November 2006 - Exam C

Exam C, Fall 2006

PRELIMINARY ANSWER KEY

Question # Answer Question # Answer

1 E 19 B 2 D 20 D 3 B 21 A 4 C 22 A 5 A 23 E 6 D 24 E 7 B 25 D 8 C 26 A 9 E 27 C

10 D 28 C 11 E 29 C 12 B 30 B 13 C 31 C 14 A 32 A 15 B 33 B 16 E 34 A 17 D 35 A 18 D

Exam C: Fall 2006 - 1 - GO ON TO NEXT PAGE

**BEGINNING OF EXAMINATION**

1. You are given: (i) Losses follow a Burr distribution with α = 2. (ii) A random sample of 15 losses is:

195 255 270 280 350 360 365 380 415 450 490 550 575 590 615 (iii) The parameters γ and θ are estimated by percentile matching using the smoothed empirical

estimates of the 30th and 65th percentiles. Calculate the estimate of γ. (A) Less than 2.9 (B) At least 2.9, but less than 3.2 (C) At least 3.2, but less than 3.5 (D) At least 3.5, but less than 3.8 (E) At least 3.8


2. An insurance company sells three types of policies with the following characteristics:

Type of Policy Proportion of Total Policies

Annual Claim Frequency

I 5% Poisson with 0.25λ = II 20% Poisson with 0.50λ = III 75% Poisson with 1.00λ =

A randomly selected policyholder is observed to have a total of one claim for Year 1 through Year 4. For the same policyholder, determine the Bayesian estimate of the expected number of claims in Year 5. (A) Less than 0.4

(B) At least 0.4, but less than 0.5

(C) At least 0.5, but less than 0.6

(D) At least 0.6, but less than 0.7

(E) At least 0.7


3. You are given a random sample of 10 claims consisting of two claims of 400, seven claims of 800, and one claim of 1600. Determine the empirical skewness coefficient. (A) Less than 1.0 (B) At least 1.0, but less than 1.5 (C) At least 1.5, but less than 2.0 (D) At least 2.0, but less than 2.5 (E) At least 2.5


4. You are given: (i) The cumulative distribution for the annual number of losses for a policyholder is:

n ( )NF n 0 0.125 1 0.312 2 0.500 3 0.656 4 0.773 5 0.855 M M

(ii) The loss amounts follow the Weibull distribution with θ = 200 and τ = 2.

(iii) There is a deductible of 150 for each claim subject to an annual maximum out-of-

pocket of 500 per policy. The inversion method is used to simulate the number of losses and loss amounts for a policyholder.

(a) For the number of losses use the random number 0.7654.

(b) For loss amounts use the random numbers: 0.2738 0.5152 0.7537 0.6481 0.3153

Use the random numbers in order and only as needed. Based on the simulation, calculate the insurer’s aggregate payments for this policyholder. (A) 106.93 (B) 161.32 (C) 224.44 (D) 347.53 (E) 520.05


5. You have observed the following three loss amounts: 186 91 66 Seven other amounts are known to be less than or equal to 60. Losses follow an inverse exponential with distribution function ( ) / , 0xF x e xθ−= > Calculate the maximum likelihood estimate of the population mode. (A) Less than 11 (B) At least 11, but less than 16 (C) At least 16, but less than 21 (D) At least 21, but less than 26 (E) At least 26


6. For a group of policies, you are given: (i) The annual loss on an individual policy follows a gamma distribution with parameters

α = 4 and θ. (ii) The prior distribution of θ has mean 600. (iii) A randomly selected policy had losses of 1400 in Year 1 and 1900 in Year 2. (iv) Loss data for Year 3 was misfiled and unavailable. (v) Based on the data in (iii), the Bühlmann credibility estimate of the loss on the selected

policy in Year 4 is 1800. (vi) After the estimate in (v) was calculated, the data for Year 3 was located. The loss on

the selected policy in Year 3 was 2763. Calculate the Bühlmann credibility estimate of the loss on the selected policy in Year 4 based on the data for Years 1, 2 and 3. (A) Less than 1850 (B) At least 1850, but less than 1950 (C) At least 1950, but less than 2050 (D) At least 2050, but less than 2150 (E) At least 2150


