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For Exam # 3 Study Purposes – Math 478 / 568, Spring 2013

UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Actuarial Science Program

DEPARTMENT OF MATHEMATICS Math 478 / 568 Prof. Rick Gorvett Actuarial Modeling Spring, 2009

Exam # 3 (17 Problems – Max possible points = 40)

Wednesday, May 13, 2009

You have 2 hours (from 8 to 10 am) to complete this exam. The exam is open-book, open-note. A standard normal distribution table and other distribution information is provided at the end of this exam (feel free to tear it off the back of the exam for convenience). Problems (1) through (14) are multiple choice, each worth two points. Circle the letter associated with the best answer to each question. Problems (15) through (17) are each worth four points. Please provide dollar answers to the nearest cent ($x,xxx.xx), and proportion and probability answers either as percentages to two decimal places (xx.xx%) or as numbers to four decimal places (0.xxxx). When using the normal distribution, there is no need to interpolate in the standard normal table – just use the value in the table closest to the value you need to do your problem. No clarification questions may be asked during the exam. Good luck – and have a great summer! (1) Consider a random variable X , which has a continuous uniform distribution from 0 to 500.

Determine the mean excess loss value of X above 400.

(a) 20 (b) 50 (c) 100 (d) 450 (e) None of (a) through (d) is correct.

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(2) Suppose ground-up loss variable X follows a two-parameter Pareto distribution with =

2,000, and = 5. Let d be the deductible level at which the loss elimination ratio is 0.35. Find d.

(a) d ≤ 200 (b) 200 < d ≤ 250 (c) 250 < d ≤ 300 (d) 300 < d ≤ 350 (e) 350 < d

(3) Three independent random variables X1, X2, and X3 have the following probability functions:

x f1(x) f2(x) f3(x) 0 0.40 0.60 0.70 1 0.30 0.20 0.20 2 0.20 0.10 0.10 3 0.10 0.10 - Let S = X1 + X2 + X3 , and let p = Prob[S = 2]. Find p. (a) p ≤ 0.13 (b) 0.13 < p ≤ 0.16 (c) 0.16 < p ≤ 0.19 (d) 0.19 < p ≤ 0.22 (e) 0.22 < p

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(4) An insurance policy has a deductible of 10 and a “policy limit” (maximum payment per loss) of 70. The three loss payments under this policy have been 20, 50, and 70. The insurer is only aware of those losses on which a payment is made. Assuming the ground-up losses

were generated by a distribution of the form

x

exf

1

)( , x > 0, find the maximum

likelihood estimate of .

(a) 55 (b) 65 (c) 75 (d) 85 (e) 95

(5) You are given the following observations from an exponential distribution with mean :

1, 1, 2, 3, 3, 5, 7, 8, 10, 15, 19

Estimate by percentile matching, using the empirical smoothed estimate of the 80th percentile. Using that result, find the limited expected value of this distribution at 5.

(a) 3.64 (b) 3.73 (c) 3.96 (d) 4.15 (e) 4.29

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(6) Based on a sample of observations, the sample mean is 120, and the sample standard deviation is 50. Assuming that this sample is from a Gamma distribution with unknown parameters and , determine the method of moments estimate of .

(a) ̂ ≤ 20

(b) 20 < ̂ ≤ 22

(c) 22 < ̂ ≤ 24

(d) 24 < ̂ ≤ 26

(e) 26 < ̂

(7) A loss random variable X has a probability density function .100 ,02.020.0)( xxxf (f(x) = 0 for other values of x.) Determine Q, the 95%

Value-at-Risk (VaR) of loss random variable X.

(a) Q ≤ 8.0 (b) 8.0 < Q ≤ 8.5 (c) 8.5 < Q ≤ 9.0 (d) 9.0 < Q ≤ 9.5 (e) 9.5 < Q

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(8) Consider the following loss distribution:

Size of Loss Probability $ 10 0.40 50 0.25 100 0.15 250 0.10 500 0.05 1,000 0.05

Determine the 80% Tail Value at Risk (i.e., the 80% Conditional Tail Expectation).

(a) 200 (b) 350 (c) 500 (d) 750 (e) 900

(9) You are given a sample of claim payments as follows: 5, 10, 10, 20, 40. Claim sizes are assumed to follow an exponential distribution, with the mean estimated using the method of moments. Determine the value of the Kolmogorov-Smirnov test statistic, .

(a) ≤ 0.19 (b) 0.19 < ≤ 0.21 (c) 0.21 < ≤ 0.23 (d) 0.23 < ≤ 0.25 (e) 0.25 <

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(10) You are given the following observations: 1, 1, 3, 3, 3, 5, 7, 8, 12. You believe that these observations come from a model distribution which is exponential with a mean of 5. Which of the following p-p plots correctly reflects these data and model?

