TABLE OF CONTENTS - VOLUME 2 - University of Torontoutstat.utoronto.ca/sam/coorses/act466/ACT466 Notes.pdf · ©ACTEX 2007 SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial

© ACTEX 2007 SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models

TABLE OF CONTENTS - VOLUME 2

CREDIBILITY

SECTION 1 - LIMITED FLUCTUATION CREDIBILITY CR-1PROBLEM SET 1 CR-17

SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR CR-27PROBLEM SET 2 CR-37

SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR CR-47PROBLEM SET 3 CR-59

SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR CR-79PROBLEM SET 4 CR-89

SECTION 5 - BAYESIAN CREDIBILITY APPLIED TO THE EXAM C TABLE DISTRIBUTIONS CR-99PROBLEM SET 5 CR-111

SECTION - BUHLMANN CREDIBILITY CR-127PROBLEM SET 6 CR-137

SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS CR-159PROBLEM SET 7 CR-169

SIMULATION

SECTION 1 - THE INVERSE TRANSFORMATION METHOD SI-1PROBLEM SET 1 SI-9

SECTION 2 - THE BOOTSTRAP METHOD SI-23PROBLEM SET 2 SI-35

SECTION 3 - THE LOGNORMAL DISTRIBUTION AND ASSET PRICES SI-41PROBLEM SET 3 SI-49

SECTION 4 - MONTE CARLO SIMULATION SI-53PROBLEM SET 4 SI-61

SECTION 5 - RISK MEASURES STUDY NOTE SI-63PROBLEM SET 5 SI-71

CREDIBILITY

CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY CR-1


CREDIBILITY - SECTION 1, LIMITED FLUCTUATION CREDIBILITY

The material in this section relates to Section 16.3 of "Loss Models". The suggested time framefor this section is 3-4 hours.

CR-1.1 Introductory Comments on Credibility Theory

The objective of credibility theory as it is covered on Exam C is to estimate the mean of a randomvariable from a random sample . Credibility estimation can be applied to anyrandom variable. There are several approaches to credibility estimation that are considered. Mostof the approaches that we look at assume that is part of a Bayesian framework. Theseapproaches are the Bayesian approach to credibility, the Buhlmann approach to credibility, andthe non-parametric empirical approach and the semi-parametric approach. These will beconsidered later. The first approach to credibility estimation that we consider is "LimitedFluctuation Credibility", also called "classical credibility", and does not involve any Bayesiancomponent.

The Mahler-Dean study note introduces some notation to refer to the mean and variance ofvarious random variables, and the "Loss Models" book also has some of its own notation. Forinstance, the M-D study note uses for the mean and for the variance in some cases.A "frequency distribution" is a non-negative integer valued random variable that represents thenumber of claims occurring in a specific period of time (claims per month, for instance). TheM-D study note uses and to denote the mean and variance of a frequency distribution.A "severity distribution" is a non-negative random variable that represents the size of anindividual loss. The M-D study note uses and to denote the mean and variance of a severitydistribution. In the M-D study note, the aggregate amount of claims (or losses) in a period of timeis referred to as the "pure premium" with mean and variance denoted and . Theaggregate amount of claim in a period might be a compound distribution (a combination of aclaim number random variable , and a severity per claim random variable ; compounddistributions were reviewed in the Modeling unit of this study guide).

In these notes the mean of a random variable will be denoted or or , and thevariance will be denoted or or . An observed value of may be referred to asan exposure (or exposure unit) of .

If is the number of claims on an insurance policy occurring in one month (the frequency permonth), then an observation of is a non-negative integer of the number of claims for aparticular month. " exposures of " can refer to either of the following two interpretations:(i) a single insurance policy is observed for separate months; each month results in a number ofclaims, and would be the numbers of claims in month 1 would be the numbers of claimsin month 2 would be the numbers of claims in month , (one policy, months), or(ii) separate insurance policies, all of the same type, are observed for one month;then would be the numbers of claims in that month forpolicy 1, policy 2, ..., policy ( policies, one month).

If is the amount of one loss on an insurance policy, then " exposures" of is observedlosses. These might be separate losses for a particular policy, or these might be one loss amountfor each of separate policies of the same type.

CR-2 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY


If is the aggregate loss occurring in one month, then " exposures" can refer to the aggregateloss amounts that occur over the months (one aggregate loss amount each month,

). " exposures" can also refer to aggregate losses from separate policiesduring one month with an aggregate loss for each policy.

In general, an exposure for a particular random variable is one observation of that randomvariable.

CR-1.2 Limited Fluctuation Credibility Theory - The Standard for Full Credibility

Suppose that the random variable being analyzed is , with mean (usually isunknown) and variance , and suppose that a sample of independent observations

is available. Suppose that there is a of amount for themanual premiumclaim distribution The manual premium is an amount which has been determined by some pastexperience and underwriting expertise. is not necessarily equal to , but it is somepreconceived estimate since we usually don't know the value of .

Two Equivalent Full Credibility StandardsUnder the full credibility approach, the estimate of (also called the premium estimate for

) is chosen as one of the following two possible values:(a) (the sample mean of ) , and we say that , orfull credibility is applied(b) .

The decision as to which of or is chosen to use as the estimate of is based on how "close" is to . If is "small enough" then is chosen as the estimate of . Note that

may be referred to as the pure premium, and may be referred to as the expected purepremium. "Small enough" in this context means that is less than some fraction of ;we want to have . Since the 's are random variables, we will never know whenthis closeness criterion is satisfied. What we can do is try to get a probability that the closenessrequirement is satisfied. The meaning of "close" is based on two quantities

usually , but other values of are possible, such as .02 or .1range parameter :

usually , but other values of are possible, such as .95 .probability level :

We say that the if the probability relationfull credibility standard is satisfied is satisfied . (1.1)This means that the probability is at least that the absolute deviation of from is lessthan the fraction of the mean.We saw earlier in the estimation review that the sample mean of a random sample satisfies thefollowing relationships: and (1.2)



As gets larger, gets smaller, and becomes more likely to be "close to" . It is notsurprising, therefore, that the application of full credibility is related to a "large enough" , thenumber of sample values (or exposures) of . If we have a large enough number of samplevalues of , we can be confident that the sample mean is "close enough" to sothat can be used as a "credible" estimate of . Thus, the full credibility requirement that" is close enough to " is translated into a requirement that " , the number of sampleobservations of is large enough". The comments that follow make the notion of "large enough

" more precise algebraically.

After the factors and have been chosen, we define the quantities and as follows:

(i) : . This quantity may be difficult to find for a general

random variable . However, if is assumed to be (approximately) normal, then is the percentile of the standard normal distribution. (1.3)

A way of describing this in a little less technical way is: given probability , we find thevalue such that the probability , where has a standard normaldistribution.

For instance, with , we know that , so that (1.645 is the 95-th percentile of the standard normal distribution so that 5% probability is in thetail to the right of 1.645 and 5% probability is in the tail to the left of 1.645). If we choose

, then , which is the . percentile of the standard normal distribution, and ifwe choose then .

(ii) (1.4)If is assumed to be (approximately) normal, and and (the two mostcommonly used values for a full credibility standard), then . (1.5)

The reason for this is as follows. We are assuming that is approximately normal, and we want.

This probability can be written in the form .

Since has a distribution which is approximately standard normal, in order for the

probability inequality to be satisfied, it must be true that . This inequality can be

rewritten in the form . This is the number of observations of needed for theprobability inequality to be satisfied. If the number of observations of is , where

, then the probability inequality is satisfied, and we regard the sample mean to be a "fully credible" estimate of .



This is restated in the following way. For a random variable , full credibility is given to (meaning that , the number of observations of is large enough so that is chosen as the

_

estimated premium), if the following condition is satisfied

The Standard For Full Credibility

(1) (square of coefficient of variation) where is the number of observations of (1.6)

This condition is equivalent to the following condition

(2) the sum of all observed -values (1.7)

To get some insight into the meaning of full credibility, and the equivalence of conditions (1) and(2), we consider the following simple situation. Suppose the random variable has mean

and variance , and suppose that we have chosen the .05 "closeness"criterion and the 90% probability criterion. Then the value of is (the approximatevalue for 1082.41, the typical value mentioned above which corresponds to and

). The two inequalities ((1) and (2)) in the equivalent standards for full credibility areinterpreted in the following way.

Using condition (1), we see that the requirement for full credibility is .This tells us that 271 or more observations of the random variable will be sufficient for fullcredibility, and the value of for that sample of 271 observations will be regarded as being"close enough" to the true value of . There is a 90% probability that the inequality

is satisfied (this is the inequality ).

Using condition (2), we have total amount of all observed 's, or (rounded) .This says that when we have a sample of observations so that the of the observed values is atsumleast 542, the sample will be sufficiently large for full credibility and again, the value of forthat sample of observations will be regarded as being "close enough" to the true value of .

The difference between conditions (1) and (2) is that in the first inequality we get the minimumnumber of observations of needed for full credibility (271 in this case). From the secondinequality we get the minimum sum of the observed values of needed (542). Since wasgiven to be 2 in this simple example, we can see that 271 observations should, on average, sum toabout (an average value of 2 per observation). The inequality in condition (2)however, doesn't specify the minimum number of observations needed, but rather specifies theminimum needed sum of observed values. Conditions (1) and (2) are statistically equivalent, andthe one that would be used might be implied by the form in which data is available. Fullcredibility standards (1) and (2) can be applied to any random variable.



In questions involving full credibility, it is important to first identify the random variable whose credibility estimate is being found. Then, in condition (1) for full credibility,

is the minimum number of occurrences, or observations, or exposures of neededto satisfy the standard for full credibility.Each represents a single exposure of , and is the sum of all observations.

is approximately equal to , so that in condition (2) for full credibility, the right handside represents the that must be reached in order tosum of all observed valuesattain full credibility.

Example CR-1: Suppose that represents the size of a claim amount, and has as exponentialdistribution with mean 5. Therefore and (the variance of an exponentialrandom variable is the square of the mean). With and , condition (1) for thestandard for full credibility can be formulated as

. Full credibility is reached as soon as 1083 are observed. An alternative equivalent formulation for full credibility is condition (2),

. This indicates that full credibility isreached when the sum of all observed s is at least 5412.05; this may occur a little before orafter 1083 claims are observed, but should occur when about 1083 claims have occurred. Forinstance, the first claim might be , the second claim might be for a total amountof claim of 11, etc. We keep a cumulative total continuing until that cumulative total reaches5412.05. At that point we have enough claims for full credibility.

Note that for condition (1) we did not need to know the mean of if we are assuming it hasan exponential distribution, because , so condition (1) becomes

. To apply condition (2) we do need to know .

CR-1.3 Full credibility applied to a frequency distribution

Suppose that the random variable to which we are applying the full credibility method isnumber of claims in a single period, also called the frequency (often is Poisson)

Then is the expected number of claims in a single period, and is the expected totalnumber of claims in claim periods. Also in this case, an "exposure" is one period of claims(one observation of ), and an observation is the .number of claims for that period

Condition (1) gives us the minimum number of periods (not the minimum number of observedclaims) needed for full credibility. Condition (2) gives the minimum of observations needed;sumthe sum of all observations in this case is the total number of claims observed in all periodsobserved (i.e. number claims in period 1 + number claims in period 2 + ). The differencebetween conditions (1) and (2) is that with condition (1) we count the number of periods ofobservation, in condition (2) we count the total number of observed claims.

Suppose we use the simple numerical example above in which and , andsuppose we interpret as the number of claims in one week. Then full credibility is met with



271 observed values of (271 weeks, with one for each week). Full credibility is also met assoon as a total of 542 claims have been observed (which could occur sooner or later than 271weeks).

Example CR-2: Suppose that represents the number of claims that occur in one day, andsuppose that the mean and variance of are . Each day represents oneexposure of , because each day we get a value of , an integer number of claims for that day.We will continue to use the and values for full credibility. Applying fullcredibility standard (1), we see that the number of observations of needed (the number of daysneeded) for full credibility is If we apply credibilitystandard (2), we see that the sum of -values that we need for full credibility is

. As soon as the total number of claims observedreaches 2,165, we have satisfied the standard for full credibility. The should occur in about 433days (5 claims per day, on average) but we might reach a total of 2,165 claims some days beforeor after 433 days.

Example CR-3: Suppose that represents the number of claims that occur in one day, andsuppose that has a Poisson distribution with mean . The variance of the Poisson distributionis the same as the mean, so also. If we apply credibility standard (1) (again with

), then full credibility is reach when the number of exposures of (the number ofdays) is at least . We would need to know inorder to know when full credibility is reached with standard (1).

If we apply credibility standard (2), we see that the total number of claims we need to observe(the sum of the -values) must be at least . As soonas we have observed a total of 1083 claims, we have satisfied the full credibility standard.

Note that we do not need to know the value of in order to apply standard (2) to thePoisson distribution.

Example CR-4: Suppose that a single die is being tossed in groups of tosses. is thenumber of 1's tossed in the -th group of tosses, and the probability of tossing a 1 on any giventoss is (assumed to be unknown). Then has a binomial distribution with mean

and variance . Each is an exposure, or singleobservation of and is a number from 0 to 10 (the number of 1's in 10 tosses). The randomvariable to which we are applying the full credibility method is the binomial random variable .The number of groups of tosses that is required to meet the standard for full credibility is foundfrom (each group of 10 tosses is an exposure and results in an between 0and 10). Using the usual values of , the standard for full credibility usingcondition (1) becomes .

This is the number of observations of that are needed for full credibility; it is the number of"groups of 10 tosses" that are needed.



The equivalent condition (2) tells us that full credibility is reached as soon as the total number of1's observed is .Since is unknown, the right-hand side using condition (2) could be as large as 1082.41. Thestandard for full credibility based on condition (2) would be regarded as having been satisfied if

. This means that the total observed number of 1's for all combined groups of 10tosses must be at least 1083 (which insures that the number of 1's observedis ). Using condition (2) the standard is based on the sum of the observations(the total number of 1's in all the groups of tosses), rather than the number of observations (eachobservation is based on a group of 10 tosses).

Note that if we know that the die was fair, then , and full credibility standard (1) would tell

us that we need groups of 10 tosses (5,412 tosses). Full credibility

standard (2) tells us that we need to observe a total of 1's in all thetosses.

CR-1.4 Review of Compound Distributions

Compound distributions were reviewed in some detail in Section 17 of the Modeling unit of thisstudy guide. A compound distribution has two component distributions :(i) is the "frequency" distribution, which is a non-negative integer random variable, and(ii) is the "severity" distribution which is a non-negative random variable (may be continuousor discrete).

The compound distribution random variable is . The usualinterpretation is that represents either the number of claims that occur in one period or thenumber of claims experienced by one policyholder, and is the size of a claim. It is generallyunderstood that and , ... are mutually independent. is the aggregate of all claimsoccurring in the period.

The mean and variance of the compound distribution are and . (1.8)

These can be found using conditioning rules of probability. If and are any two randomvariables, then it is always true that and (1.9)

. (1.10)

In the case of a compound distribution, we can find the mean and variance of by conditioningover : , and

.



CR-1.5 The Full Standard of Credibility Applied to Compound Distributions

Credibility methods can be applied to any random variable, including a compound distribution.It was pointed out earlier that for any random variable there are two equivalent full credibilitystandards. These two full credibility standards applied to the compound distribution are

(1) number of observations of needed(1.11)

where is the number of observations (exposures) of needed , and

(2) sum of all observed 's(1.12)

The interpretation applied earlier to applies here as well. Condition (1) gives the minimumnumber of observations of needed (we get one per period, or perhaps we have separatepolicies of the same type, and we get one observed value of from each policy).Condition (2) gives us the value of needed for full credibility.

When applying the full credibility method to a compound distribution, there is a thirdequivalent full credibility standard that can be used.

(3) total number of observed claims . (1.13)

Suppose that our interpretation of the compound distribution is that represents the number ofclaims in one month (frequency), and is the amount (severity) of one claim. Each month therewill be some claims, and is the number of claims for month . This third equivalent fullcredibility standard tells us the minimum needed. This third standard only applies in thecase of a compound distribution . Keep in mind that although full credibility can be reached inthree equivalent ways, we are still ultimately trying to estimate , and when we have satisfiedthe full credibility requirement based on any of the three conditions, we will use as ourestimate of , so we need to know how many months were observed, and the aggregate claimamount for each month.

In general for a compound distribution , an exposure is one period of claims, and an observation is the .total amount of claims for period

Condition (1) gives us the minimum number of periods (not the minimum number of observedclaims) needed for full credibility.

Condition (2) gives the minimum sum , the sum needed of the observed values of ; the sumof all observations in this case is the total amount of claims observed in all periods.

Condition (3) , which applies only to compound distributions, is based on the total number ofclaims needed in all periods; this is .



Example CR-5: Suppose we have a compound distribution in which and , and suppose that is the number of claims in one

month. Then , and .

Using the .05 "closeness" and .90 probability criteria, condition (1) for full credibility tells us thatwe need at least observations of . That is a minimum of 68 months forfull credibility.

Condition (2) tells us that we need at least in total claim amount forfull credibility. This means we continue observing successive months until , andwhen we reach that total we have enough observations for full credibility.

Condition (3) tells us that we need at least claims to occur forfull credibility. This means that we continue observing successive months until (total number of observed claims is at least 1354), and when we reach that point we have enoughobservations for full credibility.

Equivalent conditions (1) and (2) for full credibility appear to require knowing the value of in order to determine the full credibility standard. For certain distributions (particularly

the exponential, the Poisson and the compound Poisson, as described below), algebraicsimplification of the conditions will result in the value of not being required. Thosedistributions for (Poisson and compound Poisson) are the ones most likely to arise in aquestion involving full credibility.

CR-1.6 Full Credibility Standard Applied to a Poisson Random Variable

Suppose that the with parameternumber of claims (per period) has a Poisson distribution, so that (the random variable being considered for credibility is in

this case). If is a sample of the number of claims for the past periods, thenusing factors , we have .

The standard for full credibility based on condition (1) isnumber of observed values of needed

number of periods needed . (1.14)

The standard for full credibility based on condition (2) istotal number of claims needed . (1.15)

In order to apply condition (1) we would need to know the value of . Condition (2) can beapplied without knowing the value of . This tells us that the standard for full credibility for aPoisson random variable simplifies to the following criterion: if the total number of observedclaims (for all exposure periods combined) is at least 1083, then full credibility is satisfied. Withdifferent values of or , would be some value other than 1082.41. Note that we areapplying the full credibility method to the random variable (not ).



CR-1.7 Full Credibility Standard for When has aCompound Poisson distribution

Suppose that the claim frequency per period has a Poisson distribution with mean , and theclaim severity distribution is , with mean and variance and , and suppose that wemake the usual assumption of mutual independence of and the 's. The aggregate claim perperiod is , and has a compound Poisson distribution with mean ,and with variance

. (1.16)

The standard for full credibility for using condition (1) is(1) number of observations of needed .This minimum number of observations needed can be written as

Coeff. of Variation of (1.17)

This requires knowing both and the coefficient of variation of .

The standard for full credibility of using condition (2) is(2) total amount of all observed 's needed .This total amount needed can be written as

. (1.18)

Note that this requires that we know the mean and variance of , but we do not need to know thatvalue of .

The standard for full credibility of using condition (3) is(3) total number of observed claims needed .This total number of observed claims needed can be written as

. (1.18)

Note that this requires knowing only the coefficient of variation of but we do not need to knowthe value of (keep in mind that is the square of ) .

These equivalent full credibility standards for the can becompound Poisson distributionsummarized as follows (for and ). Full credibility is satisfied if any one of thefollowing conditions is satisfied.

(1) if the total is at least , ornumber of exposure periods

(2) if the total (for all periods or exposures combined) is at leastamount of observed claims , or



(3) if the total (for all periods or exposures combined) is at leastnumber of observed claims2 .

Since is usually unknown, but and are more likely to be known, condition (1) isusually not possible to check, so (2) or (3) is used. Again, might not be 1082.41 if differentvalues of or are used.

Example CR-6: Assume that claim frequency, , per week has Poisson distribution with mean20, and claim severity, , has an exponential distribution with a mean of 5. Frequency andseverity are assumed to be independent. In this example we investigate the various equivalentforms of the full credibility standard for the frequency ( ), severity ( ) and aggregate claim ( )distributions. We will use the values and , so that .The aggregate claim per week has mean and variance

.

(a) . The number of weeks needed for full credibility isFrequency for frequency ; this is condition (1) applied to .

The number of claims needed for is full credibility of (this is condition (2)) applied to ).

Therefore, either 55 weeks or 1083 claims will satisfy the full credibility standard for .

Note that we didn't need the value of to apply condition (2), because for thePoisson.

(b) . The number of observations of needed for full credibility isSeverity for severity (condition (1) applied to ).

The total amount (not number) of claims needed for full credibility of is (condition (2) applied to ).

Therefore, either 1083 observed values of , or a total amount of claims of 5412 will satisfy thefull credibility standard for .

Note that we did not need to know the mean and variance to apply condition (1). This is true forthe exponential distribution since , and so we get cancellation if theexpression for condition (1).

(c) . The number of weeks needed for full credibility for isAggregate Claim (condition (1) applied to ).

Since has a compound Poisson distribution, this can also be written as



Example CR-6 continuedThe total amount of claims needed for full credibility for is

(this is condition (2) for ). Since has a compound Poisson distribution, this can also be written as

.

The number of claims needed for full credibility for is

(this is condition (3) for ) . Since has a compound Poisson distribution, this can also be

written as .

Therefore, either 109 weeks, or a total amount of claims of 10,825, or a total of 2165 claims, willsatisfy the full credibility standard for .

Note that since has a Poisson distribution and has an exponential distribution, in applyingcondition (3) we do not actually need the value of for and we don't need the mean of .Since is Poisson, condition (3) reduces to 2 , and since is

exponential, , so that condition (3) becomes is the totalnumber of claims needed for full credibility.

Note: When claims follows a compound Poisson distribution (claim frequency Poisson, claimseverity independent of frequency), the (or periods) needed for fullnumber of exposurescredibility for (orclaim frequency random variable number of exposures is , and the periods) needed for full credibility for the aggregate claims random variable (or aggregateclaim amount) is . It is important to be clear about which random variable isbeing considered for full credibility. If a situation involves a compound distribution , then thatis usually the random variable to which full credibility is being applied.

Also, language that is often used in a full credibility context is something like"the observed pure premium should be within 2% of the expected pure premium 90%of the time", or "the full credibility standard is for aggregate losses to be within 5% of theexpected with probability 0.90".

Example CR-7: You are given the following: - The number of claims follows a Poisson distribution. - Claim sizes follow a lognormal distribution with parameters and . - The number of claims and claim sizes are independent. - 13,000 claims are needed for full credibility. - The full credibility standard has been selected so that the actual aggregate claims costs will be within 5% of expected aggregate claim costs 90% of the time.Determine .



Solution: It is implied by the wording of the question that 13,000 claims is the full credibilitystandard for aggregate claims . From the information given we see that has a compoundPoisson distribution. Three possible equivalent standards for full credibility for compoundPoisson (labeled (1), (2) and (3)) were summarized above. The one that we are able to usedepends on the information available. We are told that 13,000 claims are needed for fullcredibility of . This suggests that we must use condition (3), which is based on the total numberobserved claims. The full credibility standard (3) for compound Poisson aggregate claims isgiven above as claims, where is the claim size distribution. We are giventhat (90%) and (5%), so that

For the lognormal distribution with parameters and , we have , and

Then and the full credibility standard based on condition (3) is

, which we are given is . Therefore, .

Example CR-8: Suppose that for a compound Poisson claims distribution with severity , thestandard for full credibility for based on expected total amount of claims is 1800. If theseverity distribution was changed to be a constant equal to the original , the standard for fullcredibility for based on expected total amount of claims would be 1200. Assuming

, find the variance of the severity distribution for the original .Solution: We are given , where is the claim amount random variable.The value of 1800 is the total claim amount needed for full credibility for , the random variableof aggregate claim per period when . This is full credibility standard (2) for therandom variable . Thus, .The value of 1200 is the total claim amount needed for full credibility for when .Thus, 2 when .

Then,

Example CR-9: A compound Poisson random variable has severity random variable . If theexpected number of exposures of required for severity to be fully credible is and if the totalnumber of claims required for full credibility for the compound distribution is ,calculate the expected number of claims required for the frequency random variable to be fullycredible. Assume the same and for all full credibility standards.Solution: If represents the severity distribution, then

and Since frequency is Poisson, the standard for full credibility for frequency based on number ofclaims needed is claims.



Example CR-10: The distribution of aggregate claims per period is a compound Poissondistribution for which the coefficient of variation of the loss severity distribution is .894. A fullcredibility standard for claim frequency is applied in which the total number of exposures of

is within 5% of the expected number with a probability of 98%. If the same number ofexposures needed in the frequency standard was applied as the number of exposures needed for afull credibility standard for total cost of claims, then the actual total cost would be within %of the expected total cost with 95% probability. Using the normal approximation to aggregateclaims, find .Solution: With Poisson parameter , and "closeness" , and probability , we have

, and . The full credibility standard for claim frequency (usingcondition (1) for claim frequency) is an exposure number of We now want the fullcredibility standard for total claims to be based on , with and closeness factor to bedetermined, so that The coefficient of variation of the severity distribution is

(given), so that the full credibility standard based on number of exposures for

the aggregate cost of claims is . If this is numerically

equal to (the standard for number of exposures for claim frequency), then

, from which we get

CR-1.8 Partial Credibility

In the full credibility approach, the premium is chosen to be either or (where is therandom variable to which the credibility estimate is being applied). The partial credibilityapproach sets the premium to be , a weighted average of the samplemean and the manual premium. is called the and is in the interval andcredibility factor

is the . The general way in which we formulate iscredibility premium . (1.19)information available

information needed for full credibilitySince there are 2 equivalent full credibility standards for any random variable , and there is athird equivalent standard for a compound distribution , the denominator in refers to the fullcredibility standard for whichever equivalent standard we are using, and the numerator refers tothe partial value of that standard available from sample information.

For any random variable , the partial credibility factor will be either

(1) if we are usinumber of observations (exposures) availablenumber of observations (exposures) needed for full credibility ng condition (1) , or

(2) if we are using condition (2) .sum of available observationstotal sum of observations needed for full credibility

Furthermore, if the random variable is , then can also be

(3) if we are using condition (3) .number of claims observedtotal number of claims needed for full credibility



Keep in mind that if the full credibility standard is satisfied then we have ( will never belarger than 1). The form in which information is given will likely force us to use a specific one ofthe formulations for given above.

Example CR-11: Past experience with a production process which produces 6-sided dice hasshown that most of the dice are fair (each face has a chance of turning up when a die is tossed).A game is played in which a single die (randomly chosen from those produced by the processmentioned) is tossed, and the player wins the number of dollars equal to the number that turns up.In 100 previous tosses, there were 18-1's , 16-2's , 21-3's , 15-4's , 15-5's and 15-6's .We wish to find the credibility premium for playing this game using the partial credibilityapproach.

If is the number turning up when a die is tossed, then assuming a fair die, is the manual premium and

, so that

With the usual assumptions of , , , the full credibility standard for

the number of tosses needed (assuming the die is fair) is .

With 100 tosses, the partial credibility factor is .

From the given data, . The credibilitypremium for playing this game is .

Example CR-12: For a particular group of insureds, the prior estimate of total losses perexposure period is 10,000,000 (the manual premium). The standard for number of exposures forfull credibility of total claim amount has been established at 175 exposures. Over the course of100 exposure periods it is found that the observed average total claim amount per exposure periodis 12,500,000. Find the partial credibility premium for the next exposure period.Solution: and , where is the aggregate claim randomvariable for one exposure. The partial credibility factor is

number of observations (exposures) availablenumber of observations (exposures) needed for full credibility .

The partial credibility premium is .

Note that we must use this form of since we are not given any information about the totalamount of loss observed or needed for full credibility.

When applying the limited fluctuation method, it must be clear which random variable thestandard is applied to, and which of the 2 (or 3 in the compound distribution case) full credibilitystandards is being applied.



As a final comment on this limited fluctuation credibility topic, it should be noted that in the pastexams there have been about 1 question per exam on this topic. The coverage provided in thispast section may be a little disproportionate to the weight the topic has received on those exams.

CREDIBILITY - PROBLEM SET 1 CR-17


CREDIBILITY - PROBLEM SET 1Limited Fluctuation Credibility

1. The criterion for the number of exposures needed for full credibility is changed from requiring to be within with probability , to requiring to be within with probability . Find the value of that results in no change in the standard for full credibility for number of

exposures of .A) .0524 B) .0548 C) .0572 D) .0596 E) .0620

2. Total claim amount per period follows a compound Poisson claims distribution. Thestandard for full credibility for total claims in a period based on number of claims is 1500claims. It is then discovered that an incorrect value of the coefficient of variation for the severitydistribution was used to determine the full credibility standard. The original coefficient ofvariation used was , but the corrected coefficient of variation for is .Find the corrected standard for full credibility for based on number of claims.A) 1300 B) 1325 C) 1350 D) 1375 E) 1400

3. The partial credibility factor for random variable based on exposures of is .How many additional exposures are needed to increase the partial credibility factor to atleast ?A) 55 B) 56 C) 57 D) 58 E) 59

4. Total claims per period follows a compound Poisson distribution and claim severity has thepdf , for . A full credibility standard based on number of exposures of needed has been determined so that the total cost of claims per period is within 5% of theexpected cost with a probability of 90%. If the same number of exposures for full credibility oftotal cost is applied to the number of exposures needed for the frequency variable , the actualnumber of claims per exposure period would be within 100 % of the expected number of claimsper exposure period with probability 95%. Find .A) .054 B) .058 C) .062 D) .066 E) .070

5. An analysis of credibility premiums is being done for a particular compound Poisson claimsdistribution, where the criterion is that the total cost of claims is within 5% of the expected cost ofclaims with a probability of 90%. It is found that with exposures (periods) and

, the credibility premium is . After 20 more exposures (for a total of 80) andrevised , the credibility premium is . After 20 more exposures (for a total of100) the revised is . Assuming that the manual premium remains unchanged in all cases,and assuming that full credibility has not been reached in any of the cases, find the credibilitypremium for the 100 exposure case.A) 191.5 B) 192.5 C) 193.5 D) 194.5 E) 196.5

CR-18 CREDIBILITY - PROBLEM SET 1


6. Total claims per period has a compound Poisson distribution. You have determined that asample size of 2670 claims is necessary for full credibility for total claims per period if theseverity distribution is constant. If the severity distribution is lognormal with mean 1000 andvariance 1,500,000, find the number of claims needed for full credibility of total claims perperiod.A) 6650 B) 6675 C) 6700 D) 6725 E) 6750

7. You are given the following: - The number of claims follows a Poisson distribution. - The variance of the number of claims is 10. - The variance of the claim size distribution is 10. - The variance of aggregate claim costs is 500. - The number of claims and claim sizes are independent. - The full credibility standard has been selected so that actual aggregate claim costs per period will be within 5% of expected aggregate claim costs 95% of the time.Using the methods of limited fluctuation credibility determine the number of claims required forfull credibility of aggregate claim costs per period.

8. You are given the following: - The number of claims per period follows a Poisson distribution. - Claim sizes follow a lognormal distribution with parameters (unknown) and - The number of claims and claim sizes are independent. - 6,600 expected claims are needed for full credibility of aggregate claims per period. - The full credibility standard has been selected so that actual aggregate claim costs per period will be within 10% of expected aggregate claim costs per period % of the time.Using the methods of limited fluctuation credibility to determine the value of .

9. has a compound distribution with frequency and severity . and all claim amounts areindependent of one another.

Limited fluctuation credibility is being applied to , with the full credibility standard based on the samplemean of being within 5% of the true mean of with probability 90%.The following information is given regarding the three equivalent full credibility standards for .

The expected number of exposures of needed for full credibility is 108.24 .The expected aggregate amount of claim needed for full credibility is 10,824.The expected total number of claims needed for full credibility is 541.2 .

Find all of the following quantities: and

Show that cannot have a Poisson distribution and cannot have an exponential distribution.



10. is the distribution of the number of claims occuring per week. has a Poisson distribution with anunknown mean. The standard for full credibility for is based on the sample mean of being within5% of the true mean of with probability 90%.With 400 observed claims in 20 weeks, the credibility premium based on partial credibilityis . With 500 observed claims in 30 weeks, the credibility premium based on partial credibility is

. Find the credibility premium based on partial credibility if there are 550 observed claims in35 weeks. Assume that the same manual premium is used in all cases.

11. You are given:(i) pure premium calculated from partially credible data(ii)(iii) Fluctuations are limited to of the mean with probability (iv) credibility factorWhich of the following is equal to ?(A) (B) Z(C) Z (D) Z(E) Z

12. (SOA) You are given:(i) Claim counts follow a Poisson distribution.(ii) Claim sizes follow a lognormal distribution with coefficient of variation 3.(iii) Claim sizes and claim counts are independent.(iv) The number of claims in the first year was 1000.(v) The aggregate loss in the first year was 6.75 million.(vi) The manual premium for the first year was 5.00 million.(vii) The exposure in the second year is identical to the exposure in the first year.(viii) The full credibility standard is to be within 5% of the expected aggregate loss 95% of thetime.Determine the limited fluctuation credibility net premium (in millions) for the second year.(A) Less than 5.5 (B) At least 5.5, but less than 5.7 (C) At least 5.7, but less than 5.9(D) At least 5.9, but less than 6.1 (E) At least 6.1

13. An insurer has two separate classes of policies. The characteristics of the loss per insured ineach of the two classes during a one year period are as follows:Class I: Expected claim per insured is 100. To be within 5% of expected loss 90% of the time,the standard for number of insureds needed for full credibility is 1082.4 .Class II: Expected claim per insured is 200. To be within 5% of expected loss 90% of the time,the standard for number of insureds needed for full credibility is 1082.4 .Class I has twice the number of insureds as Class II. The two classes of insureds are combinedand regarded as a single class with the appropriate adjusted loss per insured during a one yearperiod. Find the full credibility standard for the minimum number of insureds required in thecombined portfolio, where the full credibility is to be within 5% of expected loss 90% of the time.(A) 1082.4 (B) 1190.7 (C) 1244.8 (D) 1298.9 (E) 1353.0



14. The partial credibility approach is applied to a data set of 50 claim amounts. It is assumedthat the claim amount distribution is uniform on the interval . The full credibility standardis to be within 5% of the expected claim amount 90% of the time. The partial credibility factor is found. After 25 additional claim amounts are recorded, the claim amount distribution is revisedto be uniform on the interval . The revised partial credibility factor is found. Find theratio .(A) (B) (C) 1 (D) (E)

15. Number of claims per year follows a Poisson distribution. A number of claims are recordedover a specified period of time. A full credibility standard is set so as to be within 5% ofexpected claims per year 90% of the time. Based on the observed number of claims, the fullcredibility standard is not met, but the partial credibility factor is .Find the maximum value of so that this same number of claims satisfies a full credibilitystandard within 10% of expected claims per year % of the time.(A) 99.5 (B) 99 (C) 98 (D) 97.5 (E) 96

16. has a compound distribution with frequency and severity . and all claim amounts areindependent of one another.

Limited fluctuation credibility is being applied to , with the full credibility standard based on the samplemean of being within 5% of the true mean of with probability 90%.The following information is given regarding the three equivalent full credibility standards for .

The expected number of exposures of needed for full credibility is 108.24 .The expected aggregate amount of claim needed for full credibility is 10,824.The expected total number of claims needed for full credibility is 541.2 .

Find all of the following quantities: and

17. The aggregate loss in one week, , follows a compound negative binomial distribution, andthe severity distribution is exponential. Limited fluctuation credibility is being applied to sothat the full credulity standard is to be within 5% of expected aggregate losses 95% of the time.It is found that the expected number of claims needed for full credibility is 5,412 .

Suppose that the frequency distribution is modified (but still negative binomial) so that mean andvariance of the frequency both increase by 20%. Find the full credibility standard for the numberof claims needed for the new compound negative binomial distribution (severity is the sameexponential distribution as before).



CREDIBILITY - PROBLEM SET 1 SOLUTIONS

1. Since and , are unchanged, the standard for full credibility will beunchanged if is unchanged. With we have , and

. With , is the percentile of the standard normaldistribution, so that . In order for to remain unchanged, we must have

. Answer: D

2. For the compound Poisson distribution with Poisson parameter (frequency distribution ornumber of claims per period) and claim amount distribution (severity distribution or amountper claim), the standard for full credibility for expected number of claims is

. Thus, with . With the coefficient of variation of changed to

5200 , we have . Answer: D

3. The partial credibility factor with is , where is the fullcredibility standard for number of exposures needed.The credibility factor with is .

Then, . Answer: C

4. The severity distribution has mean and so that

With and , we have The full credibility standard for

number of exposures needed for the compound Poisson distribution is .The full credibility standard for the number of exposures needed for the Poisson frequencydistribution only is . Since we are considering the same Poisson frequency distribution,the value of (which is not known) stays the same. If the same value of for full credibilityfrom the aggregate compound Poisson distribution is applied to the Poisson frequencydistribution alone, then we set and the " " for the Poisson frequencycredibility standard must change, which is why it has been denoted .Then .With , , and then in order for this to be the proper for , we musthave . Answer: B



5. For the 60 exposure case, the credibility premium is , and for the 80 exposure case,. We wish to find .

In going from 60 to 80 exposures, the credibility factor changes from

to (where is the severity distribution).

Thus, , and the two credibility premium equations become , .6

Juggling these equations results inwhich results in the quadratic equation

.Using and substituting into the equations above, we get ,and using , we get .With , we get , and the new credibility premium is

With , we get , and the new credibility premium is Answer: A

6. If the severity distribution has variance , then .If then the standard for full credibility of aggregateclaims based on number of claims is Answer: B

7. If the claim number distribution is Poisson, the full credibility standard for aggregate claimcosts based on number of claims is , where is the claim size distribution, and

96 We are given . For the compound Poisson aggregateclaims distribution,



8. When the claim number distribution is Poisson, the standard for full credibility foraggregate claims per period based on number of claims is ,where is the claim amount random variable.For the lognormal, and .

Therefore, .

Since , we have . But is the percentile of the standard normaldistribution. From the normal table, we have , where

. Therefore, .

9

Then, , and then .

Since , we have .If is Poisson, then .Then

, which is not possible.

If has an exponential distribution, then Then

, which is not possible.

10. Since is Poisson, the full credibility standard for estimating the mean of is either(i) as the expected number of exposures of (weeks) needed,or(ii) as the total expected number of claims needed.Since we do not know the value of , the only standard we can apply is (ii).

With 400 claims in 20 weeks, the average number of claims per week (sample mean) is .Using credibility standard (ii) above, the partial credibility factor is

, and the partial credibility premium is

, where is the manual premium .

With 500 claims in 30 weeks, the average number of claims per week (sample mean) is . Using credibility standard (ii) above, the partial credibility factor is


, where is the manual premium.