7. The following is a sample of 10 payments: 4 4 5+ 5+ 5+ 8 10+ 10+ 12 15 where + indicates that a loss exceeded the policy limit. Determine Greenwood’s approximation to the variance of the product-limit estimate S (11). (A) 0.016 (B) 0.031 (C) 0.048 (D) 0.064 (E) 0.075


8. Determine ( )3f using the second degree polynomial that interpolates the points: (2, 25) (4, 20) (5, 30) (A) Less than 15 (B) At least 15, but less than 18 (C) At least 18, but less than 21 (D) At least 21, but less than 23 (E) At least 23


9. You are given: (i) For Q = q, 1 2, , , mX X XK are independent, identically distributed Bernoulli

random variables with parameter q. (ii) 1 2m mS X X X= + + +L (iii) The prior distribution of Q is beta with 1, 99,a b= = and 1θ = . Determine the smallest value of m such that the mean of the marginal distribution of mS is greater than or equal to 50. (A) 1082 (B) 2164 (C) 3246 (D) 4950 (E) 5000


10. You are given: (i) A portfolio consists of 100 identically and independently distributed risks. (ii) The number of claims for each risk follows a Poisson distribution with mean λ . (iii) The prior distribution of λ is:

4 50(50 )( ) , 0

6e λλπ λ λλ

−= >

During Year 1, the following loss experience is observed:

Number of Claims Number of Risks 0 90 1 7 2 2 3 1

Total 100 Determine the Bayesian expected number of claims for the portfolio in Year 2. (A) 8 (B) 10 (C) 11 (D) 12 (E) 14


11. You are planning a simulation to estimate the mean of a non-negative random variable. It is known that the population standard deviation is 20% larger than the population mean. Use the central limit theorem to estimate the smallest number of trials needed so that you will be at least 95% confident that the simulated mean is within 5% of the population mean. (A) 944

(B) 1299

(C) 1559

(D) 1844

(E) 2213


12. You are given: (i) The distribution of the number of claims per policy during a one-year period for

10,000 insurance policies is:

Number of Claims per Policy Number of Policies 0 5000 1 5000

2 or more 0

(ii) You fit a binomial model with parameters m and q using the method of maximum likelihood.

Determine the maximum value of the loglikelihood function when m = 2. (A) −10,397 (B) −7,781 (C) −7,750 (D) −6,931 (E) −6,730


13. You are given: (i) Over a three-year period, the following claim experience was observed for two

insureds who own delivery vans:

Year Insured 1 2 3

A Number of Vehicles 2 2 1 Number of Claims 1 1 0

B Number of Vehicles N/A 3 2 Number of Claims N/A 2 3

(ii) The number of claims for each insured each year follows a Poisson distribution. Determine the semiparametric empirical Bayes estimate of the claim frequency per vehicle for Insured A in Year 4. (A) Less than 0.55 (B) At least 0.55, but less than 0.60 (C) At least 0.60, but less than 0.65 (D) At least 0.65, but less than 0.70 (E) At least 0.70


14. For the data set 200 300 100 400 X you are given: (i) k = 4 (ii) 2s = 1 (iii) 4r = 1 (iv) The Nelson-Åalen Estimate ˆ (410)H > 2.15 Determine X. (A) 100 (B) 200 (C) 300 (D) 400 (E) 500


15. You are given: (i) A hospital liability policy has experienced the following numbers of claims over a

10-year period:

10 2 4 0 6 2 4 5 4 2 (ii) Numbers of claims are independent from year to year. (iii) You use the method of maximum likelihood to fit a Poisson model. Determine the estimated coefficient of variation of the estimator of the Poisson parameter. (A) 0.10 (B) 0.16 (C) 0.22 (D) 0.26 (E) 1.00


16. You are given: (i) Claim sizes follow an exponential distribution with mean θ . (ii) For 80% of the policies, θ = 8.

(iii) For 20% of the policies, 2θ = . A randomly selected policy had one claim in Year 1 of size 5. Calculate the Bayesian expected claim size for this policy in Year 2. (A) Less than 5.8 (B) At least 5.8, but less than 6.2 (C) At least 6.2, but less than 6.6 (D) At least 6.6, but less than 7.0 (E) At least 7.0


17. For a double-decrement study, you are given: (i) The following survival data for individuals affected by both decrements (1)

and (2):

j cj ( )Tjq

0 0 0.100 1 20 0.182 2 40 0.600 3 60 1.000

(ii) ( )2 0.05jq ′ = for all j (iii) Group A consists of 1000 individuals observed at age 0. (iv) Group A is affected by only decrement (1). Determine the Kaplan-Meier multiple-decrement estimate of the number of individuals in Group A that survive to be at least 40 years old. (A) 343 (B) 664 (C) 736 (D) 816 (E) 861


18. You are given: (i) At time 4 hours, there are 5 working light bulbs.

(ii) The 5 bulbs are observed for p more hours.

