PRACTICE EXAMS

PRACTICE EXAM 1

PE-1

ACTEX EXAM C/4 - PRACTICE EXAM 1 1. ? has a Weibull distribution with parameters  and . Find the density function ²'³ of the  random variable A ~  c c²?° ³ . A)  for   '  B)  for   '   C) c' for ' €   D)  c'° for ' € 

E)  for   '  

2. A portfolio of risks models the annual loss of an individual risk as having an exponential distribution with a mean of $. For a randomly selected risk from the portfolio, the value of $ has an inverse gamma distribution with a mean of 40 and a standard deviation of 20. For a randomly chosen risk, find the probability that the annual loss for that risk is greater than 20. A) .524 B) .544 C) .564 D) .584 E) .604

3. You are given the following: - Losses follow a distribution (prior to the application of any deductible) with mean 2000. - The loss elimination ratio (LER) at a deductible of 1000 is 0.3. - 60 percent of the losses (in number) are less than the deductible of 1000. Determine the average size of a loss that is less than the deductible of 1000. A) Less than 300 B) At least 300 nut less than 320 C) At least 320 but less than 340 D) At least 340 but less than 360 E) At least 360

4. A casino has a game that makes payouts at a Poisson rate of 5 per hour and the payout amounts are 1,2,3,... without limit. The probability that any given payout is equal to  is  . Payouts are independent. Calculate the probability that there are no payouts of 1, 2, or 3 in a given 20 minute period. A) 0.08 B) 0.13 C) 0.18 D) 0.23 E) 0.28

5. Zoom Buy Tire Store, a nationwide chain of retail tire stores, sells 2,000,000 tires per year of various sizes and models. Zoom Buy offers the following road hazard warranty: "If a tire sold by us is irreparably damaged in the first year after purchase, we'll replace it free, regardless of the cause." The average annual cost of honoring this warranty is $10,000,000, with a standard deviation of $40,000. Individual claim counts follow a binomial distribution, and the average cost to replace a tire is $100. All tires are equally likely to fail in the first year, and tire failures are independent. Calculate the standard deviation of the placement cost per tire. A) Less than $60 B) At least $60, but less than $65 C) At least $65, but less than $70 D) At least $70, but less than $75 E) At least $75

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

PE-2

PRACTICE EXAM 1

6. You are given the following random sample: 7 , 12 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53 Determine the method of percentile matching estimate of T Ò\  $!Ó using the 25-th and " 75-th smoothed empirical percentiles for a distribution with cdf J ÐBÑ œ "  "Ð B Ñα . )

A) Less than .30 B) At least .30 but less than .32 D) At least .34 but less than .36 E) At least .36

C) At least .32 but less than .34

7. Suppose a 3-year data set is divided into a year-by-year count of new entrants, deaths and right-censored observations: .! œ "!!! ß B! œ #! ß ?! œ $! ß ." œ #!! ß B" œ "! ß ?" œ #! ß .# œ #!! ß B# œ "& ß ?# œ $! Þ E is the estimate of WÐ#Ñ using the approximation for large data sets if α œ " and " œ ", and F is the estimate of WÐ#Ñ using the approximation for large data sets if α œ Þ& and " œ !. Find EÎF . A) Less than 1.000 B) At least 1.000 but less than 1.025 C) At least 1.025 but less than 1.050 D) At least 1.050 but less than 1.075 E) At least 1.075

8. Claim sizes of 10 or greater are described by a single parameter Pareto distribution, with parameter α. A sample of claim sizes is as follows: 10 12 14 18 21 25 Calculate the method of moments estimate for α for this sample. A) Less than 2.0 B) At least 2.0, but less than 2.1 C) At least 2.1, but less than 2.2 D) At least 2.2, but less than 2.3 E) At least 2.3

9. Let \1 , \2 , \3 be independent Poisson random variables with means ), 2), and 3) respectively. What is the maximum likelihood estimator of ) based on sample values B" ß B# ß and B$ from the distributions of \1 , \2 and \3 , respectively, _ _ 3B 2B B 6B 3B 2B3 B 2B 3B3 A) 12 B B) B C) 1 62 D) 1 6 2 3 E) 1 112

10. You are given the following random sample of 12 data points from a population distribution \: 7 , 15 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53 Suppose that the distribution variance is 100. Determine the bias in the biased form of the sample variance as an estimator of the distribution variance. A) Less than  & B) At least  & but less than ! C) At least ! but less than & D) At least & but less than "! E) At least "!

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

PRACTICE EXAM 1

PE-3

11. The following table was obtained by fitting both a Poisson distribution and a binomial distribution to a data set of 100,000 integer-valued observations. V ~ À    V ~  Á V ~ À  Fitted Poisson Fitted Binomial  expected  expected    , ,À

À Á À À   ,  Á À À  Á À À    À À  À À  À À 

À À  ‚   Totals Á  Á  À  Á  À  Degrees of freedom cc~ cc~   À À    À -value You are given that the negative loglikelihood of the fitted Binomial model is 36,787 . Use Schwarz Bayesian Criterion to choose between the Poisson and Binomial models. A) Choose Poisson B) Choose Binomial C) They are equally preferable D) Not enough information is available to determine the Poisson likelihood function. E) None of A, B, C are correct

12. 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 7 ´# ~ µ ~ À Á 7 ´# ~ µ ~ À Á 7 ´# ~ µ ~ À Find the mean of the posterior distribution. A) 2.5 B) 2.6 C) 2.7 D) 2.8 E) 2.9

13. 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) 

14. In a portfolio of risks, each risk has an exponential claim amount distribution. The mean of the claim amount distribution for a randomly chosen risk is , where  has a Gamma distribution with parameters  ~ À and ~  . A single claim amount of 2 is observed for a randomly chosen risk. Find the Buhlmann credibility premium for the next claim amount for the same risk. A) Less than 1.0 B) At least 1.0, but less than 1.25 C) At least 1.25, but less than 1.5 D) At least 1.5, but less than 1.75 E) At least 1.75

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

PE-4

PRACTICE EXAM 1

15. 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 ° % ~  7 ²claim of amount %OType A and a claim occurs³ ~ D Á ° % ~  ° % ~  7 ²claim of amount %OType B and a claim occurs³ ~ D Á ° % ~  ° % ~  7 ²claim of amount %OType C and a claim occurs³ ~ D À ° % ~  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) À ? b À

 B) À ? b À  C) À

? b À

 D) À

? b À  E) À? b À

16. 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): 0.15 , 0.62 , 0.37, 0.78 . Generate the total number of trials until three successes result. A) 3 B) 4 C) 5 D) 6 E) 7

17. You are given the following information on towing losses for two classes of insureds, adults and youths: Exposures Year Adult Youth Total 1996 2000 450 2450 1997 1000 250 1250 1998 1000 175 1175 1999 1000 125 1125 Total 5000 1000 6000 Pure Premium Year 1996 1997 1998 1999 Weighted Average

Adult 0 5 6 4 3

Youth 15 2 15 1 10

Total 2.755 4.400 7.340 3.667 4.167

You are also given that the estimated variance of the hypothetical means is 17.125. Determine the nonparametric empirical Bayes credibility premium for the youth class. A) Less than 5 B) At least 5, but less than 6 C) At least 6, but less than 7 D) At least 7, but less than 8 E) At least 8

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

PRACTICE EXAM 1

PE-5

18. \ and ] are independent random variables with IÒ\Ó œ Z +  ² c - ²& ³³ h ² c- i ²&  ³ ³ b  - ²& ³ h ² - i ²&b³ ³?. ~ i

The model cdf is - ²%³ ~  c 

b

c%°

c- i ²& ³ h ² c- i ²&  ³ ³ b



~

. There are 11 distinct %-values. - i ²&

³

,  ~ - ²& ³ h ² - i ²&b³ ³  & - ²& ³ - i ²& ³      À   À  À  À À  À

 À  À

À  À À  À  À  À À

 À

À 

À  À 

À À  À À  À  À  À

 À   À À  À À  À  À  À  À

 À

À   À   À  À À  À À    À  The Anderson-Darling statistic is c  b ´À b Ä b À b  b À b Ä b À µ ~ À  À Answer: E 

 ~ ² c - ²& ³³

© ACTEX 2009

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

PE-38

PRACTICE EXAM 2

28. 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 = ´@ µ

 ´ b ²,´@ µ³ µ ~  ´ b ² @ ³ µ . Thus, with @ ~ À  Á @ @  ´ b ² @ ³ µ ~ À  ~   . With the coefficient of variation of @ changed to @ @

@

~ À5200 , we have  ´ b ² @ ³ µ ~ À ~  . @

29. 7 ´$ ~ O? ~ µ ~ 7 ´$ ~ O? ~ µ ~ 7 ´$ ~ O? ~ µ ~

7 ´?~O$~µh7 ´$~µ 7 ´?~µ

~

c h [ h²À³ c h c h c h  h²À ³b [ [ h²À³b [ h²À³ c h  [ h²À³ c   h c h c h  [ h²À ³b [ h²À³b [ h²À³

Answer: D

c h [ h²À ³ c h c h c h  [ h²À ³b [ h²À³b [ h²À³

~ À  ,

~ À , and ~ À .

Then, since ,´? O$ ~ O ~  Á ,´? O$ ~ O ~  and ,´? O$ ~ O ~ , we have ,´? O? ~ µ ~ ²³²À ³ b ²³²À³ b ² ³²À ³ ~ À  . Answer: D

30. ?O has an exponential distribution, and  has an inverse gamma distribution with  ~  Á ~  . ²³ ~ ,´?Oµ ~  and #²³ ~ = ´?Oµ ~  . 

# ~ ,´#²³µ ~ ,´ µ ~ ²c³² c³ ~  .   ~ = ´²³µ ~ = ´µ ~ ,´ µ c ²,´µ³ ~  c ² c ³ ~  .

A ~ b # ~ b  ~  .   

Answer: B

~ -coefficients: 31. According to Buhlmann's approach, we solve the normal equations for the  ~ ~ ~ ,´? µ ~  b  ,´? µ b  ,´? µ ~  = ´? µ b  ~  *#´? Á ? µ *#´? Á ? µ ~  ~ ~  = ´? µ . *#´? Á ? µ ~  *#´? Á ? µ b  Substituting the given values, these equation become ~ b  ~  b  ~ ~ ~ b  ~ ~ ~ ~  ~  b   ~ ~ ~  ~  . The credibility premium is with solution  ~  Á  ~  Á   ~  b  ~  ? ~  b ? b ? , with  ~  .  Answer: E ~

© ACTEX 2009

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

PRACTICE EXAM 2

PE-39

32. The average number of claims per insured is c ? ~ ²³²À  ³ b ²³²À ³ b ²³²À ³ b ²³²À ³ b ²³²À³ ~ À  À This is the estimate of ,´?µ ~ . Since the conditional distribution of ? given # is Poisson with parameter #, we have ,´?O#µ ~ = ´?O#µ ~ # . Then, since ,´,´?O#µµ ~ ,´?µ , our estimate for # ~ ,´= ´?O#µµ is also À  (since = ´?O#µ ~ ,´?O#µ ). We estimate = ´?µ, the variance of the relative claim frequency per insured; the relative frequency of claims forms the estimated probability distribution of ? . =V ´?µ ~ ² c À ³ ²À  ³ b ² c À ³ ²À ³ b ² c À ³ ²À ³ b ² c À ³ ²À ³ b ² c À ³ ²À³ ~ À  À Since = ´?µ ~ # b  , we use the estimated variance of ? along with the estimate of # to get an estimate of ; V  ~ =V ´?µ c V# ~ À  c À ~ À À À  V The estimate of  is  ~ VV# ~ À ~ À , and the estimated credibility factor for one  V individual is A ~ ~ À À The credibility premium for the next period for an individual bÀ

V b ² c A³ V  who had  claims in the current period is A V ~ À b À . Answer: C

 33. ? has a mixed distribution with cdf - ²%³ ~

H

%   %b 



% % %~ . % %‚

According to the inversion method of simulation, given a uniform random number " from ´Á µ, the simulated value of ? is % ~ " if   "   , % ~  if   "   , % ~ " c  if   "   . Then, " ~  S % ~  Á " ~  S % ~  Á " ~  S % ~  Á " ~  S % ~  . The sample mean is bbbÀ ~  Answer: D  .

34. For an Inverse Exponential distribution with parameter , the mode is  . We are given that the mode is 10,000, so that ~ Á  . The median, say , is the point for which 7 ´?  µ ~ - ²³ ~ À . For the Inverse Exponential, - ²%³ ~ c °% . We solve for  from cÁ cÁ° ~ À . Taking the natural log of both sides results in ~ ²À ³ , so that   ~  Á  . Answer: E 35. The conditional density function at the uncensored ground up loss % is 

  ²%O? € ³ ~ c ~ c À The conditional probability at a censored loss is c 

7 ´? €  O? € µ ~ c ~ c c À The likelihood function for the given data is ² c ³

 c  3² ³ ~ ² c ³ ² c ³ ~ ² c ³ À The loglikelihood is  3² ³ ~  ² c  ³ c  ² c ³ , and setting the derivative equal to 0 results in    V   3² ³ ~ c c c ~  , from which we get ~  . Answer: C

© ACTEX 2009

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

PE-40

© ACTEX 2009

PRACTICE EXAM 2

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

PRACTICE EXAM 3

PE-41

ACTEX EXAM C/4 - PRACTICE EXAM 3 1. A Mars probe has two batteries. Once a battery is activated, its future lifetime is exponential with mean 1 year. The first battery is activated when the probe lands on Mars. The second battery is activated when the first fails. Battery lifetimes after activation are independent. The probe transmits data until both batteries have failed. Calculate the probability the probe is transmitting data three years after landing. A) 0.05 B) 0.10 C) 0.15 D) 0.20 E) 0.25

2. An insurance company sells hospitalization reimbursement insurance. You are given: • Benefit payment for a standard hospital stay follows a lognormal distribution with  ~  and  ~ . • Benefit payment for a hospital stay due to an accident is twice as much as a standard benefit. • 25% of all hospitalizations are for accidentally causes. Calculate the probability that a benefit payment will exceed $15,000. A) Less than 0.12 B) At least 0.12, but less than 0.14 C) At least 0.14, but less than 0.16 D) At least 0.16, but less than 0.18 E) At least 0.18

3. Losses during the current year follow a Pareto distribution with  ~  and ~ Á . Annual inflation is 10%. Calculate the ratio of the expected proportion of claims that will exceed $750,000 next year to the proportion of claims that exceed $750,000 this year. A) Less than 1.105 B) At least 1.105, but less than 1.115 C) At least 1.115, but less than 1.125 D) At least 1.125, but less than 1.135 E) At least 1.135

4. Customers arrive at a bank according to a Poisson process at the rate of 100 per hour. 20% of them make only a deposit, 30% make only a withdrawal and the remaining 50% are there only to complain. Deposit amounts are distributed with mean 8000 and standard deviation 1000. Withdrawal amounts have mean 5000 and standard deviation 2000. The number of customers and their activities are mutually independent. Using the normal approximation, calculate the probability that for an 8-hour day the total withdrawals of the bank will exceed the total deposits. A) 0.27 B) 0.30 C) 0.33 D) 0.36 E) 0.39

5. An insurance company increases the per claim deductible of all automobile policies from $300 to $500. The mean payment and standard deviation of claim severity are shown below: Mean Payment Standard Deviation Deductible $300 1,000 256 $500 1,500 678 The claims frequency is Poisson distributed both before and after the change of deductible. The probability of no claims increases by 30%, and the probability of having exactly one claim decreases by 10%. Calculate the percentage increase in the variance of the aggregate claims. A) Less than 30% B) At least 30%, but less than 50% C) At least 50%, but less than 70% D) At least 70%, but less than 90% E) 90% or more

© ACTEX 2009

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

PE-42

PRACTICE EXAM 3

6. You are given the following for a sample of five observations from a bivariate distribution: % & 1 4 2 2 4 3 5 6 6 4 (ii) % ~ 3.6, & ~ 3.8 ( is the covariance of the empirical distribution - as defined by these five observations. ) is the maximum possible covariance of an empirical distribution with identical marginal distributions to - . Determine ) c (. A) 0.9 B) 1.0 C) 1.1 D) 1.2 E) 1.3 (i)

7. When data is in interval grouped form, the usual assumption that is made is that within each interval the data points for that interval are uniformly distributed on the interval. This means that  for interval ²Á µ , a data point within that interval would have pdf  ²%³ ~ c .

For an interval grouping with interval endpoints  ~       Ä  c   , and with  data points in interval ²c Á  µ , this results in an empirical estimate of the first     b moment being   h c  , where  ~   (total number of data points). ~ ~

²%c³

Suppose that the following pdf is assumed for data in the interval ²Á µ :  ²%³ ~ ²c³ . Find the estimate of the first moment based on the grouped data set.    b A)   h c  ~

   c D)   h c  ~

   b B)   h c  ~

   c E)   h   c ~

   b C)   h c  ~

8. An inverse gamma distribution is fit to a data set using maximum likelihood estimation. The estimates of  and are  V ~ À and V ~ À . À  À The information matrix that results from the estimation of  and is > . À À  ? Apply the delta method to find a 95% confidence interval for the mean of the distribution. What is the upper limit of the interval? A) 8.0 B) 8.1 C) 8.2 D) 8.3 E) 8.4

© ACTEX 2009

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

PRACTICE EXAM 3

PE-43

9. 1000 workers insured under a workers compensation policy were observed for one year. The number of work days missed is given below: Number of Days of Work Number of Workers Missed 0 818 1 153 2 25 3 or more 4 Total 1000 Total Number of Days Missed 230 The chi-square goodness-of-fit test is used to test the hypothesis that the number of work days missed follows a geometric distribution where: (i) The geometric parameter is estimated by the average number of work days missed. (ii) Any interval in which the expected number is less than one is combined with the previous interval. Determine the results of the test (use at least one decimal accuracy in all calculations). A) The hypothesis is not rejected at the 0.10 significance level. B) The hypothesis is rejected at the 0.10 significance level, but is not rejected at the 0.05 significance level. C) The hypothesis is rejected at the 0.05 significance level, but is not rejected at the 0.025 significance level. D) The hypothesis is rejected at the 0.025 significance level, but is not rejected at the 0.01 significance level. E) The hypothesis is rejected at the 0.01 significance level.

10. A loss distribution is being analyzed using the Bayesian credibility approach. The parameter  has a prior gamma distribution with  ~  and ~  . The model distribution ? is Poisson with a mean of . A sample of 6 observations of ? results in a Bayesian premium of 37.48 . A 7-th observation of X is obtained and the Bayesian premium is recalculated to be 37.84. Find the value of the 7-th observation. A) 40 B) 41 C) 42 D) 43 E) 44

11. 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: () = (0.5)5c  b ²À ³  c° À (iii) In the first policy year, no claims were observed for the insured. Determine the expected number of claims in the second policy year. A) 0.3 B) 0.4 C) 0.5 D) 0.6 E) 0.7

© ACTEX 2009

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

PE-44

PRACTICE EXAM 3

12. Type A risks have each year's losses uniformly distributed on the interval ²Á ³ . Type B risks have each year's losses uniformly distributed on the interval ²Á ³ . A risk is selected at random, with each type being equally likely. The first year's losses equal 3. Find the Buhlmann credibility premium for the second year's losses in terms of 3. A) À3 b À B) À 3 b À C) À 3 b À  D) À 3 b À E) À 3 b À

13. You are given the following table of data for three policyholders over a three year period. Policy Year S Policyholder ¨ 1

1

2

3

Number of Claims Average Claim Size

40 200

50 220

2

Number of Claims Average Claim Size

100 200

120 200

3

Number of Claims Average Claim Size

50 200

60 250

120 150

Apply the nonparametric empirical Bayes credibility method to find the credibility premium per claim in the 4-th year for Policyholder 2. A) Less than 165 B) At least 165 but less than 175 C) At least 175 but less than 185 D) At least 185 but less than 195 E) At least 195

14. 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) 4.0 B) 4.1 C) 4.2 D) 4.3 E) 4.4

15. You are given a random sample of 3 values from a distribution - : 4 , 5 , 9  You estimate the median of ? using the estimator  ²smallest ? b largest ? ³ . Determine the bootstrap approximation to the mean square error. A) 2.2 B) 2.4 C) 2.6 D) 2.8 E) 3.0

16. For a particular data set, the product limit estimator results in the following estimates: (i) : ²³ ~ .94 (ii) With data truncated at 1, the estimated conditional probability of surviving to time 2 is .88 . (iii) With data truncated at 2, the estimated conditional probability of surviving to time 3 is .75 . Find : ²³ . A) .58 B) .62 C) .66 D) .70 E) .74

© ACTEX 2009

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

PRACTICE EXAM 3

PE-45

17. For an insurance portfolio, you are given: (i) For each individual insured, the number of claims follows a Poisson distribution. (ii) The mean claim count varies by insured, and the distribution of mean claim counts follows a gamma distribution. (iii) For a random sample of 1000 insureds, the observed claim counts are as follows: Number of Claims,  0 1 2 3 4 5 Number of Insureds,  512 307 123 41 11 6  ~ 750 2  ~ 1494 (iv) Claim sizes follow a Pareto distribution with mean 1500 and variance 6,750,000. (v) Claim sizes and claim counts are independent. (vi) The full credibility standard is to be within 5% of the expected aggregate loss 95% of the time. Determine the minimum number of insureds needed for the aggregate loss to be fully credible. A) Less than 8300 B) At least 8300, but less than 8400 C) At least 8400, but less than 8500 D) At least 8500, but less than 8600 E) At least 8600

18. For a group of lives you are given that the time until death random variable for each member of the group has a constant hazard rate that is drawn from the uniform distribution on ´ À Á À µ . For someone selected at random from this group, calculate the probability that the individual dies between one and 3 years from now. A) .026 B) .029 C) .032 D) .035 E) .038

19. Losses follow and exponential distribution with parameter . For a deductible of 100, the expected payment per loss is 2,000. Which of the following represents the expected payment per loss for a deductible of 500? A) B) ² c c ° ³ C) Á c° c ° D) Á  E) Á ² c c ° ³² c c° ³

20. 5 has a geometric distribution with a mean of 2. Determine the mean of the zero-modified distribution with 4 ~  . A) 2.1 B) 2.2 C) 2.3 D) 2.4 E) 2.5

21. For a portfolio of 2,500 policies, claim frequency is 10% per year, and severity is distributed uniformly between 0 and 1,000. Each policy is independent and has no deductible. Calculate the reduction in expected annual aggregate payments, if a deductible of $200 per claim is imposed on the portfolio of policies. A) Less than $46,000 B) At least $46,000, but less than $47,000 C) At least $47,000, but less than $48,000 D) At least $48,000, but less than $49,000 E) $49,000 or more

© ACTEX 2009

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

PE-46

PRACTICE EXAM 3

22. In a given week, the number of projects that require you to work overtime has a Poisson distribution with a mean of 2. For each project, the distribution of the number of overtime hours in the week is the following: %  ²%³ 5 0.2 10 0.3 20 0.5 The number of projects and the number of overtime hours are independent. You will get paid for overtime hours in excess of 15 hours in the week. Calculate the expected number of overtime hours for which you will get paid in the week. A) 15 B) 16 C) 17 D) 18 E) 19

23. A study of the time until failure, ? , of an electronic device is based on observing 20 of the devices. One failure and one right-censoring is observed at each of the integer time points 1, 2, . . , 10. The probability 7 ´  ?  O ?  µ is to be estimated. Find the absolute difference between the Kaplan-Meier product limit estimate and the Nelson-Aalen estimate. A) .0020 B) .0022 C) .0024 D) .0026 E) .0028

24. You are given the following random sample of 12 data points from a population distribution ? : 7 , 12 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53 V ²³. Using the triangle kernel with bandwidth 3, find A)  B)  C)  D)  E)      

25. The following sample is taken from the distribution  ²%Á ³ ~ 2  3c%° Observation 1 2 3 4 5 6 7 % 0.49 1.00 0.47 0.91 2.47 5.03 16.09 Determine the maximum likelihood estimator of  , where 7 ²? € ³ ~ À . A) Less than 1.0 B) At least 1.0, but less than 1.2 C) At least 1.2, but less than 1.4 D) At least 1.4, but less than 1.6 E) 1.6 or more

26. Call center response times are described by the cumulative distribution function - ²%³ ~ % b . where   %   and € c . A random sample of response times is as follows: 0.56 0.83 0.74 0.68 0.75 Calculate the maximum likelihood estimate of . A) Less than 1.4 B) At least 1.4, but less than 1.6 C) At least 1.6, but less than 1.8 D) At least 1.8, but less than 2.0 E) At least 2.0

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27. 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: % ~  Á % ~ . 4 :,²- ³ is the mean square error of the estimator when the empirical distribution is used. Find the exact value of 4 :,²- ³. A) .125 B) .250 C) .375 D) .500 E) .625

28. You are given the following sample of five claims: 43 , 145 , 233 , 396 , 775  The distribution with density function  ²%³ ~ ²b%³  Á % €  is being tested with the Kolmogorov-Smirnov test. Find the value of the Kolmogorov-Smirnov statistic. A) Less than .05 B) At least .05, but less than .140 C) At least .140, but less than .230 D) At least .230, but less than .320 E) At least .320

29. Total claims per period : follows a compound Poisson distribution and claim severity has the pdf  ²&³ ~ &c , 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 5 , 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) .070

30. 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. A) Less than .50 B) At least .50 but less than .52 C) At least .52 but less than .54 D) At least .54 but less than .56 E) At least .56

31. 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  ² ³ ~ ² b ³  ,    . 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. 7 is the Buhlmann credibility premium for Risk 1 for the next period and 7 is the Buhlmann credibility premium for Risk 2 for 7 the next period. Find lim 7 . A) 

 B)

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¦

C) 





D) 

E) B

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

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PRACTICE EXAM 3

32. 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 7 ²@ ~ ³ ~ 7 ²@ ~ ³ ~ À . 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 : ~  Á : ~  and : ~  . Find the Buhlmann credibility premium for the 4-th year for this policy. A) Less than 0.7 B) At least 0.7 but less than 0.8 C) At least 0.8 but less than 0.9 D) At least 0.9 but less than 1.0 E) At least 1.0

33. The prior distribution of  is a gamma distribution with parameters  ~  and ~ . The conditional distribution of ? given  is Poisson with a mean of . The unconditional distribution of ? is to be simulated in two steps. Step 1: simulate a value of  Step 2: simulate a value of ? given the value of  simulated in Step 1. To apply step 1, we use the fact that the gamma distribution with parameters  ~  and is the sum of three independent exponential random variables each with mean , and simulate three independent exponentials using the inverse transformation method and add the simulated values to get the simulated gamma distribution value. To apply step 2 we use the product algorithm for the Poisson. The sequence of uniform ²Á ³ numbers to be used in the overall simulation are À Á À Á À Á À , À Á À Á À These numbers are used in the order given, with the first three used to simulate the gamma distribution and the remaining numbers used in step 2. Each number is used once until the simulation is complete. Determine the value of ? simulated. A) 0 B) 1 C) 2 D) 3 E) 4

34. According to the Loss Models book, which of the following is/are true, based on the existence of moments test? I. The Loglogistic Distribution has a heavier tail than the Gamma Distribution. II. The Paralogistic Distribution has a heavier tail than the Lognormal Distribution. III. The Inverse Exponential has a heavier tail than the Exponential Distribution. A) I only B) II only C) I and III only D) II and III only E) I, II and III

V ~ À , and 35. For these data, the maximum likelihood estimate for the Poisson distribution is  for the negative binomial distribution, it is V  ~ À and  ~ À. The Poisson has a negative loglikelihood value of 385.9, and the negative binomial has a negative loglikelihood value of 382.4. Determine the likelihood ratio test statistic, treating the Poisson distribution as the null hypothesis. A) c 1 B) 1 C) 3 D) 5 E) 7

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PRACTICE EXAM 3

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ACTEX EXAM C/4 - PRACTICE EXAM 3 SOLUTIONS 1. There are a few ways to approach this problem. One approach is the convolution approach to finding the distribution function of the sum of random variables ? and @ . If ? and @ are continuous independent non-negative random variables, the ! -?b@ ²!³ ~  ? ²%³ h -@ ²! c %³ % . In this case, ? and @ are both exponential with mean 1,    and ! ~ , this becomes  c% h ´ c c²c%³ µ % ~  c% % c c   % ~  c c c c ~ À  . This is the probability that the total time until failure of both batteries is  . The probability that total time until failure is €  is À . An alternative solution is based on the observation that if ? and @ are independent exponential random variables both with mean 1, then ? b @ has a gamma distribution with  ~  and ~  , and the pdf of ? b @ is ?b@ ²!³!c! . B B Then 7 ´? b @ € µ ~  !c! ! ~ c !c! c c! c ~ c ~ À . !~

Yet another approach is to note that with exponential inter-event time with mean 1 year, the number of failures forms a Poisson process with a rate of 1 per year. The probe will be transmitting in 3 years if there is at most one battery failure in the 3 year period. The number of failures in a 3-year period, 5 ²³, is Poisson with mean 3, so the probability is 7 ´5 ²³  µ ~ c b c h  ~ c . Answer: D

2. The benefit payment ? is a mixture of standard benefit payment ? with weight .75 and accident benefit payment ? with weight .25. 7 ²? €  Á ³ ~ À 7 ²? €  Á ³ b À 7 ²? €  Á ³ . ? has a lognormal distribution, so   Á c  7 ²? €  Á ³ ~  c )² ³ ~  c )²À³ ~  c À  ~ À .  ? ~ ? , so  Á  c  7 ²? €  Á ³ ~ 7 ²? €  Á ³ ~ 7 ²? € Á ³ ~  c )² ³  ~  c )²À ³ ~  c À  ~ À . Then 7 ²? €  Á ³ ~ À ²À ³ b À ²À ³ ~ À . Answer: A

3. Let ? denote the claim random variable for this year. ? has a Pareto distribution with  ~  and ~ Á . The expected proportion of claims that exceed 750,000 is the same as 7 ´? €  Á µ, which is the probability that a claim exceeds 750,000. This probability is (from Á the distribution table)  c - ² Á ³ ~ ²  ÁbÁ ³ ~ À  .

