TABLE OF CONTENTS - VOLUME 2

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1 TABLE OF CONTENTS - VOLUME 2 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY PROBLEM SET 1 SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR PROBLEM SET 2 SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR PROBLEM SET 3 SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR PROBLEM SET 4 SECTION 5 - BAYESIAN CREDIBILITY APPLIED TO THE EXAM C TABLE DISTRIBUTIONS PROBLEM SET 5 CR-1 CR-17 CR-27 CR-37 CR-47 CR-59 CR-79 CR-89 CR-99 CR-111 SECTION - BUHLMANN CREDIBILITY CR-127 PROBLEM SET 6 CR-137 SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS PROBLEM SET 7 CR-159 CR-169 SIMULATION SECTION 1 - THE INVERSE TRANSFORMATION METHOD PROBLEM SET 1 SECTION 2 - THE BOOTSTRAP METHOD PROBLEM SET 2 SI-1 SI-9 SI-23 SI-35 SECTION 3 - THE LOGNORMAL DISTRIBUTION AND ASSET PRICES PROBLEM SET 3 SECTION 4 - MONTE CARLO SIMULATION PROBLEM SET 4 SECTION 5 - RISK MEASURES STUDY NOTE PROBLEM SET 5 SI-41 SI-49 SI-53 SI-61 SI-63 SI-71

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4 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY CR-1 CREDIBILITY - SECTION 1, LIMITED FLUCTUATION CREDIBILITY The material in this section relates to Section 163 of "Loss Models" The suggested time frame for this section is 3-4 hours CR-11 Introductory Comments on Credibility Theory The objective of credibility theory as it is covered on Exam C is to estimate the mean of a random variable from a random sample Credibility estimation can be applied to any random variable There are several approaches to credibility estimation that are considered Most of the approaches that we look at assume that is part of a Bayesian framework These approaches are the Bayesian approach to credibility, the Buhlmann approach to credibility, and the non-parametric empirical approach and the semi-parametric approach These will be considered later The first approach to credibility estimation that we consider is "Limited Fluctuation Credibility", also called "classical credibility", and does not involve any Bayesian component The Mahler-Dean study note introduces some notation to refer to the mean and variance of various random variables, and the "Loss Models" book also has some of its own notation For instance, 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 the number of claims occurring in a specific period of time (claims per month, for instance) The M-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 an individual loss The M-D study note uses and to denote the mean and variance of a severity distribution In the M-D study note, the aggregate amount of claims (or losses) in a period of time is referred to as the "pure premium" with mean and variance denoted and The aggregate amount of claim in a period might be a compound distribution (a combination of a claim number random variable, and a severity per claim random variable ; compound distributions 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 the variance will be denoted or or An observed value of may be referred to as an exposure (or exposure unit) of If is the number of claims on an insurance policy occurring in one month (the frequency per month), then an observation of is a non-negative integer of the number of claims for a particular 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 of claims, and would be the numbers of claims in month 1 would be the numbers of claims in 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 for policy 1, policy 2,, policy ( policies, one month) If is the amount of one loss on an insurance policy, then " exposures" of is observed losses These might be separate losses for a particular policy, or these might be one loss amount for each of separate policies of the same type

5 CR-2 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY If is the aggregate loss occurring in one month, then " exposures" can refer to the aggregate loss amounts that occur over the months (one aggregate loss amount each month, ) " exposures" can also refer to aggregate losses from separate policies during one month with an aggregate loss for each policy In general, an exposure for a particular random variable is one observation of that random variable CR-12 Limited Fluctuation Credibility Theory - The Standard for Full Credibility Suppose that the random variable being analyzed is, with mean (usually is unknown) and variance, and suppose that a sample of independent observations is available Suppose that there is a manual premium of amount for the claim distribution The manual premium is an amount which has been determined by some past experience and underwriting expertise is not necessarily equal to, but it is some preconceived estimate since we usually don't know the value of Two Equivalent Full Credibility Standards Under 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 full credibility is applied, or (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 pure premium "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 when this closeness criterion is satisfied What we can do is try to get a probability that the closeness requirement is satisfied The meaning of "close" is based on two quantities range parameter : usually, but other values of are possible, such as 02 or 1 probability level : usually, but other values of are possible, such as 95 We say that the full credibility standard is satisfied if the probability relation is satisfied (11) This means that the probability is at least that the absolute deviation of from is less than the fraction of the mean We saw earlier in the estimation review that the sample mean of a random sample satisfies the following relationships: and (12)

6 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY CR-3 As gets larger, gets smaller, and becomes more likely to be "close to" It is not surprising, therefore, that the application of full credibility is related to a "large enough", the number of sample values (or exposures) of If we have a large enough number of sample values of, we can be confident that the sample mean is "close enough" to so that 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 sample observations 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 (13) A way of describing this in a little less technical way is: given probability, we find the value such that the probability, where has a standard normal distribution For instance, with, we know that, so that (1645 is the 95-th percentile of the standard normal distribution so that 5% probability is in the tail to the right of 1645 and 5% probability is in the tail to the left of 1645) If we choose, then, which is the percentile of the standard normal distribution, and if we choose then (ii) (14) If is assumed to be (approximately) normal, and and (the two most commonly used values for a full credibility standard), then (15) 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 the probability 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

7 CR-4 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY 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 (16) This condition is equivalent to the following condition (2) the sum of all observed -values (17) 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 approximate value for , the typical value mentioned above which corresponds to and ) The two inequalities ((1) and (2)) in the equivalent standards for full credibility are interpreted 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 full credibility, 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 sum of the observed values is at least 542, the sample will be sufficiently large for full credibility and again, the value of for that 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 minimum number of observations of needed for full credibility (271 in this case) From the second inequality we get the minimum sum of the observed values of needed (542) Since was given to be 2 in this simple example, we can see that 271 observations should, on average, sum to about (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 the minimum needed sum of observed values Conditions (1) and (2) are statistically equivalent, and the one that would be used might be implied by the form in which data is available Full credibility standards (1) and (2) can be applied to any random variable

8 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY CR-5 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 needed to 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 hand side represents the sum of all observed values that must be reached in order to attain full credibility Example CR-1: Suppose that represents the size of a claim amount, and has as exponential distribution with mean 5 Therefore and (the variance of an exponential random variable is the square of the mean) With and, condition (1) for the standard 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 is reached when the sum of all observed s is at least ; this may occur a little before or after 1083 claims are observed, but should occur when about 1083 claims have occurred For instance, the first claim might be, the second claim might be for a total amount of claim of 11, etc We keep a cumulative total continuing until that cumulative total reaches 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 has an exponential distribution, because, so condition (1) becomes To apply condition (2) we do need to know CR-13 Full credibility applied to a frequency distribution Suppose that the random variable to which we are applying the full credibility method is number 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 total number 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 observed claims) needed for full credibility Condition (2) gives the minimum sum of observations needed; the sum of all observations in this case is the total number of claims observed in all periods observed (ie number claims in period 1 + number claims in period 2 + ) The difference between conditions (1) and (2) is that with condition (1) we count the number of periods of observation, in condition (2) we count the total number of observed claims Suppose we use the simple numerical example above in which and, and suppose we interpret as the number of claims in one week Then full credibility is met with

9 CR-6 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY 271 observed values of (271 weeks, with one for each week) Full credibility is also met as soon as a total of 542 claims have been observed (which could occur sooner or later than 271 weeks) Example CR-2: Suppose that represents the number of claims that occur in one day, and suppose that the mean and variance of are Each day represents one exposure 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 full credibility standard (1), we see that the number of observations of needed (the number of days needed) for full credibility is If we apply credibility standard (2), we see that the sum of -values that we need for full credibility is As soon as the total number of claims observed reaches 2,165, we have satisfied the standard for full credibility The should occur in about 433 days (5 claims per day, on average) but we might reach a total of 2,165 claims some days before or after 433 days Example CR-3: Suppose that represents the number of claims that occur in one day, and suppose that has a Poisson distribution with mean The variance of the Poisson distribution is 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 of days) is at least We would need to know in order 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 soon as 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 the Poisson distribution Example CR-4: Suppose that a single die is being tossed in groups of tosses is the number of 1's tossed in the -th group of tosses, and the probability of tossing a 1 on any given toss is (assumed to be unknown) Then has a binomial distribution with mean and variance Each is an exposure, or single observation of and is a number from 0 to 10 (the number of 1's in 10 tosses) The random variable 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 found from (each group of 10 tosses is an exposure and results in an between 0 and 10) Using the usual values of, the standard for full credibility using condition (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

10 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY CR-7 The equivalent condition (2) tells us that full credibility is reached as soon as the total number of 1's observed is Since is unknown, the right-hand side using condition (2) could be as large as The standard 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 10 tosses must be at least 1083 (which insures that the number of 1's observed is ) 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 (each observation 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 tosses 1's in all the CR-14 Review of Compound Distributions Compound distributions were reviewed in some detail in Section 17 of the Modeling unit of this study 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 continuous or discrete) The compound distribution random variable is The usual interpretation is that represents either the number of claims that occur in one period or the number of claims experienced by one policyholder, and is the size of a claim It is generally understood that and, are mutually independent is the aggregate of all claims occurring in the period The mean and variance of the compound distribution are and (18) These can be found using conditioning rules of probability If and are any two random variables, then it is always true that and (19) (110) In the case of a compound distribution, we can find the mean and variance of by conditioning over :, and

11 CR-8 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY CR-15 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 credibility standards These two full credibility standards applied to the compound distribution are (1) number of observations of needed where is the number of observations (exposures) of needed, and (111) (2) sum of all observed 's (112) The interpretation applied earlier to applies here as well Condition (1) gives the minimum number of observations of needed (we get one per period, or perhaps we have separate policies 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 third equivalent full credibility standard that can be used (3) total number of observed claims (113) Suppose that our interpretation of the compound distribution is that represents the number of claims in one month (frequency), and is the amount (severity) of one claim Each month there will be some claims, and is the number of claims for month This third equivalent full credibility standard tells us the minimum needed This third standard only applies in the case of a compound distribution Keep in mind that although full credibility can be reached in three equivalent ways, we are still ultimately trying to estimate, and when we have satisfied the full credibility requirement based on any of the three conditions, we will use as our estimate of, so we need to know how many months were observed, and the aggregate claim amount 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 observed claims) needed for full credibility Condition (2) gives the minimum sum, the sum needed of the observed values of ; the sum of 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 of claims needed in all periods; this is

12 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY CR-9 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 that we need at least observations of That is a minimum of 68 months for full credibility Condition (2) tells us that we need at least in total claim amount for full credibility This means we continue observing successive months until, and when we reach that total we have enough observations for full credibility Condition (3) tells us that we need at least claims to occur for full 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 enough observations 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), algebraic simplification of the conditions will result in the value of not being required Those distributions for (Poisson and compound Poisson) are the ones most likely to arise in a question involving full credibility CR-16 Full Credibility Standard Applied to a Poisson Random Variable Suppose that the number of claims (per period) has a Poisson distribution with parameter, 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, then using factors, we have The standard for full credibility based on condition (1) is number of observed values of needed number of periods needed (114) The standard for full credibility based on condition (2) is total number of claims needed (115) In order to apply condition (1) we would need to know the value of Condition (2) can be applied without knowing the value of This tells us that the standard for full credibility for a Poisson random variable simplifies to the following criterion: if the total number of observed claims (for all exposure periods combined) is at least 1083, then full credibility is satisfied With different values of or, would be some value other than Note that we are applying the full credibility method to the random variable (not )

13 CR-10 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY CR-17 Full Credibility Standard for When has a Compound Poisson distribution Suppose that the claim frequency per period has a Poisson distribution with mean, and the claim severity distribution is, with mean and variance and, and suppose that we make the usual assumption of mutual independence of and the 's The aggregate claim per period is, and has a compound Poisson distribution with mean, and with variance (116) 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 (117) 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 (118) Note that this requires that we know the mean and variance of, but we do not need to know that value 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 (118) Note that this requires knowing only the coefficient of variation of the value of (keep in mind that is the square of ) but we do not need to know These equivalent full credibility standards for the compound Poisson distribution can be summarized as follows (for and ) Full credibility is satisfied if any one of the following conditions is satisfied (1) if the total number of exposure periods is at least, or (2) if the total amount of observed claims (for all periods or exposures combined) is at least, or

14 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY CR-11 (3) if the total number of observed claims (for all periods or exposures combined) is at least 2 Since is usually unknown, but and are more likely to be known, condition (1) is usually not possible to check, so (2) or (3) is used Again, might not be if different values of or are used Example CR-6: Assume that claim frequency,, per week has Poisson distribution with mean 20, and claim severity,, has an exponential distribution with a mean of 5 Frequency and severity are assumed to be independent In this example we investigate the various equivalent forms 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) Frequency The number of weeks needed for full credibility for frequency is ; this is condition (1) applied to The number of claims needed for full credibility of is (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 the Poisson (b) Severity The number of observations of needed for full credibility for severity is (condition (1) applied to ) The total amount (not number) of claims needed for full credibility of (condition (2) applied to ) is Therefore, either 1083 observed values of, or a total amount of claims of 5412 will satisfy the full credibility standard for Note that we did not need to know the mean and variance to apply condition (1) This is true for the exponential distribution since, and so we get cancellation if the expression for condition (1) (c) Aggregate Claim The number of weeks needed for full credibility for is (condition (1) applied to ) Since has a compound Poisson distribution, this can also be written as

15 CR-12 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY Example CR-6 continued The 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, will satisfy the full credibility standard for Note that since has a Poisson distribution and has an exponential distribution, in applying condition (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 total number of claims needed for full credibility Note: When claims follows a compound Poisson distribution (claim frequency Poisson, claim severity independent of frequency), the number of exposures (or periods) needed for full credibility for claim frequency random variable is, and the number of exposures (or periods) needed for full credibility for the aggregate claims random variable (or aggregate claim amount) is It is important to be clear about which random variable is being considered for full credibility If a situation involves a compound distribution, then that is 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 the expected with probability 090" 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

16 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY CR-13 Solution: It is implied by the wording of the question that 13,000 claims is the full credibility standard for aggregate claims From the information given we see that has a compound Poisson distribution Three possible equivalent standards for full credibility for compound Poisson (labeled (1), (2) and (3)) were summarized above The one that we are able to use depends on the information available We are told that 13,000 claims are needed for full credibility of This suggests that we must use condition (3), which is based on the total number observed claims The full credibility standard (3) for compound Poisson aggregate claims is given above as claims, where is the claim size distribution We are given that (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, the standard for full credibility for based on expected total amount of claims is 1800 If the severity distribution was changed to be a constant equal to the original, the standard for full credibility 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 variable of aggregate claim per period when This is full credibility standard (2) for the random 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 the expected number of exposures of required for severity to be fully credible is and if the total number 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 fully credible 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 of claims needed is claims

17 CR-14 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY Example CR-10: The distribution of aggregate claims per period is a compound Poisson distribution for which the coefficient of variation of the loss severity distribution is 894 A full credibility 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 of exposures needed in the frequency standard was applied as the number of exposures needed for a full 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 aggregate claims, find Solution: With Poisson parameter, and "closeness", and probability, we have, and The full credibility standard for claim frequency (using condition (1) for claim frequency) is an exposure number of We now want the full credibility standard for total claims to be based on, with and closeness factor to be determined, 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-18 Partial Credibility In the full credibility approach, the premium is chosen to be either or (where is the random variable to which the credibility estimate is being applied) The partial credibility approach sets the premium to be, a weighted average of the sample mean and the manual premium is called the credibility factor and is in the interval and is the credibility premium The general way in which we formulate is information available information needed for full credibility (119) Since there are 2 equivalent full credibility standards for any random variable, and there is a third equivalent standard for a compound distribution, the denominator in refers to the full credibility standard for whichever equivalent standard we are using, and the numerator refers to the partial value of that standard available from sample information For any random variable, the partial credibility factor will be either number of observations (exposures) available (1) number of observations (exposures) needed for full credibility if we are using condition (1), or sum of available observations (2) total sum of observations needed for full credibility if we are using condition (2) Furthermore, if the random variable is, then can also be number of claims observed (3) total number of claims needed for full credibility if we are using condition (3)

18 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY CR-15 Keep in mind that if the full credibility standard is satisfied then we have ( will never be larger than 1) The form in which information is given will likely force us to use a specific one of the formulations for given above Example CR-11: Past experience with a production process which produces 6-sided dice has shown 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 process mentioned) 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 credibility approach 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 credibility premium for playing this game is Example CR-12: For a particular group of insureds, the prior estimate of total losses per exposure period is 10,000,000 (the manual premium) The standard for number of exposures for full credibility of total claim amount has been established at 175 exposures Over the course of 100 exposure periods it is found that the observed average total claim amount per exposure period is 12,500,000 Find the partial credibility premium for the next exposure period Solution: and, where is the aggregate claim random variable for one exposure The partial credibility factor is number of observations (exposures) available number 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 total amount of loss observed or needed for full credibility When applying the limited fluctuation method, it must be clear which random variable the standard is applied to, and which of the 2 (or 3 in the compound distribution case) full credibility standards is being applied

19 CR-16 CREDIBILITY SECTION 1 - LIMITED FLUCTUATION CREDIBILITY As a final comment on this limited fluctuation credibility topic, it should be noted that in the past exams there have been about 1 question per exam on this topic The coverage provided in this past section may be a little disproportionate to the weight the topic has received on those exams

20 CREDIBILITY - PROBLEM SET 1 CR-17 CREDIBILITY - PROBLEM SET 1 Limited 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) Total claim amount per period follows a compound Poisson claims distribution The standard for full credibility for total claims in a period based on number of claims is 1500 claims It is then discovered that an incorrect value of the coefficient of variation for the severity distribution was used to determine the full credibility standard The original coefficient of variation 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) 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 at least? A) 55 B) 56 C) 57 D) 58 E) 59 4 Total claims per period follows a compound Poisson distribution and claim severity has the pdf, 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 the expected cost with a probability of 90% If the same number of exposures for full credibility of total cost is applied to the number of exposures needed for the frequency variable, the actual number of claims per exposure period would be within 100 % of the expected number of claims per exposure period with probability 95% Find A) 054 B) 058 C) 062 D) 066 E) An analysis of credibility premiums is being done for a particular compound Poisson claims distribution, where the criterion is that the total cost of claims is within 5% of the expected cost of claims 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) and revised, the credibility premium is After 20 more exposures (for a total of 100) 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 credibility premium for the 100 exposure case A) 1915 B) 1925 C) 1935 D) 1945 E) 1965

21 CR-18 CREDIBILITY - PROBLEM SET 1 6 Total claims per period has a compound Poisson distribution You have determined that a sample size of 2670 claims is necessary for full credibility for total claims per period if the severity distribution is constant If the severity distribution is lognormal with mean 1000 and variance 1,500,000, find the number of claims needed for full credibility of total claims per period A) 6650 B) 6675 C) 6700 D) 6725 E) 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 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 for full 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 are independent of one another Limited fluctuation credibility is being applied to, with the full credibility standard based on the sample mean 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 The expected aggregate amount of claim needed for full credibility is 10,824 The expected total number of claims needed for full credibility is 5412 Find all of the following quantities: and Show that cannot have a Poisson distribution and cannot have an exponential distribution

