!l\ œ #!Ó œ œ œ Þ"%)# "
0 ÒÐ\ $!Ñ∩Ð\ œ#!ÑÓ0Ð\ œ#!Ñ Þ!!(%"
Þ!!""!# "
" . Answer: C
20. Since we are given one observation (total annual losses of 500), the Bayesian\ œ"
premium is . Since the "parameter" in this Bayesian situation is the class A orIÒ\ l\ œ &!!Ó# "
B, the prior distribution is (risks A and B are equally likely). For each ofTÐEÑ œ TÐFÑ œ Þ&the two classes, total annual loss has a compound distribution.The Bayesian premium can be formulated asIÒ\ l\ œ &!!Ó œ IÒ\ lEÓ † T ÒEl\ œ &!!Ó IÒ\ lFÓ † T ÒFl\ œ &!!Ó# " # " # " .Within class A, .IÒ\ lEÓ œ Ð! † " † # † ÑÐ&!! † "#$& † Ñ œ ''!#
% % " " #* * * $ $
Within class B, .IÒ\ lFÓ œ Ð! † " † # † ÑÐ#&! † $#) † Ñ œ $')#" % % # "* * * $ $
T ÒEl\ œ &!!Ó œ œ Þ"T ÒE∩Ð\ œ&!!ÑÓ T Ò\ œ&!!lEÓ†T ÒEÓ
T Ò\ œ&!!Ó T Ò\ œ&!!lEÓ†T ÒEÓT Ò\ œ&!!lFÓ†T ÒFÓ" "
" " "
T Ò\ œ &!!lEÓ œ T Ò Ó † T Ò Ó œ œ" 1 claim claim of 500 ,Ð ÑÐ Ñ% " %* $ #(
T Ò\ œ &!!lFÓ œ T Ò Ó † ÐT Ò ÓÑ œ œ"# #2 claims claim of 250 .Ð ÑÐ Ñ% # "'
* $ )"
Then, .T ÒEl\ œ &!!Ó œ œ"Ð ÑÐ Ñ
Ð ÑÐ ÑÐ ÑÐ Ñ
% "#( #
% " "' "#( # )" #
$(
In a similar way, .T ÒFl\ œ &!!Ó œ"%(
Then, . Answer: AIÒ\ l\ œ &!!Ó œ Ð''!Ñ Ð$')Ñ œ %*$Þ"# " Ð Ñ Ð Ñ$ %( (
21. There are 4 possible die/spinner pairs, each with prior probability of :"%
T ÐEß\Ñ œ TÐEß ] Ñ œ TÐFß\Ñ œ TÐFß ] Ñ œ"% . The associated expected losses are
IÒPlEß\Ó œ Ð ÑÐ ÑÐ"# -Ñ ß IÒPlEß ] Ó œ Ð ÑÐ"#Ñ ß$ "% #
$%
IÒPlFß\Ó œ Ð ÑÐ ÑÐ"# -Ñ ß IÒPlFß ] Ó œ Ð ÑÐ"#Ñ" " "% # % .
In order to find the posterior expectation we use the relationshipIÒP lP œ "#Ó# "
IÒP lP œ "#Ó œ IÒP lEß\Ó † T ÒEß\lP œ "#Ó IÒP lEß ] Ó † T ÒEß ] lP œ "#Ó# " # " # "
IÒP lFß\Ó † T ÒFß\lP œ "#Ó IÒP lFß ] Ó † T ÒFß ] lP œ "#Ó# " # "
œ Ð ÑÐ"# -Ñ † T ÒEß\lP œ "#Ó Ð*Ñ † T ÒEß ] lP œ "#Ó$) " "
Ð ÑÐ"# -Ñ † T ÒFß\lP œ "#Ó Ð$Ñ † T ÒFß ] lP œ "#Ó") " " .
To find the posterior probabilities, we first findT ÒP œ "#Ó œ T ÒEß\ß "#Ó T ÒEß ] ß "#Ó T ÒFß\ß "#Ó T ÒFß ] ß "#Ó"
œ T Ò"#lEß\Ó † T ÐEß\Ñ T Ò"#lEß ] Ó † T ÐEß ] Ñ T Ò"#lFß\Ó † T ÐFß\Ñ T Ò"#lFß ] Ó † T ÐFß ] Ñ
œ ÒÐ ÑÐ Ñ Ð ÑÐ"Ñ Ð ÑÐ Ñ Ð ÑÐ"ÑÓÐ Ñ œ Þ$ " $ " " " " $% # % % # % % )
The posterior probabilities are then
T ÒEß\lP œ "#Ó œ œ œ ß"T ÒP œ"#lEß\Ó†T ÐEß\Ñ
T ÒP œ"#Ó %
Ð ÑÐ Ñ†Ð Ñ ""
"
$ " "% # %
$)
T ÒEß ] lP œ "#Ó œ œ œ ß"T ÒP œ"#lEß] Ó†T ÐEß] Ñ
T ÒP œ"#Ó"
"
Ð ÑÐ"Ñ†Ð Ñ "#
$ "% %
$)
T ÒFß\lP œ "#Ó œ œ œ ß"T ÒP œ"#lFß\Ó†T ÐFß\Ñ
T ÒP œ"#Ó "#
Ð ÑÐ Ñ†Ð Ñ ""
"
" " "% # %
$)
T ÒFß ] lP œ "#Ó œ œ œ"T ÒP œ"#lFß] Ó†T ÐFß] Ñ
T ÒP œ"#Ó '
Ð ÑÐ"Ñ†Ð Ñ ""
"
" "% %
$)
.