10 continuedFrom the two equations, and ,we get and . Then, with 550 claims in 35 weeks, we have

. Using credibility standard (ii) above, the partial credibility factor is


.

11. (E) is correct. This is the formula at the bottom of page 514 in the Mahler-Dean Credibilitystudy note. Answer: E

12. When considering the compound Poisson aggregate claims distribution with claim sizedistribution , we have three ways of setting the standard for full credibility:(i) the number of exposures (periods of ) is

(ii) the number of claims is , or

(iii) the aggregate amount of claims is .We do not know or in this case, and therefore (i) and (iii) cannot be used. We are given

that has coefficient of variation , so that , and , and we can use

standard (ii). The credibility criterion has , since and . The standard for full credibility is as the number of

claims needed. Since only 1000 claims occurred in the first year, we have not met the standardfor full credibility, and we apply the method of partial credibility.The credibility factor is .We are trying to determine the credibility premium for aggregate claims for the second year. Wehave only one exposure for aggregate claims, that being the first year, so million . Themanual premium is given to be millionThe credibility premium for the second year is

. Answer: A

13. Class I: .

Class II: .

For the combined portfolio of policies, the loss per insured is , is a mixture of and ,with and weighting applied, respectively. Then and

, so that .

The full credibility standard for is .Answer: E



14. The partial credibility approach sets the credibility factor to be

. In this example .

Based on the original assumption of being uniform on , and ,

we get .

Based on the revised assumption of being uniform on , and , we get

.

. Answer: D

15. The full credibility standard within 5% of expected number claims per year 90% of the timehas (5%) and (95-th percentile of the standard normal distribution). For theannual claim number distribution being Poisson, the number of claims (not exposures) needed forfull credibility is .

The partial credibility factor (when less than 1) is .actual number of observationsnumber needed for full credibility

We are told that , from which we getactual number of observations

number needed for full credibility , so thatactual number of observations . 500.5 is the full credibility standardwith and , where , so that .From reference to the standard normal table, 2. is the 98.74-th percentile. Therefore, theprobability that the 500 observed claims will satisfy the full credibility standard within 10% ofexpected number of claims per year is . Answer: D

16.

Then, , and then .

Since , we have .



17. We will denote the frequency by and the severity will be .

The full credibility standard for the expected number of claims needed is .

Since the severity is exponential, we have and .

The full credibility standard for expected number of claims needed becomes . Since both the mean and variance of are increasing by 20%,

this full credibility standard is unchanged at 5412.

CREDIBILITY SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR CR-27


CREDIBILITY - SECTION 2, BAYESIAN ESTIMATION, DISCRETE PRIOR

The material in this section relates to Section 12.4 of "Loss Models". The suggested time framefor this section is 3 hours.

Bayesian analysis involves making an initial, or estimate of a quantity, and then updatingpriorthat estimate after some sample information becomes available. This results in the posteriorestimate of that quantity. The quantity being estimated can be a probability, or a probabilitydistribution, or a distribution parameter, or an expected claim amount, etc.

We will start by considering the Bayesian approach to updating the estimate of a probability sinceit is simple and captures the main concepts which are used throughout Bayesian analysis.Bayesian analysis is related to conditional probabilities and conditional distributions, and wereview some probability relationships.

Bayesian Probability Estimates Based on a Discrete Prior Distribution

For any events and , the conditional probability that event occurs given that event hasoccurred is . This can be reformulated as .It is always true that , where is the complement of event .This is illustrated in the Venn diagram below.

Using the relationship above, we have

, and (2.1)

. (2.2)

This last expression is a simple version of Bayes Rule.

Suppose that we are given and . With this information, the usualobjective is to calculate or . We are trying to "reverse the conditioning". Thisis a typical situation to which we apply Bayesian analysis. The probability formulations abovethat allow us to reverse the conditioning can be summarized in the following table. Thenumbered items indicate the order in which calculations can be made.

CR-28 CREDIBILITY SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR


(known) and (known) (known) and (known)

3 1. 2.

4.

or or

5. 6.

7. 8.

An alternative diagram to indicate the order of calculations is as follows. Again, it is assumedthat we are given and , and

and are all known also.

The expression is Bayes Theorem, and it can

be extended to any partition of a probability space . For any event , we have

. (2.3)

This is illustrated in the Venn diagram below.

We can then reverse the conditioning using Bayes rule:

(2.4)



Example CR-13: Two similar bowls each contain 10 similarly shaped numbered balls. Bowl 1contains 5 balls with the number 1 and 5 balls with the number 2 (equally likely to be chosen).Bowl 2 contains 3 balls with the number 1 and 7 balls with the number 2 (equally likely to bechosen). A bowl is chosen at random, and a ball is chosen from that bowl.Find the probability that the ball chosen has number 1, and find the (posterior) probability thatbowl 1 was chosen given that the ball chosen was had number 1.Solution: We identify the following events: ball chosen has number 1 ,

ball chosen has number 2, Bowl 1 is chosen, and Bowl 2 was chosen.We are given that the (prior) probability of is , since we are told that a bowl ischosen at random. In general, in this sort of situation, without any additional information abouthow the bowl is chosen, the phrase "a bowl is chosen at random" is interpreted to mean that eachbowl has the same chance of being chosen.

It then follows that (given that bowl 1 was chosen, there is a probability thatthe ball chosen is a "1"), and (although these probabilities are not explicitlystated, they are implied by the nature of the situation) Then using the probability rules reviewedabove, the "overall" probability that the ball is a "1" is

.

We are also asked to find the probability that bowl 1 was chosen given that the ball chosen was a"1". This is The "reversed conditioning" probability . Using Bayes rule, we have

.

These calculations can be summarized in the following table. Ball Number is 1 Ball Number is 2

Bowl 1 ( ) ones twos

Bowl 2 ( ) ones twos

The (posterior) probability of having chosen bowl 1 given that the ball chosen was a "1" is.

Note that in this very simple situation, we could have anticipated this answer without so muchtechnical machinery. There are a total of 8 "1"s in the two bowls combined. Therefore, knowingthat the ball chosen was a "1", it must be one of those 8. Since 5 of the 8 "1"s are in bowl 1 itappears by general reasoning that there is a probability that the ball with a 1 came from bowl 1,with a similar comment for bowl 2.



Example CR-13 continuedWe must be careful with applying "general reasoning" in this way. It is important that there wasa 50-50 chance of choosing bowl 1/bowl 2 initially. Algebraically what happens is the following:since it follows that

.

We can generalize this example to more than 2 bowls. Suppose there was a 3rd bowl with 1 "1"and "2"s, and suppose that in the initial choice of a bowl, each bowl is equally likely to bechosen (prior probability for each bowl).

If we are given that a "1" was chosen, then the (posterior) probability that the ball came frombowl 1 is from bowl 2 is , and from bowl 3 is . A keyassumption again was that initially each bowl had the same chance of being chosen.

Going back to the original situation with 2 bowls, if the probabilities for the initial choice ofbowls 1 and 2 had not been each, we would have to apply the appropriate "weight"(probability) to the 5 balls numbered "1" in bowl 1 and the appropriate "weight" tothe 3 balls numbered "1" in bowl 2. This would essentially result in the same technicalcalculations we just went through earlier.

For instance, suppose that we were told that, initially we are twice as likely to choose bowl 1 aswe are to choose bowl 2. so that and (we are still assuming that bowl 1has 5 "1"s and 5 "2"s, and bowl 2 has 3 "1"s and 7 "2"s). Then

.

So now when we are given that the ball chosen was numbered 1 (and bowl 1 has 5 "1"s and bowl2 has 3 "1"s), it is no longer true that there is a probability that bowl 1 was chosen.

The previous example can be put into the framework of the initial steps of a Bayesian analysis forevent probabilities. Bayesian analysis for probabilities begins with the identification of the priorevent probabilities model event probabilities and the . A model event probability is aconditional probability given that a particular prior event has occurred. One objective ofBayesian analysis for event probabilities is to "reverse the conditioning" to find the posteriorprobability of a prior event occurring given that a particular model event has occurred. InExample CR-13, the prior events are the events of choosing Bowl 1 or Bowl 2, and the modelevents are the conditional events of choosing a ball numbered "1" given Bowl 1, a ball numbered"1" given Bowl 2, a ball numbered "2" given Bowl 1 and a ball numbered "2" given Bowl 2. Thesequence of calculations for Example CR-13 can be summarized in this Bayesian framework asfollows.



Prior Events Bowl 1 , given Bowl 2 , given(These are marginal probabilities of the prior parameter distribution; the bowl type is the"parameter")

Model Events Choose number 1 from Bowl 1 Choose number 1 from Bowl 2(Conditional , given , givenprobabilities) Choose number 2 from Bowl 1 Choose number 2 from Bowl 2 , given , given7

Joint Events (Intersection probabilities) These are probabilities of the type

Bowl 1 was chosen a ball numbered "1" was chosenand

Model Events (Marginal Probabilities) These are unconditional probabilities of the type the ball chosen was numbered "1"

Posterior Events

and

(Posterior Probabilities)

These are "reverse conditioning" probabilities of the typeBowl 1 was chosen the ball chosen was numbered "1"

This is an example of a discrete prior distribution. The prior distribution can be thought of as atwo-point random variable , where indicates that bowl 1 is chosen and indicatesthat bowl 2 is chosen.

The posterior probabilities are the "reversed conditioning" probabilities.

The following two examples also illustrate Bayesian analysis applied to event probabilities. Theprior distribution is discrete, which is usually indicated by a finite number of "categories" for theprior outcome. There are two "categories" in Example CR-13, those being bowl 1 and bowl 2.



Example CR-14: Suppose that an insured population consists of 1500 youthful drivers and 8500adult drivers. Based on experience, suppose that we have derived the following probabilities thatan individual driver will have claims in a one year period. Youth Adult 0 0.50 0.80 1 0.30 0.15 2 0.15 0.05 3 0.05 0.00It is found that a particular randomly chosen policy has exactly one claim on it in the past year.Find the probability that the policy is for a driver who is a youthful driver.Solution: We use the Bayes rule ,

where 1 claim , and youth . Then

youth 1 claim 1 claim youth 1 claim youth youth1 claim 1 claim youth youth 1 claim adult adult

The probability calculations can be summarized in the following table:

Prior Prob. youth (given) and adult (given)Model Prob. 1 claim youth (given) 1 claim adult (given) Joint Prob. 1 claim youth 1 claim adult 1 claim youth youth 1 claim adult adult

Marginal Prob. 1 claim 1 claim youth 1 claim adult

Posterior Prob. youth 1 claim .

In this example, the (discrete) prior distribution is driver type (youth or adult).

One aspect of the Bayesian analysis presented here can also be looked at from the point of viewof mixtures of distributions. For instance, in Example CR-14 the number of claims for a randomlyselected driver is a mixture of the number of claims distributions for Youths and Adults. Let denote the number of claims that a Youth will have in one year and let denote the number ofclaims that an Adult will have. The number of claims experienced by a randomly selected driverwill be the mixture of with mixing weight and with mixing weight . We can thinkof as the conditional distribution of given that the driver chosen is a Youth, and as theconditional distribution of given that the driver chosen is a Adult, and Youth and

Adult . The probability function of can the be written in asYouth Youth Adult Adult

.



Other probabilities for can be found in a similar way. Note that in Example CR-14 we found (called a marginal probability of ), and the calculations applied to find it are the

same as those in the mixture of distributions approach, but with different names for thecomponents.

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 risk is chosen at random from the portfolio and is observed to have 2 claims in the year. Findthe probability that the risk was chosen from Class I.Solution: We can describe the prior Poisson parameter in terms the random variable , where

and . We can describe the model distribution asthe conditional Poisson random variable given : . We are askedto find the posterior probability Class . Since Class corresponds to , thisprobability can written as . We will use the usual Bayesian formulation ofconditional probability: 2 .

The components of this probability can be found within the Bayesian structure as follows.

Prior Dist. of (given) Model Dist. of (given)

Joint Dist. of and

and

Marginal Dist. of

Posterior Dist. of Joint Dist. of and Prob.Marginal Dist. of Prob.



Example CR-15 continuedFrom this structure, we see that

.

Note that the numerator in this posterior probability is one of the terms that are being added up inthe denominator. This is a general feature of a posterior probability when the prior distribution isdiscrete.

Examples CR-13, CR-14 and CR-15 all have discrete prior distributions as well as discrete modeldistributions (ball number in CR-13, number of claims in CR-14 and CR-15). We now consideran example in which the prior distribution is discrete, but the model distribution is continuous.The analysis is essentially the same, except that the model distribution is described using adensity function rather than a probability function that is used for a discrete model distribution.

Example CR-16: A portfolio of risks is divided into two classes. The characteristics of the lossamount distributions for the two risk classes is as follows: Class I Class II Loss Amount Exponential ParetoDistribution mean 1000 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 theprobability that the risk was chosen from Class I.(b) A risk is chosen at random from the portfolio and is observed to have a loss greater than 2000.Find the probability that the risk was chosen from Class I.Solution: We can put this in the context of the Bayesian analysis we have seen in previousexamples. The risk class can be described as the random variable which is either I or II.Since the model distribution is either exponential or Pareto, it is described using a pdf.

Prior Dist. of (given) Model Dist. of (given) ,

(a) The joint distribution of and is and

.

The marginal distribution of is .

The posterior probabilities for are

(this is the same form as a conditional probability, but in the numerator we use the joint densitynotation for and , instead of the intersection of probability notation),



Example CR-16 continuedand

(we can cancel the .5).

Then, .

(b) We are asked to find . It is generally the case that to find a conditionalprobability we use the basic definition of conditional probability.

.

We apply the usual probability rules to find the numerator and denominator.

(since has an exponential distribution with mean 1000 if ).

.We have already found the first term on the right hand side of this expression.The second term on the right hand side is found in a similar way, except we use the Paretodistribution probability

.

Then

.

The main objective of Bayesian analysis usually goes beyond finding the (marginal) distributionof , with a further objective of finding posterior probabilities (reversing the conditioning as wehave done in the previous examples) and also predictive probabilities (which will be considered alittle later). As we have seen in the previous examples, the marginal distribution of may befound in one of the steps within the framework of Bayesian analysis. It can also be found in thecontext of a mixture of distributions as was seen in Example CR-14..





CREDIBILITY - PROBLEM SET 2Bayesian Analysis - Discrete Prior

Questions 1 and 2 relate to the following situation. Two bowls each contain 10 similarly shapedballs. Bowl 1 contains 5 red and 5 white balls (equally likely to be chosen). Bowl 2 contains 2red and 8 white balls (equally likely to be chosen). A bowl is chosen at random with each bowlhaving the chance of being chosen. A ball is chosen from that bowl.

1. Find the probability that the ball chosen is red.A) B) C) D) E)

2. Suppose that the ball chosen is red. Find the probability that bowl 1 was the chosen bowl.A) B) C) D) E)

Questions 3 and 4 relate to the following situation. A portfolio of insurance policies consists oftwo types of policies. Policies of type 1 each have a Poisson claim number per month with mean2 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 portfolioand the number of claims generated by that policy in the following is the random variable .

3. Find .A) .0321 B) .0482 C) .0642 D) .0803 E) .0963

4. Suppose that a policy is chosen at random and the number of claims is observed to be 1 forthat period. Find the probability that the policy is of type 1.A) .85 B) .88 C) .91 D) .94 E) .97

5. A risk class is made up of three equally sized groups of individuals. Groups are classified asType A, Type B and Type C. Any individual of any type has probability of .5 of having no claimin the coming year and has a probability of .5 of having exactly 1 claim in the coming year. Eachclaim is for amount 1 or 2 when a claim occurs. Suppose that the claim distributions given that aclaim occurs, for the three types of individuals are

claim of amount Type A and a claim occurs

claim of amount Type B and a claim occurs

claim of amount Type C and a claim occurs

An insured is chosen at random from the risk class and is found to have a claim of amount 2.Find the probability that the insured is Type A.A) B) C) D) E)



6. 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. - Loss sizes for property risks follow a Pareto distribution 1,000 and .(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. Determinethe limit of the posterior probability that this risk is a liability risk as goes to zero.

7. A portfolio consists of 100 independent risks. 25 of the risks have a policy with a $5,000 perclaim policy limit, 25 of the risks have a policy with a $10,000 per claim policy limit, and 50 ofthe risks have a policy with a $20,000 per claim policy limit. The risks have identical claimcount distributions. to censoring by policy limits, claim sizes for each risk follow a ParetoPriordistribution with parameters 5,000 and 2. A claims report is available which shows thenumber of claims in various claim size ranges for each policy censoring by policy limits,afterbut does not identify the policy limit associated with each policy.The claims report shows exactly one claim for a policy selected at random. This claim falls in theclaim size range of $9,000 - $11,000. Determine the probability that this policy has a $10,000policy limit.

8. (CAS May 05) An insurer selects risks from a population that consists of three independentgroups.• The claims generation process for each group is Poisson.• The first group consists of 50% of the population. These individuals are expected to generate one claim per year.• The second group consists of 35% of the population. These individuals are expected to generate two claims per year.• Individuals in the third group are expected to generate three claims per year.A certain insured has two claims in year 1. What is the probability that this insured has more thantwo claims in year 2?A) Less than 21% B) At least 21%, but less than 25%C) At least 25%, but less than 29% D) At least 29%, but less than 33% E) 33% or more

9. (CAS May 06) Claim counts for each policyholder are independent and follow a commonNegative Binomial distribution. A priori, the parameters for this distribution are or . Each parameter set is considered equally likely. Policy files are sampled atrandom. The first two files samples do not contain any claims. The third policy file contains asingle claim. Based on this information, calculate the probability that .A) Less than 0.3 B) At least 0.3, but less than 0.45 C) At least 0.45, but less than 0.6D) At least 0.6, but less than 0.75 E) At least 0.75



10. A portfolio of insurance policies consists of three types of policies. The loss distribution for each typeof policy is summarized as follows:Policy Type Type I Type II Type IIILoss Distribution Exponential Exponential Exponential With Mean 2 With Mean 4 With Mean 8Half 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 loss amount is .(a) Find .(b) Find each of the following two ways: (i) (ii) , where is the random variable describing the Type of policy chosen.

11. (SOA) You are given:(i) A portfolio of independent risks is divided into two classes, Class A and Class B.(ii) There are twice as many risks in Class A as in Class B.(iii) The number of claims for each insured during a single year follows a Bernoulli distribution.(iv) Classes A and B have claim size distributions as follows:

Claim Size Class A Class B 50,000 .60 .36

100,000 .40 .64(v) The expected number of claims per year is 0.22 for Class A and 0.11 for Class B.One insured is chosen at random. The insured’s loss for two years combined is 100,000.Calculate the probability that the selected insured belongs to Class A.(A) 0.55 (B) 0.57 (C) 0.67 (D) 0.71 (E) 0.73

12. Prior to tossing a coin, it is believed that the chance of tossing a head is equally likely to be or . The coin is tossed twice, and both tosses result in a head. Determine the posterior

probability of tossing a head.(A) (B) (C) (D) (E)

13. A portfolio of insureds consists of two types of insureds. Losses from the three types are:Type 1 insured loss: exponential with a mean of 1 ,Type 2 insured loss: exponential with a mean of 2 ,In the portfolio, of the insureds are of Type 1 and of the insureds are of Type 2.

Two insureds are chosen at random and one loss is observed from each insured.The first insured is observed to have a loss of 1 and the second insured is observed to have aloss of 2. Find the probability that the two insured are of the same Type. It is assumed that thelosses of the two insureds are independent of one another.




1. red red bowl 1 bowl 1 red bowl bowl 2 . Answer: C

2. bowl 1 red .bowl 1 red red bowl 1 bowl 1red

Answer: D

3. type

This is a mixture of two Poisson distributions. .For this is .Answer: E

4. Type 1 Type 1 Type 1

Type 1 Type 1Type 1 Type 1 Type Type

. Answer: B

5. Type A A A

. Answer: B

6. (a) The unconditional claim model is a mixture of two Paretos. The moments will be themixture of the corresponding Pareto moments, with weights .75 and .25.We use the Pareto moments (first moment) and (second moment).

.

. .

(b)

As the limit is .



7. , and will denote the event that the risk has limit 5, 10 and 20 thousand,respectively. The prior probabilities for policy limits for a randomly selected policy are

and .We wish to find the posterior probability .The distribution function of the Pareto distribution with 5,000 and 2is . The table of calculations is as follows.

Prior Prob.

Model Prob. Note that if a claim payment is between 9,000 and 11,000, it is not possible that the policy had alimit of 5,000. That is the reasoning behind . If a policy has a limit of10,000, then no payment will be over 10,000 so to say that the payment is between 9,000 and11,000 is the same as saying that the payment is over 9,000. Therefore,

.

Joint Prob.

Marginal Prob. .

Posterior Prob. .

We have used the following conditional probability rules to solve this problem.We define the following events: 5,000 limit , 10,000 limit , 20,000 limitand is the event . Then the expression above for is

, and , which is

.



8. Let us label the groups as Group 1 (50%), Group 2 (35%) and Group 3 (15%),and let denote the annual number of claims of an individual chosen at random for the threegroups combined. The conditional distribution of given that the individual is chosen fromGroup 1 is Poisson with a mean of 1, given that the individual is chosen from Group 2 isPoisson with a mean of 2, and for Group 3 is Poisson with mean 3.

We wish to find , where and are the numbers of claims years 1 and 2,respectively. By the definition of conditional probability, this probability is

.We use the rule that if is a partition of a probability space, then

.

With event as " " and as the events "individual is from Group 1, 2, 3",the denominator is Group 1 Group 1 Group 2 Group 2 Group 3 Group 3

The numerator isGroup 1 Group 1

Group 2 Group 2 Group 3 Group 3 .

For each particular group, the numbers of events in successive years are independent, so thatGroup 1 Group 1 Group 1

, andGroup 2 Group 2 Group 2

, andGroup 3 Group 3 Group 3

.Then ,and . Answer: C

9. We are asked to find .This is .

Since the conditional distribution of given is negative binomial, we get, and ,

so that .Then, , since .In a similar way, we get

,where , so that

.



9 continuedThen

,

and . Answer: E2

10.(a)

(b)(i)

(ii) and .prob. prob.

prob. prob.

prob. prob.

.

11. This is a fairly standard application of Bayesian analysis, using conditional probability rules.Let and denote the events that the chosen risk is from Class A or Class B, respectively, andlet and denote the claim amounts in years 1 and 2, respectively.

.

We are given since there are twice as many risks in Class A as in ClassB. In Class A there is a .22 chance of a claim in each year, and in Class B there is a .11 chance ofa claim in each year. In order for the total claim in two years to be 100,000, one of threepossibilities must occur: and , or

and , or and . Therefore,

.



11 continuedIn order for the annual claim to be 100,000 in Class A, there must be a claim, and it must be for100,000; this has probability . In order for the annual claim in Class A to be 0, theremust be no claim; this has probability . In order for the annual claim to be 50,000 in Class A,there must be a claim, and it must be for 50,000; this has probability .

Then, .

In a similar way, we get .

Then, .

The order of calculations can be summarized as follows. , given , given

. Answer: D

12. Posterior probability of H is 3H 1H and 2H .3H,1H,2H1H,2H

We will denote by " H" the event that the probability of tossing a head with the coin is , with a

similar definition for " H".

1H,2H 1H,2H H H 1H,2H H H .

3H,1H,2H 3H,1H,2H H H 3H,1H,2H H H

.

Then 3H 1H and 2H . Answer: C



13. We will define and to be the loss amounts from the first and second insured, respectively.We are given that and .

We will define to be the random variable that identifies Type, and and are the types ofinsured 1 and insured 2.

We wish to find . Using the definition of conditional probability, this is

Since must be 1 or 2, the numerator can be formulated as

Since both policies are chosen at random, and .Also, since the insureds are independent of one another,

(the notation really means density of , not probability, in this situation).Similarly,

.

Then, .

Also, because of independence of and .Each of the 's is a mixture of two exponenetials with pdf , so

, and , and .

Finally, .



CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR CR-47


CREDIBILITY - SECTION 3BAYESIAN CREDIBILITY, DISCRETE PRIOR

The material in this section relates to Section 16.4.2 of "Loss Models". The suggested time framefor this section is 3-4 hours.

The Bayesian approach to credibility has the same initial steps as the Bayesian analysis presentedin Section 2, we just take the analysis a few steps further. We begin with the basic components ofBayesian analysis, which are the (parameter ) distribution and the distribution (aprior modeldistribution that is conditional on the value of the prior parameter ). We then find the jointdistribution of and (joint probabilities in the case of discrete and ), and the marginaldistribution of (unconditional probabilities). We then find the distribution of ,posteriorwhich is the conditional of given the outcome that has occurred. The Bayesian credibilityprocedure extends this analysis to the distribution, which is the distribution of apredictivesecond outcome of given the value of the first outcome of . An extension of Example CR-13illustrates this procedure.

CR-3.1 A Basic Example of Bayesian Analysis With a Discrete Prior

Example CR-13 continued: Bowl 1 has 5 balls numbered "1" and 5 balls numbered "2", andBowl 2 has 3 balls numbered "1" and 7 balls numbered "2". A bowl is chosen at random (witheach bowl having the same chance of being chosen) and a ball is randomly chosen from thatbowl. Suppose that after the first ball is chosen, it is replaced into the bowl from which it waschosen, and a second ball is chosen . Find the probability that the secondfrom that same bowlball chosen is a "1" given that the first ball chosen was a "1".Solution: We had identified the following events: ball chosen has number 1 ,

ball chosen has number 2, Bowl 1 is chosen, and Bowl 2 was chosen.

Along with these events, we identify the following events:second ball chosen is a "1" , second ball chosen is a "2".

We are being asked to find second ball is a "1" first ball is a "1" .Using the definition of conditional probability, we have .

From Example CR-13 we have . We find using the same rule as was usedfor finding in Example CR-13: .

Suppose that it is known that bowl 1 was chosen (event ). Then since the first ball picked isreplaced, the number of the second ball picked is independent of that of the first ball, and since itis being picked from the same bowl as the first ball it has the same probability of being a "1" asthe first. Therefore, .By the same reasoning, .

Then , and

.Example CR-13 continued

CR-48 CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR


This is the that the second ball chosen is a "1" given that the first ballpredictive probabilitychosen is a "1".

There is an alternative way of finding the predictive probabilitysecond ball is a "1" first ball is a "1" . This alternative uses a modification of the

probability rule mentioned earlier, . (3.1)The modification to the rule is . (3.2)In this example is the event that bowl 1 was initially chosen, and is the event that bowl 2was initially chosen. From the makeup of the two bowls, we know that

and . We found that and .Therefore, .

In this example two methods have been presented for finding the predictive probability .

Method 1:

.

This uses the basic definition of the conditional probability .Note that we are assuming that and are conditionally independent given (given thatwe know which bowl was chosen, the number of the second ball chosen from that bowl isindependent of the number of the first ball chosen from that bowl). This will always be the casefor the Bayesian credibility situations that arise on Exam C, so we will always be able to use therelationships and . (3.3)

Method 2: (3.4)(this is a more general form of the relationship ) .It is not clear from this formulation, but Method 2 requires the same assumption of conditionalindependence of and given .

It will almost always be the case that Method 1 is preferable to Method 2 when calculating apredictive probability (although they require the same calculations in different orders). In theBayesian framework, a predictive probability is a conditional probability of a model-type eventoccurring on the next trial given information on the model-type event that occurred on the firsttrial (or trials). In this example, a "trial" is the choice of a ball after a bowl has been picked. Asubsequent "trial" is the next choice of ball from the same bowl (after the first ball is replaced).A "model" event is the ball number chosen (the prior event is the bowl type chosen).

A point to mention again is the assumption of conditional independence that always arises in theBayesian credibility context. Keep in mind that even though events and are conditionallyindependent given event , it is not necessarily true that and are unconditionallyindependent. In fact, they usually will not be unconditionally independent.



CR-3.2 The Framework for Finding Predictive Probabilities

Example CR-13 illustrates the framework for formulating and calculating predictive probabilities(and densities). This can be summarized in the following way.

Prior Distribution CR-13 - bowl type, CR-14 - age type, CR-15 and CR-16 - distribution typeModel Distribution CR-13 - ball number, CR-14 and CR-15 - number of claims, CR-16 - loss size

A random choice is made from the prior distribution, but we do not know what that choice is. Anobservation is obtained from the model distribution for that prior type.

Posterior Distribution CR-13 - bowl type given ball number, CR-14 - age type given number of claims CR-15 - Poisson parameter given number of claims CR-16 - distribution type given loss size

Predictive DistributionA second observation is obtained from that same (still unknown) prior type. CR-13 - a second ball is chosen from the same bowl CR-14 - number of claims in the 2nd year are observed for the same driver CR-15 - number of claims observed in the second year for the same Poisson CR-16 - loss size of 2nd loss is observed for the same loss distribution

In this section we are considering discrete prior distributions. Examples CR-13, CR-14 and CR-15 also had discrete model distributions, and CR-16 had a continuous model distribution. If theprior distribution is discrete, then the posterior distribution will also be discrete, with updated(posterior) probabilities. If the model distribution is discrete, then the predictive distribution willbe discrete, and if the model distribution is continuous, the predictive distribution will also becontinuous. Along with identifying the posterior and predictive distributions, we may beinterested in the mean of the posterior distribution, and also the mean of the predictivedistribution.



CR-3.3 Predictive Expectation - The Bayesian Premium

We now review a couple of important principles involving the expected value of a randomvariable. We have already used the following probability rules.(i) and(ii) .Rule (ii) requires that events and are conditionally independent given (and ).The corresponding rules for the expected value of a random variable are:

(i) and (3.5)(ii) . (3.6)Again, rule (ii) requires that events and are conditionally independent given (and ).These rules can be extended to any partition of a probability space .

(i) and (ii) (3.7)

Note that these rules are versions of the "double expectation" rule which states that for any tworandom variables and , . In the case of rule (i), can be thought ofas a two-point random variable indicating whether or not event has occurred,

if occursif occurs .

Then .if occurs, prob. if occurs, prob.

So is itself a two-point random variable. The expected value of this two-point randomvariable is ,which was seen to be equal to in rule (i) above, so we have .

As an illustration of rule (i) above for expectations we can use Example CR-14. Suppose that adriver is chosen at random from the population of 10,000 drivers. The expected number of claimsfor the randomly selected driver will be

youth youth adult adult

.

As another illustration of these concepts we have the following continuation of Example CR-13,the two-bowl example with numbered balls.



Example CR-13 continued: Bowl 1 contains 5 balls with the number 1 and 5 balls with thenumber 2. Bowl 2 contains 3 balls with the number 1 and 7 balls with the number 2. A bowl ischosen at random, and a ball is chosen at random from that bowl. The ball is replaced in the bowlfrom which it was drawn and another ball is drawn at random from that same bowl.(a) Find the expected value of the number of the first ball drawn.(b) Find the expected value of the number of the second ball drawn given that the first ball wasnumbered with a 1.Solution: (a) Suppose that denotes the number on the first ball drawn. We are asked to find

. We have two methods that can be applied.

Method 1: We could formulate this expectation using expectation rule (i) above. Recall thatevent is the event that bowl 1 was chosen. .Note that , and is found in a similar way.

Method 2: We can use the unconditional (marginal) probabilities for and find its expectationdirectly as follows. The event was denoted event in Example CR-13 when it was firstpresented in Section 2.

, and in a similar way. Then, .

This method requires the marginal probabilities for .

(b) We are asked to find , where denotes the number of the ball on draw .The two approaches taken in part (a) for the unconditional expectation of can be adapted tofind the predictive (conditional) expectation .

Method 1: We use expectation rule (ii) above. We condition over the bowl from which the firstball was drawn, . The expressiongiven that the first draw was a 1

which was used to find the (unconditional)expectation of the number on the first ball picked can be modified to

. and (once it is known that we chose bowl 1, the expected draw

will be , with similar reasoning if it is known thatthe bowl chosen is bowl 2). Then in order to find the conditional probabilities and

, we must use the probability rules for conditioning that were used when thisexample was first presented in Section 2.

Recall that

and .

Putting these probabilities together, we get.



Example CR-13 continuedThese calculations can be summarized in a table similar to that presented in Example CR-13.

Bowl 1 ( )

Bowl 2 ( )

Then , and .

Method 2: A second approach to find is to find the conditional distribution of given that . We wish to find and .

We have two methods that can be applied to find these probabilities.

(i)

We could find in a similar way, or note that .

(ii) , and in a similar way, .A careful inspection of (i) and (ii) will show that we are really doing the same calculations in adifferent order.

Then, .

In a Bayesian credibility situation, it is almost always more efficient to use Method 1 ratherthan Method 2 to find a predictive expectation (particularly if the prior distribution iscontinuous).



Example CR-13 continuedThe components of this example can be put into the general context of Bayesian estimation.

The is the two point random outcome of which bowl is chosen. prior distribution of or , with ).

The is the conditional distribution of the number on the ball drawnmodel distribution of given the bowl from which it is drawn: number on ball drawn (what was previouslycalled ) ; , etc .

The describes probabilities of the typejoint distribution of and that were calculated earlier.

The is the unconditional distribution of ball number:marginal distribution of and .

The is the conditional distribution of bowl type given the numberposterior distribution of of first ball. For instance and are posteriorprobabilities.

The is the conditional distribution of the second ball numberpredictive distribution of given the number on the first ball: are predictive probabilities.

In this example, alternative methods were presented to calculate various factors. The method thatis most efficient may depend on the data available and the way in which a situationis presented.

The following example is another illustration of the Bayesian method applied to a discrete priordistribution.

Example CR-17: You are given the following information.The number of claims per year for Risk A follows a Poisson distribution with mean .The number of claims per year for Risk B follows a Poisson distribution with mean .The probability of selecting Risk A is the same as the probability of selecting Risk B.One of the risks is randomly selected and zero claims are observed for this risk during one year.Determine the probability that the selected risk will have at least one claim during the next year.Solution: We denote by and the number of claims in the first and second years,respectively. We are asked to find ] ]

]We interpret the statement "One of the risks is randomly selected" as meaning that the priorprobabilities for risk type are . The numerator and denominator of thepredictive probability can both be found by conditioning over Risk Classes A and B.

The denominator is]



Example CR-17 continuedThe numerator is ]

.

As mentioned earlier, we are implicitly assuming that given risk , the numbers of claims inyears 1 and 2 are independent of each other, and the same is true for risk . It follows that

and .

Note that we use the relationship and Poisson probabilities(and the same for )

Then ] ,

and ]

We can also find this probability using the following method, but the method presented abovemay be more efficient..

]As an exercise, you can check that this will result in the same probability as found above.

The Bayesian context of this example is as follows.

The is the 2-point distribution consisting of the choice of risk or risk ,prior distribution with and as the prior probabilities.

The is , the number of claims per year. The conditional distribution of model distributiongiven is Poisson with mean , and the conditional distribution of given isPoisson with mean .

The of and has probabilities of the form . Thisjoint distributionprobability can be found from

.

After a risk is chosen at random (not knowing whether it is from group A or B), the number ofclaims for the year is found to be 0. We find the probabilitymarginal

.

The of given becomes posterior distribution

no claim in yr 1 , and

no claim in yr 1 .

The is the distribution of number of claims in the 2nd year ( ) given nopredictive distributionclaims in the first year ( ). We were asked to find the predictive probability

].



Suppose that in Example CR-17 we were asked to find the predictive expectation .The understanding would be that a risk class was chosen and an individual from that risk classwas observed to have 0 claims for the year. We wish to find the expected number of claims forthe same individual in the following year (we don't know the risk class).

The most efficient solution would be to use the relationship .

We know that and . The main work involved would be to findthe conditional probabilities and (actually, since this example has a2-point prior, we get ). We saw in Example CR-17 that

, so that

CR-3.4 Exam C Bayesian Credibility Question Types Based on Discrete Prior

The techniques discussed so far have related to a discrete 2- or 3-point prior distribution.This is usually apparent in a question if we are told something like "one-third of a portfolio ofrisks is in Class and two-thirds are in Class ", or "a portfolio of risks has 50 risks of Type and 100 risks of Type ". The prior distribution would be , with prior probabilities

and . The ideas can be extended to a discrete prior distribution with anynumber of points. Bayesian problems involving a finite point discrete prior distributions havecome up regularly on Exam C. The following is a summary of the main quantities that we mightbe asked to find in exam questions.

Suppose that the prior distribution is a 2-point distribution which takes on one of the two values or with probabilities and , and the model distribution may be a discrete or

continuous random variable with conditional probability/density function and .

(i) Marginal probability of an event involving the model distribution If is an event involving the model distribution , then the marginal probability of event is

.

(ii) Posterior probabilitiesSuppose an individual is chosen at random from the population (type or is unknown) and anobserved value of the model distribution is obtained. The posterior probability that theindividual is of type is . (3.8)

Suppose that event involving the model distribution has been observed to occur (for instance,suppose that we are given that , or might be the event that ).The posterior probability that the individual is of type is .

(3.9)



(iii) Predictive probabilitiesSuppose an individual is chosen at random from the population (type or is unknown) and anobserved event for the model distribution is known to have occurred. Suppose that is anevent involving the model distribution for the next occurrence of for the same individual. Thenthe predictive probability that event occurs given that has occurred is

. (3.10)Event might be in which case , and event might be ,so that . (3.11)

(iv) Predictive expectationSuppose an individual is chosen at random from the population (type or is unknown) and anobserved event for the model distribution is known to have occurred. The predictiveexpectation for the next occurrence of is

.If event is , then .

(3.12) might be the event , and we might wish to find .

The following examples illustrate these points.

Example CR-17 Continued: Calculate the following quantities.(a) The probability that the risk class is for a randomly chosen individual given that thenumber of claims for that individual in the first year is at least 1.(b) The probability that the number of second year claims for a randomly chosen individual is atleast 1 given that the number of claims for that individual in the first year is at least 1.(c) The expected number of claims in the second year for a randomly chosen individual giventhat the number of claims for that individual in the first year is at least 1.Solution: (a)

.

(b)

.

(c) .From part (a) we have . In a similar way, we get

. Then, since and B 22 2

.



Example CR-14 Continued: Recall that 15% of the drivers are youths and 85% are adults.The table of claim number probabilities is on page CR-27 where the example was first presented.A driver is chosen at random from the insured population and it is found the driver had 1 claim ina two-year period. Calculate each of the following.(a) The probability that the driver is a youth.(b) The probability that the same driver will have at least one claim in the third year.(c) The expected number of claims for the same driver in the third year.Solution: Let and denote the number of claims in years 1 and 2, respectively,(a) We can do calculations in the following order. Youth , Adult ,

Note that to find we consider the two combinations .

We always assume conditional independence, so that ,

and , so that .

(b) .From (a) we have . We calculate the numerator using the sameapproach that was used to find in (a). This is summarized in the followingseries of calculations. Again, it is assumed that and are conditionally independentgiven either or .