22 CREDIBILITY - PROBLEM SET 1 CR is the distribution of the number of claims occuring per week has a Poisson distribution with an unknown mean The standard for full credibility for is based on the sample mean of being within 5% of the true mean of with probability 90% With 400 observed claims in 20 weeks, the credibility premium based on partial credibility is 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 in 35 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 factor Which 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 675 million (vi) The manual premium for the first year was 500 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 the time Determine the limited fluctuation credibility net premium (in millions) for the second year (A) Less than 55 (B) At least 55, but less than 57 (C) At least 57, but less than 59 (D) At least 59, but less than 61 (E) At least An insurer has two separate classes of policies The characteristics of the loss per insured in each 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 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 Class I has twice the number of insureds as Class II The two classes of insureds are combined and regarded as a single class with the appropriate adjusted loss per insured during a one year period Find the full credibility standard for the minimum number of insureds required in the combined portfolio, where the full credibility is to be within 5% of expected loss 90% of the time (A) (B) (C) (D) (E) 13530

23 CR-20 CREDIBILITY - PROBLEM SET 1 14 The partial credibility approach is applied to a data set of 50 claim amounts It is assumed that the claim amount distribution is uniform on the interval The full credibility standard is 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 revised to be uniform on the interval The revised partial credibility factor is found Find the ratio (A) (B) (C) 1 (D) (E) 15 Number of claims per year follows a Poisson distribution A number of claims are recorded over a specified period of time A full credibility standard is set so as to be within 5% of expected claims per year 90% of the time Based on the observed number of claims, the full credibility 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 credibility standard within 10% of expected claims per year % of the time (A) 995 (B) 99 (C) 98 (D) 975 (E) has a compound distribution with frequency and severity and all claim amounts are independent of one another Limited fluctuation credibility is being applied to, with the full credibility standard based on the sample mean 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 The expected aggregate amount of claim needed for full credibility is 10,824 The expected total number of claims needed for full credibility is 5412 Find all of the following quantities: and 17 The aggregate loss in one week,, follows a compound negative binomial distribution, and the severity distribution is exponential Limited fluctuation credibility is being applied to so that 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 and variance of the frequency both increase by 20% Find the full credibility standard for the number of claims needed for the new compound negative binomial distribution (severity is the same exponential distribution as before)

24 CREDIBILITY - PROBLEM SET 1 CR-21 CREDIBILITY - PROBLEM SET 1 SOLUTIONS 1 Since and, are unchanged, the standard for full credibility will be unchanged if is unchanged With we have, and With, is the percentile of the standard normal distribution, so that In order for to remain unchanged, we must have Answer: D 2 For the compound Poisson distribution with Poisson parameter (frequency distribution or number of claims per period) and claim amount distribution (severity distribution or amount per 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 full credibility 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 frequency distribution 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 credibility from the aggregate compound Poisson distribution is applied to the Poisson frequency distribution alone, then we set and the " " for the Poisson frequency credibility standard must change, which is why it has been denoted Then With,, and then in order for this to be the proper for, we must have Answer: B

25 CR-22 CREDIBILITY - PROBLEM SET 1 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 in which 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 aggregate claims based on number of claims is Answer: B 7 If the claim number distribution is Poisson, the full credibility standard for aggregate claim costs based on number of claims is, where is the claim size distribution, and 96 We are given For the compound Poisson aggregate claims distribution,

26 CREDIBILITY - PROBLEM SET 1 CR-23 8 When the claim number distribution is Poisson, the standard for full credibility for aggregate 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 normal distribution 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, and the partial credibility premium is, where is the manual premium

27 CR-24 CREDIBILITY - PROBLEM SET 1 10 continued From 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, and the partial credibility premium is 11 (E) is correct This is the formula at the bottom of page 514 in the Mahler-Dean Credibility study note Answer: E 12 When considering the compound Poisson aggregate claims distribution with claim size distribution, 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 standard for 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 We have only one exposure for aggregate claims, that being the first year, so million The manual premium is given to be million The 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

28 CREDIBILITY - PROBLEM SET 1 CR 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 time has (5%) and (95-th percentile of the standard normal distribution) For the annual claim number distribution being Poisson, the number of claims (not exposures) needed for full credibility is actual number of observations The partial credibility factor (when less than 1) is number needed for full credibility We are told that, from which we get actual number of observations number needed for full credibility, so that actual number of observations 5005 is the full credibility standard with and, where, so that From reference to the standard normal table, 2 is the 9874-th percentile Therefore, the probability that the 500 observed claims will satisfy the full credibility standard within 10% of expected number of claims per year is Answer: D 16 Then,, and then Since, we have

29 CR-26 CREDIBILITY - PROBLEM SET 1 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

30 CREDIBILITY SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR CR-27 CREDIBILITY - SECTION 2, BAYESIAN ESTIMATION, DISCRETE PRIOR The material in this section relates to Section 124 of "Loss Models" The suggested time frame for this section is 3 hours Bayesian analysis involves making an initial, or prior estimate of a quantity, and then updating that estimate after some sample information becomes available This results in the posterior estimate of that quantity The quantity being estimated can be a probability, or a probability distribution, 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 since it is simple and captures the main concepts which are used throughout Bayesian analysis Bayesian analysis is related to conditional probabilities and conditional distributions, and we review 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 has occurred 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 (21) (22) This last expression is a simple version of Bayes Rule Suppose that we are given and With this information, the usual objective is to calculate or We are trying to "reverse the conditioning" This is a typical situation to which we apply Bayesian analysis The probability formulations above that allow us to reverse the conditioning can be summarized in the following table The numbered items indicate the order in which calculations can be made

31 CR-28 CREDIBILITY SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR (known) and (known) (known) and (known) or or An alternative diagram to indicate the order of calculations is as follows Again, it is assumed that 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 (23) This is illustrated in the Venn diagram below We can then reverse the conditioning using Bayes rule: (24)

32 CREDIBILITY SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR CR-29 Example CR-13: Two similar bowls each contain 10 similarly shaped numbered balls Bowl 1 contains 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 be chosen) 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 that bowl 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 is chosen at random In general, in this sort of situation, without any additional information about how the bowl is chosen, the phrase "a bowl is chosen at random" is interpreted to mean that each bowl has the same chance of being chosen It then follows that (given that bowl 1 was chosen, there is a probability that the ball chosen is a "1"), and (although these probabilities are not explicitly stated, they are implied by the nature of the situation) Then using the probability rules reviewed above, 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 much technical machinery There are a total of 8 "1"s in the two bowls combined Therefore, knowing that the ball chosen was a "1", it must be one of those 8 Since 5 of the 8 "1"s are in bowl 1 it appears 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

33 CR-30 CREDIBILITY SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR Example CR-13 continued We must be careful with applying "general reasoning" in this way It is important that there was a 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 be chosen (prior probability for each bowl) If we are given that a "1" was chosen, then the (posterior) probability that the ball came from bowl 1 is from bowl 2 is, and from bowl 3 is A key assumption 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 of bowls 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" to the 3 balls numbered "1" in bowl 2 This would essentially result in the same technical calculations we just went through earlier For instance, suppose that we were told that, initially we are twice as likely to choose bowl 1 as we are to choose bowl 2 so that and (we are still assuming that bowl 1 has 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 bowl 2 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 for event probabilities Bayesian analysis for probabilities begins with the identification of the prior event probabilities and the model event probabilities A model event probability is a conditional probability given that a particular prior event has occurred One objective of Bayesian analysis for event probabilities is to "reverse the conditioning" to find the posterior probability of a prior event occurring given that a particular model event has occurred In Example CR-13, the prior events are the events of choosing Bowl 1 or Bowl 2, and the model events 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 The sequence of calculations for Example CR-13 can be summarized in this Bayesian framework as follows

34 CREDIBILITY SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR CR-31 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, given probabilities) Choose number 2 from Bowl 1 Choose number 2 from Bowl 2 7, given, given Joint Events (Intersection probabilities) These are probabilities of the type Bowl 1 was chosen and a ball numbered "1" was chosen 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 type Bowl 1 was chosenthe ball chosen was numbered "1" This is an example of a discrete prior distribution The prior distribution can be thought of as a two-point random variable, where indicates that bowl 1 is chosen and indicates that bowl 2 is chosen The posterior probabilities are the "reversed conditioning" probabilities The following two examples also illustrate Bayesian analysis applied to event probabilities The prior distribution is discrete, which is usually indicated by a finite number of "categories" for the prior outcome There are two "categories" in Example CR-13, those being bowl 1 and bowl 2

35 CR-32 CREDIBILITY SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR Example CR-14: Suppose that an insured population consists of 1500 youthful drivers and 8500 adult drivers Based on experience, suppose that we have derived the following probabilities that an individual driver will have claims in a one year period Youth Adult It 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 1 claimyouth 1 claimyouthyouth youth1 claim 1 claim 1 claimyouth youth 1 claimadult adult The probability calculations can be summarized in the following table: Prior Prob youth (given) and adult (given) Model Prob 1 claimyouth (given) 1 claimadult (given) Joint Prob 1 claim youth 1 claim adult 1 claimyouthyouth 1 claimadultadult Marginal Prob 1 claim 1 claim youth 1 claim adult Posterior Prob youth1 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 view of mixtures of distributions For instance, in Example CR-14 the number of claims for a randomly selected 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 of claims that an Adult will have The number of claims experienced by a randomly selected driver will be the mixture of with mixing weight and with mixing weight We can think of as the conditional distribution of given that the driver chosen is a Youth, and as the conditional distribution of given that the driver chosen is a Adult, and Youth and Adult The probability function of can the be written in as YouthYouthAdultAdult

36 CREDIBILITY SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR CR-33 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 the components Example CR-15: A portfolio of risks is divided into three classes The characteristics of the annual claim distributions for the three risk classes is as follows: Class I Class II Class III Annual Claim Poisson Poisson Poisson Number Distribution mean 1 mean 2 mean 5 50% of the risks are in Class I, 30% are in Class II, and 20% are in Class III A risk is chosen at random from the portfolio and is observed to have 2 claims in the year Find the 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 as the conditional Poisson random variable given : We are asked to find the posterior probability Class Since Class corresponds to, this probability can written as We will use the usual Bayesian formulation of 2 conditional probability: 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

37 CR-34 CREDIBILITY SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR Example CR-15 continued From this structure, we see that Note that the numerator in this posterior probability is one of the terms that are being added up in the denominator This is a general feature of a posterior probability when the prior distribution is discrete Examples CR-13, CR-14 and CR-15 all have discrete prior distributions as well as discrete model distributions (ball number in CR-13, number of claims in CR-14 and CR-15) We now consider an 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 a density 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 loss amount distributions for the two risk classes is as follows: Class I Class II Loss Amount Exponential Pareto Distribution 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 the probability 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 previous examples 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 density notation for and, instead of the intersection of probability notation),

38 CREDIBILITY SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR CR-35 Example CR-16 continued and Then, (we can cancel the 5) (b) We are asked to find It is generally the case that to find a conditional probability 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 Pareto distribution probability Then The main objective of Bayesian analysis usually goes beyond finding the (marginal) distribution of, with a further objective of finding posterior probabilities (reversing the conditioning as we have done in the previous examples) and also predictive probabilities (which will be considered a little later) As we have seen in the previous examples, the marginal distribution of may be found in one of the steps within the framework of Bayesian analysis It can also be found in the context of a mixture of distributions as was seen in Example CR-14

39 CR-36 CREDIBILITY SECTION 2 - BAYESIAN ESTIMATION, DISCRETE PRIOR

40 CREDIBILITY - PROBLEM SET 2 CR-37 CREDIBILITY - PROBLEM SET 2 Bayesian Analysis - Discrete Prior Questions 1 and 2 relate to the following situation Two bowls each contain 10 similarly shaped balls Bowl 1 contains 5 red and 5 white balls (equally likely to be chosen) Bowl 2 contains 2 red and 8 white balls (equally likely to be chosen) A bowl is chosen at random with each bowl having the chance of being chosen A ball is chosen from that bowl 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 of two types of policies Policies of type 1 each have a Poisson claim number per month with mean 2 per period and policies of type 2 each have a Poisson claim number with mean 4 per period of the policies are of type 1 and are of type 2 A policy is chosen at random from the portfolio and the number of claims generated by that policy in the following is the random variable 3 Find A) 0321 B) 0482 C) 0642 D) 0803 E) Suppose that a policy is chosen at random and the number of claims is observed to be 1 for that 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 as Type A, Type B and Type C Any individual of any type has probability of 5 of having no claim in the coming year and has a probability of 5 of having exactly 1 claim in the coming year Each claim is for amount 1 or 2 when a claim occurs Suppose that the claim distributions given that a claim occurs, for the three types of individuals are 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)

41 CR-38 CREDIBILITY - PROBLEM SET 2 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 Determine the 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 per claim policy limit, 25 of the risks have a policy with a $10,000 per claim policy limit, and 50 of the risks have a policy with a $20,000 per claim policy limit The risks have identical claim count distributions Prior to censoring by policy limits, claim sizes for each risk follow a Pareto distribution with parameters 5,000 and 2 A claims report is available which shows the number of claims in various claim size ranges for each policy after censoring by policy limits, but 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 the claim size range of $9,000 - $11,000 Determine the probability that this policy has a $10,000 policy limit 8 (CAS May 05) An insurer selects risks from a population that consists of three independent groups 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 than two 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 common Negative Binomial distribution A priori, the parameters for this distribution are or Each parameter set is considered equally likely Policy files are sampled at random The first two files samples do not contain any claims The third policy file contains a single claim Based on this information, calculate the probability that A) Less than 03 B) At least 03, but less than 045 C) At least 045, but less than 06 D) At least 06, but less than 075 E) At least 075

42 CREDIBILITY - PROBLEM SET 2 CR A portfolio of insurance policies consists of three types of policies The loss distribution for each type of policy is summarized as follows: Policy Type Type I Type II Type III Loss Distribution Exponential Exponential Exponential With Mean 2 With Mean 4 With Mean 8 Half of the policies are of Type I, one-quarter of the policies are of Type II and one-quarter are Type III A policy is chosen at random, and the 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, , (v) The expected number of claims per year is 022 for Class A and 011 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) 055 (B) 057 (C) 067 (D) 071 (E) 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 a loss of 2 Find the probability that the two insured are of the same Type It is assumed that the losses of the two insureds are independent of one another

43 CR-40 CREDIBILITY - PROBLEM SET 2 CREDIBILITY - PROBLEM SET 2 SOLUTIONS 1 redredbowl 1bowl 1redbowl bowl 2 Answer: C bowl 1red redbowl 1bowl 1 2 bowl 1red red 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 1 Type 1 Type 1 Type Type Answer: B A A 5 Type A Answer: B 6 (a) The unconditional claim model is a mixture of two Paretos The moments will be the mixture 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

44 CREDIBILITY - PROBLEM SET 2 CR-41 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 2 is 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 a limit of 5,000 That is the reasoning behind If a policy has a limit of 10,000, then no payment will be over 10,000 so to say that the payment is between 9,000 and 11,000 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 limit and is the event Then the expression above for is, and, which is

45 CR-42 CREDIBILITY - PROBLEM SET 2 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 three groups combined The conditional distribution of given that the individual is chosen from Group 1 is Poisson with a mean of 1, given that the individual is chosen from Group 2 is Poisson 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 1Group 1 Group 2Group 2 Group 3Group 3 The numerator is Group 1Group 1 Group 2Group 2 Group 3Group 3 For each particular group, the numbers of events in successive years are independent, so that Group 1 Group 1 Group 1, and Group 2 Group 2 Group 2, and Group 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

46 CREDIBILITY - PROBLEM SET 2 CR-43 9 continued Then 2 and Answer: E, 10(a) (b)(i) prob prob (ii) prob and 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, and let 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 Class B In Class A there is a 22 chance of a claim in each year, and in Class B there is a 11 chance of a claim in each year In order for the total claim in two years to be 100,000, one of three possibilities must occur: and, or and, or and Therefore,

47 CR-44 CREDIBILITY - PROBLEM SET 2 11 continued In order for the annual claim to be 100,000 in Class A, there must be a claim, and it must be for 100,000; this has probability In order for the annual claim in Class A to be 0, there must 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 3H,1H,2H 12 Posterior probability of H is 3H1H and 2H 1H,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,2H1H,2H H H1H,2H H H 3H,1H,2H3H,1H,2H H H3H,1H,2H H H Then 3H1H and 2H Answer: C

48 CREDIBILITY - PROBLEM SET 2 CR 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 of insured 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,

49 CR-46 CREDIBILITY - PROBLEM SET 2

50 CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR CR-47 CREDIBILITY - SECTION 3 BAYESIAN CREDIBILITY, DISCRETE PRIOR The material in this section relates to Section 1642 of "Loss Models" The suggested time frame for this section is 3-4 hours The Bayesian approach to credibility has the same initial steps as the Bayesian analysis presented in Section 2, we just take the analysis a few steps further We begin with the basic components of Bayesian analysis, which are the prior (parameter ) distribution and the model distribution (a distribution that is conditional on the value of the prior parameter ) We then find the joint distribution of and (joint probabilities in the case of discrete and ), and the marginal distribution of (unconditional probabilities) We then find the posterior distribution of, which is the conditional of given the outcome that has occurred The Bayesian credibility procedure extends this analysis to the predictive distribution, which is the distribution of a second outcome of given the value of the first outcome of An extension of Example CR-13 illustrates this procedure CR-31 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", and Bowl 2 has 3 balls numbered "1" and 7 balls numbered "2" A bowl is chosen at random (with each bowl having the same chance of being chosen) and a ball is randomly chosen from that bowl Suppose that after the first ball is chosen, it is replaced into the bowl from which it was chosen, and a second ball is chosen from that same bowl Find the probability that the second ball 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 used for finding in Example CR-13: Suppose that it is known that bowl 1 was chosen (event ) Then since the first ball picked is replaced, the number of the second ball picked is independent of that of the first ball, and since it is being picked from the same bowl as the first ball it has the same probability of being a "1" as the first Therefore, By the same reasoning, Then, and Example CR-13 continued

51 CR-48 CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR This is the predictive probability that the second ball chosen is a "1" given that the first ball chosen is a "1" There is an alternative way of finding the predictive probability second ball is a "1" first ball is a "1" This alternative uses a modification of the probability rule mentioned earlier, (31) The modification to the rule is (32) In this example is the event that bowl 1 was initially chosen, and is the event that bowl 2 was 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 that we know which bowl was chosen, the number of the second ball chosen from that bowl is independent of the number of the first ball chosen from that bowl) This will always be the case for the Bayesian credibility situations that arise on Exam C, so we will always be able to use the relationships and (33) Method 2: (34) (this is a more general form of the relationship ) It is not clear from this formulation, but Method 2 requires the same assumption of conditional independence of and given It will almost always be the case that Method 1 is preferable to Method 2 when calculating a predictive probability (although they require the same calculations in different orders) In the Bayesian framework, a predictive probability is a conditional probability of a model-type event occurring on the next trial given information on the model-type event that occurred on the first trial (or trials) In this example, a "trial" is the choice of a ball after a bowl has been picked A subsequent "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 the Bayesian credibility context Keep in mind that even though events and are conditionally independent given event, it is not necessarily true that and are unconditionally independent In fact, they usually will not be unconditionally independent

52 CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR CR-49 CR-32 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 type Model 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 An observation 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 Distribution A 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 the prior 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 will be discrete, and if the model distribution is continuous, the predictive distribution will also be continuous Along with identifying the posterior and predictive distributions, we may be interested in the mean of the posterior distribution, and also the mean of the predictive distribution