Next year after inflation of 10%, the claim random variable will be @ ~ À? . The Pareto distribution is a "scale distribution" with scale parameter . This means that if ? is Pareto with parameters  and , and if @ ~ ? ( € ), then @ also has a Pareto distribution with the same , and with Z ~  . In this case, @ ~ À? will have a Pareto distribution with Á  ~  and ~ Á  , and 7 ´@ €  Á µ ~ ²  ÁbÁ ³ ~ À  . The ratio is À  À  ~ À .

Answer: D

© ACTEX 2009

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

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PRACTICE EXAM 3

4. Because of independence, we can separate the deposit process and the withdrawal process as two independent processes. The rate per hour at which depositors arrive is ²À³ ~  , and the rate per hour of withdrawers arriving is 30. The number of depositors arriving in an 8-hour day has a Poisson distribution with a mean of ²³ ~   , and the number of withdrawers arriving in an 8-hour day has a Poisson distribution with a mean of ²³ ~ . The total amount deposited in a day has a compound Poisson distribution with Poisson parameter 160, and individual deposit amount (severity distribution) with mean 8000 and standard deviation 1000. The mean and variance of the total deposit in an 8-hour day is ,´:+ µ ~ ,´5+ µ h ,´?+ µ ~ ² ³² ³ ~ Á  Á  and = ´:+ µ ~ ,´5+ µ h = ´?+ µ b = ´5+ µ h ²,´?+ µ³ ~ ² ³² ³ b ² ³² ³ ~ Á Á Á  . In a similar way, we get the mean and variance of the total withdrawals in an 8-hour day: ,´:> µ ~ ,´5> µ h ,´?> µ ~ ²³² ³ ~ Á Á  and = ´:> µ ~ ,´5> µ h = ´?> µ b = ´5> µ h ²,´?> µ³ ~ ²³² ³ b ²³² ³ ~ Á Á Á  . We wish to find 7 ´:> € :+ µ using the normal approximation. ,´:> c :+ µ ~ c Á , and since :> and :+ are independent, = ´:> c :+ µ ~ Á Á Á  b Á Á Á  ~ Á  Á Á  . Then, using the normal approximation, : c: c²c Á³

c²c Á³

+ 7 ´:> € :+ µ ~ 7 ´:> c :+ € µ ~ 7 ´ > jÁ ÁÁ € jÁ ÁÁ µ ~  c )²À ³ ~  c À  ~ À . Answer: A

5. Let ,´5 µ ~  be the Poisson expected claim frequency before the change in deductible and let ,´5 Z µ ~ Z be the Poisson expected claim frequency after the change in deductible. We are given Z that 7 ´5 Z ~ µ ~ c ~ Àc ~ À7 ´5 ~ µ and Z 7 ´5 Z ~ µ ~ Z c ~ À c ~ À 7 ´5 ~ µ . Dividing the second equation by the first, we get Z ~  À The second moment of the claim severity @ before the change in deductible is ,´@  µ ~ = ´@ µ b ²,´@ µ³ ~   b  ~ Á  Á  . The second moment of the claim severity @ after the change in deductible is ,´²@ Z ³ µ ~ = ´@ Z µ b ²,´@ Z µ³ ~   b   ~ Á  Á  . Since aggregate claims follow a compound Poisson distribution, the variance of : , aggregate claims before deductible is = ´:µ ~ ,´@  µ ~ Á  Á   , and the variance of : Z , aggregate claims before deductible is = ´: Z µ ~ Z ,´²@ Z ³ µ ~ Á  Á ²  ³ . The proportional increase in the variance of aggregate claims is Á  Á ²  ³°Á  Á   ~ À . This is a 76% increase. Answer: D 

6. The empirical distribution assigns probability of  ~ À for each ²%Á &³ pair. The covariance of the empirical distribution is ,´²? c ? ³²@ c @ ³µ ~ ²À³´² c À ³² c À ³ b ² c À ³² c À ³ b ² c À ³² c À ³ b ² c À ³² c À ³ b ² c À ³² c À ³µ ~ À , or ,´?@ µ c ? h @ ~ bbbb c ²À ³²À ³ ~ À ~ ( . Using the same marginal distributions (the same ? 's and the same @ 's, perhaps in different pairings), the covariance will be maximized if ,´?@ µ is maximized. Maximization will occur if the largest @ 's are paired with the largest ? 's. The following bivariate distribution has the same marginal distributions as the original bivariate distribution:

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

PRACTICE EXAM 3

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6. continued ?   

@    

The covariance of ? and @ in this bivariate distribution is b b bb

c ²À ³²À ³ ~ À ~ ) . This is the maximum covariance with identical marginal distributions. ) c ( ~ À c À ~ À . Answer: D

7. An underlying relationship upon which the estimate is based is 

,´?µ ~  ,´? O c  ?   µ h 7 ²c  ?   ³ . ~

When the uniform distribution is assumed for each interval, this results in  b ,´? O c  ?   µ ~ c  . Also, the estimate of 7 ²c  ?   ³  is  (fraction of total set of data points that lie in the interval). 

²%c³

With the new pdf, we get ,´? O  ?  µ ~  % h ²c³ % . A somewhat simplified approach is to find ²%c³ ²c³   ²%c³ ,´? c  O  ?  µ ~  ²% c ³ h ²c³ % ~  ²c³ % ~  , so that ,´? O  ?  µ ~ ,´? c  O  ?  µ b  ~ Then, ,´? O c  ?   µ ~

c b 

²c³ 

b  ~ b  .

   b , and the estimated mean of ? is   h c  ~

Answer: C

8. The covariance matrix is the inverse of the information matrix. This will be c À  À À   ~ ²À ³²À  ³c²À ³²À ³ h > *#² VÁ V ³ ~ > ? À À  c À À  c À  =V ² V³ *#² V  VÁ V ³ . ~ > ~ @ ? c À  À  *#² =V ²V ³ A V  VÁ V ³

c À À  ?

À The mean of the inverse gamma is c . The estimate of this is Àc ~ À  . According to the delta method, the variance of the mle estimate of ²Á ³ ~ c is C C C C ² ²Á ³³ h = ² V³ b ² ²Á ³³² ²Á ³³ h *#² VÁ V ³ b ² ²Á ³³ h = ²V ³

C

C

evaluated at the estimate values. This is

C

C

    ² c ²c³  ³ ²À ³ b ² c ²c³ ³² c ³² c À ³ b ² c ³ ²À ³ À À     ~ ² c ²Àc³  ³ ²À ³ b ² c ²Àc³ ³² Àc ³² c À ³ b ² Àc ³ ²À ³ ~ À À

The 95% confidence interval for the mean is À  f À jÀ ~ ²À Á À³ . Answer: E

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PRACTICE EXAM 3

 9. The estimate of the geometric mean  is  ~ À .

 Of the 1000 workers, the expected number who miss 0 days of work is  h À ~ À, À the expected number who miss 1 day of work is  h ²À³ ~  À, ²À³

the expected number who miss 2 days of work is  h ²À³ ~  À, and the expected number who miss 3 or more days of work is  c ² À b  À b  À³ ~ À . Each "interval", "0", "1", "2" and "3 or more" has an expected number of at least 1, so we keep all four intervals. The chi-squared statistic is ² Àc  ³

² Àc ³

² Àc ³

² À c³

8~ b b b À ~ À . À  À  À The test has 2 degrees of freedom (there are 4 intervals and one parameter estimated). From the chi-square table with 2 degrees of freedom we see that the 90-th percentile is 4.605. Therefore, the hypothesis is not rejected at the 10 per cent level of significance. Answer: A





~

~

 % ³²  ³ ~ À 10. The original Bayesian premium is ² b  % ³² b

³ ~ ² b 

from which it follows that  % ~  . ~





~

~

 % ³²  ³ ~ À  The updated Bayesian premium is ² b  % ³² b ³ ~ ² b 

from which it follows that  % ~  . 

~

~

~

Therefore, % ~  % c  % ~  . Answer: A

11. The problem involves determining the Bayesian premium. The prior distribution of the parameter  has density ²³ ~ ²À ³ c  b ²À ³²  ³c° (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  ~  Á  ~  Á the second with  ~  Á  ~ ). 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 ,´? O? ~ µ , which is the Bayesian premium. B We can write the expectation as  ,´? Oµ h ²O? ~ ³  . Since ? given  has a Poisson distribution with mean , this integral becomes B  h ²O? ~ ³  Á 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 %   ²Á ³ ~  ²O³ h ²³ ~ c h %[ h ´²À ³ c  b ²À ³² ³c° µ 

~ ²À ³´ c  b ² ³cÀ µ , and the marginal probability that ? ~  is B





 ²³ ~   ²Á ³   ~ ²À ³´ b µ ~ À .

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11. continued The posterior density of  given ? ~  is then  ²Á³

²À ³´ c  b²  ³cÀ µ



²O? ~ ³ ~ 7 ´?~µ ~ ~ c  b ² ³cÀ . À  This can be written in the form ²O? ~ ³ ~ ² ³² c  ³ b ² ³²ÀcÀ ³ . 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

    À . The mean of the posterior distribution is then ² ³² ³ b ² ³² À ³ ~ À . Answer: A

12. Credibility is being applied to aggregate losses : , which has a compound distribution. The frequency 5 depends on the parameter , and the severity @ depends on the parameter , so : depends on both parameters. In general, for a compound distribution, the mean is ,´:µ ~ ,´5 µ h ,´@ µ and the variance is = ´:µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ . The hypothetical mean in this example is ,´:O Á µ ~ ,´5 Oµ h ,´@ O µ ~  h and the process variance is = ´:OÁ µ ~ ,´5 Oµ h = ´@ O µ b = ´5 Oµ h ²,´@ O µ³ ~   b   ~   . The variance of the hypothetical mean is  ~ = ´ ,´:O Á µ µ ~ = ´ µ ~ ,´² ³ µ c ²,´ µ³ . Since  and are independent, we have ,´ µ ~ ,´µ h ,´ µ ~  d  , and ,´² ³ µ ~ ,´  µ ~ ,´ µ h ,´  µ ~ ²³²³ ~  (since  has an exponential distribution with mean 1, the second moment of  is ²mean³ ~ , and since has a Poisson distribution with mean 1,  ~ = ´ µ ~ ,´  µ c ²,´ µ³ ~ ,´  µ c ²³ , so that ,´  µ ~ ). Therefore,  ~  c  ~  . The expected process variance is # ~ ,´ = ´:OÁ µ µ ~ ,´  µ ~ ,´µ h ,´  µ ~ ²³²³ ~  . Then  ~ # ~  . Answer: B ²³²³b² ³²³ c 13. ?  ~ ~ À Á  ~  b  ²³²³b²³²³b²³² ³ c ? ~ ~  À Á  ~  bb c ² ³²³b² ³² ³ ? ~ ~ À Á  ~  b  ²³²³b² ³²³b²³²³b²³²³b²³² ³b² ³²³b² ³² ³ c  ~  À V~?~

bb  ~  .

V# ~ V# ~ V# ~

 c  c  c

h ´² c À³ b ² c À³ µ ~ À Á h ´² c  À ³ b ² c  À ³ b ²  c  À ³ µ ~ Á  À  Á h ´ ² c À³ b ²  c À³ µ ~ Á  À 

V# b# V b# V ~ Á  À . V# ~ bb  c c  ~ h ´   ²?  c ? ³ c V#² c ³ µ V 

c   

~

~

~ c  ² b b ³  d ´ ´ ²À c  À³ b ² À c  À³ b ²À c  À³ µ c Á  À²³µ ~  À .

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PRACTICE EXAM 3

13. continued The credibility premium for policyholder 2 is  V c  b ² c A³ V  A? V~²

Á À ³ h ² À ³ b ² c b

 À

 ³² À³ b Á À  À

~  À .

Answer: D

c c  ?  b? 14. The sample mean based on the first 19 points is ?  ~ ~    , c c  ?  b?  and the sample mean based on all 20 points is ?  ~ ~  À  We use the relationship b c c c    :b ~   ²? c ? b ³ ~ ² c  ³: b ² b ³²? b c ?  ³ to get ~

c c    : ~ ²   ³: b ² ³²?  c ?  ³ , and then c c    : ~ ²   ³: b ²³²?  c ?  ³ c c  c c   ~ ²   ³: b ² ³²?  c ?  ³ b ²³²?  c ?  ³

     ~ ²   ³²³ b ² ³²  c ³ b ²³²  c  ³ ~ À .

Answer: B

15. 4 :, is approximated (estimated) by using the corresponding quantity in the empirical distributionÀ The empirical distribution consists of the three data points, so the median of the empirical distribution is . The mean square error of the estimator is ,´²V c ³ µ . Since the empirical distribution consists of three points, there are  ~  possible samples of size 3 that  can be drawn from the empirical distribution, each with probability  . The samples and values of the estimator V are Sample # 1 2 3 4 5 6 7 8 9 Sample 4,4,4 4,4,5 4,4,9 4,5,4 4,5,5 4,5,9 4,9,4 4,9,5 4,9,9 V ~ Sample median 4 4.5 6.5 4.5 4.5 6.5 6.5 6.5 6.5 Sample # Sample Sample median

10 5,4,4 4.5

11 5,4,5 4.5

12 5,4,9 6.5

13 5,5,4 4.5

14 5,5,5 5

15 5,5,9 7

16 5,9,4 6.5

17 5,9,5 7

18 5,9,9 7

Sample # Sample Sample median

19 9,4,4 6.5

20 9,4,5 6.5

21 9,4,9 6.5

22 9,5,4 6.5

23 9,5,5 7

24 9,5,9 7

25 9,9,4 6.5

26 9,9,5 7

27 9,9,9 9







4 :, ~  ² c ³ b  ²À c ³ b  ² c ³ 

 b  ² À c ³ b  ² c ³ b  ² c ³ ~ À  .

: ²³

Answer: C

: ²³

16. We are given : ²³ ~ .94 , : ²³ ~ À and : ²³ ~ À .   : ²³

: ²³

: ²³ ~ : ²³ h : ²³ h : ²³ ~ À  .  

© ACTEX 2009

Answer: B

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17. The general standard for the number of exposures (or insureds in this case) needed for full = ´:µ

credibility for the random variable : is

 ²,´:µ³ , since : is the aggregate loss per insured.

The distribution of : is a compound distribution (but not compound Poisson, since the claim count parameter has a gamma distribution). We use the more general relationships for compound distributions for which claim frequency is 5 and claim severity is @ . ,´:µ ~ ,´5 µ h ,´@ µ and = ´:µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ . In this case, we are given that @ has a Pareto distribution with mean ,´@ µ ~   , and variance = ´@ µ ~ Á  Á  . Since the full credibility standard is to be within 5% of the  expected aggregate loss 95% of the time, we have  ~ ² À

À ³ ~   À  . From the given observations about the claim counts of 1000 insureds we can estimate the mean of c    V µ~5 5 , ,´5 ~  h ' ~  ~ À , and we can estimate the variance of 5 , c  c   =V ´5 µ ~ c h '² c 5 ³ ~

h ´' c ²5 ³ µ ~ À  . The estimated mean of : is ²À ³² ³ ~  , and the estimated variance is ²À ³² Á  Á ³ b ²À ³² ³ ~ Á  Á  . = ´:µ

Á Á

The full credibility standard is then  ²,´:µ³ ~ ²  À ³´ ² ³ µ ~

 , which is the number of insureds needed for full credibility. Note that for a compound aggregate loss distribution : , with frequency 5 and severity @ , there are three alternative equivalent standards for full credibility. This is reviewed in Section 1 of the notes on credibility in Volume 1 of this study guide. The standard for full credibility used above is the standard for the number of insureds (or exposures) needed. The other two standards are: = ´:µ

(i) standard based on aggregate losses; this is  ,´:µ , which in this example would be 9,780,750 in aggregate losses needed for full credibility; and (ii) standard based on total number of claims; this is  would be

 d ²À ³ ~ À .

= ´:µh,´5 µ ²,´:µ³ ,

which in this example

Answer: E

18. Constant hazard rate  means that time until death has an exponential distribution with mean  . The time until death for a randomly chosen individual has a continuous mixture distribution. The probability 7 ²  ;  ³ is found by conditioning over .  The pdf of  is  ²³ ~ À for À    À. À À 7 ²  ;  ³ ~ À 7 ²  ;  O³ h  ²³   ~ À ´c c c µ h   ~ ´²

cÀ

c

cÀ

³c

c²À³ cc²À³ µ 

À

~ À .

Answer: B

19. With deductible , the expected payment per loss is ,´²? c ³b µ ~ ,´?µ c ,´? w µ . For the exponential with parameter , from the table of distributions, we have ,´?µ ~ and ,´? w µ ~ ² c c° ³ , so that ,´²? c ³b µ ~ c° . We are given that ,´²? c ³b µ ~ Á  , and therefore, c° ~ Á  . We are asked to find ,´²? c ³b µ ~ c ° . ,´²?c ³ µ

c °

We see that ,´²?c³b µ ~ c° ~ c° , and therefore b ,´²? c ³b µ ~ ,´²? c ³b µ h c° ~ Á c° .

© ACTEX 2009

Answer: C

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

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PRACTICE EXAM 3

20. The probability function of the geometric distribution with mean ,´5 µ ~  ~  is of the   form  ~ ²b ³b ~ ²³b ~ ²  ³²  ³ , so that  ~  . The zero-modified distribution has probability function

c4

c 

4 ~ c h  ~ ° h  ~   for  ~ Á Á Á ÀÀÀ 

c4 The mean of the zero-modified distribution is ,´54 µ ~ c h ,´5 µ ~  h  ~  .  Note that since the probability at 0 is reduced by an amount of  ²reduced from  to  ), the total °

probability ‚ 1 must be increased from  to , which is a proportional increase of ° ~ À . It follows that the mean (or any moment of distribution) is increased proportionally by 1.25 from 2 to 2.5. Answer: E

21. With no deductible the aggregate annual payment : has a compound distribution with frequency 5 that is binomial with  ~   and  ~ À, and severity @ that is uniform on the interval ´Á µ . The expected aggregate payment with no deductible is ,´:µ ~ ,´5 µ h ,´@ µ ~ ² ³²À³² ³ ~  Á  . If a deductible of 200 is applied to each claim, the severity becomes @ Z ~ ²@ c ³b  and the expected severity is ,´@ Z µ ~ ,´²@ c ³b µ ~  ²& c ³²À³ & ~  . With deductible of 200 per claim, the aggregate annual payment : Z has a compound distribution with frequency 5 that is still binomial with  ~   and  ~ À, and severity @ Z . The expected aggregate payment with a deductible of 200 per claim is ,´: Z µ ~ ,´5 µ h ,´@ Z µ ~ ² ³²À³²³ ~ Á  . The reduction in expected annual claims is  Á  c Á  ~  Á  . Alternatively, the expected reduction per claim when the deductible of 200 is imposed is  ,´? w µ ~  &²À³ & b ´ c -@ ²³µ ~  b ´ c Àµ ~   . The expected number of claims is  ²À³ ~  , so the expected reduction in aggregate claims is ² ³² ³ ~  Á  . Answer: A 22. This is a stop-loss problem where : is the aggregate number of overtime hours worked in the week and the deductible is 15. : has a compound distribution with frequency 5 that is Poisson with mean 2 and severity ? that is 5 (prob. .2), 10 (prob. .3) or 20 (prob. .5). We wish to find ,´²: c  ³b µ ~ ,´:µ c ,´: w  µ . The mean of : is ,´:µ ~ ,´5 µ h ,´?µ ~ ´ ²À³ b ²À³ b ²À ³µ ~  . Note that : must be a multiple of 5, with 7 ²: ~ ³ ~ 7 ´5 ~ µ ~ c ~ À  (the only way that : ~  is if 5 ~ ), 7 ´: ~ µ ~ 7 ´5 ~ µ h 7 ´? ~ µ ~ ²À³²À³ ~ À  , and 7 ´: ~ µ ~ ²7 ´5 ~ µ h 7 ´? ~ µ b 7 ´5 ~ µ h ²7 ´? ~ µ³ ³ ~ ²À³²À³ b ²À³²À³ ~ À  . Then, 7 ²: ‚  ³ ~  c 7 ²: ~ Á Á ³ ~ À   .  : ~ , prob. À  : ~ , prob. À  , : w  ~  : ~ , prob. À   : ‚ , prob. .  so ,´: w  µ ~ ²À ³ b ²À ³ b  ²À ³ ~ À . Then, ,´²: c  ³b µ ~  c À ~  À . Answer: B

H

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

PRACTICE EXAM 3

PE-57

7 ´? µ

23. 7 ´  ?  O ?  µ ~ 7 ´? µ Since the observed failure times are all integers, the estimate of 7 ´  ?  µ is the same as the estimate of 7 ´  ?  µ ~ :²³ c :² ³ . :²³c:² ³

Therefore, we wish to estimate c:² ³ . The numbers at risk and the numbers of failures at each failure time are: & ¢    



 ¢   

  

           ¢ The product limit estimate of :²³ is ² c

  ³²

c

  ³²

c

  ³

  

~ À  .

     The product limit estimate of :² ³ is ² c  ³² c  ³² c 

³² c  ³² c  ³ ~ À  . The product limit estimate of :² ³ is       ² c  ³² c  ³² c 

³² c  ³² c  ³² c  ³² c  ³² c  ³² c  ³ ~ À   .

The product limit estimate of

:²³c:² ³ c:² ³

is À  cÀ  ~ À  À cÀ  

 V The Nelson-Aalen estimate of /²³ is /²³ ~  b cÀ 

so the N-A estimate of :²³ is  ~ À   .

 

b

 

~ À  ,

 V The Nelson-Aalen estimate of /² ³ is /² ³ ~  b cÀ  so the N-A estimate of :²³ is  ~ À .

 

b

 

b

 

b

 

~ À  ,

The Nelson-Aalen estimate of /² ³ is       V /² ³ ~  b  b 

b  b  b  b  b  b  ~ À   , so the N-A estimate of :²³ is cÀ   ~ À   . The Nelson-Aalen estimate of

:²³c:² ³ c:² ³

is À   cÀ

~ À À cÀ  

The absolute difference between the two estimates is .0026 . Answer: D

24. Triangle kernel with bandwidth  ~ . For the point % ~  there is one & value within the band from  c  ~  to 0c ~  ; this is the data value & ~  . .

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

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PRACTICE EXAM 3

24. continued The intervals centered at points & ~  Á & ~  Á and & ~  all lie completely to the left of % ~ , so that 2 ²³ ~ 2 ²³ ~ 2 ²³ ~ . For the &'s from 26 to 53, the intervals all lie completely to the right of 20, so 2& ²³ ~  for each of them. From the triangle diagram above, it can be seen that the area in the triangle to the right of 20 is  d  d  ~  , so that the area to the left of 20 in that triangle is 2 ²³ ~  c  ~  À Therefore, V ²³ ~ ²  ³²³ b ²  ³²³ b ²  ³²³ b ²  ³²  ³ ~  . Answer: D     

V 25. The estimate of 7 ´? € µ is c° , where V is the mle of . Since ? has an exponential distribution with mean , the mle of is the sample mean of the data set, which is 3.78 . The estimate of 7 ´? € µ is c°À , which we are given as 0.75 .  Therefore c À ~ ²À ³ and  ~ À  . Answer: B

26. The pdf is  ²%³ ~ ² b ³% and   ²%³ ~ ² b ³ b %  and    ²%³ ~ b b  % . Setting the derivative of the loglikelihood function to 0, we get  

 3 ~ '    ²% ³ ~ b b ' % ~  S ~ 'c  % c  ~ À  .

Answer: D

27. 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, % b% . Thus, for the 4 pairs  ²% Á % ³, we have estimator values  Á À Á À Á  , and the estimate of the MSE is      Answer: A  ´² c À ³ b ²À c À ³ b ²À c À ³ b ² c À ³ µ ~ À .

28. The distribution function of the hypothesized distribution is % %   - ²%³ ~   ²!³ ! ~  ²b!³  ! ~  c b! À

Alternatively, it can be noted that ? has a Pareto distribution with parameters  ~  and c  ~ , so that - ²%³ ~  c ² %b ~  c %b À ³ The empirical and model distribution function values are % - ²%³ - ²%c ³ - ²%³ +  À   À À  À

 À À À  À  À À

À 

À  À

À À  À

À  À The Kolmogorov-Smirnov statistic is the maximum from column +, .3404. Answer: E

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

PRACTICE EXAM 3

PE-59

B 29. The severity distribution has mean ,´@ µ ~  & h &c & ~  Á and B ,´@  µ ~  & h &c & ~  Á so that = ´@ µ ~  c ²  ³ ~  À

 With  ~ À and  ~ À , we have  ~ ² À  À ³ ~  À À The full credibility standard for



= ´@ µ

number of exposures needed for the compound Poisson distribution is  ´ b ²,´@ µ³ µ . The full credibility standard for the number of exposures needed for the Poisson frequency Z

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 Z



= ´@ µ

distribution alone, then we set  ~  ´ b ²,´@ µ³ µ and the " " for the Poisson frequency credibility standard must change, which is why it has been denoted Z . Z

= ´@ µ



°

Then  ~  ´ b ²,´@ µ³ µ ~  À ´ b ² °³ µ S Z ~  À .  With 7 ~ À , & ~ À , and then in order for this to be the proper Z for 7 ~ À , we must  have  À  ~ ² À

Answer: B  ³ S  ~ À  .

30. ,´?Oµ ~  Á = ´?Oµ ~  . Since  is a mixture, ,´µ ~ ²À ³²³ b ²À ³²³ ~  and ,´ µ ~ ²À ³²³ b ²À ³² d  ³ ~ À (the first component is constant at 1). Then  ~ ,´²³µ ~ ,´µ ~  Á # ~ ,´#²³µ ~ ,´µ ~  Á  ~ = ´²³µ ~ = ´µ ~ ,´ µ c ²,´µ³ ~ À . For a single observation of ? ,  ~ , the Buhlmann credibility factor is A ~ b  ~ À . With ? ~ , the Buhlmann credibility estimate for the next year is c A? b ² c A³ ~  b ²À ³²³ ~ À  . Answer: C

À

31. Prior distribution is 7 ²(³ ~ 7 ²)³ ~  À

Hypothetical means are ²(³ ~ ,´?O(µ ~  Á ²)³ ~ ,´?O)µ ~  .

   Process variances are #²(³ ~ = ´?O(µ ~  ~  ~  Á #²)³ ~  .     ~ ,´?µ ~ expected hypothetical mean ~ ²³²  ³ b ²À ³²  ³ ~  .  # ~ expected process variance ~ ²  ³²  ³ b ²  ³²  ³ ~  .

  ~ variance of hypothetical mean ~ ² c À ³ ²  ³²  ³ ~ 

.