(iii) Three light bulbs burn out at times 5, 9, and 13 hours, while the remaining light bulbs

are still working at time 4 + p hours.

(iv) The distribution of failure times is uniform on ( )0,ω .

(v) The maximum likelihood estimate of ω is 29. Determine p. (A) Less than 10 (B) At least 10, but less than 12 (C) At least 12, but less than 14 (D) At least 14, but less than 16 (E) At least 16


19. You are given: (i) The number of claims incurred in a month by any insured follows a Poisson

distribution with mean λ . (ii) The claim frequencies of different insureds are independent. (iii) The prior distribution of λ is Weibull with 0.1θ = and 2τ = .

(iv) Some values of the gamma function are

( ) ( ) ( ) ( )0.5 1.77245, 1 1, 1.5 0.88623, 2 1Γ = Γ = Γ = Γ =

(v) Month Number of Insureds Number of Claims

1 100 10 2 150 11 3 250 14

Determine the Bühlmann-Straub credibility estimate of the number of claims in the next 12 months for 300 insureds. (A) Less than 255 (B) At least 255, but less than 275 (C) At least 275, but less than 295 (D) At least 295, but less than 315 (E) At least 315


20. You are given: (i) The following data set:

2500 2500 2500 3617 3662 4517 5000 5000 6010 6932 7500 7500

(ii) ^

1(7000)H is the Nelson-Åalen estimate of the cumulative hazard rate function calculated under the assumption that all of the observations in (i) are uncensored.

(iii) ^

2 (7000)H is the Nelson-Åalen estimate of the cumulative hazard rate function calculated under the assumption that all occurrences of the values 2500, 5000 and 7500 in (i) reflect right-censored observations and that the remaining observed values are uncensored.

Calculate|^ ^

1 2(7000) (7000)H H− |. (A) Less than 0.1 (B) At least 0.1, but less than 0.3 (C) At least 0.3, but less than 0.5 (D) At least 0.5, but less than 0.7 (E) At least 0.7


21. For a warranty product you are given: (i) Paid losses follow the lognormal distribution with 13.294μ = and 0.494σ = .

(ii) The ratio of estimated unpaid losses to paid losses, y, is modeled by

0.851 0.7470.801 xy x e−= where x = 2006 − contract purchase year

The inversion method is used to simulate four paid losses with the following four uniform (0,1) random numbers: 0.2877 0.1210 0.8238 0.6179 Using the simulated values, calculate the empirical estimate of the average unpaid losses for purchase year 2005. (A) Less than 300,000 (B) At least 300,000, but less than 400,000 (C) At least 400,000, but less than 500,000 (D) At least 500,000, but less than 600,000 (E) At least 600,000


22. Five models are fitted to a sample of n = 260 observations with the following results:

Model Number of Parameters Loglikelihood I 1 −414 II 2 −412 III 3 −411 IV 4 −409 V 6 −409

Determine the model favored by the Schwarz Bayesian criterion. (A) I (B) II (C) III (D) IV (E) V


23. You are given: (i) The annual number of claims for an individual risk follows a Poisson distribution

with mean λ . (ii) For 75% of the risks, 1=λ . (iii) For 25% of the risks, 3=λ . A randomly selected risk had r claims in Year 1. The Bayesian estimate of this risk’s expected number of claims in Year 2 is 2.98. Determine the Bühlmann credibility estimate of the expected number of claims for this risk in Year 2. (A) Less than 1.9 (B) At least 1.9, but less than 2.3 (C) At least 2.3, but less than 2.7 (D) At least 2.7, but less than 3.1 (E) At least 3.1