Youth , Adult ,

.

We can also find from the relationship

.



Example CR-14 continuedSince the prior distribution consists only of the two classes and ,

.From the calculations above, this becomes

.In most cases, the first method of finding will generally be moreefficient.

(c) .We have already found in (a), so that

(as in (b)).From the description of the model distributions, we have and .Then, .

Note: Bayesian premium Exam C questions may ask for the calculation of the . This is thepredictive expectation of the next occurrence of the model random variable given whateverinformation is available about observations of that have already been made. The general formof the Bayesian premium given information about the first occurrences of is

. For a Bayesian situation with a two-point priordistribution based on categories and for the prior, we would calculate this predictiveexpectation as

. (3.13)



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 andthe 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'sclaims for that policy). Find .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 withparameter , where the prior distribution of is a discrete uniform distribution on the integers

3 . A single observation of is made.

3. Find the joint pf .A) B) C) D) E)

4. Find the mean of the posterior distribution given that .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




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 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. - Loss sizes for property risks follow a Pareto distribution 1,000 and .(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. Determinethe limit of the posterior probability that this risk is a liability risk as 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

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 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 ( isnon-random), and the prior distribution of is Poisson with parameter . Find the posteriordistribution of based on one observation of .

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) (ii) (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 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 thenobserve 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

Two spinners, and , are used to determine claim cost. Spinner has two areas marked 12and . Spinner has only one area marked 12.

Spinner1/2 1/2

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, thesame 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.




1. 2nd red 1st red 2nd red 1st red1st red

2nd red 1st red bowl 1 bowl 1 2nd red 1st red bowl 2 bowl 2

. Note also that

2nd red 1st red 2nd red bowl 1 bowl 1 1st red 2nd red bowl 2 bowl 2 1st red

. Answer: E

2. Type 1 Type 1 Type Type

.

Alternatively, Type 1 Type 1 Type Type 2

. Answer: C

3. Answer: D

4. The marginal pf for is

and the posterior distribution of has pf

, and

222

Answer: C



5. (a)

.

.

.

.In Example CR-14 we found

.

In a similar way, we get , and

.

Then, since and , we have .

(b) We are to find and ..

In part (a) we found .We find in a similar way.

.Then .

.Then .

6. Type A A A

. Answer: B



7. Type TypeType

A A B B C C .As in Problem 6, we find B and C .Also, A , and similarly,

B and C Then, . Answer: A

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 ).Therefore Then

both shots land in same Areaboth 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 hitthe same Area, and if the two shooters have one Area in common, then there is probabilitythat both hit the same area. The probability in question is then .

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.

.

. .

(b)

As the limit is .

10. The model distribution is .

The joint distribution of and is

.

The marginal distribution of is

.The posterior distribution of is

,

,



10. continued

For the given vector of values, this becomes

The posterior mean is .Answer: A

11.(a) Class ] Class ] Class

Class ] , and Class ] .In Example CR-15 it was found that Class , and

Class . Thus,

(b) We wish to find Class , . This probability is,

, . The numerator is

, .The denominator is

, ,

Then Class , , and

We are also asked to find , . This predictive expectation isClass Class ,

Class Class , .

12. We wish to find . We condition over the combinations ofFront/Side air bags , Wall/Lake accident , so that

We immediately get that since with side air bags there will be 2 or 4dummies and with a lake crash the minimum damage per dummy is 1, so that the minimumoverall damage with is 2. Therefore,

We find (2 or 4 dummies equally likely, average

damage of for each dummy) ,



12. continuedand

.

We then calculate ,

,

and , where , and

since there must be 2 dummies (prob. .5) and each sustainsdamage of .5 (prob. .5 each),

(either 1 dummy and damage of 1, or twodummies and damage of .5 each), and

.

Then.

Then

and .

We can summarize these conditional probability calculations as follows.

Finally, .Answer: C



13. The model distribution is ,

and the prior distribution of is Poisson with , so that the joint distribution of

and is .

The marginal distribution of is (the summation

starts at because as a binomial distribution with parameter , it must be the case that). Applying the change of variable to the summation results in

.

The marginal distribution of is Poisson with parameter .

The posterior distribution of has probability function

for (this is a translated Poisson; a Poisson with parameter translated by ).

14. Class 1 Class 1 Class 2 Class 2 .

Class 1 ,Class 2 .

Given Class 1 Class 2Given Class 1 Class 2 Class 1 Class 2 Class 1 Class 1 Class 2 Class 2

Class 1 Class 2

Class 1 ,Class 1

Class 2 .

Then . Answer: B



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 (15months) between claims, so that the expected number of claims per year for Class 1 is

. 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 .

Let us denote the number of claims in year 1 by and the number of claims in year 2 is denotedby . We wish to find , and then, since each claim cost is 1,000 in all classes,the expected loss in year 2 is .

We find by conditioning over the Class type.Class I Class I

Class II Class II Class III Class III .The conditional expectations are the expected number of claims in a year for a given class:

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 , III

I II III

I II III I I II II III III

Then, I II III , and

Class I .Class I

Class II .Class II

Class III .Class III

Then , and the expected lossfor year 2 is . Answer: A

16. The prior parameter has distribution .prob.

2 prob.The model distribution has a compound distribution with Poisson frequency with mean , andthe stated severity distribution.(a)

1 claim for amount 2 2 claims for amount 1 each .



16. continued


.

and .

(b) .

(c)(i) (from part (b) ).

.Since has a compound Poisson distribution with , the mean is 2,and the variance is , where is the severity distribution.From the distribution of we have , and then .Similarly,

.

From part (a), we know the posterior distribution of given , so that

.

Then .

(ii) To find , we first note that .Then, .We found the posterior in part (a),

and .Then, ,and .

To find , we note that .Then, From the posterior distribution of found in part (a), we have

, so that .

Then,



17. Suppose we use the following notation:aggregate claims in second year for the selected risknumber of claims in first year for the selected riskamount of first claim in first year for the selected riskamount of second claim in first year for the selected risk

We wish to find .This can be found by conditioning over the risk class.

selected risk is from Class A Class A selected risk is from Class B Class B .If the selected risk is from Class A, then has a compound Poisson distribution with Poissonparameter (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 that

Class A (Poisson parameter)(expected claim amount) .In a similar way, we get Class B .

Once the conditioning relationship is set up, most of the work in this problem is in finding theconditional probabilities Class Aand Class B .This requires the use of rules for conditional probability.

Class A andClass A

Class A A A .

If it is known that the risk is from Class A then we know the distributions of and , so we cancalculate A . We are told that a risk is selected randomly. Thismeans that risk Classes A and B are equally likely to be chosen, so that A . Then

Class A A A

From independence of and the 's, it follows thatA A Two Claim amounts are 1, 3 A .

Since claim amount has a continuous (exponential distribution), we use the density 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 theexponential distribution with mean is .

Then AA Two Claim amounts are 1, 3 A .

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 A(this is the numerator of the probability Class A that we are tryingto find).



17 continuedIn order to find the denominator we use the fact that the selectedrisk must be either from Class A or from Class B, so that

Class A Class BWe find Class B in the same way we found

Class A .

Class B B B .Using the Poisson frequency (mean 3) and exponential severity (mean 3) from risk Class B, we

get B ,

and then Class B .

We then have

Now we can findClass A Class A

.

We can use the same reasoning to find Class B , which will turn

out to be .

Alternatively, and , once Class A is known,more efficientlyClass B is equal to its complementClass B Class A .

Finally, the posterior expected value we are looking for is

selected risk is from Class A Class A selected risk is from Class B Class B . The main work in this problem was in finding Class A .The following "table" summarizes the steps outlined to find that probability.

, given , given

Aand

Bare both know from the given distributions

Class A and

Class B .(this calculation uses the compound Poisson distribution)



17 continued

Class A Class B

Class A Class A

and

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 aregiven that is a two point random variable that can take on the values 2 and 4 with

and (this means that any randomly chosen individual has anequal chance of being 2 or 4). We are given that for a randomly chosenindividual ( and 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 forthat insured in year 3 given the information about years 1 and 2, so we are asked to find

. The typical way to find this expectation is to condition over ; we canwrite the expectation as

.(Note that if the distribution of had been continuous, say uniform on the interval ,then we could write as ,and we would have to find the conditional density of given ; this conditionaldistribution of would not necessarily be uniform).

Since given has a Poisson distribution, we have and .We must find the conditional probabilities and

. We now use standard methods of conditional and joint probability (ordensity in the continuous case). We use the relationship

,with events and defined as follows:

: , : , and (complement of ): . Then

|| | .

We are given that , and the distribution of given isPoisson with parameter . It must be implicitly assumed that for a randomly chosen individualgiven the value of , 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 conditionaldistributions and are independent, but the unconditional distributions of and are not generally independent.Therefore

, and

.

Then,

.Since is either 2 or 4, is the complement of so that .Finally,

. .

The main work in the solution of this problem was to find and . The calculations needed to find these probabilities can be summarized

as follows. and and The probabilities for were given explicitly, and the conditional probabilities follow since isPoisson given .

and

.

.

and

.

Answer: C

19. We wish to find . This can be written as.

The numerator is

,



19 continued

and the denominator is .

The Pareto distribution with has pdf ,

and cdf .

,so that is the denominator.

From conditional independence of the 's given , we have ,

and the same for , so that,

and and3 .3

Then, 3

is the numerator, and then . Answer: C

20. Since we are given one observation (total annual losses of 500), the Bayesianpremium is . Since the "parameter" in this Bayesian situation is the class A orB, the prior distribution is (risks A and B are equally likely). For each ofthe two classes, total annual loss has a compound distribution.The Bayesian premium can be formulated as

.Within class A, .Within class B, .

1 claim claim of 500 ,2 claims claim of 250 .

Then, .

In a similar way, .Then, . Answer: A



21. There are 4 possible die/spinner pairs, each with prior probability of :

. The associated expected losses are

.In order to find the posterior expectation we use the relationship

.To find the posterior probabilities, we first find

The posterior probabilities are then

.

The posterior expectation is .

We are given that this is equal to 10. Solving for results in . Answer: E

22. We first fine the posterior probabilities for the parameter :



.

and .The Bayesian premium is

.

CREDIBILITY SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR CR-79


CREDIBILITY - SECTION 4BAYESIAN CREDIBILITY, CONTINUOUS PRIOR

The material in this section relates to Section 16.4.2 of "Loss Models". The suggested time framefor this section is 3 hours.

The application of Bayesian analysis to credibility estimation can be summarized as follows. Therandom variable (usually loss frequency or severity, or perhaps a compound aggregate claimsdistribution) has a pf/pdf that depends on a parameter , where is a random variable aswell. The initially assumed distribution of is referred to as the , and hasprior distributionpf/pdf . The distribution of based on a value is the , but it is really amodel distributionconditional distribution of given , with pf/pdf . The prior distribution of can be discrete or continuous, and the model distribution can be discrete or continuous. Underthe Bayesian analysis approach, an observation (or several observations) are made from thedistribution of , and then an updated form of the distribution of is found; this is called theposterior distribution.

The posterior distribution of is a conditional distribution given the value of (or values of )observed. If one data value of is available, say , then the posterior distribution haspf/pdf . If several data values of are available then the posterior distributionhas pf/pdf . It is then possible to extend this analysis tofind the of the next occurrence of . If one data point of has beenpredictive distributionobserved, then the predictive distribution is a conditional distribution of (the next occurrenceof ) given , with pf/pdf . A predictive distribution can also befound if several data values of are available; we would find the predictive density

. This is the conditional density of the nextoccurrence of ( ) given the observed data points .

Up to now in these notes on Bayesian credibility, we have considered only the situation in whichthe prior distribution is discrete (2- or 3-points usually). When the prior distribution of iscontinuous, the Bayesian analysis leading to the posterior and predictive distributions can becomequite complicated. There are a few specific combinations of prior and model distributions whichlead to recognizable forms for the posterior (and perhaps the predictive) distribution. We firstsummarize the general Bayesian analysis procedure and then consider a number of examples inwhich the prior parameter has a continuous distribution.

CR-80 CREDIBILITY SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR


CR-4.1 The Bayesian Structure

General Bayesian estimation has the following components.

The initial assumption for the distribution of the parameter is called the , andprior distributionit has a pdf/pf (continuous/discrete) often denoted ( ). A typical interpretation of is that thereis a population of individuals (also called "risks", or "policyholders") and each one has his ownvalue , and these values are distributed over the population according to the prior density of .

The (usually frequency or severity of loss or aggregate loss) is amodel distributionconditional distribution given , with pf/pdf (or more simply ).For a data set of random observed values from the distribution of for a particular member ofthe population, say , and a specific possible value of , the ismodel distribution

. (4.1)

We are implicitly assuming that are conditionally independent given .In Exam C questions is often 1 (or a small integer)

The of and has pf/pdf ( ) .joint distribution In the multiple data-point situation, and have joint pdf/pf

(4.2)(note that only one factor of ) appears in this joint density).

The of is .marginal distribution In the multiple data-point situation, have joint pdf

. (4.3)The limits of the integration are the original limits of the prior distribution of .

The of given has pf/pdfposterior distribution (for continuous ), (4.4)

where the integration is over the range of the prior distribution of .Also, as we have seen earlier for discrete .

allIn the case of multiple data points, the posterior pdf/pf is

(for continuous ), (4.5)

integrated over the range of 's distribution; and

all (for discrete )

Given the data , the of (a new observation) has pdf/pfpredictive distribution (4.6)

(in the continuous case) (in the discrete case)

Note that both the prior and model distributions can be continuous or discrete in any combination,but in this section of the notes we will consider continuous prior distributions.



In a Bayesian credibility situation we may be given a single data value for , or we may be givenseveral data values , or we may be given the sum of several data values . Typically weare asked to find the density function or mean of the posterior distribution of , or we may beasked to calculate a probability related to the posterior distribution. We also may be asked tocalculate a probability or expectation for the predictive distribution. It is less likely that we willneed to find the marginal density of or that we will be asked to find the density of thepredictive distribution.

A useful point to note regarding Bayesian analysis is that the posterior density of is generallyproportional to (a constant multiple of) the joint density of and (or and ).This is true since the posterior density is ( is constant with respect to ). Therefore, once we have formulated the joint density of and , we might know what the form of the posterior density must be in terms of . Some priorand model distribution pairs combine in a convenient way to give a joint density that has a fairlyrecognizable posterior density, and some do not. The following example illustrates a prior-modelpair that results in a posterior distribution which is in the Table of Distributions. The examplealso illustrates a few different ways in which information can be presented.

Another point note is the similarity between the formulation of the density of marginaldistribution of and the density of the predictive distribution of given .The density of the marginal distribution of is , and the densityof the predictive distribution is

.Both integrals involve the model density, in the marginal case, and in thepredictive case. The difference is that in the marginal case the integral uses , the prior densityof , and the predictive case the integral uses , the posterior density of .

CR-4.2 Examples To Illustrate A Continuous Prior

Example CR-18: The distribution of (the ) is assumed to be exponentialmodel distributionwith parameter , so that (we use instead of so as not to get confused withthe prior distribution that we will use in this example) The of is assumed toprior distributionbe an inverse gamma distribution with parameters and , and has pdf

for .

The of and isjoint distribution .

(a) Suppose that we are given a single data point with value . The joint density becomes .

The of will be , and for the specific value ofmarginal density

this becomes , which is some numerical value.



Example CR-18 continuedWe don't actually need to find the numerical value of the integral in order to identify the posteriordistribution. In the next section (Section 5) of these notes there will be a few comments about thetypes of integrals than can arise in the Bayesian credibility context, and comments made therewill show how to find the integral above in a fairly efficient way.

The of will be , which for becomesposterior density

.

Even though we do not know the value of , we know that the posterior density, , is some constant multiplied by (the constant factor can be ignored

for the time being).If we search the table of distributions in the Exam C tables, we see that the general form of theinverse gamma distribution with parameters and has pdf (we use instead of as thevariable) , which is proportional to +1 +1( )

( is multiplied by some constant not involving ).+1

Since the posterior distribution that we are considering is proportional to , it follows thatthe posterior distribution must have an inverse gamma distribution with parameters and

(since ). This is true for the following reason. We know that the posteriordensity is , where is a constant not involving (since the posterior

density is proportional to ). In order to be a properly defined pdf, it must be true that

. But we know that the inverse gamma distribution with parameters

and has pdf , and we know that( ) ( )

. Therefore is follows that must be the same as( )

( ) , since they both integrate to 1, and we see that the posterior density is the density ofthe inverse gamma with and .

In other words, once we recognize that the algebraic form of the variable in the posteriordensity is the same as the inverse gamma with and , we can conclude that theposterior distribution is that inverse gamma. The posterior density will be

+1 ( ) . Therefore we have found the posterior density withoutdirectly actually having to find the numerical value of (in fact, the numerical value of

must satisfy the relationship , so that , although( )this may be an unnecessary calculation).

We can also quickly find the mean of the posterior distribution, since the mean of the inversegamma random variable is . The mean of this posterior distribution is



Example CR-18 continued(b) Suppose that we are given several data points, say a random sample of 8 values fromthe distribution of all based on the same unknown value of . Suppose that the random sampleis .The joint distribution now involves and . We will continue to assume that themodel distribution of is exponential with parameter , so that and that theprior distribution of is an inverse gamma distribution with parameters and , andhas pdf for .

The joint distribution of and has density

.

The same analysis applied to part (a) of this example shows that the posterior distribution must beproportional to , which implies that the posterior distribution must have an inversegamma distribution with and (since ). The density function of

the posterior would be .+1 ( )The mean of the posterior distribution is

(c) Suppose that we are told that there are 8 data points, , but we are only told that

the sum of the 8 data values is (without knowing the individual values). The model

and prior distributions involved in this example allow us to proceed just as in part (b). In part (b)the only information we really used regarding the 8 data values was that the sum of them was225. The joint density of and was , which only needed

. The same posterior distribution would result.

There are a few important observations to make from Example CR-18. First, it is useful to befamiliar with the distribution types in the Exam C Tables. Second, an inverse gamma priordistribution combines with an exponential model distribution to result in an inverse gammaposterior distribution. In the next section several prior-model pair combinations are listed thatresult in a posterior distribution that is of the same type as the prior distribution. In such a casethe prior distribution is said to be a .conjugate prior for the model distribution

The third observation is related to part (c) of the example. It is possible that the sum ofobservations is given rather than the individual values. For some prior-model distribution pairsthis is all that is needed since in the expression for the joint distribution, only the sum of the 's isneeded, as was the case in part (c) of Example CR-18. A fourth point to note is that it may not benecessary to find the marginal distribution of , although the marginal distribution of may alsohave a recognizable form. In part (a) of the example, it could have been shown that the marginaldistribution of is Pareto with the same values of and as are in the inverse gamma priordistribution of . In other words, with integration by parts we get

, where and .



Therefore, the marginal distribution of is Pareto with and . For Exam Cquestions. it is not usually necessary to find the marginal distribution of in a Bayesiancredibility situation.

It is also possible that the prior-model pair do not combine in such a recognizable way. Thefollowing example illustrates this.

Example CR-19: A random sample of 8 values from the distribution of is given:

The distribution of is assumed to be exponential with parameter , so that The prior distribution of is assumed to be uniform on the interval , so that for In the expressions below, (boldface) denotes the vector .

(i) the for the given data points ismodel distribution

(ii) the of the 's and is ; for the given data pointsjoint density

this is

(iii) the of a set of observations ismarginal distribution

(iv) the of is ; evaluated at the given dataposterior distribution

points, this is for 20

(it is possible to integrate the denominator exactly, but the number is a computer-generated approximation of ). The mean of the posterior distribution of is 14.9 .

(v) the of a new observation has densitypredictive distribution

(which can be integrated, but is a rather nasty function). Using a computer integration package,the probability that a predicted value is is .If we use the Bayesian estimate , then the distribution of given is exponential withparameter , and .

Example CR-19 shows that even when the prior and model distributions are fairly elementary(uniform, exponential in this case) the marginal and posterior distributions can be complicated.A main objective of Bayesian credibility is to determine the Bayesian credibility premium. Thiswas mentioned earlier, and is equal to the predictive expectation. Suppose the prior parameter is

and the model distribution has density . Suppose we are given a data set of observations from a particular member of the population ( unknown), .



The usual way to find the predictive expectation is to first find the posterior distribution,which has density , and then

. (4.7)The integration is over the range of values of .

It is often the case that the mean of the model distribution is (for instance, ifthe model distribution is Poisson or exponential). In that case, the predictive mean becomes

which is same as the posterior mean so that we can find the,predictive mean if we can identify the posterior distribution and find its mean. This is apoint worth noting.

In Example CR-18, the model distribution is exponential with density function .Therefore the mean of the model distribution is , the prior parameter.From the previous paragraph, it follows that in Example CR-18, the predictive mean will be themean of the posterior distribution. For part (a) of that example, given the single data value

, it was found that the posterior distribution had an inverse gamma distribution withparameters and . The mean of this inverse gamma posterior distribution is

. Therefore, for part (a) of Example CR-18 the predictive mean is equal to theposterior mean, which is .

Example CR-20: You are given the following: - Claim sizes for a given policyholder follow a distribution with density function . - The prior distribution of has density function .The policyholder experiences a claim size of 2. Determine the expected value of a second claimfrom this policyholder.Solution: The approach to finding a predictive expectation that is usually most efficient is to usethe relationship d .This requires that we find the conditional mean , and also the posterior pdf

.

In this example, is .

The posterior distribution has pdf, where and

(the actual lower limit of the integral must be 2, since if , then it is impossible for to be 2,since ). This is one of the occasional cases in which the marginal density of must becalculated.Then,

Thus, , for , and

for .



Example CR-20 continuedThen, .

There is an alternative way of finding the predictive expectation that first requires finding thepredictive density function. In general, this is not the most efficient method for finding apredictive expectation.

, where .

As above, we find the posterior pdf to be

, .

Then, if andif , .

Then we have .This is a more complicated way of finding the predictive expectation because of the complicatednature of the predictive density. The predictive density has two different forms

depending upon whether or not .

Example CR-21: The prior parameter has a uniform distribution on the interval . Themodel distribution has a Poisson distribution with a mean of . A single observation of ismade, and the value of is 0. Find the posterior probability .Solution: Since must be between 2 and 4, the posterior probability can be formulated as

. The posterior density is , where is the jointsdensity and is the marginal probability the .The joint density is (because is exponential, and is uniform on ).The marginal probability that is .

The posterior density is then for .

Then, .2

This posterior probability can be expressed as

.

Note that the numerator and denominator are both integrals of the joint density , but thenumerator is over a subinterval of the denominator. The denominator is an integral over the entireregion for , and the numerator is over the interval for the probability in question. Questions of the type in Example CR-21 have come up a few times on Exam C. We generallymust find the joint density (if there is one sample value ), and theposterior probability is a ratio of two integrals of , with the numerator integration overa subinterval of the full integration in the denominator. If there are multiple sample values

, then the joint density is .

This is integrated in the numerator and denominator over intervals for .



CR-4.3 The Double Expectation Rule Applied to Bayesian Credibility

The "double expectation" rule has been mentioned a few times in the context of Bayesianestimation. The basic form of the double expectation rule is that for any random variables and

, . This rule can be used to find the mean of the marginal distribution of in a Bayesian credibility situation. For instance, suppose that we consider part (a) of ExampleCR-18, where the parameter had an inverse gamma distribution with parameters and

, and the model variable had an exponential distribution with mean . The mean ofthe marginal distribution of is .

A version of the double expectation rule also applies to the predictive distribution. In general, forrandom variable and , the double expectation rule can be stated as

. Applying this to Example CR-18(a), we see that . Once we have identified the posterior

distribution of given , we see that is the mean of the posterior distribution.In the case of Example CR-18(a), with , we saw that the posterior distribution wasinverse gamma with and . Therefore,

.

A conditional approach to finding variance can also be applied. The following rule forformulating variance was reviewed earlier in the Modeling unit of this study guide. Given anytwo random variables and , .If we apply this to Example CR-18(a) to find the variance of the marginal distribution of , wehave .Since has an exponential distribution with mean , we have .Then, . Since has an inverse exponential distribution with , thevariance of is infinite (the second moment of is infinite), so the variance of the marginaldistribution of will be infinite. This could have also been anticipated because in ExampleCR-18(a) it was noted that the marginal distribution of had a Pareto distribution with ,which has an infinite variance.

We can also apply a version of this rule to find the variance of the predictive distribution, but itgets a little more complex. The general rule for conditional variance is the following. .If we apply this to the predictive distribution in Example CR-18(a), we get

.

First we have , since is exponential with mean .Then . The posterior distribution of is inversegamma with parameters and , and the variance of the inverse gamma is

, so that .

Second, we have since is exponential with mean .Then , which is the second moment of the posteriordistribution. The second moment of the inverse gamma with and is

, so that .



Putting these together, we get .

These comments, particularly about variance of the marginal and predictive distributions areprobably taking the Bayesian credibility analysis somewhat past what might be expected on anexam question.



CREDIBILITY - PROBLEM SET 4Bayesian Credibility - Continuous Prior

Problems 1 and 2 are based on the following situation. Assume that the number of claims eachyear for an individual insured has a Poisson distribution. Assume also that the expected annualclaim frequencies (the Poisson parameters ) of the members of the population of insureds areuniformly distributed over the interval . Finally, we assume that an individual'sexpected annual claim frequency is constant over time. An individual is chosen at random fromthe population and is found to have 0 claims for the year.

1. Find the pdf of the posterior distribution of , .

A) for B) for C) for

D) for E) 1 for

2. Find the Bayesian Premium.A) .40 B) .42 C) .44 D) .46 E) .48

3. The parameter has a prior distribution with pdf for .The conditional distribution of given is uniform on the interval .(a) Find the joint density , the marginal density of , and the posterior density , and describe the appropriate region or intervals over which they are defined.(b) Find the Bayesian premium .(c) Find the Bayesian premium , .

4. Annual aggregate claims for a particular policy are modeled as a compound Poisson distribution withPoisson parameter for the frequency (number of claims per year), and a severity (individual claim size)

that is either 1 or 2 with . An insurer has a large portfolio of policies,and each policy has its own value of . For a randomly chosen policy from the portfolio, the distributionof is exponential with a mean of 1.The claim sizes and the numbers of claims are independent of one another given .A policy is chosen at random from the portfolio, and denotes the aggregate claim for that policy for oneyear.(a) Find .(b) Find the posterior density of if the observed aggregate loss for the year is 1.(c) Find the Bayesian premium for the next year if the observed aggregate loss this year is 1.



Problems 5 to 15 relate to the severity distribution with pdf for ,where has pdf e for .

5. Find the conditional mean of given .A) B) C) D) E)

6. Find the mean of the marginal distribution of .A) 1 B) 2 C) 3 D) 4 E) 8

7. Find the pdf of the marginal distribution of A) B) C) D) E)

8. Find the pdf of the posterior distribution of given .

A) B) C)

D) E)

9. For a random sample , find the pdf of the joint distribution of and .

A) B)

C) D)

E)

10. Find the pdf of the marginal distribution of .

A) B) C)

D) E)

11. Find the pdf of the posterior distribution of given .

A) B)

C) D)

E)



12. Find the pdf for the predictive distribution of given a single sample value .A) B) C)

D) E)

13. Find the Bayesian premium .A) B) C) D) E)

14. Find the pdf of the predictive distribution of given the sample .A) B)

C) D)

E)

15. Find the Bayesian premium .A) B) C) D) E)

16. The number of claims in one exposure period follows a Bernoulli distribution with mean .The prior density function of is assumed to be .

Hint: and .

(a) Determine the expected value of the conditional variance .A) B) C) D) E)

(b) The claims experience is observed for one exposure period and no claims are observed.Determine the posterior density function of .A) B) C)

D) E)

17. Find the mean and variance of the posterior distribution in Example CR-18(b).Use the double expectation rule to find the mean and variance of the predictive distribution of

in Example CR-18(b).

18. Find the mean and the variance of the marginal distribution of in Example CR-19.



19. (SOA) The size of a claim for an individual insured follows an inverse exponential

distribution with the following probability density function: ( ) , 02

The parameter has a prior distribution with the following probability density function:

( ) , 04

4One claim of size 2 has been observed for a particular insured.Which of the following is proportional to the posterior distribution of ?(A) (B) (C) (D) (E) 2 3 4 2 2 2 9 4

20. You are given:(i) An individual automobile insured has annual claim frequencies that follow a Poissondistribution with mean .(ii) An actuary’s prior distribution for the parameter has probability density function: ( ) = (0.5)5(iii) In the first policy year, no claims were observed for the insured.Determine the expected number of claims in the second policy year.(A) 0.3 (B) 0.4 (C) 0.5 (D) 0.6 (E) 0.7

21. The number of claims per period has a Bernoulli distribution with mean .The parameter has a uniform distribution on the interval .The parameter has pdf , . Find .(A) (B) (C) (D) (E)




1. We wish to findSince the model distribution is Poisson, we have , and since the priordistribution is uniform on we have 1 for . The marginal probability

is .

Then for . Answer: B

2.

Note that the antiderivative of is . Answer: B

3.(a) The model density is for .

The joint density is on the (triangular) region

.

The marginal density of is for .

The posterior density is

for (this is uniform on the interval ).

(b) .

(c) The joint density for and is for .

The marginal density for is .

The posterior density is

for .

The Bayesian premium is

.



4.(a) .

(b) .

, and .

Then , and8

84 .

(c) .

5.Answer: B

6. .Answer: B

7 e for Note that we can find from the marginal distribution of , by

finding . Applying the change of variable , this

integral becomes . The method applied inproblem 4 is more efficient. Answer: B

8. . Notice the prior and posterior distributions are both gamma distributions, so that the gammadistribution is a conjugate prior for the model distribution. Answer: C

9. The joint pdf is

e .

Answer: E



10.

We have used the rule

if is an integer and . Answer: D

11.

Notice that the algebraic form of the posterior distribution involves the parameter in the sameform as the joint distribution. Answer: C

12. . From Problem 5,

From Problem 7, .

Then, . We can also find from the relationship

.Answer: D

13. To find the Bayesian premium we use the relationship

With from Problem 3, and from Problem 6,we get

3

.Alternatively, using the pdf of the predictive distribution from Problem 10, we have

(this is the Bayesian premium). This integration can be done by a symbolic computing program.Answer: D



14. .

From Problem 8, we have

and

.

Then . Answer: C

15. The Bayesian premium can be found in two ways:(i) , or(ii) .Both methods were used to solve Problem 11 above when the sample was a single value. In thegeneral -sample case, approach (i) can become quite complicated due to the complicated natureof the pdf of the predictive distribution, .Approach (ii) is often more straightforward. From Problem 3 we have and

from Problem 8 we have .

Then

(in this integral, the following general integration form was used -if is an integer and , then , in this case and

). Note that this Bayesian premium can be written in the form ,

where is . Answer: B

16. (a) If is Bernoulli given , then the variance of is .The expected value of this is , and from the hint, we have

, and .

Then, . Answer: B

(b)

.

Answer: E



17. It was seen in Example CR-18(b) that the posterior distribution has an inverse gammadistribution with and . The mean is , and the

variance is .

The mean of the predictive distribution is . This is the same as the mean of

the posterior distribution, which is 14.11.

The variance of the predictive distribution is .

, so that , which is the variance ofthe posterior distribution, .

, since has an exponential distribution given , and the variance of theexponential is the square of the mean. Then, ,which is the second moment of the posterior distribution, .Then, .

18. has a uniform distribution on the interval , and the conditional distribution of has an exponential distribution with mean . Using the double expectation rule, we have

Using the conditioning rule for variance, we have .

(variance of the uniform distribution).

(2nd moment of the uniform distribution).Then .

19. Given the prior distribution of , say with pdf , and given the "model distribution" of given , say , we construct the joint distribution of and :

. The marginal distribution of is found by integrating (or summing inthe case of discrete ) the joint density over the range of -values.

. Then, the posterior distribution of given has pdf

, which is proportional to , since it is equal to divided by thefactor which does not depend on .From the given information, , and with one claim size of

, this becomes . From the comments above, the posteriordensity of is proportional to . The only one of the answers which is proportional to thisexpression is B. Answer: B



20. The problem involves determining the Bayesian premium. The prior distribution of theparameter has density (a mixture of two exponentialdistributions, one with mean and on with mean 5, which can also be regarded as a mixture oftwo gamma distributions, the first with the second with ).The model distribution, given , is Poisson with mean .We are given one observation of , which is in the first policy year.We wish to find , which is the Bayesian premium.We can write the expectation as .Since given has a Poisson distribution with mean , this integral becomes

which is the mean of the posterior distribution.Therefore, if we can identify the posterior distribution, it may be easy to determine its mean.

The joint density of and at is

, and the marginal probability that is

.The posterior density of given is then

.

This can be written in the form .Therefore the posterior distribution is a mixture of two exponential distributions, with mixingweight for the exponential with mean , and mixing weight for the exponential with mean

. The mean of the posterior distribution is then . Answer: A

21. For a Bernoulli distribution with mean , and . is found by conditioning over the parameter ;

.The pdf is found by conditioning over the parameter ;

We are given that given has a uniform distribution on and therefore for and for .It follows that in the integral , the conditional density isequal to 0 if . Therefore, the integral becomes

Then, Answer: D

CREDIBILITY SECTION 5 - BAYESIAN CREDIBILITY, DISTRIBUTION TABLE CR-99


CREDIBILITY - SECTION 5, BAYESIAN CREDIBILITY APPLIED TODISTRIBUTIONS IN EXAM C TABLE

The material in this section relates to Section 16.4.2 of "Loss Models". The suggested time framefor this section is 3-4 hours.

The first important step in proceeding with Bayesian credibility is to identify the prior and modeldistributions. There are a number of combinations of prior-model distribution pairs that result inrecognizable marginal, posterior and predictive distributions. These are summarized a few pageslater in this section of notes. Before that summary, we consider in detail one particularlyimportant prior-model combination (it has come up on exam questions more often than others).The following example involves some integrations that you would not likely be expected to findunder exam conditions, but there are a couple of more general points in the example that areworth noting.

CR-5.1 The Gamma-Poisson Credibility Model

Example CR-22: A portfolio of insurance policies has the following characteristics. Eachpolicy has a Poisson claim frequency distribution , but the mean of the Poisson distribution (say

) varies from one policy to another. For a randomly chosen policy from the portfolio ofpolicies, the parameter has a gamma distribution with known values and (we use thenotation to avoid confusion with the conventional use of in the gamma distribution). Apolicy is chosen at random from the portfolio.(i) Find the joint distribution of and , and find the marginal distribution of .(ii) The observed value (number of claims) of in one period is . Find the posteriordistribution of .Find the predictive distribution of given (for that same randomly chosen policy).Find the predictive expectation of given .(iii) The observed values of in three successive periods are (it is assumed that oncethe policy is chosen, the same policy is observed for three periods).Find the posterior distribution of .Find the predictive distribution of given .Find the predictive expectation of given .Solution: The prior distribution of has pdf , and

the model distribution of claim frequency given has pf ( is an integer ). The joint distribution of and has pdf

, which can be written as .

(i) The marginal distribution of is found by integrating out ; .This can be a complicated integral (it will involve the gamma function, but we will see someintegration "shortcuts" a little later), but it reduces to the probability function for a negativebinomial distribution with and , which is

; this is the marginal distribution of .

CR-100 CREDIBILITY SECTION 5 - BAYESIAN CREDIBILITY, DISTRIBUTION TABLE


Example CR-22 continued(ii) The posterior distribution of has pdf

,

where is a factor that does not involve . The expression occurs in the pdf of/ 1+

the gamma distribution with parameters and . Therefore the posteriordistribution of is a gamma distribution with new parameters, and .

The predictive distribution has pf .This will be a complicated integral when the pdf's are written down. The integral will reduceshowing that the pf for the predictive distribution is negative binomial, with new parameters,

and .

The predictive expectation will be

(this last equality is true because given has a Poisson distribution with mean ).This is equal to the mean of the posterior distribution, which is .Notice that once we identified the predictive distribution as negative binomial with parameters and , the predictive mean can also be written as .

(iii) The posterior distribution of has pdf

.

where is a factor that does not involve . The expression occurs inthe pdf of the gamma distribution with parameters and .Therefore, the posterior distribution of is a gamma distribution with new parameters,

and .As in (ii), the predictive distribution will be negative binomial with and . The predictive mean will be , which is the same as the mean of theposterior distribution in this example.

There are few points to note about Example CR-22.1. The posterior distribution has pdf of the form ,where is a factor which does not involve . Therefore, the distribution family type for theposterior distribution can be identified by the way the parameter occurs in the expression for

. This is true for many combinations of prior and modeldistributions. This means that it is usually not necessary to actually find in order toidentify the posterior distribution (finding can be a tedious exercise, as in Example CR-22). For example, suppose that a pdf is given as , where does notinvolve ; then this must be the pdf for a gamma distribution with parameters and

. This comment also applies when there is more than one -value given, as in part (iii) ofCR-21.



2. When there are multiple -values given, it may not be necessary to know the individual-values, we may only need to know the sum of the -values. This is the case in (iii) of CR-21,

where the posterior and predictive distributions depend on .

3. For a Poisson distribution with parameter , the mean of the distribution is the parameter .This is also true for the exponential distribution (the mean is the distribution parameter).If the model distribution satisfies the relationship , then the predictive mean is equalto the posterior mean.

4. The prior distribution of in Example CR-22 is a gamma distribution, and the posteriordistribution in parts (ii) and (iii) are also gamma distributions with modified parameters. Whenthe posterior distribution is from the as the prior distribution (such assame family of distributionsin Examples CR-18 and CR-22), that distribution is referred to as being a conjugate prior forthe model distribution. From Example CR-22 we see that the gamma distribution is a conjugateprior for the Poisson distribution. The gamma prior/Poisson model combination has come up inExam C questions on a regular basis, and it is worthwhile being familiar with it.

CR-5.2 A Few Useful Integration Relationships

One of the requirements for a function to be a valid density function is that the integral of over the entire probability region must be 1. From this observation we can derive a number

of integration relationships from the distributions that are in the Exam C table.

Gamma Distribution, parameters , : The pdf is .

Since , it follows that . (5.1)

We can now simplify the integral that appeared in Example CR-22(i) .

.A close look at this expression reveals that it is the probability function for a negative binomialrandom variable with and .

Inverse Gamma Distribution, parameters , : The pdf is .

Since , it follows that (5.2)(we must have in order for the integral to exist).

These integration relationships can be adapted to any of the distributions found in the Exam Ctable. Note also that if is an integer , .

We now summarize a number of prior-model distribution pairs that result in recognizableposterior distributions. In each case the model distribution is formulated with parameter .When looking at the joint distribution, we will isolate the part involving to try to determine theform of the posterior.



CR-5.3 Summary of Prior/Model Combinations

1. Model: Poisson dist. , , pf

Prior: gamma dist., , , pdf .( )This is the prior-model distribution pair in Example CR-20, and it is the most important of allcombinations presented here (in terms of how often it has occurred on Exam C).