 A ~ b ~ À . # ~ b °  °

Buhlmann credibility premium is A3 b ² c A³ ~ À 3 b À

²  ³ ~ À 3 b À 

. Answer: A

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

PE-60

PRACTICE EXAM 3

32. Hypothetical mean is ,²:O³ ~  h ,²@ ³ ~ À  À Process variance is = ²:O³ ~  h ,²@  ³ ~ À  . Expected hypothetical mean is  ~ ,²,²:O³³ ~ ,²À ³ ~ À . Expected process variance is # ~ ,²= ²:O³³ ~ ,²À ³ ~ À . Variance of hypothetical mean is  ~ = ²,²:O³³ ~ = ²À ³ ~ À . A ~ b # ~ bÀ ~ À  À  À c Buhlmann credibility premium is A: b ² c A³ ~ ²À ³² bb  ³ b ²À³²À ³ ~ À . Answer: E

33. The simulation of an exponential variable @ with mean using the inverse transform method is & ~ c ² c "³ . In this case, ~ , so the three simulated exponential values are @ ~ c  ² c À³ ~ À   Á @ ~ c  ² c À ³ ~ À   Á and @ = c  ² c À³ ~ À  . The simulated value of the gamma is  ~ @ b @ b @ ~ À

 . This completes Step 1. According to the product algorithm we multiply the successive uniform ²Á ³ values until the product is first less then cÀ

 ~ À  . The remaining uniform ²Á ³ numbers are .1 , .7 , .7 , .1 . We see that À d À € À   € À d À d À . Since the product of the first two uniform numbers was greater than cÀ

 , but the product of the first three was less than cÀ

 , the simulated value of ? is 2. Answer: C

34. According to the existence of moments test, a distribution has a light right tail if all positive moments exists; this means that ,´?  µ  B if  € . If not all positive moments exist then the distribution has a heavy right tail. From the Exam C Tables we see that all positive moments exist for the Gamma, Lognormal and Exponential distributions and we also see that not all positive moments exist for the LogLogistic, Paralogistic and Inverse Exponential distributions. For instance, for the Loglogisitic distribution with parameters  and , ,´?  µ exists only for c      (not for all  ). Therefore, according to the existence of moments test, the LogLogistic, Paralogistic and Inverse Exponential distributions all have heavier tails than the Gamma, Lognormal and Exponential distributions. All three statements are true. Answer: E

35. The likelihood ratio test is an approximate test of the null hypothesis that model B is preferable to model A, where model B has more parameters than model A. The test statistic is ²) c ( ³ , where  represents the log of the likelihood at the estimated parameter values. This statistic is approximately chi-square with degrees of freedom equal to #parameters in model ) c #parameters in model (. In this case the Poisson is model A (one parameter) and the negative binomial is model B (two parameters). The test statistic is ² c  À c ² c  À ³³ ~  . Answer: E

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

PRACTICE EXAM 4

PE-61

ACTEX EXAM C/4 - PRACTICE EXAM 4 1. ? has a Pareto distribution with parameters  and . The random variable @ defined as follows: @ ~ ² ?b ³. Identify the type of distribution that @ has. A) Exponential with mean  B) Exponential with mean  C) Gamma with parameters  and

E) Inverse Pareto with parameters  ~

D) Gamma with parameters  and

 

and

2. ? has a normal distribution with a mean of $ and variance of 1. $ has a normal distribution with a mean of 1 and variance of 1. Find the 95-th percentile of ? . A) 3.00 B) 3.33 C) 3.67 D) 4.00 E) 4.33

3. A loss random variable ? has the following characteristics. There is a 90% chance of no loss occurring, ? ~ , and there is a 10% chance that a positive loss occurs. If a positive loss occurs, it is uniformly distributed between 1000 and 5000. An insurance policy on this loss has an ordinary deductible of 2000 applied. Find the expected cost per loss (including when ? ~ ) and the expected cost per payment for this policy. Find the variance of the cost per payment random variable. A) 600,000 B) 650,000 C) 700,000 D) 750,000 E) 800,000

4. Insurance losses are a compound Poisson process where: (i) The approvals of insurance applications arise in accordance with a Poisson process at a rate of 1000 per day. (ii) Each approved application has a 20% chance of being from a smoker and an 80% chance of being from a non-smoker. (iii) The insurances are priced so that the expected loss on each approval is c 100. (iv) The variance of the loss amount is 5000 for a smoker and is 8000 for a non-smoker. Calculate the variance for the total losses on one day's approvals. A) 13,000,000 B) 14,100,000 C) 15,200,000 D) 16,300,000 E) 17,400,000

5. The random variable ? has an exponential distribution with mean . The loss random variable @ is defined to be a mixture of ? with mixing weight  , and 0 with mixing weight  c  , where     . A random sample of  losses is observed: @ Á À À À Á @ . Suppose that  of the losses are 0 and the remaining losses are € . Find the maximum likelihood estimators of  and in terms of Á  and @ Á À À À Á @ . '@ '@ '@ '@ A)   B) c C)   D) b E) None of A, B, C or D is correct 

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

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PRACTICE EXAM 4

6. XYZ Re provides reinsurance to Bigskew Insurance Company. XYZ agrees to pay Bigskew for all losses resulting from "events", subject to: a $500 deductible per event and a $100 annual aggregate deductible For providing this coverage, XYZ receives a premium of $150. Use a Poisson distribution with mean equal to 0.15 for the frequency of events. Event severity is from the following distribution. Probability Loss 250 0.10 500 0.25 800 0.30 1,000 0.25 1,250 0.05 1,500 0.05  ~ % What is the actual probability that XYZ will pay out more than it receives? A) 8.9% B) 9.0% C) 9.1% D) 9.2% E) 9.3%

7. You are given the following grouped data set of 80 random data points taken from the distribution of ? : Interval Number of Data Points ²Á µ  ² Á µ  ²Á µ

²Á µ 

²Á µ  ²Á µ  ²Á µ  Find the empirical estimate of the -th percentile of ? . A) 640 B) 644 C) 648 D) 652 E) 656

8. Loss data for 925 policies with deductibles of 300 and 500 and policy limits of 5,000 and 10,000 were collected. The results are given below: Deductible Range 300 500 Total ²Á µ  c  ² Á µ    ²Á µ      ² Á Á µ    At        At Á  Total  

 Using the Kaplan-Meier approximation for large data sets with  ~ À and  ~ À , estimate - ² ³. A) 0.25 B) 0.36 C) 0.47 D) 0.58 E) 0.69

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

PRACTICE EXAM 4

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Questions 9 and 10 are based on the following random sample of 12 data points from a random variable ? 7 , 12 , 15 , 19 , 26 , 27 , 29 , 29 , 30 , 33 , 38 , 53 9. Assuming a uniform distribution on interval ²Á ³ , maximum likelihood estimation is applied to estimate . Find the estimated variance of the parameter estimate. A) 10 B) 12 C) 14 D) 16 E) 18

10. The mle found in question 5 is used with the delta method to construct an approximate 95% confidence interval for the probability 7 ´? € µ . Find the upper limit of the interval. A) .47 B) .49 C) .51 D) .53 E) .55

11. The random variable ? has pdf  ²%³ ~ % for   %   . ? Á ÀÀÀÁ ? is a random sample from the distribution of the continuous random variable ? . 

V ~   ? is taken as an estimate of the distribution mean . Find 4 :, ²V ³ . c ~  A) 

b B) ²c³ 

 C) 

b D)  ²c³ 

b E) ²c³ 

12. You are given the following random sample of size 4: 1 , 2 , 5 , 9 An exponential distribution with a mean of is used as a model for the data. In the -plot for this data set and model distribution, it is found that the right tails of the exponential distribution are thinner than the right tails of the (smoothed) empirical distribution at each of the data points. Determine the maximum value of (rounded to the nearest .1) that is consistent with this -plot. A) 3.5 B) 3.9 C) 4.5 D) 4.9 E) 5.5

13. The prior distribution of the parameter has pdf ² ³ ~  for € . The model distribution has a uniform distribution on the interval ´ Á  µ . Find the posterior density of given % if % €  , and indicate the region of density for the posterior. A) % B) % C) % D) % E) %  

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

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PRACTICE EXAM 4

Questions 14 and 15 relate to the following information. A portfolio of independent risks is divided into two classes of equal size. All of the risks in Class 1 have identical claim count and claim size distributions as follows: Class 1 Class 1 Number of Claims Probability Claim Size Probability 1 1/2 50 2/3 2 1/2 100 1/3 Class 2 Number of Claims 1 2

Probability 2/3 1/3

Class 2 Claim Size Probability 50 1/2 100 1/2

The number of claims and claim size(s) for each risk are independent. A risk is selected at random from the portfolio, and a pure premium of 100 is observed for the first exposure period. 14. Determine the Bayesian estimate of the expected number of claims for this same risk for the second exposure period. A) 1.0 B) 1.1 C) 1.2 D) 1.3 E) 1.4

15. An aggregate claim of 150 is observed for this risk for the second exposure period. Determine the Buhlmann credibility estimate of the expected aggregate claim for this same risk for the third exposure period. A) 100 B) 125 C)  D) 150 E)   

16. An actuary applies empirical Bayes credibility analysis to a portfolio consisting of  ~  policyholders. Each policyholder has  ~  exposure observations. The data summary is as c c c follows: ?  ~ ?  ~ À Á ?  ~ À Á    c  c  c   ²? c ?  ³ ~ À Á  ²? c ?  ³ ~ À Á  ²? c ?  ³ ~ À . ~

~

~

The estimated process variance V #  is found based on this data. Another actuary combines policyholders 1 and 2 together as a single policy holder, and applies empirical Bayes credibility analysis to the newly grouped portfolio of 2 policyholders (previous policyholders 1 and 2 combined are now "policyholder 1" and previous policyholder 3 is now "policyholder 2"). The estimated process variance V# is found by the second actuary. Find V# c V# . A) c À B) c À C)  D) À E) À

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

PRACTICE EXAM 4

PE-65

17. You are given a sample of losses from an exponential distribution. However, if a loss is 1000 or greater, it is reported as 1000. The summarized sample is: Reported Loss Number Total Amount Less than 1000 62 28,140 38,000 1000 38 Total 100 66,140 Determine the maximum likelihood estimate of , the mean of the exponential distribution. A) Less than 650 B) At least 650, but less than 850 C) At least 850, but less than 1050 D) At least 1050, but less than 1250 E) At least 1250

18. 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 first year and 2 claims in the second year. Find the probability that the risk will have 2 claims in the third year. A) Less than .20 B) At least .20 but less than .21 C) At least .21 but less than .22 D) At least .22 but less than .23 E) At least .23

19. You are given: (i) claim count 5 has a binomial distribution with parameters 4 and  ~ À . (ii) 4 has a discrete uniform distribution on the integers Á Á Á ÀÀÀÁ  . Find = ´5 µ . A) 1.21 B) 1.24 C) 1.27 D) 1.30 E) 1.33

20. For a particular loss random variable ? , if an ordinary deductible of 1000 is applied, the mean excess loss is 3500. If the deductible of 1000 is applied to ? as a franchise deductible, then the expected cost per loss is 3600. Find the expected cost per loss if the deductible of 1000 is applied as an ordinary deductible. A) 2000 B) 2200 C) 2400 D) 2600 E) 2800

21. A compound distribution : has frequency 5 and severity ? , both of which are members of the ²Á Á ³ class. You are given the following: ,²5 ³ ~ À Á = ²5 ³ ~ À  Á ,²:³ ~ À Á = ²:³ ~  À Find 7 ²: ~ ³ . A) Less than .025 B) At least .025 but less than .050 D) At least .075 but less than .100 E) At least .100

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C) At least .050 but less than .075

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

PE-66

PRACTICE EXAM 4

22. For a collective risk model: (i) The number of losses has a geometric distribution with a mean of 2. (ii) The common distribution of the individual losses is: % ? ²%³  À

 À An insurance covers aggregate losses subject to a deductible of 3. Calculate the expected aggregate payments of the insurance. A) Less than 1.0 B) At least 1.0 but less than 1.2 C) At least 1.2 but less than 1.4 D) At least 1.4 but less than 1.6 E) At least 1.6

23. PQR Re provides reinsurance to Telecom Insurance Company. PQR agrees to pay Telecom for all losses resulting from "events", subject to a $500 per event deductible. For providing this coverage, PQR receives a premium of $250. Use a Poisson distribution with mean equal to 0.15 for the frequency of events. Event severity is from the following distribution: Probability Loss 250 0.10 500 0.25 750 0.30 1,000 0.25 1,250 0.05 1,500 0.05  ~ % Using the normal approximation to PQR's annual aggregate losses on this contract, what is the probability that PQR will pay out more than it receives? A) Less than 12% B) At least 12%, but less than 13% C) At least 13%, but less than 14% D) At least 14%, but less than 15% E) 15% or more

24. The "redistribute to the right" procedure is a method of dealing with censored data when  estimating :²!³. All  data points start out with "probability mass" of  . When an observation(s) is (are) censored, its (their) probability mass(es) at the time of censoring is (are) divided among the remaining individuals still alive at that time. The redistribute to the right method is applied to a group of 10 individuals, one of whom had a censored observation. The other 9 deaths were observed to occur at all different times. We are given that :  ²! ³ ~ À (! is the time of the 8-th observed death). The censored observation occurred A) before the first death B) between the first and third deaths C) between the third and fifth deaths D) between the fifth and seventh deaths E) between the seventh and ninth deaths

25. The random variable ? has the density function  ²%³ ~ ²% b ³cc Á   %  B Á  €  Assuming  € , determine the method of moments estimator of . c c c c c ? ? ?b c c c c A) ? B) ?c C) ?b D) ?c E) ? ?

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

PRACTICE EXAM 4

PE-67

26. Which of the following statements is true? A) For a null hypothesis that the population follows a particular distribution, using sample data to estimate the parameters of the distribution tends to decrease the probability of a Type II error. B) The Kolmogorov-Smirnov test can be used on individual or grouped data. C) The Anderson-Darling test tends to place more emphasis on a good fit in the middle rather than in the tails of the distribution. D) For a given number of cells, the critical value for a chi-square goodness-of-fit test becomes larger with increased sample size. E) None of A, B, C or D are true.

27. Annual claim counts follow a Negative Binomial distribution. The following claim count observations are available: Year: 2005 2004 2003 Claim Count: 0 3 5 Assuming each year is independent, calculate the likelihood function of this sample. 

  A) ²  b ³ ² b ³ 

  B) ²  b ³ ² b ³ 

  C) ²  b ³ ² b ³ 

  D) ²  b ³ ² b ³ 

  E) ²  b ³ ² b ³

 ²b³ ²b³ [ [  ²b³ ²b³ [ [  ²b³ ²b³ ²b3³ [ [  ²b³ ²b³ ²b3³²b³ [ [  ²b³ ²b³ ²b3³²b³ [ [

28. 40 observed losses have been recorded in thousands of dollars and are grouped as follows: Interval Number of Total Losses Losses ($000) ($000) ² Á  µ 

  ²  Á µ   ²Á µ   ²Á B³   The null hypothesis, / , is that the random variable ? underlying the observed losses, in thousands, has the density function   ²%³ ~ % , % €  . Calculate the value of the chi-square goodness-of-fit statistic used to test the null hypothesis. A) 7.0 B) 7.2 C) 7.4 D) 7.6 E) 7.8

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

PE-68

PRACTICE EXAM 4

29. The ABC Insurance Company has decided to establish its full credibility requirements for an individual state rate filing using a standard under which the observed total cost of claims should be within 5% of the true value with probability .95 . The claim frequency follows a Poisson  distribution and the claim severity is distributed according to the distribution  ²%³ ~ Á for   %  Á  . What is the expected number of claims necessary to obtain full credibility? A) Less than 1500 B) At least 1500, but less than 1800 C) At least 1800, but less than 2100 D) At least 2100, but less than 2400 E) At least 2400

 30. ? has the following spliced distribution:  ²%³ ~ H   The mean of ? is 1.25. Find the variance of ? . A)  B)  C)  D)  E)      

% % otherwise

31. 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 7 ´? ~ O? ~ µ. A) .15 B) .20 C) .25 D) .30 E) .35

32. You are given the following: - Partial Credibility Formula A is based on the methods of limited fluctuation credibility, with 1600 expected claims needed for full credibility. - Partial Credibility Formula B is based on Buhlmann's credibility formula with a  of 391. - One claim is expected during each period of observation. Determine the largest number of periods of observation for which Partial Credibility Formula B yields a larger credibility factor than Partial Credibility Formula A. A) 524 B) 526 C) 528 D) 530 E 532

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PRACTICE EXAM 4

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33. Semi-parametric empirical Bayesian credibility is being applied in the following situation. The distribution of annual losses ? on an insurance policy has pdf  ²%O ³ ~ %  for   %  , 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 4-th year for this policy. A) Less than 1.1 B) At least 1.1 but less than 1.3 C) At least 1.3 but less than 1.5 D) At least 1.5 but less than 1.7 E) At least 1.7

34. The following random sample of size 7 is drawn from a distribution: 1 , 1 , 2 , 4 , 5 , 6 , 10 The distribution median is estimated, and the estimator used is the sample median. The following five bootstrap samples, each of size 7 have been simulated from the empirical distribution of the original random sample. Sample 1 : 1 , 1 , 1 , 5 , 6 , 6 , 10 Sample 2 : 1 , 2 , 2 , 4 , 6 , 6 , 10 Sample 3 : 2 , 4 , 4 , 5 , 5 , 6 , 6 Sample 4 : 1 , 1 , 4 , 4 , 4 , 10 , 10 Sample 5 : 1 , 2 , 5 , 6 , 6 , 6 , 10 Use these five bootstrap samples to determine the bootstrap approximation to the mean square error of the sample median estimator. A) 1.0 B) 1.2 C) 1.4 D) 1.6 E) 1.8

35. The parameter  has prior distribution ²³ ~  c b  ²  c° ³ (mixture of two exponentials). The model distribution ? has a conditional distribution given  that is Poisson with mean . Find the Buhlmann credibility premium if there is a single observation of 0.



A)  B)  C)  D)  E) 

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PRACTICE EXAM 4

ACTEX EXAM C/4 - PRACTICE EXAM 4 SOLUTIONS 

1. ? ²%³ ~ ²%b ³b for % €  . @ ~ ² ?b S ? ~ ²@ c ³ ~ ²@ ³. ³ ~ ²?³ 



  & & c& @ ²&³ ~ ? ²²&³³ h  Z ²&³ ~ ²²&³b . ³b h  ~ ² & ³b h  ~   This is the pdf of the exponential distribution with mean  . Answer: A

2. The conditional distribution of ? given $ has ,´?O$µ ~ $ and = ´?O$µ ~ . We are given that ,´$µ ~ = ´$µ ~  . ? is a continuous mixture with mean ,´?µ ~ ,´ ,´?O$µ µ ~ ,´$µ ~  and = ´?µ ~ = ´ ,´?O$µ µ b ,´ = ´?O$µ µ ~ = ´$µ b ,´µ ~  b  ~  . The continuous mixture of a normal "over" a normal distribution is also normal, so the unconditional distribution of ? is normal with mean 1 and variance 2. c c The 95-th percentile of ? is  , where 7 ²?  ³ ~ 7 ² ?c j  j ³ ~ )² j ³ ~ À . 



From the standard normal table we get c j ~ À  , so that  ~ À . 



Answer: B

3. The distribution of ? is a mixture of a discrete point at ? ~ , with mixing weight 7 ²? ~ ³ ~ À , and a continuous uniform distribution on then interval ´Á µ with mixing weight .1. The pdf of ? on ´Á µ is ? ²%³ ~ ²À³²À ³ ~ À for   %   .  if ?   . ? c  if ? €    ,´@3 µ ~  ²% c ³ h ? ²%³ % ~  ²% c ³ h ²À ³ % ~ À À  ,´@3 µ ~  ²% c ³ h ²À ³ % ~  Á  À The cost per loss variable is @3 ~ ²? c ³b ~ D

,´@ µ

3 The cost per payment variable @7 has expected value ,´@7 µ ~ 7 ²?€³ .   7 ²? € ³ ~  ? ²%³ % ~  À % ~ À .

,´@ µ

3 ,´@7 µ ~ 7 ²?€³ ~ À À ~   .

,´@  µ

 Á

3 ,´@7 µ ~ 7 ²?€³ ~ À ~ Á Á  . = ´@7 µ ~ Á Á  c   ~  Á  .

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Answer: D

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

PRACTICE EXAM 4

PE-71

4. The total losses for the day can be formulated as : ~ : b : , where : is the total losses from smokers and : is the total losses from non-smokers. The number of approvals in a day has a Poisson mean of 1000. Since 20% are smokers and 80% non-smokers, the number of approvals of smokers per day is Poisson with mean 200 and for nonsmokers it is Poisson with mean 800. We must assume that application approvals arise independently, and : is independent of : , so that = ´:µ ~ = ´: µ b = ´: µ . : has a compound Poisson distribution with  ~  (frequency) and severity @ with ,´@ µ ~ c  and = ´@ µ ~  . Similarly, : has a compound Poisson distribution with  ~  (frequency) and severity @ with ,´@ µ ~ c  and = ´@ µ ~  . Then using the general form of variance of a compound distribution, we have = ´: µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ ~ ² ³ b ² c ³ ~ Á Á  , and = ´: µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ ~ ² ³ b ² c ³ ~ Á Á . = ´:µ ~ Á Á  b Á Á  ~ Á Á  . Note that since : has a compound Poisson distribution, it is also true that = ´: µ ~  h ,´@ µ ~  h ´= ²@ ³ b ²,´@ µ³ µ ~ ²³´  b ² c ³ µ ~ Á Á  (and a similar comment applies to : ). Answer: E

5. @ has a mixed distribution with a probability mass at @ ~  which has probability 7 ²@ ~ ³ ~  and density function @ ²&³ ~ ² c ³ h  c&° for & €  . The likelihood function will be the product of a factor of  for each loss that is 0, and a factor of ² c ³ h  c&° for each loss that is € . The loglikelihood will be a sum of factors of   for & each loss that is 0, and ² c ³ c  c for each loss that is €  . There are  ~  c  losses that are € . Let us denote them ' Á ÀÀÀÁ ' (there are the  @ values that are € ). The loglikelihood function is ''  3 ~    b  ² c ³ c   c  . C C The mle's of  and are found by solving the two equations C  3 ~  and C  3 ~  .   Because of the additive separation of  and , it can be seen that the mle of  is V ~  b ~   '' ''  (the usual binomial parameter mle) and V ~  ~ . Since the 'Z s are the non-zero @ 's, it '@ is true that '' ~ '@ , so that V ~ c . 



c

Answer: B

6. We denote by : the aggregate amount of payment made by XYZ before the aggregate deductible of 100 is applied, but after the individual claim deductible of 500 has been applied to each individual claim for the year. : has a compound Poisson distribution with mean frequency 0.15 (per year) and severity distribution @ (after the individual claim deductible of 500): Probability Individual Claim Payment @ 0 0.35 (if original loss is 250 or 500) 300 0.30 500 0.25 750 0.05 1,000 0.05

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PRACTICE EXAM 4

6. continued XYZ receives 150. With an aggregate deductible of 100, XYZ will pay out more than it receives if : €   . This will happen if there is at least one individual loss ? of 800 or more, which is equivalent to an individual claim @ (after deductible of 500) of 300 or more. Alternatively, 7 ´: €  µ ~  c 7 ´:   µ , and :    in one of two cases: (i) there are 5 ~  claims, or (ii) there are more than 0 claims, but each one is of amount 0 (after 500 deductible). B 7 ´:   µ ~  7 ´:   O5 ~ µ h 7 ´5 ~ µ . ~

We note that 7 ´:   O5 ~ µ ~  and 7 ´:   O5 ~ µ ~ ²À ³ (since there is a .35 probability of each loss being  500 and thus having a claim payment of 0). B B ²À ³ 7 ´:   µ ~  ²À ³ h 7 ´5 ~ µ ~  ²À ³ h cÀ h B

~ cÀ h 

~

~

²À dÀ ³ [

[

~

~ cÀ h ²À dÀ ³ ~ cÀ  ~ À  .

Then, 7 ´: €  µ ~  c À  ~ À  .

Answer: E

7. The empirical cdf has -  ²³ ~ À Á -  ²³ ~ À  . The estimated 60-th percentile  À cÀ VÀ c  ~ À  cÀ (linear interpolation in the ²c Á µ for which VÀ is found from c -  ²c ³  À  -  ² ³ ).  Answer: C VÀ ~  À .

8. Using the notation associated with the Kaplan-Meier approximation for large data sets, we have  ~  Á  ~  Á  ~  Á  ~  Á  ~ Á  .  ~  (there are 400 observations with deductible of 300) ,  ~ 25 (observations with deductible of 500) ,  ~  ~  ~  (there are no observations with deductible higher than 5000). % ~  , % ~  Á % ~  Á % ~  Á " ~  Á " ~  À c

c

~

~

With  ~ À and  ~ À , the number at risk at  is  ~   b À  c  ²% b " ³ c À " , V ² ³ ~  c 4 c and we use the estimator -

%c %  5Ä4 c c 5

.

Then with  ~ , we have  ~ À  c À " ~ À ²³ c À ²³ ~  Á  ~ ² b À  ³ c ²% b " ³ c À " ~ ´ b À ²  ³µ c ²  b ³ c À ²³ ~ À Á  ~ ² b  b À  ³ c ²% b % b " b " ³ c À " ~ ² b  b ³ c ²  b  b  b ³ c À ²³ ~   . V ² ³ ~  c 4 c % 54 c % 54 c % 5 Then ~  c 4 c

       54 c À 54 c   5

~ À .

Note that " ~  is the number of censorings in the interval from  ~  to  ~ . There is censoring at 5000, but that is counted in the interval from 5000 to 10,000. Answer: B

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

PRACTICE EXAM 4

PE-73

9. The pdf of the uniform distribution is  , so the likelihood function is 3´ µ ~  . This is maximized at the minimum feasible value for . Since each % must be in the interval ²Á ³ it must be true that € % for each % . Therefore € %¸% ¹ . The minimum feasible value for is the largest of the %'s, which is V ~ 4 %¸? ¹ . To find the variance of V we must find the distribution of @ ~ 4 %¸? ¹ and then the variance of V is = ´@ µ . Since there are 12 %'s we have the cdf for @ is -@ ²&³ ~ 7 ´@  &µ ~ 7 ´4 %¸? Á ÀÀÀÁ ? ¹  &µ ~ 7 ´²?  &³ q ²?  &³ q Ä q ²?  &³µ & ~ 7 ´?  &µ h 7 ´?  &µÄ7 ´?  &µ ~ ² ³ . &

The pdf of @ is @ ²&³ ~ -@Z ²&³ ~ 

for   &  .

The first and second moments of @ are & ,´@ µ ~  & h & ²&³ & ~   & ~   , &



,´@  µ ~  & h & ²&³ & ~   & ~  .  

 Then, = ´@ µ ~ ,´@  µ c ²,´@ µ³ ~  c ²  ³ ~    . The estimated variance is found using the mle of , which is 53.

² ³

The estimate variance is   ~ À .

Answer: C

10. For the uniform distribution on the interval ²Á ³ , we have 7 ´? € µ ~  c  ~ ² ³ À Using the maximum likelihood estimate for , the estimated variance of the estimated probability is =V ´²V ³µ ~ ´Z ²V ³µ h = ´V µ ~ ²  ³ h = ´V µ . Using the mle V ~  and = V ´V µ ~ À V

from question 5, we get the estimated variance of the estimate of the probability 7 ´? € µ to be   ²   ³ h ²À³ ~ À  . The estimate of the probability using the mle is  7 ´? € µ ~  c  V V ~  c  ~ À . The approximate 95% confidence interval for 7 ´? € µ is À f À jÀ  ~ ²À Á À ³ . Answer: C

11. ~ ,´?µ ~  %²%³% ~  ,   = ´?µ ~ ,´²? c  ³ µ ~  ²% c  ³ ²%³% ~  4 :, ²V ³ ~ = ´V µ b ´ ²V ³µ . 

  °     = ´V µ ~ = ´ c ? µ ~ ²c³ = ´? µ ~ ²c³ Á  ~ ~

      ²V ³ ~ ,´V µ c ~ ,´ c ? µ c  ~ ²c³ c  ~ ²c³ ~

°



 b 4 :, ²V ³ ~ ²c³ b 4 ²c³ 5 ~  ²c³  . Answer: D

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PRACTICE EXAM 4

12. The -plot uses the following smoothed empirical distribution function values, and model probabilities: %  

Empirical À À À

À Dist. Fn. (left tail) Model  c c°  c c°  c c °  c c ° Dist. Fn. (left tail) In order for the right tails of the exponential model to be thinner than those of the (smoothed) empirical distribution at each data point, we must have c°  À , and c2°  À , and and c °  À , and and c °  À . This translates into the following required inequalities for that must all be satisfied:  À Á  À  Á  À Á  À . Maximum value of consistent with all inequalities is 3.9 . Answer: B

13.  ²%O ³ ~  for  %   .