53 CR-50 CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR CR-33 Predictive Expectation - The Bayesian Premium We now review a couple of important principles involving the expected value of a random variable 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 (35) (ii) (36) 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) (37) Note that these rules are versions of the "double expectation" rule which states that for any two random variables and, In the case of rule (i), can be thought of as a two-point random variable indicating whether or not event has occurred, if occurs if occurs if occurs, prob Then if occurs, prob So is itself a two-point random variable The expected value of this two-point random variable 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 a driver is chosen at random from the population of 10,000 drivers The expected number of claims for the randomly selected driver will be youthyouthadultadult As another illustration of these concepts we have the following continuation of Example CR-13, the two-bowl example with numbered balls

54 CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR CR-51 Example CR-13 continued: Bowl 1 contains 5 balls with the number 1 and 5 balls with the number 2 Bowl 2 contains 3 balls with the number 1 and 7 balls with the number 2 A bowl is chosen at random, and a ball is chosen at random from that bowl The ball is replaced in the bowl from 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 was numbered 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 that event 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 expectation directly as follows The event was denoted event in Example CR-13 when it was first presented 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 to find the predictive (conditional) expectation Method 1: We use expectation rule (ii) above We condition over the bowl from which the first ball was drawn, given that the first draw was a 1 The expression 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 that the 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 this example was first presented in Section 2 Recall that and Putting these probabilities together, we get

55 CR-52 CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR Example CR-13 continued These 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 a different order Then, In a Bayesian credibility situation, it is almost always more efficient to use Method 1 rather than Method 2 to find a predictive expectation (particularly if the prior distribution is continuous)

56 CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR CR-53 Example CR-13 continued The components of this example can be put into the general context of Bayesian estimation The prior distribution of is the two point random outcome of which bowl is chosen or, with ) The model distribution of is the conditional distribution of the number on the ball drawn given the bowl from which it is drawn: number on ball drawn (what was previously called ) ;, etc The joint distribution of and describes probabilities of the type that were calculated earlier The marginal distribution of is the unconditional distribution of ball number: and The posterior distribution of is the conditional distribution of bowl type given the number of first ball For instance and are posterior probabilities The predictive distribution of is the conditional distribution of the second ball number given the number on the first ball: are predictive probabilities In this example, alternative methods were presented to calculate various factors The method that is most efficient may depend on the data available and the way in which a situation is presented The following example is another illustration of the Bayesian method applied to a discrete prior distribution 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 prior probabilities for risk type are The numerator and denominator of the predictive probability can both be found by conditioning over Risk Classes A and B The denominator is ]

57 CR-54 CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR Example CR-17 continued The numerator is ] As mentioned earlier, we are implicitly assuming that given risk, the numbers of claims in years 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 above may 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 prior distribution is the 2-point distribution consisting of the choice of risk or risk, with and as the prior probabilities The model distribution is, the number of claims per year The conditional distribution of given is Poisson with mean, and the conditional distribution of given is Poisson with mean The joint distribution of and has probabilities of the form This probability can be found from After a risk is chosen at random (not knowing whether it is from group A or B), the number of claims for the year is found to be 0 We find the marginal probability The posterior distribution of given becomes no claim in yr 1, and no claim in yr 1 The predictive distribution is the distribution of number of claims in the 2nd year ( ) given no claims in the first year ( ) We were asked to find the predictive probability ]

58 CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR CR-55 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 class was observed to have 0 claims for the year We wish to find the expected number of claims for the 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 find the conditional probabilities and (actually, since this example has a 2-point prior, we get ) We saw in Example CR-17 that, so that CR-34 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 of risks 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 any number of points Bayesian problems involving a finite point discrete prior distributions have come up regularly on Exam C The following is a summary of the main quantities that we might be 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 probabilities Suppose an individual is chosen at random from the population (type or is unknown) and an observed value of the model distribution is obtained The posterior probability that the individual is of type is (38) 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 (39)

59 CR-56 CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR (iii) Predictive probabilities Suppose an individual is chosen at random from the population (type or is unknown) and an observed event for the model distribution is known to have occurred Suppose that is an event involving the model distribution for the next occurrence of for the same individual Then the predictive probability that event occurs given that has occurred is (310) Event might be in which case, and event might be, so that (311) (iv) Predictive expectation Suppose an individual is chosen at random from the population (type or is unknown) and an observed event for the model distribution is known to have occurred The predictive expectation for the next occurrence of is If event is, then (312) 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 the number 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 at least 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 given that 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 2 and 2B 2

60 CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR CR-57 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 in a 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 same approach that was used to find in (a) This is summarized in the following series of calculations Again, it is assumed that and are conditionally independent given either or Youth, Adult, We can also find from the relationship

61 CR-58 CREDIBILITY SECTION 3 - BAYESIAN CREDIBILITY, DISCRETE PRIOR Example CR-14 continued Since 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 more efficient (c) We have already found in (a), so that (as in (b)) From the description of the model distributions, we have and Then, Note: Exam C questions may ask for the calculation of the Bayesian premium This is the predictive expectation of the next occurrence of the model random variable given whatever information is available about observations of that have already been made The general form of the Bayesian premium given information about the first occurrences of is For a Bayesian situation with a two-point prior distribution based on categories and for the prior, we would calculate this predictive expectation as (313)

62 CREDIBILITY - PROBLEM SET 3 CR-59 CREDIBILITY - PROBLEM SET 3 Bayesian Credibility - Discrete Prior 1 Two bowls each contain 10 similarly shaped balls Bowl 1 contains 5 red and 5 white balls (equally likely to be chosen) Bowl 2 contains 2 red and 8 white balls (equally likely to be chosen) A bowl is chosen at random with each bowl having the chance of being chosen A ball is chosen from that bowl After the ball is chosen it is returned to its bowl and another ball is chosen at random from the same bowl Suppose that first ball chosen is red Find the probability that second ball chosen is red A) B) C) D) E) 2 A portfolio of insurance policies consists of two types of policies Policies of type 1 each have a Poisson claim number per month with mean 2 per period and policies of type 2 each have a Poisson claim number with mean 4 per period of the policies are of type 1 and are of type 2 A policy is chosen at random from the portfolio and the number of claims generated by that policy in the following is the random variable Suppose that a policy is chosen at random and the number of claims is observed to be 1 for that month The same policy is observed the following month and the number of claims is (assumed to be independent of the first month's claims for that policy) Find A) 15 B) 20 C) 25 D) 30 E) 35 Problems 3 and 4 refer to the distribution of, which is a Poisson random variable with parameter, where the prior distribution of is a discrete uniform distribution on the 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) 15 B) 16 C) 17 D) 18 E) 19

63 CR-60 CREDIBILITY - PROBLEM SET 3 5 (Example CR-15) A portfolio of risks is divided into three classes The characteristics of the annual claim distributions for the three risk classes is as follows: Class I Class II Class III Annual Claim Poisson Poisson Poisson Number Distribution mean 1 mean 2 mean 5 50% of the risks are in Class I, 30% are in Class II, and 20% are in Class III (a) A risk is chosen at random from the portfolio and is observed to have 2 claims in the year Find the probability that the risk will have 2 claims next year, and find the expected number of claims for the risk next year (b) A risk is chosen at random from the portfolio and is observed to have 2 claims in the first year and 2 claims in the second year Find the probability that the risk will have 2 claims in the third year, and find the expected number of claims for the risk in the third year Problems 6 and 7 are based on the following situation A risk class is made up of three equally sized groups of individuals Groups are classified as Type A, Type B and Type C Any individual of any type has probability of 5 of having no claim in the coming year and has a probability of 5 of having exactly 1 claim in the coming year Each claim is for amount 1 or 2 when a claim occurs Suppose that the claim distributions given that a claim occurs, for the three types of individuals are 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 6 Find the probability that the insured is Type A A) B) C) D) E) 7 Find the Bayesian premium A) B) C) D) E) 8 You are given the following: - Four shooters are available to shoot at a target some distance away that has the following design: R S T U V W X Y Z - Shooter A hits areas R,S,U,V, each with probability 1/4 - Shooter B hits areas S,T,V,W each with probability 1/4 - Shooter C hits areas U,V,X,Y each with probability 1/4 - Shooter D hits areas V,W,Y,Z each with probability 1/4 Two distinct shooters are randomly selected and each fires one shot Determine the probability that both shots land in the same Area

64 CREDIBILITY - PROBLEM SET 3 CR-61 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 Determine the 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) 25 B) 26 C) 27 D) 28 E) (Example CR-16) A portfolio of risks is divided into two classes The characteristics of the loss amount distributions for the two risk classes is as follows: Class I Class II Loss Amount Exponential Pareto Distribution mean 1000 The portfolio is evenly divided between Class I and Class II risks (a) A risk is chosen at random from the portfolio and is observed to have a loss of 2000 Find the expected value of the next loss from the same risk (b) A risk is chosen at random from the portfolio and is observed to have a first loss of 2000 and a second loss of 1000 Find the probability that the risk was chosen from Class I Find the expected value of the third loss from the same risk 12 A car manufacturer is testing the ability of safety devices to limit damages in car accidents You are given: (i) A test car has either front air bags or side air bags (not both), each type being equally likely (ii) The test car will be driven into either a wall or a lake, with each accident type equally likely (iii) The manufacturer randomly selects 1, 2, 3 or 4 crash test dummies to put into a car with front air bags (iv) The manufacturer randomly selects 2 or 4 crash test dummies to put into a car with side air bags (v) Each crash test dummy in a wall-impact accident suffers damage randomly equal to either 05 or 1, with damage to each dummy being independent of damage to the others (vi) Each crash test dummy in a lake-impact accident suffers damage randomly equal to either 1 or 2, with damage to each dummy being independent of damage to the others One test car is selected at random, and a test accident produces total damage of 1 Determine the expected value of the total damage for the next test accident, given that the kind of safety device (front or side air bags) and accident type (wall or lake) remain the same (A) 244 (B) 246 (C) 252 (D) 263 (E) 309

65 CR-62 CREDIBILITY - PROBLEM SET 3 13 Suppose that the distribution of given is binomial with parameters and ( is non-random), and the prior distribution of is Poisson with parameter Find the posterior distribution 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 Claim Amount for Class 1 Amount for Class , , (ii) Class 1 has twice as many claims as Class 2 A claim of 250 is observed Determine the Bayesian estimate of the expected value of a second claims from the same policyholder (A) Less than 10,200 (B) At least 10,200, but less than 10,400 (C) At least 10,400, but less than 10,600 (D) At least 10,600, but less than 10,800 (E) At least 10, (CAS) The Allerton Insurance Company insures 3 indistinguishable populations The claims frequency of each insured follows a Poisson process Given: Population Expected Probability Claim (class) time between of being in cost claims class I 12 months 1/3 1,000 II 15 months 1/3 1,000 III 18 months 1/3 1,000 Calculate the expected loss in year 2 for an insured that had no claims in year 1 A) Less than 810 B) At least 810, but less than 910 C) At least 910, but less than 1,010 D) At least 1,010, but less than 1,110 E) At least 1, A portfolio of insurance policies consists of two types of policies The annual aggregate loss distribution for each type of policy is is a compound Poisson distribution Policies of Type I have a Poisson parameter of 1 and policies of Type 2 have a Poisson parameter of 2 For both policy types, the claim size (severity) distribution is uniformly distributed on the integers 1, 2 and 3 Half of the policies are of Type I and half areof Type II A policy is chosen at random and an aggregate annual claim of 2 is observed (a) Find the posterior distribution of the Poisson parameter for the two policy types (b) Find the mean of the aggregate claim next year for the same policy (given aggregate claim this year was 2) (c) Find the variance of the aggregate claim next year for the same policy (given aggregate claim this year was 2) in two ways: (i) (ii) (this requires using the posterior distribution of found in part (a))

66 CREDIBILITY - PROBLEM SET 3 CR (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 10 per year (ii) Risks in Class B have a Poisson claim count distribution with a mean of 30 per year (iii) Risks in Class A have an exponential severity distribution with a mean of 10 (iv) Risks in Class B have an exponential severity distribution with a mean of 30 (v) Each class has the same number of risks (vi) Within each class, severities and claim counts are independent A risk is randomly selected and observed to have two claims during one year The observed claim amounts were 10 and 30 Calculate the posterior expected value of the aggregate loss for this risk during the next year (A) Less than 20 (B) At least 20, but less than 40 (C) At least 40, but less than 60 (D) At least 60, but less than 80 (E) At least (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 20 claims per year (iii) The other half of the insureds are expected to have 40 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) 32 (B) 34 (C) 36 (D) 38 (E) (SOA) Prior to observing any claims, you believed that claim sizes followed a Pareto distribution with parameters 10 and or 1, 2 or 3, with each value being equally likely You then observe one claim of 20 for a randomly selected risk Determine the posterior probability that the next claim for this risk will be greater than 30 (A) 006 (B) 011 (C) 015 (D) 019 (E) (SOA) The claim count and claim size distributions for risks of type A are: Number of Claims Probabilities Claim Size Probabilities 0 4/ /3 1 4/ /3 2 1/9 The claim count and claim size distributions for risks of type B are: Number of Claims Probabilities Claim Size Probabilities 0 1/ /3 1 4/ /3 2 4/9 Risks are equally likely to be type A or type B Claim counts and claim sizes are independent within each risk type The variance of the total losses is 296,962 A randomly selected risk is observed to have total annual losses of 500 Determine the Bayesian premium for the next year for this same risk (A) 493 (B) 500 (C) 510 (D) 513 (E) 514

67 CR-64 CREDIBILITY - PROBLEM SET 3 21 Two eight-sided dice, A and B, are used to determine the number of claims for an insured The faces of each die are marked with either 0 or 1, representing the number of claims for that insured for the year Die A 1/4 3/4 B 3/4 1/4 Two spinners, and, are used to determine claim cost Spinner has two areas marked 12 and Spinner has only one area marked 12 Spinner 1/2 1/2 To determine the losses for the year, a die is randomly selected from A and B and rolled If a claim occurs, a spinner is randomly selected from and and spun For subsequent years, the same die and spinner are used to determine losses Losses for the first year are 12 Based upon the results of the first year, you determine that the expected losses for the second year are 10 Calculate (A) 4 (B) 8 (C) 12 (D) 24 (E) The annual aggregate loss A 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 a Poisson parameter of 1 and policies of Type 2 have a Poisson parameter of 2 For both policy types, the claim size (severity) distribution is uniformly distributed on the integers 1, 2 and 3 Half of the policies are of Type I and half are of Type II A policy is chosen at random and an aggregate annual claim of 2 is observed Find the Bayesian premium for the same policy for next year

68 CREDIBILITY - PROBLEM SET 3 CR-65 CREDIBILITY - PROBLEM SET 3 SOLUTIONS 2nd red1st red 1 2nd red1st red 1st red 2nd red1st redbowl 1 bowl 1 2nd red1st redbowl 2 bowl 2 Note also that 2nd red1st red 2nd redbowl 1bowl 11st red2nd redbowl 2bowl 21st red Answer: E 2 Type 1 Type 1 Type Type Alternatively, Type 1Type 1 Type Type 2 Answer: C 3 Answer: D 4 The marginal pf for is and the posterior distribution of has pf, and Answer: C

69 CR-66 CREDIBILITY - PROBLEM SET 3 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 A A 6 Type A Answer: B

70 CREDIBILITY - PROBLEM SET 3 CR-67 7 TypeType Type AA BB CC 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 in common (call this event ), and the last has 1 target Area in common (call this event ) Therefore Then both shots land in same Area both shots land in same Areaboth shots land in same Area If the two shooters have two Areas in common, then there is probability that both hit the same Area, and if the two shooters have one Area in common, then there is probability that both hit the same area The probability in question is then 9 (a) The unconditional claim model is a mixture of two Paretos The moments will be the mix of the corresponding Pareto moments, with weights 75 and 25 (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,,

71 CR-68 CREDIBILITY - PROBLEM SET 3 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 is Class Class, Class Class, 12 We wish to find We condition over the combinations of Front/Side air bags, Wall/Lake accident, so that We immediately get that since with side air bags there will be 2 or 4 dummies and with a lake crash the minimum damage per dummy is 1, so that the minimum overall damage with is 2 Therefore, We find (2 or 4 dummies equally likely, average damage of for each dummy),

72 CREDIBILITY - PROBLEM SET 3 CR continued and We then calculate,, and, where, and since there must be 2 dummies (prob 5) and each sustains damage of 5 (prob 5 each), (either 1 dummy and damage of 1, or two dummies and damage of 5 each), and Then Then and We can summarize these conditional probability calculations as follows Finally, Answer: C

73 CR-70 CREDIBILITY - PROBLEM SET 3 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 2 Given 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

74 CREDIBILITY - PROBLEM SET 3 CR For each Class the claims follow a Poisson process For a particular Class, if the expected amount of time between claims is (in years), then the expected number of claims per year is Class I has an expected amount of time of 1 year between claims, so that the expected number of claims per year for Class 1 is Class II has an expected amount of time of 125 years (15 months) between claims, so that the expected number of claims per year for Class 1 is Class III has an expected amount of time of 15 years (15 months) between claims, 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 denoted by 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 IClass I Class IIClass II Class IIIClass 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 loss for year 2 is Answer: A prob 16 The prior parameter has distribution 2 prob The model distribution has a compound distribution with Poisson frequency with mean, and the stated severity distribution (a) 1 claim for amount 2 2 claims for amount 1 each

75 CR-72 CREDIBILITY - PROBLEM SET 3 16 continued 1 claim for amount 2 2 claims for amount 1 each 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,

76 CREDIBILITY - PROBLEM SET 3 CR Suppose we use the following notation: aggregate claims in second year for the selected risk number of claims in first year for the selected risk amount of first claim in first year for the selected risk amount 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 AClass A selected risk is from Class BClass B If the selected risk is from Class A, then has a compound Poisson distribution with Poisson parameter (frequency) 10 and has a claim amount distribution (severity) with a mean of 10 (we are 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 the conditional probabilities Class A and Class B This requires the use of rules for conditional probability Class A Class A and Class A AA If it is known that the risk is from Class A then we know the distributions of and, so we can calculate A We are told that a risk is selected randomly This means that risk Classes A and B are equally likely to be chosen, so that A Then Class A AA From independence of and the 's, it follows that A ATwo Claim amounts are 1, 3A 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 for the Poisson distribution with mean is, and the density function for the exponential distribution with mean is Then A A Two Claim amounts are 1, 3A The factor of 2 arises from the two combinations of claims of amounts 1 and 3 (first claim is amount 1 and second claims is amount 3, and first claim is amount 3 and second claim is amount 1) It follows that Class A (this is the numerator of the probability Class A that we are trying to find)

77 CR-74 CREDIBILITY - PROBLEM SET 3 17 continued In order to find the denominator we use the fact that the selected risk must be either from Class A or from Class B, so that Class A Class B We find Class B in the same way we found Class A Class B BB 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 find Class A Class A We can use the same reasoning to find Class B, which will turn out to be Alternatively, and more efficiently, once Class A is known, Class B is equal to its complement Class B Class A Finally, the posterior expected value we are looking for is selected risk is from Class AClass A selected risk is from Class BClass 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 A and B are both know from the given distributions Class A and Class B (this calculation uses the compound Poisson distribution)