(a) With a single data value available, the joint density of and is

( ) , which is proportional to .

The and .posterior distribution must be gamma with parameters

An important related property is that the marginal distribution of can be shown to benegative binomial with and . We can also find by the double expectationrule,

(b) With data values and the individual values or the sum available, theeitherjoint density of and is


The and .posterior distribution must be gamma with parameters

Since both prior and posterior distributions are gamma distributions, the gamma distribution is aconjugate prior for the Poisson distribution. Note that the variance of the posterior distribution isthe gamma distribution variance .

In both (a) and (b) the withpredictive distribution will be negative binomial and , and the predictive mean is the same as the mean of

the posterior distribution, . (5.3)Note that the variance of the predictive distribution will be the variance of the negative binomial

. (5.4)

A special case arises when in the prior distribution. The prior becomes an exponentialdistribution and the marginal distribution of becomes a geometric distribution. Theposterior will be a gamma distribution. If the observation is 0 (or all are 0) then theposterior will have and will still be exponential.



2. Model: exponential dist. , , pdf

Prior: inverse gamma dist., , , pdf .+1 ( )This is the prior-model distribution pair in Example CR-18.


+1(

+( ) , which is proportional to .

We can find the mean of the marginal distribution of by the double expectation rule, (5.5)

The with , and . (5.6)posterior distribution must be inverse gammaSince both prior and posterior distributions are inverse gamma, the inverse gamma is a conjugateprior for the exponential distribution.

Although usually of less importance, it is possible to show that the marginal distribution of isPareto with the same and as in the prior distribution, and the predictive distribution of given is a Pareto distribution with the same and as the posterior distribution.T (5.7)he predictive mean is

(b) With data values and either the individual values or the sum available, thejoint density of and is

+1(

+( ) , which is proportional to .

The with , and .posterior distribution must be inverse gammaIn both (a) and (b) an exponential model distribution combines with an inverse gamma priordistribution to produce an inverse gamma posterior distribution. The inverse gamma is aconjugate prior for the exponential distribution.

The predictive mean is (5.8)

A variation on this example would have the model distribution being a gamma distribution with known and unknown. For instance, suppose that in the gamma model

distribution . Then the model density is . We are still assuming that the prior( )

density is inverse gamma with parameters that are still called and , so that .+1 ( )

For a single observed data point , the joint density of and is

( ) ( )+1 + , which is proportional (in ) to . From this expressionwe see that the posterior distribution must have an inverse gamma distribution with and . We see that a gamma model distribution combines with an inverse gamma priorto produce an inverse gamma posterior distribution. The case considered above in which themodel distribution is exponential is really just a special case of this one, since an exponentialdistribution is a special case of the gamma distribution.



3. Model: binomial , pf

Prior: beta , pdf Note that the mean of this beta distribution is .

(a) With a single data value of available, the joint density of and is ,

which is proportional to .The posterior distribution must be a beta with parameters and . (5.9)Since both prior and posterior distributions are beta distributions, we see that the beta distributionis a conjugate prior for the binomial distribution.


,

which is proportional to .The posterior distribution must be a beta with parameters and . (5.10)

The predictive expectation will be ,

since has a binomial distribution with parameters and . Therefore, the predictive meanis posterior mean . (5.11)

4. Model: inverse exponential dist. , pdf

Prior: gamma dist., , , pdf .( )


( ) , which is proportional to .The with , and posterior distribution must have a gamma distribution

. (5.12)

Although usually of less importance, it is possible to show that the marginal distribution of isinverse Pareto with and the same as in the prior distribution, and the predictivedistribution of given is inverse Pareto with and the same as the posteriordistribution.





The with , and . (5.13)posterior distribution must be gamma Since both prior and posterior distributions are gamma distributions, the gamma distribution is aconjugate prior for the inverse exponential distribution.

A generalization similar to that in Combination 1 applies here. Suppose that instead of beinginverse exponential, the model distribution is inverse gamma with known and unknown.For instance, suppose that in the inverse gamma model distribution. Then the modeldensity is . We are still assuming that the prior density is gamma with( )

parameters that are still called and , so that (it is a little confusing that the( )Greek letters and are used as the conventional parameters for both the gamma and inversegamma distributions). For a single observed data point , the joint density of and is

( ) ( ) , which is proportional to . We see that the

posterior distribution has a gamma distribution with and .Since prior and posterior both have gamma distributions, we see that the gamma distribution is aconjugate prior for the inverse gamma distribution.

5. Model: normal with mean and known variance , pdf

Prior: normal with mean and variance , pdf

(a) With a single data value of available, the joint density of and is , which is proportional to

( ) ( ) .

Therefore the posterior distribution is normal with mean ( ) ( ) (5.14)

and variance . (5.15)

The normal distribution is a conjugate prior for the normal distribution.

(b) With data values and either the individual values or the sum available, theposterior distribution of can be shown to be normal with mean ( ) ( ) (5.16)

and variance . (5.17)

The predictive mean will be .Since is normal with mean , it follows that the predictive mean is the same as theposterior mean, which is ( ) ( ) . (5.18)



6. distribution on the interval , pdf Model: uniformPrior: single parameter Pareto, , pdf , With observations , suppose that .The posterior distribution of is a single parameter Pareto with and , andpdf , . The Bayesian premium is (5.19)

As an example, suppose that the prior density is , , and the model density isfor Note that the prior density is a single parameter Pareto with

and .

Suppose that two observations of are made, and they are and . Since thelargest observed value of is 25, it must be true that is at least , since each observed mustbe in the interval . The joint density of and is

, and the region of joint density is and . The posterior density of is proportional to on the interval .

Therefore the posterior must be a single parameter Pareto with and .

The pdf of the posterior distribution of is , .The Bayesian premium will be

(since the posterior density is non-zero only for ).This integral is (posterior mean) .

We now consider a few examples to illustrate these prior-model combinations.

Example CR-23: You are given the following: - A portfolio of 10 identical and independent risks. - The number of claims per year for each risk follows a Poisson distribution with mean . - The prior distribution of is assumed to be a gamma distribution with mean . 5, variance .01. - During the latest year a total of claims are observed for the entire portfolio. - Variance of the posterior distribution of is equal to the variance of the prior dist. of .Determine .Solution: We reviewed the gamma prior-Poisson model case in Combination 1 above. If thegamma prior has parameters and , and if sample values are obtained, and the total of those

observations is , then the posterior distribution of is also gamma with parameters

, and . In this case, , so there would be 10 observed's (claim numbers), . The distributions in this example correspond to Combination 1

in the summary above, but we are using instead of in the notation in this example.From the prior distribution of , we have , and , so that and .The variance of the posterior distribution of is

, which we are told is equal to the prior varianceof .01 . Therefore, , from which it follows that .



Example CR-24: The claim size distribution is exponential with mean , where has the pdf, . Three claims are observed, of amounts 40, 50 and 75.

Find the Bayesian premium and the variance of the predictive distribution.Solution: We first note that the prior distribution of is an inverse gamma with and

. We identify this from the exponential exponent (100) and the exponent of in thedenominator (which is ). This is Combination 2 described above. We see that the posterior

density will be proportional to . Therefore, the

posterior density is proportional to , and so the posterior distribution is inverse gammawith and . The Bayesian premium is the predictive mean. Since ( is exponential given ), it follows that the predictive mean is equal to the posterior mean,which is

The variance of the predictive distribution is .

We have found the predictive mean to be . .

Since is exponential given , we have (the second moment of an exponential distribution is 2 times the square of the mean).Therefore,

,which is (2nd moment of the posterior distribution). Since the posterior is inverse gamma, its2nd moment is . The predictive variance is

.

Example CR-25: Tom has a coin, but he doesn't know the probability of tossing a head. Heassumes a prior distribution for the probability of tossing a head to be

for . Tom tosses the coin 10 times and observes 4 heads.Find the posterior distribution of the probability of tossing a head. Suppose that Tom tosses thecoin another 5 times. Find the expected number of heads in those next 5 tosses.Solution: We first observe that , which is a betadistribution with , . The number of heads tossed in 10 tosses has a binomialdistribution with parameters and , and probability function

. The joint density of and is

, which is proportionalto . Therefore, the posterior is also a beta distribution, with and .The predictive mean for the number of heads in 5 more tosses is

no. heads in 5 more tosses 4 heads no. heads in 5 more tosses 4 heads4 heads (posterior mean of ) .



A variation on the Beta prior and Negative Binomial model distributionsThe parametrization of the negative binomial distribution in the Exam C table has parameters and , both . A variation on this parametrization is to keep the parameter , but use theparameter , where so that . Suppose that we use this parametrization, andthat we assume that is a prior parameter with a beta distribution with parameters and , withprior density . Assume that the conditional distribution of given is binomial with known parameter , and with parameter , so that the model distributionhas probability function for .

The joint distribution of and has joint density

.This is proportional, in , to , which implies that the posterior distributionof also has a beta distribution, with parameters and .

If there are observations of available, say , then the joint density of is proportional to . Again this implies that

the posterior distribution of is beta, with parameters and .

Under this parametrization, the beta distribution is a conjugate prior for the negative binomial.

CR-5.4 Some Additional Comments on Bayesian Estimation

There are a couple of other points to mention regarding Bayesian estimation. One of theobjectives of a Bayesian estimation situation may be to obtain an estimate of the parameter . Anapproach that can be taken is to define a "loss function" , and the estimate of is that whichminimizes the expected value of the loss function using the posterior distribution. The followingis a summary of the estimates obtained for three standard loss functions. We wish to find thevalue of that minimizes the loss function.(i) Squared-error loss: ; the estimate is the mean of the posterior distribution.(ii) Absolute loss: ; the estimate is the median of the posterior distribution.

(iii) Zero-one loss: ; the estimate is the mode of the posterior. if if

Another point to note about Bayesian estimation is the concept of the highest posterior densitycredibility set. We will use the example given above for the single parameter Pareto prior anduniform distribution model combination. In that example the posterior distribution of was foundto be single parameter Pareto with and , with pdf

, .



Suppose that we wish to find an interval such that for thatinterval. There are many different possible intervals that satisfy the equation. For instance

, and also , and also ,are all examples of such intervals. If we consider each of these intervals, ,

and , and if we look at the posterior density on each of theseintervals, we note that the interval over which the pdf has the largest values is , sincethe pdf for the single parameter Pareto is a decreasing function. The interval is thehighest posterior density (HPD) credibility interval for probability .75.

In general, the % HPD credibility interval for a given is an interval over which theposterior probability is %, and the numerical values of the posterior density are higheron that interval than any other. As in the example above, all three intervals (and many others)have posterior probability .75, but has the largest numerical values on

.

There is some additional terminology that sometimes arises in the Bayesian credibility context(and also in the Buhlmann credibility context, which will be reviewed in the next section of thesenotes).

1. The model distribution is the conditional distribution of given . The conditionalmean of is referred to as the . This may be denoted and is equal tohypothetical mean

.

2. The conditional variance of in the model distribution is called the .process variance .

3. The expected value of the hypothetical mean is , the mean of the marginal distribution of .

This may also be referred to as the .pure premium4. Note that the marginal -distribution variance is

.When the predictive distribution is found for given the observations , it may bepossible to formulate the Bayesian premium in terms of the observed sample mean and in acredibility formulation: ,where may depend upon .





CREDIBILITY - PROBLEM SET 5Bayesian Credibility - Exam C Table Distributions

1. An individual insured has a frequency distribution per year that follows a Poisson distributionwith mean . The prior distribution for is exponential with a mean of 2. An individual isobserved to have 2 claims in a year. Find the Bayesian premium for the same individual for thefollowing year.A) B) C) D) E)

Questions 2 to 6 are based on a single observation, an integer from a Poisson distributionwith parameter , where has a prior distribution which is exponential with parameter , so thepdf of the prior distribution of is .

2. Find the pf of the model distribution, .

A) B) C) D) E)

3. Find the pdf of the joint distribution, .

A) B) C)

D) E)

4. Find the marginal pf of , .

A) B) C)

D) B)

5. Find the pdf of the posterior distribution of , .

A) B)

C) D)

E)



6. Find the pf of the predictive distribution, .

A) B)

C) D)

E)

7. For a coin chosen at random from a large collection of coins, the probability of tossing a headwith that randomly chosen coin is , where has pdf defined on the interval

(it is assumed that ) . Suppose that a coin is chosen at random from the collection ofcoins. The solutions to the following questions should be expressed in terms of .(a) The coin is tossed twice. denotes the number of heads that occur. Find the marginalprobability function of , and .(b) Suppose that there is one head and one tail observed on the first two tosses of the coin.(i) Find the posterior density of .(ii) Find the probability that the next toss will be a head.

8. You are given the following: - Claim size for a given risk follows a distribution with density function ( ) , 0 , 01 )

The prior distribution of is assumed to follow a distribution with mean 50 and density function ( ) , 0500,000

4)100

(a) Determine the variance of the conditional mean.(b) Determine the mean and variance of the marginal distribution of .(c) Determine the density function of the posterior distribution of after 1 claim of size 50has been observed for this risk.

Problems 9 and 10 are based on the following situation. Suppose that the claim severity for anindividual in a risk class has a binomial distribution with parameters and . Suppose alsothat a randomly chosen individual in the risk class has parameter where has a betadistribution with pdf for ( are the beta parameters).Then the claim severity for an individual with parameter is

for . An individual is chosen at random from thepopulation and is found to have a claim of size 1.

9. Find the pdf of the posterior distribution of , 1 .A) beta with 1 6 B) beta with C) beta with D) beta with E) beta with



10. Find the Bayesian premium.A) B) C) D) E)

11. has a prior distribution with pdf , for and the conditional distribution of given has pdf , for .For the following problem, you can use the identities .(a) Identify the type of prior distribution and the parameter values, and identify the type of modeldistribution and the parameter values (refer to the table of distributions if necessary).(b) Find the marginal distribution of , Refer to the attached table and verify that has aGeneralized 3-Parameter Pareto distribution (A.2.3.1 in the table), and indicate the values of ,

, and for this generalized Pareto.(c) For a single observed value of , find the posterior density of . Show that the givenprior is a conjugate prior for the given model distribution.(d) For a single observed value of , find the Bayesian premium.

12. An individual insured has a frequency distribution per year that follows a Poissondistribution with mean . The prior distribution for is a mixture of two exponentialdistributions with means of 1 and 3, and with mixing weights both equal to .5 . An individual isobserved to have 0 claims in a year. Find the Bayesian premium for the same individual for thefollowing year.A) B) C) D) E)

13. An individual insured has a frequency distribution per year that follows a Poissondistribution with mean . The prior distribution for is a mixture of two distributions.Distribution 1 is constant with value 1, and distribution 2 is exponential with a mean of 3, and themixing weights are both .5. An individual is observed to have 0 claims in a year. Find theBayesian premium for the same individual for the following year.A) .7 B) .9 C) 1.1 D) 1.3 E) 1.5

14. (SOA) You are given:(i) An individual automobile insured has an annual claim frequency distribution that follows aPoisson distribution with mean .(ii) follows a gamma distribution with parameters and .(iii) The first actuary assumes that 1 and 1/6.(iv) The second actuary assumes the same mean for the gamma distribution, but only half thevariance.(v) A total of one claim is observed for the insured over a three year period.(vi) Both actuaries determine the Bayesian premium for the expected number of claims in thenext year using their model assumptions.Determine the ratio of the Bayesian premium that the first actuary calculates to the Bayesianpremium that the second actuary calculates.(A) 3/4 (B) 9/11 (C) 10/9 (D) 11/9 (E) 4/3



15. (SOA) For a risk, you are given:(i) The number of claims during a single year follows a Bernoulli distribution with mean .(ii) The prior distribution for is uniform on the interval [0, 1].(iii) The claims experience is observed for a number of years.(iv) The Bayesian premium is calculated as 1/5 based on the observed claims.Which of the following observed claims data could have yielded this calculation?(A) 0 claims during 3 years (B) 0 claims during 4 years (C) 0 claims during 5 years(D) 1 claim during 4 years (E) 1 claim during 5 years

16. You are given:(i) Annual claim counts follow a Poisson distribution with mean .(ii) The parameter has a prior distribution with probability density function:

Two claims were observed during the first year. Determine the variance of the posterior distribution of .(A) 9/16 (B) 27/16 (C) 9/4 (D) 16/3 (E) 27/4

17. (SOA) You are given:(i) The number of claims per auto insured follows a Poisson distribution with mean (ii) The prior distribution for has the following probability density function:

A company observes the following claims experience:Year 1 Year 2

Number of claims 75 210Number of autos insured 600 900

The company expects to insure 1100 autos in Year 3.Determine the expected number of claims in Year 3.(A) 178 (B) 184 (C) 193 (D) 209 (E) 224

18. For a group of insureds, you are given:(i) The amount of a claim is uniformly distributed but will not exceed a certain unknown limit .

(ii) The prior distribution of is ) = .(iii) Two independent claims of 400 and 600 are observed.Determine the probability that the next claim will exceed 550.(A) 0.19 (B) 0.22 (C) 0.25 (D) 0.28 (E) 0.31



19. (SOA) You are given:(i) The annual number of claims for each policyholder has a Poisson distribution with mean .(ii) The distribution of across all policyholders has probability density function: ,

(iii)

A randomly selected policyholder is known to have had at least one claim last year. Determinethe posterior probability that this same policyholder will have at least one claim this year.(A) 0.70 (B) 0.75 (C) 0.78 (D) 0.81 (E) 0.86

20. For a coin chosen at random from a large collection of coins, the probability of tossing a headwith that randomly chosen coin is , where has pdf defined on the interval

(it is assumed that ) . Suppose that a coin is chosen at random from the collection ofcoins. Suppose that the coin is tossed twice and there is one head and one tail observed.Find the posterior density of in terms of .

21. A Bayesian model has a model distribution which is negative binomial with parameters and . The parameter has a prior distribution which is exponential with a mean of 1.

A single sample value of is observed to be .Find the posterior density of .

22. A loss distribution is being analyzed using the Bayesian credibility approach.The parameter has a prior gamma distribution with and .The model distribution is Poisson with a mean of .A sample of 6 observations of results in a Bayesian premium of 37.48 .A 7th observation of X is obtained and the Bayesian premium is recalculated to be 37.84.Find the value of the 7th observation.




1. A gamma prior with parameters and combines with a Poisson model distribution to resultin a gamma posterior distribution and a negative binomial predictive distribution. If observations are available, then the posterior distribution is gamma with parameters

and , and the predictive distribution is negative binomial with and . In this problem there is observation of size . The prior distributionis exponential with mean 2, which is the same as a gamma distribution with and .The posterior distribution is gamma with parameters , and ,and the predictive distribution is negative binomial with , so that the Bayesianpremium (predictive mean) is . Answer: D

2. The pf of the model distribution is for integer values Answer: A

3. The joint pdf isfor integers and real numbers . Answer: D

4.

for In simplifying this integral, we use the general rule

for integer , and - in the integral above, is the integer, and. Notice that has a geometric distribution with (using the

parametrization of distributions in the Exam C Table). In the summary of prior-model pairs inSection 5 of the notes, it was noted that an exponential prior distribution combines with a Poissonmodel distribution to result in a geometric marginal distribution of with . Answer: B

5. Since has a continuous distribution, the posterior distribution of is continuous with pdf

.

Note that the posterior distribution is a gamma distribution with and In the summary of prior-model pairs in the Bayesian analysis notes, it was noted that anexponential prior distribution combines with a Poisson model distribution to result in a posteriordistribution of which is a gamma distribution with and Answer: C



6.

Note that this can be written as , which is

the pf for the negative binomial distribution with , and .It was mentioned in the summary of prior-model pairs in the Section 5 notes that the combinationof an exponential prior distribution with a Poisson model distribution results in a negativebinomial predictive distribution. Answer: E

The situation in questions 2 to 6 is one of the standard Bayesian model-prior distributioncombinations summarized in the notes. If the prior is exponential with parameter (or gammawith parameters and ) and if the model distribution is Poisson with mean , then(i) the marginal dist. of is geometric with parameter (negative binomial with

) ;(ii) the posterior dist. of is gamma with and ;(iii) the predictive distribution of given is negative binomial with

7. (a) The distribution of given is binomial with and .

.

Alternatively, is the (the Beta function), which is

.

.

Alternatively,

.

.



7.(b)(i) We wish to find .

This can be formulated as .

Alternatively, the prior is a beta distribution with and , and since the model distribution of is binomial with and , and since we have observed ,

the posterior distribution of is also beta with and , with pdf

.

(ii) next toss next toss

.

Alternatively, this probability is the mean of the posterior, and the mean of the posterior is.

8. (a) Given , the expected claim (hypothetical mean) is the expected value of the exponentialwith mean . The variance of is .Noting that the distribution of is inverse gamma with and , we have

, and

and .

(b) as in part (a). .

(c) The posterior distribution of is proportional to the joint distribution of and , which isequal to , and with , this is1 500,000) )

4100

1 . This is not a pdf, but the posterior pdf is some) ) )

constant multiplied by this function. Again, we identify this as an inverse gamma that must have

and . The pdf is .) )

This is Model 2 in the prior-model distribution pair summary in Section 5.



9. Then the joint distribution of and is for and .

With , the joint density becomes .The posterior density of given must be proportional to the joint density, and thereforethe posterior distribution of must also be a beta distribution with (the exponent of is and the exponent of is ). This result also follows from the prior-model

combination of beta prior and binomial model distribution; for that combination, if the priordistribution of is a beta distribution with parameters and , and if the model distribution isbinomial with parameters and , then given observation , the posterior distribution isalso beta with parameters and . The prior of in this case is a betadistribution with , and the model distribution is binomial with and ; sincethe observation is , the posterior distribution is also beta with parameters and

. Answer: C

10. . In Problem 8, the posterior distribution wasfound to be a beta with parameters and . The conditional expectation is

, since the conditional distribution of given is binomial with and .Then (mean of theposterior). The mean of the beta distribution with parameters and is . Therefore, themean of this posterior is , and the Bayesian premium is . Answer: E

11.(a) The prior has an inverse gamma distribution with .The model distribution is gamma with .(b) The joint distribution isThe marginal distribution of has pdf

.

This is a generalized Pareto with , and .

(c) The posterior density is .

This is an inverse gamma distribution with .Alternatively, since , and we know that the posterior density is proportional with

respect to to , we see that the posterior density is proportional to This is the from the density function of an inverse gamma distribution with .

(d) The Bayesian premium is .The model distribution of given is gamma with , so .The Bayesian premium is .Since is the mean of the posterior distribution, we see that the Bayesian premium is

posterior mean . The posterior distribution in inverse gamma, and has mean .



12. The prior distribution has pdf , and the model distribution is The joint distribution of and is

. With this becomes

.The posterior distribution of will have pdf that is proportional in to . The form ofthe components of indicate that the posterior must be a mixture of two exponentialdistributions: exp has mean .5, and exp has mean .75.The posterior pdf is , where

Therefore, .The posterior distribution is a mixture of two exponentials, a mixing weight of on theexponential with mean .5 and a mixing weight of on the exponential with a mean of .75.The Bayesian premium (the predictive mean) is

(the last equality is true since is Poisson with mean if is given). Therefore, the predictivemean is the same as the mean of the posterior distribution, which is the mixture of the means ofthe two exponential distributions: . Answer: E

13. The prior distribution has pdf, probability mass

and the model distribution has pf .

The joint distribution of and is , prob. mass

With this becomes ., prob. mass

The posterior pdf is , where

Then, the posterior distribution is

., probability mass

The Bayesian premium is the predictive premium. By the same reasoning given in Problem 10,this will be the mean of the posterior distribution. The posterior distribution is a mixture of aconstant 1 with mixing weight and an exponential with mean .75 and mixing weight

The mean of the posterior is .Answer: B



14. The distribution being considered is the number of claims per year (annual claimfrequency). We are given that the total number of claims observed in the first three years is 1( ). The Bayesian premium for the fourth year is the conditional expectednumber of claims in the fourth year given the information about the first three years

. The distribution of is Poisson with mean , and is assumed tohave a prior gamma distribution, .

We find the Bayesian premium by conditioning over : .

Since has a Poisson distribution, for both actuaries. is the pdf of the posterior distribution of given that there is one claim

in the first three years. The Bayesian premium will be the mean of the posterior distribution of .

We use the following relationship: if given is Poisson with parameter and if the prior distribution of is a gamma distribution with parameters and , and if we have a sample of values, say , then the posterior distribution of also is a gamma distribution, with parameters and , where is the sample mean of the sample of -values

In this problem is the discrete random variable for of claims in one period, andnumber is a sample of separate periods; is the number of claims for period .for

Therefore is the total number of claims observed in the periods.

Actuary 1 assumes , , and we are given , and periods(or exposures), so that the posterior distribution of given is also gammawith parameters and .The mean of a gamma distribution with parameters and is . Actuary 1's Bayesianpremium estimate is the mean of the posterior distribution of , which is .

The variance of the gamma distribution with parameters and is . We are told thatActuary 2 assumes that has a gamma distribution, say with and , with the same mean asActuary 1's gamma distribution, so . We are also assuming that the variance ofthe second actuary's gamma distribution half of the variance of Actuary 1's gamma distribution,so that . From the two equations and , we get

and .

The posterior distribution of for Actuary 2 will be a gamma distribution with and .

Actuary 2's Bayesian premium is the mean of the posterior distribution of , which is .

The ratio of Actuary 1's Bayesian premium to that of Actuary 2 is . Answer: C



15. Let be the number of claims, which has a Bernoulli distribution with mean (a Bernoulli distribution is a binomial distribution with 1 trial).The prior distribution of is uniform on , and has density .The probability function for is .The Bayesian premium given observations is ,which can be formulated by conditioning over the parameter ,

. , since is Bernoulli with mean .

The Bayesian premium becomes .

Also .Since , it follows that

.

Therefore, ,so that must be the density for a Beta distribution with

and .Then, is , which is the expected value of theBeta distribution with .The mean of this beta distribution is The beta-prior/Bernouilli-model pair is one of the Bayesian pairs considered in the notes inVolume I of this study guide.We are given that the Bayesian premium, which is this expected value , is equal to .The 5 possible answers are A - ; B - C - ; D - ; E - .The only combination of and that result in is A. Answer: A

16. This problem involves Bayesian estimation of a posterior distribution.The prior distribution of is exponential with (mean ) and the model distribution (of given ) is Poisson with mean . This combination of prior and model distributions is a specialcase of the more general situation in which the prior distribution of is a gamma distributionwith parameters and , and the model distribution (of given ) is Poisson with mean . Inthis general case, if sample data values are obtained, then the posterior distributionof given is also a gamma distribution, with parameters

, and .

The exponential distribution is a special case of the gamma distribution in which .In this problem, has a gamma distribution with and . , the annual claim countdistribution, has a Poisson distribution with mean . The data available is the claim count duringthe first year, a single observation of . Therefore, sample value, and the value of isgiven to be 2. The posterior distribution of is also a gamma distribution, with and . The first moment of this gamma distribution is (from the distributiontable) ' , and the second moment is . The variance is thesecond moment minus the square of the first . Answer: B



17. The prior distribution is gamma with and .This is a Bayesian problem with a Poisson model distribution with mean , and aprior distribution that is gamma, with parameters and . In this problem is theexpected number of claims for one auto in a year.

With data values and either the individual values known or the sum known,

the posterior distribution must be gamma with parameters and . Inthis problem there are observations (1500 's) in the first two years, and

the sum of the 's is claims for the 1500 insured autos during the two

years. The posterior distribution of is gamma with , and . The predictive expected number of claims per auto is the

expected value of in the posterior distribution. This is the mean of the (posterior) gammadistribution, which is . With 1100 autos insured in the 3rd year, theexpected number of claims from all 1100 autos is . Answer: B

18. We are asked to find the conditional probability .

This probability can be formulated as .

The denominator and numerator are found by conditioning over the prior parameter .First the denominator:

.A crucial observation that must be made before continuing with this integral calculation isthat, since given has a uniform distribution on the interval it follows that

forfor , or

.

Therefore, andforfor , or

forfor , or

.

We are given a prior distribution for on the interval .However, for , so that

.



18 continuedThe numerator is found in a similar way:

(if is uniform on the interval , then )

.

Then .

Again, note that a key point in this problem is that the lower limit of the integration is 600 (not500). This is true since one of the observations was 600 implies that must be greater than 600(since any observation must be between 0 and ).

Alternatively, can be formulated as ,

where is the posterior distribution of .Since one of the observation is 600, we must have .

.

As before, ,

and for ,

and .

Then, , and

.Answer: E

19. Usually the most efficient way to find a predictive probability is to condition the both thenumerator and denominator over the prior parameter. We are asked to find

which is written asWe find by conditioning over : .Since given has a Poisson distribution, .Then,(we have used identity (iii) given in the statement of the problem).

In a similar way, we get

1 (as always, in a Bayesian situation, we implicitlyassume that and are conditionally independent given )

. Then, .



19 continuedAn alterative approach is to "factor" the predictive probability through the posterior distributionof , so that As before,

.When the prior distribution of is a gamma distribution with parameters and and the modelhas a Poisson distribution with mean , then the marginal distribution of is negative binomialwith and . In this case, the prior has a gamma distribution with and so that the marginal distribution of is negative binomial with and , and therefore

Then, , and

.

One other point to note about this problem is the following. The identity given in (iii) of thestatement of the problem is . This is an example of a more general rule thathas been stated a few places elsewhere in this study guide. The general rule is that for any integer

and any number , . In the integral ,the variable of integration is instead of , and , so the value of the integralis . Note that this rule is only valid for , not for finite intervals. Answer: D

20. .

Alternatively,

.

.

We wish to find .

This can be formulated as .

Alternatively, the prior is a beta distribution with and , and since the model distribution of is binomial with and , and since we have observed ,

the posterior distribution of is also beta with and , with pdf

.



21. The posterior density of is ,where is the joint density of and , and is the marginal probability of .

, where is the prior density of .We are given that (exponential with a mean of 1).We are also given that (negative binomial with ) .The marginal probability for is found from

We use the integration rule , with , so that .

The posterior density is .

22. The original Bayesian premium is

from which it follows that .

The updated Bayesian premium is

from which it follows that .

Therefore, .

CREDIBILITY SECTION 6 - BUHLMANN CREDIBILITY CR-127


CREDIBILITY -SECTION 6, BUHLMANN CREDIBILITY

The material in this section relates to Section 16.4.3 to 16.4.5 of "Loss Models". The suggestedtime frame for this section is 3-4 hours.

CR-6.1 The Buhlmann Credibility Structure

The Buhlmann model has the same initial structure as the Bayesian credibility model. There is apopulation or portfolio of risks. Each member of the population has an associated randomvariable (the model distribution which is loss amount, or claim number usually). The modeldistribution random variable depends on a parameter (the prior distribution). Each memberof the population has an associated parameter, and the model distribution is a conditional randomvariable given the value of . The following is a repeat of the prior and model distributioncomponents of a Bayesian credibility model.

The initial assumption for the distribution of the parameter is called the (ofprior distribution), and it has a pdf/pf (continuous/discrete) often denoted ( ); a typical interpretation of is

that there is a population of individuals (also called "risks", or "policyholders") and eachindividual has their own value , and these values are distributed over the population according tothe prior density of .

The (usually frequency or severity of loss) is a conditional distributionmodel distributiongiven , with pf/pdf (or more simply ).

Under the Buhlmann credibility model, the conditional distributions of the 's given areassumed to be independent and identically distributed (iid), with the following characteristics: is the (6.1)hypothetical mean is the (6.2)process variance(and since the conditional distributions of given and given areindependent, for ; but the unconditional distributions of and

will most likely not be independent).

Using the conditional expectation rule, we have . (6.3)

is called the (also called the ).pure premium collective premiumWe also define the following factors. (the ) , and (6.4)expected process variance, EPV (the ) (6.5)variance of the hypothetical mean, VHM(Keep in mind that since is a random variable, the hypothetical mean and process variance arealso random variables, and have means and variances themselves.)Then the unconditional variance of is , (6.6)(and the unconditional covariance between and for is ) . (6.7)

This situation can be described as(and , where ).

CR-128 CREDIBILITY SECTION 6 - BUHLMANN CREDIBILITY


A member of the population is chosen at random, and the value of is not known. A number ofindependent observations of are collected for that member of the population, .The 's are independent (but all come from a distribution with the same unknown ).If we apply Buhlmann's credibility approach to this (Buhlmann's) model, then the credibilitypremium is

Buhlmann Credibility Premium

, (6.8)

where (6.9)

The Buhlmann credibility premium is the estimate of the next occurrence of for that samevalue of .

The factor is the ( ) . (6.10)Buhlmann credibility factor

Note also that since , the credibility premium can be written in the form

. (6.11)sum of observed 's

Keep in mind that is the number of observations or exposures of the random variable ,and, as in the Bayesian credibility situation, it is understood that these observations are all froman with the same value of .

Buhlmann Credibility questions on Exam C have tended to be fairly mechanical, with one of themain objectives being to find either the Buhlmann credibility factor or the credibility premium.This requires finding , which in turn, requires finding and .Once the distribution of and the conditional distribution of given are identified, and can be found as described above. In the special case that (this occurs if is aconstant for all values of ), the credibility factor is .

In both the Bayesian and Buhlmann credibility approach, we are trying to estimate , butwe don't know the specific value of . The Buhlmann credibility premium is weighted average ofthe sample mean from the data set and the mean of the marginal distribution of from theBayesian structure . The weight applied to the sample mean is , and we can see thatas gets larger, approaches 1. This is reasonable, since the more data we have for , the better

is as an estimate of the mean of . When we calculate from the Bayesian structure, we aretaking into account all possible values of in the model distribution of . We do not know thevalue of , but as we increase the number of sample values of for the unknown distribution of

, we should be more willing to rely on the sample mean as a credible estimate of .This is reflected in the weight getting closer to 1 as gets larger.



CR-6.2 Examples of the Buhlmann Credibility Method

Example CR-26: You are given the following: - The number of claims follow a Poisson distribution with mean . - Claim sizes follow a distribution with density function . - The number of claims and claim sizes are independent. - The prior distribution of has density function Determine the value of Buhlmann's for aggregate losses.Hint:Solution: Aggregate losses are , the random variable to which we are applying Buhlmann'smethod. The conditional distribution of given is a compound Poisson distribution withPoisson mean and exponential claim size distribution also with mean . We recall that acompound Poisson distribution with Poisson parameter and severity distribution has mean

and variance .In this case, the hypothetical mean is .The process variance is

( has an exponential distributionwith mean , and the 2nd moment of the exponential is 2 times the square of the mean).The variance of the hypothetical mean is

,since and .The expected process variance is .The value of Buhlmann's is .Note that we can also find , so that if one or moreobservations are given, we can find the Buhlmann credibility premium to be

, where .

Example CR-27: You are given the following: - A portfolio of independent risks is divided into two classes. - Each class contains the same number of risks. - The claim count distribution for each risk in Class A is a mixture of a Poisson distribution with mean 1/6 and a Poisson distribution with mean 1/3, with each distribution in the mixture having a weight of 0.5. - The claim count distribution for each risk in Class B is a mixture of a Poisson distribution with mean 2/3 and a Poisson distribution with mean 5/6, with each distribution in the mixture having a weight of 0.5.A risk is selected at random from the portfolio. Determine the Buhlmann credibility factor forone observation from this risk.Solution: The Buhlmann credibility (factor) is . In this case, there is one observation,so that . The process variance is . In this case, refers to the Class. Sinceeach Class is a mixture, the moments within each class can be found using the mixing weights.We also use the fact that for a Poisson distribution with mean , the second moment is .Then, , and

.Then .Similarly, , and



Example CR-27 continuedThe process variance is one of two values, either or depending upon whichclass of Class A or B was picked. We are told that a risk is selected at random, which we interpretas meaning that there is a .5 chance of either class being picked.The expected process variance

Class .The hypothetical mean is , which is and for the two classes(with and each having probability .5 of occurring). The variance of the hypothetical mean is

.Then,

Example CR-28: You are given the following: - An urn contains six dice. - Three of the dice have two sides marked 1, two sides marked 2, and two sides marked 3. - two of the dice have two sides marked 1, two sides marked 3, and two sides marked 5. - One die has all six sides marked 6.One die is randomly selected from the urn and rolled. A 6 is observed.(a) Determine the Buhlmann credibility estimate of the expected value of the second roll of thissame die.(b) The selected die is placed back into the urn. A seventh die is then added to the urn. Theseventh die is one of the following three types:1. Two sides marked 1, two sides marked 3, two sides marked 5.2. All six sides marked 3.3. All six sides marked 6.One die is randomly selected from the urn and rolled. An estimate is to be made of the expectedvalue of the second roll of this same die. Determine which of the three types for the seventh diewould increase the Buhlmann credibility factor of the first roll of the selected die (compared tothe Buhlmann credibility factor in part (a)).Solution: (a) With observation, .

The hypothetical means are type 1 die type 2 type type 3 type .

The variance of the hypothetical mean is .

The process variances aretype 1 ,type 2 ,type 3 .

Then, the expected value of the process variance is

Then, , and .

The prior mean is .With a first roll of 6, the expected value of the second roll is

.



Example CR-28 continued(b) In order for to increase, the ratio must decrease.Die 1 is a die of type 2 and die 3 is of type 3 from part (a) .The hypothetical mean for die 2 is 3 and the hypothetical variance is 0.Adding die 1: ,7

.

Adding die 2: ,

.

Adding die 3: ,

.The reduction of by adding Die 3 could have been anticipated. All faces are 6 so die 6 has noprocess variance, and there would be a decrease in expected process variance . Since the meanfor die 3 is 6 and the prior mean was 3 without the extra die, the variance of the hypotheticalmean will increase. Thus will decrease.

Example CR-29: You are given the following: - The number of claims follows a Poisson distribution with mean . - Claim sizes follow a distribution with mean 20 and variance 400 .2

- is a gamma random variable with density function ( ) , 022

- For any value of , the number of claims and claim sizes are independent.Determine the expected value of the process variance of the aggregate losses.Solution: For a given value of , the process variance is the variance of a compound Poissondistribution; the process variance is

.2

2 .

The Buhlmann model is somewhat simple and also restricted. It requires that from one period tothe next, the characteristics of the loss-related random variables (claim number or total claimamount) remain the same. No allowance is made for increase (or any change) in the number ofpolicyholders from one period to the next. The next model still applies the Buhlmann approachto credibility (the credibility premium modeled as a linear function of the 's), but allows forchanges in loss characteristics from one period to the next. In fact, the following model isactually equivalent to the original Buhlmann model when the process variance is known, but itmay be more convenient for dealing with changing numbers of exposures from one period to thenext.