 ²%Á ³ ~  ²%O ³ h ² ³ ~  h  ~  (on the appropriately defined region for ? and ) .  % If   %   , then    % , and ? ²%³ ~    ~  c %  ~ %%c  . %  %    If % € , then   % , and ? ²%³ ~ %°   ~  c  ~  . 

Since % € , ? ²%³ ~

 %

and ² O%³ ~

 % % ,´ O%µ ~ %° h %    ~  . Answer: B

 ²%Á ³ ? ²%³

~

% °  °%

~

% %  

%

on the region %   % .

14. We wish to find ,´number of 2nd period claimO1st period claim ~ 100µ . The Bayesian approach shows that this expectation can be "mixed" or "factored" through the two classes: ,´number of 2nd period claimsO1st period claim ~ 100µ ~ ,´number of 2nd period claimsOfrom class 1µ h 7 ´from class 1O1st period claim ~ 100µ b ,´number of 2nd period claim 2Ofrom class 2µ h 7 ´from class 2O1st period claim ~ 100µ From the description of classes 1 and 2, we have ,´number of 2nd period claimsOfrom class 1µ ~ ²  ³ b ²  ³ ~  Á and ,´number of 2nd period claimsOfrom class 2µ ~ ²  ³ b ²  ³ ~  À Using Bayes theorem, we have

7 ´Oclass 1µh7 ´class 1µ

7 ´from class 1O1st period claim ~ 100µ ~ 7 ´Oclass 1µh7 ´class 1µb7 ´Oclass 2µh7 ´class 2µ ~

´²  ³²  ³b²  ³²  ³ µ²  ³       ´²  ³²  ³b²  ³²  ³ µ²  ³b´²  ³²  ³b²  ³²  ³ µ²  ³

~  

(in the numerator, 7 ´Oclass 1µ is 7 ´1 claimµ h 7 ´µ b 7 ´ claimsµ h ²7 ´ µ³ , similarly for class 2 in the denominator). Also, 7 ´Oclass 2µh7 ´class 2µ

7 ´from class 2O1st period claim ~ 100µ ~ 7 ´Oclass 1µh7 ´class 1µb7 ´Oclass 2µh7 ´class 2µ ~

´²  ³²  ³b²  ³²  ³ µ²  ³ ~   À ´²  ³²  ³b²  ³²  ³ µ²  ³b´²  ³²  ³b²  ³²  ³ µ²  ³

Then, ,´number of 2nd period claimsO1st period claim ~ 100µ    ~ ²  ³²  Answer: E  ³ b ²  ³²  ³ ~  ~ À .

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

PRACTICE EXAM 4

PE-75

15. We are given ? ~  and ? ~   for the chosen risk. c The Buhlmann credibility estimate is A? b ² c A³ , where   ~ ,´²#³µ and A ~ b # . In this case, # takes on two "values", class 1 and class 2. 

²class 1³ ~ (expected frequency)(expected severity) ~ ²  ³²   ³ ~  , and

²class 2³ ~ ²  ³² ³ ~  . Since ² ³ is a constant for all , we have  ~ = ´²#³µ ~  , and this case A ~  . Thus, the Buhlmann credibility estimate is  ~ ,´²#³µ ~  . Answer: A 16. Applying empirical Bayes in the equal sample size case with  ~  Á  ~  ,   c   V# ~ ²c³   ²? c ?  ³ ~ ²³²³ h ´À b À b Àµ ~ À

 . ~ ~

When policyholders 1 and 2 are combined, the resulting model has  ~  Á  ~  ,  ~  for  ~ Á Á ÀÀÀÁ  (combined old policyholders 1 and 2), c c  ~  Á  ~  for  ~ Á Á ÀÀÀÁ  . ?  ~ À as before, and ?  ~ À .   c   c   ²? c ? ²? c ?  ³ ~ À b À ~ À Á ³ ~

~ 

~

~

~

 c   c   ²? c ? ²? c ?  ³ ~ À . ³ ~

In this unequal sample size case, V# ~

  c  h    ²? c ?  ³ ~ b h ´À b Àµ ~ À  .  ² c³ ~ ~ 



~

V# c V# ~ À  c À

 ~ c À .

Answer: B

17. This problem involves maximum likelihood estimation with right-censored data.  The likelihood function to be maximized is 3 ~   ²% ³ h ´ c - ²"³µ , ~

where % Á ÀÀÀÁ % are the non-censored observations, and  is the number of right-censored observations, censored at limit ". In this case, the limit is " ~  , and there are  ~  right-censored observations (observations that are 1000 or greater). For the exponential distribution with mean , the density function is  ²%³ ~  h c%° and the distribution function is - ²%³ ~  c c%° . The likelihood function becomes



3 ~  ²  h c% ° ³ h ´c° µ ~   h c²'% ³° h c Á° ~

~   h c Á° h c Á° ~   h c

Á° . It is generally easier to maximize the log of the likelihood. The resulting value of is still the maximum likelihood estimate. The natural log of the likelihood is

Á

Á  3 ~ c   c Á and   3 ~ c  b  . Setting this equal to 0 results in ~   . This is the maximum likelihood estimate of . The general (and simple) rule for maximum likelihood estimation for the exponential distribution is that given a data set that may be include left-truncated (after deductible) payments, and rightcensored (limit) payments, the mle of the ground up loss mean is total amount paid

Á V ~ number of non-censored observations . in This example that will be  ~   . Answer: D

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PRACTICE EXAM 4

18. We are to find 7 ´? ~ O? ~ Á ? ~ µ and ,´? O? ~ Á ? ~ µ . 7 ´? ~ O? ~ Á ? ~ µ ~

7 ´²? ~³q²? ~³q²? ~³µ 7 ´²? ~³q²? ~³µ

.

In part (a) we found 7 ´²? ~ ³ q ²? ~ ³µ ~ À . We find 7 ´²? ~ ³ q ²? ~ ³ q ²? ~ ³µ in a similar way. 7 ´²? ~ ³ q ²? ~ ³ q ²? ~ ³µ c   c   c   ~ 2  h 3 h ²À ³ b 2  h 3 h ²À³ b 2  h 3 h ²À³ ~ À  . [

[

Then 7 ´? ~ O? ~ Á ? ~ µ ~

[ À  À ~

À .

Answer: D

19. = ´5 µ ~ = ´,´5 O4 µµ b ,´= ´5 O4 µµ ,´5 O4 µ ~ 4  ~ À4 , = ´5 O4 µ ~ 4 ² c ³ ~ À 4 . ,´4 µ ~ ²À³² b Ä b ³ ~ À Á = ´4 µ ~ ,´4  µc²,´4 µ³ ~ ²À³² b Ä b  ³ c ² À ³ ~ À . = ´,´5 O4 µµ ~ = ´À4 µ ~ À= ´4 µ ~ À , ,´= ´5 O4 µµ ~ ,´À 4 µ ~ À ,´4 µ ~ À À = ´5 µ ~ À b À ~ À . Answer: A

20. For an ordinary deductible of 1000, the mean excess loss is

,´²?c³b µ :²³

~   ,

where :²³ ~ 7 ²? € ³ . For a franchise deductible, the expected cost per loss is ,´²? c ³b µ b :²³ ~   . Therefore, ,´²? c ³b µ ~   :²³ , and it follows that   :²³ b  :²³ ~   , from which we get :²³ ~ À . Then, ,´²? c ³b µ ~  ²À ³ ~   is the expected cost per loss if the deductible of 1000 is applied as an ordinary deductible. Answer: E 21. The probability generating functions of :Á 5 and ? satisfy the relationship 7: ²!³ ~ 75 ²7? ²!³³ . Then, 7 ²: ~ ³ ~ 7: ²³ ~ 75 ²7? ²³³ . An ²Á Á ³ distribution must be either Poisson, Negative Binomial or Binomial. Binomial is the only one of the three with expected value greater than variance. Therefore, 5 is binomial, say with parameters  and . ,´5 µ ~  ~ À and = ´5 µ ~ ² c ³ ~ À  , and it follows that  ~ À and  ~  . The probability generating function of 5 is 75 ²'³ ~ ´ b ²' c ³µ ~ ´ b ²À³²' c ³µ . The mean of : is ,´:µ ~ ,´5 µ h ,´?µ , so that À ~ À,´?µ and we get ,´?µ ~ À . The variance of : is = ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ , so that  À ~ À= ´?µ b À ² À³ , and we get = ´?µ ~  . Since ? is an ²Á Á ³ distribution, it is either Poisson, Negative Binomial or Binomial. The Negative Binomial distribution is the only one of the three whose mean is less than its variance. Therefore ? has a negative binomial distribution, say with parameters  and  . ,´?µ ~  ~ and = ´?µ ~  ² b  ³ ~  , and it follows that  ~  and  ~  .   The probability generating function of ? is 7? ²!³ ~ ´c ²!c³µ  ~ ´c²!c³µ .   Then, 7? ²³ ~ ´c²c³µ  ~  , and   7 ²: ~ ³ ~ 7: ²³ ~ 75 ²7? ²³³ ~ 75 ² 

³ ~ ´ b ²À³² 

c ³µ ~ À  . Answer: D

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PRACTICE EXAM 4

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22. The amount paid is a stop-loss insurance with a deductible of 3 applied to aggregate losses. The aggregate loss has a compound distribution with an integer-valued severity. Therefore, : is integer-valued. ,´:µ ~ ,´5 µ h ,´?µ ~ ´²³²À ³ b ²³²À³µ ~ À . ,´²:c³b µ ~ ,´:µ c ,´: w µ . The geometric distribution with mean  ~  has probability function 



7 ²5 ~ ³ ~ ²b ³b ~ b . : is integer valued with probability values 7 ²: ~ ³ ~ 7 ²5 ~ ³ ~  Á  7 ²: ~ ³ ~ 7 ²5 ~ ³ h 7 ²? ~ ³ ~  d ²À ³ ~  Á 7 ²: ~ ³ ~ 7 ²5 ~ ³ h 7 ²? ~ ³ b 7 ²5 ~ ³ h ²7 ²? ~ ³³  ~  d ²À³ b  d ²À ³ ~  ,

 

and 7 ²: ‚ ³ ~  c 7 ²: ~ Á Á ³ ~  c ²  b  b  ³ ~  .   prob. 

Then,

: w~



 prob. 



prob. 



H

, prob. 

 so that ,´: w µ ~ ²³²  ³ b ²³²  ³ b ²³² 

 ³ ~ À . ,´²: c ³b µ ~ À c À ~ À . Answer: C

23. The amount paid out by PQR, say > (after deductible) has a compound Poisson distribution with a mean frequency of ,´5 µ ~ À and the following severity distribution (after deductible): Individual Claim Payment @ Probability 0 0.35 (if original loss is 250 or 500) 250 0.30 500 0.25 750 0.05 1,000 0.05 The first and second moments of the claim payment @ are ,´@ µ ~ ² ³²À³ b ² ³²À ³ b ² ³²À ³ b ²³²À ³ ~  À , and ,´@  µ ~ ´²  ³²À³ b ²  ³²À ³ b ²  ³²À ³ b ² ³²À ³µ ~  Á  À Then ,´> µ ~ ,´5 µ h ,´@ µ ~  À and = ´> µ ~ ,´5 µ h ,´@  µ ~ Á  À (this is a form of the variance for a compound Poisson distribution). We wish to find 7 ´> €  µ using the normal approximation with continuity correction. This is 7 ´> €  ´ ~  c 7 ´>   À µ ~  c )²  À cÀ ³ ~  c )²À³ ~  c À 

j ~ À  .

Á  À

Answer: A

24. If the withdrawal is before the first death, then the amount of failure mass redistributed to À each survivor is ~ À , and the estimated : values will decrease by steps of size À b À ~ À so that : ²! ³ ~ À . If the withdrawal is between the first and third À deaths then the amount of failure mass redistributed is either ~ À with : ²! ³ ~ À or À V ³ ~ .1149 . If the withdrawal is between the third and fifth deaths then the ~ À with :²! 

À

À

amount of failure mass redistributed is either ~ À  with : ²! ³ ~ À  or ~ À with : ²! ³ ~ .1 . Answer: C

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

PE-78

PRACTICE EXAM 4

25. This is a Pareto distribution with ~  . From the Exam C distribution tables, we have ,´?  µ ~

 !²b³!²c³ !²³

!²³!²c³  ~ c . !²³ c c  ?b c set c ~ ? S  ~ ?

Á so that ,´?µ ~

According to the method of moments, we

. Answer: E

26. A) False. See the bottom of page 427 of the Loss Models book. B) False. The K-S test is applied to individual data. C) False. See page 430 of the Loss Models text. D) False. The critical value depends on the number of cells, the number of estimated parameters, and the level of significance. Answer: E

27. The likelihood function is  ²³ h  ²³ h  ² ³ .

For the Negative Binomial,  ²³ ~ ²b ³ ,  ²³ ~ ²b³²b³²b³²b³  h [ ²b ³ b .  ²b³ ²b³ ²b³²b³ likelihood function is [ [

²b³²b³ [



h ²b ³b ,

and  ² ³ ~ The



h ²b ³ b .

Answer; E



²6 c, ³ , where there  interval groupings for the dataÀ , ~

28. The chi-square statistic is  ~ 

 is the observed number of occurrences in interval grouping . , ~ ´- ² ³ c - ²c ³µ is the expected number of occurrences of a total sample of size  in interval grouping  given that the hypothesized distribution is correct ²- is the distribution function of the hypothesized distribution, and the  's are the interval endpoints). In this case, for the  ~  interval groupings, we have  ~ , with 6 ~  Á 6 ~  Á 6 ~  Á 6 ~  . The distribution function of the hypothesized distribution is % %  - ²%³ ~ cB  ²!³ ! ~  ! ! ~  c % for % €  . 



Then , ~ ´- ²  ³ c - ²³µ ~ ´  c µ ~  Á 





, ~ ´- ²³ c - ²  ³µ ~ ´  c  µ ~ ,   , ~ ´- ²³ c - ²³µ ~ ´  c  µ ~ , and 

, ~ ´- ²B³ c - ²³µ ~ ´ c  µ ~  . The chi-square statistic is  ~ Answer: B

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² c³ ²c³ ²c³ ²c³ b b b ~ À.    

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

PRACTICE EXAM 4

PE-79

29. For a compound Poisson total claims distribution, the standard for number of exposures for &

= ´@ µ

full credibility for total claims is ²  ³ h  , the number of people leaving by cab has a mean of ² ³ ~   and variance ²³ ~   . If > is assumed to be approximately normal, then since > is an integer 7 ´> ‚  µ ~ 7 ´> ‚  À µ ~ 7 ´ >jc  ‚  À c  µ j  

~ 7 ´A ‚ c À

µ ~ 7 ´A  À

µ ~ À (A has a standard normal distribution). Answer: D

 

V  ³ ~  ~ À Á /²! V  ³ ~  b  ~ À  S  ~ À Á 24. /²!     V  ³ ~  b  b  ~ À  S  ~ À Á /²!     V  ³ ~  b  b  b  ~ À  S  ~ À  . /²! 









: ²! ³ ~ ² c  ³² c  ³² c  ³² c  ³ ~ ²À  ³²À ³²À ³²À  ³ ~ À  .     Answer: D

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

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PRACTICE EXAM 7

25. The kernel density estimator of the cumulative distribution function is 

V ²%³ ~  ²& ³ h 2& ²%³ , and the uniform kernel with bandwidth  is ~

 2& ²%³ ~ H

%c&b 

%&c

&c %&b .  %€&b There are 15 data points and the empirical probability function is     ²³ ~  Á ²³ ~  Á ²³ ~  Á ² ³ ~  Á ²³ ~  Á   ²³ ~  Á ²³ ~  .

For % ~ , we have 2& ² ³ ~ H



c&



&

& .  & V - ² ³ ~ ²³2 ² ³ b ²³ ² ³ b ²³2 ² ³ b ² ³2 ² ³ b ²³2 ² ³ b ²³2 ² ³ b ²³2 ² ³    

c   ~ ²  ³²³ b ²  ³²³ b ²  ³²³ b ²  ³² c

³ b ²  ³² ³ b ²  ³²³ b ²  ³²³ ~  .

Answer: D

26. Since the data are grouped, the likelihood function is

   3² ³ ~ - ²³ ´ c - ²³µ ~ ´ c b µ ´ b µ ~ ² b³  .  3² ³ ~   c  ² b ³ À    V Answer: E   3² ³ ~ c b ~  S ~  À

27. The covariance matrix of the estimates is the matrix inverse ´0 ² ³µc (recall that in the covariance matrix, the diagonal entries are the variances of the estimates of the  's, and the entries off the diagonal are the covariances between the estimates of the 's).    c  The inverse of > is c . > c  ?  ? À À  À À  Therefore, [0 ²V ³µc ~ ²À ³²À³c²cÀ ³ ~  h> À  À ? > À À ? Answer: C and the estimated variance of V  is 27.4 .

28.  ²%³ ~ % ,   %   S - ²%³ ~ % ,   %   .  % - ²%c ³ - ²%³ - ²%³ O- ²%³ c - ²%c ³O O- ²%³ c - ²%c ³O  À  À À À À  À À À À

À

À   À À À

À À  À   À À

À À À  À  À  À À À  À À + ~ the maximum of the final two columns ~ À . Answer: E

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

PRACTICE EXAM 7

PE-137

29. The claim amount distribution is irrelevant in establishing the standard for full credibility for expected number of claims. With a Poisson claim number distribution, the standard for full & credibility of expected number of claims is  ~ ²  ³ .  In this question, with  ~ À and  ~ À ,  ~ ² À

À ³ ~   À  À

 The number of expected claims needed for 60% credibility is  where k   À  ~ À , so that  ~ À . Answer: B

30. The prior distribution is exponential (gamma) with ~  . With a gamma prior distribution of  (exponential is a special case of the gamma distribution) and a Poisson model distribution  whose mean is , the posterior distribution is a gamma distribution as well, with Z ~  b  % and Z ~

 b



~

, where  is the number of observations (exposure periods), and  % is the 

~

total number of claims observed during those  periods. We are given that  % ~  . ~

    b

 ~ b  . For a Poisson model distribution the predictive mean is the same as the expected value of the posterior. Therefore the predicted number of claims in the  b -st period is  b Z Z ~ ² b ³² b Answer: E  ³ ~ b . Z

Z

Therefore, the posterior distribution is gamma with  ~  b  and ~

31. ? is continuous on   %  , with - ²%³ ~ À % , which rises from 0 to .25. ? has a probability mass at % ~ , with probability ²³ ~ À . Then ? is continuous on   %  , with - ²%³ ~ À b À ²% c ³ (this is just the straight line with slope .25 on the interval   %   and has - ²³ ~ À ). Then ? has a probability mass at ? ~ , with ²³ ~ À (this is true because as %S from the left, - ²%³SÀ , but then - ²³ ~ , so there is a jump in - ²%³ at % ~ ). The inverse transform simulation method is applied as follows: - if   "  À , then " ~ - ²%³ ~ À %, so % ~ " is the simulated value, - if À  "  À , then the simulated value of ? is % ~ , - if À  "  À , then " ~ - ²%³ ~ À b À ²% c ³ ~ À b À %, so that % ~ " c , - if À  "   , then % ~  . From the given uniform values, the simulated values are % ~  Á % ~  Á % ~  Á % ~ . The sample mean is 1.5 . Answer: C

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

PE-138

PRACTICE EXAM 7

32.  ~  groups, and  ~  ~  ~  ~  exposure periods. Since no mention is made of different numbers of policies from group to group, we can regard this situation as one exposure unit per exposure period, which is the basic Buhlmann empirical Bayesian credibility approach. V  ~  ~  V# , where  ~  ~  (one exposure unit per exposure period for two A V  b  b V c c c c exposure periods). ?  ~ Á ?  ~ À Á ?  ~ Á ? ~ À À   c  V# ~ ²c³   ²? c ?  ³ ~ ~

 ~ ²c³ .3

.3 20 B) > c30 .2 ?

c30 .2 C) > 50 ? .3

.3 .5 ?

5 D) > 3

3 2?

2 E) > c3

V 1 . V 2 A

c3 5 ?

19. 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 ²Á ³ , 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 3. Find the semi-parametric estimate of the losses in the 4-th year for this policy. A) Less than 1.5 B) At least 1.5, but less than 1.7 C) At least 1.7, but less than 1.9 D) At least 1.9, but less than 2.1 E) At least 2.1

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PRACTICE EXAM 9

Questions 20 and 21 relate to the following situation. Claims arrive for processing according to a Poisson process with mean rate  ~  per hour. Claim processing takes either  or  hour, with any given claim having a .5 probability of taking  hour. 20. Use the inverse transform method to simulate the number of claims in each of the first two hours. The uniform random numbers to be used in sequence to simulate the numbers of claims in hours 1 and 2 are: À Á À5 . Find the number of claims simulated in each hour. A) 1 in the first hour and 3 in the second hour B) 2 in the first hour and 2 in the second hour C) 3 in the first hour and 2 in the second hour D) 3 in the first hour and 1 in the second hour E) None of A, B, C or D is correct

21. The simulated arrival times (in hours) of the claims during the first 2 hours are .2 , .8 , 1.1 , 1.3 , 1.7 There is only one claims processor. The claim processing times of the successive claims are simulated using the inversion method, using the following uniform random numbers, where small random numbers correspond to small processing times: .8 , .6 , .1 , .1 , .7 . Determine the state of the claims processing system at the end of 2 hours. A) No claims are being processed, no claims are waiting B) A claim is being processed and no claims are waiting to be processed. C) A claim is being processed and one claim are waiting to be processed. D) A claim is being processed and two claims are waiting to be processed. E) A claim is being processed and three claims are waiting to be processed.

22. ? has a uniform distribution on the interval ´Á j$µ , and $ has a uniform distribution on the interval ´Á µ . Find the mean of the unconditional distribution of ? . A)  B)  C)  D)  E) 

23. A loss random variable has a continuous uniform distribution between 0 and $100 . An insurer will insure the loss amount above a deductible  . The variance of the amount that the insurer will pay is 69.75 . Find  . A) 65 B) 70 C) 75 D) 80 E) 85

24. The times of death in a mortality study are ! Á ! Á ! Á ÀÀÀ The following information is given. There was one death at time ! , two deaths at time ! and one death at time ! . The Product-Limit estimate of survival probability for those times are :  ²! ³ ~ À Á :  ²! ³ ~ À  and :  ²! ³ ~ À  . Determine the number of right-censorings that took place in the interval ´! Á ! ³ . A) 0 B) 1 C) 2 D) 3 E) 4

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PRACTICE EXAM 9

PE-169

25. On Time Shuttle Service has one plane that travels from Appleton to Zebrashire and back and each day. Flights are delayed at a Poisson rate of two per month. Each passenger on a delayed flight is compensated $100. The numbers of passengers on each flight are independent and distributed with mean 30 and standard deviation 50. (You may assume that all months are 30 days long and that years are 360 days long). Calculate the standard deviation of the annual compensation for the delayed flights. A) Less than $25,000 B) At least $25,000, but less than $50,000 C) At least $50,000, but less than $75,000 D) At least $75,000, but less than $100,000 E) At least $100,000

26. A farmer develops a model for his seeding season for a particular crop. The number of days in which crops can be seeded during the season has a Poisson distribution with a mean of 20. On a day suitable for seeding, the number of acres than can be seeded is either 1 or 2, each with probability .5. The farmer wishes to insure against a poor seeding season. The farmer purchases insurance which will pay if the number of acres seeded during the season is under 20. For each acre under 20 that is not seeded the insurance will 5000. : represents that number of acres that will be seeded in the season. The farmer has determined ,´²: c ³b µ ~ 1À . Find the expected insurance payment. A) 4000 B) 5000 C) 6000 D) 7000 E) 8000

27. You are given the following random sample of 6 observations from the distribution of the random variable ? : 2 , 4 , 4 , 5 , 7 , 10 Kernel smoothing is applied to estimate the density function of ? . The kernel function used for the data point & is the pdf of the normal distribution with mean & and variance 1. Use kernel V ²³ . smoothing to estimate the distribution function of ? at the point % ~  , A) Less then .06 B) At least .06, but less than .12 C) At least .12 but less than .18 D) At least .18, but less than .24 E) At least .24

28. You are given: (i) A sample of losses is: 600 700 ‚ 900 (the third loss is known only to be at least 900) (ii) No information is available about losses of 500 or less. (iii) Losses are assumed to follow an exponential distribution with mean . Determine the maximum likelihood estimate of . A) Less than 500 B) At least 500 but less than 600 C) At least 600 but less than 700 D) At least 700 but less than 800 E) At least 800

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PRACTICE EXAM 9

29. If the proposed model is appropriate, which of the following tends to zero as the sample size goes to infinity? A) Kolmogorov-Smirnov test statistic B) Anderson-Darling test statistic C) Chi-square goodness-of-fit test statistic D) Schwarz Bayesian adjustment E) None of A), B), C), or D)

30. When applying the method of limited fluctuation credibility to a certain compound Poisson distribution of total claims cost so that total claims cost will be within % of expected total claims cost % of the time, the full credibility standard is 1200 expected claims. We also know that the coefficient of variation of the severity distribution is 2. Suppose the following changes are made in our assumptions: (i) the coefficient of variation of the severity distribution is doubled to 4, and (ii) the standard for full credibility is based on total claims cost being within % of expected claims cost % of the time ² is unchanged). Find the new standard for full credibility. A) 500 B) 1000 C) 1020 D) 1200 E) 2040

31. In a portfolio of insureds, each insured will have either 0 or 1 claim in a year, with independence from one year to another. The probability that an individual insured will have a claim in a given year is %. The portfolio of insureds is such that for a randomly chosen individual from the portfolio, the probability % is uniformly distributed on ²Á ³ . A randomly chosen individual is found to have no claims in  consecutive years, where  ‚ . Determine the expected number of claims that the individual will have in the  b -st year.     A) c B) c C)  D) b E) b

32. Four machines are in a shop. The number needing repair in each week has a binomial distribution with  = 0.5. For each machine, the repair time, in hours, is uniformly distributed on [0,10]. You are to estimate ? , the total repair time (in hours) for a three-week period, using the inverse transformation method of simulation. Use the following numbers from the uniform distribution on [0Á 1] to simulate the number of machines needing repair during each of three weeks: 0.3, 0.6, 0.7. Use the following numbers from the uniform distribution on [0Á 1] to simulate repair times: 0.3 0.1 0.7 0.6 0.5 0.8 0.1 0.3 Determine ? . A) 17 B) 22 C) 23 D) 30 E) 34

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PE-171

33. ? has the following distribution : 7 ´? ~ µ ~ À Á 7 ´? ~ µ ~ À . The distribution of @ is conditional on the value of ? : if ? ~  then the distribution of @ is 7 ´@ ~ µ ~ À Á 7 ´@ ~ µ ~ À Á 7 ´@ ~ µ ~ À , and if ? ~  then the distribution of @ is 7 ´@ ~ µ ~ ÀÁ 7 ´@ ~ µ ~ À Á 7 ´@ ~ µ ~ À . A is the sum of @ independent normal random variables, each with mean and variance 2. What is = ´Aµ ? A) 5.0 B) 6.0 C) 7.0 D) 8.0 E) 9.0

34. Random sampling from the distribution of ? results in the three sample values 1, 2 and 4. A uniform distribution on the interval ´Á µ is fitted to the data set by finding the value of that minimizes the Kolmogorov-Smirnov goodness-of-fit statistic. Find . A) 4.4 B) 4.5 C) 4.6 D) 4.7 E) 4.8

35. 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 .80 100 .10 200 .10 400 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 .04

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PRACTICE EXAM 9

ACTEX EXAM C/4 - PRACTICE EXAM 9 SOLUTIONS

1. I. The mean residual lifetime with deductible  is B

For the single parameter Pareto, :²!³ ~ !  ²%³ % B B  Á  Á Then  :²!³ ! ~  ! ~ .  !

The mean residual lifetime is

  Á°  Á°  ~

 B :²!³ !  . :²³ B Á ~ ! %

% ~

 Á !

.

 , which is increasing. I is true.