The posterior expectation isIÒP lP œ "#Ó œ Ð ÑÐ"# -Ñ † Ð Ñ Ð*Ñ † Ð Ñ Ð ÑÐ"# -Ñ † Ð Ñ Ð$Ñ † Ð Ñ# "
$ " " " " ") % # ) "# ' .
We are given that this is equal to 10. Solving for results in . Answer: E- - œ $'
22. We first fine the posterior probabilities for the parameter :-TÐ œ "lW œ #Ñ œ- "
T ÐW œ#∩ œ"ÑT ÐW œ#Ñ"
"
-
1 claim for amount 2 2 claims for amount 1 eachTÐW œ #l œ "Ñ œ TÐ l œ "Ñ TÐ l œ "Ñ" - - -
œ / † † † œ" " / " "$ # $ $
" (/")
"
.
TÐW œ # ∩ œ "Ñ œ TÐW œ #l œ "Ñ † T Ð œ "Ñ œ † œ Þ" ""#- - - (/ (/
") $'
" "
T ÐW œ #l œ #Ñ œ TÐ l œ #Ñ TÐ l œ #Ñ" - - -1 claim for amount 2 2 claims for amount 1 eachœ / † # † † † œ# " " "
$ $ $/ †# )/
# *
# # #
.
TÐW œ # ∩ œ #Ñ œ TÐW œ #l œ #Ñ † T Ð œ #Ñ œ † œ" ""#- - - )/ %/
* *
# #
Þ
T ÐW œ #Ñ œ TÐW œ # ∩ œ "Ñ TÐW œ # ∩ œ #Ñ œ " " "- - (/ %/$' *
" #
.
TÐ œ "lW œ #Ñ œ œ œ Þ&%$#- "T ÐW œ#∩ œ"Ñ
T ÐW œ#Ñ $' $' *(/ (/ %/"
"
" " #- Ð Ñ Ð Ñ‚and .TÐ œ #lW œ #Ñ œ " TÐ œ "lW œ #Ñ œ Þ%&')- -" "
The Bayesian premium isIÒW lW œ #Ó œ IÒW l œ "Ó † T Ð œ "lW œ #Ñ IÒW l œ #Ó † T Ð œ #lW œ #Ñ# " # " # "- - - - .œ Ð"ÑÐ#ÑÐÞ&%$#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"Ñ Ð#Ñ Ð%Ñ œ #Ð Ñ Ð Ñ Ð Ñ" " "# % %
(b)(i) IÒ\ Ó œ IÒ\ lMÓ † T ÐMÑ IÒ\ lMMÓ † T ÐMMÑ IÒ\ lMMMÓ † T ÐMMMÑ# # # #
œ Ð" " Ñ Ð# # Ñ Ð% % Ñ œ (Þ&# # #Ð Ñ Ð Ñ Ð Ñ" " "# % %
Z +<Ò\Ó œ (Þ& # œ $Þ&#
(ii) and prob probprob. prob.prob. prob.
IÒ\lX Ó œ Z +<Ò\lX Ó œ
" X œ M ß " X œ M ß
# X œ MM ß # X œ MM ß
% X œ MMM ß % X œ MMM ß
. .
.
" "# #" "% %" "% %
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) Type ITÐ l\ œ "Ñ œTÐ\œ"l †T Ð Ñ
T Ð\œ"ÑType I) Type I .
Type I, Type II , Type III, " " "# % %
T Ð\ œ "l œ / TÐ\ œ "l œ #/ T Ð\ œ "l œ %/Type I) Type II) Type I)" # %
T Ð\ œ " ∩ Ñ T Ð\ œ " ∩ Ñ T Ð\ œ " ∩ ÑType I Type II Type IIIœ / œ / œ /" "
# #" # %
TÐ\ œ "Ñ œ / / /" "# #
" # %
T Ð l\ œ "Ñ œ œ Þ')" ßType I "#
"
" "# #
" # %
/
/ / /
T Ð l\ œ "Ñ œ œ Þ#&" ßType II "#
#
" "# #
" # %
/
/ / /
T Ð l\ œ "Ñ œ œ Þ!')Type III .// / /
%
" "# #
" # %
(d) Type I Type I Type II Type IIIÒ\ l\ œ "Ó œ IÒ\ l Ó † T Ð l\ œ "Ñ IÒ\ l Ó † T Ð l\ œ "Ñ# " # " # "
Type III Type IIIIÒ\ l Ó † T Ð l\ 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\ l Ó † T Ð l\ œ "Ñ IÒ\ l Ó † T Ð l\ œ "Ñ# # ## # #
" " "Type I Type I Type II Type II Type III Type IIIIÒ\ l Ó † T Ð l\ 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\ l Ó l\ œ "Ó Z +<Ò IÒ\ l Ó l\ œ "Ó# " # " # "Type Type .
Z +<Ò\ l Ó œZ +<Ò\ l Ó œ " TÐ l\ œ "Ñ œ Þ')"Z +<Ò\ l Ó œ # TÐ l\ œ "Ñ œ Þ#&"Z +<Ò\ l Ó œ TÐ
#
# "
# "
#
TypeType I prob. Type IType II prob. Type IIType III 4 prob. Type IIIl\ œ "Ñ œ Þ!')"
Type .p IÒ Z +<Ò\ l Ó l\ œ "Ó œ Ð"ÑÐÞ')"Ñ Ð#ÑÐÞ#&"Ñ Ð%ÑÐÞ!')Ñ œ "Þ%&&# "
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TypeType I prob. Type IType II prob. Type IIType III 4 prob. Type III "Ñ œ Þ!')
Z +<Ò IÒ\ l Ó l\ œ "Ó# "Typeœ 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""# " .