CR-6.3 The Buhlmann-Straub model

This is a generalization of the Buhlmann model just considered. It is still assumed that theconditional distributions of the given are independent, and the hypothetical mean isthe same for all : for all .The difference from the original Buhlmann model is that the conditional variances of the given (the process variance) might not all be the same. It is assumed that there are(positive) factors for , and a factor such that for ,

. (6.12)

As in the basic Buhlmann model, we define the factors , and : .

We also define to be . (6.13)The credibility factor is , (6.14)

and (6.15)credibility premium , where

( is a weighted average of the 's).Also, it can be shown that , and for . (6.16)

Note that, as in the original Buhlmann model case, this credibility premium can be written astotal claims for observed values of . (6.17)

A standard interpretation of the Buhlmann-Straub model is that is the average claim for independent exposure units in period (or group) , where, given , each exposure unit hasmean claim amount (or number) and variance ; , where each

has mean and variance . Since is the average of the claims, it has the samehypothetical mean as any individual exposure unit, but has process variance . It is stillassumed that there is independence from one period (or group) to another. The credibilitypremium that is calculated is the credibility (i.e., per occurrence ofpremium per exposure unitan individual ) for the next period. If there will be exposure units (policies) in period(or group) , then the credibility premium for period for all exposures combined is

(the number of exposure units multiplied by the credibility premium perexposure unit).

In fact, if we consider the total collection of policies as occurring in periods

with one policy per period, the credibility premium using the original Buhlmann modelwould be the same as that found from this Buhlmann-Straub model.

The credibility premium per policy in either the original Buhlmann model or this Buhlmann-Straub model is , where total claims all periods

total number of exposures for all periodsand . This shows the equivalence between the original Buhlmannmodel and the Buhlmann-Straub model. The following example illustrates this.



Example CR-30: You are given the following data on large business policyholders:(i) Losses for each employee of a given policyholder are independent and have a common meanand variance.(ii) The overall average loss per employee for all policyholders is 25.(iii) The variance of the hypothetical means is 50.(iv) The expected value of the process variance is 10,000.(v) The following experience is observed for a randomly selected policyholder: Year Average Loss per Employee Number of Employees

1 20 10002 15 7503 10 600

Determine the Buhlmann-Straub credibility premium per employee for this policyholder.¨Solution: This problem can be viewed as a standard Buhlmann credibility problem. The randomvariable being considered is loss per employee. There are a total of observations(number of losses in all 3 years). We are given , and forthe loss per employee distribution. The sample mean loss per employee for this policyholder is

.The Buhlmann credibility factor is .

The Buhlmann credibility premium per employee is .

The fact that there are 3 years is not relevant, what is important is that we have 2350observations. We don't know individual loss amounts for each of them, but we don't need them.We need the sum of the losses for the 2350 employees, which we have (we are given year by yearsample means, but from that we can determine the total loss for each year, and add them).

A variation on the Buhlmann-Straub modelThere is a variation to the Buhlmann-Straub model that we note (derivations can be found inExamples 16.29 and 16.30 of the "Loss Models" book). This variation allows for thedifferentiation between the variances of the claim-related distribution; this is represented as amore general form for . The characteristics of this variation to the Buhlmann-Straubmodel are as follows.

is the same for all , but for (instead of in the original Buhlmann-Straub

model). We still have the factors ,but we now also have a new factor, .The unconditional mean, variance and covariances of the 's are

, , . (6.18)

We define the factor as follows . (6.19)

The solution of the normal equations results in credibility premium , where

, and . (6.20)

A further variation has (where ) instead of .

Now in the credibility premium, (also, and are found by

replacing in the expressions above with ).



CR-6.4 Exact Credibility

The Bayesian credibility method and the Buhlmann credibility method both require the sameBayesian structure of a parameter distribution for and model distribution for given .When the conditional distribution of given and the distribution of satisfy certainconditions, the Bayesian premium will be equal to the (Buhlmann) credibility premium. Thissituation is referred to as exact credibility. One way of checking whether or not exact credibilityis satisfied is to look at the Bayesian premium . If it is a linear function of

then exact credibility is satisfied. There is a general class of distributions for whichexact credibility is satisfied.

If the model distribution of is from the linear exponential family with ,

and if the distribution of had pf/pdf for ,and if (for the situations that are likely to occur on Exam C, it is likely that

), then exact credibility is satisfied.) Also, and .

It may be a tedious exercise to verify that a distribution is from the linear exponential family.

Example CR-31: Show that the gamma prior, Poisson model combination satisfies exactcredibility.Solution: The prior distribution of the parameter is a gamma distribution with parameters and , and the model distribution of given is Poisson with mean . Given observations of , say , the predictive distribution is negative binomial with mean

; this will be the Bayesian premium. This Bayesian premium is a linearfunction of the 's.

We can find the Buhlmann credibility premium in the usual way; it is .For the given distributions, .

(from the gamma distribution for ) , , .

. The Buhlmann credibility premium is

.This is the same as the Bayesian premium. Since the Bayesian premium is a linear function of the

's, it could have been anticipated that exact credibility would be satisfied.

Along with the gamma prior and Poisson model combination just considered in Example CR-31,it can be shown that the following prior , model combinations satisfy exact credibility. TheBayesian premium is also given and can be seen to be a linear function of the .

Prior/Model Bayesian Premium is Inverse Gamma , /Exponential with mean

is beta /binomial

normal normal with mean ( ) ( )



CE-16.5 Theoretical Background to Buhlmann Credibility

The following is a theoretical preamble to Buhlmann's approach to credibility and is not crucialbackground. We continue to assume that the distribution of depends upon a parameter , andthat has a distribution as well (the prior distribution). Given the sample values , wecan find the conditional distribution of given . In this approach tocredibility, we estimate ( ) (also denoted when there are -sample values available) asa linear function of the 's in the form . Given the sample 's, we find the values of the -coefficients thatminimize the expectation . This least squaresestimation is done in the usual way, taking partial derivatives of the expectation with respect toeach , and solving a system of equations. This results in the set of

, with the 's being the least squares estimates:~normal equations

for we have the ,~ ~unbiasedness equation

for we have the remaining normal equations

.~

Note that in the normal equations, all expectations and covariances are unconditional (ormarginal) expectations and covariances of the 's. Also note that for any ,

.

The is then ~ ~ ~ ~ ~credibility premium

This credibility premium is not only the least squares linear estimator of the hypothetical mean, but it is also the least squares linear estimator of the Bayesian premium

, and of .

The information needed to apply this credibility method (to solve the normal equations to find the~-values) is the mean and variance of each for , and the covariance betweenany pair and for . Keep in mind that initial information is usually in theform of the conditional distribution of given (along with the prior distribution of ), and itis usually assumed that for , the conditional distributions of and given areindependent. This does not mean that the unconditional (marginal) distributions of and areindependent; the main reason that credibility methods are meaningful is that does dependupon ; if the unconditional distribution of was independent of , thenthe Bayesian premium is just the pure premium - , and pastexperience would be of no value in determining the premium.In Example 16.24 of the "Loss Models" book it is shown that if for all and , and if for all , , then and for ~ ~

and also the credibility premium can be written as , where~ ~

and

It is unlikely that an exam question will go as deeply into the background of the Buhlmanncredibility model as was done just now.





CREDIBILITY - PROBLEM SET 6Buhlmann Credibility

1. Assume that the number of claims each year for an individual insured has a Poissondistribution. Assume also that the expected annual claim frequencies (the Poisson parameters )of the members of the population of insureds are uniformly distributed over the interval

. Finally, we assume that an individual's expected annual claim frequency is constantover time. Find the Buhlmann credibility factor if observations are available.A) B) C) D) E)

Problems 2 and 3 refer to the following situation. Two urns contain balls with each ball marked 0or 1 in the proportions described below: Percentage of Balls in Urn Marked 0 Marked 1 Urn A 20% 80% Urn B 70% 30%An urn is randomly selected and two balls are drawn from the urn with replacement. The sum ofthe values on the selected balls is 1. Two more balls are selected from the same urn withreplacement. Assume that each selected ball has been returned to the urn before the next ball isdrawn.

2. Determine the Bayesian analysis estimate of the expected value of the sum of the values on thesecond pair of selected balls.A) .99 B) 1.01 C) 1.03 D) 1.05 E) 1.07

3. Determine the Buhlmann credibility estimate of the expected value of the sum of the values onthe second pair of balls chosen.A) 1.00 B) 1.03 C) 1.06 D) 1.09 E) 1.12

Problems 4 and 5 refer to the following situation. The aggregate loss distribution for two risksfor one exposure period is: Aggregate losses Risk $0 $50 $1000 A .80 .16 .04 B .60 .24 .16A risk is selected at random and observed to have 0 losses in the first two exposure periods.

4. Determine the Bayesian analysis estimate (Bayesian premium) of the expected value of theaggregate losses for the same risk's third exposure period.A) 92.00 B) 92.16 C) 92.32 D) 92.48 E) 92.64

5. Determine the credibility premium (Buhlmann's credibility estimate of the expected value ofthe aggregate losses for the same risk's third exposure period).A) 99.50 B) 99.83 C) 100.17 D) 100.50 E) 100.83



Problems 6 to 8 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




6. Find the expected hypothetical mean .A) B) C) D) E)

7. Find the value of that appears in the credibility factor .A) 111 B) 115 C) 119 D) 134 E) 145

8. If an individual is chosen at random from the risk population and observation isavailable for that individual, find the credibility premium for the next exposure period for thisindividual.A) B) C) D) E)

9. For an individual risk in a population, the number of claims per month follows a Poissondistribution with mean . For the population, is distributed according to the exponentialdistribution, with pdf . A randomly selected risk in the population is found tohave had 1 claim in the final 6 months of 2000 and 1 claim for all 2001. Find the Buhlmann-Straub credibility premium for this individual for the first three months of 2002.A) B) C) D) E)

10. Aggregate claims for an exposure period for an individual in a risk class with parameter has a binomial distribution with parameters and . The parameter has pdf

for . A portfolio of insurance policies has 10 individuals in eachrisk class in 1997, 12 individuals in each risk class in 1998 and 15 individuals in each risk class in1999. An individual from a randomly chosen risk class is observed for the three years, and it isfound that aggregate claims for the three consecutive years are 18 (1997), 20 (1998) , 27 (1999).Using the Buhlmann-Straub model find the credibility premium in 2000 for aggregate claims forthis risk class, if there are 20 individuals in the risk class in 2000.A) 33 B) 34 C) 35 D) 36 E) 37



11. You are given the following: - The number of claims for a single insured follows a Poisson distribution with mean . - varies by insured and follows a Poisson distribution with mean .Determine the value of Buhlmann's .

12. You are given the following: - Partial Credibility Formula A is based on the methods of limited fluctuation credibility, with 1600 expected claims needed for full credibility. - Partial Credibility Formula B is based on Buhlmann's credibility formula with a of 391. - One claim is expected during each period of observation.Determine the largest number of periods of observation for which Partial Credibility Formula Byields a larger credibility factor than Partial Credibility Formula A.

13. You are given the following: - Claim sizes follow a gamma distribution with parameters and . - The prior distribution of is assumed to be uniform on the interval (0, 4).Determine the value of Buhlmann’s for estimating the expected value of a claim.

14. is a Poisson random variable with parameter , where the prior distribution of is adiscrete uniform distribution on the integers 3 . A single observation of is made. Find theBuhlmann factor .

15. Find the Buhlmann credibility premium in Example CR-19.

16. You are given the following: - Claim size for a given risk follows a distribution with density function ( ) , 0 , 01 )

The prior distribution of is assumed to follow a distribution with mean 50 and density function ( ) , 0500,000

4)100

Find the Buhlmann credibility factor for a single observation of .

17. An individual insured has a frequency distribution per year that follows a Poissondistribution with mean . The prior distribution for is a mixture of two exponentialdistributions with means of 1 and 3, and with mixing weights both equal to .5 . An individual isobserved to have 0 claims in a year. Find the Buhlmann credibility estimated number of claimsfor the same individual for the following year.



18. An individual insured has a frequency distribution per year that follows a Poissondistribution with mean . The prior distribution for is a mixture of two distributions.Distribution 1 is constant with value 1, and distribution 2 is exponential with a mean of 3, and themixing weights are both .5. An individual is observed to have 0 claims in a year. Find theBuhlmann credibility premium for the same individual for the following year.

19. The distribution of in three consecutive periods has the following characteristics:

Find the credibility premium for period 3 in terms of and using Buhlmann's credibilityapproach.A) B) C) D) E)

Problems 20 to 23 refer to the aggregate claim distribution per period with pdf for , where has pdf e for .

20. Find the process variance.A) B) C) D) E)

21. Find for A) B) C) D) E)

22. Suppose that there are periods of claims. Find the value of coefficient that arises in the~credibility premium.A) B) C) D) E)

23. Find the credibility factor if periods of claim information is available.A) B) C) D) E)

24. An insurance company writes a book of business that contains several classes ofpolicyholders. You are given:(i) The average claim frequency for a policyholder over the entire book is 0.425.(ii) The variance of the hypothetical means is 0.370.(iii) The expected value of the process variance is 1.793.One class of policyholders is selected at random from the book. Nine policyholders are selected atrandom from this class and are observed to have produced a total of seven claims. Five additionalpolicyholders are selected at random from the same class. Determine the Buhlmann credibilityestimate for the total number of claims for these five policyholders.A) 2.5 B) 2.8 C) 3.0 D) 3.3 E) 3.9



25. You are given:(i) The number of claims incurred in a month by any insured has a Poisson distribution withmean .(ii) The claim frequencies of different insureds are independent.(iii) The prior distribution is gamma with probability density function:

(iv) Month Number of Insureds Number of Claims 1 100 6 2 150 8 3 200 11 4 300 ?Determine the Buhlmann-Straub credibility estimate of the number of claims in Month 4.(A) 16.7 (B) 16.9 (C) 17.3 (D) 17.6 (E) 18.0

26. Type A risks have each year's losses uniformly distributed on the interval .Type B risks have each year's losses uniformly distributed on the interval .A risk is selected at random, with each type being equally likely. The first year's losses equal .Find the Buhlmann credibility premium for the second year's lossesin terms of .

27. Annual aggregate claims for a particular policy are modeled as a compound Poissondistribution with Poisson parameter for the frequency (number of claims per year), and a severity(individual claim size) that is either 1 or 2 with . An insurer has a largeportfolio of policies, and each policy has its own value of . For a randomly chosen policy from theportfolio, the distribution of is exponential with a mean of 1.The claim sizes and the numbers of claims are independent of one another given .A policy is chosen at random from the portfolio, and denotes the aggregate claim for that policy for oneyear. The policy is observed for three years and the observed aggregate losses for the 3 years are

and . Find the Buhlmann credibility premium for the 4th year for this policy.

28. (SOA) You are given:(i) is the claim count observed for driver for one year.(ii) has a negative binomial distribution with parameters 0.5 and .(iii) is the expected claim count for driver for one year.(iv) The ’s have an exponential distribution with mean 0.2.Determine the Buhlmann credibility factor for an individual driver for one year.(A) Less than 0.05 (B) At least 0.05, but less than 0.10 (C) At least 0.10, but less than 0.15 (D) At least 0.15, but less than 0.20 (E) At least 0.20



29. For a portfolio of independent risks, the number of claims per period for a randomly chosenrisk has a Poisson distribution with a mean of , where has pdf , .Two risks are chosen at random and observed for one period, and it is found that Risk 1 has noclaims for the period and Risk 2 has 2 claims for the period. is the Buhlmann credibilitypremium for Risk 1 for the next period and is the Buhlmann credibility premium for Risk 2 forthe next period. Find .lim

(A) (B) (C) (D) (E)

30. It is known that there are two groups of drivers in an insured population. One group has a 20percent accident probability per year and the other group has a 40 percent accident probability peryear. Two or more accidents per year per insured are not possible. The two groups compriseequal proportions of the population and each has the following accident severity distribution: Probability Size of Loss .80 100 .10 200 .10 400A merit rating plan is based on the pure premium experience of individual insureds for the prioryear. Calculate the credibility of an insured's experience.(A) Less than .01 (B) At least .01, but less than .02 (C) At least .02, but less than .03(D) At least .03, but less than .04 (E) At least .04

31. For a portfolio of independent risks, you are given:(i) The risks are divided into two classes, Class A and Class B.(ii) Equal numbers of risks are in Class A and Class B.(iii) For each risk, the probability of having exactly 1 claim during the year is 20% and theprobability of having 0 claims is 80%.(iv) All claims for Class A are of size 2.(v) All claims for Class B are of size , an unknown but fixed quantity.One risk is chosen at random, and the total loss for one year for that risk is observed. You wish toestimate the expected loss for that same risk in the following year.Determine the limit of the Buhlmann credibility factor as goes to infinity.(A) 0 (B) 1/9 (C) 4/5 (D) 8/9 (E) 1

32. The Buhlmann credibility factor based on exposures of a single risk is . Based on

exposures, the Buhlmann credibility factor is . Find the Buhlmann credibility factorbased on exposures.(A) (B) (C) (D) (E)

33. The Buhlmann credibility model is being applied to the loss variable (per exposure). It isfound that after exposures, the Buhlmann credibility factor is .How many additional exposures are needed to increase the factor to 0.5?(A) 10 (B) 15 (C) 20 (D) 25 (E) 30



34. (SOA) You are given:(i) 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

(ii) 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

(iii) Risks are equally likely to be type A or type B.(iv) Claim counts and claim sizes are independent within each risk type.(v) The variance of the total losses is 296,962.A randomly selected risk is observed to have total annual losses of 500.Determine the Buhlmann credibility premium for the next year for this same risk.¨(A) 493 (B) 500 (C) 510 (D) 513 (E) 514

35. (SOA) You are given the following information about a single risk:(i) The risk has exposures in each year.(ii) The risk is observed for years.(iii) The variance of the hypothetical means is .(iv) The expected value of the annual process variance is Determine the limit of the Buhlmann-Straub credibility factor as approaches infinity.¨(A) (B) (C) (D) (E) 1

(SOA) Use the following information for questions 36 and 37.You are given the following information about workers' compensation coverage:(i) The number of claims for an employee during the year follows a Poisson distribution with

mean , where is the salary (in thousands) for the employee.

(ii) The distribution of is uniform on the interval .

36. An employee is selected at random. No claims were observed for this employee during theyear. Determine the posterior probability that the selected employee has salary greater than 50thousand.(A) 0.5 (B) 0.6 (C) 0.7 (D) 0.8 (E) 0.9

37. An employee is selected at random. During the last 4 years, the employee has had a total of 5claims. Determine the Buhlmann credibility estimate for the expected number of claims theëmployee will have next year.(A) 0.6 (B) 0.8 (C) 1.0 (D) 1.1 (E) 1.2



38. (SOA) You are given:(i) The full credibility standard is 100 expected claims.(ii) The square-root rule is used for partial credibility.You approximate the partial credibility formula with a Buhlmann credibility formula by selectingä Buhlmann value that matches the partial credibility formula when 25 claims are expected.¨Determine the credibility factor for the Buhlmann credibility formula when 100 claims areëxpected.(A) 0.44 (B) 0.50 (C) 0.80 (D) 0.95 (E) 1.00

39. (SOA) You are given:(i) Claim size, , has mean and variance 500.(ii) The random variable has a mean of 1000 and variance of 50.(iii) The following three claims were observed: 750, 1075, 2000Calculate the expected size of the next claim using Buhlmann credibility.(A) 1025 (B) 1063 (C) 1115 (D) 1181 (E) 1266

40. Losses for the year for a risk are uniformly distributed on , where is uniformlydistributed on the interval .

The first year loss for a risk is . Find the Buhlmann credibility premium for the second year'sloss for the same risk in terms of .




1. Hypothetical mean , Process variance .

(variance of the uniform distribution on interval is ). . Answer: C

2. This is a standard Bayesian analysis. The formal algebraic explanation is more long-windedthan the calculations needed.

(urn A is chosen) (sum of two balls is 1)sum of two balls is 1 urn A is chosen urn A is chosen

balls are or urn A is chosenSimilarly, (urn B is chosen) (sum of two balls is 1)Then sum of two balls is 1

(urn A is chosen) (sum of two balls is 1) (urn B is chosen) (sum of two balls is 1) .We can find conditional (posterior) probabilities of urn chosen given sum of balls:

urn A is chosen sum of two balls is 1 (urn A is chosen) (sum of two balls is 1)sum of two balls is 1

, and similarlyurn B is chosen sum of two balls is 1 .

The expected value on a ball if chosen from urn A is ,and the expected value on a ball if chosen from urn B is 7 3Using the posterior probabilities for urns A and B, we can find the conditional expected value ona ball chosen from the same urn as the first two balls - this conditional expectation is

3rd ball value sum of first two balls is 1 ball value urn A urn A sum of first two balls is 1 ball value urn B urn B sum of first two balls is 1

The expected value of the sum of two more balls under these circumstances would be .

Another way to describe these calculations is the following. We know the expected value of thesum of two balls if urn A is chosen (1.6) and also if urn B is chosen (.6). If we had no priorinformation about the sum of the first two balls, then the expected value of the sum of two ballsfrom a randomly chosen urn would be

sum urn A urn A sum urn B urn B .Although each of urns A and B have the same chance of being chosen initially, we now haveadditional information about which urn might have been chosen. We use Bayesian analysis tofind the probability that the urn was A (or B) given that the two balls chosen from the unknownurn added up to 1. These conditional probabilities were found to be

A sum is 1 and B sum is 1 . We use these Bayesian updated probabilitiesto find the expected sum of the next two balls.



2. continuedsum urn A urn A sum of first 2 balls is 1

sum urn B urn B sum of first 2 balls is 1In this last formulation, we have replaced urn A with

urn A sum of first 2 balls is 1 , and similarly for B.

Note that we could have found the conditional probabilities of the next ball being either 0 or 1given that the sum of the first two balls was 1 -

next ball is 0 sum of first two balls is 1ball is 0 (urn A) (sum of first two balls is 1) urn A sum of first two balls is 1ball is 0 (urn B) (sum of first two balls is 1) urn B sum of first two balls is 1

.Algebraically, this relationship is

next ball is 0 sum of first two balls is 1 (sum is 1)sum is 1

A (sum is 1) B (sum is 1)sum is 1 sum is 1

A (sum is 1) A (sum is 1) B (sum is 1) B (sum is 1)A sum is 1 sum is 1 B sum is 1 sum is 1

ball is 0 (urn A) (sum of first two balls is 1) urn A sum of first two balls is 1ball is 0 (urn B) (sum of first two balls is 1) urn B sum of first two balls is 1 .

In a similar way, next ball is 1 sum of first two balls is 1

The comments that were made above regarding the expectation apply to probabilities as well.The probability of choosing a 0 from urn A is .2, and the probability of choosing a 0 from urn Bis .7. If we had no information about the sum of the first two balls, then each of urns A and Bhave a chance of being chosen, and the probability of choosing a 0from a randomly chosen urn would be , and the probability of choosing a 1from a randomly chosen urn would be The Bayesian analysis above gaveus the conditional probabilities of choosing urn A or urn B given that the first two balls had a sumof 1 - A sum is 1 and B sum is 1 .We use these Bayesian updated probabilities (instead of ) to find the probability of choosing a 0or a 1 given that the sum of the first two balls was 1 -

next ball is 0 sum of first two balls is 1 andnext ball is 1 sum of first two balls is 1

Then ball value sum of first two balls is 1 ( ) (1)( ) ,as before. The two conditional probabilities

next ball is 0 sum of first two balls is 1 andnext ball is 1 sum of first two balls is 1 form the posterior distribution of the value of

the next ball chosen (from the same urn) given that the sum of the first 2 balls was 1. Answer: C



3. We now put this situation in the context of the Buhlmann credibility approach. The parameter

describes the urn chosen: To find the Buhlmann credibility estimateA prob. B prob.

(premium) for the sum of the next two balls we must identify the random variable , and theconditional distribution of given . In this case, is the number on a ball chosen from anurn.

The conditional distribution of given A is A with,,

A , and B , with B (these conditional

expectations were found in Problem 2).

The credibility premium is , where3. continuedThe values of the various components are found as follows.

The hypothetical mean and its expected value (the collective or pure premium) - since, in the marginal distribution of , there is chance of choosing

urn A or urn B, we have A B .Note that the hypothetical mean is , which in this case is

A A A , and B B B , and thenA A B B ,

as before.

The process variance, and its expected value , where

A A A A ,and since A wehave A A , and similarly,

B B Then,A A B B .

The variance of the hypothetical mean

A A B B

There are observations, so that .Also, so that , andthe credibility premium is pernew ball chosen. There will be two balls chosen, so the expected total of the next two balls (thecredibility premium for two more claims) is Answer: C



4. The solution requires a Bayesian analysis similar to that in Problem 7 above.We find unconditional probabilities:

( claims in first period and claims in second period) claims in first period and claims in second period

(this is the unconditionalprobability of no claims in either of two periods for a randomly chosen risk)We now find conditional probabilities:

, andThe conditional expectation of aggregate claims per period for each risk:

,

The conditional expected aggregate claims per period given no claims in first two periods isfound using the updated (Bayesian) probabilities of and :

Answer: E

5. The collective premium is ;in this case, is either risk A or risk B, each with prob. .5, and the hypothetical means are

. The process variances for risks A and B are

and the expected process variance isThe variance of the hypothetical mean is

The credibility factor is Since there are no claims in each of the

first two periods, , so that .The credibility premium is then Notice the strong similarity between Problems 2,3 and 4,5. Urn becomes risk, and number onball becomes aggregate claim for one exposure period. Otherwise, the analysis is the same. Answer: E

6. each with probability , and :

Therefore Answer: D



7. .

,

, and

.

Then

Also,

Then . Answer: C

8. The credibility premium is In this case, and so that , and

. From Problem 6, we have . The credibility premium is Answer: A

9. Let represent the number of claims in a month for a randomly chosen individual. Then (mean of the Poisson distribution with parameter ),

(mean of the exponential),(variance of the Poisson distribution with parameter ),

, and (variance of the exponential isthe square of the mean).

Let denote the number of claims in each of the final 6 months of 2000 for theindividual, and let denote the number of claims in each of the 12 months of2001. We have 18 monthly claim numbers and we can apply the Buhlmann method to .

,

and .Then the Buhlmann(-Straub) credibility premium is

per month.The credibility premium for the first three months of 2001 is Answer: A



10. Let denote the claims for an individual in a single year. Then, and

(binomial mean and variance with 3 trials).

Then, using the distribution of , we have;

; and

.

With 10 policyholders per risk class in 1997, the average aggregate claim per policyholder in therandomly selected risk class is , .Similarly, , , and , are the observed average aggregateclaim amounts per individual for 1998 and 1999.

Applying the Buhlmann-Straub model, , and the credibility factor is . Also,

, so that the credibility

premium per individual in the risk class is With 20 individuals in the risk class

in 2000, the credibility premium is Note that we have a total of 37 exposures ( , with an of ,and we have applied the basic Buhlmann method, without really needing to refer toBuhlmann-Straub. Answer: C

11. Expected value of process varianceVariance of hypothetical mean .

Process variance .Hypothetical mean (since has a Poisson distribution,its mean and variance are equal). Then .

12. The "credibility factor" is the value of . In the partial credibility context, , where . We are told that full credibility, , is

reached with Therefore, , and the partial credibility is if. The Buhlmann credibility factor is . In order to have ,

we must have , or equivalently, , or equivalently, . The roots of the quadratic equation are

and . Any between 289 and 529 will result in .



13. The hypothetical mean given is , and the variance of the hypothetical mean is .

The process variance is , the variance of the gamma distribution, which is , and the expected value of the process variance is

. Then .14. ,

. .

.(one observation of ), so .

15. has an exponential distribution, and is uniformly distributed on . .

, , .

For data points, the Buhlmann credibility factor is .

For the data points , with , the Buhlmanncredibility premium is .

16. has an exponential distribution, and has an inverse gamma distributionwith . and .

.

. .

17. . Since is a mixture, and (each component of the mixture has anexponential distribution, and the second moment of an exponential is two times the square of themean). Then

.For a single observation of , , the Buhlmann credibility factor is .

With , the Buhlmann credibility estimate for the next year is .



18. . Since is a mixture, and (the first component is constant at 1). Then

.For a single observation of , , the Buhlmann credibility factor is .

With , the Buhlmann credibility estimate for the next year is .

19. According to Buhlmann's approach, we solve the normal equations for the -coefficients: ~~ ~ ~

~ ~ .~ ~Substituting the given values, these equation become

~ ~ ~~ ~~ ~

with solution . The credibility premium is~ ~ ~

~ ~ , with . Answer: E

20. Answer: D

21. This model satisfies the requirements of Buhlmann's model (common mean, variance andcovariances for the 's). .From Problem 20 above , so that

e e Answer: C

22. , where and (from Problem 21),~

and e (from Problem 20).Thus, (in fact, all ). Answer: D~ ~

23. Answer: B

24. The Buhlmann credibility factor is , and the Buhlmann credibility estimate is

. We are given and . For the policyholders, the sample mean for number of claims is . The credibility factor is

. The Buhlmann credibility estimate for number of claims for another

policyholder of the same class is . The credibility estimate for5 policyholders of the same class is . Answer: D



25. This is a combination of a gamma prior distribution and Poisson model distribution. For thiscombination, the Buhlmann credibility premium is the same as the Bayesian credibility premiumestimate (called exact credibility). The Buhlmann-Straub estimate in this case is the same as theBuhlmann estimate (which is the same as the Bayesian estimate). The first three months of dataare combined so that we have a total of insureds with a total number of claims of

. Using the Bayesian method, the predictive mean for the next policy of

the same type is (actually, the predictive distribution for the number ofclaims on the next policy of the same type is a negative binomial distribution with parameters

and ).

In this problem where and are the gamma parameters from the priordistribution. Therefore, and .The predictive expectation is , This is the expected number of claims for a singleinsured in Month 4. For 300 insureds in Month 4 we would expect .If this had not involved the specific combination of the gamma prior and Poisson modeldistributions we would have had to use the Buhlmann method.

Note that we are implicitly assuming that all 450 insureds have the same (unknown) , and wewish to find the expected number of claims from 300 more policies with the same . That is thestandard Bayesian approach. Answer: B

26. Prior distribution isHypothetical means are .Process variances are .

expected hypothetical mean .expected process variance .variance of hypothetical mean .

.

Buhlmann credibility premium is .

27. Hypothetical mean is Process variance is .Expected hypothetical mean is .Expected process variance is .Variance of hypothetical mean is .




28. The conditional claim count given is negative binomial with parameters and .Therefore, and For driver , the expected claim count is , which we are told has an exponentialdistribution with mean 0.2 . Under the Buhlmann model,

(since we are told that has anexponential distribution with mean 0.2). Also,

(variance of the exponential is the square of the mean). Thecredibility factor for a single ( ) driver for one year is

. Answer: C

29. The Buhlmann credibility premium is .In this case, and . Then

,

, and

.

Each risk has data for period so that .

As , , and .

Risk 1 has no claims for the period, so that and therefore as , .

Risk 2 has two claims for the period, so that and therefore as ,

. Then Answer: Dlim

30. is the group number. Hypothetical mean for group 1 is (.2 is expected frequency, and is the expected severityfor group 1) and for group 2 is .Since the groups are of equal sizes, .Then, the expected hypothetical mean is .The variance of the hypothetical means is

(also equal to ) .The process variance for group 1 is

, andfor group 2 it is

.Then, . The credibility for 1 driver for 1 year is

. Answer: C



31. is the total loss for one year on the randomly chosen claim.is either A or B, denoting the two classes. .

.The credibility factor is ,

where , and

. Credibility is based on year.Then . As , ,

and . Answer: B

32. . .

With exposures, we have . Answer: B

33. .In order to have , we must have , which is an additional 25exposures. Answer: D

34. The Buhlmann credibility premium is , where is the sample mean ofthe observed values. There is one ) observation, so that .Also, , and , .

is the class, A or B. , (from problem 11),so that , and

.We can find and directly, or we can use the fact that

, and since we are given that so that .Then, , and the Buhlmann premium is

. Answer: D

35. For this model, the factor is , which becomes

when for all . The credibility factor is .

As , , and . Answer: B



36. We wish to find .

.

.

Then . Alternatively, the prior distribution of is , and the model distribution

for (claim number) is for The marginal probability function for is , and for ,this marginal probability is as given

above. The posterior density of is .The posterior probability in question is

.Answer: B

37. The parameter is . There are data points, and we are given

is the total number of claims for the 4 years. Therefore, .

The hypothetical mean is , where is uniformly distributed from 0to 100. Then .The process variance is (since given has a Poissondistribution, the mean and variance are the same).The expected process variance is .The variance of the hypothetical mean is

.The Buhlmann credibility factor is .

The Buhlmann credibility estimate for next year's claim number is . Answer: B

38. Using the partial credibility approach, the credibility factor with 25 claims is . The Buhlmann credibility factor is . In order for this to be .5 when

, we must have , so that . Then, if (and is still 25) the

Buhlmann credibility factor becomes . Answer: C



39. Hypothetical mean is process variance is . .

Expected process variance is .Variance of hypothetical mean is .There are claims, so that the Buhlmann credibility factor is .

The sample mean of the given data is , so that the expected size of the next claim is. Answer: B

40. Prior distribution is .

The hypothetical mean is , since is uniformly distributed on .

Process variance is .

Expected hypothetical mean is

Expected process variance is .

Variance of the hypothetical mean is

There is one observation, so




CREDIBILITY SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS CR-159


CREDIBILITY - SECTION 7EMPIRICAL BAYES CREDIBILITY METHODS


CR-7.1 Non-parametric Empirical Bayes Credibility

In the Bayesian and Buhlmann credibility approaches that have been considered, it was assumedthat the conditional distribution of given was known to be of a certain form (usually aparametric distribution such as Poisson with parameter ). It was also assumed that thedistribution of the risk parameter was known (such as a gamma distribution with parameters and for Combination 1 of the prior-model distribution pairs that were summarized earlier). Wenow consider the situation in which the conditional distribution of given might not beparametric, and the distribution of is unknown. The unknown quantities (such as mean andvariance) related to the distribution of are called structural parameters.

Our objective is still to apply the Buhlmann or Buhlmann-Straub models to determine acredibility premium based on observed claim data only. We still wish to find , which requiresfinding and , as before, and then is needed to find the credibility premium. Since the pf/pdfof the conditional distribution of given and of the distribution of are not known, wecannot directly find or or . Theempirical Bayes approach uses the experience data to estimate , and . The analysis can beplaced in the following general setting.

- an insurance portfolio consists of policyholders, (each "policyholder" might represent a group of insureds)

- for policy holder , data on "exposure periods" is available,( might represent the number of years of observations available for group ,or policyholder might represent a group with policies in the group )

- for policyholder and exposure period , there are exposure units, with an average observed claim (amount or number) of per exposure unit , so that the total claim observed for policyholder in exposure period is , and the total claim observed

for policyholder for all exposure periods is (an interpretation of this is (7.1)

that in group , for individual there might be years of data available, with average claim of over those years)

- the total number of exposure units for policyholder is (7.2)

- the average observed claim per exposure unit for policyholder is (7.3)

- the total number of exposure units for all policyholders is (7.4)

CR-160 CREDIBILITY SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS


- the average claim per exposure period for all policyholders istotal observed claim for all policyholders in all exposure periods

total number of exposure periods for all policyholders (7.5)

- policyholder has risk parameter random variable , and are assumed to be independent and identically distributed

- for policyholder and exposure period , the conditional distribution ofaverage claim per exposure period given has mean and variance

and (7.6)

- the structural parameters are , , (7.7) (these are the same for each , since the 's are assumed to be identically distributed)

- once the values of and are found (or estimated), then the credibility premium for the next exposure period for policyholder is , for , where ; this is the credibility premium for policyholder per unit (7.8) of exposure; if there will be exposure units for the next period for policyholder , then the credibility premium for policyholder for all exposure units combined for the next period is (7.9)

We can apply the empirical Bayes approach to the Buhlmann model and to the Buhlmann-Straubmodel.

The notation described above is a little burdensome, but a few examples may help.

Example CR-32: As a simple illustration of the structure of the data for a non-parametricempirical Bayesian analysis, suppose that there are policyholders. Policy holder 1 isobserved for years (exposure periods) , policyholder 2 is observed for years,and policyholder 3 is observed for years. The data is summarized below:

Policyholder 1, Policyholder 2, Policyholder 3,

,

,

.



Example CR-32 continuedThe 3 years of observations for policyholder 1 resulted in 4 claims in the first year withaverage claim size of 2500, 6 claims in the second year with average claim size of 1800,and 2 claims in the third year with average claim size of 2900. Similar interpretations are appliedto policyholders 2 and 3. Since is the average claim size of the claimsin the first year of observation for policyholder 1, it follows that the total claim for policyholder 1in the first year is . Similarly, the total claim amounts for the second andthird years for policyholder 1 are 10,800 and 5,800 , for an overall total of all claims amounts forpolicyholder 1 for all 12 exposures (claims) in the 3 exposure periods (years) of 26,600 .The average claim amount for policyholder 1 per exposure is . Similarly,the total claim amount for policyholder 2 for all exposures in all 4 exposure periods is16,600 and the average claim amount per exposure is . For policyholder 3there are a total of exposures, and an average claim of .The aggregate claims for all policyholders in all exposure periods is

.

CR-7.2 Empirical Bayes Estimation for the Buhlmann model (Equal Sample Size)

Under the Buhlmann model, there are the same number of exposure periods for eachpolicyholder, which means that , and there is one exposure unit for eachexposure period, which means that for . There are two commoninterpretations to this situation.

(1) is the number of years (or months) that each policyholder is observed, and there is oneobservation per year. For policyholder in year the claim amount (or claim number, if that iswhat is being represented by ) is (since there is only one observation for each year(exposure period), is a single observed value (a sample mean of a sample of size 1).

(2) Each policyholder consists of insureds. The observations are the claim amounts for each of the insureds for policyholder 1. In general, is the claimamount for the -th insured individual of the insureds for policyholder .

Also, this situation can be interpreted as saying that each of the groups has the same number ofindividuals ( ) and all are observed for 1 ( ) period. This can be considered the"equal sample size" case. Then for each we have

, and . (7.10)

The unbiased estimates of the structural parameters that we will use (there are other unbiasedestimates) are- , (7.11)the estimated prior mean:

- the estimated expected process variance:

( ) , and (7.12)where

- . (7.13)the estimated variance of the hypothetical means:



(The derivations that show that the following estimates are unbiased can be found in the "LossModels book).

If , then we set . The credibility factors will be equal for all groups sincefor all . The estimated credibility factor for each is . (7.14)

Example CR-33: An insurance company has two group policies. The aggregate claim amounts(in millions of dollars) for the first four policy years are summarized in the table below. Assumethat the two groups have the same number of insureds. Use Buhlmann's model with empiricalBayesian estimation to estimate the credibility premium for each of the two groups during thenext (5th) policy year. Aggregate Claim Amounts Group Policy Year Policy Solution: In this case, (groups) , and (exposure periods per group,and exposure unit per group/year combination), and , and

. Then .The estimates of the structural parameters are

andSince , the credibility factors for the two groups are equal:

The credibility premium for the 5th year for group is .For group 1 the credibility premium is ( )and for group 2 it is ( ) .