Alternatively, if ? has an decreasing hazard rate, then it has an increasing mean residual lifetime. The hazard rate for ? is

c: Z ²%³ :²%³

 ²%³

Á°%

~ :²%³ ~  Á°% ~ % , which is a decreasing function of %.

II. It can be shown that the gamma distribution with  €  has an increasing hazard rate. B B &c&°  We would have to find :²!³ ~ ! ²&³ & ~ ! & . Using integration by parts, The hazard rate is

²!³ :²!³

~

Á !c!°  c!°  this is :²!³ ~ Ád  b Ád   . c!°  ! ° Á ~ ! !  , which is an !c!°  c!°   b  b

Ád 

Ád 

increasing function.



II is true. III. According to the hazard rate test, a random variable with a decreasing hazard rate has a heavy right tail, and a if the hazard rate is increasing the right tail is light. ? has a decreasing hazard rate so it has a heavy right tail, and @ has an increasing hazard rate so it has a light right tail. III is true. Answer: E

2. The randomly chosen policy is a mixture of the three policy types. The expected number of claims per year for a randomly chosen policy is ²À ³²À³ b ²À³²³ b ²À³ ~ À b À . The second moment of a Poisson random variable is ,´5  µ ~ = ´5 µ b ²,´5 µ³ ~  b  À The second moment of number of claims for a low risk policy is À b ²À³ ~ À. The second moment of number of claims for a medium risk policy is  b ²³ ~ . The second moment of number of claims for a high risk policy is  b ²³ . The second moment of the number of claims per year for a randomly chosen policy is ²À ³²À³ b ²À³²³ b ²À³² b  ³ ~ À b À² b  ³ . The variance of the number of claims in a year for a randomly chosen policy is À b À² b  ³ c ²À b À³ ~ À  b À  b À  . We are given that this is 1. , so that À  b À  b À  ~ À . This is the quadratic equation À  b À  c À  ~  . The two roots of the equation are 2.5 and c 2.67. We ignore the negative root.  ~ 2.5 . Answer: C

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

PRACTICE EXAM 9

PE-173

3. The loss elimination ration with a deductible of amount  is

,´?wµ ,´?µ

À

The distribution function of the loss distribution is - ²%³ ~  c c%° Á and ,´?µ ~  À With a deductible of 500, the limited expected value is  ,´? w µ ~  %  ²%³ % b ´ c - ² ³µ   c%° ~  % h   % b  h c ° ~ ² c %c%° c c%° ³c

%~  %~

b c° ~  À À

The LER is  À  ~ À  À With a deductible of amount , the limited expected value is  ,´? w µ ~  %  ²%³ % b ´ c - ²³µ   c%° ~  % h   % b  h c° ~ ² c %c%° c c%° ³c

%~ %~

b c° ~ ² c c° ³ .

²cc° ³

The LER is ~  c c° .  In order for this to be twice as large as .39347, we must have  c c° ~ À

 S  ~   À Answer: E

4. The probability of a given loss exceeding 500 is c ° ~ c° ~ À   . If there are  exposures, then the expected number of losses exceeding the deductible will be c° ~ À   . We are told that this is 10, so that  ~ ° . If all loss amounts doubled, the loss distribution will be exponential with mean 2000, so that - ²%³ ~  c c%° , and the expected number of losses exceeding 500 will be ´ c -$ ² ³µ ~ ° c° ~ ° ~ À  À Answer: C

5. Consider an employee with salary * and probability of becoming disabled  . Let 0 be the indicator random variable indicating whether or not disability will take place. Let ? denote the disability claim random variable for this employee. Then ,´?µ ~ ,´,´? “ 0µµ . If 0 ~  then there is no claim so that ,´? “ 0 ~ µ ~  . If 0 ~  (probability  ) then the expected number of weeks of disability is  h ²À ³ b  h ²À³ b  h ²À ³ b  h ²À ³ ~ À , so the expected disability claims will be À * . Thus, ,´?µ ~  h ²À *³ . We can find the variance of ? from = ´?µ ~ = ´,´? “ 0µµ b ,´= ´? “ 0µµ ~ ² c ³²À *³ b ,´= ´? “ 0µµ . If 0 ~  then = ´? “ 0 ~ µ ~  , and if 0 ~  then = ´? “ 0 ~ µ ~ *  h ²À ³ b ²*³ h ²À³ b ²*³ h ²À ³ b ²*³ h ²À ³ c ²À *³ ~ À  *  , and thus, ,´= ´? “ 0µµ ~ ² c ³ h  b  h ²À  *  ³ . Then, = ´?µ ~ ² c ³²À *³ b  h ²À  *  ³ . The disability claim variance for each of the category 1 employees is ²À ³²À  ³²  ³ b ²À ³²Á ³ ~ Á À ,

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

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PRACTICE EXAM 9

5. continued for category 2 employees it is ²À³²À ³²  ³ b ²À³² Á ³ ~ Á À , and for category 3 employees it is ²À³²À

³²  ³ b ²À³²Á ³ ~  Á À . The expected aggregate claim for disability is  h ²À ³² ³ b  h ²À³² ³ b  h ²À³² ³ ~ Á  . The variance for the aggregate disability claim random variable for all 1000 employees is ²³² Á À ³ b ² ³²Á À ³ b ²³² Á À ³ ~ Á Á  . The premium required to ensure with 95% probability that claims will be covered is 7 where 7 cÁ Á Á

7 ´7 ‚ :µ ~ 7 ´ j

:c,´:µ

7 cÁ

‚ j µ ~ À ¦ j ~ À  ¦ 7 ~ Á  . Á Á = ´:µ Total weekly payroll is  h  b  h  b  h  ~ Á  . The fraction of weekly Á  payroll is Á ~ À , or À % . Answer: D.

6. With the policy limit raised to 7000, the 30 policies between 6000 and 7000 will add Á  c  Á  ~ Á  in losses (total of those 30 claims minus what was paid under the policy limit of 6000). The 270 claims above 7000 will add 270,000 (and additional 1000 is paid on each of these 270 policies). So the total amount added to losses paid is 290,000, or an  Á average of  ~  , which brings the LEV to   b  ~  À An alternative approach is the following. With a policy limit of 6000, the empirical LEV is V w µ ~ (Total claim amount from claims  6000)b²³² ³ ~   . ,´? 

It follows that the total of the 1700 claims that were each no greater than 6000 is ² ³²³ c ²³² ³ ~ Á Á  À V w µ ~ (Total claim amount from claims  7000)b²³²³ À Then ,´? 

Total claim amount from claims  7000 is equal to Total claim amt from claims  6000 b Total claim amt from claims between 6000 and 7000 ~ Á Á  b Á  ~ Á Á  . V w µ ~ ÁÁb²³²³ ~  À Then ,´? Answer: D 

7. À   ~ c c ~  c  c  S  ~ À  c  ~ À  c À V V  ³ ~  b  ~ c ²À   ³ ~ À  À À   ~ c/²! ³ S /²!  c Since

 °  b c ~  b c²   °³   ~  ~ À  or À  À

 



~  b À  c ~ À  Á we solve the quadratic for  c

But if  ~ À  Á then  ~ À   À Thus, V  ³ Á and the Nelson-Aalen estimate of :²! ³ is cÀ  ~ À  À  ~  ~ À  ~ /²! Answer: D and get

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

PRACTICE EXAM 9

PE-175

8. After the deduction of 1 from the losses, the data grouping is Interval Number of Losses ($000) ² Á  µ 

²  Á µ  ²Á µ  ²Á B³  The 40-th and 90-th percentiles from the empirical distribution are  and 3, since 



~ À of the after-deductible payments are   , and 

 ~ À of the after-deductible  payments are   . We use  and 3 as the 40-th and 90-th percentiles in the Weibull & distribution. With Weibull distribution function - ²&³ ~  c %´ c ² ³ µ , we have the following percentile matching equations: °

 c %´ c ² ³ µ ~ À Á  c %´ c ²  ³ µ ~ À . ° Then, ² ³ ~ c  À ~ À   , and ²  ³ ~ c  À ~ À . ²° ³ ²  ° ³

À

~ À   ~ À  , which simplifies to  ~ À  , so that V ~ À (and then V ~ À ). Answer: D Dividing the second equation by the first, we have

9. According to the method of moments for a one parameter distribution, we set the sample mean equal to the distribution mean and solve for the parameter. The distribution mean is B % h ²% b ³cc % . This can be simplified using integration by parts B % h ²% b ³cc % ~ B % ´ c ²% b ³c µ %~B B ~ c %²% b ³c c c  ´ c ²% b ³c µ % ~ c  b  b  À c

%~

Integration by parts can be avoided if it is noticed that the distribution of ? is Pareto with  parameters  and ~  , and density  ²%³ ~ ²%b ³b ~ ²% b ³cc .

 Then the mean of the Pareto is ,´?µ ~ c ~ c . c c  ~ ~ ?b c . Applying the method of moments, we have ? ~ c , so that  ? We could also avoid integration by parts by using the substitution " ~ % b  in the integral. Answer: E

10. Given the sample values % Á ÀÀÀÁ % the likelihood function is 





~

~

~

cc

3²% Á ÀÀÀÁ % O³ ~   ²% Á ³ ~  ´²% b ³cc µ ~  <  ²% b ³ = The loglikelihood is





M²³ ~  3 ~    c ² b ³ <  ²% b ³ = ~    c ² b ³  ²% b ³ ~

The maximum likelihood equation is  V~

   ²% b³

 



~

 M²³ ~  c  ²% b ³ ~  , so that the mle is ~

. If the sample mean goes to infinity, then some of the sample elements

~

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PRACTICE EXAM 9

10. continued approach infinity (we are not necessarily assuming that the number of sample points goes to infinity). In that case, the denominator in the expression for  V goes to infinity and  V goes to 0. The significance of this is as follows, if the true value of  is 1 or less, then distribution has an infinite mean, so that sample means could be extremely large. Answer: A

11. For claims reported in 1997, only those settled in 1 or 2 years can be observed. The likelihood function will be formulated using conditional probabilities related to those events. The likelihood factor for a claim reported in 1997 and settled in 1 year is 3 ²³ 3 ²³b3 ²³

²c³

 ~ ²c³b²c³ ~ b and the likelihood factor for a claim reported in 1997 and ²c³

 ²³



3 settled in 2 years  ²³b ~ ²c³b²c³ ~ b . 3 3 ²³ The denominator 3 ²³ b 3 ²³ is the probability that a claim was observed in years 1 or 2, so

 ²³

3 7 ´claim observed in year 1Oclaim observed in year 1 or year 2µ ~  ²³b , and 3 3 ²³

 ²³

3 7 ´claim observed in year 2Oclaim observed in year 1 or year 2µ ~  ²³b is how the 3 3 ²³ conditional probabilities above are formulated. For claims reported in 1998, only those settled in 0 or 1 years are observed. The likelihood factor

 ²³

²c³

 3 for a claim reported in 1998 and settled in 0 years is  ²³b ~ ²c³b²c³ ~ b , and 3 3 ²³ the likelihood factor for a claim reported in 1998 and settled in 1 year is 3 ²³ 3 ²³b3 ²³

²c³



~ ²c³b²c³ ~ b .

For claims reported in 1999, only those settled in 0 years are observed, and the likelihood factor  ²³

is 3 ²³ ~ . 3 The likelihood function for the data set given is 





   3 ~ ´ b µ h ´ b µ h ´ b µ h ´ b µ h ´µ ~ ²b³ . The log of the likelihood function is  3 ~    c  ² b ³ . The maximum is found   where   3 ~  S  c b ~  S  ~  is the maximum likelihood estimate.

Answer: D

12. The two classes have the same number of risks, and so are equally likely to be chosen. 7 ²ÁOClass 1³h7 ²Class 1³

7 ²Class 1OÁ ³ ~ 7 ²ÁOClass 1³h7 ²Class 1³b7 ²ÁOClass 2³h7 ²Class 2³ ~

c h c h  [ ³² [ ³²  ³ c  c  c h c h  ² [ ³² [ ³²  ³b²  [h ³²  [h ³²  ³

²

~

c2 

c2 c  b

 ~ b c ~ À  À

Answer: D

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PRACTICE EXAM 9

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13. The hypothetical means are the expected number of claims from each class: /4 ~  Á /4 ~  . Then  ~ ,´/4 µ ~ ²³²  ³ b ²³²  ³ ~  , and  ~ = ´/4 µ ~ ²  ³´² c  ³ b ² c  ³ µ ~  . The process variances are the variance of the number of claims from each class: 7 = ~  Á 7 = ~  À Then # ~ ,´7 = µ ~ ²³²  ³ b ²³²  ³ ~  . With  ~  exposures, the Buhlmann credibility factor is     A ~ b ~ b # ~ ° ~  . The Buhlmann credibility premium is 

b °

c    A? b ² c A³ ~ ²  ³² b  ³ b ²  ³²  ³ ~ ~ À À

Answer: D

14. The prior density is ²³ ~ c ,  € . The joint density of ? and  at ? ~  is  ²Á ³ ~ ²O³ h ²³ ~ ²À ³´c b c µ h c ~ ²À ³´c b c µ . The marginal probability that ? ~  is B B 7 ´? ~ µ ~   ²Á ³   ~  ²À ³´c b c µ  ~ ²À ³´  b  µ ~  .  ²Á³

²À ³´c bc µ

The posterior density of  is ²O³ ~ 7 ´?~µ ~ ~ ² ³´c b c µ . ° The Bayesian estimate for the expected number of claims next year is B ,´? O? ~ µ ~  ,´?Oµ h ²O³  . Since the conditional distribution of ? given  is a mixture of a Poisson with mean  and a Poisson with mean 2, with mixing weights of .5 for each part of the mixture, it follows that ,´?Oµ ~ ²À ³´ b µ ~ À  . Then B ,´?Oµ h ²O³  ~ B À  h ² ³´c b c µ  ~ ² ³B ´c b c µ  ~ ² ³´  b  µ ~   .

Answer: D

c 15. 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 c V c b ² c A³ V  V ~  . We are given that with  ~  and @ A@ ~  , we have V , where A V b V c    V h ²³ b ² c A³ V  A V ~ ² c V ³ V~ V ~  , and with  ~  and @ ~ , we have V h b b V    V h ²³ b ² c A³ V  A V ~ ² c V ³ V~ V ~ . It follows that V h b b V V V c   b V ² V h V³,² V h  V³ ~ V ~  , so that  ~ À . Then, with  ~  and @ ~ , we have b b b V V V   b  À   V h ²³ b ² c A³ V  A h V ~ ² c V ³ V~ V ~ ² V ³² V h  V³ ~ ² À ³² ³ ~  . V b b b b

Answer: C

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PRACTICE EXAM 9

16. 7 ´ € O ~ µ ~

7 ´²~³ q² €³µ 7 ´~µ

À

c   7 ´² ~ ³ q ² € ³µ ~  7 ´ ~ Oµ  ²³  ~  c²c³ h   ~ c ~ À   c   7 ´ ~ µ ~  7 ´ ~ Oµ  ²³  ~  c²c³ h   ~ c ~ À  À 

7 ´ € O ~ µ ~ À  À  ~ À  .

Answer: D

17. The hypothetical mean for a driver with gradepoint  is /4 ~  c  , so that  ~ ,´/4 µ ~ ,´ c µ ~  c  ~  , since  is uniformly distributed on ´Á µ . The process variance of number of claims for a driver with gradepoint  is also 7 = ~  c  (since Poisson mean is equal to variance), and therefore, # ~ ,´7 = µ ~  . The variance of the  hypothetical mean is  ~ = ´/4 µ ~ = ´ c µ ~ = ´µ ~ 

 ~  (this is the variance of  the uniform distribution on ´Á µ ). The Buhlmann factor  is  ~ # ~ ° ~  , and the  credibility factor is A ~ b ~ bÀ ~   for the driver with five years of no claims. Then the Buhlmann credibility estimate for expected number of claims for this driver is c 

A? b ² c A³ ~ ²  Answer: C  ³²³ b ²  ³²³ ~  ~ À À

18. If maximum likelihood estimation is applied to a distribution with parameters  and  , then  C v C  M² ³ M² ³ y C C  C   x { the information matrix 0 ² ³ is the following  d  matrix c , C C M² ³ z w C  C  M² ³ C  where M² ³ is  3²  Á  ³. The covariance matrix of the estimates is the matrix inverse ´0 ² ³µc (recall that in the covariance matrix, the diagonal entries are the variances of the estimates of the  's, and the entries off the diagonal are the covariances between the estimates of 

the 's). The 1,1 entry of 0 ² ³ is c ,´ CC   3²  Á  ³µ ~ c ,² c ³µ ~ À 

The 1,2 entry and the 2,1 entry are the same (0 ² ³ is always a symmetric matrix) and they are 

c ,´ C C C  3²  Á  ³µ ~ c ,´² c ³µ ~ . 





The 2,2, entry is c ,´ CC  M² ³µ ~ c ,´ c µ ~ À Then 0 ² ³ ~ > 

  matrix is ´ 0 ² ³µc ~ hch h> c



c  c ~ . ? > c ?     Note that the inverse of the  d  matrix > is hch h> ?   c

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 , and the covariance ? c .  ?

Answer: E

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PRACTICE EXAM 9

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19. Hypothetical mean is ,´?O µ ~  .  Process variance is = ´?O µ ~  .

Expected hypothetical mean is  ~ ,´?µ ~ ,´ ,´?O µ µ ~ , ´  µ ~  ,´ µ .   Expected process variance is # ~ ,´ = ´?O µ µ ~ , ´  µ ~  ,´  µ .

Variance of hypothetical mean is  ~ = ´ ,´?O µ µ ~ = ´  µ ~  = ´ µ ~  ´ ,´  µ c ²,´ µ³ µ À

c From the sample, we can estimate ,´?µ as ? ~ , so this is also the estimate  ,´ µ . The estimate of ,´ µ is 4. From the sample we can estimate = ´?µ using the unbiased sample variance, c    

´ '? c ? µ ~

´  c ²³ µ ~ À .

 But = ´?µ ~  b # ~  ,´  µ b  ´ ,´  µ c ²,´ µ³ µ ~  ,´  µ c  ²,´ µ³ . Using the estimated variance of ? and the estimated mean of , we have À ~  ,´  µ c  ²³ , so that the estimate of ,´  µ is 18.06 .  Then, # ~  ,´  µ is estimated to be 1.505, and

 ~  ´ ,´  µ c ²,´ µ³ µ is estimated to be .515 .

V c b ² c A³ V  The estimate of losses in the 4th year is A@ V, c   V where A ~ ~ ~ À , and  V ~ ? ~ , b À  b VV# À  c V b ² c A³ V  so that A@ V ~ ²À ³²³ b ²À ³²³ ~ À .

Answer: A

20. The Poisson probability function for  ~  is ²³ ~ c ~ À  Á c

c



c



²³ ~  [h ~ À  Á ²³ ~  [h ~ À  Á ²³ ~  [h ~ À  Á ... The probability function and distribution function are: %     ÀÀÀ ²%³ À  À  À  À  ÀÀÀ - ²%³ À  À 

À 



À  ÀÀÀ The uniform number .7 results in 3 simulated claims in the first hour, since À  À  À , and the uniform number .5 results in 2 simulated claims in the second hour. Answer: C

21. Processing time is  hour if uniform random number is  À , and is  hour if € .5 .

The simulated processing times are  Á  Á  Á  Á  . Claim 1 arrives at .2 and is processed at .7 . Claim 2 arrives at .8 and is processed at 1.3, and there are 2 claims waiting (claims 3 and 4 have arrived). Claim 3 is processed at time 1.55, and there is 1 claim waiting. Claim 4 is processed at time 1.8 and there is one claim waiting (Claim 5 has arrived). Claim 5 is processed at time 2.3 . At time 2, a claim is being processed and no claims are waiting to be processed. Answer: B

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PRACTICE EXAM 9

22. The uniform distribution on ´Á j$µ has pdf ?O$ ²%O³ ~ j or equivalently,  ‚ % . Also, the pdf of $ is $ ²³ ~ .



for   %  j ,

  Then ? ²%³ ~  ?O$ ²%O³ h $ ²³  ~ % j h 1 d ~ ² c %³ for   %   .



 The mean of % is ,´?µ ~  % h ² c %³ % ~  .

j$

Alternatively, since ,´?O$µ ~  , we have j$  j  j ,´?µ ~ ,´ ,´?O$µ µ ~ ,´  µ ~   h $ ²³  ~    ~  . , if %

23. The insurer pays @ ~ D  %c

Then ,´@ µ ~ 

 % ²c³ 

%c , if %

Answer: B

with constant density .01 .

²c³ ²c³  ²%c³ and ,´@  µ ~    % ~  ²c³  c ´  µ . Substituting the possible answers, we

~

so that = ´@ µ ~  ~ , the variance of @ is 69.75 .

see that with

Answer: B

24. À ~ :  ²! ³ ~ :  ²! ³´ c  µ ~ ²À ³´ c  µ S  ~  S  ~ is the number alive just before the death at time ! . A similar analysis at time ! shows that À ~ :  ²! ³ ~ :  ²! ³´ c  µ ~ ²À³´ c  µ S  ~  S  ~  is the number alive just    before the 2 deaths at time ! . Therefore there are 10 survivors just after the 2 deaths at time ! , and 6 of them are still alive at the next death point ! . This implies that 4 observations were censored between ! and ! . Answer: E 25. The number of passengers compensated in one year : has a compound distribution. The frequency is 5 , the number of delayed flights in one year, which is Poisson with mean 24 (2 per month for 12 months). The severity @ is the number of passengers on a delayed flight. We have ,´5 µ ~ = ´5 µ ~  Á ,´@ µ ~  and = ´@ µ ~  ~   . Since the frequency has a Poisson distribution, it follows that = ´:µ ~ ,´5 µ h ,´@  µ À From the given information we have ,´@  µ ~ = ´@ µ b ²,´@ µ³ ~   b  ~  À Then, = ´:µ ~ ²³²³ ~ Á  . Alternatively, = ´:µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ ~ Á  À Each passenger on a delayed flight receives $100 in compensation. The total compensation paid in one year is : , and the variance is = ´:µ ~ ² ³² Á ³ À The standard deviation of annual compensation is j Á  ~  Á

À Answer: B

26. Insurance pays @ ~  d ² c :³b ~  d F

 c : 

:   : € 

~  d ² c ,´: w µ³ . : ~ ²: w ³ b ²: c ³b S  ~ ,´:µ ~ ,´: w µ b ,´²: c ³b µ ~ ,´: w µ b À S ,´: w µ ~  À S ,´@ µ ~ ² c  À ³ ~  . Answer: C

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PRACTICE EXAM 9

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27. The data points are & ~  Á & ~  Á & ~ Á & ~  Á & ~  . The empirical distribution is ²³ ~  , ²³ ~  , ² ³ ~  , ²³ ~  , ²³ ~ 

The kernel smoothed estimate of - ²³ is 

V ²³ ~  ²& ³ h 2& ²³ 

~

~ h 2 ²³ b  h 2 ²³ b  h 2 ²³ b  h 2 ²³ b  h 2 ²³ , where 2& ²%³ is the cdf of the normal distribution with mean & and variance 1. %c& 2& ²%³ ~ )²  ³ ~ )²% c &³ . 2 ²³ ~ )² c ³ ~ )²³ ~ À  Á 2 ²³ ~ )² c ³ ~ )² c ³ ~ À  Á 2 ²³ ~ )² c ³ ~ )² c ³ ~ À Á 2 ²³ ~ )² c ³ ~ )² c ³ ~  Á 2 ²³ ~ )² c ³ ~ )² c ³ ~  . V ²³ ~  h 2 ²³ b  h 2 ²³ b  h 2 ²³ b  h 2 ²³ b  h 2 ²³ Then,







~  h ²À ³ b  h ²À ³ b  h ²À ³ b  h ²³ b  h ²³ ~ À

. Answer: D

28. When we have truncated and censored losses for an exponential distribution, the mle of the total insurance payment ground up exponential mean is V ~ number of losses (non-censored) . In this case, the insurance payments are 100, 200 and 400. The mle of is bb ~   . 

Answer: A

29. The Kolmogorov-Smirnov test statistic is the maximum deviation of the fitted model distribution function from the empirical distribution function. If the model is appropriate, as the sample size increases, the empirical distribution function should approach the true model distribution function (which is the fitted model if it is appropriate), so the maximum deviation will go to 0. Answer: A

30. For a compound Poisson distribution of total claims cost, the standard for full credibility = ´@ µ

based on number of claims needed is  µ . This does not depend on the ? 's having exponential distributions. Answer: C

3. 1. True. This can be verified by looking at the moment generating function of the sum of two Poisson random variables and seeing that it is also Poisson. 2. False. The sum is negative binomial if  ~  Z (in that case the sum is negative binomial with parameters  b Z and  . This can be thought of in another way. If  is an integer, then the negative binomial is the number of failures until the -th success, where the probability of success on any given trial is b  . The sum of the two negative binomial random variables with  ~  Z is the number of failures until the -th success, plus the number of failures after that until the Z -th success; this total is the number of failures until the  b Z -th success). 3. True. The sum is binomial with parameters  b Z and  . This can be thought as follows. The binomial random variable with parameters  and  is the number of successes in  trials of an experiment, where  is the probability of success in any given trial. If the success probability is the same for two binomial random variable, ( ~  Z ), then the sum of the two independent binomial random variables is the number of successes in the first  trials, plus the number of successes in the next Z trials, which is the number of successes in  b Z trials. Answer: C

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PRACTICE EXAM 11

4. This problem involves a compound distribution. The frequency (number of prizes) is 5 and the severity (prize amount) is ? . The aggregate prize amount is : ~ ? b ? b Ä b ?5 , with mean ,´:µ ~ ,´5 µ h ,´?µ ~ ²À³²³ ~  and variance = ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ À In this case, = ´5 µ ~ ,´5  µ c ²,´5 µ³ ~ À c ²À³ ~ À , and = ´?µ ~ ,´?  µ c ²,´?µ³ ~ Á  c ²³ ~  Á  . Then, = ´:µ ~ ²À³² Á ³ b ²À ³²³ ~ Á  À The budget is ,´:µ b À j= ´:µ ~  b À j Á  ~ . Answer: E

5. This problem involves stop-loss insurance. The aggregate amount of prizes, : , is integervalued, with multiples of 100. The cost of the stop-loss insurance is ² b À ³,´²: c ³b µ, where ,´²: c ³b µ ~ ,´:µ c ,´: w µ . From Problem 29, we have ,´:µ ~  . We can find probabilities for : by looking at combinations of 5 and : . 7 ´: ~ µ ~ 7 ´5 ~ µ h 7 ´? ~ µ b 7 ´5 ~ µ h 7 ´? ~ µ h 7 ´? ~ µ ~ ²À ³²À³ b ²À³²À³²À³ ~ À , 7 ´: ~ µ ~ 7 ´5 ~ µ h 7 ´? ~ µ b 7 ´5 ~ µ h 7 ´? ~ µ h 7 ´? ~ µ b 7 ´5 ~ µ h 7 ´? ~ µ h 7 ´? ~ µ ~ ²À ³²À³ b ²À³²À³²À³ ~ À  . Then, ,´: w µ ~ 7 ²: ~ ³ b 7 ²: ‚ ³³ ~ ²À  ³ b ² c À c À  ³ ~ À , and ,´²: c ³b µ ~  c À ~

À . Therefore, the cost of the insurance is ² b À ³²

À³ ~ À . Answer: D 6. : ~ number of acres seeded has a compound Poisson distribution with mean ,´:µ ~ ,´?µ ~ ²³²À ³ ~  , and variance = ´:µ ~ ,´?  µ ~ ²³²  ³ ~  . : is an integer, so :c,´:µ = ´:µ

7 ´: ‚ µ ~ 7 ´: ‚ À µ ~ 7 ´ j

‚ À c µ ~ 7 ´A ‚ À µ ~  c )²À ³ ~ À

. j 

Answer: A

7.  ~  Á  ~  Á  ~  Á  ~  Á  ~  Á  ~ Á  ~  Á  ~ Á  ~   ~ Á  ~ Á  ~ Á  ~ Á ~ Á ~ Á  ~ Á ~ À         ² c ³ ~ ² ³²³ b ²³²³ b ²³²³ b ² ³²³ b ²³²³ !  





   b ² ³²³ b ²³² ³ b ² ³² ³ ~ À  . Greenwood's approximation is  =V ´:  ²³µ ~ ´:  ²³µ ²À ³ ~ ²À ³ ²À ³ ~ À . Answer: C

8.





 ² c  ³ !   