78 CREDIBILITY - PROBLEM SET 3 CR 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 conditional distribution of (claim count for an individual) given is Poisson with mean ; and we are given 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 an equal chance of being 2 or 4) We are given that for a randomly chosen individual ( 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 for that 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 can write 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 conditional distribution 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 (or density 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 is Poisson with parameter It must be implicitly assumed that for a randomly chosen individual given the value of, the (conditional) distributions of number of claims in separate years are independent of one another also

79 CR-76 CREDIBILITY - PROBLEM SET 3 18 continued This assumption is always made in the Bayesian estimation context, but note that it is not generally true that the unconditional distributions of numbers of claims in separate years are independent What is meant here is that if the value of is known, then the conditional distributions 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 is Poisson given and and Answer: C 19 We wish to find This can be written as The numerator is,

80 CREDIBILITY - PROBLEM SET 3 CR 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 and 3 3, 3 Then, is the numerator, and then Answer: C 20 Since we are given one observation (total annual losses of 500), the Bayesian premium is Since the "parameter" in this Bayesian situation is the class A or B, the prior distribution is (risks A and B are equally likely) For each of the two classes, total annual loss has a compound distribution The Bayesian premium can be formulated as Within class A, Within class B, 1 claim claim of 500, 2 claims claim of 250 Then, In a similar way, Then, Answer: A

81 CR-78 CREDIBILITY - PROBLEM SET 3 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 : 1 claim for amount 2 2 claims for amount 1 each 1 claim for amount 2 2 claims for amount 1 each and The Bayesian premium is

82 CREDIBILITY SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR CR-79 CREDIBILITY - SECTION 4 BAYESIAN CREDIBILITY, CONTINUOUS PRIOR The material in this section relates to Section 1642 of "Loss Models" The suggested time frame for this section is 3 hours The application of Bayesian analysis to credibility estimation can be summarized as follows The random variable (usually loss frequency or severity, or perhaps a compound aggregate claims distribution) has a pf/pdf that depends on a parameter, where is a random variable as well The initially assumed distribution of is referred to as the prior distribution, and has pf/pdf The distribution of based on a value is the model distribution, but it is really a conditional distribution of given, with pf/pdf The prior distribution of can be discrete or continuous, and the model distribution can be discrete or continuous Under the Bayesian analysis approach, an observation (or several observations) are made from the distribution of, and then an updated form of the distribution of is found; this is called the posterior 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 has pf/pdf If several data values of are available then the posterior distribution has pf/pdf It is then possible to extend this analysis to find the predictive distribution of the next occurrence of If one data point of has been observed, then the predictive distribution is a conditional distribution of (the next occurrence of ) given, with pf/pdf A predictive distribution can also be found if several data values of are available; we would find the predictive density This is the conditional density of the next occurrence of ( ) given the observed data points Up to now in these notes on Bayesian credibility, we have considered only the situation in which the prior distribution is discrete (2- or 3-points usually) When the prior distribution of is continuous, the Bayesian analysis leading to the posterior and predictive distributions can become quite complicated There are a few specific combinations of prior and model distributions which lead to recognizable forms for the posterior (and perhaps the predictive) distribution We first summarize the general Bayesian analysis procedure and then consider a number of examples in which the prior parameter has a continuous distribution

83 CR-80 CREDIBILITY SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR CR-41 The Bayesian Structure General Bayesian estimation has the following components The initial assumption for the distribution of the parameter is called the prior 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 each one has his own value, and these values are distributed over the population according to the prior density of The model distribution (usually frequency or severity of loss or aggregate loss) is a conditional distribution given, with pf/pdf (or more simply ) For a data set of random observed values from the distribution of for a particular member of the population, say, and a specific possible value of, the model distribution is (41) We are implicitly assuming that are conditionally independent given In Exam C questions is often 1 (or a small integer) The joint distribution of and has pf/pdf ( ) In the multiple data-point situation, and have joint pdf/pf (42) (note that only one factor of ) appears in this joint density) The marginal distribution of is In the multiple data-point situation, have joint pdf (43) The limits of the integration are the original limits of the prior distribution of The posterior distribution of given has pf/pdf (for continuous ), (44) where the integration is over the range of the prior distribution of Also, as we have seen earlier for discrete all In the case of multiple data points, the posterior pdf/pf is (for continuous ), (45) integrated over the range of 's distribution; and (for discrete ) all Given the data, the predictive distribution of (a new observation) has pdf/pf (46) (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

84 CREDIBILITY SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR CR-81 In a Bayesian credibility situation we may be given a single data value for, or we may be given several data values, or we may be given the sum of several data values Typically we are asked to find the density function or mean of the posterior distribution of, or we may be asked to calculate a probability related to the posterior distribution We also may be asked to calculate a probability or expectation for the predictive distribution It is less likely that we will need to find the marginal density of or that we will be asked to find the density of the predictive distribution A useful point to note regarding Bayesian analysis is that the posterior density of is generally proportional 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 prior and model distribution pairs combine in a convenient way to give a joint density that has a fairly recognizable posterior density, and some do not The following example illustrates a prior-model pair that results in a posterior distribution which is in the Table of Distributions The example also illustrates a few different ways in which information can be presented Another point note is the similarity between the formulation of the density of marginal distribution of and the density of the predictive distribution of given The density of the marginal distribution of is, and the density of the predictive distribution is Both integrals involve the model density, in the marginal case, and in the predictive case The difference is that in the marginal case the integral uses, the prior density of, and the predictive case the integral uses, the posterior density of CR-42 Examples To Illustrate A Continuous Prior Example CR-18: The distribution of (the model distribution) is assumed to be exponential with parameter, so that (we use instead of so as not to get confused with the prior distribution that we will use in this example) The prior distribution of is assumed to be an inverse gamma distribution with parameters and, and has pdf for The joint distribution of and is (a) Suppose that we are given a single data point with value The joint density becomes The marginal density of will be, and for the specific value of this becomes, which is some numerical value

85 CR-82 CREDIBILITY SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR Example CR-18 continued We don't actually need to find the numerical value of the integral in order to identify the posterior distribution In the next section (Section 5) of these notes there will be a few comments about the types of integrals than can arise in the Bayesian credibility context, and comments made there will show how to find the integral above in a fairly efficient way The posterior density of will be, which for becomes 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 the inverse gamma distribution with parameters and has pdf (we use instead of as the variable), 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 that the posterior distribution must have an inverse gamma distribution with parameters and (since ) This is true for the following reason We know that the posterior density 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 of the inverse gamma with and In other words, once we recognize that the algebraic form of the variable in the posterior density is the same as the inverse gamma with and, we can conclude that the posterior distribution is that inverse gamma The posterior density will be +1 Therefore we have found the posterior density without ( ) directly 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 inverse gamma random variable is The mean of this posterior distribution is

86 CREDIBILITY SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR CR-83 Example CR-18 continued (b) Suppose that we are given several data points, say a random sample of 8 values from the distribution of all based on the same unknown value of Suppose that the random sample is The joint distribution now involves and We will continue to assume that the model distribution of is exponential with parameter, so that and that the prior distribution of is an inverse gamma distribution with parameters and, and has 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 be proportional to, which implies that the posterior distribution must have an inverse gamma 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 was 225 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 be familiar with the distribution types in the Exam C Tables Second, an inverse gamma prior distribution combines with an exponential model distribution to result in an inverse gamma posterior distribution In the next section several prior-model pair combinations are listed that result in a posterior distribution that is of the same type as the prior distribution In such a case the 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 of observations is given rather than the individual values For some prior-model distribution pairs this is all that is needed since in the expression for the joint distribution, only the sum of the 's is needed, as was the case in part (c) of Example CR-18 A fourth point to note is that it may not be necessary to find the marginal distribution of, although the marginal distribution of may also have a recognizable form In part (a) of the example, it could have been shown that the marginal distribution of is Pareto with the same values of and as are in the inverse gamma prior distribution of In other words, with integration by parts we get, where and

87 CR-84 CREDIBILITY SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR Therefore, the marginal distribution of is Pareto with and For Exam C questions it is not usually necessary to find the marginal distribution of in a Bayesian credibility situation It is also possible that the prior-model pair do not combine in such a recognizable way The following 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 model distribution for the given data points is (ii) the joint density of the 's and is ; for the given data points this is (iii) the marginal distribution of a set of observations is (iv) the posterior distribution of is ; evaluated at the given data 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 149 (v) the predictive distribution of a new observation has density (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 with parameter, 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 This was 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),

88 CREDIBILITY SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR CR-85 The usual way to find the predictive expectation is to first find the posterior distribution, which has density, and then (47) The integration is over the range of values of It is often the case that the mean of the model distribution is (for instance, if the 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 a point 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 the mean 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 with parameters and The mean of this inverse gamma posterior distribution is Therefore, for part (a) of Example CR-18 the predictive mean is equal to the posterior 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 claim from this policyholder Solution: The approach to finding a predictive expectation that is usually most efficient is to use the 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 be calculated Then, Thus,, for, and for

89 CR-86 CREDIBILITY SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR Example CR-20 continued Then, There is an alternative way of finding the predictive expectation that first requires finding the predictive density function In general, this is not the most efficient method for finding a predictive expectation, where As above, we find the posterior pdf to be, Then, if and if, Then we have This is a more complicated way of finding the predictive expectation because of the complicated nature 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 The model distribution has a Poisson distribution with a mean of A single observation of is made, 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 joints density 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 2 Then, This posterior probability can be expressed as Note that the numerator and denominator are both integrals of the joint density, but the numerator is over a subinterval of the denominator The denominator is an integral over the entire region 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 generally must find the joint density (if there is one sample value ), and the posterior probability is a ratio of two integrals of, with the numerator integration over a 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

90 CREDIBILITY SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR CR-87 CR-43 The Double Expectation Rule Applied to Bayesian Credibility The "double expectation" rule has been mentioned a few times in the context of Bayesian estimation 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 Example CR-18, where the parameter had an inverse gamma distribution with parameters and, and the model variable had an exponential distribution with mean The mean of the marginal distribution of is A version of the double expectation rule also applies to the predictive distribution In general, for random 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 was inverse gamma with and Therefore, A conditional approach to finding variance can also be applied The following rule for formulating variance was reviewed earlier in the Modeling unit of this study guide Given any two random variables and, If we apply this to Example CR-18(a) to find the variance of the marginal distribution of, we have Since has an exponential distribution with mean, we have Then, Since has an inverse exponential distribution with, the variance of is infinite (the second moment of is infinite), so the variance of the marginal distribution of will be infinite This could have also been anticipated because in Example CR-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 it gets 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 inverse gamma 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 posterior distribution The second moment of the inverse gamma with and is, so that

91 CR-88 CREDIBILITY SECTION 4 - BAYESIAN CREDIBILITY, CONTINUOUS PRIOR Putting these together, we get These comments, particularly about variance of the marginal and predictive distributions are probably taking the Bayesian credibility analysis somewhat past what might be expected on an exam question

92 CREDIBILITY - PROBLEM SET 4 CR-89 CREDIBILITY - PROBLEM SET 4 Bayesian Credibility - Continuous Prior Problems 1 and 2 are based on the following situation Assume that the number of claims each year for an individual insured has a Poisson distribution 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 constant over time An individual is chosen at random from the 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 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 large portfolio of policies, and each policy has its own value of For a randomly chosen policy from the portfolio, 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 one year (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

93 CR-90 CREDIBILITY - PROBLEM SET 4 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)

94 CREDIBILITY - PROBLEM SET 4 CR 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

95 CR-92 CREDIBILITY - PROBLEM SET 4 19 (SOA) The size of a claim for an individual insured follows an inverse exponential distribution with the following probability density function: ( ) 2, 0 The parameter has a prior distribution with the following probability density function: () 4, 0 4 One 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) 20 You are given: (i) An individual automobile insured has annual claim frequencies that follow a Poisson distribution with mean (ii) An actuary s prior distribution for the parameter has probability density function: ( ) = (05)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) 03 (B) 04 (C) 05 (D) 06 (E) 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)

96 CREDIBILITY - PROBLEM SET 4 CR-93 CREDIBILITY - PROBLEM SET 4 SOLUTIONS 1 We wish to find Since the model distribution is Poisson, we have, and since the prior distribution 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

97 CR-94 CREDIBILITY - PROBLEM SET 4 4(a) (b), and Then 8, and 4 8 (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 in problem 4 is more efficient Answer: B 8 Notice the prior and posterior distributions are both gamma distributions, so that the gamma distribution is a conjugate prior for the model distribution Answer: C 9 The joint pdf is e Answer: E

98 CREDIBILITY - PROBLEM SET 4 CR 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 same form 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

99 CR-96 CREDIBILITY - PROBLEM SET 4 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 the general -sample case, approach (i) can become quite complicated due to the complicated nature of the pdf of the predictive distribution, Approach (ii) is often more straightforward From Problem 3 we have 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 and 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

100 CREDIBILITY - PROBLEM SET 4 CR It was seen in Example CR-18(b) that the posterior distribution has an inverse gamma distribution 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 1411 The variance of the predictive distribution is, so that, which is the variance of the posterior distribution,, since has an exponential distribution given, and the variance of the exponential 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 in the 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 the factor which does not depend on From the given information,, and with one claim size of, this becomes From the comments above, the posterior density of is proportional to The only one of the answers which is proportional to this expression is B Answer: B

101 CR-98 CREDIBILITY - PROBLEM SET 4 20 The problem involves determining the Bayesian premium The prior distribution of the parameter has density (a mixture of two exponential distributions, one with mean and on with mean 5, which can also be regarded as a mixture of two 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 mixing weight 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 is equal to 0 if Therefore, the integral becomes Then, Answer: D

102 CREDIBILITY SECTION 5 - BAYESIAN CREDIBILITY, DISTRIBUTION TABLE CR-99 CREDIBILITY - SECTION 5, BAYESIAN CREDIBILITY APPLIED TO DISTRIBUTIONS IN EXAM C TABLE The material in this section relates to Section 1642 of "Loss Models" The suggested time frame for this section is 3-4 hours The first important step in proceeding with Bayesian credibility is to identify the prior and model distributions There are a number of combinations of prior-model distribution pairs that result in recognizable marginal, posterior and predictive distributions These are summarized a few pages later in this section of notes Before that summary, we consider in detail one particularly important 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 find under exam conditions, but there are a couple of more general points in the example that are worth noting CR-51 The Gamma-Poisson Credibility Model Example CR-22: A portfolio of insurance policies has the following characteristics Each policy 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 of policies, the parameter has a gamma distribution with known values and (we use the notation to avoid confusion with the conventional use of in the gamma distribution) A policy 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 posterior distribution 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 once the 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 some integration "shortcuts" a little later), but it reduces to the probability function for a negative binomial distribution with and, which is ; this is the marginal distribution of

103 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 / 1+ occurs in the pdf of the gamma distribution with parameters and Therefore the posterior distribution 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 reduce showing 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 in the 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 the posterior 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 the posterior distribution can be identified by the way the parameter occurs in the expression for This is true for many combinations of prior and model distributions This means that it is usually not necessary to actually find in order to identify 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 not involve ; 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) of CR-21

104 CREDIBILITY SECTION 5 - BAYESIAN CREDIBILITY, DISTRIBUTION TABLE CR 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 equal to the posterior mean 4 The prior distribution of in Example CR-22 is a gamma distribution, and the posterior distribution in parts (ii) and (iii) are also gamma distributions with modified parameters When the posterior distribution is from the same family of distributions as the prior distribution (such as in Examples CR-18 and CR-22), that distribution is referred to as being a conjugate prior for the model distribution From Example CR-22 we see that the gamma distribution is a conjugate prior for the Poisson distribution The gamma prior/poisson model combination has come up in Exam C questions on a regular basis, and it is worthwhile being familiar with it CR-52 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 (51) 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 binomial random variable with and Inverse Gamma Distribution, parameters, : The pdf is Since, it follows that (52) (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 C table Note also that if is an integer, We now summarize a number of prior-model distribution pairs that result in recognizable posterior 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 the form of the posterior

105 CR-102 CREDIBILITY SECTION 5 - BAYESIAN CREDIBILITY, DISTRIBUTION TABLE CR-53 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 all combinations 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 posterior distribution must be gamma with parameters and An important related property is that the marginal distribution of can be shown to be negative binomial with and We can also find by the double expectation rule, (b) With data values and either the individual values or the sum available, the joint density of and is ( ), which is proportional to The posterior distribution must be gamma with parameters and Since both prior and posterior distributions are gamma distributions, the gamma distribution is a conjugate prior for the Poisson distribution Note that the variance of the posterior distribution is the gamma distribution variance In both (a) and (b) the predictive distribution will be negative binomial with and, and the predictive mean is the same as the mean of the posterior distribution, (53) Note that the variance of the predictive distribution will be the variance of the negative binomial (54) A special case arises when in the prior distribution The prior becomes an exponential distribution and the marginal distribution of becomes a geometric distribution The posterior will be a gamma distribution If the observation is 0 (or all are 0) then the posterior will have and will still be exponential

106 CREDIBILITY SECTION 5 - BAYESIAN CREDIBILITY, DISTRIBUTION TABLE CR Model: exponential dist,, pdf Prior: inverse gamma dist,,, pdf +1 ( ) This is the prior-model distribution pair in Example CR-18 (a) With a single data value available, the joint density of and is ( +1, which is proportional to ( ) + We can find the mean of the marginal distribution of by the double expectation rule, (55) The posterior distribution must be inverse gamma with, and (56) Since both prior and posterior distributions are inverse gamma, the inverse gamma is a conjugate prior for the exponential distribution Although usually of less importance, it is possible to show that the marginal distribution of is Pareto 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 he predictive mean is (57) (b) With data values and either the individual values or the sum available, the joint density of and is ( +1, which is proportional to ( ) + The posterior distribution must be inverse gamma with, and In both (a) and (b) an exponential model distribution combines with an inverse gamma prior distribution to produce an inverse gamma posterior distribution The inverse gamma is a conjugate prior for the exponential distribution The predictive mean is (58) 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 expression we 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 prior to produce an inverse gamma posterior distribution The case considered above in which the model distribution is exponential is really just a special case of this one, since an exponential distribution is a special case of the gamma distribution

107 CR-104 CREDIBILITY SECTION 5 - BAYESIAN CREDIBILITY, DISTRIBUTION TABLE 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 (59) Since both prior and posterior distributions are beta distributions, we see that the beta distribution is a conjugate prior for the binomial distribution (b) With data values and either the individual values or the sum available, the joint density of and is, which is proportional to The posterior distribution must be a beta with parameters and (510) The predictive expectation will be, since has a binomial distribution with parameters and Therefore, the predictive mean is posterior mean (511) 4 Model: inverse exponential dist, pdf Prior: gamma dist,,, pdf ( ) (a) With a single data value available, the joint density of and is ( ), which is proportional to The posterior distribution must have a gamma distribution with, and (512) Although usually of less importance, it is possible to show that the marginal distribution of is inverse Pareto with and the same as in the prior distribution, and the predictive distribution of given is inverse Pareto with and the same as the posterior distribution (b) With data values and either the individual values or the sum available, the joint density of and is ( ), which is proportional to

108 CREDIBILITY SECTION 5 - BAYESIAN CREDIBILITY, DISTRIBUTION TABLE CR-105 The posterior distribution must be gamma with, and (513) Since both prior and posterior distributions are gamma distributions, the gamma distribution is a conjugate prior for the inverse exponential distribution A generalization similar to that in Combination 1 applies here Suppose that instead of being inverse exponential, the model distribution is inverse gamma with known and unknown For instance, suppose that in the inverse gamma model distribution Then the model density 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 inverse gamma 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 a conjugate 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 ( ) ( ) (514) and variance (515) The normal distribution is a conjugate prior for the normal distribution (b) With data values and either the individual values or the sum available, the posterior distribution of can be shown to be normal with mean ( ) ( ) (516) and variance (517) The predictive mean will be Since is normal with mean, it follows that the predictive mean is the same as the posterior mean, which is ( ) ( ) (518)