CR-7.3 Empirical Bayes Estimation for the Buhlmann-Straub Model

The Buhlmann-Straub model is the general form with data in the form presented at the start ofthis section. There are exposure periods for policy holder , .For policyholder and exposure period , , , there are exposure units, and represents the observed average claim per exposure unit (for "cell" ).This can be considered the "unequal sample size" case.

The unbiased estimates that are used for the structural parameters are

, (7.15)

. (7.16)

As in the Buhlmann model, if , then set .

The estimated credibility factor for group is . (7.16)

Note that the factors are not necessarily equal, since may vary from one policyholder toanother.

An alternative to the unbiased estimate just described for the Buhlmann-Straub model is the (also called themethod that preserves total losses credibility-weighted average) estimate of . This is found by first estimating

in the way just described for the Buhlmann-Straub model, and

then . If we use this estimate of to find the credibility premiums

for groups , and then calculate total credibility premiums for pastexposures, that total will equal to the actual total past claims. There have beenseveral questions on older exams that referred to the "method that preserves totallosses". It appears that this method has not been covered on recent exams, but itis part of the material covered in the "Loss Models" book according to the examcatalog.

Example CR-34: Compute the estimated credibility premium for the fourth policy year for eachof the two groups of insureds whose claims experience for the first three year is presented in thefollowing table.Group Policy YearPolicyholder 1 2 3 4 Aggregate Claim Amount 8,000 11,000 15,000 - 1 Size of Group 40 50 70 75 Aggregate Claim Amount 20,000 24,000 19,000 - 2 Size of Group 100 120 115 95



Solution: There are policyholders, with exposure periods for eachpolicyholder (group).

For policyholder 1, so that , andfor policyholder 2 so that .

The average claim amounts per exposure unit for each of the groups and each of the exposureperiods are and

and

and

Overall number of exposure units is The estimate of the overall mean is

The estimated expected process variance using the Buhlmann-Straub model:

The estimated variance of the hypothetical means is

The estimated value of is , and the credibility factors for the twopolicyholders are .

The credibility premiums for the two policyholders per exposure unit for the fourth year arefor policyholder 1, andfor policyholder 2 For the 75

exposure units for policyholder 1 in the fourth year, the credibility premium is, and for the 95 exposure units for policyholder 2 in the fourth year, the

credibility premium is .

The credibility-weighted average estimate of is



Example CR-34 continuedUsing this value of , for policyholder 1, the credibility premium per unit of exposure is

and for policyholder 2, the credibility premium perunit of exposure is (note that the values of and

do not depend on , so they are unchanged even though a different estimate is being used)Policyholder 1 has total past exposure of 160, and policyholder 2 has total past exposure of 335,so the total credibility premium for all past exposure for both policyholders 1 and 2 combined is

. The total past claims observed for all exposures is .

It is no coincidence that the two values (97,001.5 and 97,000) are almost the same (they differbecause of roundoff error). This will always be the case if the credibility-weighted estimate of (instead of ) is used for to calculate credibility premiums.

A couple of variations on the Buhlmann/Buhlmann-Straub models can be considered:

(i) if the manual rate is known (not necessarily an unbiased estimate) then it would be used inthe credibility premium equation instead of ; the estimates of and would be formulated the

same as before - and

(ii) if the actual value of is known (possibly approximated by using the manualpremium), then an alternative unbiased estimate for is

, and is the same as in (i) above (7.17)

(iii) if is known (again, possibly the manual premium) and if there is data available for onlypolicyholder (although there are other policyholders), then the following estimates can be used

for and : and (7.18)



CR-7.4 Semiparametric Empirical Bayesian Credibility

The model for the portfolio may have a parametric distribution for given , but anunspecified non-parametric distribution for . In this case, we may be able to use therelationships linking and and the fact that

in order to get estimates for and to use in the credibility premiumformulation.

We will consider the case in which represents the number of claims for a period of time (sayone month) and the conditional distribution of claim number given is Poisson withparameter . In this model and .Then and , so that .Also, .

A data set will consist of a random sample of observations from the entire population.Therefore, each for is a claim number from a different randomly selectedmember of the population. The 's each come from population members with differentvalues of .

We wish to focus on one particular member of the population, and apply credibility estimation toestimate the number of claims that individual will have . Suppose for that individual, we have observations, . The difference between the 's and the 's is that the 's areobservations that range over all possible values of in the population, but the 's areobservations from a single individual with an unknown value of .

We wish to apply the Buhlmann credibility method to estimate the mean ( ) for that individual.Note now that the data points labeled are not all from the same individual as the were inthe Bayesian approach and the parametric Buhlmann approach considered earlier. It is the

's that come from one individual.

The semiparametric credibility estimate will be of the form , where is thesample mean of the observations for the particular individual. will have a similar form to theusual Buhlmann method, ; note that we use in , since is the number of

observations ( ) we have for the particular individual. The 's will be used to estimate and and .

For this semiparametric Poisson model, we have .The sample of 's is drawn from the entire range of possible values of in the population, so wecan use as an estimate for , this is the usual estimate for .

Since for Poisson it is also true that , it follows that , and can also be estimated by .

For any model it is always true that , and this leads us to the estimate of . Wefirst use the data set of the -values, , to find the (unbiased) sample variance of the 'sand use this as the estimate the variance of . Then sample variance sample mean.As before, if , we set .This is how we get the Buhlmann quantities that we need.



Example CR-35: A group of 500 insureds submitted the following 290 claims during a one yearperiod of observation. Number of Claims Number of Insureds 0 300 1 120 2 70 3 10For each individual, the number of claims is assumed to follow a Poisson distribution, but themean of each distribution may vary among insureds. Suppose that an individual from this groupof insureds has experienced two claims during the one year period. Using semi-parametricestimation, find the Buhlmann credibility estimate of the expected number of claims in the nextyear for this insured. Suppose that the insured had 4 claims in a two-year period. Find the semi-parametric credibility estimate of the expected number of claims in the next year for this insured.Solution: Let denote the number of claims for a randomly chosen individual. Then theconditional distribution of given is Poisson with parameter (mean and variance) , so that

.

From the given data, we find the estimate of the mean of ..

Since , our estimate for is .

Since also, and , our estimate for isthe same as our estimate for , .

To find the estimate of , we use the general variance relationship .

Since we have an estimate for , if we can find an estimate for , then the estimate of will be .

We use the standard unbiased sample variance expression to estimate .

, ,

.

The estimate of is .

The estimate of the factor is .

Since (one individual insured is being considered for one year; note that the sample sizeof 500 includes individuals with values of that range over the full unknown distribution of

), the estimate of the credibility factor is .We denote by the number of claims of a particular insured for one year. We are told that for the individual chosen.

The credibility estimate of the expected number of claims for this individual for the next year is .



Example CR-35 continuedIf the insured has 4 claims in two years, there would be observations (number of claims in thefirst year) and (number of claims in the second year), and , and , so that

. and and are the same as before (they are based on the entire sample of 500observations).

is now , and the credibility estimate for the number of claims in

the 3rd year is .

Note that although the conditional distribution of given is Poisson (and so the mean andvariance of the distribution of given are the same), the unconditional distributionconditionalof is not Poisson and will likely have mean and variance not equal. That is why we estimatethe variance of the unconditional as well as estimating the mean of the unconditional .

The crucial point in applying semiparametric empirical Bayesian estimation when the conditionaldistribution of given is Poisson with parameter is that the sample mean is used for both

and . Then the estimated variance of is found from the data and is used for , so that can be found.



CREDIBILITY - PROBLEM SET 7Empirical Bayes Credibility Methods

Problems 1 to 3 refer to the following situation. An insurance company has two group policies.The aggregate claim amounts (in millions of dollars) for the first three policy years aresummarized in the table below. Assume that the two groups have the same number of insureds. Aggregate Claim Amounts Group Policy Year Policy 5 8 11 1 3 2

1. Find the Buhlmann credibility premium for group 1 for the fourth year.A) 8.0 B) 8.2 C) 8.4 D) 8.6 E) 8.8

2. Using the credibility-weighted average estimate of (also referred to as the method thatpreserves total losses), find the Buhlmann credibility premium for group 1 for the fourth year.A) 8.0 B) 8.2 C) 8.4 D) 8.6 E) 8.8

3. Suppose that the aggregate claim amounts for group 2 for policy years 1, 2 and 3 are2 , 8 and 14 (instead of 11, 13 and 12). Find the estimated Buhlmann credibility premium forgroup 1 for the fourth year.A) 8.0 B) 8.2 C) 8.4 D) 8.6 E) 8.8

Problems 4 and 5 refer to the data of Example CR-34 in the notes, with the followingmodification. Assume that there is a third group with the following experience for the threeyears: Policy Year 1 2 3 Aggregate Claim 10,000 15,000 13,500 Size of Group 50 60 60

4. Find the credibility premium per exposure unit for policyholder 1 for the fourth year(nearest 1).A) 203 B) 205 C) 207 D) 209 E) 211

5. Using the credibility-weighted average estimate of , find the Buhlmann credibility premiumper unit of exposure for group 1 for the fourth year (nearest 1).A) 203 B) 205 C) 207 D) 209 E) 211



Problems 6 and 7 refer to the data of Example CR-34 in the notes, with the followingmodification. For group 1, assume that there is no data for the first policy year (but all threeyears of data are still available for group 2).

6. Find the credibility premium per exposure unit for policyholder 1 for the fourth year(nearest 1).A) 206 B) 208 C) 210 D) 212 E) 214

7. Using the credibility-weighted average estimate of , find the Buhlmann credibility premiumper unit of exposure for group 1 for the fourth year (nearest 1).A) 206 B) 208 C) 210 D) 212 E) 214

8. For a large sample of insureds, the observed relative frequency of claims during anobservation period is as follows: Number Relative Frequency of Claims of Claims 0 61.9% 1 28.4% 2 7.8% 3 1.6% 4 .3% 5 or more 0Assume that for a randomly chosen insured, the underlying conditional distribution of number ofclaims per period given the parameter is Poisson with parameter .Given an individual who had claims in the observation period, use semiparametric empiricalBayesian estimation to find expected number claims that the individual will have in the nextperiod.A) B) C) D) E)

9. You are given the following:- The number of losses arising from 500 individual insureds over a single period of observation is distributed as follows: Number of Losses Number of Insureds 0 450 1 30 2 10 3 5 4 5 5 or more 0 - The number of losses for each insured follows a Poisson distribution, but the mean of eachdistribution may be different for individual insureds.Determine the Buhlmann credibility of the experience of an individual insured over a singleperiod.



10. You are given the following table of data for three policyholders over a three year period.

Policy Year 1 2 3

Policyholder

1 Number of Claims 40 50 Average Claim Size 200 220

2 Number of Claims 100 120 120 Average Claim Size 200 200 150


Apply the nonparametric empirical Bayes credibility method to find the credibility premium perclaim in the 4th year for Policyholder 2 using the standard method for (not the method thatpreserves total losses).

11. Semi-parametric empirical Bayesian credibility is being applied in the following situation.The distribution of annual losses on an insurance policy is uniform on the interval , where hasan unknown distribution. A sample of annual losses for 100 separate insurance policies is available. It is

found that and .

For a particular insurance policy, it is found that the total losses over a 3 year period is 4.Find the semi-parametric estimate of the losses in the 4th year for this policy.

12. (SOA) An insurer has data on losses for four policyholders for seven years. is the lossfrom the policyholder for year . You are given:

( ) 33.60 and ( ) 3.301 1 1

4 7 42 2

Calculate the Buhlmann credibility factor for an individual policyholder using nonparametricempirical Bayes estimation.(A) Less than 0.74 (B) At least 0.74, but less than 0.77(C) At least 0.77, but less than 0.80 (D) At least 0.80, but less than 0.83 (E) At least 0.83



13. (SOA) The number of claims a driver has during the year is assumed to be Poissondistributed with an unknown mean that varies by driver. The experience for 100 drivers is asfollows: Number of Claims during the Year Number of Drivers 0 54 1 33 2 10 3 2 4 1Determine the credibility of one year’s experience for a single driver using semiparametricempirical Bayes estimation.(A) 0.046 (B) 0.055 (C) 0.061 (D) 0.068 (E) 0.073

14. (SOA) The following information comes from a study of robberies of convenience storesover the course of a year:(i) is the number of robberies of the store, with = 1, 2, , 500.(ii) 50 (iii) 2202

(iv) The number of robberies of a given store during the year is assumed to be Poissondistributed with an unknown mean that varies by store.Determine the semiparametric empirical Bayes estimate of the expected number of robberies nextyear of a store that reported no robberies during the studied year.(A) Less than 0.02 (B) At least 0.02, but less than 0.04(C) At least 0.04, but less than 0.06 (D) At least 0.06, but less than 0.08 (E) At least 0.08

15. Survival times are available for four insureds, two from Class A and two from Class B. Thetwo from Class A died at times 1 and 9. The two from Class B died at times 2 and

4. Nonparametric Empirical Bayes estimation is used to estimate the mean survival time foreach class. Unbiased estimators of the expected value of the process variance and the variance ofthe hypothetical means are used. Estimate , the Buhlmann credibility factor.(A) 0 (B) 2/19 (C) 4/21 (D) 8/25 (E) 1

16. You are given the following experience for two insured groups:Year

Group 1 2 3 Total1

2

Number of members 8 12 5 25Average loss per member 96 91 113 97

Number of members 25Average loss per member 113 111 116 113

30 20 75

Number of members 100Average loss per member 109Total

_

4800_ _

Determine the nonparametric Empirical Bayes credibility premium for group 1, using the methodthat preserves total losses.(A) 98 (B) 99 (C) 101 (D) 103 (E) 104



17. The number of claims per month for a given risk is assumed to be Poisson distributed with anunknown mean that varies by risk. It is found that for a risk that has reported no claims for thepast month, the semiparametric empirical Bayes estimate of the expected number of claims nextmonth is , and it is found that for a risk that has reported no claims for the past two months,

the semiparametric empirical Bayes estimate of the expected number of claims next month is .Find the semiparametric empirical Bayes estimate of the expected number of claims next monthfor a risk that has reported no claims for the past three months.(A) (B) (C) (D) (E)

18. You are given the following table of data for three policyholders over a three year period.

Policy Year 1 2 3

Policyholder


2 Number of Claims 100 120 120 Average Claim Size 200 200 150


Apply the nonparametric empirical Bayes credibility method to find the credibility premium perclaim in the 4th year for Policyholder 2.

19. A particular type of individual health insurance policy models the annual loss per policy asan exponential distribution with a mean that varies with individual insured. A sample of 1000randomly selected policies results in the following data regarding annual loss amounts in intervalgrouped form.

Interval Number of Losses It is assumed that the loss amounts are uniformly distributed within each interval.

Apply semiparametric empirical Bayes credibility to estimate the loss in the 3rd year for aparticular individual who had annual policy losses of 150 in the first year and 0 in the secondyear.




1. policyholders (groups) and exposure periods (years)for each group, and exposure unit for combination of group and year.

, , ,

, and

.

Then , and the estimated credibility factor for group 1 is

. The credibility premium for group 1 for the fourth year is

Answer: C

2. The credibility-weighted average estimate of is .

From Problem 1, we have . We find .

The credibility-weighted average estimate of is .This is the same as the original sample mean estimate of , so the resulting credibility premiumwill be the same as in Problem 1. It is not a coincidence that the credibility-weighted estimate of

is the same as the sample mean estimate. When we have an "equal sample size" data set( and for all ) then the two will always be equal. Answer: C

3. , , ,

, and

Since , the we assign a value of 0 to . The credibility premium is . Answer: A



4. There are policyholders, with exposure periods for eachpolicyholder. Exposure units are so that

, and so that and so that . As found in Example CR-34,

and

and

and , and with the additional

policyholder, we have and

Overall number of exposure units is The estimate of the overallmean is

The estimated variance of the hypothetical means is

The estimated value of is , and the credibility factors for the threepolicyholders are , and

.

The credibility premium per exposure unit for policyholder 1 for the fourth year is. Answer: D

5. From Problem 4, we have . The credibility- weighted

estimate of is

.The credibility premium for group 1 per unit of exposure using this estimate of is

. Answer: E



6. There are policyholders, with exposure periods for the twopolicyholders. Exposure units are (second policy year) and (third policyyear) so that , and as in the original Example CR-34,

so that . The average claim amounts per exposure unit are and and

and

.

The overall number of exposure units is .

,

The estimated value of is , and the credibility factors for the twopolicyholders are

The credibility premium for group 1 per unit of exposure using this estimate of is. Answer: A

7. From Problem 6, we have and . The credibility- weighted estimate

of is .

The credibility premium for group 1 per unit of exposure using this estimate of is. Answer: B

8. The average number of claims per insured is

This is the estimate of . Since the conditional distribution of given is Poissonwith parameter , we have . Then, since

, our estimate for is also (since ). We estimate , the variance of the relative claim frequency per

insured; the relative frequency of claims forms the estimated probability distribution of .



8. continuedSince , we use the estimated variance of along with the estimate of to getan estimate of ; The estimate of is , and the estimated credibility factor for oneindividual is The credibility premium for the next period for an individualwho had claims in the current period is .Answer: C

9. Since each insured has a Poisson claim number, .We use the semiparametric Buhlmann credibility estimate for one observation, .

is the estimate of , whose estimate from the data is .

We can also estimate from the data .

Since , we can estimate as theestimate of . This estimate is .Then and .

10.

.

.

.

The credibility premium for policyholder 2 is .



11. Hypothetical mean is .

Process variance is .Expected hypothetical mean is ,

Expected process variance .Variance of hypothetical mean

.

From the sample, we can estimate as , so this is also the estimate of .The estimate of is 4.From the sample we can estimate using the unbiased sample estimate,

.But .Using the estimated variance of and the estimated mean of , we have

, so that the estimate of is .Then, is estimated to be , and

is estimated to be .515 .

The estimate of losses in the 4th year is where , and ,

so that .

12. Under the nonparametric empirical Bayes method applied to the Buhlmann credibility model,

the estimated expected process variance is ,

where is the number of policyholders and is the number of exposures per policyholder,

and the estimated variance of the hypothetical means: .

The estimated credibility factor for each policyholder is .

In this problem we have policyholders and exposure periods (years) perpolicyholder. From the given values we get and

. Then, . Answer: D

13. "Credibility" refers to the factor found in the semiparametric empirical Bayes approach.The following comments review semiparametric estimation. The model for the portfolio mayhave a parametric distribution for given , but an unspecified non-parametric distributionfor . In this case, we may be able to use the fact that relationships linking and

and the fact that in order to get estimates for and touse in the credibility premium formulation.



In this problem (the typical example) the conditional distribution of claim number given isPoisson with parameter . Then , so that

. We then use to estimate , and also use this as theestimate of . We find the (unbiased) sample variance of the 's and set that equal to , sothat the estimate of is . Then, as usual . From the

given data, we have , and

.Then

We are asked for the "credibility of one year’s experience for a single driver". This is the valueof when (one driver's experience for one year). . Answer: E

14. The semiparametric estimate is , where ,and sample mean of the claims for the risk being considered. In this case there is a singleobservation of (the number of robberies in one year) and , since there were no robberiesin the year for the store being considered. Then , where exposure period one

year) for the risk being considered.

For the semiparametric approach in which is Poisson, we have and , and .

In this example .

Then , and , and finally, the semiparametric

estimate is . Answer: B

15. For empirical Bayes estimation in the equal sample size case, the estimated credibility factorfor each is In this formulation, is the number of observations for each

group (or policyholder, or sample), and ,

where ) , and is the number of policyholders.

Also, .

In this example , , (Class A death times) , (Class B death times) ,

.When , the credibility factor is set equal to 0. Answer: A



16. Since the numbers of exposures differs among exposure periods, we use non-parametricempirical Bayes estimation for the Buhlmann-Straub model. Under the Buhlmann-Straub model,there are exposure periods for policy holder (group) , for . For policyholder andexposure period (year) , there are exposure units (8 members for group 1 in year 1, etc.), and

represents the observed average claim per exposure unit (member) (for "cell" ) (96 forgroup 1 in year 1). In this case, policyholders (groups 1 and 2), and exposure periods (years) for each policyholder. The usual unbiased estimates that are used for thestructural parameters

are , (estimated mean of the process variances),

and (estimated variance of the hyp. means).

From the given values we get . , and then

.

The estimated credibility factor for group is ;

.

An alternative to the unbiased estimate just described for the Buhlmann-Straub model isthe credibility-weighted average estimate of , which is found by first estimating

in the way just described for the Buhlmann-Straub model, and then .

If we use this estimate of to find the credibility premiums for groups , and then calculatetotal credibility premiums for past exposures, that total will equal to the actual total past claims.This is what is meant by using the method that "preserves total losses". In this case,

.

Credibility premium for the group 1 is .Answer: A

17. For a risk with sample mean of claims for the past months, the semiparametric empiricalBayes estimate of the expected number of claims next month is

, where . We are given that with and , we have

, and with and , we have

. It follows that

, so that . Then, with and , we have

.Answer: C



18.

.

.

.

The credibility premium for policyholder 2 is .

19. is the random variable for annual loss. We are given that the conditional distribution of given is exponential with a mean of , where has an unspecified distribution.

Therefore, the hypothetical mean is and the process variance is .

The expected hypothetical mean is (using the double expectation rule ).The expected process variance is .The variance of the hypothetical mean is

.From this we see that .

In general, .

From the data set we can use empirical estimation to estimate :

.

From the data set we can use empirical estimation to estimate :

.



19 continued

The empirical estimate of the variance of is .

In the semiparametric empirical Bayes credibility model, we use the empirical estimate of for ,so that . We also know that , so using the empirical estimate of givesis . But we also know that, for this model, , so using our sample estimateof , we have . We can then solve the two equations

and to get and .

We can now find the estimated loss in the 3rd year for a policy that had losses of in the first year and in the second year. The estimate is ,

where and . The credibility premium is

.

SIMULATION

SIMULATION SECTION 1 - THE INVERSE TRANSFORM METHOD SI-1


SIMULATION - SECTION 1, THE INVERSE TRANSFORM METHOD

The material in this section relates to Section 17.1 of "Loss Models". The suggested time framefor this section is 2-3 hours.

The objective in performing a simulation is to reproduce the behavior of a random variable bygenerating observations from another random variable which has the same distribution as therandom variable being simulated. For instance, to simulate the flip of a fair coin, which has

, we can toss a fair die. The simulation can be defined as follows:the event is simulated by tossing a 1, 2 or 3 on the die, and the event is simulated by tossinga 4, 5 or 6 on the die. To see if this is a valid simulation we must check of the simulated eventsreplicate the original probability distribution. In this example, there is .5 chance of tossing a 1, 2or 3, so the simulation replicates the correct .5 probability of getting a when a fair coin istossed. Same for

Example SI-1: You are given a random number generator that produces sample observations

from the following probability density function: .forotherwise

You use this random number generator to simulate the color of a traffic light facing a randomly-arriving car. The light is green 36% of the time, yellow 13% of the time, and red 51% of thetime. The following table gives the correspondence between the values of and the color of thetraffic light: Value of Color of light Green Yellow RedDetermine .Solution: In for the simulation to be valid it must reproduce the probabilities for the color of thetraffic light. In the simulating distribution , the probability of Green is

. This must be in the traffic lightcolor distribution. Thus, . In a similar way,

.

One of the general requirements for simulation is to have a way of obtaining independent uniformrandom numbers from the interval . There are various ways of generating such randomnumbers, but we will simply assume that they will be available when needed.

SI-2 SIMULATION SECTION 1 - THE INVERSE TRANSFORM METHOD


SI -1.1 Simulation of A Discrete Random Variable

Example SI-2: We wish to simulate the number of heads in 3 tosses of a fair coin.The actual distribution and distribution function are

We consider the following partition of the unit interval

Using the uniform distribution on we simulate as follows. Let denote the uniform value that will be used for simulation.

If simulate , if simulate ,if simulate , and if simulate .

We must check that the simulated distribution replicates the original probabilities. This isguaranteed since the probability of a subinterval of is simply the length of the subinterval.Therefore, , ,etc. This is a properly defined simulation.

We can generalize the procedure used in Example SI-2. Suppose that a discrete random variable has probability function for where .

We can simulate a value of in the following way. Given a random uniform number from

, find the integer such that . Then the simulated value of is .

For instance, if then is the simulated value of ,if then is the simulated value of ,if then is the simulated value of , etc.

Note that , and , so that

if , then the simulated value of is . This method is referred to as theinverse transform method of simulation. This procedure is sometimes referred to with thephrase .small uniform random numbers correspond to small simulated values



Example SI-3: Of a group of three independent lives with medical expense insurance, thenumber having medical expenses during a year is distributed according to a binomial distributionwith and . The amount, , of medical expenses for any person, once expensesoccur, has the following distribution: Each year, the insurance company pays the total medical expenses for the group in excess of5,000. Use the uniform random numbers, and , in the order given, to generate thenumber of claims for each of two years. Use the following uniform random numbers, inthe order given, to generate the amount of each claim: .Calculate the total amount that the insurance company pays for the two years.Solution: The probabilities for the given binomial distribution are

, and

The uniform number .01 satisfies .Therefore, the simulated number of medical expenses for the first year is 1. The uniform number.20 satisfies , so the simulated number of medicalexpenses for the second year is 2.

For the claim amount distribution, . In the first year, there isone medical expense. Since the first uniform number to be used for simulating expenses is.8 , the simulated expense in the first year is 100. There are two expenses in thesecond year. The first is simulated using the uniform number .95, and since

, the simulated expense is 10,000. The second expensein the second year is simulated from the uniform number .7, and since , thissimulated expense is 100. The total expense for the second year is 10,100. Each year the insurerpays total medical expenses in excess of 5000. The insurer pays 0 in the first year, since the totalmedical expense was 100, and the insurer pays in the second year.



SI-1.2 Simulation of a Continuous Random Variable

The inverse transformation method for a continuous random variableGiven the continuous random variable with cdf , a value of can be simulated from arandom number from the uniform distribution on in the following way. Solve theequation for ; the solution is the simulated value of . The solution of the equationis sometimes written . This is illustrated in the graph below.

Example SI-4: You are given a random variable with the following probability density

function: .ifotherwise

Let be a uniform random number between 0 and 1. Use the inverse transformation method todetermine a random observation from the distribution of , given .Solution: The cdf of is for , and

for . Given uniform on , the simulated value of using the inversetransformation method is the value of in the solution of , so that , fromwhich we get .

It is understood when applying the inverse transformation method, that given a uniform number, we solve for in the equation to get the simulated value . This may also be

referred to with the phrase . It issmall random numbers correspond to small simulated valuesalso possible the may see the phrase "small random numbers correspond to large simulatedvalues. If that is the case, then we solve for in the equation . This will result in adifferent simulated numerical value , but it will be a statistically valid simulation.



Example SI-5: You are to use the inversion method to simulate three values of a randomvariable with the following distribution function:

is linear in the interval and in the interval .You are to use the following random numbers from the uniform distribution on : .Calculate the mean of the three simulated numbers.

Solution: The cdf is .

We have constructed the cdf by finding the equation of the lines between successive point in thedistribution. For instance, since and , and since we know that is a linearfunction for . it must be true that for (we can use the "two-point" formula for finding the equation of a straight line; in this case the two -points are

and ). We can use the same reasoning to find for .

For , ranges from 0 to .4, and for ranges from .4 to 1. Inorder to apply the inversion method, and solve for from , we must use theappropriate "piece" of from this piecewise formulation of . If , we use

, and if , we use . The first uniform number is , therefore, the first simulated value of is , the solution of

.Then, , and

.The mean of is .The graph below illustrates the relationship between the uniform values and thecorresponding simulated -values.



SI-1.3 Simulation of Some Specific Random Variables

Simulation of the exponential distributionThe exponential distribution with mean has cdf .Given the value from the uniform distribution over , the simulated value of based on the standard form of the inverse transform method (for which small random numberscorrespond to small simulated values) is the solution of , so that

.

If we want to apply a simulation method in which small random numbers correspond to largesimulated values, we would solve the equation , so that .This would result in a different numerical simulated value, but it would be a statistically validsimulation (and would be slightly simpler to apply algebraically).

Simulation of the Poisson distributionThe relationship linking the exponential and Poisson distributions results in an alternative methodfor simulating the Poisson distribution. If the time between successive events has an exponentialdistribution with mean , and successive event times are independent, then the number of eventsin a unit of time has a Poisson distribution with mean The Poisson with mean can besimulated in the following way.

From successive independent uniform numbers find the successive products .Find . Then has a Poisson distribution with mean .This is called the "cumulative product algorithm"

Example SI-6: We wish to simulate the Poisson distribution with a mean of 3 using thecumulative product algorithm. We first calculate . We then multiply successiveuniform numbers until the product is less than .049787 . Apply the algorithm with each ofthe following sequences of uniform random numbers.(i) .45 , .63 , .30 , .81 , .66 , .15 ,(ii) .23 , .20 , ,88 ,(iii) .03 , .58 ,Solution: (i) , go to the next number ;

, go to the next number ; , go to the next number ;

, go to the next number ; , so we stop.

The simulated value of the Poisson is , since .

(ii) , go to the next number ; , so we stop.

The simulated value of the Poisson is , since .

(iii) , so we stop. The simulated value is



Simulation of the gamma distributionThe gamma distribution with parameters and , where is an integer can be simulated inthe following way (in the Exam C Tables, this distribution would be described as the gammadistribution with parameters and ). With being an integer , this gammarandom variable is the sum of independent exponential random variables, each with mean .This gamma is simulated from uniform variables as follows:

. When is an integer 1, the gamma distribution is also referred toas an Erlang distribution.

Example SI-7: Let be random numbers from the uniform distribution on .

Define for and .

Which of the following are true?I. The numbers and are random numbers from the same exponential distribution.II. The average of the numbers is equal to the average of the numbers .III. is a random number from a gamma distribution.Solution: I. Each and each is a simulated value from an exponential distribution withmean 1. True.II. Just because both the 's and the 's are valid simulated values from an exponentialdistribution with mean 1, there is no guarantee that the numerical values are the same. Forinstance, if , there is no guarantee that . False.III. The sum of independent exponential random variables all with same mean is a gammarandom variable. True.

Example SI-8: You have generated a random value from the uniform distributionon . Use the Normal Distribution Table, applying linear interpolation for values not in thetable. Using the inversion method, determine the random observation (corresponding to )that is from a normal distribution having mean 1 and variance 4.Solution: We solve for from the equation . Since is normal, we canstandardize ,

,

where has a standard normal distribution ( ). From the normal table, and . Using linear interpolation,

.



SI-1.4 Simulation of a Mixed Distribution

If a distribution is partly continuous and partly discrete, then the cdf increasescontinuously on the continuous region of , and there are discrete jumps in at the pointsof probability of , with the size of the jump being the amount of probability at the point. Givenuniform number , if is in a region where is continuous, then we solve toget the simulated value of . If is inside a "jump" interval, then the simulated value of is thediscrete point where the "jump" occurs.

Example SI-9: A random variable has distribution function with:(i) for (ii) (iii) for , where is the derivative of Three sample values are simulated from the distribution of by applying theinversion method to the three observations from the uniform distribution on . Theobservations are and . Determine the sample mean .Solution: From the definition of , we see that has a discrete point of probability at

, with . Then, (antiderivative of ).Since , it follows that for to continue from to as goes from 0 to 1.The inversion method is applied as follows. If , then the simulated value of is 0,and if , then the simulated value of is the solution of .From the three given uniform values, results in a simulated value of ,

and .

SI-1.5 Using Simulation to Estimate a Mean or a Probability

Suppose that we are simulating a random variable which has mean and variance . Supposewe are trying to estimate , and we want to determine the number of simulated values of needed, say so that . This is the same idea that was considered inlimited fluctuation credibility. We saw that the number of (simulated) observations of neededis . We can change the " " factor to " ", and if we do so, then wechange .05 to .01 in the denominator of the right side of the inequality. We can change theprobability to .95 from .9, and if we do so, then we change 1.645 to 1.96 on the right side of theinequality. We usually do not know or , so as the successive values of are simulated, wecalculate updated estimates of and , and continue to simulate 's until the inequality

is satisfied.

We can apply a similar sort of idea to estimating a probability. Suppose that is the unknownprobability of success for an experiment. Our objective is to estimate to within 1% of the actualvalue of with a probability of 95%. We will estimate with ,#successes in simulated trials

which is the proportion of the simulated experiments out of that are successes.We want . With a relatively large , is approximately normal, sowe can apply a similar approach as in the previous paragraph. This results in the followinginequality

. could be any probability, for instance , in which case,#simulated 's that are in simulated trials .

SIMULATION - PROBLEM SET SI-9


SIMULATION - PROBLEM SET 1

1. The random variable X has probability density function .ifif otherwise

Using the inverse transform method of simulation, find the random observation generated by the(uniform decimal) random number r .716 .A) .358 B) .608 C) .716 D) .858 E) 1.216

2. A binomial distribution with n 3 and p .4 is simulated by the inverse transform methodwith the uniform random numbers .31 , .71 , .66 , .48 , .19 . How many of the generated randomvariables are equal to 2?A) 1 B) 2 C) 3 D) 4 E) 5

3. A “mixed exponential" distribution has P[ X 0 ] p (0 p 1) and density functionf(x) (1 p) e for x 0. The distribution can be simulated by the inverse transformx

method as follows:if U is a random uniform (0,1) value and U p , then X 0 ; if p U 1 then X A) ln(1 U) B) ln(1 U) C) ln D) ln E) ln1 1 1 U1 p 1 p

1 U 1 U 1 p

4. X has density function f(x) = , 0 x , and f(x) = 0 elsewhere. Use the inversesin x2

transform method to simulate three values of X based on uniform [0 , 1] values

u = , u = .5 , u = . Find the mean of the simulated values x , x , x .1 2 3 1 2 32 1

2 22 + 3

4

A) B) C) D) E) 19 19 17 17 536 24 36 24 12

5. You are given a probability distribution with the following probability density function:

The inverse transformation method was used to generate a random observation fromthis distribution by using , a uniform random number between 0 and 1.Determine the value of .A) B) C) D) E)

SI-10 SIMULATION - PROBLEM SET 1


6. You are to use the inverse transform method to generate two random observations from the

distribution with probability density function .elsewhere

You are to use the following random numbers from the uniform distribution on :

Calculate the sum of the resulting random observations.A) B) C) D) E)

7. The random variable has distribution function . You are given: for , , for ,

You simulate by using a function of , the uniform distribution on the interval , andobtain the following sequence of values from : .Determine , the sample mean.A) B) 1 C) 2 D) E) 3

8. A company is insured against liability suits. The number of suits for a given year isdistributed as follows:Number of suits: 0 1 2 3Probability: 0.5 0.2 0.2 0.1For each of three years, you simulate the number of suits per year by generating one uniformrandom number from the interval and then applying the inverse transformation method.Your three random numbers are 0.83 , 0.59 , 0.19 .

The liability loss from an individual suit is a random variable with the following cumulative

distribution function .

forforfor

You generate uniform random numbers from the interval and apply the inversetransformation method to simulate the amounts of the individual losses. The first nine uniformrandom numbers, in order, are: 0.51 , 0.01 , 0.78 , 0.74 , 0.03 , 0.69 , 0.17 , 0.86 , 0.82Use, in order, as many of these random numbers as needed to simulate the company's total lossdue to liability suits over the three year period.A) 0.0 B) 0.2 C) 5.6 D) 5.8 E) 14.4



9. Your company insures a risk that is modeled as a surplus process as follows:(i) Interarrival times for claims are independent and exponentially distributed withmean 1/3.(ii) Claim size equals , where t equals the time the claim occurs.(iii) Initial surplus equals 5.(iv) Premium is collected continuously at rate .(v)You simulate the interarrival times for the first three claims by using 0.5, 0.2, and 0.1,respectively, from the uniform distribution on , where small random numberscorrespond to long interarrival times. Of the following, which is the smallest such that yourcompany does not become insolvent from any of these three claims?A) 22 B) 35 C) 49 D) 113 E) 141

10. When generating random variables, it is important to consider how much time it takes tocomplete the process. Consider a discrete random variable with the following distribution:

Of the following algorithms, which is the most efficient way to simulate ?

A) If set and stop.If set and stop.If set and stop.If set and stop.Otherwise set and stop.

B) If set and stop.If set and stop.If set and stop.If set and stop.Otherwise set and stop.

C) If set and stop.If set and stop.If set and stop.If set and stop.Otherwise set and stop.

D) If set and stop.If set and stop.If set and stop.If set and stop.Otherwise set and stop.

E) If set and stop.If set and stop.If set and stop.If set and stop.Otherwise set and stop.



Problems 11 and 12 relate to the following situation. A random sample of size 30 is taken fromthe distribution of the random variable . The sample mean is , and the sample

_

variance is 10 additional random values are drawn in the following order:

11. It is decided that sampling will stop when the estimated standard deviation of is first lessthan .2 . How many samples values after the 30th are needed to satisfy this criterion?A) 2 B) 4 C) 6 D) 8 E) 10

12. It is decided that sampling will stop when the width of the 95% confidence interval for isfirst less than (the 97.5 percentile of the standard normal distribution is 1.96).Ho many sample values after the 30th are needed to satisfy this criterion?A) 2 B) 4 C) 6 D) 8 E) 10

13. An estimate is being made of the probability of flipping a head with a particular coin. Thecoin will be flipped until the estimated standard deviation of the estimated value of is less than.075. In the first 40 coin flips there are 25 heads and 15 tails. The next 10 coin flips areT H H H T H H T H H . With which flip is the stopping criterion reached?A) 41 B) 42 C) 43 D) 44 E) 45

14. To estimate , you have simulated and with the following results:1 2 3 4 5

You want the standard deviation of the estimator of to be less than 0.05.Estimate the total number of simulations needed.A) Less than 150 B) At least 150, but less than 400 C) At least 400, but less than 650 D) At least 650, but less than 900 E) At least 900

15. A random sample of 18 data points has a sample mean of 8 and an unbiased sample varianceof 4. and are added to the sample. Find the updated unbiased samplevariance based on all 20 data points.A) 4.0 B) 4.1 C) 4.2 D) 4.3 E) 4.4

16. (CAS) A scientist perform experiments, each with a 60% success rate. Let represent thenumber of trials until the first success. Use the inverse transform method to simulate the randomvariable, , and the following random numbers (where low numbers correspond to a highnumber of trials): 0.15 , 0.62 , 0.37, 0.78 . Generate the total number of trials until threesuccesses result.A) 3 B) 4 C) 5 D) 6 E) 7



17. (CAS) Two actuaries are simulating the number of automobile claims for a book of business.For the population that are studying:i) The claim frequency for each individual driver has a Poisson distribution.ii) The means of the Poisson distributions are distributed as a random variable, .iii) has a gamma distribution.In the first actuary's simulation, a driver is selected and one year's experience is generated. Thisprocess of selecting a driver and simulating one year is repeated times. In the second actuary'ssimulation, a driver is selected an years of experience are generated for that driver.Which of the following is/are true?I. The ratio of the number of claims the first actuary simulates to the number of claims thesecond actuary simulates should tend towards 1 as tends to infinity.II. The ratio of the number of claims the first actuary simulates to the number of claims thesecond actuary simulates will equal 1, provided that the same uniform random numbers are used.III. When the variances of the two sequences of claim counts are compared the first actuary'ssequence will have a smaller variance because more random numbers are used in computing it.A) I only B) I and II only C) I and III onlyD) II and II only E) None of I, II, or III is true

18. (SOA) Insurance for a city’s snow removal costs covers four winter months.(i) There is a deductible of 10,000 per month.(ii) The insurer assumes that the city’s monthly costs are independent and normally distributedwith mean 15,000 and standard deviation 2,000.(iii) To simulate four months of claim costs, the insurer uses the Inverse Transform Method(where small random numbers correspond to low costs).(iv) The four numbers drawn from the uniform distribution on [0,1] are: 0.5398 0.1151 0.0062 0.7881Calculate the insurer’s simulated claim cost.(A) 13,400 (B) 14,400 (C) 17,800 (D) 20,000 (E) 26,600

19. (SOA) You are simulating a continuous surplus process, where claims occur according to aPoisson process with frequency 2, and severity is given by a Pareto distribution with parameters

and . The initial surplus is 2000, and the relative security loading is 0.1. Premiumis collected continuously, and the process terminates if surplus is ever negative.