~ À  (from Problem 4 above), :  ²³ ~ À

À j= V ´: ²³µ

²À ³hjÀ

< ~ %> : ²³h´: ²³µ ? ~ %< ²À ³h´: ²³µ = ~ À  À    * ~ :  ²³ c ²À ³ h j= V ´:  ²³µ ~ À c ²À ³j À ~ À . °< °À  V + ~ ´:²³µ ~ ²À ³ ~ À À * c + ~ c À À Answer: D

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PRACTICE EXAM 11

PE-215

9. Given the likelihood function 3² ³ for a parameter based on an observed random sample or random experiment, the maximum likelihood estimate for is that value of that maximizes 3² ³. We first determine the likelihood function for the parameter . Let coin 1 be the one with probability  of heads, and let coin 2 be the one with probability  of heads. Let ( denote the event that the coin chosen at random and then flipped turns up as a head. Let ) and ) denote the events that a coin chosen at random is coin 1 () ) or coin 2 () ), respectively. Then, since ) and ) are mutually exclusive and exhaustive events (i.e., exactly one of ) and ) must occur), it follows that 7 ´(µ ~ 7 ´( q ²) r ) ³µ ~ 7 ´²( q ) ³ r ²( q ) ³µ ~ 7 ´( q ) µ b [( q ) µ ~ 7 ´( “ ) µ h 7 ´) µ b 7 ´( “ ) ] h 7 ) µ ~ 12 h 12 b  h  ~  ²b). The likelihood function for  based on the occurrence of event ( on two independent trials of the procedure outlines above is [  ²b³µ . Since 0    , this is maximized at the point  ~ . Answer: D 10. The sample has  ~  data points, so the smoothed empirical percentiles for the sample are %       ÀÀÀ ...   

  Smoothed Perc.       À À  À À

 À

À 

Since À d  ~ À , the smoothed empirical estimate of the 20-th percentile is 60% of the way from the 2nd to the 3rd data point, which is 13.8 . Similarly, since À d  ~ À, the smoothed empirical estimate of the 80-th percentile is 40% of the way from the 10-th to the 11-th data  % c  point, which is 35. The cdf of the lognormal with parameters  and  is - ²%³ ~ )²  ³ . The 20-th and 80-th percentiles of the standard normal distribution are 'À ~ c À  and 'À ~ À  .  À c    c  The percentile matching equations are ~ c À  and ~ À  À   Solving these two equations results in  V ~ À Á  ~ À  . c À Then, 7 ´? € µ ~  c - ²³ ~  c )²  À  ³ ~  c )²À ³ ~  c À ~ À  . Answer: C

11. The mle of the mean of an exponential random variable is the sample mean. c²% b% b% ³°

The likelihood function is 3²³ ~ ²  c% ° ³²  c% ° ³²  c% ° ³ ~ .  % b% b% The loglikelihood is M²³ ~  3²³ ~ c c    , and the maximum likelihood  c % b% b%   V ~ % b% b% ~ ? equation is M²³ ~  S c ~S .  







This is just the mle for the exponential distribution, which is the sample mean.

Answer: A

= ´?µ c c V~? Vµ ~ = ´?µ 12. Since  , it follows that = ´ ~ À For the exponential  distribution with mean  , the variance is the square of the mean, = ´?µ ~  . In this case, c V~? V is  ~ , and the estimate of  is  ~ bb ~  , so the estimated variance of    

~   .

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Answer: B

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

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PRACTICE EXAM 11

13. Let * represent the number of claims in a month for a randomly chosen individual. Then ² ³ ~ ,´* O# ~ µ ~ (mean of the Poisson distribution with parameter ),  ~ ,´*µ ~ ,´²#³µ ~ ,´#µ ~ À (mean of the exponential), #² ³ ~ = ´*O# ~ µ ~ (variance of the Poisson distribution with parameter ), # ~ ,´#²#³µ ~ ,´#µ ~ À, and  ~ = ´²#³µ ~ = ´#µ ~ À (variance of the exponential is the square of the mean). Let *Á Á *Á Á ÀÀÀÁ *Á denote the number of claims in each of the final 6 months of 2000 for the individual, and let *Á Á *Á Á ÀÀÀÁ * Á 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 * . À    ~ # ~ À ~  Á A ~ b ~  b ~   , c *Á b*Á bÄb*Á b*Á b*Á bÄÀb* Á  and * ~ ~  .  Then the Buhlmann(-Straub) credibility premium is c    A* b ² c A³ ~ ²   ³²  ³ b ²  ³²À³ ~  per month. 

The credibility premium for the first three months of 2001 is ²  ³ ~  À Answer: A 

²6 c, ³ À 6 ~   Á 6 ~   Á 6 ~  Á 6 ~  Á 6 ~   , ~

14.  ~ 



, ~ - ²³ ~ ´ c ²  ³À µ ~  À Á 



, ~ ´- ²À ³ c - ²³µ ~ ´²  ³À c ² À ³À µ ~ À Á 



, ~ ´- ² ³ c - ²À ³µ ~ ´² À ³À c ² ³À µ ~ À Á 



, ~ ´- ²³ c - ² ³µ ~ ´² ³À c ²  ³À µ ~ À Á 

, ~ ´- ²B³ c - ²³µ ~ ´²  ³À µ ~ À À Á

² c À ³ ² cÀ³ ² cÀ ³ b b  À À À ² cÀ³ ² cÀ ³ b b ~ À À Answer: D À À

 ~

15. 5 ~ number of claims in one year. For Actuary ? , ,´5 µ ~ ,´5 O(µ h 7 ²(³ b ,´5 O)µ h 7 ´)µ À ,´5 O(µ ~ ,´5 O(µ h 7 ²(O(³ b ,´5 O(µ h 7 ²(O(³ ~ ²À³²À ³ b ²À³²À ³ ~ À À Also, ,´5 O)µ ~ ²³²À³ ~ À . We are given 7 ²(³ ~ À Á 7 ²)³ ~ À , so that ,´5 µ ~ ²À ³²À ³ b ²À³²À³ ~ À . For Actuary @ , we wish to find ,´5 O5 ~ µ , where 5 is the number of claims from someone drawn randomly from the same class as for 5 . We condition over the class: ,´5 O5 ~ µ ~ ,´5 O(µ h 7 ²(O5 ~ ³ b ,´5 O)µ h 7 ²)O5 ~ ³ . Since Class A has two types of policyholder, we condition over the two types to get ,´5 O(µ ~ ,´5 O(µ h 7 ²(O(³ b ,´5 O(µ h 7 ²(O(³ ~ ²À³²À ³ b ²À³²À ³ ~ À À Class B has one type of policyholder, ,´5 O)µ ~ ²³²À³ ~ À . We must now find 7 ²(O5 ~ ³ and 7 ²)O5 ~ ³ . Since the Class is selected at random and two-thirds are in Class A, 7 ²(³ ~ À and 7 ²)³ ~ À , and since Class A is divided equally into the two types, 7 ²(³ ~ 7 ²(³ ~ À (and 7 ²(³ ~ 7 ²(³ b 7 ²(³ ~ À ).

© ACTEX 2009

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

PRACTICE EXAM 11

PE-217

15. continued We find 7 ²(O5 ~ ³ in the usual Bayesian way: 7 ´²5 ~³q(µ 7 ²5 ~³

7 ²5 ~|(³h7 ²(³

 ~ 7 ²5 ~|(³h7 ²(³b7 ²5 ~|)³h7 ²)³ .  7 ²5 ~ |(³ ~ 7 ²5 ~ O( ³ h 7 ²( O(³ b 7 ²5 ~ O( ³ h 7 ²( O(³ ~ ²À ³²À ³ b ²À ³²À ³ ~ À Á and 7 ²5 ~ |)³ ~ ²À ³ ~ À  À

7 ²(O5 ~ ³ ~

²À ³²À ³

Then, 7 ²(O5 ~ ³ ~ ²À ³²À ³b²À ³²À³ ~ À  Á 7 ´)O5 ~ µ ~ À  . Finally, ,´5 O5 ~ µ ~ ,´5 O(µ h 7 ²(O5 ~ ³ b ,´5 O)µ h 7 ²)O5 ~ ³ ~ ²À ³²À  ³ b ²À³²À ³ ~ À  For Actuary A , we wish to find ,´5 O5 ~ µ , where 5 is the number of claims from the same policyholder. We condition over policyholder type. ,´5 O5 ~ µ ~ ,´5 O(µ h 7 ²(O5 ~ ³ b ,´5 O(µ h 7 ²(O5 ~ ³ b ,´5 O)µ h 7 ²)O5 ~ ³ . As above, ,´5 O(µ ~ À Á ,´5 O(µ ~ À Á ,´5 O)µ ~ À , 7 ´(µ ~ 7 ´(µ ~ À Á 7 ²5 ~ O(³ ~ À , 7 ²5 ~ O(³ ~ À and 7 ²5 ~ ³ ~ ²À ³²À ³ b ²À ³²À³ ~ À  . Then, 7 ²(O5 ~ ³ ~ ²À ³²À³

7 ´²5 ~³q(µ 7 ²5 ~³

7 ²5 ~|(³h7 ²(³

~ 7 ²5 ~|(³h7²(³b7 ²5 ~|)³h7 ²)³  

~ À  ~ À , and similarly 7 ²(O5 ~ ³ ~ À  . As above, 7 ²)O5 ~ ³ ~ À  . Then ,´5 O5 ~ µ ~ ²À³²À ³ b ²À³²À ³ b ²À³²À ³ ~ À  . Then A  @  ? . Answer: E

16. 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 ¸Á ¹ random variable that identifies Type, and $ and $ are the types of insured 1 and insured 2. We wish to find 7 ²$ ~ $ O? ~ Á ? ~ ³ . Using the definition of conditional probability, this is 7 ´²$ ~$ ³q²? ~Á? ~³µ 7 ²? ~Á? ~³

À

Since $ must be 1 or 2, the numerator can be formulated as 7 ´²$ ~ $ ³ q ²? ~ Á ? ~ ³µ ~ 7 ´²$ ~ $ ~ ³ q ²? ~ Á ? ~ ³µ b 7 ´²$ ~ $ ~ ³ q ²? ~ Á ? ~ ³µ ~ 7 ²? ~ Á ? ~ O$ ~ $ ~ ³ h 7 ²$ ~ $ ~ ³ b 7 ²? ~ Á ? ~ O$ ~ $ ~ ³ h 7 ²$ ~ $ ~ ³ Since both policies are chosen at random, 7 ²$ ~ $ ~ ³ ~  and 7 ²$ ~ $ ~ ³ ~ Also, since the insureds are independent of one another, 7 ²? ~ Á ? ~ O$ ~ $ ~ ³ ~ 7 ²? ~ O$ ~ $ ~ ³ h 7 ²? ~ O$ ~ $ ~ ³ ~ c h c ~ c (the notation 7 really means density of ? , not probability, in this situation).

© ACTEX 2009

 

.

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16. continued Similarly, 7 ²? ~ Á ? ~ O$ ~ $ ~ ³ ~ 7 ²? ~ O$ ~ $ ~ ³ h 7 ²? ~ O$ ~ $ ~ ³ ~  c° h  c° ~  c° . Then, 7 ´²$ ~ $ ³ q ²? ~ Á ? ~ ³µ ~ c h  b  c° h  ~ À   . Also, 7 ²? ~ Á ? ~ ³ ~ 7 ²? ~ ³ h 7 ²? ~ ³ because of independence of ? and ? . Each of the ? 's is a mixture of two exponentials with pdf  ²c% b  c%° ³ , so 7 ²? ~ ³ ~  ²c b  c° ³ , and 7 ²? ~ ³ ~  ²c b  c° ³ , and 7 ²? ~ Á ? ~ ³ ~  ²c b  c° ³ h  ²c b  c° ³ ~ À   . Finally, 7 ²$ ~ $ O? ~ Á ? ~ ³ ~ À   Answer: E À   ~ À  .

c 17.  V ~ V# ~ ? ~  ~ À  Á     =V ´?µ ~  ´² c  ³ b ² c  ³ b ² c  ³ µ ~ À Á  V  ~ À c À  ~ À À A ~ V À  ~ À b À

The credibility premium is ²À³² b  ³ b ²À ³²À ³ ~ À .

Answer: C

18. 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 5 trials is 5 h ,´$µ . The second actuary selects a driver with Poisson parameter  and the average number of claims in 5 years for that driver will be 5 . The ratio

5 ,´$µ 5

tends to 1 only if the second actuary's driver's  is equal to ,´$µ. 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

19. With  ~  ordered numbers, we find the empirical estimate of the 90-th percentile as  b follows. First identify the integer value of  such that b  À  b . For this example, we 

b

find  so that   À   , from which it follows that   À   b , and therefore,  ~ À We then perform a linear interpolation between the 9-th (-th) and 10-th ( b -st) numbers in the series. The 9-th and 10-th numbers (in order) are 2199 and 3207, and we interpolate to the "9.9-th number". Since 9.9 is .9 of the way from 9 to 10, the smoothed empirical estimate is ² c À ³²

³ b ²À ³²³ ~  À . Answer: D

© ACTEX 2009

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20. 5 is the number of coins found per minute today. We are told that the distribution of 5 given  is Poisson with mean , and the distribution of  is gamma with mean  ~  and variance  ~   . It follows that ~  and  ~  À The unconditional distribution of 5 is negative binomial with parameters  ~  ~  and  ~ ~  .  The probability 7 ´5 ~ µ ~ ²b ³ ~  . Answer: A 21. In 2000, with no policy limit and with a deductible of 100, the expected amount paid per loss is ,´²? c ³b µ . The density function of ? is ? ²%³ ~ À for   %   (0 otherwise) and the distribution function -? ²%³ ~ À% for   %  . The expectation can be formulated in each of the following ways:  (i)  ²% c ³²À³ % ~   (ii) ,´?µ c ,´? w µ ~  c ´ %²À³ % b ² c -? ²³³µ ~  c ´ b ² c À³µ ~  .   (iii)  ´ c -? ²%³µ % ~  ´ c À%µ % ~  . The variance of amount paid is ,´²? c ³b µ c ²,´²? c ³b µ³ .  ,´²? c ³b µ ~  ²% c ³ ²À³ % ~ Á  , so that the standard deviation of amount paid per loss in 2000 is jÁ  c ² ³ ~  .0 . In 2001, the loss @ is uniform on ²  Á   ³ . The variance of amount paid per loss is ,´²@ c ³b µ c ²,´²@ c ³b µ³       ~  ²% c ³ ²   ³ % c ´ ²% c ³ ²   ³ %µ ~ Á   c ² À ³ ~ Á  , and the standard deviation is 295. The percentage increase in standard deviation from 2000 to 2001 is  À Answer: A  À c  ~ À  . 22. This question can be solved by referring to the ²Á Á ³ class of distributions. A discrete non negative integer-valued random variable with probability function  ²³ ~  Á  ~ Á Á Á Á ÀÀÀ is a member of the ²Á Á ³ class is there are constants  and  such that for all   ~ Á Á Á ÀÀÀ, the probability function satisfies the relationship   ~  b  . The Poisson with parameter  has

 c

~

²c  ³°[ ²c c ³°²c³[

~

 

c

, so that  ~  Á  ~ À

Therefore, the distribution in this problem is Poisson with  ~  . The probability the ? ~  is c . The zero-truncated distribution has probabilities  7 ²? A ~ ³ ~ c c d 7 ²? ~ ³ for  ~ Á Á ÀÀÀ .

  

 Then ,´? A µ ~ c ,´²? A ³ µ ~ c c d ,´?µ ~ cc and c d ,´? µ ~ cc (note that ,´?  µ ~ = ´?µ b ²,´?µ³ ~  b  ~ ³. c

 c   = ´? A µ ~ c c c ² cc ³ ~ ²cc ³ ~ À .

Answer: D

23. The overall rate at which coins are found is 30 per hour (.5 per minute). The rate at which coins worth 1 are found is ²³²À ³ ~  per hour, for coins worth 5 the rate is ²³²À³ ~ per hour, and for coins worth 10 the rate is ²³²À³ ~ per hour. Since the overall way in which coins are found is a Poisson process, and since the denominations are randomly distributed (and therefore independent) it is also true that for each denomination of coin amount, coins are found according to a Poisson process.

© ACTEX 2009

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

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PRACTICE EXAM 11

23. continued Suppose that 5 is the number of coins worth 1 found in one hour, 5 is the number of coins worth found in one hour, 5 is the number of coins worth 10 found in one hour. Then = ´5 µ ~  Á = ´5 µ ~ and = ´5 µ ~ , and 5 Á 5 and 5 are mutually independent. The total value of all coins found in the hour is 5 b 5 b 5 , and the variance is = ´5 b 5 b 5 µ ~ = ´5 µ b  = ´5 µ b = ´5 µ ~  b  ² ³ b ² ³ ~  . Answer: E

24. The average sample cost per payment is bbb bb b b  ~ . For a uniform distribution on the interval ²Á ³, the expected cost per loss, with deductible 20 and maximum covered loss 170 is     ,´? w µ c ,´? w µ ~  ´ c - ²%³µ % ~  ´ c % µ % ~   c  c À The expected cost per payment is

,´?wµc,´?wµ 7 ²?€³

~

We set this equal to 96 and solve for , which results in

 c Á 

c  c Á  c 

~

  c Á  À c 

~ , so that ~  À .

Answer: D

25. The log of the density of the Pareto distribution is   ²%³ ~   b   c ² b ³ ²% b ³ and C C

b   ²%³ ~  c  %b

,

C C

  ²%³ ~  b  c ²% b ³

.

Given  ~  , with 2 sample values, to find the mle of , we solve C   C  3 ~  b   c '²% b ³ ~  b    c ²  À b   À ³ ~  . Solving for  results in  ~ À . Now, with  ~ À , to find the mle of , we solve

²À³ À À  3 ~   c ' %b ~ c  Àb c À b ~  .  b

À ² Àb ³² À b ³cÀ ² À b ³cÀ ² Àb ³ This expression becomes ~. ² Àb ³² À b ³ C C

The numerator is À b 

À  c   . Setting this equal to 0 results in two solutions for , 113.5 and c 28.9 . We ignore the negative root and use  ~ À as the new estimate for . The next estimate of  is found from   b   À c ² À b   À³ ~  . Solving for  results in  ~ À  . Answer: E

© ACTEX 2009

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26. The variance/covariance matrix of the maximum likelihood estimates is the inverse of the    c  information matrix. The inverse of > is c . ? >   c  ?

 À À  The inverse of the given matrix is ² ³² ³c²c³²c³ >   ? ~ > À À ? . V . Therefore, the variance of  This is the variance/covariance matrix of  V, V is .12 . Answer: A

27. We will denote the frequency by 5 and the severity will be @ . The full credibility standard for the expected number of claims needed is  À h

= ´5 µh²,´@ µ³ b,´5 µh= ´@ µ ,´5 µh²,´@ µ³

~ À .

Since the severity is exponential, we have ,´@ µ ~ and = ´@ µ ~  ~ ²,´@ µ³ . The full credibility standard for expected number of claims needed becomes  À h

= ´5 µb,´5 µ ,´5 µ

. Since both the mean and variance of 5 are increasing by 20%,

this full credibility standard is unchanged at 5412.

Answer: B

   28. 7 ²? ~ ³ ~   ²Á ³  ~  7 ²? ~ O³ ²³  ~  ² c ³ h c      ~  ² c b ³   ~  h ´ b c b µ ~ ²b³² b³ .

  Alternatively,  ² c ³ h c  ~   h ² c ³  ~  h )² b Á ³

~  h

!²b³h!²³ !²b³

h!²b³h

 ~ ²b³²b³h!²b³ ~ ²b³² b³ .

   7 ²? ~ ³ ~   ²Á ³  ~  7 ²? ~ O³ ²³  ~   h c    ~  b   ~ b .

We wish to find ²O? ~ ³ . This can be formulated as

 ²?~O³h²³ 7 ²?~³

~

²c³hc  ²b³²b³

~ ² b ³² b ³ ² c ³ .

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 Z ~  b % ~  b  and Z ~  b  c % ~  b  c  ~  , with pdf !²Z bZ ³

!²b³

²O% ~ ³ ~ !²Z ³h!²Z ³ h  c ² c ³ c ~ !²b³h!²³ h  ² c ³ ~ ² b ³² b ³ ² c ³ . Answer: B

© ACTEX 2009

Z

Z

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PRACTICE EXAM 11

29. ,´?Oµ ~  Á = ´?Oµ ~  . Since  is a mixture, ,´µ ~ ²À ³²³ b ²À ³²³ ~  and ,´ µ ~ ²À ³² d  ³ b ²À ³² d  ³ ~  (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  ~ ,´²³µ ~ ,´µ ~  Á # ~ ,´#²³µ ~ ,´µ ~  Á  ~ = ´²³µ ~ = ´µ ~ ,´ µ c ²,´µ³ ~ . For a single observation of ? ,  ~ , the Buhlmann credibility factor is A ~ b  ~  .

With ? ~ , the Buhlmann credibility estimate for the next year is c A? b ² c A³ ~  b ²  ³²³ ~  . Answer: D

30. 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 -? ² c ³    -? ²³ , and the simulated value of ? is . In this problem we are asked to apply the inverse transform method in the form where low numbers correspond to a high number of trials, so that given a uniform random number , we find the integer  such that -? ² c ³   c   -? ²³ . The random variable ? is the number of trials until the first success. This is a version of the geometric distribution with probability function ²%³ and distribution function - ²%³ as follows: ?~     ÀÀÀ   ²³ À

²À³²À ³ ~ À ²À³ ²À ³ ~ À

²À³ ²À ³ ~ À  ÀÀÀ - ²³ À

À  À 

À  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 ? b ? b ? . The first random number is  ~ À , so that  c  ~ À . We see that - ²³ ~ À   À  À  ~ - ²³ so that the simulated value of ? is 3. This is ? , the simulated number of trials until the first success. The second random number is  ~ À , with  c  ~ À . We see that À  À ~ - ²³ , so that the simulated value of ? is 1.  ~ À , so that  c  ~ À  . We see that - ²³ ~ À  À   À  ~ - ²³, so that the simulated value of ? is 2. Then ? b ? b ? ~  b  b  ~ . Answer: D

31. ? is the total loss for one year on the randomly chosen claim. # is either A or B, denoting the two classes. 7 ´# ~ (µ ~ 7 ´# ~ )µ ~  . ² ³ ~ ,´?O µ S ²(³ ~ ,´?O(µ ~ ²À³ ~ À Á ²)³ ~ ,´?O)µ ~ À Á #² ³µ ~ = ´?O µ S #²(³ ~ = ´?O(µ ~  ²À³²À ³ ~ À  Á  #²)³ ~ = ´?O)µ ~   ²À³²À ³ ~ À   . The credibility factor is A ~ b # , 





where # ~ ,´#²#³µ ~  h #²(³ b  h #²)³ ~ À b À   , and      ~ = ´,´?O#µµ ~ = ´²#³µ ~ ´  h ²À³ b  h ²À³ µ c ´  h ²À³ b  h ²À³µ ~ À c À b À  . Credibility is based on  ~  year. 

ÀbÀ  



Then A ~ b # ~ . As SB , ÀcÀbÀ  S , ÀbÀ   b ÀcÀbÀ   



and A S b ~ .

© ACTEX 2009

Answer: B

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PRACTICE EXAM 11

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32. Suppose that a random sample from a distribution is given: ? Á ? Á ÀÀÀÁ ? , and suppose that the sample is used to estimate some parameter of the distribution. If the estimator is V , then the bootstrap approximation to the mean square error of this estimator is , denote the claims for an individual in a single year. Then ² ³ ~ ,´> O# ~ µ ~  , and #² ³ ~ = ´> O# ~ µ ~  ² c ³ (binomial mean and variance with 3 trials). Then, using the distribution of #, we have   ~ ,´?µ ~ ,´²#³µ ~ ,´#µ ~   h ² c ³  ~  ;  # ~ ,´#²#³µ ~ ,´#² c #³µ ~   ² c ³ h ² c ³  ~  ; and  ~ = ´²#³µ ~ = ´#µ ~ = ´#µ ~ ´ ,´# µ c ²,´#µ³ µ 

~ ´   h ² c ³  c ²  ³ µ ~  . With 10 policyholders per risk class in 1997, the average aggregate claim per policyholder in the randomly selected risk class is ? ~

© ACTEX 2009

>Á b>Á bÄb>Á 

~   ,  ~  .

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

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PRACTICE EXAM 12

30. continued c  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  A ~ b ~ bb  ~ À . Also, 

bb b 

c       

? ~   h ? ~ ²  ³²  ³ b ²  ³²  ³ b ²  ³²  ³ ~  ~ À  , so that the credibility ~ premium per individual in the risk class is c A? b ² c A³ ~ ²À ³²À ³ b ²À ³²À ³ ~ À À With 20 individuals in the risk class in 2000, the credibility premium is ²À ³ ~ À À c Note that we have a total of 37 exposures ( b  b  ³, with an ? of  bb ~   ~ À , and we have applied the basic Buhlmann method, without really needing to refer to Buhlmann-Straub. Answer: C

31. To apply the Inverse Transform Method we must formulate the cumulative distribution function. First we find the probability function for this binomial distribution.   The probability function is  ~ 7 ²? ~ ³ ~ 2  3 ² c ³c ~ 2  3²À³ ²À ³c . ?~  - ²³

 À À

 À  À

 À À 

 À À  

 ÀÀÀ À  À 

The simulated value of ? is  , where - ² c ³  <  - ²³ . We see that À ~ - ²³  À  À  ~ - ²³ . Therefore, the simulated value of ? is 2. Answer: C

32. The probability for ? ~  is .4, the probability for ? ~  is À c À ~ À , the probability for ? ~  is À c À ~ À , and the probability for ? ~  is  c À ~ À . Answer: B



33. The conditional distribution of 5 given  is geometric, with  ~ c , so the conditional   mean and variance of 5 are ,´5 Oµ ~  ~ c and = ´5 Oµ ~  ² b  ³ ~ ²c³ . 





Then, ,´5 µ ~ ,´ ,´5 Oµ µ ~ , ´ c µ ~  c h ² c  b  ³     ~  c h  ² c ³  ~   ² c ³  ~  , and b

 b

,´5  µ ~ ,´ ,´5  Oµ µ ~ , ´ ²c³ µ ~  ²c³ h  ² c ³   ~  ² b  ³  ~  À = ´5 µ ~  c  ~  . The coefficient of variation of 5 is

© ACTEX 2009

j= ´5 µ ,´5 µ

~

j 

~ j  .

Answer: D

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PRACTICE EXAM 12

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34. The likelihood function for Phil's mle is 37 ² 7 ³ ~  c% °  ~  c'% °  ,  ~  





and the likelihood function for Sylvia's mle is 3: ² : ³ ~  c& ° : ~  c'& ° S . S ~ : The mle of the exponential distribution mean is the sample mean of the data points. Therefore, '% ~ V 7 ~ Á  , and '& ~ V : ~  Á  . The combined likelihood function, using the restriction : ~  7 is     3 ~  c'% °  h ( ³ c'& ° 7 ~  cÁ°  h ( ³ c Á° 7 



7

  ~  h  cÁ °  . 

7

Á  7 , and Á     . Setting   3 ~  and solving for 7 results in  7  3 ~ c 7 b 7 7

Then,  3 ~ c    c   7 c V 7 ~  À .



Note that once the function we wish to maximize is in the form  c° , the maximum occurs    at  ~  . Thus, when we have 3 ~  h  cÁ °  , we see that the maximum will occur at 

Á   ~  ~  À . Answer: B

35. We wish to find ,´? O? b ? ~ µ , where ? is the size of the 3rd claim and ? , ? are the first two claim sizes. Since the claim size distribution is described as a conditional distribution given $, we can find the expectation by conditioning over $. B ,´? O? b ? ~ µ ~  ,´? O$ ~ µ h ²O? b ? ~ ³  . Since ? has an exponential distribution with mean , we have ,´? O$ ~ µ ~  . B Therefore, ,´? O? b ? ~ µ ~   h ²O? b ? ~ ³  , which is the mean of the posterior distribution, which has pdf ²O? b ? ~ ³ . We see that the prior distribution has density () ~ 500c4 c10/ , which is the pdf of an 29 inverse gamma distribution with  ~  and ~ . In this study guide's notes on Bayesian estimation, the combination of the inverse gamma prior and exponential model distribution was one of the various prior/model combinations considered. It was seen there that this combination results in a posterior distribution that is also inverse gamma. If there are  sample values available % Á % Á ÀÀÀÁ % , then the posterior distribution has Z ~  b  and Z ~ b '% . This can be seen from the joint distribution of % Á % Á ÀÀÀÁ % and which has pdf  ²% Á % Á ÀÀÀÁ % Á ³ ~  ²% O³ h  ²% O³Ä ²% O³ h ²³ 







c² b'% ³°

~  c% ° h  c% ° Ä  c% ° h b !²³ c ° which is proportional to  bb . The posterior density is proportional to this, and therefore must have an inverse gamma distribution with Z ~  b  and Z ~ b '% .