109 CR-106 CREDIBILITY SECTION 5 - BAYESIAN CREDIBILITY, DISTRIBUTION TABLE 6 Model: uniform distribution on the interval, pdf Prior: single parameter Pareto,, pdf, With observations, suppose that The posterior distribution of is a single parameter Pareto with and, and pdf, The Bayesian premium is (519) As an example, suppose that the prior density is,, and the model density is for Note that the prior density is a single parameter Pareto with and Suppose that two observations of are made, and they are and Since the largest observed value of is 25, it must be true that is at least, since each observed must be 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 the gamma 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 variance of 01 Therefore,, from which it follows that

110 CREDIBILITY SECTION 5 - BAYESIAN CREDIBILITY, DISTRIBUTION TABLE CR-107 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 the denominator (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 gamma with 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, its 2nd moment is The predictive variance is Example CR-25: Tom has a coin, but he doesn't know the probability of tossing a head He assumes 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 the coin another 5 times Find the expected number of heads in those next 5 tosses Solution: We first observe that, which is a beta distribution with, The number of heads tossed in 10 tosses has a binomial distribution with parameters and, and probability function The joint density of and is, which is proportional to 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 tosses4 heads no heads in 5 more tosses 4 heads 4 heads (posterior mean of )

111 CR-108 CREDIBILITY SECTION 5 - BAYESIAN CREDIBILITY, DISTRIBUTION TABLE A variation on the Beta prior and Negative Binomial model distributions The 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 the parameter, where so that Suppose that we use this parametrization, and that we assume that is a prior parameter with a beta distribution with parameters and, with prior density Assume that the conditional distribution of given is binomial with known parameter, and with parameter, so that the model distribution has probability function for The joint distribution of and has joint density This is proportional, in, to, which implies that the posterior distribution of 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-54 Some Additional Comments on Bayesian Estimation There are a couple of other points to mention regarding Bayesian estimation One of the objectives of a Bayesian estimation situation may be to obtain an estimate of the parameter An approach that can be taken is to define a "loss function", and the estimate of is that which minimizes the expected value of the loss function using the posterior distribution The following is a summary of the estimates obtained for three standard loss functions We wish to find the value 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 if (iii) Zero-one loss: ; the estimate is the mode of the posterior if Another point to note about Bayesian estimation is the concept of the highest posterior density credibility set We will use the example given above for the single parameter Pareto prior and uniform distribution model combination In that example the posterior distribution of was found to be single parameter Pareto with and, with pdf,

112 CREDIBILITY SECTION 5 - BAYESIAN CREDIBILITY, DISTRIBUTION TABLE CR-109 Suppose that we wish to find an interval such that for that interval 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 these intervals, we note that the interval over which the pdf has the largest values is, since the pdf for the single parameter Pareto is a decreasing function The interval is the highest posterior density (HPD) credibility interval for probability 75 In general, the % HPD credibility interval for a given is an interval over which the posterior probability is %, and the numerical values of the posterior density are higher on 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 these notes) 1 The model distribution is the conditional distribution of given The conditional mean of is referred to as the hypothetical mean This may be denoted and is equal to 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 premium 4 Note that the marginal -distribution variance is When the predictive distribution is found for given the observations, it may be possible to formulate the Bayesian premium in terms of the observed sample mean and in a credibility formulation:, where may depend upon

113 CR-110 CREDIBILITY SECTION 5 - BAYESIAN CREDIBILITY, DISTRIBUTION TABLE

114 CREDIBILITY - PROBLEM SET 5 CR-111 CREDIBILITY - PROBLEM SET 5 Bayesian Credibility - Exam C Table Distributions 1 An individual insured has a frequency distribution per year that follows a Poisson distribution with mean The prior distribution for is exponential with a mean of 2 An individual is observed to have 2 claims in a year Find the Bayesian premium for the same individual for the following year A) B) C) D) E) Questions 2 to 6 are based on a single observation, an integer from a Poisson distribution with parameter, where has a prior distribution which is exponential with parameter, so the pdf 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)

115 CR-112 CREDIBILITY - PROBLEM SET 5 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 head with 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 of coins 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 marginal probability 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 ( ) 1 ), 0, 0 The prior distribution of is assumed to follow a distribution with mean 50 and density function ( ) 500, , 0 4 ) (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 50 has been observed for this risk Problems 9 and 10 are based on the following situation Suppose that the claim severity for an individual in a risk class has a binomial distribution with parameters and Suppose also that a randomly chosen individual in the risk class has parameter where has a beta distribution 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 the population 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

116 CREDIBILITY - PROBLEM SET 5 CR 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 model distribution 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 a Generalized 3-Parameter Pareto distribution (A231 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 given prior 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 Poisson distribution with mean The prior distribution for is a mixture of two exponential distributions with means of 1 and 3, and with mixing weights both equal to 5 An individual is observed to have 0 claims in a year Find the Bayesian premium for the same individual for the following year A) B) C) D) E) 13 An individual insured has a frequency distribution per year that follows a Poisson distribution 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 the mixing weights are both 5 An individual is observed to have 0 claims in a year Find the Bayesian premium for the same individual for the following year A) 7 B) 9 C) 11 D) 13 E) (SOA) You are given: (i) An individual automobile insured has an annual claim frequency distribution that follows a Poisson 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 the variance (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 the next year using their model assumptions Determine the ratio of the Bayesian premium that the first actuary calculates to the Bayesian premium that the second actuary calculates (A) 3/4 (B) 9/11 (C) 10/9 (D) 11/9 (E) 4/3

117 CR-114 CREDIBILITY - PROBLEM SET 5 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 Number of autos insured 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) 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) 019 (B) 022 (C) 025 (D) 028 (E) 031

118 CREDIBILITY - PROBLEM SET 5 CR (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 Determine the posterior probability that this same policyholder will have at least one claim this year (A) 070 (B) 075 (C) 078 (D) 081 (E) For a coin chosen at random from a large collection of coins, the probability of tossing a head with 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 of coins 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 3748 A 7th observation of X is obtained and the Bayesian premium is recalculated to be 3784 Find the value of the 7th observation

119 CR-116 CREDIBILITY - PROBLEM SET 5 CREDIBILITY - PROBLEM SET 5 SOLUTIONS 1 A gamma prior with parameters and combines with a Poisson model distribution to result in 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 distribution is 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 Bayesian premium (predictive mean) is Answer: D 2 The pf of the model distribution is for integer values Answer: A 3 The joint pdf is for 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 in Section 5 of the notes, it was noted that an exponential prior distribution combines with a Poisson model 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 an exponential prior distribution combines with a Poisson model distribution to result in a posterior distribution of which is a gamma distribution with and Answer: C

120 CREDIBILITY - PROBLEM SET 5 CR 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 combination of an exponential prior distribution with a Poisson model distribution results in a negative binomial predictive distribution Answer: E The situation in questions 2 to 6 is one of the standard Bayesian model-prior distribution combinations summarized in the notes If the prior is exponential with parameter (or gamma with 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,

121 CR-118 CREDIBILITY - PROBLEM SET 5 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 exponential with 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 is equal to 1 500, ) ) 4, and with, this is 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

122 CREDIBILITY - PROBLEM SET 5 CR 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 therefore the 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 prior distribution of is a beta distribution with parameters and, and if the model distribution is binomial with parameters and, then given observation, the posterior distribution is also beta with parameters and The prior of in this case is a beta distribution with, and the model distribution is binomial with and ; since the observation is, the posterior distribution is also beta with parameters and Answer: C 10 In Problem 8, the posterior distribution was found to be a beta with parameters and The conditional expectation is, since the conditional distribution of given is binomial with and Then (mean of the posterior) The mean of the beta distribution with parameters and is Therefore, the mean 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 is The 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

123 CR-120 CREDIBILITY - PROBLEM SET 5 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 of the components of indicate that the posterior must be a mixture of two exponential distributions: 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 the exponential 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 predictive mean is the same as the mean of the posterior distribution, which is the mixture of the means of the two exponential distributions: Answer: E, probability mass 13 The prior distribution has pdf and the model distribution has pf, prob mass The joint distribution of and is, prob mass With this becomes 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 a constant 1 with mixing weight and an exponential with mean 75 and mixing weight The mean of the posterior is Answer: B

124 CREDIBILITY - PROBLEM SET 5 CR The distribution being considered is the number of claims per year (annual claim frequency) 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 expected number of claims in the fourth year given the information about the first three years The distribution of is Poisson with mean, and is assumed to have 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 number of claims in one period, and is a sample of for separate periods; is the number of claims for period 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 gamma with parameters and The mean of a gamma distribution with parameters and is Actuary 1's Bayesian premium estimate is the mean of the posterior distribution of, which is The variance of the gamma distribution with parameters and is We are told that Actuary 2 assumes that has a gamma distribution, say with and, with the same mean as Actuary 1's gamma distribution, so We are also assuming that the variance of the 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

125 CR-122 CREDIBILITY - PROBLEM SET 5 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 the Beta 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 in Volume 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 special case of the more general situation in which the prior distribution of is a gamma distribution with parameters and, and the model distribution (of given ) is Poisson with mean In this general case, if sample data values are obtained, then the posterior distribution of 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 count distribution, has a Poisson distribution with mean The data available is the claim count during the first year, a single observation of Therefore, sample value, and the value of is given to be 2 The posterior distribution of is also a gamma distribution, with and The first moment of this gamma distribution is (from the distribution table) ', and the second moment is The variance is the second moment minus the square of the first Answer: B

126 CREDIBILITY - PROBLEM SET 5 CR The prior distribution is gamma with and This is a Bayesian problem with a Poisson model distribution with mean, and a prior distribution that is gamma, with parameters and In this problem is the expected 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 In this 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) gamma distribution, which is With 1100 autos insured in the 3rd year, the expected 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 is that, since given has a uniform distribution on the interval it follows that for for, or Therefore, for for, or and for for, or We are given a prior distribution for on the interval However, for, so that

127 CR-124 CREDIBILITY - PROBLEM SET 5 18 continued The 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 (not 500) 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 the numerator and denominator over the prior parameter We are asked to find which is written as We 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 implicitly assume that and are conditionally independent given ) Then,

128 CREDIBILITY - PROBLEM SET 5 CR continued An alterative approach is to "factor" the predictive probability through the posterior distribution of, so that As before, When the prior distribution of is a gamma distribution with parameters and and the model has a Poisson distribution with mean, then the marginal distribution of is negative binomial with 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 the statement of the problem is This is an example of a more general rule that has 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 integral is 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

129 CR-126 CREDIBILITY - PROBLEM SET 5 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,

130 CREDIBILITY SECTION 6 - BUHLMANN CREDIBILITY CR-127 CREDIBILITY -SECTION 6, BUHLMANN CREDIBILITY The material in this section relates to Section 1643 to 1645 of "Loss Models" The suggested time frame for this section is 3-4 hours CR-61 The Buhlmann Credibility Structure The Buhlmann model has the same initial structure as the Bayesian credibility model There is a population or portfolio of risks Each member of the population has an associated random variable (the model distribution which is loss amount, or claim number usually) The model distribution random variable depends on a parameter (the prior distribution) Each member of the population has an associated parameter, and the model distribution is a conditional random variable given the value of The following is a repeat of the prior and model distribution components of a Bayesian credibility model The initial assumption for the distribution of the parameter is called the prior distribution (of ), 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 each individual has their own value, and these values are distributed over the population according to the prior density of The model distribution (usually frequency or severity of loss) is a conditional distribution given, with pf/pdf (or more simply ) Under the Buhlmann credibility model, the conditional distributions of the 's given are assumed to be independent and identically distributed (iid), with the following characteristics: is the hypothetical mean (61) is the process variance (62) (and since the conditional distributions of given and given are independent, for ; but the unconditional distributions of and will most likely not be independent) Using the conditional expectation rule, we have (63) is called the pure premium (also called the collective premium) We also define the following factors (the expected process variance, EPV), and (64) (the variance of the hypothetical mean, VHM) (65) (Keep in mind that since is a random variable, the hypothetical mean and process variance are also random variables, and have means and variances themselves) Then the unconditional variance of is, (66) (and the unconditional covariance between and for is ) (67) This situation can be described as (and, where )

131 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 of independent 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 credibility premium is Buhlmann Credibility Premium, (68) where (69) The Buhlmann credibility premium is the estimate of the next occurrence of for that same value of The factor is the Buhlmann credibility factor ( ) (610) Note also that since, the credibility premium can be written in the form sum of observed 's (611) 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 from an with the same value of Buhlmann Credibility questions on Exam C have tended to be fairly mechanical, with one of the main 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 a constant for all values of ), the credibility factor is In both the Bayesian and Buhlmann credibility approach, we are trying to estimate, but we don't know the specific value of The Buhlmann credibility premium is weighted average of the sample mean from the data set and the mean of the marginal distribution of from the Bayesian structure The weight applied to the sample mean is, and we can see that as 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 are taking into account all possible values of in the model distribution of We do not know the value 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

132 CREDIBILITY SECTION 6 - BUHLMANN CREDIBILITY CR-129 CR-62 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's method The conditional distribution of given is a compound Poisson distribution with Poisson mean and exponential claim size distribution also with mean We recall that a compound 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 distribution with 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 more observations 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 05 - 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 05 A risk is selected at random from the portfolio Determine the Buhlmann credibility factor for one 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 Since each 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

133 CR-130 CREDIBILITY SECTION 6 - BUHLMANN CREDIBILITY Example CR-27 continued The process variance is one of two values, either or depending upon which class of Class A or B was picked We are told that a risk is selected at random, which we interpret as 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 this same die (b) The selected die is placed back into the urn A seventh die is then added to the urn The seventh 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 expected value of the second roll of this same die Determine which of the three types for the seventh die would increase the Buhlmann credibility factor of the first roll of the selected die (compared to the 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 are type 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

134 CREDIBILITY SECTION 6 - BUHLMANN CREDIBILITY CR-131 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 no process variance, and there would be a decrease in expected process variance Since the mean for die 3 is 6 and the prior mean was 3 without the extra die, the variance of the hypothetical mean 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 is a gamma random variable with density function ( ) 2 2, 0 - 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 Poisson distribution; the process variance is 2 2 The Buhlmann model is somewhat simple and also restricted It requires that from one period to the next, the characteristics of the loss-related random variables (claim number or total claim amount) remain the same No allowance is made for increase (or any change) in the number of policyholders from one period to the next The next model still applies the Buhlmann approach to credibility (the credibility premium modeled as a linear function of the 's), but allows for changes in loss characteristics from one period to the next In fact, the following model is actually equivalent to the original Buhlmann model when the process variance is known, but it may be more convenient for dealing with changing numbers of exposures from one period to the next

135 CR-132 CREDIBILITY SECTION 6 - BUHLMANN CREDIBILITY CR-63 The Buhlmann-Straub model This is a generalization of the Buhlmann model just considered It is still assumed that the conditional distributions of the given are independent, and the hypothetical mean is the 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, (612) As in the basic Buhlmann model, we define the factors, and : We also define to be (613) The credibility factor is, (614) and credibility premium, where (615) ( is a weighted average of the 's) Also, it can be shown that, and for (616) Note that, as in the original Buhlmann model case, this credibility premium can be written as total claims for observed values of (617) 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 has mean claim amount (or number) and variance ;, where each has mean and variance Since is the average of the claims, it has the same hypothetical mean as any individual exposure unit, but has process variance It is still assumed that there is independence from one period (or group) to another The credibility premium that is calculated is the credibility premium per exposure unit (ie, per occurrence of an 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 per exposure 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 model would be the same as that found from this Buhlmann-Straub model The credibility premium per policy in either the original Buhlmann model or this Buhlmanntotal claims all periods Straub model is total number of exposures for all periods, where and This shows the equivalence between the original Buhlmann model and the Buhlmann-Straub model The following example illustrates this

136 CREDIBILITY SECTION 6 - BUHLMANN CREDIBILITY CR-133 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 mean and 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 Determine the Buhlmann-Straub credibility premium per employee for this policyholder Solution: This problem can be viewed as a standard Buhlmann credibility problem The random variable being considered is loss per employee There are a total of observations (number of losses in all 3 years) We are given, and for the 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 2350 observations 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 year sample means, but from that we can determine the total loss for each year, and add them) A variation on the Buhlmann-Straub model There is a variation to the Buhlmann-Straub model that we note (derivations can be found in Examples 1629 and 1630 of the "Loss Models" book) This variation allows for the differentiation between the variances of the claim-related distribution; this is represented as a more general form for The characteristics of this variation to the Buhlmann-Straub model 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,, (618) We define the factor as follows (619) The solution of the normal equations results in credibility premium, where, and (620) A further variation has (where ) instead of Now in the credibility premium, (also, and are found by replacing in the expressions above with )

137 CR-134 CREDIBILITY SECTION 6 - BUHLMANN CREDIBILITY CR-64 Exact Credibility The Bayesian credibility method and the Buhlmann credibility method both require the same Bayesian structure of a parameter distribution for and model distribution for given When the conditional distribution of given and the distribution of satisfy certain conditions, the Bayesian premium will be equal to the (Buhlmann) credibility premium This situation is referred to as exact credibility One way of checking whether or not exact credibility is 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 which exact 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 exact credibility 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 linear function 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 The Bayesian 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 ( ) ( )

138 CREDIBILITY SECTION 6 - BUHLMANN CREDIBILITY CR-135 CE-165 Theoretical Background to Buhlmann Credibility The following is a theoretical preamble to Buhlmann's approach to credibility and is not crucial background We continue to assume that the distribution of depends upon a parameter, and that has a distribution as well (the prior distribution) Given the sample values, we can find the conditional distribution of given In this approach to credibility, we estimate ( ) (also denoted when there are -sample values available) as a linear function of the 's in the form Given the sample 's, we find the values of the -coefficients that minimize the expectation This least squares estimation is done in the usual way, taking partial derivatives of the expectation with respect to each, and solving a system of equations This results in the set of normal equations, with the ~ 's being the least squares estimates: 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 (or marginal) expectations and covariances of the 's Also note that for any, The credibility premium is then ~ ~ ~ ~ ~ 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 between any pair and for Keep in mind that initial information is usually in the form of the conditional distribution of given (along with the prior distribution of ), and it is usually assumed that for, the conditional distributions of and given are independent This does not mean that the unconditional (marginal) distributions of and are independent; the main reason that credibility methods are meaningful is that does depend upon ; if the unconditional distribution of was independent of, then the Bayesian premium is just the pure premium -, and past experience would be of no value in determining the premium In Example 1624 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 Buhlmann credibility model as was done just now