You simulate the time between claims using the inverse transform method (where small numberscorrespond to small times between claims) using the following values from the uniformdistribution on [0,1]: 0.83, 0.54, 0.48, 0.14.

You simulate the severities of the claims using the inverse transform method (where smallnumbers correspond to small claim sizes) using the following values from the uniformdistribution on [0,1]: 0.89, 0.36, 0.70, 0.61.

Calculate the simulated surplus at time 1.(A) 1109 (B) 1935 (C) 2185 (D) 4200(E) Surplus becomes negative at some time in [0,1].



(SOA) Use the following information for Questions 20 and 21.Lucky Tom finds coins on his 60 minute walk to work at a Poisson rate of 2 per minute. 60% ofthe coins are worth 1 each; 20% are worth 5 each; 20% are worth 10 each. The denominationsof the coins found are independent. Two actuaries are simulating Tom’s 60 minute walk to work.

(i) The first actuary begins by simulating the number of coins found, using the procedure ofrepeatedly simulating the time until the next coin is found, until the length of the walk has beenexceeded. For each coin found, he simulates its denomination, using the inverse transformalgorithm. The expected number of random numbers he needs for one simulation of the walkis .

(ii) The second actuary uses the same algorithm for simulating the times between events.However, she first simulates finding coins worth 1, then simulates finding coins worth 5, thensimulates finding coins worth 10, in each case simulating until the 60 minutes are exceeded. Theexpected number of random numbers she needs for one complete simulation of Tom’s walk is

20. Which of the following statements is true?(A) Neither is a valid method for simulating the process.(B) The first actuary’s method is valid; the second actuary’s is not.(C) The second actuary’s method is valid; the first actuary’s is not.(D) Both methods are valid, but they may produce different results from the same sequence ofrandom numbers.(E) Both methods are valid, and will produce identical results from the same sequence of randomnumbers.

21. Determine which of the following ranges contains the ratio / .(A) 0.0 / 0.4(B) 0.4 / 0.8(C) 0.8 / 1.2(D) 1.2 / 1.6(E) 1.6 /

22. (SOA) You wish to simulate a value, , from a two point mixture.With probability 0.3, is exponentially distributed with mean 0.5.With probability 0.7, is uniformly distributed on .You simulate the mixing variable where low values correspond to the exponential distribution.Then you simulate the value of , where low random numbers correspond to low values of .Your uniform random numbers from are 0.25 and 0.69 in that order.Calculate the simulated value of Y .(A) 0.19 (B) 0.38 (C) 0.59 (D) 0.77 (E) 0.95



23. (SOA) An actuary is evaluating two methods for simulating , the future lifetime of thejoint-life status of independent lives and :(i) Mortality for and follows De Moivre’s law with ., which states that

for and .(ii)(iii) Both methods select random numbers , and independently from the uniformdistribution on .(iv) Method 1 sets: (a) (b) (c) smaller of and (v) Method 2 first determines which lifetime is shorter: (a) If , it chooses that is the first to die, and sets . (b) If R , it chooses that is the first to die, and sets .Which of the following is correct?(A) Method 1 is valid for but not for ; Method 2 is never valid.(B) Method 1 is valid for but not for ; Method 2 is valid for but not for .(C) Method 1 is valid for but not for ; Method 2 is valid for all and .(D) Method 1 is valid for all and ; Method 2 is never valid.(E) Method 1 is valid for all and ; Method 2 is valid for but not for .

24. (SOA) You are simulating a compound claims distribution.(i) The number of claims, , is binomial with and mean 1.8.(ii) Claim amounts are uniformly distributed on .(iii) Claim amounts are independent and are independent of the number of claims.(iv) You simulate the number of claims, , and the amounts of each of those claims, . Then you repeat another , its claim amounts, and so on until you have performed the desired number of simulations.(v) When the simulated number of claims is , you do not simulate any claims amounts.(vi) All simulations use the inverse transform method, with low random numbers corresponding to few claims or small claim amounts.(vii) Your random numbers from are 0.7 , 0.1 , 0.3 , 0.1 , 0.9 , 0.5 , 0.5 , 0.7 , 0.3, and 0.1.Calculate the aggregate claim amount associated with your third simulated value of .(A) 3 (B) 5 (C) 7 (D) 9 (E) 11

25. is a mixture of two exponential distributions.Distribution 1 has a mean of 1 and a mixing weight of .25 and distribution 2 has a mean of 2 anda mixing weight of .75. is simulated using the inverse transformation method with a uniform

value of .7. Find the simulated value of .



SIMULATION - PROBLEM SET 1 SOLUTIONS

1. F(x) = 0 if x 0 , F(x) = 1 if x , and F(x) = .34

x if 0 x

2x if x

12

1 1 32 2 4

r = .716 = 2x x = .608 . Answer: B.12

2. s : 0 1 2 3 (s) .216 .432 .288 .064 (s) .216 .648 .936 1.00A binomial value of 2 will be simulated by a uniform (0,1) value that is both greater than or equalto .648 and less than .936 . Thus, the uniform numbers .71 and .66 result in simulated binomialvalues of 2. Answer: B.

3. For x 0 , F(x) = P[ X x ] = P[ X = 0 ] + P[ 0 X x] = p + (1 p) e dtx – t

= p + (1 p)(1 e ) . Thus, F(x) = ififif

– x

Thus, for U p , the inversion method implies thatU = F(x) = p + (1 p)(1 e ) x = ln . Answer: D.– x 1 1 p

1 U

4. The c.d.f. is F(x) = dt = for 0 x .sin t 1 cos x2 2

By the inversion method, for uniform u we have x = F (u) , or equivalently, u = .–1 1 cos x2

u = = cos x = x = ,1 1 12 1

2 21 cos x

2 41

21

u = .5 cos x = 0 x = ,2 2 2 2

u = cos x = x = . = . Answer: A.3 3 32 + 3 3

4 2 6 3 365 19x + x + x1 2 3

5. From the diagram of the density function we must have ,since the total probability is 1. From the inverse transformation method,

(this is the integral of from 2 to 3.5).Answer: D

6. . We solve for .

, . . Answer: A



7. has a mixed distribution with cdf .

According to the inversion method of simulation, given a uniform random number from ,the simulated value of is

if , if , if .Then, .The sample mean is . Answer: D

8. The cdf of (number of suits per year) is

The simulated numbers of suits in the three years are

.The liability loss from a suit has a mixed distribution. If then , and if

then The simulated liability losses from the 3 suits are .

Total liability losses in 3 years . Answer: D

9. The interarrival time for claims is exponential. We must simulate the three claim times. Weare told that where small random numbers correspond to long interarrival times. This is theopposite of the usual inverse transform method. Instead of solving for interarrival claim time from (the simulation method that has small random number corresponding to smallinterarrival time ) , we solve for from the equation . Since interarrival time forclaims is exponential with mean , .The simulated interarrival times are and , where ,

, and .These are interarrival times, so the actual successive claim times are at times .231 , .767 , and1.535 . The claim amounts are , , and .The total premium received up to time .231 is .

The total premium received up to time .767 is .

The total premium received up to time 1.535 is .Surplus at time each of the claim times is:- at time .231 surplus is ,- at time .767 surplus is , and- at time 1.535 surplus is .In order for the company to be solvent at each claim time it must be true that

, and and . Answer: C



10. In order to generate values as quickly as possible, we want the -value with the largestprobability to be the first possible one generated, and second largest probability to be the secondpossible one generated, etc. We place the -values in order of largest to smallest probability:

Cumulative

The simulation procedure with generate if ,it would generate if ,it would generate if ,it would generate if , andit would generate if .

The reason this method results in the least time required for simulation is that the probability thatthe stopping time occurs on the first step is higher than any of the other simulation methods.

For instance, with the stopping points for each of the 5 methods given areA. so set (stop on the 4th step)B. so set (stop on the 3rd step)C. so set (stop on the 4th step)D. so set (stop on the 2nd step)E. so set (stop on the 4th step)D stops sooner than the other methods. In general, method D stops at the same time or soonerthan any of the other four methods, but never later.

Note that all five answers result in a valid simulation method since they reproduce theprobabilities correctly. Answer: D

11. The stopping rule is

Successive sample means can be found from the relationship :

Successive sample variances can be found from the relationship :

We see that the stopping rule is reached at the 34-th data point. Answer: B



12. The width of the 95% confidence interval is .

From the calculations in Problem 9, we see that interval width is

interval width is

interval width is

interval width is

interval width isContinuing with the calculations in Problem 9,

interval width is ,

interval width isWe see that the stopping rule is reached at the 36-th data point. Answer: C

13. . .if headif tail

number of heads in flips

The stopping criterion is .

After the first 40 flips, we have so that

Answer: B

14. The estimator of is the sample mean , which has variance

, if there are simulated observations.

The standard deviation of is . From the data given, the estimated variance of is

.

In order for the standard deviation of to be less than .05 we must have ,

which translates to . Answer: E

15. The sample mean based on the first 19 points is ,

and the sample mean based on all 20 points is We use the relationship

to get

, and then

. Answer: B



16. To say that low numbers correspond to a low number of trials is the standard form of theinverse transform method: given a uniform random number we find the integer such that

, and the simulated value of is . If we apply the inverse transformmethod in the form where low numbers correspond to a high number of trials, then given auniform random number , we find the integer such that .The random variable is the number of success until the first trial. This is a geometricdistribution with probability function and distribution function as follows:

Each uniform number simulates the number of trials until the next success. We must simulate three times to get (the simulated number of trials until the first success, (the additional

number of trials until the second success) and (the additional number of trials until the thirdsuccess). Then the total number of trials until the third success is .The first random number is , so that . We see that

so that the simulated value of is 3. This is , thesimulated number of trials until the first success. The second random number is , with

. We see that , so that the simulated value of is 1. , so that . We see that , so that the

simulated value of is 2. Then . Answer: D

17. I. The combination of a Poisson claim count with mean and a gamma distribution for results in a negative binomial distribution being simulated by actuary 1. The average number ofclaims simulated by actuary 1 in trials is . The second actuary selects a driver withPoisson parameter and the average number of claims in years for that driver will be .The ratio tends to 1 only if the second actuary's driver's is equal to . False

II. This is false for the same reason as I. False

III. For actuary 1, the variance of the sequence generated is the variance of a negative binomialdistribution. For actuary 2, the variance of the sequence generated is the variance of the Poissondistribution with parameter (the for the driver chosen by actuary 2). Either variance could belarger than the other depending on the value of for actuary 2's driver. FalseAnswer: E

18. According to the inverse transform method, given uniform [0,1] number , the simulatedstandard normal value is , where .From the given uniform values, we get

The loss random variable has a mean of 15,000 and standard deviation of 2000.The simulated values of are

.After monthly deductibles, the insurer pays

in the four winter months, for total claim cost of 14,400.Answer: B



19. The time between successive claims has an exponential distribution with mean .The simulated inter-claim times are where , or equivalently,

. We use the relationship because we are told that "smallnumbers correspond to small times between claims"; this is the standard form of the inversetransform method. The simulated inter-claim times are

, .The second simulated claim occurs after time 1, so it is irrelevant.The simulated claim amount of the first claim is , where

, or equivalently, .

The simulated claim amount is .

The expected claim amount per claim is , so that the expected aggregateclaim per period is . With relative security loading of .1, the premium rate is

. With initial surplus of 2000, the simulated surplus at time 1 is . Answer: C

20. D.

21. Actuary 1: Average number of times that must be simulated in one hour is .For each one of those times, a simulation of coin denomination must be made. Average numberof simulations in total is .Actuary 2: Denomination 1 coins are found at the rate of per minute, so that in onehour, an average of times must be simulated.For denomination 5, coins are found at rate per minute, and an average of must be simulated.Similarly, for denomination 10, an average of 24 simulations are needed per hour.Average total is . . Answer: E

22. The mixing probability for the exponential distribution is .3 . We are told that low valuescorrespond to the exponential distribution when simulating the mixing variable. Therefore theexponential distribution is used if the uniform random number for simulating the mixing variableis . Since the first uniform random number is , we use the exponentialdistribution for . We are also told that low random numbers correspond to low values of .This just means that we are applying the usual inverse transformation method. The cdf of theexponential distribution with mean .5 is .The second uniform random number is .69, so that the simulated value of is the value of which satisfies . Answer: C



23. Under DeMovire's Law, the future lifetime of someone at age is uniformly distributed onthe interval . Therefore Method 1 is a valid simulation of both times of death. The twoindividuals are independent, so if then there is an equal chance that either is first to die.However, given that is first to die, the conditional distribution of 's time of death is no longeruniform (same for ). Therefore Method 2 is never valid. Answer: D

24. The distribution of binomial with mean , so that is the probability of aclaim. Then the distribution of number of claims is

The first uniform number .7 simulates 2 claims, since .We then use .1 and .3 to simulate claim amounts.The next uniform number is .1 and is used to simulated the second . This will be 1, since

. We use .9 to simulate the one claim amount.The next uniform number is .5, which simulates 2 claims.The uniform numbers .5 and .7 simulate the claim amounts. The claim amount distribution is

The uniform number .5 simulates a claim amount of 3, and the uniform number .7 simulatesa claim amount of 4. Aggregate claim amount simulated associated with the 3rd simulated valueof is 7. Answer: C

25. .We must solve for from the equation .The equation can be written as .Substituting results in the quadratic equation .

Solving for we get .We ignore the negative root, since must be .Solving for , we get .

SIMULATION SECTION 2 - THE BOOTSTRAP METHOD SI-23


SIMULATION - SECTION 2, THE BOOTSTRAP METHODAND STATISTICAL ANALYSIS USING SIMULATION


SI-2.1 The Bootstrap Estimate of the Mean Square Error of an Estimator

If is a quantity related to a probability distribution and is an estimator of , then wegenerally regard as a random variable. The is defined to bemean square error of

. This is not necessarily the variance of . If is an unbiased estimator,then , but in general ,where .

The bootstrap method is a way of estimating the mean square error of an estimator. The estimatoris generally calculated from a random sample of observations from the underlying distribution ofthe random variable . The bootstrap estimate is calculated by first constructing the empiricaldistribution of the random sample, and then "resampling" from the empirical distribution.The following example illustrates the bootstrap method based on a sample of size 2. Exam Cbootstrap questions have generally been based on a sample of size 2 or 3.

Example SI-10: A sample from the distribution of consists of the two values and . Apply the bootstrap technique to estimate the mean square error of the sample mean as

an estimator of the distribution mean.Solution: The empirical distribution has probabilities of assigned to each of the two values 3and 7. The mean of the empirical distribution is . The estimator of the mean thatwe are using is .

The bootstrap method is based on resampling from the original sample, with the new sampleshaving the same size as the original sample. Since the original sample size was 2, we choosesamples of size 2 from the original sample . There 4 samples of size 2 that we can take from

, and these are . For each of these samples, we calculate theestimator based on that sample, and we calculate the squared deviation of the estimator from thequantity being estimated, which is 5, the actual mean of the empirical distribution. This issummarized in the table below.

Sample Estimator Based on the Sample Squared Deviation of Estimator From Quantity Being Estimated (5)

The bootstrap estimate of the mean square error of the sample mean estimator is the average ofthe squared deviations in the last column, which is .

SI-24 SIMULATION SECTION 2 - THE BOOTSTRAP METHOD


The main idea with the bootstrap method is that we use the empirical distribution as if it actuallywas the distribution of . Our original objective was to find the mean square error of the samplemean estimator based on a sample of size 2 for the original random variable . We don't knowanything more about the distribution of than the sample that we have taken, so we use thatsample as a substitute for the distribution of and we work toward our objective again. Ourobjective is to find the mean square error of the sample mean estimator for a sample of size 2, butnow we are using the empirical distribution. This means that we take samples of size 2 from theempirical distribution. Since the empirical distribution is a 2-point discrete random variable, weare taking samples of size 2 from that 2-point distribution. As was seen in Example SI-10, thereare only different samples of size 2. For each of them, we calculate the squared deviationof the estimated mean from the "true" mean. What "true" mean refers to now is the mean ofnotthe original random variable (we don't know that value); it refers to the mean of the empiricaldistribution (which was 5 in Example SI-10).

In a more realistic situation, the sample size is (a lot) larger than 2, and complete resampling canbecome impossible. For instance, with a sample of size 10, there would be possible samplesof size 10 from the original sample of size 10, because each in a resampling can be any one ofthese original 10, with repeats. For a sample of size , there would be ways of taking a sampleof size from the original sample.

The way in which we can deal with the impossibly large number of resamples that can be takenfrom the original sample is to just take some resamples from the original sample. If the samplesize was 10, then just (randomly) choose say 1000 resamplings out the possible resamplings.Then, for each of the 1000 resamples, we calculate the estimate, and then calculate the squareddeviation of that estimate from the quantity in the empirical distribution being estimated, and thenaverage out those 1000 squared deviations. The following example illustrates the idea.

Example SI-11: The following random sample of size is drawn from the distribution ofa random variable with unknown mean and unknown variance (the sample values are givenin numerical order but were not necessarily sampled in that order).4, 5, 5, 6, 8, 9, 9, 9, 12, 14, 17, 17,18, 18, 23, 27, 31, 36, 44, 57Formulate the empirical distribution, and simulate 5 bootstrap samples of size 20 from theempirical distribution.Solution: With , the probability function and distribution function of the empiricaldistribution are



Example SI-11 continuedNotice that the mean of the empirical distribution is

,which is the sample mean of the original sample.

The variance of the empirical distribution is

(this is not to be confused with the sample variance of the random sample, which is

; in general, for a sample of size from the

original distribution of , the sample variance will be , and the variance of the

empirical distribution based on that sample will be ).

A simulated bootstrap sample is a sample of size 20 from the original sample, and will be asample of 20 's simulated from this empirical distribution just described (the connection tosimulation with the bootstrap method is that we simulate some of the possible resamplings).The 20 's will come from the 15 possible values

, using their empirical distributionprobabilities. A table of random uniform numbers, or a computer routine for generating randomuniform numbers can be used along with the inversion method for simulating a discretedistribution (the empirical distribution in this case is a discrete 15 point distribution). Thefollowing 5 bootstrap samples were simulated in that way (again the sampled values have beenplaced in increasing numerical order).Sample 1: 4, 8, 8, 9, 9, 9, 12, 14, 14, 14, 14, 14, 17, 18, 27, 31, 36, 44, 44, 57Sample 2: 5, 6, 8, 9, 9, 12, 14, 14, 17, 17, 17, 18, 23, 27, 31, 31, 44, 57, 57, 57Sample 3: 5, 5, 6, 6, 9, 9, 9, 9, 9, 9, 17, 18, 27, 27, 27, 27, 36, 44, 44, 57Sample 4: 4, 5, 5, 5, 5, 5, 6, 8, 9, 9, 9, 12, 12, 14, 17, 18, 23, 31, 36, 44Sample 5: 4, 4, 5, 8, 8, 8, 9, 12, 14, 17, 18, 18, 18, 18, 18, 23, 31, 31, 36, 44

In Example SI-11, the parameter being estimated in the distribution of might be the mean,median or variance of , or it might be , etc. We can estimate any quantityrelated to the distribution of . Our objective is then to estimate the mean square error of theestimator. If is the quantity being estimated and is the estimate, our objective is to find themean square error of , which is . Whether is the mean or median or variance of

, or the probability , etc., we don't know the actual value of . This is where the bootstrapapproximation to the mean square error of the estimator comes in. We do know the values of themean, median, variance, , etc, for the empirical distribution. Instead of finding in the original distribution of (which is impossible, since we don't know the originaldistribution of ), we try to find in the empirical distribution (which is possible,since we know the empirical distribution).



In the bootstrap approximation, now refers to the corresponding quantity in the empiricaldistribution. This is what was done in Example SI-10. We had a sample of size 2, , and wewere estimating the mean of , so is the mean of . We don't know the mean of , but we doknow the mean of the (2-point) empirical distribution, it is 5. We are using the sample meanestimator to estimate the distribution mean, so refers to the sample mean, and we are trying tofind . Here is where we have to be careful aboutin the empirical distributionunderstanding the application of the bootstrap approximation. For the 2-point empiricaldistribution of SI-10, we look at all possible samples of size that can be drawn (withreplacement).

As pointed out after Example SI-10, there are 4 possible "resamplings" of size 2. Each of themhas the same chance of being drawn. For each of those samples we calculate , the samplemean , and the calculate the squared deviation . Wefor that sample for that samplethen take the average of those squared deviations; there are 4 of them in this case, since there are

resamplings. If the original sample was of size 3, then there would be resamplings, still small enough so that we could consider all of them (Example 17.11 in the LossModels book is a bootstrap example based on a sample of size 3).

If our original sample was larger, say size 10, then we would have to consider resamplings tofind the bootstrap approximation. We would just consider a subset of those resamplings,with the subset of resamples chosen randomly by simulation. If we are estimating the quantity using the estimator , we would find the numerical value of in the empirical distribution, andthen for each resample considered, we would calculate the numerical value of , and then thevalue of (using the numerical values just mentioned). The bootstrap estimate of themean square error of the estimator would be the average of those values of for theresamplings that were taken.

The following continuation of Example SI-11 illustrates how we find the bootstrap estimate if theoriginal sample is large.

Example SI-11 (continued): Use the 5 bootstrap samples to estimate the mean square error of(a) the sample mean estimator,(b) the unbiased sample variance estimator, and(c) the following estimator of , .number of 's

Solution: (a) The parameter being estimated is the mean of the distribution of , so that in thiscase, . The estimator is the sample mean . The mean of the empirical distribution is

(which is the sample mean of the original sample). For each of the 5 resamplings, wecalculate the sample mean of that sample and then calculate the squared deviation from 18.45.

Sample Estimator Based on the Sample Squared Deviation of Estimator FromNumber Quantity Being Estimated ( ) 1 2 3 4 5



Example SI-11 continuedThe bootstrap estimate of the mean square error of the sample mean estimator is the average ofthe squared deviations in the last column, which is

.

Note that we know that the actual variance of the sample mean is , and so its sample

estimate is . From the sample of size and calculations in Example SI-11, we have

. Since the sample mean is an unbiased estimator, its mean square error is equal toits variance, so it is reasonable that the bootstrap estimate of the MSE of the sample meanestimator is approximately equal to the estimated variance of the sample mean estimator.

(b) The distribution parameter being estimated in this case is , the distribution variance.

The sample estimator is , the unbiased form of sample variance.

The variance of the empirical distribution was found earlier to be . For each of the 5resamplings, we calculate the unbiased sample variance of that sample and then calculate thesquared deviation from 193.35.

Sample Estimator Based on the Sample Squared Deviation of Estimator FromNumber Quantity Being Estimated ( ) 1 2 3

Note that for each sample we calculate the unbiased sample variance , so that forfor that samplesample 1, the estimator is

,and for sample 2, the estimator is

, etc.

The bootstrap estimate of the mean square error of the sample variance estimator is the average ofthe squared deviations in the last column, which is .

(c) The quantity being estimated now is . The sample estimator is .number of 's

Since 8 out of the original 20 sample points are , it follows that in the empirical distributionthe probability being estimated is ; this is . For each of the 5 bootstrap samples, wecalculate , which is the proportion of 20 values in the sample that are . For sample 1 this is

, etc. The bootstrap estimate calculations are summarized in the following table.



Example SI-11 continuedSample Estimator Based on the Sample Squared Deviation of Estimator FromNumber Quantity Being Estimated ( ) 1 2 3

Then the bootstrap estimate of the mean square error of the estimator is .

(Note that for a sample of size , the estimator of the probability isnumber of 's , where can be regarded as having a binomial distribution withparameters and . The actual variance of this binomial proportion estimate is

, which has a maximum of , that occurs when .Thus with , the actual should be no more that .0125 .

There has usually been a bootstrap problem on each Exam C for the past several years.

SI-2.2 Applications of Simulation to Statistical Testing

The material covered in this section is mentioned briefly in Section 17.2 of the LossModels book. There haven't been any exam questions on this material on the releasedexams.

At several places in these notes, various statistical tests have been reviewed, such as the -test,chi-square goodness-of-fit test, etc. In general, a statistical test has a null hypothesis, and analternative hypothesis . Data is collected, often in the form of a random sample (or samples).The null hypothesis may be a statement about the characteristics of the distribution from whichthe random sample has been drawn. may be a statement about the value of a parameter orsome parameters in a distribution, or it may be a statement that the sample has been drawn from aparticular distribution. The objective of the test is to analyze the sample data and determine(make an inference) whether to accept or reject . The test is performed by using the randomsample of data to calculate a test statistic . The test statistic itself is a random variable with aprobability distribution. For the typical types of statistical tests that we have considered, when

is true, the test statistic will have a distribution which is approximately equal to a well-known distribution (chi-square distribution for chi-square goodness-of-fit test, for instance).

An important component of a hypothesis test is the level of significance of the test. The level ofsignificance is usually set to be .05 (5% level of significance) or .01 (1%). Once the level ofsignificance is chosen, the critical value for the test is determined. The critical value is usuallythe 95-th percentile (for 5% significance test) of the distribution of the test statistic (thisrequires reference to a -distribution table, or a chi-square table, etc.); if a test is a two-sided test,



the critical value would be the 97.5 percentile. The following summarizes some of the main teststhat have been considered earlier.(i) Two-sided normal test. In this test, the test statistic is assumed to be normal, so there is noreference to degrees of freedom. For a 5% significance test, the critical value is ,and the null hypothesis is rejected if .(ii) One-sided normal test on the right. For a 5% significance test, the critical value is

. The null hypothesis is rejected if .(iii) One-sided normal test on the left. For a 5% significance test, the critical value is

. The null hypothesis is rejected if .(iv) Chi square test with 10 degrees of freedom. For a 5% significance test, the critical value is

. The null hypothesis is rejected if , where is the test statistic.

The typical mechanical application of the hypothesis test is as follows: if the numerical value ofthe test statistic for the given data set is , then reject if . It can happen that eventhough is true, some data sets result in a test statistic value , in which case we reject

; since we are rejecting even though is true (although we would not know is true)we have made an error. This is referred to as a Type I error. If the null hypothesis is true, thenthe probability of making a Type I error is is true . Once the level ofsignificance desired for the test is chosen, say .05, the critical value is the number whichsatisfies the probability relationship is true .

Assuming that is true, the distribution of may be known (possibly we will know theapproximate distribution of ), and from the probability table for the distribution of we canfind (in this case is the 95-th percentile of ). It follows that the level of significance of ahypothesis test is the probability of making a Type I error.

Reducing the level of significance from .05 to .01 will change (usually increase) the critical value,and will be less likely to be rejected. Reducing the level of significance from .05 to .01 doesnot mean the that test becomes more accurate; it means that the chance of making a Type I errorhas been reduced, but the chance of making a Type II error will be increased. A Type II erroroccurs if is false but the mechanical application of the test given the data results in acceptanceof .

Another way of looking at the hypothesis test it to consider the -value of the observed teststatistic. The -value is a probability defined in the following way: assume that is true andthat the test statistic has distribution ; then the -value of is . For example,suppose a chi-square test is being done and has chi-square distribution with 9 degrees offreedom. Suppose that the test statistic value is . Then the -value of is

(from a chi-square table).

If the test had a significance level of 5%, then would be rejected, because the critical valuewould be and (16.92 is the 95-th percentile of the chi-square distribution with 9 degrees of freedom). would not be rejected in a 1% test, because inthat case the critical value is 21.67.



An equivalent way of determining whether or not is rejected can be set up using the -valueof ; if the -value of is less than .05 then is rejected at the 5% significance level, otherwise is not rejected.Since the -value is .025 in the example just mentioned, would be rejected at the 5% level( would be rejected at any level of significance over 2.5%) but would not be rejected at 1%( would not be rejected at any level of significance at or below 2.5%).

A crucial component of the hypothesis test is the conditional distribution of the test statistic given that is true. For instance, in the chi-square test, given that the sample comes from thespecified distribution of , the distribution of is usually only approximately chi-square (withthe approximation becoming more exact as the sample size increases, so that is asymptoticallychi-square). It may be possible to simulate a more accurate -value of by simulating the actualdistribution of rather than using the approximate chi-square distribution for . We reviewsome of the standard statistical tests and how simulation can be applied to obtain a -value.

The Chi-Square Goodness-of-Fit TestThis test was discussed earlier in the notes for estimation and testing of loss and survival models.The test is usually applied to discrete or grouped data.

Suppose that the random variable has a discrete distribution on the integers withprobability function . Suppose that we have observations, andeach observation is one of the numbers , with being the number of observations thatare equal to . We want to test whether or not the observed values behave as if they have beendrawn from (or fit) the distribution of .

The standard test for this is the chi-square goodness-of-fit test. The test statistic is

. As the sample size gets larger, the statistic has a distribution which

approaches a chi-square distribution with degrees of freedom.

The hypothesis test has null hypothesis the sample is drawn from the distribution of and alternative hypothesis the sample is not drawn from the distribution of .For a 5% significance level, we conduct the test in the following way. If the test statistic numerical value is and if then reject .

This is equivalent to saying that (as mentioned above, the probability is called the -value of ). The null hypothesis is rejected at the 5% significance

level if the -value of the test statistic , is less than .05 (similar statements apply for significancelevel .01). The interpretation is that if the -value of the test statistic is small then it is unlikelythat the observations were drawn from the distribution of . The typical approximation appliedto determine the -value is to assume that has a chi-square distribution with degrees offreedom, and referring to such a table will give the -value .

Since does not have a chi-square distribution exactly, it may be possible to get a more accurateestimate of the -value of by simulation. We can simulate the distribution of and get anestimate of the -value of as follows:



(i) generate (simulate) independent values from the distribution of , and let be the number of 1's, 2's,..., 's respectively (the superscript indicates that

this is the first simulated sample)

(ii) calculate

(iii) repeat steps (i) and (ii) times to get .The simulated -value of is . As gets larger this proportion approachesnumber of 's

the true -value.

The Kolmogorov-Smirnov TestThis test was also discussed earlier in the notes for estimation and testing of loss and survivalmodels. The test is usually applied to data sampled from a continuous distribution. Suppose that

has a continuous distribution with known distribution function . Suppose that we have arandom sample of observations . Our objective is determine whether or not thesample values have been drawn from the distribution of . The null hypothesis being tested is the distribution from which the sample -values have been drawn is the same as the distribution of .The alternative hypothesis is the sample is not from the distribution of .

The Kolmogorov-Smirnov test requires construction of the empirical distribution of the datasample. The empirical distribution is constructed in the same way as described earlier in thesenotes, but now we focus on the distribution function of the empirical distribution. Thedistribution function of the empirical distribution is .number of 's

The Kolmogorov-Smirnov test looks at how closely fits , the distribution function of . Ifthe fit is "close enough", then will be accepted. Recall that the empirical distribution functionis a step-function. The Kolmogorov-Smirnov test statistic is the "maximum distance" that isaway from over the range of the distribution of . This is found by first calculating thefollowing quantities: for each in the sample data set find and

. Then .

In other words, at each sample point , there is a jump in the empirical distribution function, andwe find the absolute difference between the hypothesized distribution function of and both thelower and upper jump points in the empirical distribution function. is the maximum deviationof from upper and lower jump points of over all the 's in the data set. The test statistic has a distribution that depends on , the number of data points.

Some critical values for the Kolmogorov-Smirnov test are Significance level .20 .10 .05 .01 Critical Value

If the numerical value of the test statistic is , then the Kolmogorov-Smirnov test with a 5% levelof significance is performed as follows: if then is rejected. In other words, if the

maximum deviation of the empirical distribution function from the model distribution function is large enough, the inference is that the sample was not drawn from the distribution of .



As an alternative to performing the test with the table of critical values given above, this test canbe formulated in terms of the -value of . If the -value of is less than .05 then is rejectedat the 5% level of significance. The -value of can be estimated by simulation. This can bedone in the following way.(i) simulate values from the hypothesized distribution of , say (ii) calculate the Kolmogorov-Smirnov test statistic for that sample, say (iii) repeat steps (i) and (ii) times to get .The simulated -value of is . As gets larger this proportion approaches thenumber of 's

true -value.

The Chi-Square Test When There are Unknown Parameters

The standard test for the chi-square goodness-of-fit test is .

This statistic is based on a sample of size , in which the underlying distribution of ispartitioned into pieces. If the distribution of is discrete and integer-valued with , thenthe first of the " pieces" could be the integers and the -th piece is theset . If the distribution of is continuous, the pieces are usually intervals

. In any case, represents the distribution probability that " is in the -th piece", so that in the discrete

case, for and , and in the continuous case, . denotes the actual number of data points in piece or interval . The

test statistic has a (approximate) chi-square distribution with degrees of freedom.

Suppose that the distribution of has unspecified parameters. The goodness-of-fit teststatistic is found in much the same way as that just described. The difference is that the unknown distribution parameters are estimated using the data set. The estimated parameters areused to get an estimated value of the probability , the estimate being denoted by . The test

statistic is now , and has a distribution which is (approximately) chi-

square with degrees of freedom. If the numerical value of calculated from aparticular data set is , then the -value of can be simulated as follows:(i) from the data set estimate the parameters unknown parameters of ,

(ii) compute the test statistic , where is the distribution probability found

from the estimated parameters in (i),(iii) using the estimated parameters of 's distribution from (i), simulate a new sample of size

from the distribution of ; use the new sample to get estimates of the parameters of newand new estimates of the values; find the new values for the using the new simulated 's inthis step; the simulated value of the test statistic is

, which is found using 's simulated and 's estimated in this step

(iv) step (iii) is repeated many times (say times), and the simulated -value of isnumber of 's .



The Kolmogorov-Smirnov Test When There are Unknown ParametersIf a distribution has unknown parameters and a random sample of size is available for thedistribution, then the parameters can be estimated from the data to get , the distributionfunction of based on the parameter estimates. The empirical distribution function based on therandom sample is denoted by (as before). We calculate the Kolmogorov-Smirnov teststatistic value as before: , wherethe maximum is taken over the sample values. The -value of can be simulated as follows:(i) simulate new values of using the estimated distribution based on the estimatedparameters, and denote the empirical distribution function of these values sim(ii) using the simulated values of from (i), find estimates of the distributionnew newparameters, and denote by the distribution function of based on these new estimated(sim)parameters,(iii) the simulated value of the K-S statistic is , where,sim (sim)this maximum is taken over the simulated -values in (i),(iv) repeat steps (i) to (iii) times; the simulated -value of is .number of 's

The last couple of sections are somewhat more detailed than what is presented in the Loss Modelsbook on this topic.



SIMULATION - PROBLEM SET 2 SI-35



Problems 1 to 3 refer the following random sample of 15 data points:8.0 , 5.1 , 2.2 , 8.6 , 4.5 , 5.6 , 8.1 , 6.4 , 3.3 , 7.3 , 8.0 , 4.0 , 6.5 , 6.3 , 9.1The following three bootstrap samples of the empirical distribution have been simulated. Eachsample is of size 15.Sample 1: 3.3, 3.3, 3.3, 4.0, 4.0, 4.0, 6.5, 7.3, 8.1, 8.6, 8.8, 9.1, 9.1, 9.1, 9.1Sample 2: 3.3, 3.3, 5.1, 5.1, 5.1, 5.6, 6.3, 6.4, 6.5, 7.3, 7.3, 8.0, 8.0, 8.0, 8.1Sample 3: 2.2, 3.3, 3.3, 4.0, 4.5, 4.5, 5.1, 6.5, 7.3, 8.0, 8.1, 8.1, 8.1, 9.1, 9.1

1. Find , where sample variance (unbiased form), and variance of the empirical distribution

A) .20 B) .22 C) .24 D) .26 E) .28

2. Suppose that is the mean of the distribution, and it is being estimated by the sample mean.Use the three bootstrap samples to estimate the of the estimator.A) .01 B) .02 C) .03 D) .04 E) .05

3. Suppose that is the probability , and it is being estimated by the proportion of datapoints in the random sample that are . Use the three bootstrap samples to estimate the for this estimator.A) .018 B) .020 C) .022 D) .024 E) .026

4. The mean of a distribution is being estimated, using the sample mean of a random sample asan estimator. A sample of size 2 is drawn: . is the mean square error of the estimator when the empirical distribution is used. Find the exactvalue of .A) .125 B) .250 C) .375 D) .500 E) .625

5. With the bootstrapping technique, the underlying distribution function is estimated by whichof the following?(A) The empirical distribution function(B) A normal distribution function(C) A parametric distribution function selected by the modeler(D) Any of (A), (B) or (C)(E) None of (A), (B) or (C)

6. You are given a random sample of two values from a distribution function : 1 3

You estimate ( ) using the estimator ( ) , where .12 2

2

1

2 1 2

Determine the bootstrap approximation to the mean square error.(A) 0.0 (B) 0.5 (C) 1.0 (D) 2.0 (E) 2.5



7. You are given a random sample of two values from a distribution function : 1 3

You estimate ( ) using the estimator ( , ) ( ) , where .21 21

22 1 2

Determine the bootstrap approximation to the mean square error.(A) 0.0 (B) 0.5 (C) 1.0 (D) 2.0 (E) 2.5

8. The following random sample of size 5 is taken from the distribution of : 1 , 3 , 4 , 7 , 10

Bootstrap approximation of the mean square error of estimators is to be based on the following 6resamplings of size 5 from the empirical distribution: Resample 1 : 1 , 1 , 4 , 7 , 7 Resample 2 : 3 , 4 , 4 , 7 , 10 Resample 3 : 1 , 4 , 4 , 10 , 10 Resample 4 : 3 , 3 , 3 , 4 , 10 Resample 5 : 4 , 4 , 7 , 7 , 10 Resample 6 : 1 , 7 , 7 , 10 , 10

The median of is estimated by the third order statistic of a sample.Find the bootstrap approximation to the estimator of the median using the 6 resamplings.