© ACTEX 2009

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PRACTICE EXAM 12

35. continued In this case we have an inverse gamma prior with  ~  and ~ , and we are given  ~  sample values of % with % b % ~  . Therefore, the posterior distribution is inverse gamma with Z ~  b  ~ and Z ~  b  ~  . As noted above, the predictive expectation for the next claim is equal to the posterior mean, which is the mean of the inverse gamma, which is

 Z Z c ~ c ~  . Notice that we only needed the sum of the %-values, not the individual values. Answer: C

© ACTEX 2009

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PRACTICE EXAM 13

PE-243

ACTEX EXAM C/4 - PRACTICE EXAM 13 1. Two independent random variables, ? and ? , follow the negative binomial distribution with parameters ² Á  ³ and ² Á  ³, respectively. Under which of the following circumstances will ? b ? always be negative binomial? 1.  ~  2.  ~  3. The coefficients of variation ? and ? are equal. A) 1 only B) 2 only C) 3 only D) 1 and 3 only E) 2 and 3 only

2. A frailty model has a base age-at-death distribution that follows an exponential distribution with mean 100, and associated hazard rate function ²%³. The conditional hazard rate for the ageat-death random variable ? for an individual with parameter  is ?O ²%O³ ~ ²%³. For a new-born individual in the frailty model group, the value of  is uniformly distributed between .8 and 1.5. Find the probability that a randomly selected new-born from the frailty group will survive to at least age 80. A) Less than .1 B) At least .1, but less than .3 C) At least .3, but less than .5 D) At least .5, but less than .7 E) At least .7 Questions 3 and 4 are based on a continuous loss random variable ? is uniformly distributed on the interval ²Á ³. 3. If a loss is over 25 but less than 75, an insurance policy pays 60% of the loss amount that is over 25. If a loss is over 75, the insurance pays the full loss amount over 25. Find the expected cost per payment for this insurance. A) Less than 25 B) At least 25, but less than 28 C) At least 28, but less than 31 D) At least 31, but less than 34 E) At least 34 4. If ?  , a risk manager is paid a bonus equal to 75% of the difference between ? and 60. Find variance of the bonus received by the risk manager. A) 10.5 B) 11.5 C) 12.5 D) 13.5 E) 14.5

5. The claims department of an insurance company receives envelopes with claims for insurance coverage at a Poisson rate of  ~  envelopes per week. For any period of time, the number of envelopes and the numbers of claims in the envelopes are independent. The numbers of claims in the envelopes have the following distribution: Number of Claims Probability 1 0.20 2 0.25 3 0.40 4 0.15 ! Using the normal approximation, calculate the 90 percentile of the number of claims received in 13 weeks. A) Less than 1700 B) At least 1700, but less than 1720 C) At least 1720, but less than 1740 D) At least 1740, but less than 1760 E) At least 1760

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

PE-244

PRACTICE EXAM 13

6. The 95% log-transformed confidence interval for /²! ³ is ²ÀÁ À³ . Find the upper limit of the 90% linear confidence interval for /²! ³ A) Less than 3.00 B) At least 3.00 but less than 3.04 C) At least 3.04 but less than 3.08 D) At least 3.08 but less than 3.12 E) At least 3.12

7. A mortality study of 100 individuals with impaired health has resulted in the following observations (time measured in years). There is no censoring of any data points. Time .1 .3 .4 .6 .7 .8 .9 1 Number of Deaths 2 1 3 2 4 4 5 3 Determine the kernel-smoothed estimate of the density function of time at death at time .7 using a  %&c  bandwidth of  ~ À years and the uniform kernel & ²%³ ~ H  & c   %  & b  .  %€&b A) .1 B) .2 C) .3 D) .4 E) .5

8. You are given: (i) Losses follow a Pareto distribution with  ~ . (ii) A random sample of losses is distributed as follows: Loss Range Number of Losses ²  Á µ  ²  Á  µ  ²  Á  µ  ²  Á   µ 

²   Á  µ  ²  Á   µ  Total  Estimate by matching at the 80-th percentile. Determine the mean of the estimated distribution. A) Less than 250 B) At least 250 but less than 260 C) At least 260 but less than 270 D) At least 270 but less than 280 E) At least 280

9. You observe the following five ground-up claims from a data set that is truncated from below at 100: 125 150 165 175 250 You fit a ground-up two Parameter Pareto distribution with ~   using maximum likelihood estimation. Determine the mean of the fitted distribution. A) Less than 25 (B) At least 25 but less than 50 C) At least 50 but less than 75 D) At least 75 but less than 100 E) At least 100

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

PRACTICE EXAM 13

PE-245

10. An urn has a large number of balls numbered from 1 to 10. The proportion of balls with the number 1 on them is not known. You wish to draw a sample (with replacement) of balls from the urn to estimate the proportion of balls in the urn that have the number 1. You want to choose enough balls so that the standard deviation of the estimated proportion is no more than .01 . Find the minimum number of balls that must be drawn in general to satisfy this requirement. A) 1500 B) 2000 C) 2500 D) 3000 E) 3500

11. You are given the following: (i) the claims frequency rate for a group of insureds is believed to be an exponential distribution with unknown parameter; - ²%³ ~  c c% , % €  (ii) ten random observations of ? yield the following sample in ascending order: .001 , .003 , .053 , .062 , .127 , .131 , .377 , .382 , .462 . .481 (iii) summary statistics for the sample data are: 

 % ~ À Á

~





~

~

 % ~ À Á  ²% ³ ~ c  À 

(iv)  is the maximum likelihood estimator for  Use the normal distribution to determine a 95% confidence interval for  based upon the sample data. A) (.20 , .22) B) (.27, 9.36) C) (1.83 , 7.79) D) (2.50 , 7.12) E) (3.20 , 6.33)

12. > is a random variable with mean ,´> µ and variance = ´> µ. In a partial credibility analysis of > , the manual premium used is 4 ~ . A sample of 350 observations of > is available and the sum of the observed values is 300,000. Partial credibility is applied to determine a credibility premium based on the "5% closeness" and "90% probability" criteria. If the credibility standard used is the one based on the expected number of observations of > needed, then the partial readability premium is 884.40. If the credibility standard used is the one based on the expected sum of the observed values of > needed, then the partial credibility premium is 887.19. Using this information, determine the mean of > . A) Less than 750 B) At least 750 but less than 850 D) At least 850 but less than 950 D) At least 950 but less than 1050 E) At least 1050

13. The prior distribution of the parameter has pdf ² ³ ~  for € . The model distribution has a uniform distribution on the interval ´ Á  µ . Find the mean of the posterior distribution. A) % B) % C) % D) % E) %   

14. The severity distribution of ? has pdf  ²%O ³ ~  %c % for % € , where # has pdf ² ³ ~ ec for € . Find the Bayesian premium ,´?b O % À À À % µ. '%

A) b

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B)

²b'% ³ b

C)

²b'% ³ b

'%

D) b

b'%

E) b

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

PE-246

PRACTICE EXAM 13

15. ? is a Poisson random variable with parameter $, where the prior distribution of $ is a discrete uniform distribution on the integers Á Á 3 . A single observation of ? is made. Find the Buhlmann factor A . A)  B)  C)  D)  E) 

16. The per year claim amount random variable has an exponential distribution with an unknown mean that varies by individual. The experience of 200 individuals for one year is summarized in 

the following information:  % ~  Á ~

   ~ 

²% c c %³ ~ À  . ~

Determine the credibility factor A of one year's experience for a single individual using semiparametric empirical Bayes estimation. A) .03 B) .04 C) .05 D) .06 E) .07

17. You are given the following random sample of three values from the distribution function - :     c c ? b? b? You are to estimate = ²?³ using the estimator   ²? c ?³ , where ? ~    . ~

Find the bootstrap approximation to the MSE of the estimator. A) .1 B) .2 C) .3 D) .4 E) .5

18. Maximum likelihood estimation is applied to a data set of 20 observations to estimate in the distribution with pdf  ²%³ ~  h c%° Á % €  . The sample mean of the data set is c % ~. The probability 7 ´  ?  µ is estimated using the mle of . Find the estimated variance of that estimated probability using the delta method. A) .0005 () .0010 C) .0015 () .0020 E) .0025

19. You are given the following information about two classes of business, where ? is the loss for an individual insured: Class 1 Class 2 Number of insureds 25 50 , (? ) 380 23 , (? 2 ) 365,000 ---You are also given that an analysis has resulted in a Buhlmann  value of 2.65. Calculate the process variance for Class 2. A) 2,280 B) 2,810 C) 7,280

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D) 28,320

E) 75,050

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

PRACTICE EXAM 13

PE-247

20. A population has 30% who are smokers with a constant hazard rate for time until death of 0.2 and 70% who are non-smokers with a constant hazard rate for time until death 0.1. Calculate the 75-th percentile of the distribution of the future lifetime of an individual selected at random from this population. A) 10.7 B) 11.0 C) 11.2 D) 11.6 E) 11.8

21. The loss random variable ? follows a two-parameter Pareto distribution with ~  and  ~ . Calculate mean excess loss based on an ordinary deductible of 20.. A) 5 B) 15 C) 25 D) 35 E) 45

22. Dental Insurance Company sells a policy that covers two types of dental procedures: root canals and fillings. There is a limit of one root canal per year and a separate limit of two fillings per year. The number of root canals a person needs in a year follows a Poisson distribution with  ~ , and the number of fillings a person needs in a year follows a Poisson distribution with  ~ . The company is considering replacing the single limits with a combined limit of 3 claims per year, regardless of the type of claim. Determine the change in the expected number of claims per year if the combined limit is adopted. A) No change B) More than 0.00 but less than 0.20 more claims C) At least 0.20 but less than 0.25 more claims D) At least 0.25 but less than 0.30 more claims E) At least 0.30 more claims

23. A compound Poisson claim distribution : has Poisson parameter  ~  and severity  prob. .4 distribution ? ~ H 2 prob. .2 . 3 prob. .4 Stop loss insurance with a deductible of 2 is applied to : . Find ,´²: c ³b µ . A) À c B)  b À c C)  b À c D)  c À c E) c À c

24. You are given the following data in grouped form for the amount of each of 100 separate claims. Interval Number of Claims ² Á µ  ²  Á µ  ² Á  µ  ²  Á µ  ² Á µ  ² Á µ  Find the empirical estimate of the standard deviation of the loss . A) 723 B) 763 C) 803 D) 843 E) 883

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

PE-248

PRACTICE EXAM 13

25. The following data is from the first five years of a cohort life study. The study began with 50 individuals. The year by year numbers of deaths and censorings are as follows: Year # deaths # censorings 1 3 2 2 5 4 3 6 3 4 6 5 5 8 4 It is assumed that censorings occur at mid-year and deaths for a particular year occur just before year-end. Find the estimated variance of the cumulative hazard rate at the end of the 4-th year. A) .02 B) .03 C) .04 D) .05 E) .06

26. You are given the following random sample of observations: 3 , 4 , 6 , 6 , 7 , 7 , 8 , 8 , 9 , 12 The data is a sample of 10 observations from a compound Poisson distribution for which the severity distribution is exponential. Apply the method of moments to estimate the Poisson parameter (mean)  and the exponential parameter (mean) . What is the estimate of ? A) 9 B) 11 C) 13 D) 15 E) 17

27. The following claim payment information is available for a collection of policies for which the ground up loss distribution is exponential with mean (all policies have the same ground up loss distribution). Claim payments for 10 policies with no deductible, and no policy limit: 22 , 28 , 31 , 38 , 38 , 45 , 49 , 50 , 55 , 73 Claim payments for 5 policies with no deductible, but with policy limit 50: 18 , 29 , 37 , 42 , 50 (limit payment) Claim payments for 5 policies with deductible of 10 and with policy limit 40 (maximum covered loss is 50): 21 , 24 , 33 , 40 (limit payment) , 40 (limit payment) c(°

The likelihood function is of the form ) A) 780 B) 783 C) 786 D) 789

. Determine ( b ) . E) 792

28. A sample of ten losses has the following statistics: 

 ? c ~ À 



 ? À ~  À 

~ 

~ 

~ 

~ 

~

~

 ? c ~ À

 ? cÀ ~ À

 ? ~ Á   ?  ~ Á  Á 

You assume that the losses come from a Weibull distribution with  ~ À Determine the maximum likelihood estimate of the Weibull parameter À A) Less than 500 B) At least 500, but less than 1500 C) At least 1500, but less than 2500 D) At least 2500, but less than 3500 E) At least 3500

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

PRACTICE EXAM 13

PE-249

29. You are given the following random sample of size 7: 1 , 2 , 4 , 5 , 7 , 9 , 14 A uniform distribution on the interval ´Á µ is the estimated distribution. Which of the following is the -plot for the data and estimated distribution? A)

B)

C)

D)

E) None of A, B, C or D is similar to the -plot for this data and estimated distribution.

30. 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 first year and 2 claims in the second year. Find the expected number of claims for the risk in the third year. A) Less than 1.5 B) At least 1.5 but less than 1.6 C) At least 1.6 but less than 1.7 D) At least 1.7 but less than 1.8 E) At least 1.8

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

PE-250

PRACTICE EXAM 13

31. You are given: (i) The number of claims made by an individual insured in a year has a Poisson distribution with mean . (ii) The prior distribution for  is gamma with parameters  ~  and ~ À . Three claims are observed in Year 1, and no claims are observed in Year 2. Determine the estimated variance of the posterior distribution of the number of claims in Year 3. A) 1.85 B) 1.86 C) 1.90 D) 1.91 E) 1.93

32. You are making credibility estimates for regional rating factors. You observe that the Buhlmann-Straub nonparametric empirical Bayes method can be applied, with rating factor playing the role of pure premium. ? denotes the rating factor for region  and year , where  ~ Á Á  and  ~ Á Á Á . Corresponding to each rating factor is the number of reported claims,  , measuring exposure. You are given:  _ _ _ _     ~   ?  ~   ? V #  ~    ²? c?  ³  ²?  c?³ ~



~

~

  À

À 

  À À    À À Determine the credibility estimate of the rating factor for region 1. A) 1.31 B) 1.33 C) 1.35 D) 1.37 E) 1.39

À  À  À

33. The inverse transform method is used to simulate the number of failures before a success in a series of independent trials each with success probability  ~ À, low values of uniform random numbers correspond to small numbers of failures. What is the simulated number of failures that corresponds to a uniform random number of 0.78 . A) 0 B) 1 C) 2 D) 3 E) 4

34. Bob is an overworked underwriter. Applications arrive at his desk at a Poisson rate of 60 per day. Each application has a 1/3 chance of being a “bad” risk and a 2/3 chance of being a “good” risk. Since Bob is overworked, each time he gets an application he flips a fair coin. If it comes up heads, he accepts the application without looking at it. If the coin comes up tails, he accepts the application if and only if it is a “good” risk. The expected profit on a “good” risk is 300 with variance 10,000. The expected profit on a “bad” risk is –100 with variance 90,000. Calculate the variance of the profit on the applications he accepts today. A) 4,000,000 B) 4,500,000 C) 5,000,000 D) 5,500,000 E) 6,000,000

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

PRACTICE EXAM 13

PE-251

35. A sample of size  is used to estimate the parameters in two possible models for the data. The maximized log-likelihood for the 3-parameter generalized Pareto model is M( , and the maximized log-likelihood for the exponential model is M) . You are given that according to the Schwarz Bayesian Criterion, model A is preferred to model B. You are also given that according to the likelihood ratio test, in which the null hypothesis is that model B is acceptable, and the alternative hypothesis is that model A is preferable to model B, the null hypothesis is rejected at the 5% level of significance but not at the 1% level of significance. Find the maximum value of  that is compatible with these results. A) 99 B) 101 C) 103 D) 105 E) 107

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

PE-252

PRACTICE EXAM 13

ACTEX EXAM C/4 - PRACTICE EXAM 13 SOLUTIONS 1. Answer: B

2. The base hazard rate function is ²%³ ~ À. The conditional survival probability to age 80 for an individual with parameter  is   :² O³ ~ c  ²%³ % ~ ´:² ³µ ~ ²c À % ³ ~ cÀ  .  The pdf of the parameter  is $ ²³ ~ À (uniform distribution on ²À Á À ³). The survival probability to age 80 for a randomly chosen individual is ²cÀ ³À c²cÀ ³À À À  :² ³ ~ À :² O³ $ ²³  ~ À cÀ  h À  ~ ~ À . Answer: C ²À ³²À³





3.  À ²% c  ³²À³ % b  ²% c  ³²À³ % ~ À b  À  ~ À is expected cost À per loss. Expected cost per payment is 7À Answer: C ´?€ µ ~ À ~ À  .

 c ? ?   ~ À ´  c ²? w ³µ  ? € 

 ,´Bonusµ ~ À ´  c ,´? w µµ ~ À ´  c ´ %²À³ % b 7 ´? € µµµ ~ À ´  c ´ b µµ ~ À or

 ,´Bonusµ ~ À d  ²  c %³²À³ % ~ À Answer: D 4. Bonus ~ À d F

5. The number of claims received in one week has a compound Poisson distribution with ,´5 µ ~  , and @ ~ Á Á Á  with the given probabilities. The expected amount of claims received in one week is ,´5 µ h ,´@ µ ~ ² ³´²³²À³ b ²³²À ³ b ²³²À³ b ²³²À ³µ ~  . The variance of the amount of claims received in one week is ,´5 µ h ,´@  µ ~ ² ³´² ³²À³ b ² ³²À ³ b ² ³²À³ b ² ³²À ³µ ~   . The expected number of > claims in 13 weeks is ²³² ³ ~ Á  , and the variance of the number of claims in 13 weeks is ²³² ³ ~ Á  . The 90-th percentile is  , where 7 ´>  µ ~ 7 ´ >jc   c  µ ~ )² c  ³ ~ À  . j j Using the normal approximation

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c  j 

 

 

 

~ À  S  ~  . Answer: B

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

PRACTICE EXAM 13

PE-253

6. The log-transformed interval for /²! ³ has lower and upper limits of V  ³ ~ À . Therefore, < h /²!

V ³ /²!
µ

12. The full credibility standard based on the number of observations needed is  À ²,´> µ³ = ´> µ

and based on the sum of the observed values it is  À ,´> µ . Á c From the given information, the sample mean of the observed values is > ~   ~ À À c The credibility premium based on partial credibility using is A> b ² c A³4 , c where > ~ À Á 4 ~  and A is the partial credibility factor. Using the credibility standard based on the expected number of observations needed, A~

  = ´> µ , so that ÀA b ² c A³ ~ À , from which we get l  À ²,´> µ³

A ~ À  , and therefore

  = ´> µ  À ²,´> µ³

= ´> µ

~ À  , so that ²,´> µ³ ~ À  .

Using the credibility standard based on the expected sum of the observed values needed, Á

A ~l ´> µ , so that ÀA b ² c A³ ~ À , from which we get  À =,´> µ Á ´> µ ~ À   À =,´> µ = ´> µ°,´> µ  = ´> µ°²,´> µ³ ~ À  ~ 

A ~ À  , and therefore

, so that

Then, ,´> µ ~

.

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= ´> µ ,´> µ

~  .

Answer: C

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

PRACTICE EXAM 13

PE-255

13.  ²%O ³ ~  for  %   .

 ²%Á ³ ~  ²%O ³ h ² ³ ~  h  ~  (on the appropriately defined region for ? and ) .  % If   %   , then    % , and ? ²%³ ~    ~  c %  ~ %%c  . %  %     If % € , then   % , and ? ²%³ ~ %°   ~  c  ~  . 

 %

Since % € , ? ²%³ ~

and ² O%³ ~

 % % ,´ O%µ ~ %° h %    ~  . Answer: B

 ²%Á ³ ? ²%³

~

% °  °%

~

% %  

%

on the region %   % .

14. The Bayesian premium can be found in two ways: (i) ,´?b O? ~ %µ ~  %b ?b O? ²%b O%³ %b , or (ii) ,´?b O? ~ %µ ~  ,´?b O µ h #O? ² O%³  . In the general -sample case, approach (i) can become quite complicated due to the complicated nature of the pdf of the predictive distribution, ?b O? ²%b O%³. Approach (ii) is often more straightforward. B B ,´?O µ ~  % h  %c % % ~   % c % % ~  h  ~  . 

Also, ? Á# ²% Á ÀÀÀÁ % Á ³ ~ ? O# ²% Á ÀÀÀÁ % O ³ h ² ³ ~ <  ? O# ²% O ³=² ³ 

~ ´  ² %  

c %

~

c

³µ h e

~

b

~



h ²  % ³ h 

c ²b'% ³

~



c

~ b h ²  % ³ h c ²b%³ ~



B

and ? ²%³ ~  ? Á# ²% Á ÀÀÀÁ % Á ³  ~  b h ²  % ³ h c ²b'% ³  ~

B b



h ²  % ³ h 



c c ²b%³

~



 ~ ²  % ³ h ~

²b³[

~ c B b  h c ²b%³



~ ²  % ³ h ²b%³ c b À We have used the rule ~

B ! h c! ! ~ [ if  is an integer ‚  and  € . b Then,



#O? ² O%³ ~

? Á# ²%Á ³ ? ²%³

~

c

b h²  % ³hc ²b% ³ 

~

²b³[ ²  % ³h ²b% c³b

~

c b h b hc ²b%c³ ²b%³ ²b³[

.

~

Then ~

c b h b hc ²b%c³ ²b%³ B ,´?b O? ~ %µ ~   h  ²b³[

c b ²b%³ ²b³[

B

c

h   h c ²b%³  ~

c b ²b%³ ²b³[

²³[

c ²b%³

h ²b%³ c b ~ b (in this integral, the following general integration form was used B if  is an integer ‚  and  € , then  & c& & ~ [ b , in this case  ~  and c  ~  b % ). Answer: B

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

PE-256

PRACTICE EXAM 13

15. ²$³ ~ ,´?O$µ ~ $ , #²$³ ~ = ´?O$µ ~ $ Á  ~ ,´²$³µ ~ ,´$µ ~ ²  ³² b  b ³ ~  .

 ~ = ´²$³µ ~ = ´$µ ~ ²  ³² b  b  ³ c  ~  . # ~ ,´#²$³µ ~ ,´$µ ~  .    ~ (one observation of ? ), so A ~ b ~  . # ~ b  

Answer: B

°

16. 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 relationships linking ²#³ ~ ,´?O#µ and #²#³ ~ = ´?O#µ and the fact that = ´?µ ~ # b  in order to get estimates for Á # and  to use in the credibility premium formulation. In this problem, ? given # is exponential, so that ²#³ ~ ,´?O#µ ~ # and #²#³ ~ = ´?O#µ ~ # . Then, ,´?µ ~ ,´,´?O#µµ ~ ,´#µ is c estimated by ? ~   ~ À , so our estimate of ,´#µ is 2.2. Also, = ´?µ ~ ,´= ´?O#µµ b = ´,´?O#µµ ~ ,´# µ b = ´#µ ~ ,´# µ b ,´# µ c ²,´#µ³ ~ ,´# µ c ²,´#µ³ . = ´?µ is estimated by  ~ À , and this becomes our estimate for ,´# µ c ²,´#µ³ . The estimate for ,´# µ then comes from À  ~ ,´# µ c ²,´#µ³ ~ ,´# µ c ²À³ , 

so that # ~ ,´# µ is estimated to be V# ~ À bÀ ~ À.  Finally, we estimate  ~ = ´?µ c # using V  ~  c V# ~ À c À ~ À . V ~  V# ~  À ~ À . The credibility factor for a single individual is A Answer: E b V

b

À

17. The variance of the empirical distribution is ~ = ²? ³ ~  ´² c  ³ b ² c  ³ b ² c  ³ µ ~ . There are  ~  possible bootstrap samples (of size 3) from the original sample. These 27 samples can be described as follows: c - all 1's, 8 samples , ? ~  Á V ~  Á ²V c ³ ~ ² c ³ ~   Â c  V  - two 1's and one 3, 12 samples , ? ~ Á ~ ´² c ³ b ² c ³ b ² c ³ µ ~  





²V c ³ ~ ²  c ³ ~ 

 Â c - one 1 and two 3's, 6 samples , ? ~  Á V ~  ´² c  ³ b ² c ²V c ³ ~ ²  c ³ ~ 

 Â c - all 3's, 1 sample , ? ~  Á V ~  Á ²V c ³ ~ ² c ³ ~  



  ³





b ² c  ³ µ ~  Á

 .

The bootstrap estimate to the MSE of the estimator is  





  ²  ³²   ³ b ²  ³²  ³ b ²  ³²  ³ b ²  ³²  ³ ~   ~ À . An example of the count of the number of samples given above is the number of samples with one 1 and two 3's: ²Á Á ³ Á ²Á Á ³ Á ²Á Á ³ Á ²Á Á ³ Á ²Á Á ³ Á ²Á Á ³ . The reason that there are two samples of the form ²Á Á ³ is that there are two sample points equal to 1. If we label them 1a and 1b, then ²Á Á ³ and ²Á Á ) should be regarded as separate samples. The same applies to the other samples. Answer: D

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

PRACTICE EXAM 13

PE-257

18. 7 ´  ?  µ ~ - ²³ c - ²³ ~ ´ c c° µ c ´ c c° µ ~ c° c c° ~ ² ³ . The pdf is for the exponential distribution with mean , and the mle of is V ~ c % ~. Using the delta method, the variance of ²V ³ is approximately equal to ´Z ² ³µ h = ´V µ ~ ´  h c° c  h c° µ h = ´V µ . Since V ~  '? , which is the sample mean, the variance of the mle is the variance of the 5

sample mean, which is

= ´?µ 5

. For the exponential distribution, the variance is the square of  the mean, so that = ´?µ ~  , and therefore, = ´V µ ~ 5 À In this example, we know that the mle of is 1. The estimated variance of the estimate of the V V probability 7 ´  ?  µ is ´  h c° c  h c° µ h = ´V µ . V 

V

V

 

The estimate of = ´V µ is  ~ , and the estimated variance of the probability is    c° c°  ´  h  c  h  µ h ²  ³ ~ À . Answer: A

19. This is an exercise in Buhlmann's credibility model. We can put this situation in the Buhlmann context, with parameter random variable #, and conditional distribution of loss amount ? given #. The loss random variable ? depends on the class of business. There are two classes of business. Class of business is represented by the parameter #. Since there are 75 insureds of which 25 are Class 1 and 50 are class 2, the probability distribution for the class parameter # is 7 ´# ~ µ ~  and 7 ´# ~ µ ~  (when a claim occurs, the chance is  

~  that the policy was from class 1, or in other words, that # ~ ). The Buhlmann  value is  ~ # , where # is the expected process variance, and  is the variance of the hypothetical mean. The process variance for Class 1 is #²³ ~ = ´?O# ~ µ ~ ,´?  O# ~ µ c ²,´? O# ~ µ³ ~  Á  c ² ³ ~ Á  , and for Class 2 it is #²³ ~ = ´?O# ~ µ . In general for the Buhlmann credibility approach, #² ³ ~ = ´?O# ~ µ and # ~ ,´#²#³µ ~ ,´ = ´?O# ~ µ µ , ² ³ ~ ,´?O# ~ µ and  ~ = ´²#³µ ~ = ´ ,´?O# ~ µ µ .