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(11) You are asked to determine next year’s (calendar year 2010) premium for a particular large commercial risk. You have the following actuarial model that applies to similar risks:

i.) Claim counts in 2010 will follow a Poisson distribution with a mean of 80. ii.) Claim sizes in 2010 will follow a two-parameter Pareto distribution with = 3 and

= 20,000. iii.) Claim sizes and claim counts are independent. You also have the following data for the particular risk: iv.) You have loss data for this risk for just one year (calendar year 2008). During that

year, there were 55 claims, resulting in total aggregate losses of 625,000. v.) Use the expected mean from items (i) through (iii) above as your “manual premium.” vi.) The risk’s exposure for 2010 is identical to the exposure during 2008. vii.) Ignore expenses, profit provisions, and inflation. viii.) The full credibility standard is defined as total claim costs being within 5% of the

expected aggregate losses 90% of the time. Determine P, the limited fluctuation (classical) credibility premium for 2010 (based upon a full-credibility standard expressed in terms of number of claims).

(a) P ≤ 730,000 (b) 730,000 < P ≤ 750,000 (c) 750,000 < P ≤ 770,000 (d) 770,000 < P ≤ 790,000 (e) 790,000 < P

(12) Suppose that the full-credibility standard is defined such that aggregate claim dollars will be within 8% of its true value 95% of the time. The number of claims follows a Poisson distribution, and the claim severity distribution is gamma with = 4 and = 1,000. Find the number of claims, n, that would produce a credibility of 0.50.

(a) 168 (b) 178 (c) 188 (d) 198 (e) 208

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(13) You are given the following:

i.) The number of claims incurred in a month by any insured has a Poisson distribution with mean λ.

ii.) The claim frequencies of different insureds are independent. iii.) The prior distribution is gamma with = 5 and = 0.04. iv.)

Month

Number of Insureds

Number of Claims

1 400 100 2 500 100 3 600 250 4 700 ?

Determine the Bühlmann-Straub credibility estimate of the number of claims in Month 4.

(a) P ≤ 150 (b) 150 < P ≤ 170 (c) 170 < P ≤ 190 (d) 190 < P ≤ 210 (e) 210 < P

(14) For a portfolio of 100 policies in effect during 2008, you have the following claim count data, which expresses the number of policies experiencing different numbers of claims during that year:

Number of Claims Number of Policies 0 70 1 20 2 10

Assume that a Poisson distribution is appropriate for modeling the number of claims in a year. Determine a semiparametric empirical Bayes credibility estimate of the number of annual claims expected on a policy which had two claims in 2008.

(a) 0.48 (b) 0.52 (c) 0.56 (d) 0.60 (e) 0.64

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(15) You have the following 30 observations: 1 1 2 2 2 2 3 3 4 5

5 6 7 7 8 8 8 10 11 12 14 14 15 16 16 19 20 22 25 32

You hypothesize that these data came from an exponential distribution with a mean of 10. Calculate the chi-square goodness-of-fit test statistic using the following four intervals: (0,3], (3,8], (8,15], (15,infinity).

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(16) Suppose that policyholders all fall into one of two homogeneous risk groups. Policyholders

in each risk level group have the following size-of-loss distributions per period: Size of Loss Group 1 Group 2 $ 0 50% 20% 100 20 30 500 20 30 1,000 10 20

There are twice as many Group 2 policyholders as Group 1 policyholders. An insurer randomly selects a policyholder, and observes two periods of loss experience for that policyholder. Suppose that the insurer observes a $ 0 loss the first year, and a $ 1,000 loss the second year. Calculate the expected value of the policyholder’s loss in the next period using Bühlmann credibility.

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(17) Suppose that the number of claims per year for a policyholder follows a Poisson distribution

with parameter , and the prior distribution of is exponential with a mean of 2. Empirically, there were 1, 4, 3, and 4 claims, respectively, in years 1 through 4 for this policyholder. Determine the Bayesian credibility estimate for this policyholder’s expected claim frequency in year 5.

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Standard Normal Distribution Values

x 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.53590.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57530.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.61410.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.65170.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.68790.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.72240.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.75490.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.78520.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.81330.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.83891.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.86211.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.88301.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.90151.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.91771.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.93191.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.94411.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.96331.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.97061.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.97672.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.98172.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.98572.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.98902.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.99162.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.99362.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.99522.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.99642.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.99742.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.99812.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.99863.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.99903.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.99933.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.99953.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.99973.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.99983.5 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.99983.6 0.9998 0.9998 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.99993.7 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.99993.8 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.99993.9 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.00004.0 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

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Distribution Information

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