9. is the random variable denoting the number of claims in one day. The following is a sampleof the number of claims occurring on 5 randomly chosen days: 2 0 4 5 7

The following estimator from a sample of days is used to estimate , the probabilityof 4 or less claims in one day: # days with 4 claims or less

The bootstrap approximation is applied to estimate the mean square error of usingthe following 8 samples simulated from the empirical distribution of the original sample:

Sample 1 0 2 0 7 7Sample 2 2 2 4 2 5Sample 3 7 2 5 4 5Sample 4 4 0 7 2 5Sample 5 2 7 2 5 5Sample 6 4 5 4 2 7Sample 7 0 0 2 4 7Sample 8 5 0 4 7 5

Find the bootstrap approximation to the mean square error of the estimator.




1. Sample variance

Variance of the empirical distribution is

.

Then Answer: E

2. The estimate of the distribution mean is the sample mean . In this example, is the number of bootstrap samples that are available. Also, is the distribution mean of

the empirical distribution, which is equal to the sample mean of the original random sample. The values of the estimator for the three samples are the sample means of the bootstrap

samples - and . Then the estimated MSE is[ Answer: D

3. The value of is the probability in the empirical distribution that . This is ,since 4 of the 15 original sample points are . The values of the estimator for arethe proportions in each bootstrap sample of the points that are . These are6 6and . Then, the estimated MSE is[ Answer: A

4. The parameter being estimated is the distribution mean, and the mean of the empiricaldistribution is .5 . The estimator being used is the sample mean, . Thus, for the 4 pairs

, we have estimator values , and the estimate of the MSE is . Answer: A

5. Answer: A.

6. Suppose that a random sample from a distribution is given: , and suppose thatthe sample is used to estimate some parameter of the distribution. If the estimator is , then thebootstrap approximation to the mean square error of this estimator is .In this expression is the value in the empirical distribution of the parameter being estimated, andthe expected value is taken within the empirical distribution. In this case, the parameter beingestimated is the distribution variance. The mean of the empirical distribution is , and

the variance of the empirical distribution is .



6 continuedTo find the expectation, we must consider all samples of size 2 from the empirical distribution(since the empirical distribution was based on a sample of size 2 itself).

The mean square error of the estimator is .From the empirical distribution on the set , there are four possible samples, which are

and . For each sample, we must calculate 12 1

22

( )

for that sample. For instance, for the sample , we have , and 1

22

( ) .

The bootstrap approximation to the mean square error of the estimator is the average of the 4

values of ( ) that we get. This is summarized in the following table.2

1

2

Sample

Since each of the four possible samples is equally likely to occur, the bootstrap estimate of . Answer: B

7. Since the sample consists of only 2 sample points, there are only 4 possible, equally likelybootstrap samples: (1) (2) (3) (4) .We are trying to estimate , so we find the variance of the empirical distribution.The empirical distribution is a 2-point random variable and its variance is

.The bootstrap estimate of the MSE of the estimator is .

To find this expectation we first find for each of the samples:1

22( )

(1) (2) (3) (4) .The bootstrap estimate is the average of the values of for the samples:

. Answer: C

8. The median of the empirical distribution is .

Resample 1 , 1 , 4 , 7 , 7 4 3 , 4 , 4 , 7 , 10 1 , 4 , 4 , , 3 , 3 , 3 , 4 , 10 4 , 4 , 7 , 7 , 10 1 , 7 , 7 , 10 , 10

The bootstrap estimate of MSE( is .



9. In the empirical distribution, the actual value of the parameter being estimated is ,since three of the values in the 5-point empirical distribution are .

We calculate and for each simulated sample.

Sample 1 0 2 0 7 7 Sample 2 2 2 4 2 5 Sample 3 7 2 5 4 5 Sample 4 4 0 7 2 5 Sample 5 2 7 2 5 5 Sample 6 4 5 4 2 7 Sample 7 0 0 2 4 7 Sample 8 5 0 4 7 5

The average value of is ; this is the bootstrap approximation to themean square error of based on the 8 samples.



SIMULATION SECTION 3 - LOGNORMAL MODEL FOR STOCK PRICES SI-41


SIMULATION - SECTION 3THE LOGNORMAL DISTRIBUTION APPLIED TO ASSET PRICES

The material in this section relates to Chapter 18 of "Derivatives Markets". The suggested timeframe for this section is 2 hours.

SI-3.1 Review of the Normal Distribution

The normal distribution was reviewed in Section 3 of the Modeling unit of this study guide.

The distribution has a mean of 0 and a variance of 1.standard normal is notation used to indicate that has a standard normal distribution.

The conventional notation for the pdf of the standard normal is , (3.1)and the region of density is . Conventional notation for the cdf of standardnormal is . Notation used for this cdf in the "Derivatives Markets" book is

instead of . The following is a repeat of the excerpt given in Section 3 of theModeling unit for the Exam C standard normal distribution table.

We use the symmetry of the distribution to find for negative values of . For instance, since the two regions have the same area

(probability). Notice also that , since, and this area is deleted from both ends of the curve.

SI-42 SIMULATION SECTION 3 - LOGNORMAL MODEL FOR STOCK PRICES


The general form of the normal distribution has mean and variance . This is a continuousdistribution with a "bell-shaped" density function similar to that of the standard normal, butsymmetric around the mean . The median and the mode are also .

Given any normal random variable , it is possible to find by first"standardizing" , and then

. (3.2)

As an example, suppose that has a normal distribution with mean 1 and variance 4. Then (interpolation in the first

table above). The 95-th percentile of is , where .

Sums of Normal Random VariablesOne measure of the relationship between random variables and is the covariance betweenthem. Covariance between and is denoted or , and is defined to be equal to

. It is unlikely that an Exam C question would directly ask for thecalculation of a covariance if the joint distribution of and is given (that is more like an ExamP type of question).

A related measure of the link between random variables and is the coefficient of correlationbetween them. It is denoted (or just ) and is defined to be , (3.3)where is the standard deviation of and is the standard deviation of .

For any two random variables and , the mean and variance of the sum are and . (3.4)

If and are any two numbers, then and (3.5)

. (3.6)

These relationships involving sums of random variables can be extended to any number ofrandom variables. Suppose that are any random variables with means andvariances and for . Suppose that the covariancebetween and is denoted , and we will use the notation ( .

Now suppose that are numbers, and .

Then and . (3.7)

As a special case, if and , and and , then

(3.8)

. (3.9)Note that we have used the fact that .



These were general comments about the sums of any random variables. If the 's are all normalrandom variables, then the sum of the 's will also have a normal distribution.

In Section 3 of the Modeling Unit we briefly mentioned the Central Limit Theorem. If are independent and identically distributed random variables with common mean and common

variance , then the Central Limit Theorem of probability states that has a

distribution which is approximately normal with mean and variance . As getslarger, approaches a normal distribution. In practice, a value of 30 or so is regarded as "largeenough" for the normal distribution to be a reasonable approximation to the distribution of .

SI-3.2 Application of the Lognormal Distribution to Stock Prices

Review of the lognormal distributionIf has a normal distribution, , and if , then we say that has a lognormaldistribution with parameters and . The lognormal distribution is included in the Exam C tableof distributions. Note that .

The mean of is , the moment generating function of evaluated at1 for the standard normal random variable . This is . Alternatively,from the Exam C table, we see that the -th moment of is . (3.10)Then , and . (3.11)

Suppose that are lognormal random variables.Then .

If we multiply the 's, we get . (3.12)

Since the 's are normal, so is , and therefore, has a lognormal distribution.

Lognormal stock pricesSuppose that we consider an interval of time from time 0 to time , and we break up that timeinterval into subintervals of equal length , .Suppose that we consider and investment whose value at time is , so the values ofthe investment are at the intermediate time points.The proportional growth factors for the investment from one time point to the next are

, and the natural logs of those growth factors are

.

Now suppose that each of those natural log of growth factors are random variables all with thesame mean and variance, say has mean and variance , so does , etc.

From the central limit theorem, the sum has a distribution that is approximately

normal with a mean of . If the growth factors in successive intervals of time are independentof one another, then the variance of the sum is .



We also note that .

(3.13)

Therefore, has a distribution that is approximately normal with mean and

variance (because ), so the mean and variance of are proportional to the

time length for the investment. If has a distribution that is normal, then has alognormal distribution. This is the motivation behind using the lognormal distribution to modelstock prices. Another minor point is that is the continuously compounded growth fromtime 0 to time , which is like the force of interest that came up in Exam FM.

We assume that a stock pays dividends continuously at annual rate , and that the average annualcontinuously compounded growth before dividend payment is . Therefore the average annualcontinuously compounded growth rate measured after the dividend is paid is . We alsoassume that and the annual variance of the continuously compounded growth is .

The lognormal model for the stock price at time , based on stock price at time 0 is that has a normal distribution with mean and variance , where is

measured in years. The factor is often referred to as the volatility of the stock.We can write this as , (3.14)where has a standard normal distribution. An alternative equivalent way to formulate the stockprice at time is . (3.15)

We can write Equation 3.15 as .Note that is not random, but is random, and , because

is standard normal. Then . (3.16)

Example SI-12: A stock based on the lognormal model has a current price of $100. Theexpected price of the stock in one year is $110. The stock pays no dividends and the volatility ofis 20% per year. Find the median price of the stock in one year,, and find the expected stock pricein two years.Solution: The mean of the annual continuously compounded return is . We aregiven that and . Then, , so that .The expected stock price in 2 years is .The median stock price in one year is , where . We can write this probability as

. Since is standard normal, it follows that, from which we get

.

A one-standard deviation move up for the stock price in one year ( ) would be based on, so the stock price would be .



Probabilities of stock prices based on the lognormal distributionWe continue to use the lognormal distribution for stock prices, so that

has a normal distribution with mean and variance, where is measured in years, and has a normal distribution with mean

and variance , so that has a lognormal distribution.

From the Exam C table, we know that if has a lognormal distribution with parameters and ,then . Applying this to the lognormal random variable, we get

.

We can use the chain rule to identify the behavior of as one of the variables , , , of changes. For instance,

. (pdf of standard normal) , and

.

Example SI-12 (continued): A stock based on the lognormal model has a current price of $100.The expected price of the stock in one year is $110. The stock pays no dividends and thevolatility of is 20% per year.(a) Find the probability that the stock price is less than or equal to 80 at the end of 3 years.(b) Find how changes as a function of , and as a function of .(c) Find the 5-th and 95-th percentiles of the stock price at the end of 3 years.Solution: (a) In the first part of the example, we found that , so that .

has a lognormal distribution with parameters and .

.

(b) .

We see that for , and for .The probability increases as increases up to 3 and then decreases for (as a function of ).

. .

. for all and .

The probability increases as the volatility increases.



Example SI-12 continued(c) The 5th percentile will be , where .Therefore, , so that .The 95-th percentile will be , where . Therefore,

, so that .Note that , where , and can be formulated in the sameway. Note also that , so that is a 90%confidence interval for that has 5% probability outside each end of the interval.

Conditional Expectations for Lognormal Stock PricesWe continue to use the lognormal model for stock prices, .The following conditional expectations can be derived from the normal distribution.

,

and .

Example SI-12 (continued): A stock based on the lognormal model has a current price of $100.The expected price of the stock in one year is $110. The stock pays no dividends and thevolatility of is 20% per year. Find the following conditional expectations:

.Solution: , , , and .

.

.



Black-Scholes Option Price FormulaThe Black-Scholes option pricing formula can be expressed using conditional stock priceexpectations and the risk free rate of return.

Suppose that a non-dividend paying stock price follows the lognormal model with annualvolatility . Suppose that the current stock price is and the risk free, continuouslycompounded rate of interest is .

The price of a European call option expiring at time with strike price is ,

where and .

The price of a European put option expiring at time with strike price is .

Example SI-12 (continued): A stock based on the lognormal model has a current price of $100.The expected price of the stock in one year is $110. The stock pays no dividends and thevolatility of is 20% per year. The continuously compounded risk free rate of return is 5%.Find the price of European call and put options that expire in 3 years with a strike price of 120.Solution: We are given .

Then, and .

The price of the call option is ,and the price of the put option is .

SI-3.3 Normal Probability Plots

Given a random sample of data, a way of assessing whether or not the data might be from anormal distribution is to construct a normal probability plot for the data. Normal probability plotsmake use of distribution quantiles. If , the -th quantile of the standard normaldistribution is just the -th percentile. For instance, the 80-th quantile of the standard normaldistribution is the point satisfying . From the normal table, we see that (using interpolation). The quantiles of a random sample are found by first ordering the samplevalues from smallest to largest. If there are observations, then we assign the smallest samplevalue to be the -the quantile, the 2nd smallest sample value is the -th quantile, etc.The normal probability plot for the data set plots the data quantile on the horizontal axis and thecorresponding normal quantile on the vertical axis. The following example illustrates the idea. Ifthe plot is on a straight line, this indicates that the data is from a normal distribution.



Example SI-13: The following random sample of size 10 is given from smallest to largest: Construct the normal probability for the data.Solution: Since there are 10 data points, the smallest value, 1.0 is the % quantile,1.1 is the 15% quantile, . . . , 2.9 is the 95% quantile.The corresponding quantiles from the normal distribution are5% quantile is , 15% quantile is , 25% quantile is ,35% quantile is , 45% quantile is , 55% quantile is ,65% quantile is , 75% quantile is , 85% quantile is ,and 95% quantile is .

The outlying value on the left strays from a straight line, so it does not appear that the data arefrom a normal distribution.

If the first data point in the sample in Example SI-13 was .5, the points would be in a straighterline which would be more indicative that data is from a normal distribution.




Questions 1 to 9 are based on the following information. A stock based on the lognormal modelhas a current price of $100. The expected price of the stock in one year is $110. The stock paysno dividends and the volatility is 40% per year.

1. Find the expected stock price in two years.

2. Find the median price of the stock in one year.

3. Find the 5-th and 95-th percentiles of .

4. Find , .

5. Find .

6. Find .

7. Assuming a continuously compounded risk free rate of interest of 5%, find the Black-Scholesprice of a European call option with a strike price of 125 expiring at the end of 2 years.

8. Assuming a continuously compounded risk free rate of interest of 5%, find the Black-Scholesprice of a European put option with a strike price of 80 expiring at the end of 2 years.

9. Suppose that the stock starts paying dividends continuously at the rate of 5% at the start of thesecond year, but the other parameters and remain unchanged. Find the median and expectedstock price at the end of two years.

10. Construct a normal probability plot of the following data set of 8 points:3 , 4 , 8 , 10 , 12 , 18 , 22 , 35




1. The mean of the annual continuously compounded return is . We are given that and . Then, , so that .

The expected stock price in 2 years is .

2. The median stock price in one year is , where . We can write this probabilityas

. Since is standard normal, it follows that, from which we get

.

3. The 5th percentile of is , where

.

Therefore, , so that .

The 95th percentile of is , where

.

Therefore, , so that .

4.

.

.

5.

.



.

.

.

6. ,

because and are independent. , and

,

so .

.

,

so .

7. ,

.Using the Black-Scholes formula, the price of a European call option expiring at time withstrike price is

.

8. ,

.Using the Black-Scholes formula, the price of a European put option expiring at time withstrike price is

.



9. From the start of the second year, with , and and , the model for thestock price at time is

.The expected price at the end of 2 years would be .The median price at the end of 2 years would be , where

Then, is the median stock price at the end of

2 years.

10. With data points, we find the % quantile of the standard normal,the 18.75% quantile, and the 31.25% , 43.75% , 56.25% , 68.75% , 81.25% and 93.75%quantiles. These are , , , , , , The normal probability plot is

SIMULATION SECTION 4 - MONTE CARLO SIMULATION SI-53


SIMULATION - SECTION 4MONTE CARLO SIMULATION

The material in this section relates to Chapter 19 of "Derivatives Markets". The suggested timeframe for this section is 2-3 hours.

SI-4.1 Risk Neutral Valuation of an Option in a Binomial Tree

Section 19.1 of "Derivatives Markets" refers to risk neutral valuation of an option using abinomial tree. There is a fair amount of background to this topic that is covered in the book and ispart of the Exam M material. We will briefly review the basic definitions of stock options, thebinomial stock price model and how it can be used to price an option.

OptionsA call option on an asset allows the holder of the option to purchase the asset on (European) orbefore (American) a specified date (expiry date) at a specified price (strike price). A put optionallows the holder to sell the asset (with the same terminology as the call option). The asset can bea stock, foreign currency, a market index, a futures contract, etc.

The option price is related to the current stock price, the strike price, the time to expiration, thevolatility of stock price, the risk-free interest rate, and dividends payable.

To develop a theoretical framework for valuing options, we assume that there are no transactioncosts (no commissions), no taxes, and borrowing and lending can be done at the risk-free rate.

Notation:- current stock price (at time 0) ; - stock price at time ;- strike price ; - time of expiration of option ;

- risk-free rate of interest (continuously compounded) ;- value (at time ) of American call option to buy one share ;- value (at time ) of American put option to sell one share ;- value (at time ) of European call option to buy one share ;- value (at time ) of European put option to sell one share ; (4.1)

We have the following relationships:

is the payoff on a call option at expiry. The payoff on a call is (4.2)unbounded.

is the payoff on a put option at expiry. (4.3)

On October 11, 2006, the price of IBM stock at the close of trading was $84.19. At that time, theprice of a call option on one share of IBM stock, expiring in April, 2007 with a strike price of $80was $7.70, and the price of a put expiring in April, 2007 with the same strike price was $2.75.

SI-54 SIMULATION SECTION 4 - MONTE CARLO SIMULATION


Binomial Branch Model - One PeriodIn this simplistic model, we imagine a financial market that consists of one stock and one "bond"(risk-free investing). The financial market is being considered over a 1 period time frame,starting at time , and ending at time . The analysis is carried out at time referencepoint . It is assumed that the value of the stock is known to be at , and that thevalue that the stock will have at time is a two-point random variable:

stock value at time is . (4.4) probability

probability The probability refers to the "true" probability (perhaps based on historical stock pricebehavior) of the stock price being at the end of one period.

We assume that the stock can be bought or sold in unlimited quantities. The bond is a risk-freeinvestment earning a continuously compounded rate of interest per period, so 1 dollar investedin the bond at time will accumulate to at time . We assume that money can beborrowed or loaned out in unlimited quantities at the risk free rate . There is no uncertainty inthe risk-free rate, there is uncertainty only in the stock value at time In addition, in orderfor this simple market to be rational, we must have , or equivalently

otherwise a risk-free gain greater than the risk free rate of interest would beavailable. Another of our assumptions for this simple financial market is that no arbitrageopportunities exist. This means that it is not possible to get "something for nothing", it is notpossible to get a guaranteed return that is higher than the risk free rate of return.

We consider the following example. Suppose that , and in thebinomial stock price model, and suppose that . The model for the stock price at the end

of one year is probability

probability .

The of the stock price going up is , which is the probability the makesrisk neutral probabilitythe stock price at time 0 equal to the expected present value of the possible year-end values of thestock. is the solution to the equation . Solving for results ina value of .

It is the risk neutral probability that can be used to find the value at time 0 of an option thatexpires at time 1. For example, consider a call option on this stock with a strike price of 110.If the stock rises to 125, the option value at time 1 is 15, and if the stock falls to 80, the optionvalue at time 1 is 0. The value of the option at time 0 is the expected present value of theseoutcomes using the risk neutral probabilities. The option value at time 0 is

. We do not make use of the true probabilities to find thisoption price.

Suppose we consider a put option expiring at time 1 with strike price 110. If the stock price goesto 125 the put option has value 0 at time 1, but if the stock price goes to 80, the value of the putoption is 30 at time 1. The expected present value of the put option using the risk neutralprobabilities is .

This model can be extended to two periods. Suppose that the stock either rises by 25% or falls by20% each period, and the risk free rate of interest continues to be 7% per period.



The following diagram illustrates the two-period binomial stock price model.

Note that the risk neutral probabilities of and continue toapply in the second period of the model. For instance, for the upper branch of the model.

We can continue to price (European) options expiring at time 2 by find expected present valueswith risk neutral probabilities. Suppose we consider the call with strike price 110 that nowexpires at time 2. The value of the option at the three nodes at time 2 are46.25 at 156.25 , 0 at 100 and 0 at 64. The expected present value of the option is

.

The value of a (European) put option expiring at time 2 with strike price 110 will be0 at 156.25 , 10 at 100 , and 46 at 64 . The expected present value at time 0 is

(0) (10) .Notice that in finding the expected present value from node 100 at time 2, there were two pathsback to time 0, so we had to find the combined probabilities of both paths.

This is the same idea as in Equation 19.1 of the "Derivatives Markets" book. .



SI-4.2 Simulating the Normal and Lognormal Distributions

Section 19.2 of "Derivatives Markets" presents two methods for simulating the Standard Normaldistribution. The second method is the inverse transform method that was reviewed in Section 1of this Simulation Unit. An example of it is Example SI-8 in Section 1.

The second method is an application of the central limit theorem. The central limit theorem statesthat if is a sequence of independent random values from a distribution with mean

and variance , then has a distribution which is approximately normal with a mean of

and a variance of . As increases, the sum approaches the normal distribution as a limit.

If are uniform random variables, then has a mean of (4.5)

and a variance of , and it is approximately normal (from the centrallimit theorem). Then is also approximately normal, with a mean of 0 and a variance of 1.This is a way to simulate a standard normal value.

The lognormal model for a stock price at time is . (4.6)We can choose a specific point in time and simulate the stock price at that time by simulating astandard normal and substituting into Equation 4.6 . Alternatively, we can simulate a path ofstock prices using any increment of time we wish. To simulate we use

, which can be done by substituting a simulated standard normal .Then we can use with a new standard normal to get the stockprice at time . We can continue this procedure repeatedly.

Example SI-14: A lognormal model for stock prices has , , , and .Simulate the stock price at time 1 using a standard normal value of .Simulate the stock price at successive times .25 , .5 , .75 and 1 using the standard normal values

.Solution: .

, , ,

.



SI-4.3 Simulating the Price of a European Option or an Asian Option

A European option can be exercised only on the expiry date. With a call option strike price of ,if the stock price at expiry is , the payoff of the option at expiry is max . (4.7)We can use the lognormal stock price model, . At a continuouslycompounded risk free rate , the present value of the payoff is max . (4.8)

The value of the option at time 0 can be simulated by simulating the time price of the stock,and averaging the present values of the simulated option payoffs. The -th time the time stockprice is simulated, the option payoff is max , and the time 0 option value is

max . After simulations, the average of the time 0 simulated option values is

max . (4.9)

A comment made in the subsection "Accuracy of Monte Carlo" in Section 19.4 of "DerivativesMarkets" is a restatement of a principle that we have seen earlier in the Model Estimation Unit ofthis study guide. The comment indicates that the standard deviation of the average of the simulated estimates is of the standard deviation of one simulated value. This is just

saying that for a sample mean of observations. As the number ofsimulations increases, the variance of the simulated value of decreases, and the accuracy of as an estimate of the true option price increases.

An Asian option has payoff at expiry based on an average of stock prices at specified earliertimes. An example would be that for an Asian option expiring at time 3, the payoff at time ismax , where is the stock value at time 1, etc. To simulate the option valueat time 0 we would have to simulate the stock price at times 1, 2 and 3 for each simulated payoff.

As the number of simulated values increases, the simulated estimate of an option or stock pricebecomes a more accurate estimate of the option price, so the simulated value converges to the truevalue. It may be possible to speed up the convergence of Monte Carlo simulation. One method todo this is the control variate method. Suppose there are two related options whose true prices are

and , and suppose that we know the true price but we do not know the true price . If wesimulate the price of and get the simulated estimate , and then if we use the same randomnumbers to estimate and get , then the adjusted estimate may be a more

_

accurate estimate of . An alternative is the adjusted estimate ( , where

.

Another technique to improve the accuracy of a simulation is the method of .stratified samplingUnder this method, we divide up the distribution into portions ( , which are usuallystratasubintervals), where each stratum would be sampled individually. For instance, for every 100simulations, we can ensure that each of the following subintervals of is represented oncein the simulation: . We can do this as follows. Suppose that

are 100 uniform random numbers. must be in ,and must be in , must be in , etc.

The method of complementary random numbers is yet another technique to improve theaccuracy of a simulation. This is referred to as the antithetic variate method in the "Derivatives



Markets" textbook. For each uniform random number , we use 1 to generateanother simulated value.

SI-4.4 Using the Poisson Distribution to Simulate Jumps in Stock Prices

Section 19.7 of "Derivatives Markets" reviews the Poisson distribution and Poisson process.Under a Poisson process with unit time period parameter , the expected number of events perunit time is (an event might be a stock price jump that is greater than expected for lognormalstock price distribution). The random variable is the number of events in a time interval oflength , which is Poisson with mean .

In the discussion in Section 19.8 of "Derivatives Markets" regarding the modeling of jumps instock prices, the argument is made that incorporating jumps into the model for a stock priceshould not change the overall expected return on the stock. The lognormal model for , theprice at time of a stock that does not have jumps, is ,where has a standard normal distribution (this was discussed in Section 3 of these simulationnotes).

The model proposed in Section 19.8 of "Derivatives Markets" for a stock that has jumps in its

stock price is . (4.10)This model is constructed so that this stock price which includes jumps has the same expectedreturn as the stock modeled without jumps. and are the same in both models, and and are the parameters modeling a jump in price as a lognormal random variable.

In order for the two models (with and without jumps) to be consistent, we must have thefollowing relationships satisfied: , and has a Poisson distribution with mean .

and are standard normal random variables.

In order to simulate the stock price at time for a stock that incorporates jumps, we need tosimulate the Poisson distribution to get , and we need to simulate standard normal values

(where was the simulated Poisson value).

SI-4.5 Simulating Correlated Stock Prices

Simulating the prices of two correlated stocksSuppose that and are the prices of two stocks, each based on the lognormal model,with and . (4.11)If the continuously compounded returns of the two stocks (the natural logs of the prices) arecorrelated with correlation coefficient , then it is possible to simulate the pair of prices thefollowing way. Let and be independent standard normal variables, and let , and . (4.12)The simulated values of and using these values of and will be correlated and thecoefficient of correlation of the natural logs of the stock prices will be .



The simulation of correlated stock prices can be extended to more than 2 stocks. Suppose that , and are three stock prices based on the lognormal model.

Suppose also that is the coefficient of correlation between the prices of stocks and .Note that and . We can simulate the three stock prices at time in the

following way. We first calculate entries in the matrix ;

.

Note that the matrix multiplication results in

1 ,

is referred to as the of the matrix of correlation coefficients.Cholesky decomposition

Given independent standard normal random variables , and , we find correlated standardnormal variables and from the equations

and .We use these values of in the lognormal stock price model. Note that and are the sameas and for the two stock simulation on the previous page.

Example SI-15: Three correlated stocks whose prices follow the lognormal model have thefollowing parameters:

. . Given independent standardnormal values find the simulated values of the three stockprices at the end of one year.Solution: The entries in the matrix are

.

The simulated values are , ,and .The simulated stock prices are

and .






Questions 1 to 5 are based on the following information. A stock based on the lognormal modelhas a current price of $100. The expected price of the stock in one year is $110. The stock paysno dividends and the volatility is 40% per year.

1. Use the following uniform numbers to simulate the stock price at time 2 using theinverse transformation method. Find the average simulated price.

2. Assuming a continuously compounded risk free rate of interest of 5%, use the simulatedvalues in Question 1 to estimate the value at time 0 of a European call option with a strike priceof 125 expiring at the end of 2 years.

3. Assuming a continuously compounded risk free rate of interest of 5%, use the simulatedvalues in question 1 to estimate the value at time 0 of a European put option with a strike price of80 expiring at the end of 2 years.

4. A model incorporating jumps is formulated so that it has the same expected stock price of 110in one year. The number of jumps has a Poisson distribution with a mean of jump per year, andthe jump parameters are and . Use the model

to simulate the stock price at time 2. The inverse transform method is applied with the followinguniform random numbers used in order to simulate : .61 , .65 , .89 , .17 , .51 , .38Use as many uniform numbers that are needed.

5. Another stock that follows a lognormal model is correlated with the first stock, with coefficientof correlation of . The first stock is the stock described above, and the second stock hasinitial price , and and . The price of the first stock at the end of 2years was simulated, and the simulated value is . Find the expected value of .




1. The simulated standard normal values are: .The stock price at time 2 is , where ,and , so . Using the five simulated values of , the simulated stockprices are .The average of these 5 prices is .

2. The option values at time 2 based on the simulated stock prices in Question 1 are .

The present values at time 0 using the risk free rate of interest are .

The sample mean of the option values at time 0 is 18.39 .

3. The option values at time 2 based on the simulated stock prices in Question 1 are .

The present values at time 0 using the risk free rate of interest are .

The sample mean of the option values at time 0 is 6.05 .

4. The model for the stock price at time 2 is

.For this model, , and , and .

2 .The uniform value .62 simulates a standard normal of .The cdf of the Poisson distribution with mean is

, so the uniform number .65 simulates jumps, so we need two standard normal 's . The uniform value .89 simulates

, and .17 simulates .954 .The simulated stock price is

.

5. .The simulated value of was .

, where ,where is standard normal, and .Then,

.

SIMULATION SECTION 5- RISK MEASURES SI-63


SIMULATION SECTION 5 - RISK MEASURES STUDY NOTE

The material in this section relates to the "Risk Measures" study note by Mary Hardy.The suggested time frame for this section is 2-3 hours.

A "risk measure" is a variation on a premium calculation for a loss. In the notation of the studynote, the risk measure for loss random variable is . As the name suggests, a risk measure isa way of assessing the risk of a potential loss random variable. The risk measure for loss couldbe or (expected value principle), which is the average size of theloss. Another measure of risk could involve the variability of , which is usually related to thevariance of ; for instance (standard deviation principle) . Other riskmeasures look at "worst-case" scenarios. The Value at Risk (VaR) measure and the ConditionalTail Expectation measures are of this type.

SI-5.1 Value at Risk

For a continuous loss random variable , if , the-VaR (Value at Risk) is the % percentile (or quantile) of the distribution of .

This is denoted in the study note and called % quantile; .For a continuous random variable, the value of would usually be unique. may also bereferred to as the % quantile risk measure, .

The general definition of the quantile for any random variable is . (5.1)

Example SI-15: The discrete random variable has the following distribution.

Describe the various quantiles of .Solution: If , then .If , then , and if , then .It follows that for any , , but .Therefore, for any , .

Now suppose that .Then , but .It follows that the minimum for which is ,and therefore for .

In a similar way, for , for ,and for .

SI-64 SIMULATION SECTION 5 - RISK MEASURES


SI-5.2 Conditional Tail Expectation

The Value at Risk measure gives percentiles of the loss distribution, but does not indicate howlarge the losses could be above a particular percentile. The Conditional Tail Expectation measuredoes consider losses above a percentile.

For a continuous loss random variable , if , the Conditional Tail Expectation(CTE) at confidence level is (5.2)where is the % quantile. This is the expected cost per payment with a franchisedeductible of .

is related to the cost per payment formulations that were considered in the Modeling Unitof the study guide. We review a few of the relationships developed there and look at theconnection to CTE.

We recall that for a continuous loss distribution with an ordinary deductible , the expectedcost per loss (ECL) is

,and the expected cost per payment is

.

For a franchise deductible , the expected cost per loss is .

The expected cost per payment for a franchise deductible is

.

is an expected cost per payment for a franchise deductible of which becomes

since (this is the definition of ).

For a continuous random variable, the CTE at confidence level for loss random variable is equal to the 100 % quantile plus the expected cost per payment with an

ordinary deductible of .



CTE for Some Particular DistributionsIn Section 13 of the Modeling Unit we saw some simplified representations for expected cost perloss with ordinary deductible for the uniform, Pareto and exponential distributions.• Uniform : ,• Pareto , : ( is the Pareto parameter here),• Exponential : .

For these and some other loss distributions we can get convenient representations of the CTE.

Uniform For , .

Pareto (using instead of in the Exam C table parametrization)For , satisfies the equation , so .

Exponential For , satisfies the equation . .

Lognormal , : is the solution of the equation .

From the Exam C Table, .

If , this becomes .

From the definition of is follows that , so

.Then,

, and

.

Normal , is the solution of the equation .

, where (the pdf of the standard normaldistribution). The derivation of the CTE for the normal distribution is in the SOA study note.



Example SI-16: Find (the CTE at confidence level .9) for each of the following lossdistributions.(a) Uniform distribution on .(b) Pareto distribution with and .(c) Exponential distribution with mean 1000.(d) Lognormal distribution with and .(e) Normal distribution with mean 1000 and standard deviation 100 .Solution: (a) .

(b) We find from .

Then .

(c) . .

(d) is the solution of the equation .

.

(e) is the solution of the equation ..

CTE for A Discrete DistributionThe general definition of the CTE at confidence level is defined in the following way.

, . (5.3)The following example illustrates how this definition is applied.

Example SI-17: Find and for the following loss distribution

Solution: In Example SI-15 we saw that for , so that 1 .We also saw that for .Therefore, if , so that ,and for , (the same reasoning shows that

for any in the interval ) .

.



Example SI-17 continued .

Note that for , since and , , which increases from to

as increases from to .

In Example SI-15 we saw that for , so that .Therefore, .

becomes irrelevant since , and . Note that for any in the

interval .

In Example SI-17, we saw that . This is true in general for CTE.We also saw that for , , and .This is also true in general; .if , then

SI-5.3 Distortion Risk Measures

is a non-increasing function with and .lim

If is an increasing (or non-decreasing) function (called the "distortion function) with and then and .lim

Also, , since and .Therefore, has the characteristics of a survival function, non-increasing from 1 to 0 as goes from 0 to . is a "distorted" survival function.

For a loss , we know that .If we use instead of in this integral, we get , (5.4)which is called a .distortion risk measure

Some of the risk measures considered earlier can be formulated as distortion risk measures for aparticular distortion function.

(i) The ( -Var) results from the distortion function-quantile risk measureif and if .

(ii) The results from the distortion function risk measure if and if .



Some other distortion risk measures are:

(iii) The risk measure is for . (5.5)proportional hazard transformApplying this results in .

Applying this transform to the Pareto distribution with parameters and results ina Pareto with parameters and . The risk measure would be (the mean of the

transformed Pareto).

Applying this transform to the exponential distribution with mean results in theexponential distribution with mean , which would be the risk measure .

Applying this transform to the Weibull distribution with parameters and results in theWeibull distribution with parameters and .

(iv) The is defined by the distortion function . (5.6)dual power transformApplying this results in .

Applying this transform to the Inverse Pareto distribution with parameters and results inan Inverse Pareto with parameters and .

Applying this transform to the inverse exponential distribution with parameter results inthe inverse exponential distribution with parameter .

Applying this transform to the Inverse Weibull distribution with parameters and results in the Weibull distribution with parameters and .

If is an integer, then is the survival function of the -th (largest)order statistic in a random sample of observations from . The risk measure isthe mean of the -th order statistic.

(v) is . (5.7)Wang's transformApplying this transform to the lognormal distribution with parameters and results in a lognormal distribution with parameters and .

Example SI-18: The loss random variable has a uniform distribution on . Find therisk measure in each of the following cases.(a) The proportional hazard transform with is applied .(b) The dual power transform is applied .Solution: for the uniform distribution.(a) , and

.



Example SI-18 continued(b) , and

.

Example SI-19: has a lognormal distribution with and . Apply Wang'stransform with to find .Solution: The distribution that results after the application of Wang's transform is lognormalwith , and . The risk measure is the mean of thetransformed loss variable, which is . Note that the mean of the originaldistribution of is .

SI-5.4 Coherence of Risk Measures

The study note identifies four axioms which are needed to be satisfied by a risk measure in orderfor it to be .coherent

(1) Translation Invariance: for any constant .

(2) Positive Homogeneity: for any constant .

(3) Subadditivity: For losses and , .

(4) Monotonicity: If then .

The expected value risk measure is a coherent risk measure.

SI-5.5 Variability Measures of Risk

The is . (5.8)semivariance of the random variable

The , (5.8)threshold semivariance with threshold parameter is and the threshold downside semivariance with threshold parameter is

. (5.9)

Example SI-20: Find the semivariance of an exponential random variable with mean 1.Solution: .Using the substitution , this integral becomes

.



Example SI-21: Find the estimated semi-variance and threshold semi-variance withthreshold 10 for the following sample: 3 , 4 , 8 , 10 , 12 , 18 , 22 , 35 .Solution: .

.

SI-5.6 Estimating Risk Measures

Estimation of Given a random sample of size , and , we put the sample values in increasing order,

. Three possible estimators of are(i) , (ii) , and (iii) the smoothed empirical estimate of the 100 -percentile.

An approximate confidence interval for from a sample of losses.Again, we put the sample values in increasing order, .

For a 90% confidence interval we find .Round up to the next largest integer if necessary.The approximate 90% confidence interval for is .

Example SI-22: A sample of 5000 observations are taken for a loss random variable .Find the approximate 90% confidence interval for (the 95-th percentile of the loss randomvariable) in terms of the interval of ordered sample values.Solution: , and (round to the next integer). The lower limit of the interval is , andthe upper limit is .

Estimating CTE and the Variance of the Estimated CTEWith a sample of size ordered from smallest to largest, , the sample estimate

of is ; we consider the sample observations that make up

the largest % of the sample values and find the sample mean of them.

The estimated variance of the estimate is , where

, and is the sample estimate of the %

percentile.




1. Find and for each of the following loss distributions.(a) Exponential distribution with a mean of 10,000(b) Pareto distribution with mean 10,000 and standard deviation of 12,247.(c) Lognormal with mean 10,000 and standard deviation 5000.(d) Normal with mean 10,000 and standard deviation 5000.

2. Find and for a geometric loss distribution with a mean of 2.

3. You are given the following random sample from the loss random variable : 3 , 4 , 8 , 10 , 12 , 18 , 22 , 35 .Find empirical estimates of the .8-VaR risk measure and .

4. has an exponential distribution with mean . Find the distortion risk measure in each of thefollowing cases.(a) Proportional hazard transform with .(b) Dual power transform with .

5. Find the semivariance of the uniform distribution on the interval .




1.(a) . .

(b) ,

.

and .

.

(c) . .

.

.

.

(d) .

.

2. With a mean of 2, the probability function and cdf of the geometric distribution is

. For ,

.

.

.



3. The empirical distribution function has and so .For , .

. .

4.(a) .

(b) .The distortion risk measure is .

5. The mean if . The semivariance is

.



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