We are given ,´?O# ~ µ ~   Á ,´?O# ~ µ ~  , and ,´?  O# ~ µ ~  Á  and  ~ À . Therefore, ²³ ~ ,´?O# ~ µ ~   Á ²³ ~ ,´?O# ~ µ ~  .   # ~ , prob.  ²#³ is a discrete 2-point random variable, ²#³ ~ F , # ~ , prob.   and  is the variance of ²#³,  ~ = ´²#³µ . A simple rule for the variance of a discrete 2point random variable is the following: if A is the discrete two-point random variable with outcomes ¸Á ¹ with probabilities  Á  c  , then = ´Aµ ~ ² c ³ ² c ³ . Therefore,  ~ = ´²#³µ ~ ²  c ³ ²  ³²  ³ ~  Á  . Note that we can also find = ´²#³µ in the conventional way as the 2nd moment minus the square of the first moment: = ´²#³µ ~ ²  ³²  ³ b ² ³²  ³ c ´² ³²  ³ b ²³²  ³µ ~  Á   ~ # S # ~  ~ ²À ³² Á ³ ~  Á   ~ ,´= ´?O# ~ µµ À

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19. continued Since = ´?O# ~ µ ~ ,´?  O# ~ µ c ²,´? O# ~ µ³ ~  Á  c ² ³ ~ Á  , we get  Á   ~  ~ = ´?O# ~ µ h 7 ´# ~ µ b = ´?O# ~ µ h 7 ´# ~ µ ~ ²Á ³²  ³ b = ´?O# ~ µ²  ³ , from which we get = ´?O# ~ µ ~   . This is the process variance for Class 2. Answer: A

20. The future lifetime is ; ²%³, where ; ²%³ is a mixture of two distributions. The two mixing distributions are ; ²%³, the future lifetime of a smoker, and ; ²%³, the future lifetime of a nonsmoker. The mixing weights are .3 for smokers and .7 for non-smokers. The 75-th percentile of ; ²%³ is  , where 7 ´; ²%³  µ ~ À . For a mixture distribution, 7 ´; ²%³  µ ~ ²À³ h 7 ´; ²%³  µ b ²À³ h 7 ´; ²%³  µ . Since ; ²%³ and ; ²%³ have constant forces of mortality, their survival probabilities have an exponential form, c! . 7 ´; ²%³  µ ~  c cÀ , and 7 ´; ²%³  µ ~  c cÀ . Then, 7 ´; ²%³  µ ~ ²À³² c cÀ ³ b ²À³² c cÀ ³ ~ À . This becomes a quadratic equation with @ ~ cÀ : À@  b À@ c À ~  . The solutions are @ ~ À Á c À . Since @ ~ cÀ , we ignore the negative root. Then cÀ ~ À S  ~ À . Note that once we have formulated the quadratic equation, we can substitute in each of the five possible answers for  to see which one satisfies the quadratic equation. Answer: D

21. ,´? c O? € µ is the mean excess loss with a deductible of 20. ,´²?c³ µ

,´?µc,´?wµ

This can be written as 7 ´; ?€µb ~ 7 ´; ?€µ . Using the Pareto distribution, we have    ,´?µ ~ c ~  , and ,´? w µ ~ ² c ³´ c ² b  ³c µ ~ À .  Also, 7 ´? € µ ~  c -? ²³ ~ ² b  ³ ~ À  .

The mean excess loss is  cÀ À  ~  . The quick solution to this problem uses that fact that for the Pareto distribution with parameters  and , based on an ordinary deductible of , the cost per payment random variable is also Pareto with parameters  and b  , so the expected cost per payment (which is the mean excess loss) is b b c . In this problem, that becomes c ~  À Answer: D

22. Under the current limits, with 5 denoting the number of root canals a person needs in a year, the number of root canal claims is either 0 or 1, with 7 ² root canal claims in one year³ ~ 7 ²5 ~ ³ ~ c and 7 ² root canal claim in one year³ ~ 7 ´5 ‚ µ ~  c 7 ´5 ~ µ ~  c c ~ À  . The expected number of root canal claims in one year is  c c (this is not the same as the expected number of root canals a person needs, which is Poisson with a mean of 1).

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22. continued Under the current limits, with 5 denoting the number of fillings a person needs in a year, the number of filling claims is either 0 or 1 or 2, with 7 ² filling claims in one year³ ~ 7 ²5 ~ ³ ~ c and c



7 ² filling claim in one year³ ~ 7 ´5 ~ µ ~  [h ~ c , and P²2 filling claims in one year³ ~ 7 ´5 ‚ µ ~  c 7 ´5 ~ Á µ ~  c c c c The expected number of filling claims in one year is c b ² c c c c ³ ~  c c ~ À .

Under the new limit, the number of root canals and fillings combined needed in one year is 5 , which is Poisson with a mean of 3. The number of claims in one year is 0,1,2 or 3, with probabilities 7 ² claims in one year³ ~ 7 ²5 ~ ³ ~ c Á 7 ² claim in one year³ ~ 7 ²5 ~ ³ ~ c h Á c



7 ²2 claims in one year³ ~ 7 ²5 ~ ³ ~  [h Á and 7 ²2 claims in one year³ ~ 7 ²5 ‚ ³ ~  c 7 ´5 ~ Á Á µ ~  c À c . The expected number of claims in 1 year is c



c h  b ²  [h ³ b ² c À c ³ ~  c À c ~ À  . Under the original limit scheme, the expected number of claims per year is À  b À ~ À . The new limit scheme has an expected number of claims per year that is .237 larger than the original limit scheme. Answer: C

23. ,´²: c ³b µ ~ ,´:µ c ´ c - ²³µ c ´ c - ²³µ ~  c ² c  ²³³ c ² c  ²³ c  ²³³  ²³ ~ 7 ´5 ~ µ ~ c Á  ²³ ~ 7 ´5 ~ µ h 7 ´? ~ µ ~ ²c h ³²À³ ~ À c À ,´²: c ³b µ ~  c ² c c ³ c ² c c c À c ³ ~  b À c . Alternatively, we can find ,´²: c ³b µ ~ ,´:µ c ,´: w µ ~  c ´ ²³ b ² c - ²³³µ ~  c ´À c b ² c c c À c ³µ ~  b À c . Answer: C 

 24. The empirical estimate of the mean is  4  h ~

 bc 5 

, which is

 b  b ³ b ² b  ³ b  ²  b ³  ´²  ³ b ²   

b ² b ³ b ² b  ³µ ~  À .   

²  c  ³

  c The empirical estimate of the second moment is  4  h ²b³² 5 , which is  cc ³ ~   c  c    c  c   ´² ² c³ ³ b ² ²c ³ ³ b ² ² c³ ³ b  ² ²c ³ ³ 







 c  c b ² ²c³ ³ b ² ² c³ ³µ ~ Á  Á  .

The empirical variance is Á  Á  c ² À ³ ~ Á  . The empirical estimate of the standard deviation is j Á  ~   . Answer: B 25.  ~  Á  ~  Á  ~  Á  ~ Á  ~  Á V =V ´/²³µ ~    b   b   b   ~ À .

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~ Á  ~  Á  ~ . Answer: A

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26. Let : denote the compound Poisson random variable, and let 5 denote the Poisson frequency random variable with ,´5 µ ~ , and let ? denote the exponential severity random variable with ,´?µ ~ . Then ,´:µ ~ ,´5 µ h ,´?µ ~  and = ´:µ ~ ,´5 µ h ,´?  µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ ²  ³ ~  h  b  h  ~   . The first moment of the empirical distribution based on the given sample is bb b bbb b b b ~ À .   The variance of the empirical distribution is  ´² c ³ b ² c ³ b Ä b ² c ³ µ ~ À . According to the method of moments, we have the moment equations ,´:µ ~ ,´5 µ h ,´?µ ~  ~ À and = ´:µ ~ ,´5 µ h ,´?  µ ~   ~ À .   Dividing the second equation by the first results in  ~  ~ À À , from which we get the À V~ estimate of , V ~ À . Then  Answer: D V ~  À .

27. For a claim payment of amount % that is not a limit payment from a policy with no c%°

deductible, the likelihood factor is the pdf  . For a limit payment of 50 from a policy with no deductible but with limit 50, the likelihood factor is c ° . For a claim payment of amount % that is not a limit payment from a policy with deductible 10, the likelihood factor is the c²%b³° °

c%°

conditional pdf ~  . For a limit payment of 40 from a policy with deductible 10 c° and policy limit 40 (maximum covered loss 50), the likelihood factor is the conditional c²b³° probability  c° ~ c° . The overall likelihood function will be c° c ° c° c ° c° c ° d dÄd d dÄd d

c° c° c ° d  d Ä d  d c° d c° ~   . ( ~   Á ) ~  , so that ( b ) ~   . Answer: A

28. The quick way of dealing with maximum likelihood estimation for the Weibull distribution with  given and data that includes known % 's and right-censored (limit) payments is as follows (this is reviewed in the mle section of the notes in this study guide). For each data value calculate % ( or " ) for a limit payment. Find the sample mean of those values. The mle of is that  sample mean raised to the power  . For this problem, we have  ~ À , and there are  ~    data values, so we find  '%À  . From the given information, this is  ² À ³ ~  À  . Then ~ ² À ³° ~ ² À ³ ~  .

We now solve the problem from basic principles by first setting up the likelihood function. The density function of the Weibull distribution with parameters  and is 

 ²%° ³ c²%° ³  ²%³ ~ . The likelihood function for the 10 observed losses is %    ²% ° ³ c²% ° ³     % ³ c h c  h'% . 3 ~   ²% ³ ~  ~ h ²   % ~ ~

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28. continued With  ~ À the likelihood function becomes 3~

²À ³

h ²% ³cÀ h 

3~

²À ³

c h ²% ³cÀ h 

c À h'%À 

 À  À



, and then using  %À ~  À  , we get ~

.

The natural log of the likelihood is  3 ~   À c  c À ²% ³ c  À  , and the À C maximum likelihood estimate is found by setting C  3 ~ c b À À ~  . The result is an estimate of V ~ Á   . Answer: C

29. The -plot uses the following smoothed empirical distribution function values, and model probabilities: %













Empirical















 

 



 



 

Dist. Fn. (left tail) Model

 

Dist. Fn. (left tail) Graph D is an accurate representation of the -plot.

30. Posterior probabilities are found first. 7 ´$ ~ O? ~ µ ~

7 ´?~O$~µh7 ´$~µ 7 ´?~µ

~

Answer: D

c h [ h²À ³ c h c h c h  h²À ³b [ [ h²À³b [ h²À³ c h [ h²À³ c   h c h c h  [ h²À ³b [ h²À³b [ h²À³

~ À  .

In a similar way, we get 7 ´$ ~ O? ~ µ ~ 7 ´$ ~ O? ~ µ ~

c h  [ h²À³ c h c h c h  [ h²À ³b [ h²À³b [ h²À³

~ À , and

~ À .

Then the Bayesian premium is found: ,´? O? ~ Á ? ~ µ ~ ,´? O$ ~ O h 7 ´$ ~ O? ~ Á ? ~ µ b ,´? O$ ~ O h 7 ´$ ~ O? ~ Á ? ~ µ b ,´? O$ ~ O h 7 ´$ ~ O? ~ Á ? ~ µ 7 ´$ ~ O? ~ Á ? ~ µ ~

c h  3 h²À ³ [ c h  c h  c    2 3 h²À ³b2 3 h²À³b2  h 3 h²À³ [ [ [

2

~ À Á

7 ´$ ~ O? ~ Á ? ~ µ ~ À  Á 7 ´$ ~ O? ~ Á ? ~ µ ~ À . Then ,´? O? ~ Á ? ~ µ ~ ²³²À³ b ²³²À  ³ b ² ³²À ³ ~ À . Answer: C

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31. The model distribution ? is Poisson with mean , where  has a gamma distribution with  ~  and ~ À . The posterior distribution of the number of claims in year 3 is negative binomial with  ~  b '% and  ~  b . We are given  ~  with % ~  and % ~ . Therefore, the posterior distribution of ? is negative binomial with  ~  b  b  ~  and À  ~ ²À³b À The variance of ? is  ² b  ³ ~ À  . Answer: D

V  ~  ~  V# , 32. The estimated credibility factor for group  is A V  b V  b c V V and the estimated rating factor is A  ?  b ² c A  ³ V. We are given  ~  Á  ~  Á  ~  Á  ~   and  ~  ~  ~  Á so that  ~  b  b   ~  À c c c c  ?  b ?  From the given information we can find  ~ À Á and V ~ ? ~  ? b  b b   c  h    ²? c ?  ³ ~  ²À  b À b À³ ~ À Á and V# ~   ² c³

~

~ V

     c  

~

~ ~

 c c h ´   ²?  c ? ³ c V#² c ³ µ ~

 h ´²À  b À  b À ³ c ²À³²³µ ~ À

 .  c  ²  b b  ³   V  ~  V# ~ V Then A À ~ À Á A  ~ À ~ À  and b À

 b À

  b V   V ~ A À ~ À . Then the estimated rating factor for region 1 is  b À



~

c V ? V  ³ A V ~ ²À ³²À ³ b ²À ³²À ³ ~ À À Answer: C  b ² c A

33. The distribution of the number of failures before the first success is geometric with distribution No. Failures      À ²À³²À³ ²À³ ²À³ ²À³ ²À³  ! À À  À  À

 The uniform random number is " ~ À . We see that À  À   À  , so the simulated number of failures is . Answer: B

34. The total profit : on a the applications accepted in a given day can be modeled as a compound Poisson random variable with mean frequency rate  ~  , and with severity distribution @ . @ represents the profit on an arriving application. @ can be modeled as a mixture of three component distributions. Good risks will always be accepted no matter what the result of the coin flip. Since good risks arrive  of the time, it follows that  of the accepted risks are

"good". Bad risks arrive  of the time, but half of the arriving bad risks will be rejected since the coin flip is "tails" half of the time, and in that case the bad risk is rejected. Therefore,  d  ~  of the accepted risks are "bad". The other  of the accepted risks are 0 (a rejected bad risk).

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34. continued The accepted risks are a mixture of (i) "good" risks (say @ ) with mixing weight  , (ii) "bad" risks (say @ ), with mixing weight  , and (iii) risks of amount 0 (say @ ) with mixing weight  . Since : has a compound Poisson distribution, the variance of : can be formulated as = ´:µ ~  h ,´@  µ ~  h ,´@  µ . Since @ is a mixture distribution, ,´@  µ ~ ²  ³,´@ µ b ²  ³,´@ µ b ²  ³,´@ µ . ,´@ µ ~ = ´@ µ b ²,´@ µ³ ~ Á  , and ,´@ µ ~ = ´@ µ b ²,´@ µ³ ~ Á  , and ,´@ µ ~  À Then ,´@  µ ~ ²  ³²Á ³ b ²  ³²Á ³ b ²  ³²³ ~ Á  . Finally, = ´:µ ~ ² ³² Á ³ ~ Á Á  . Answer: C

35. The Schwarz Bayesian Criterion compares M( c  ln²³ and M) c  ln²³ , and since model A is preferable to model B this means that M( c  ln²³ c ´M) c  ln²³µ €  , which can be rewritten as M( c M) € ²³ . The likelihood ratio test has test statistic ²M( c M) ³ . Since model A has 3 parameters and model B has 1 parameter, the number of degrees of freedom in the chi-square statistic is  c  ~  . The critical value for a test with significance level 5% is À ²³ ~ À

 and the critical value for a test with significance level 1% is À ²³ ~ À . Since the null hypothesis is not rejected at the 1% level, it must be true that ²M( c M) ³  À, so that M( c M)  À  . From the Schwartz Bayesian Criterion, we had ²³  M( c M) , and therefore, ²³  À  . It then follows that   À  ~

À . The maximum (integer) value for  is

. Answer: A

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ACTEX EXAM C/4 - PRACTICE EXAM 14 1. You are given: (i) Annual claim counts follow a Poisson distribution with mean -. (ii) The parameter - has a prior distribution with probability density function: 0 Ð-Ñ œ "$ /-Î$ ß -  ! Two claims were observed during the first year. Determine the variance of the posterior distribution of -. A) 9/16 () 27/16 C) 9/4 D) 16/3 E) 27/4

2. \ 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 Ð!ß "Ñ value of .7. Find the simulated value of \ . A) Less than 2.0 B) At least 2.0 but less than 2.1 C) At least 2.1 but less than 2.2 D) At least 2.2 but less than 2.3 E) At least 2.3

3. The parameter - has prior distribution 1Ð-Ñ œ "# /-  "# Ð "# /-Î# Ñ (mixture of two exponentials). The model distribution \ has a conditional distribution given - that is Poisson with mean -. Find the Buhlmann credibility premium if there is a single observation of 2. A) $" B) $$ C) $' D) $* E) %# "( "( "( "( "(

4. A random sample of eight times until failure has the following Nelson-Aalen estimates for the cumulative hazard function: s LÐ>Ñ > >$ ! $Ÿ>& !Þ"#& &Ÿ>' !Þ%"!( 'Ÿ>) !Þ'"!( )Ÿ>* !Þ)'!( * Ÿ >  "! "Þ"*%! "! Ÿ > #Þ"*%! There is no censoring or truncation of the data. Find the empirical estimate of the variance of the time until failure. A) Less than 5.0 B) At least 5.0, but less than 5.2 C) At least 5.2, but less than 5.4 D) At least 5.4, but less than 5.6 E) At least 5.6

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5. 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 0.425. (ii) The variance of the hypothetical means is 0.370. (iii) The expected value of the process variance is 1.793. One class of policyholders is selected at random from the book. Nine policyholders are selected 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) 2.5 B) 2.8 C) 3.0 D) 3.3 E) 3.9

6. Suppose that \" ß \# ß á are independent random variables, each with probability of successes : and probability of failure "  :, where !  : Ÿ ". Let R be the number of observations needed to obtain the first success. What is a maximum likelihood estimator of : ? A) R "# B) R "" C) R" D) R "" E) R "#

7. A survival study of 100 individuals with no censoring results in the following Nelson-Aalen estimates for the cumulative hazard rates at successive death points. It is known that there were 3 deaths at the third death point, >$ . Death Point >3 >" ># >$ s 3Ñ LÐ> .03 .103495 Determine the number of deaths at the second death point, ># . A) 0 B) 1 C) 2 D) 3 E) 4

8. 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 Ò!ß "!!Ó &!! Ð"!!ß #!!Ó #&! Ð#!!ß &!!Ó "&! Ð&!!ß "!!!Ó '! Ð"!!!ß #!!!Ó %! 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. A) 150 B) 152 C) 154 D) 156 E) 158

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9. You are simulating the gain/loss from insurance where: (i) Claim occurrence follows a Poisson distribution with - œ #Î$ per year. (ii) Each claim amount is 1, 2 or 3, with T Ð"Ñ œ !Þ#& ß :Ð#Ñ œ !Þ#& and :Ð$Ñ œ !Þ&! . (iii) Claim occurrences and amounts are independent. (iv) The annual premium equals expected annual claims plus 1.8 times the standard deviation of annual claims. (v) 3 œ !Þ You use 0.75, 0.60, 0.40, and 0.20 from the unit interval to simulate time between claims, where small numbers correspond to longer times. You use 0.30, 0.60, 0.20, and 0.70 from the unit interval to simulate claim size, where small numbers correspond to smaller claims. Calculate the gain or loss for the first two years from this simulation. A) loss of 5 B) loss of 4 C) 0 D) gain of 4 E) gain of 5

10. The Allerton Insurance Company insures 3 indistinguishable populations. The claims frequency of each insured follows a Poisson process. Given: Population Expected Probability Claim (class) time between of being in cost claims class I 12 months 1/3 1,000 II 15 months 1/3 1,000 III 18 months 1/3 1,000 Calculate the expected loss in year 2 for an insured that had no claims in year 1. A) Less than 810 B) At least 810, but less than 910 C) At least 910, but less than 1,010 D) At least 1,010, but less than 1,110 E) At least 1,110

11. The following table that was obtained by fitting both a Poisson distribution and a binomial distribution to a data set of 100,000 integer-valued observations. s œ Þ""&#( 7 s œ % ß s; œ Þ!#))"(& Fitted Poisson Fitted Binomial 85 expected ;# expected ;# 5 ! )*,!!! )*,""#Þ' Þ"%## ))ß *'"Þ) Þ!"'% " "!,%)( "!ß #(#Þ! %Þ&!! "!ß &&)Þ* Þ%)*( # &!! &*#Þ! "%Þ$!&! %(!Þ! "Þ*"*& #$Þ% %Þ'## *Þ$ "Þ$()(   $ "$ Totals "!!ß !!! "!!ß !!! #$Þ&( "!!ß !!! $Þ)! Degrees of freedom %""œ# %#"œ" :  Þ!!" Þ!&  :  Þ" :-value Which of the following pair of statements has both statements true regarding the chi-square goodness-of-fit test of the null hypothesis that the model is a good fit to the data? Poisson Binomial A) Accept at 1% level Reject at 1% level B) Accept at 5% level Reject at 5% level C) Reject at 10% level Reject at 10% level D) Reject at 5% level Reject at 5% level E) Reject at 1% level Reject at 1% level

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12. Non-parametric empirical Bayes estimation has been applied to a data set with three policyholders, with four exposure periods for each policyholder, and with one exposure unit for    each exposure period. You are also given that \ " œ % ß \ # œ ' ß \ $ œ ( , and s is found based on this estimated expected process variance is # . The credibility factor ^ information. It is decided that a fourth policyholder will be added to the data set and the credibility factor will be recalculated. The fourth policyholder has four exposure periods with  s one unit for each exposure period, and \ % œ $ and s@% œ " . Find the percentage change in ^ from original to recalculated value. A)  ##% B)  ""% C) !% D) ""% E) ##%

13. A spliced distribution is defined to have the following density function. + † 0" ÐBÑ !  B  "!! . 0 ÐBÑ œ œ , † 0# ÐBÑ "!! Ÿ B  #!! 0" ÐBÑ is the density function of a uniform random variable on the interval Ð!ß "!!Ñ , + œ Þ%, and 0# ÐBÑ is the density function of the uniform distribution on the interval Ò"!!ß #!!Ñ. Find the mean of the spliced distribution. A) 100 B) 110 C) 120 D) 130 E) 140

14. The loss random variable \ has an exponential distribution and an ordinary deductible is applied to all losses. You are given the following: • The variance of the cost per loss random variable is 20,480. • The variance of the cost per payment random variable (excess loss random variable) is 25,600. Find the average cost per loss. A) Less than 80 B) At least 80 but less than 83 C) At least 83 but less than 86 D) At least 86 but less than 89 E) At least 89

15. W is the mixture of two compound Poisson distributions, W" and W# , with mixing weights " prob. "# of "# each. W" has Poisson frequency R" with -" œ " and severity ]" œ œ . # probÞ "# W# has Poisson frequency R# with -# œ " and severity ]# œ œ Find Z +" is =" "!! œ Þ!$ p =" œ $ . Then, the Nelson-Aalen estimate of the cumulative hazard rate function to $ "!!

=

$

 *(#  *(= œ Þ"!$%*& Þ By trial and error (substitution) # we find that =# œ % . Answer: E time >$ is

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

PRACTICE EXAM 14

PE-275

8. \ 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 LQ œ IÒ\l)Ó œ ) and the process variance is T Z œ Z +Î"Þ& so > œ Þ(( is the time between the first and second claim, which implies that the second claim occurs at time 1.2. The time between the second and third claim is the solution of Þ% œ />Î"Þ& so > œ "Þ$( , so the third

© ACTEX 2009

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

PE-276

PRACTICE EXAM 14

9. continued claim occurs after the end of the second year. There are two claims simulated in the first two years. We now simulate claim amounts. The claim amount distribution is \ " # $ : Þ#& Þ#& Þ& J Þ#& Þ&! "Þ!! We are told that small numbers correspond to small claim amounts, so we apply the usual inverse transformation approach to simulate \ (using J ÐBÑ). The first uniform number for simulating claim amounts is .3. Since Þ#&  Þ$ Ÿ Þ& , the simulated claim amount is \ œ # . The second uniform number for simulating claim amounts is .6. Since Þ&  Þ' Ÿ "Þ! , the simulated claim amount is \ œ $. Aggregate claims for the two years is 5. The expected aggregate claim per year is IÒR Ó † IÒ\Ó œ Ð #$ ÑÒÐ"ÑÐÞ#&Ñ  Ð#ÑÐÞ#&Ñ  Ð$ÑÐÞ&ÑÓ œ "Þ& The variance of the compound Poisson annual aggregate claim distribution is IÒR Ó † IÒ\ # Ó œ Ð #$ ÑÒÐ"# ÑÐÞ#&Ñ  Ð## ÑÐÞ#&Ñ  Ð$# ÑÐÞ&ÑÓ œ $Þ)$$. The annual premium is "Þ&  "Þ)È$Þ)$$ œ &Þ!# . The gain for the two years is premium minus claims, which is #Ð&Þ!#Ñ  & œ &Þ!% . Answer: E

10. For each Class the claims follow a Poisson process. For a particular Class, if the expected amount of time between claims is >! (in years), then the expected number of claims per year is >" . ! Class I has an expected amount of time of 1 year between claims, so that the expected number of claims per year for Class 1 is -" œ " . Class II has an expected amount of time of 1.25 years (15 months) between claims, so that the expected number of claims per year for Class 1 is " -# œ "Þ#& œ Þ) . Class III has an expected amount of time of 1.5 years (15 months) between " claims, so that the expected number of claims per year for Class 1 is -$ œ "Þ& œ $# .

Let us denote the number of claims in year 1 by R" and the number of claims in year 2 is denoted by R# . We wish to find IÒR# lR" œ !Ó , and then, since each claim cost is 1,000 in all classes, the expected loss in year 2 is "!!!IÒR# lR" œ !Ó . We find IÒR# lR" œ !Ó by conditioning over the Class type. IÒR# lR" œ !Ó œ IÒR# lClass IÓ † T ÒClass IlR" œ !Ó  IÒR# lClass IIÓ † T ÒClass IIlR" œ !Ó  IÒR# lClass IIIÓ † T ÒClass IIIlR" œ !Ó . The conditional expectations are the expected number of claims in a year for a given class: IÒR# lClass IÓ œ " , IÒR# lClass IIÓ œ Þ) and IÒR# lClass IIIÓ œ #$ . The conditional probabilities can be found from the following probability table: Class I , T Ò I Ó œ "$

Class II , T Ò II Ó œ "$

" ! R" œ ! T ÒR" œ !lIÓ œ / !x†" œ Þ$'())

Þ) ! / $ †Ð $ Ñ! T ÒR" œ !lIIÓ œ / !x†Þ) T ÒR" œ !lIIIÓ œ !x œ Þ%%*$$ œ Þ&"$%#

T ÒR" œ ! ∩ IÓ œ T ÒR" œ !lIÓ † T Ò I Ó œ ÐÞ$'())ÑÐ "$ Ñ

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Class III , T Ò III Ó œ "$ #

#

T ÒR" œ ! ∩ IIÓ T ÒR" œ ! ∩ IIIÓ œ T ÒR" œ !lIIÓ † T Ò II Ó œ T ÒR" œ !lIIIÓ † T Ò III Ó œ ÐÞ%%*$$ÑÐ "$ Ñ œ ÐÞ&"%$#ÑÐ "$ Ñ

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

PRACTICE EXAM 14

PE-277

10. continued Then, T ÒR" œ !Ó œ T ÒR" œ ! ∩ IÓ  T ÒR" œ ! ∩ IIÓ  T ÒR" œ ! ∩ IIIÓ œ Þ%%$)% , and ÐÞ$'())ÑÐ "$ Ñ T ÒClass I ∩ R" œ!Ó œ œ Þ#('$ . T ÒR" œ!Ó Þ%%$)% " ÐÞ%%*$$ÑÐ Ñ T ÒClass II ∩ R" œ!Ó IIlR# œ !Ó œ œ Þ%%$)% $ œ Þ$$(& . T ÒR" œ!Ó ÐÞ&"%$#ÑÐ "$ Ñ T ÒClass III ∩ R" œ!Ó IIIlR# œ !Ó œ œ œ Þ$)'$ T ÒR" œ!Ó Þ%%$)%

T ÒClass IlR" œ !Ó œ T ÒClass T ÒClass

.

Then IÒR# lR" œ !Ó œ Ð"ÑÐÞ#('$Ñ  ÐÞ)ÑÐÞ$$(&Ñ  Ð #$ ÑÐÞ$)'$Ñ œ Þ)!% , and the expected loss for year 2 is "!!!ÐÞ)!%Ñ œ )!% . Answer: A

11. Since the :-value for the Poisson model is  .001, the Poisson model is rejected as a good fit at any significance level above .001 (.1%). Since the :-value for the Binomial model is between .05 and .1, the binomial model is rejected at the 10% level of significance or above, and is accepted at any level of significance below 5%. Statement C is the only one satisfying both requirements. Answer: C

8   12. Original data set: \ œ "< \ 3 œ $" Ò%  '  (Ó œ "( @ œ # (given), 8 œ % ß $ . s 4œ"

" + œ C:/ "Ó œ Ð"ÑÐ # Ñ ß IÒWl>C:/ #Ó œ Ð"ÑÐ $ Ñ p Z +C:/ÓÓ œ Ð #  $ Ñ Ð # ÑÐ # Ñ œ "%%

" $*( Z +