24. You are given the following ages at time of death for 10 individuals:

25 30 35 35 37 39 45 47 49 55 Using a uniform kernel with bandwidth b = 10, determine the kernel density estimate of the probability of survival to age 40. (A) 0.377 (B) 0.400 (C) 0.417 (D) 0.439 (E) 0.485


25. The following is a natural cubic spline passing through the points (0, 3), (1, 2), (3, 6):

( )( ) ( )( )( ) ( )( )

33 12 2

2 33 12 4

3 , 0 1

2 1 1 , 1 3

x x xf x

x x x

⎧ − + ≤ ≤⎪= ⎨+ − − − ≤ ≤⎪⎩

Using the method of extrapolation as given in the Loss Models text, determine ( )4 .f (A) 7.0

(B) 8.0

(C) 8.8

(D) 9.0

(E) 10.0


26. The random variables 1 2, , , nX X XK are independent and identically distributed with probability density function

/

( ) , 0xef x xθ

θ

−

= ≥

Determine 2E X⎡ ⎤⎣ ⎦ .

(A) 21nn

θ+⎛ ⎞⎜ ⎟⎝ ⎠

(B) 22

1nn

θ+⎛ ⎞⎜ ⎟⎝ ⎠

(C) 2

nθ

(D) 2

nθ

(E) 2θ


27. Three individual policyholders have the following claim amounts over four years:

Policyholder Year 1 Year 2 Year 3 Year 4 X 2 3 3 4 Y 5 5 4 6 Z 5 5 3 3

Using the nonparametric empirical Bayes procedure, calculate the estimated variance of the hypothetical means. (A) Less than 0.40 (B) At least 0.40, but less than 0.60 (C) At least 0.60, but less than 0.80 (D) At least 0.80, but less than 1.00 (E) At least 1.00


28. You are given: (i) A Cox proportional hazards model was used to compare the fuel economies of

traditional and hybrid cars. (ii) A single covariate z was used with z = 0 for a traditional car and z = 1 for a hybrid

car. (iii) The following are sample values of miles per gallon for the two types of car:

Traditional: 22 25 28 33 39 Hybrid: 27 31 35 42 45

(iv) The partial maximum likelihood estimate of the coefficient β is –1. Calculate the estimate of the baseline cumulative hazard function ( )0 32H using an analog of the Nelson-Åalen estimator which is appropriate for proportional hazard models. (A) Less than 0.7 (B) At least 0.7, but less than 0.9 (C) At least 0.9, but less than 1.1 (D) At least 1.1, but less than 1.3 (E) At least 1.3, but less than 1.5


29. You are given: (i) The number of claims made by an individual in any given year has a binomial

distribution with parameters m = 4 and q. (ii) The prior distribution of q has probability density function

( ) 6 (1 ), 0 1q q q qπ = − < < . (iii) Two claims are made in a given year. Determine the mode of the posterior distribution of q. (A) 0.17 (B) 0.33 (C) 0.50 (D) 0.67 (E) 0.83


30. A company has determined that the limited fluctuation full credibility standard is 2000 claims if: (i) The total number of claims is to be within 3% of the true value with probability p. (ii) The number of claims follows a Poisson distribution. The standard is changed so that the total cost of claims is to be within 5% of the true value with probability p, where claim severity has probability density function:

( ) 110,000

f x = , 0 000,10≤≤ x

Using limited fluctuation credibility, determine the expected number of claims necessary to obtain full credibility under the new standard. (A) 720 (B) 960 (C) 2160 (D) 2667 (E) 2880


31. For a mortality study with right censored data, you are given the following:

Time Number of Deaths Number at Risk 3 1 50 5 3 49 6 5 k 10 7 21

You are also told that the Nelson-Åalen estimate of the survival function at time 10 is 0.575. Determine k. (A) 28 (B) 31 (C) 36 (D) 44 (E) 46


32. A dental benefit is designed so that a deductible of 100 is applied to annual dental charges. The reimbursement to the insured is 80% of the remaining dental charges subject to an annual maximum reimbursement of 1000. You are given: (i) The annual dental charges for each insured are exponentially distributed with mean

1000.