139 CR-136 CREDIBILITY SECTION 6 - BUHLMANN CREDIBILITY

140 CREDIBILITY - PROBLEM SET 6 CR-137 CREDIBILITY - PROBLEM SET 6 Buhlmann Credibility 1 Assume that the number of claims each year for an individual insured has a Poisson distribution 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 constant over 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 0 or 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 of the values on the selected balls is 1 Two more balls are selected from the same urn with replacement Assume that each selected ball has been returned to the urn before the next ball is drawn 2 Determine the Bayesian analysis estimate of the expected value of the sum of the values on the second pair of selected balls A) 99 B) 101 C) 103 D) 105 E) Determine the Buhlmann credibility estimate of the expected value of the sum of the values on the second pair of balls chosen A) 100 B) 103 C) 106 D) 109 E) 112 Problems 4 and 5 refer to the following situation The aggregate loss distribution for two risks for one exposure period is: Aggregate losses Risk $0 $50 $1000 A B A 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 the aggregate losses for the same risk's third exposure period A) 9200 B) 9216 C) 9232 D) 9248 E) Determine the credibility premium (Buhlmann's credibility estimate of the expected value of the aggregate losses for the same risk's third exposure period) A) 9950 B) 9983 C) D) E) 10083

141 CR-138 CREDIBILITY - PROBLEM SET 6 Problems 6 to 8 are based on the following situation A risk class is made up of three equally sized groups of individuals Groups are classified as Type A, Type B and Type C Any individual of any type has probability of 5 of having no claim in the coming year and has a probability of 5 of having exactly 1 claim in the coming year Each claim is for amount 1 or 2 when a claim occurs Suppose that the claim distributions given that a claim occurs, for the three types of individuals are 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 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) If an individual is chosen at random from the risk population and observation is available for that individual, find the credibility premium for the next exposure period for this individual A) B) C) D) E) 9 For an individual risk in a population, the number of claims per month follows a Poisson distribution with mean For the population, is distributed according to the exponential distribution, with pdf A randomly selected risk in the population is found to have 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 each risk class in 1997, 12 individuals in each risk class in 1998 and 15 individuals in each risk class in 1999 An individual from a randomly chosen risk class is observed for the three years, and it is found 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 for this risk class, if there are 20 individuals in the risk class in 2000 A) 33 B) 34 C) 35 D) 36 E) 37

142 CREDIBILITY - PROBLEM SET 6 CR 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 One claim is expected during each period of observation Determine the largest number of periods of observation for which Partial Credibility Formula B yields 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 a discrete uniform distribution on the integers 3 A single observation of is made Find the Buhlmann factor 15 Find the Buhlmann credibility premium in Example CR You are given the following: - Claim size for a given risk follows a distribution with density function ( ) 1 ), 0, 0 The prior distribution of is assumed to follow a distribution with mean 50 and density function ( ) 500, , 0 4 ) Find the Buhlmann credibility factor for a single observation of 17 An individual insured has a frequency distribution per year that follows a Poisson distribution with mean The prior distribution for is a mixture of two exponential distributions with means of 1 and 3, and with mixing weights both equal to 5 An individual is observed to have 0 claims in a year Find the Buhlmann credibility estimated number of claims for the same individual for the following year

143 CR-140 CREDIBILITY - PROBLEM SET 6 18 An individual insured has a frequency distribution per year that follows a Poisson distribution 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 the mixing weights are both 5 An individual is observed to have 0 claims in a year Find the Buhlmann 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 credibility approach 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 of policyholders You are given: (i) The average claim frequency for a policyholder over the entire book is 0425 (ii) The variance of the hypothetical means is 0370 (iii) The expected value of the process variance is 1793 One class of policyholders is selected at random from the book Nine policyholders are selected at random from this class and are observed to have produced a total of seven claims Five additional policyholders are selected at random from the same class Determine the Buhlmann credibility estimate for the total number of claims for these five policyholders A) 25 B) 28 C) 30 D) 33 E) 39

144 CREDIBILITY - PROBLEM SET 6 CR You are given: (i) The number of claims incurred in a month by any insured has a Poisson distribution with mean (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 ? Determine the Buhlmann-Straub credibility estimate of the number of claims in Month 4 (A) 167 (B) 169 (C) 173 (D) 176 (E) 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 losses in terms of 27 Annual aggregate claims for a particular policy are modeled as a compound Poisson distribution 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 large portfolio of policies, and each policy has its own value of For a randomly chosen policy from the portfolio, 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 one year 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 05 and (iii) is the expected claim count for driver for one year (iv) The s have an exponential distribution with mean 02 Determine the Buhlmann credibility factor for an individual driver for one year (A) Less than 005 (B) At least 005, but less than 010 (C) At least 010, but less than 015 (D) At least 015, but less than 020 (E) At least 020

145 CR-142 CREDIBILITY - PROBLEM SET 6 29 For a portfolio of independent risks, the number of claims per period for a randomly chosen risk 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 no claims for the period and Risk 2 has 2 claims for the period is the Buhlmann credibility premium for Risk 1 for the next period and is the Buhlmann credibility premium for Risk 2 for the 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 20 percent accident probability per year and the other group has a 40 percent accident probability per year Two or more accidents per year per insured are not possible The two groups comprise equal proportions of the population and each has the following accident severity distribution: Probability Size of Loss A merit rating plan is based on the pure premium experience of individual insureds for the prior year 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 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 the probability 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 to estimate 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 factor based on exposures (A) (B) (C) (D) (E) 33 The Buhlmann credibility model is being applied to the loss variable (per exposure) It is found that after exposures, the Buhlmann credibility factor is How many additional exposures are needed to increase the factor to 05? (A) 10 (B) 15 (C) 20 (D) 25 (E) 30

146 CREDIBILITY - PROBLEM SET 6 CR (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/ /3 1 4/ /3 2 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/ /3 1 4/ /3 2 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) (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 the year Determine the posterior probability that the selected employee has salary greater than 50 thousand (A) 05 (B) 06 (C) 07 (D) 08 (E) An employee is selected at random During the last 4 years, the employee has had a total of 5 claims Determine the Buhlmann credibility estimate for the expected number of claims the employee will have next year (A) 06 (B) 08 (C) 10 (D) 11 (E) 12

147 CR-144 CREDIBILITY - PROBLEM SET 6 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 selecting a 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 are expected (A) 044 (B) 050 (C) 080 (D) 095 (E) (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, 2000 Calculate the expected size of the next claim using Buhlmann credibility (A) 1025 (B) 1063 (C) 1115 (D) 1181 (E) Losses for the year for a risk are uniformly distributed on, where is uniformly distributed on the interval The first year loss for a risk is Find the Buhlmann credibility premium for the second year's loss for the same risk in terms of

148 CREDIBILITY - PROBLEM SET 6 CR-145 CREDIBILITY - PROBLEM SET 6 SOLUTIONS 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-winded than the calculations needed (urn A is chosen) (sum of two balls is 1) sum of two balls is 1 urn A is chosenurn A is chosen balls are or urn A is chosen Similarly, (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 similarly urn 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 3 Using the posterior probabilities for urns A and B, we can find the conditional expected value on a ball chosen from the same urn as the first two balls - this conditional expectation is 3rd ball valuesum 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 the sum of two balls if urn A is chosen (16) and also if urn B is chosen (6) If we had no prior information about the sum of the first two balls, then the expected value of the sum of two balls from a randomly chosen urn would be sum urn A urn Asum urn B urn B Although each of urns A and B have the same chance of being chosen initially, we now have additional information about which urn might have been chosen We use Bayesian analysis to find the probability that the urn was A (or B) given that the two balls chosen from the unknown urn 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 probabilities to find the expected sum of the next two balls

149 CR-146 CREDIBILITY - PROBLEM SET 6 2 continued sum urn A urn A sum of first 2 balls is 1 sum urn B urn B sum of first 2 balls is 1 In 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 1 given that the sum of the first two balls was 1 - next ball is 0 sum of first two balls is 1 ball is 0 (urn A) (sum of first two balls is 1) urn A sum of first two balls is 1 ball is 0 (urn B) (sum of first two balls is 1) urn B sum of first two balls is 1 Algebraically, this relationship is (sum is 1) next ball is 0 sum of first two balls 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) Asum is 1 sum is 1 Bsum 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 1 ball 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 B is 7 If we had no information about the sum of the first two balls, then each of urns A and B have a chance of being chosen, and the probability of choosing a 0 from a randomly chosen urn would be, and the probability of choosing a 1 from a randomly chosen urn would be The Bayesian analysis above gave us the conditional probabilities of choosing urn A or urn B given that the first two balls had a sum of 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 0 or a 1 given that the sum of the first two balls was 1 - next ball is 0 sum of first two balls is 1 and next 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 and next 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

150 CREDIBILITY - PROBLEM SET 6 CR We now put this situation in the context of the Buhlmann credibility approach The parameter A prob describes the urn chosen: To find the Buhlmann credibility estimate B prob (premium) for the sum of the next two balls we must identify the random variable, and the conditional distribution of given In this case, is the number on a ball chosen from an urn, 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, where 3 continued The 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 AA, and B BB, and then A A B B, as before The process variance, and its expected value, where A A A A, and since A we have A A, and similarly, B B Then, A A B B The variance of the hypothetical mean A AB B There are observations, so that Also, so that, and the credibility premium is per new ball chosen There will be two balls chosen, so the expected total of the next two balls (the credibility premium for two more claims) is Answer: C

151 CR-148 CREDIBILITY - PROBLEM SET 6 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 unconditional probability of no claims in either of two periods for a randomly chosen risk) We now find conditional probabilities:, and The conditional expectation of aggregate claims per period for each risk:, The conditional expected aggregate claims per period given no claims in first two periods is found 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 is The 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 on ball becomes aggregate claim for one exposure period Otherwise, the analysis is the same Answer: E 6 each with probability, and : Therefore Answer: D

152 CREDIBILITY - PROBLEM SET 6 CR-149 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 is the square of the mean) Let denote the number of claims in each of the final 6 months of 2000 for the individual, and let denote the number of claims in each of the 12 months of 2001 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

153 CR-150 CREDIBILITY - PROBLEM SET 6 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 the randomly selected risk class is, Similarly,,, and, are the observed average aggregate claim 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 to Buhlmann-Straub Answer: C Expected value of process variance 11 Variance 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

154 CREDIBILITY - PROBLEM SET 6 CR 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 Buhlmann credibility premium is 16 has an exponential distribution, and has an inverse gamma distribution with and 17 Since is a mixture, and (each component of the mixture has an exponential distribution, and the second moment of an exponential is two times the square of the mean) Then For a single observation of,, the Buhlmann credibility factor is With, the Buhlmann credibility estimate for the next year is

155 CR-152 CREDIBILITY - PROBLEM SET 6 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 and covariances 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 for 5 policyholders of the same class is Answer: D

156 CREDIBILITY - PROBLEM SET 6 CR This is a combination of a gamma prior distribution and Poisson model distribution For this combination, the Buhlmann credibility premium is the same as the Bayesian credibility premium estimate (called exact credibility) The Buhlmann-Straub estimate in this case is the same as the Buhlmann estimate (which is the same as the Bayesian estimate) The first three months of data are 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 of claims 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 prior distribution Therefore, and The predictive expectation is, This is the expected number of claims for a single insured 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 model distributions we would have had to use the Buhlmann method Note that we are implicitly assuming that all 450 insureds have the same (unknown), and we wish to find the expected number of claims from 300 more policies with the same That is the standard Bayesian approach Answer: B 26 Prior distribution is Hypothetical 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 Buhlmann credibility premium is

157 CR-154 CREDIBILITY - PROBLEM SET 6 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 exponential distribution with mean 02 Under the Buhlmann model, (since we are told that has an exponential distribution with mean 02) Also, (variance of the exponential is the square of the mean) The credibility 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 lim Answer: D 30 is the group number Hypothetical mean for group 1 is (2 is expected frequency, and is the expected severity for 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, and for group 2 it is Then, The credibility for 1 driver for 1 year is Answer: C

158 CREDIBILITY - PROBLEM SET 6 CR 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 25 exposures Answer: D 34 The Buhlmann credibility premium is, where is the sample mean of the 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

159 CR-156 CREDIBILITY - PROBLEM SET 6 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 0 to 100 Then The process variance is (since given has a Poisson distribution, 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

160 CREDIBILITY - PROBLEM SET 6 CR 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 Buhlmann credibility premium is

161 CR-158 CREDIBILITY - PROBLEM SET 6

162 CREDIBILITY SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS CR-159 CREDIBILITY - SECTION 7 EMPIRICAL BAYES CREDIBILITY METHODS The material in this section relates to Section 165 of "Loss Models" The suggested time frame for this section is 3 hours CR-71 Non-parametric Empirical Bayes Credibility In the Bayesian and Buhlmann credibility approaches that have been considered, it was assumed that the conditional distribution of given was known to be of a certain form (usually a parametric distribution such as Poisson with parameter ) It was also assumed that the distribution 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) We now consider the situation in which the conditional distribution of given might not be parametric, and the distribution of is unknown The unknown quantities (such as mean and variance) related to the distribution of are called structural parameters Our objective is still to apply the Buhlmann or Buhlmann-Straub models to determine a credibility premium based on observed claim data only We still wish to find, which requires finding and, as before, and then is needed to find the credibility premium Since the pf/pdf of the conditional distribution of given and of the distribution of are not known, we cannot directly find or or The empirical Bayes approach uses the experience data to estimate, and The analysis can be placed 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 (71) 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 (72) - the average observed claim per exposure unit for policyholder is (73) - the total number of exposure units for all policyholders is (74)

163 CR-160 CREDIBILITY SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS - the average claim per exposure period for all policyholders is total observed claim for all policyholders in all exposure periods total number of exposure periods for all policyholders (75) - policyholder has risk parameter random variable, and are assumed to be independent and identically distributed - for policyholder and exposure period, the conditional distribution of average claim per exposure period given has mean and variance and (76) - the structural parameters are,, (77) (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 (78) 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 (79) We can apply the empirical Bayes approach to the Buhlmann model and to the Buhlmann-Straub model 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-parametric empirical Bayesian analysis, suppose that there are policyholders Policy holder 1 is observed 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,,,

164 CREDIBILITY SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS CR-161 Example CR-32 continued The 3 years of observations for policyholder 1 resulted in 4 claims in the first year with average 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 applied to policyholders 2 and 3 Since is the average claim size of the claims in the first year of observation for policyholder 1, it follows that the total claim for policyholder 1 in the first year is Similarly, the total claim amounts for the second and third years for policyholder 1 are 10,800 and 5,800, for an overall total of all claims amounts for policyholder 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 is 16,600 and the average claim amount per exposure is For policyholder 3 there are a total of exposures, and an average claim of The aggregate claims for all policyholders in all exposure periods is CR-72 Empirical Bayes Estimation for the Buhlmann model (Equal Sample Size) Under the Buhlmann model, there are the same number of exposure periods for each policyholder, which means that, and there is one exposure unit for each exposure period, which means that for There are two common interpretations to this situation (1) is the number of years (or months) that each policyholder is observed, and there is one observation per year For policyholder in year the claim amount (or claim number, if that is what 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 claim amount 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 of individuals ( ) and all are observed for 1 ( ) period This can be considered the "equal sample size" case Then for each we have, and (710) The unbiased estimates of the structural parameters that we will use (there are other unbiased estimates) are - the estimated prior mean:, (711) - the estimated expected process variance: ( where ), and (712) - the estimated variance of the hypothetical means: (713)

165 CR-162 CREDIBILITY SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS (The derivations that show that the following estimates are unbiased can be found in the "Loss Models book) If, then we set The credibility factors will be equal for all groups since for all The estimated credibility factor for each is (714) 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 Assume that the two groups have the same number of insureds Use Buhlmann's model with empirical Bayesian estimation to estimate the credibility premium for each of the two groups during the next (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 and Since, 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 ( )

166 CREDIBILITY SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS CR-163 CR-73 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 of this 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, (715) (716) As in the Buhlmann model, if, then set The estimated credibility factor for group is (716) Note that the factors are not necessarily equal, since may vary from one policyholder to another An alternative to the unbiased estimate just described for the Buhlmann- Straub model is the method that preserves total losses (also called the 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 past exposures, that total will equal to the actual total past claims There have been several questions on older exams that referred to the "method that preserves total losses" It appears that this method has not been covered on recent exams, but it is part of the material covered in the "Loss Models" book according to the exam catalog Example CR-34: Compute the estimated credibility premium for the fourth policy year for each of the two groups of insureds whose claims experience for the first three year is presented in the following table Group Policy Year Policyholder Aggregate Claim Amount 8,000 11,000 15,000-1 Size of Group Aggregate Claim Amount 20,000 24,000 19,000-2 Size of Group

167 CR-164 CREDIBILITY SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS Solution: There are policyholders, with exposure periods for each policyholder (group) For policyholder 1, so that, and for policyholder 2 so that The average claim amounts per exposure unit for each of the groups and each of the exposure periods 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 two policyholders are The credibility premiums for the two policyholders per exposure unit for the fourth year are for policyholder 1, and for 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

168 CREDIBILITY SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS CR-165 Example CR-34 continued Using this value of, for policyholder 1, the credibility premium per unit of exposure is and for policyholder 2, the credibility premium per unit 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 (97,0015 and 97,000) are almost the same (they differ because 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 in the 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 manual premium), then an alternative unbiased estimate for is, and is the same as in (i) above (717) (iii) if is known (again, possibly the manual premium) and if there is data available for only policyholder (although there are other policyholders), then the following estimates can be used for and : and (718)

169 CR-166 CREDIBILITY SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS CR-74 Semiparametric Empirical Bayesian Credibility The model for the portfolio may have a parametric distribution for given, but an unspecified non-parametric distribution for In this case, we may be able to use the relationships linking and and the fact that in order to get estimates for and to use in the credibility premium formulation We will consider the case in which represents the number of claims for a period of time (say one month) and the conditional distribution of claim number given is Poisson with parameter 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 selected member of the population The 's each come from population members with different values of We wish to focus on one particular member of the population, and apply credibility estimation to estimate 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 are observations that range over all possible values of in the population, but the 's are observations 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 in the 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 the sample mean of the observations for the particular individual will have a similar form to the usual 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 we can 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 We first use the data set of the -values,, to find the (unbiased) sample variance of the 's and 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

170 CREDIBILITY SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS CR-167 Example CR-35: A group of 500 insureds submitted the following 290 claims during a one year period of observation Number of Claims Number of Insureds For each individual, the number of claims is assumed to follow a Poisson distribution, but the mean of each distribution may vary among insureds Suppose that an individual from this group of insureds has experienced two claims during the one year period Using semi-parametric estimation, find the Buhlmann credibility estimate of the expected number of claims in the next year for this insured Suppose that the insured had 4 claims in a two-year period Find the semiparametric 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 the conditional 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 is the 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 size of 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

171 CR-168 CREDIBILITY SECTION 7 - EMPIRICAL BAYES CREDIBILITY METHODS Example CR-35 continued If the insured has 4 claims in two years, there would be observations (number of claims in the first 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 500 observations) 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 and variance of the conditional distribution of given are the same), the unconditional distribution of is not Poisson and will likely have mean and variance not equal That is why we estimate the variance of the unconditional as well as estimating the mean of the unconditional The crucial point in applying semiparametric empirical Bayesian estimation when the conditional distribution 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

172 CREDIBILITY - PROBLEM SET 7 CR-169 CREDIBILITY - PROBLEM SET 7 Empirical 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 are summarized in the table below Assume that the two groups have the same number of insureds Aggregate Claim Amounts Group Policy Year Policy Find the Buhlmann credibility premium for group 1 for the fourth year A) 80 B) 82 C) 84 D) 86 E) 88 2 Using the credibility-weighted average estimate of (also referred to as the method that preserves total losses), find the Buhlmann credibility premium for group 1 for the fourth year A) 80 B) 82 C) 84 D) 86 E) 88 3 Suppose that the aggregate claim amounts for group 2 for policy years 1, 2 and 3 are 2, 8 and 14 (instead of 11, 13 and 12) Find the estimated Buhlmann credibility premium for group 1 for the fourth year A) 80 B) 82 C) 84 D) 86 E) 88 Problems 4 and 5 refer to the data of Example CR-34 in the notes, with the following modification Assume that there is a third group with the following experience for the three years: Policy Year Aggregate Claim 10,000 15,000 13,500 Size of Group 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) Using the credibility-weighted average estimate of, find the Buhlmann credibility premium per unit of exposure for group 1 for the fourth year (nearest 1) A) 203 B) 205 C) 207 D) 209 E) 211

173 CR-170 CREDIBILITY - PROBLEM SET 7 Problems 6 and 7 refer to the data of Example CR-34 in the notes, with the following modification For group 1, assume that there is no data for the first policy year (but all three years 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) Using the credibility-weighted average estimate of, find the Buhlmann credibility premium per unit of exposure for group 1 for the fourth year (nearest 1) A) 206 B) 208 C) 210 D) 212 E) For a large sample of insureds, the observed relative frequency of claims during an observation period is as follows: Number Relative Frequency of Claims of Claims 0 619% 1 284% 2 78% 3 16% 4 3% 5 or more 0 Assume that for a randomly chosen insured, the underlying conditional distribution of number of claims per period given the parameter is Poisson with parameter Given an individual who had claims in the observation period, use semiparametric empirical Bayesian estimation to find expected number claims that the individual will have in the next period 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 or more 0 - The number of losses for each insured follows a Poisson distribution, but the mean of each distribution may be different for individual insureds Determine the Buhlmann credibility of the experience of an individual insured over a single period

174 CREDIBILITY - PROBLEM SET 7 CR You are given the following table of data for three policyholders over a three year period Policy Year Policyholder 1 Number of Claims Average Claim Size Number of Claims Average Claim Size Number of Claims Average Claim Size Apply the nonparametric empirical Bayes credibility method to find the credibility premium per claim in the 4th year for Policyholder 2 using the standard method for (not the method that preserves 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 has an 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 loss from the policyholder for year You are given: ( ) 3360 and 2 ( ) Calculate the Buhlmann credibility factor for an individual policyholder using nonparametric empirical Bayes estimation (A) Less than 074 (B) At least 074, but less than 077 (C) At least 077, but less than 080 (D) At least 080, but less than 083 (E) At least 083

175 CR-172 CREDIBILITY - PROBLEM SET 7 13 (SOA) The number of claims a driver has during the year is assumed to be Poisson distributed with an unknown mean that varies by driver The experience for 100 drivers is as follows: Number of Claims during the Year Number of Drivers Determine the credibility of one year s experience for a single driver using semiparametric empirical Bayes estimation (A) 0046 (B) 0055 (C) 0061 (D) 0068 (E) (SOA) The following information comes from a study of robberies of convenience stores over the course of a year: (i) is the number of robberies of the store, with = 1, 2,, 500 (ii) 50 (iii) (iv) The number of robberies of a given store during the year is assumed to be Poisson distributed with an unknown mean that varies by store Determine the semiparametric empirical Bayes estimate of the expected number of robberies next year of a store that reported no robberies during the studied year (A) Less than 002 (B) At least 002, but less than 004 (C) At least 004, but less than 006 (D) At least 006, but less than 008 (E) At least Survival times are available for four insureds, two from Class A and two from Class B The two 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 for each class Unbiased estimators of the expected value of the process variance and the variance of the 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 Total Number of members Average loss per member Number of members Total Average loss per member Number of members 100 Average loss per member 109 _ 4800 Determine the nonparametric Empirical Bayes credibility premium for group 1, using the method that preserves total losses (A) 98 (B) 99 (C) 101 (D) 103 (E) 104

176 CREDIBILITY - PROBLEM SET 7 CR The number of claims per month for a given risk is assumed to be Poisson distributed with an unknown mean that varies by risk It is found that for a risk that has reported no claims for the past month, the semiparametric empirical Bayes estimate of the expected number of claims next month 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 month for 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 Policyholder 1 Number of Claims Average Claim Size Number of Claims Average Claim Size Number of Claims Average Claim Size Apply the nonparametric empirical Bayes credibility method to find the credibility premium per claim in the 4th year for Policyholder 2 19 A particular type of individual health insurance policy models the annual loss per policy as an exponential distribution with a mean that varies with individual insured A sample of 1000 randomly selected policies results in the following data regarding annual loss amounts in interval grouped 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 a particular individual who had annual policy losses of 150 in the first year and 0 in the second year

177 CR-174 CREDIBILITY - PROBLEM SET 7 CREDIBILITY - PROBLEM SET 7 SOLUTIONS 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 premium will 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

178 CREDIBILITY - PROBLEM SET 7 CR There are policyholders, with exposure periods for each policyholder 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 overall mean is The estimated variance of the hypothetical means is The estimated value of is, and the credibility factors for the three policyholders 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

179 CR-176 CREDIBILITY - PROBLEM SET 7 6 There are policyholders, with exposure periods for the two policyholders Exposure units are (second policy year) and (third policy year) 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 two policyholders 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 Poisson with 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

180 CREDIBILITY - PROBLEM SET 7 CR continued Since, we use the estimated variance of along with the estimate of to get an estimate of ; The estimate of is, and the estimated credibility factor for one individual is The credibility premium for the next period for an individual who 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 the estimate of This estimate is Then and 10 The credibility premium for policyholder 2 is

181 CR-178 CREDIBILITY - PROBLEM SET 7 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) per policyholder 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 may have a parametric distribution for given, but an unspecified non-parametric distribution for 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 to use in the credibility premium formulation

182 CREDIBILITY - PROBLEM SET 7 CR-179 In this problem (the typical example) the conditional distribution of claim number given is Poisson with parameter Then, so that We then use to estimate, and also use this as the estimate of We find the (unbiased) sample variance of the 's and set that equal to, so that 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 value of 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 single observation of (the number of robberies in one year) and, since there were no robberies in 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 factor for 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

183 CR-180 CREDIBILITY - PROBLEM SET 7 16 Since the numbers of exposures differs among exposure periods, we use non-parametric empirical Bayes estimation for the Buhlmann-Straub model Under the Buhlmann-Straub model, there are exposure periods for policy holder (group), for For policyholder and exposure 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 for group 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 the structural 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 is the 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 calculate total 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 empirical Bayes 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

184 CREDIBILITY - PROBLEM SET 7 CR 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 :

185 CR-182 CREDIBILITY - PROBLEM SET 7 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 gives is But we also know that, for this model,, so using our sample estimate of, 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

186 SIMULATION

187

188 SIMULATION SECTION 1 - THE INVERSE TRANSFORM METHOD SI-1 SIMULATION - SECTION 1, THE INVERSE TRANSFORM METHOD The material in this section relates to Section 171 of "Loss Models" The suggested time frame for this section is 2-3 hours The objective in performing a simulation is to reproduce the behavior of a random variable by generating observations from another random variable which has the same distribution as the random 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 tossing a 4, 5 or 6 on the die To see if this is a valid simulation we must check of the simulated events replicate the original probability distribution In this example, there is 5 chance of tossing a 1, 2 or 3, so the simulation replicates the correct 5 probability of getting a when a fair coin is tossed Same for Example SI-1: You are given a random number generator that produces sample observations for from the following probability density function: otherwise You use this random number generator to simulate the color of a traffic light facing a randomlyarriving car The light is green 36% of the time, yellow 13% of the time, and red 51% of the time The following table gives the correspondence between the values of and the color of the traffic light: Value of Color of light Green Yellow Red Determine Solution: In for the simulation to be valid it must reproduce the probabilities for the color of the traffic light In the simulating distribution, the probability of Green is This must be in the traffic light color distribution Thus, In a similar way, One of the general requirements for simulation is to have a way of obtaining independent uniform random numbers from the interval There are various ways of generating such random numbers, but we will simply assume that they will be available when needed

189 SI-2 SIMULATION SECTION 1 - THE INVERSE TRANSFORM METHOD SI -11 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 is guaranteed 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 the inverse transform method of simulation This procedure is sometimes referred to with the phrase small uniform random numbers correspond to small simulated values

190 SIMULATION SECTION 1 - THE INVERSE TRANSFORM METHOD SI-3 Example SI-3: Of a group of three independent lives with medical expense insurance, the number having medical expenses during a year is distributed according to a binomial distribution with and The amount,, of medical expenses for any person, once expenses occur, has the following distribution: Each year, the insurance company pays the total medical expenses for the group in excess of 5,000 Use the uniform random numbers, and, in the order given, to generate the number of claims for each of two years Use the following uniform random numbers, in the 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 medical expenses for the second year is 2 For the claim amount distribution, In the first year, there is one 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 the second year The first is simulated using the uniform number 95, and since, the simulated expense is 10,000 The second expense in the second year is simulated from the uniform number 7, and since, this simulated expense is 100 The total expense for the second year is 10,100 Each year the insurer pays total medical expenses in excess of 5000 The insurer pays 0 in the first year, since the total medical expense was 100, and the insurer pays in the second year

191 SI-4 SIMULATION SECTION 1 - THE INVERSE TRANSFORM METHOD SI-12 Simulation of a Continuous Random Variable The inverse transformation method for a continuous random variable Given the continuous random variable with cdf, a value of can be simulated from a random number from the uniform distribution on in the following way Solve the equation for ; the solution is the simulated value of The solution of the equation is sometimes written This is illustrated in the graph below Example SI-4: You are given a random variable with the following probability density if function: otherwise Let be a uniform random number between 0 and 1 Use the inverse transformation method to determine 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 inverse transformation method is the value of in the solution of, so that, from which 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 small random numbers correspond to small simulated values It is also possible the may see the phrase "small random numbers correspond to large simulated values If that is the case, then we solve for in the equation This will result in a different simulated numerical value, but it will be a statistically valid simulation

192 SIMULATION SECTION 1 - THE INVERSE TRANSFORM METHOD SI-5 Example SI-5: You are to use the inversion method to simulate three values of a random variable 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 the distribution For instance, since and, and since we know that is a linear function for it must be true that for (we can use the "twopoint" 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 In order to apply the inversion method, and solve for from, we must use the appropriate "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 the corresponding simulated -values

193 SI-6 SIMULATION SECTION 1 - THE INVERSE TRANSFORM METHOD SI-13 Simulation of Some Specific Random Variables Simulation of the exponential distribution The 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 numbers correspond 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 large simulated values, we would solve the equation, so that This would result in a different numerical simulated value, but it would be a statistically valid simulation (and would be slightly simpler to apply algebraically) Simulation of the Poisson distribution The relationship linking the exponential and Poisson distributions results in an alternative method for simulating the Poisson distribution If the time between successive events has an exponential distribution with mean, and successive event times are independent, then the number of events in a unit of time has a Poisson distribution with mean The Poisson with mean can be simulated 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 the cumulative product algorithm We first calculate We then multiply successive uniform numbers until the product is less than Apply the algorithm with each of the 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

194 SIMULATION SECTION 1 - THE INVERSE TRANSFORM METHOD SI-7 Simulation of the gamma distribution The gamma distribution with parameters and, where is an integer can be simulated in the following way (in the Exam C Tables, this distribution would be described as the gamma distribution with parameters and ) With being an integer, this gamma random 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 to as 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 with mean 1 True II Just because both the 's and the 's are valid simulated values from an exponential distribution with mean 1, there is no guarantee that the numerical values are the same For instance, if, there is no guarantee that False III The sum of independent exponential random variables all with same mean is a gamma random variable True Example SI-8: You have generated a random value from the uniform distribution on Use the Normal Distribution Table, applying linear interpolation for values not in the table 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 can standardize,, where has a standard normal distribution ( ) From the normal table, and Using linear interpolation,

195 SI-8 SIMULATION SECTION 1 - THE INVERSE TRANSFORM METHOD SI-14 Simulation of a Mixed Distribution If a distribution is partly continuous and partly discrete, then the cdf increases continuously on the continuous region of, and there are discrete jumps in at the points of probability of, with the size of the jump being the amount of probability at the point Given uniform number, if is in a region where is continuous, then we solve to get the simulated value of If is inside a "jump" interval, then the simulated value of is the discrete 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 the inversion method to the three observations from the uniform distribution on The observations 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-15 Using Simulation to Estimate a Mean or a Probability Suppose that we are simulating a random variable which has mean and variance Suppose we 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 in limited fluctuation credibility We saw that the number of (simulated) observations of needed is We can change the " " factor to " ", and if we do so, then we change 05 to 01 in the denominator of the right side of the inequality We can change the probability to 95 from 9, and if we do so, then we change 1645 to 196 on the right side of the inequality We usually do not know or, so as the successive values of are simulated, we calculate 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 unknown probability of success for an experiment Our objective is to estimate to within 1% of the actual #successes in simulated trials value of with a probability of 95% We will estimate with, which is the proportion of the simulated experiments out of that are successes We want With a relatively large, is approximately normal, so we can apply a similar approach as in the previous paragraph This results in the following inequality could be any probability, for instance, in which case, #simulated 's that are in simulated trials

196 SIMULATION - PROBLEM SET SI-9 SIMULATION - PROBLEM SET 1 if 1 The random variable X has probability density function if 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) A binomial distribution with n 3 and p 4 is simulated by the inverse transform method with the uniform random numbers 31, 71, 66, 48, 19 How many of the generated random variables 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 function x f(x) (1 p) e for x 0 The distribution can be simulated by the inverse transform method as follows: if U is a random uniform (0,1) value and U p, then X 0 ; if p U 1 then X 1 1 p 1 1 p 1 U A) ln(1 U) B) ln(1 U) C) ln 1 U D) ln 1 U E) ln 1p sin x 4 X has density function f(x) = 2, 0 x, and f(x) = 0 elsewhere Use the inverse transform method to simulate three values of X based on uniform [0, 1] values u 1 =, u 2 = 5, u 3 = Find the mean of the simulated values x 1, x 2, x A) 36 B) 24 C) 36 D) 24 E) 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 this distribution by using, a uniform random number between 0 and 1 Determine the value of A) B) C) D) E) from

197 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, and obtain 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 is distributed as follows: Number of suits: Probability: For each of three years, you simulate the number of suits per year by generating one uniform random number from the interval and then applying the inverse transformation method Your three random numbers are 083, 059, 019 The liability loss from an individual suit is a random variable with the following cumulative for distribution function for for You generate uniform random numbers from the interval and apply the inverse transformation method to simulate the amounts of the individual losses The first nine uniform random numbers, in order, are: 051, 001, 078, 074, 003, 069, 017, 086, 082 Use, in order, as many of these random numbers as needed to simulate the company's total loss due to liability suits over the three year period A) 00 B) 02 C) 56 D) 58 E) 144

198 SIMULATION - PROBLEM SET SI-11 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 with mean 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 05, 02, and 01, respectively, from the uniform distribution on, where small random numbers correspond to long interarrival times Of the following, which is the smallest such that your company does not become insolvent from any of these three claims? A) 22 B) 35 C) 49 D) 113 E) When generating random variables, it is important to consider how much time it takes to complete 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

199 SI-12 SIMULATION - PROBLEM SET 1 Problems 11 and 12 relate to the following situation A random sample _ of size 30 is taken from the 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 less than 2 How many samples values after the 30th are needed to satisfy this criterion? A) 2 B) 4 C) 6 D) 8 E) It is decided that sampling will stop when the width of the 95% confidence interval for is first less than (the 975 percentile of the standard normal distribution is 196) Ho many sample values after the 30th are needed to satisfy this criterion? A) 2 B) 4 C) 6 D) 8 E) An estimate is being made of the probability of flipping a head with a particular coin The coin 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 are T 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) To estimate, you have simulated and with the following results: You want the standard deviation of the estimator of to be less than 005 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 A random sample of 18 data points has a sample mean of 8 and an unbiased sample variance of 4 and are added to the sample Find the updated unbiased sample variance based on all 20 data points A) 40 B) 41 C) 42 D) 43 E) (CAS) A scientist perform experiments, each with a 60% success rate Let represent the number of trials until the first success Use the inverse transform method to simulate the random variable,, and the following random numbers (where low numbers correspond to a high number of trials): 015, 062, 037, 078 Generate the total number of trials until three successes result A) 3 B) 4 C) 5 D) 6 E) 7

200 SIMULATION - PROBLEM SET SI (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 This process of selecting a driver and simulating one year is repeated times In the second actuary's simulation, 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 the second 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 the second 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's sequence 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 only D) 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 distributed with 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: Calculate the insurer s simulated claim cost (A) 13,400 (B) 14,400 (C) 17,800 (D) 20,000 (E) 26, (SOA) You are simulating a continuous surplus process, where claims occur according to a Poisson 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 01 Premium is collected continuously, and the process terminates if surplus is ever negative You simulate the time between claims using the inverse transform method (where small numbers correspond to small times between claims) using the following values from the uniform distribution on [0,1]: 083, 054, 048, 014 You simulate the severities of the claims using the inverse transform method (where small numbers correspond to small claim sizes) using the following values from the uniform distribution on [0,1]: 089, 036, 070, 061 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]

201 SI-14 SIMULATION - PROBLEM SET 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% of the coins are worth 1 each; 20% are worth 5 each; 20% are worth 10 each The denominations of 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 of repeatedly simulating the time until the next coin is found, until the length of the walk has been exceeded For each coin found, he simulates its denomination, using the inverse transform algorithm The expected number of random numbers he needs for one simulation of the walk is (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, then simulates finding coins worth 10, in each case simulating until the 60 minutes are exceeded The expected 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 of random numbers (E) Both methods are valid, and will produce identical results from the same sequence of random numbers 21 Determine which of the following ranges contains the ratio / (A) 00 / 04 (B) 04 / 08 (C) 08 / 12 (D) 12 / 16 (E) 16 / 22 (SOA) You wish to simulate a value,, from a two point mixture With probability 03, is exponentially distributed with mean 05 With probability 07, 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 025 and 069 in that order Calculate the simulated value of Y (A) 019 (B) 038 (C) 059 (D) 077 (E) 095

202 SIMULATION - PROBLEM SET SI (SOA) An actuary is evaluating two methods for simulating, the future lifetime of the joint-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 uniform distribution 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 18 (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 07, 01, 03, 01, 09, 05, 05, 07, 03, and 01 Calculate the aggregate claim amount associated with your third simulated value of (A) 3 (B) 5 (C) 7 (D) 9 (E) 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 and a mixing weight of 75 is simulated using the inverse transformation method with a uniform value of 7 Find the simulated value of

203 SI-16 SIMULATION - PROBLEM SET 1 SIMULATION - PROBLEM SET 1 SOLUTIONS 1 F(x) = 0 if x 0, F(x) = 1 if x 3 x if 0 x 2 4, and F(x) = x 2 if 2 x 4 1 r = 716 = 2x 2 x = 608 Answer: B 1 2 s : (s) (s) A binomial value of 2 will be simulated by a uniform (0,1) value that is both greater than or equal to 648 and less than 936 Thus, the uniform numbers 71 and 66 result in simulated binomial values of 2 Answer: B 3 For x 0, F(x) = P[ X x ] = P[ X = 0 ] + P[ 0 X x] = p + t (1 p) e dt if x = p + (1 p)(1 e ) Thus, F(x) = if if Thus, for U p, the inversion method implies that x 1 1p U = F(x) = p + (1 p)(1 e ) x = ln 1 U Answer: D x 4 The cdf is F(x) = sin t 1 cos x 2 dt = 2 for 0 x 1 1 cos x By the inversion method, for uniform u we have x = F (u), or equivalently, u = cos x1 1 u 1= = cos x 1 = x 1 =, u 2 = 5 cos x 2 = 0 x 2 = 2, x 1 + x 2 + x3 19 u 3 = 4 cos x 3 = 2 x 3 = 6 3 = 36 Answer: A 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 35) Answer: D 6 We solve for, Answer: A

204 SIMULATION - PROBLEM SET SI-17 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 We are told that where small random numbers correspond to long interarrival times This is the opposite of the usual inverse transform method Instead of solving for interarrival claim time from (the simulation method that has small random number corresponding to small interarrival time ), we solve for from the equation Since interarrival time for claims 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, and 1535 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 1535 is Surplus at time each of the claim times is: - at time 231 surplus is, - at time 767 surplus is, and - at time 1535 surplus is In order for the company to be solvent at each claim time it must be true that, and and Answer: C

205 SI-18 SIMULATION - PROBLEM SET 1 10 In order to generate values as quickly as possible, we want the -value with the largest probability to be the first possible one generated, and second largest probability to be the second possible 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, and it would generate if The reason this method results in the least time required for simulation is that the probability that the 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 are A 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 sooner than any of the other four methods, but never later Note that all five answers result in a valid simulation method since they reproduce the probabilities 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

206 SIMULATION - PROBLEM SET SI 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 is Continuing with the calculations in Problem 9, interval width is, interval width is We see that the stopping rule is reached at the 36-th data point Answer: C if head number of heads in flips 13 if tail 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

207 SI-20 SIMULATION - PROBLEM SET 1 16 To say that low numbers correspond to a low number of trials is the standard form of the inverse transform method: given a uniform random number we find the integer such that, and the simulated value of is If we apply the inverse transform method in the form where low numbers correspond to a high number of trials, then given a uniform random number, we find the integer such that The random variable is the number of success until the first trial This is a geometric distribution 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 third success) 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, the simulated 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 of claims simulated by actuary 1 in trials is The second actuary selects a driver with Poisson 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 binomial distribution For actuary 2, the variance of the sequence generated is the variance of the Poisson distribution with parameter (the for the driver chosen by actuary 2) Either variance could be larger than the other depending on the value of for actuary 2's driver False Answer: E 18 According to the inverse transform method, given uniform [0,1] number, the simulated standard 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

208 SIMULATION - PROBLEM SET SI 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 "small numbers correspond to small times between claims"; this is the standard form of the inverse transform 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 aggregate claim 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 number of simulations in total is Actuary 2: Denomination 1 coins are found at the rate of per minute, so that in one hour, 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 values correspond to the exponential distribution when simulating the mixing variable Therefore the exponential distribution is used if the uniform random number for simulating the mixing variable is Since the first uniform random number is, we use the exponential distribution 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 the exponential 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

209 SI-22 SIMULATION - PROBLEM SET 1 23 Under DeMovire's Law, the future lifetime of someone at age is uniformly distributed on the interval Therefore Method 1 is a valid simulation of both times of death The two individuals 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 longer uniform (same for ) Therefore Method 2 is never valid Answer: D 24 The distribution of binomial with mean, so that is the probability of a claim 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 simulates a claim amount of 4 Aggregate claim amount simulated associated with the 3rd simulated value of 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

210 SIMULATION SECTION 2 - THE BOOTSTRAP METHOD SI-23 SIMULATION - SECTION 2, THE BOOTSTRAP METHOD AND STATISTICAL ANALYSIS USING SIMULATION The material in this section relates to Section 172 of "Loss Models" The suggested time frame for this section is 3 hours SI-21 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 we generally regard as a random variable The mean square error of is defined to be 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 estimator is generally calculated from a random sample of observations from the underlying distribution of the random variable The bootstrap estimate is calculated by first constructing the empirical distribution 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 C bootstrap 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 3 and 7 The mean of the empirical distribution is The estimator of the mean that we are using is The bootstrap method is based on resampling from the original sample, with the new samples having the same size as the original sample Since the original sample size was 2, we choose samples 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 the estimator based on that sample, and we calculate the squared deviation of the estimator from the quantity being estimated, which is 5, the actual mean of the empirical distribution This is summarized 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 of the squared deviations in the last column, which is

211 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 actually was the distribution of Our original objective was to find the mean square error of the sample mean estimator based on a sample of size 2 for the original random variable We don't know anything more about the distribution of than the sample that we have taken, so we use that sample as a substitute for the distribution of and we work toward our objective again Our objective is to find the mean square error of the sample mean estimator for a sample of size 2, but now we are using the empirical distribution This means that we take samples of size 2 from the empirical distribution Since the empirical distribution is a 2-point discrete random variable, we are taking samples of size 2 from that 2-point distribution As was seen in Example SI-10, there are only different samples of size 2 For each of them, we calculate the squared deviation of the estimated mean from the "true" mean What "true" mean refers to now is not the mean of the original random variable (we don't know that value); it refers to the mean of the empirical distribution (which was 5 in Example SI-10) In a more realistic situation, the sample size is (a lot) larger than 2, and complete resampling can become impossible For instance, with a sample of size 10, there would be possible samples of size 10 from the original sample of size 10, because each in a resampling can be any one of these original 10, with repeats For a sample of size, there would be ways of taking a sample of size from the original sample The way in which we can deal with the impossibly large number of resamples that can be taken from the original sample is to just take some resamples from the original sample If the sample size 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 squared deviation of that estimate from the quantity in the empirical distribution being estimated, and then average 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 of a random variable with unknown mean and unknown variance (the sample values are given in 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, 57 Formulate the empirical distribution, and simulate 5 bootstrap samples of size 20 from the empirical distribution Solution: With, the probability function and distribution function of the empirical distribution are

212 SIMULATION SECTION 2 - THE BOOTSTRAP METHOD SI-25 Example SI-11 continued Notice 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 a sample of 20 's simulated from this empirical distribution just described (the connection to simulation 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 distribution probabilities A table of random uniform numbers, or a computer routine for generating random uniform numbers can be used along with the inversion method for simulating a discrete distribution (the empirical distribution in this case is a discrete 15 point distribution) The following 5 bootstrap samples were simulated in that way (again the sampled values have been placed 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, 57 Sample 2: 5, 6, 8, 9, 9, 12, 14, 14, 17, 17, 17, 18, 23, 27, 31, 31, 44, 57, 57, 57 Sample 3: 5, 5, 6, 6, 9, 9, 9, 9, 9, 9, 17, 18, 27, 27, 27, 27, 36, 44, 44, 57 Sample 4: 4, 5, 5, 5, 5, 5, 6, 8, 9, 9, 9, 12, 12, 14, 17, 18, 23, 31, 36, 44 Sample 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 quantity related to the distribution of Our objective is then to estimate the mean square error of the estimator If is the quantity being estimated and is the estimate, our objective is to find the mean 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 bootstrap approximation to the mean square error of the estimator comes in We do know the values of the mean, median, variance,, etc, for the empirical distribution Instead of finding in the original distribution of (which is impossible, since we don't know the original distribution of ), we try to find in the empirical distribution (which is possible, since we know the empirical distribution)

213 SI-26 SIMULATION SECTION 2 - THE BOOTSTRAP METHOD In the bootstrap approximation, now refers to the corresponding quantity in the empirical distribution This is what was done in Example SI-10 We had a sample of size 2,, and we were estimating the mean of, so is the mean of We don't know the mean of, but we do know the mean of the (2-point) empirical distribution, it is 5 We are using the sample mean estimator to estimate the distribution mean, so refers to the sample mean, and we are trying to find in the empirical distribution Here is where we have to be careful about understanding the application of the bootstrap approximation For the 2-point empirical distribution of SI-10, we look at all possible samples of size that can be drawn (with replacement) As pointed out after Example SI-10, there are 4 possible "resamplings" of size 2 Each of them has the same chance of being drawn For each of those samples we calculate, the sample mean for that sample, and the calculate the squared deviation for that sample We then 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 1711 in the Loss Models 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 to find 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, and then for each resample considered, we would calculate the numerical value of, and then the value of (using the numerical values just mentioned) The bootstrap estimate of the mean square error of the estimator would be the average of those values of for the resamplings that were taken The following continuation of Example SI-11 illustrates how we find the bootstrap estimate if the original 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 number of 's (c) the following estimator of, Solution: (a) The parameter being estimated is the mean of the distribution of, so that in this case, 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, we calculate the sample mean of that sample and then calculate the squared deviation from 1845 Sample Estimator Based on the Sample Squared Deviation of Estimator From Number Quantity Being Estimated ( )

214 SIMULATION SECTION 2 - THE BOOTSTRAP METHOD SI-27 Example SI-11 continued The bootstrap estimate of the mean square error of the sample mean estimator is the average of the 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 to its variance, so it is reasonable that the bootstrap estimate of the MSE of the sample mean estimator 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 5 resamplings, we calculate the unbiased sample variance of that sample and then calculate the squared deviation from Sample Estimator Based on the Sample Squared Deviation of Estimator From Number Quantity Being Estimated ( ) Note that for each sample we calculate the unbiased sample variance for that sample, so that for sample 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 of the squared deviations in the last column, which is number of 's (c) The quantity being estimated now is The sample estimator is Since 8 out of the original 20 sample points are, it follows that in the empirical distribution the probability being estimated is ; this is For each of the 5 bootstrap samples, we calculate, 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

215 SI-28 SIMULATION SECTION 2 - THE BOOTSTRAP METHOD Example SI-11 continued Sample Estimator Based on the Sample Squared Deviation of Estimator From Number Quantity Being Estimated ( ) 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 is number of 's, where can be regarded as having a binomial distribution with parameters 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-22 Applications of Simulation to Statistical Testing The material covered in this section is mentioned briefly in Section 172 of the Loss Models book There haven't been any exam questions on this material on the released exams 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 an alternative 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 which the random sample has been drawn may be a statement about the value of a parameter or some parameters in a distribution, or it may be a statement that the sample has been drawn from a particular 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 random sample of data to calculate a test statistic The test statistic itself is a random variable with a probability 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 wellknown 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 of significance is usually set to be 05 (5% level of significance) or 01 (1%) Once the level of significance is chosen, the critical value for the test is determined The critical value is usually the 95-th percentile (for 5% significance test) of the distribution of the test statistic (this requires reference to a -distribution table, or a chi-square table, etc); if a test is a two-sided test,

216 SIMULATION SECTION 2 - THE BOOTSTRAP METHOD SI-29 the critical value would be the 975 percentile The following summarizes some of the main tests that have been considered earlier (i) Two-sided normal test In this test, the test statistic is assumed to be normal, so there is no reference 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 of the test statistic for the given data set is, then reject if It can happen that even though 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, then the probability of making a Type I error is is true Once the level of significance desired for the test is chosen, say 05, the critical value is the number which satisfies the probability relationship is true Assuming that is true, the distribution of may be known (possibly we will know the approximate distribution of ), and from the probability table for the distribution of we can find (in this case is the 95-th percentile of ) It follows that the level of significance of a hypothesis 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 does not mean the that test becomes more accurate; it means that the chance of making a Type I error has been reduced, but the chance of making a Type II error will be increased A Type II error occurs if is false but the mechanical application of the test given the data results in acceptance of Another way of looking at the hypothesis test it to consider the -value of the observed test statistic The -value is a probability defined in the following way: assume that is true and that 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 of freedom 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 value would be and (1692 is the 95-th percentile of the chisquare distribution with 9 degrees of freedom) would not be rejected in a 1% test, because in that case the critical value is 2167

217 SI-30 SIMULATION SECTION 2 - THE BOOTSTRAP METHOD An equivalent way of determining whether or not is rejected can be set up using the -value of ; 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 25%) but would not be rejected at 1% ( would not be rejected at any level of significance at or below 25%) 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 the specified distribution of, the distribution of is usually only approximately chi-square (with the approximation becoming more exact as the sample size increases, so that is asymptotically chi-square) It may be possible to simulate a more accurate -value of by simulating the actual distribution of rather than using the approximate chi-square distribution for We review some of the standard statistical tests and how simulation can be applied to obtain a -value The Chi-Square Goodness-of-Fit Test This 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 with probability function Suppose that we have observations, and each observation is one of the numbers, with being the number of observations that are equal to We want to test whether or not the observed values behave as if they have been drawn 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 significance level 01) The interpretation is that if the -value of the test statistic is small then it is unlikely that the observations were drawn from the distribution of The typical approximation applied to determine the -value is to assume that has a chi-square distribution with degrees of freedom, 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 accurate estimate of the -value of by simulation We can simulate the distribution of and get an estimate of the -value of as follows:

218 SIMULATION SECTION 2 - THE BOOTSTRAP METHOD SI-31 (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 number of 's The simulated -value of is As gets larger this proportion approaches the true -value The Kolmogorov-Smirnov Test This test was also discussed earlier in the notes for estimation and testing of loss and survival models 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 a random sample of observations Our objective is determine whether or not the sample 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 data sample The empirical distribution is constructed in the same way as described earlier in these notes, but now we focus on the distribution function of the empirical distribution The number of 's distribution function of the empirical distribution is The Kolmogorov-Smirnov test looks at how closely fits, the distribution function of If the fit is "close enough", then will be accepted Recall that the empirical distribution function is a step-function The Kolmogorov-Smirnov test statistic is the "maximum distance" that is away from over the range of the distribution of This is found by first calculating the following 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, and we find the absolute difference between the hypothesized distribution function of and both the lower and upper jump points in the empirical distribution function is the maximum deviation of 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 Critical Value If the numerical value of the test statistic is, then the Kolmogorov-Smirnov test with a 5% level of 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

219 SI-32 SIMULATION SECTION 2 - THE BOOTSTRAP METHOD As an alternative to performing the test with the table of critical values given above, this test can be formulated in terms of the -value of If the -value of is less than 05 then is rejected at the 5% level of significance The -value of can be estimated by simulation This can be done 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 number of 's The simulated -value of is As gets larger this proportion approaches the 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 is partitioned into pieces If the distribution of is discrete and integer-valued with, then the first of the " pieces" could be the integers and the -th piece is the set 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 test statistic 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 are used 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 a particular 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 new estimates of the parameters of and new estimates of the values; find the new values for the using the new simulated 's in this 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 is number of 's

220 SIMULATION SECTION 2 - THE BOOTSTRAP METHOD SI-33 The Kolmogorov-Smirnov Test When There are Unknown Parameters If a distribution has unknown parameters and a random sample of size is available for the distribution, then the parameters can be estimated from the data to get, the distribution function of based on the parameter estimates The empirical distribution function based on the random sample is denoted by (as before) We calculate the Kolmogorov-Smirnov test statistic value as before:, where the 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 estimated parameters, and denote the empirical distribution function of these values sim (ii) using the new simulated values of from (i), find new estimates of the distribution parameters, and denote by (sim) the distribution function of based on these new estimated parameters, (iii) the simulated value of the K-S statistic is,sim, where (sim) this maximum is taken over the simulated -values in (i), number of 's (iv) repeat steps (i) to (iii) times; the simulated -value of is The last couple of sections are somewhat more detailed than what is presented in the Loss Models book on this topic

221 SI-34 SIMULATION SECTION 2 - THE BOOTSTRAP METHOD

222 SIMULATION - PROBLEM SET 2 SI-35 SIMULATION - PROBLEM SET 2 Problems 1 to 3 refer the following random sample of 15 data points: 80, 51, 22, 86, 45, 56, 81, 64, 33, 73, 80, 40, 65, 63, 91 The following three bootstrap samples of the empirical distribution have been simulated Each sample is of size 15 Sample 1: 33, 33, 33, 40, 40, 40, 65, 73, 81, 86, 88, 91, 91, 91, 91 Sample 2: 33, 33, 51, 51, 51, 56, 63, 64, 65, 73, 73, 80, 80, 80, 81 Sample 3: 22, 33, 33, 40, 45, 45, 51, 65, 73, 80, 81, 81, 81, 91, 91 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 data points 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) The mean of a distribution is being estimated, using the sample mean of a random sample as an estimator A sample of size 2 is drawn: is the mean square error of the estimator when the empirical distribution is used Find the exact value of A) 125 B) 250 C) 375 D) 500 E) With the bootstrapping technique, the underlying distribution function is estimated by which of 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 Determine the bootstrap approximation to the mean square error (A) 00 (B) 05 (C) 10 (D) 20 (E) 25

223 SI-36 SIMULATION - PROBLEM SET 2 7 You are given a random sample of two values from a distribution function : You estimate ( ) using the estimator (, ) ( ), where Determine the bootstrap approximation to the mean square error (A) 00 (B) 05 (C) 10 (D) 20 (E) 25 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 6 resamplings 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 sample of the number of claims occurring on 5 randomly chosen days: The following estimator from a sample of days is used to estimate, the probability # days with 4 claims or less of 4 or less claims in one day: The bootstrap approximation is applied to estimate the mean square error of using the following 8 samples simulated from the empirical distribution of the original sample: Sample Sample Sample Sample Sample Sample Sample Sample Find the bootstrap approximation to the mean square error of the estimator

224 SIMULATION - PROBLEM SET 2 SI-37 SIMULATION - PROBLEM SET 2 SOLUTIONS 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 are the proportions in each bootstrap sample of the points that are These are 6 6 and Then, the estimated MSE is [ Answer: A 4 The parameter being estimated is the distribution mean, and the mean of the empirical distribution 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 that the sample is used to estimate some parameter of the distribution If the estimator is, then the bootstrap 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, and the expected value is taken within the empirical distribution In this case, the parameter being estimated is the distribution variance The mean of the empirical distribution is, and the variance of the empirical distribution is

225 SI-38 SIMULATION - PROBLEM SET 2 6 continued To 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 2 ( ) 2 and For each sample, we must calculate 1 2 for that sample For instance, for the sample, we have, and 2 ( 2 ) 1 The bootstrap approximation to the mean square error of the estimator is the average of the 4 2 values of ( ) 2 that we get This is summarized in the following table 1 Sample Since each of the four possible samples is equally likely to occur, the bootstrap estimate of Answer: B 1 7 Since the sample consists of only 2 sample points, there are only 4 possible, equally likely bootstrap 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 ( ) 2 for each of the samples: (1) (2) (3) (4) The bootstrap estimate is the average of the values of for the samples: Answer: C 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

226 SIMULATION - PROBLEM SET 2 SI-39 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 Sample Sample Sample Sample Sample Sample Sample The average value of is ; this is the bootstrap approximation to the mean square error of based on the 8 samples

227 SI-40 SIMULATION - PROBLEM SET 2

228 SIMULATION SECTION 3 - LOGNORMAL MODEL FOR STOCK PRICES SI-41 SIMULATION - SECTION 3 THE LOGNORMAL DISTRIBUTION APPLIED TO ASSET PRICES The material in this section relates to Chapter 18 of "Derivatives Markets" The suggested time frame for this section is 2 hours SI-31 Review of the Normal Distribution The normal distribution was reviewed in Section 3 of the Modeling unit of this study guide The standard normal distribution has a mean of 0 and a variance of 1 is notation used to indicate that has a standard normal distribution The conventional notation for the pdf of the standard normal is, (31) and the region of density is Conventional notation for the cdf of standard normal 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 the Modeling 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

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