(ii) Use the following uniform (0, 1) random numbers and the inversion method to generate four values of annual dental charges:

0.30 0.92 0.70 0.08

Calculate the average annual reimbursement for this simulation. (A) 522

(B) 696

(C) 757

(D) 947

(E) 1042


33. For a group of policies, you are given: (i) Losses follow the distribution function

( ) 1 / , F x x xθ θ= − < < ∞ .

(ii) A sample of 20 losses resulted in the following:

Interval Number of Losses 10x ≤ 9

10 25x< ≤ 6 25x > 5

Calculate the maximum likelihood estimate of θ. (A) 5.00 (B) 5.50 (C) 5.75 (D) 6.00 (E) 6.25


34. You are given: (i) Loss payments for a group health policy follow an exponential distribution with

unknown mean. (ii) A sample of losses is: 100 200 400 800 1400 3100 Use the delta method to approximate the variance of the maximum likelihood estimator of ( )1500 .S

(A) 0.019 (B) 0.025 (C) 0.032 (D) 0.039 (E) 0.045

Exam C: Fall 2006 - 35 - STOP

35. You are given: (i) A random sample of payments from a portfolio of policies resulted in the following:

Interval Number of Policies (0, 50] 36

(50, 150] x (150, 250] y (250, 500] 84 (500, 1000] 80 (1000, ∞ ) 0

Total n

(ii) Two values of the ogive constructed from the data in (i) are:

( )90 0.21,nF = and ( )210 0.51nF = Calculate x. (A) 120

(B) 145

(C) 170

(D) 195

(E) 220

**END OF EXAMINATION**

FALL 2006 EXAM C SOLUTIONS

Question #1 Key: E With n + 1 = 16, we need the 0.3(16) = 4.8 and 0.65(16) = 10.4 smallest observations. They are 0.2(280) + 0.8(350) = 336 and 0.6(450) + 0.4(490) = 466. The equations to solve are:

2 2

1/ 2 1/ 2

-1/2

1/ 2

0.3 1 and 0.65 1336 466

(0.7) 1 (336 / ) and (0.35) 1 (466 / ) (0.7) 1 (336 / )

(0.35) 1 (466 / )0.282814 (336 / 466)ln(0.282814) ln(336 / 466)

3.8614.

γ γ

γ γ γ γ

γ γ

γ

γ

γ

θ θθ θ

θ θ

θθ

γγ

− −

−

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

= + = +

−=

−

==

=

Question #2 Key: D Let E be the even of having 1 claim in the first four years. In four years, the total number of claims is Poisson(4λ).

1

2

4

Pr( | ) Pr( ) (0.05) 0.01839Pr( | ) 0.14427Pr( ) Pr( ) Pr( )

(2)(0.2) 0.05413Pr( | ) 0.42465Pr( ) Pr( )(4)(0.75) 0.05495Pr( | ) 0.43108Pr( ) Pr( )

: Pr( ) 0.01839 .05413

E Type I Type I eType I EE E E

eType II EE E

eType III EE E

Note E

−

−

−

= = = =

= = =

= = =

= + .05495 0.12747+ =

The Bayesian estimate of the number of claims in Year 5 is: 0.14427(0.25) + 0.42465(0.5) + 0.43108(1) = 0.67947.

Question #3 Key: B The sample mean is 0.2(400) + 0.7(800) + 0.1(1600) = 800. The sample variance is 2 2 20.2(400 800) 0.7(800 800) 0.1(1600 800) 96,000− + − + − = . The sample third central moment is

3 3 30.2(400 800) 0.7(800 800) 0.1(1600 800) 38,400,000− + − + − = . The skewness coefficient is 1.538,400,000 / 96,000 1.29= . Question #4 Key: C Because 0.656 < 0.7654 < 0.773, the simulated number of losses is 4. To simulate a loss by inversion, use

39.204,6481.075.236,7537.018.170,5152.012.113,2738.0

))1ln((200))1ln(()/()1ln(

1

1)(

44

33

22

11

2/1/1

)/(

)/(

========

−−=−−=

−=−

=−

=−=−

−

xuxuxuxu

uuxxu

eu

uexFx

x

τ

τ

θ

θ

θ

θ

τ

τ

With a deductible of 150, the first loss produces no payments and 113.12 toward the 500 limit. The second loss produces a payment of 20.18 and the insured is now out-of-pocket 263.12. The third loss produces a payment of 86.75 and the insured is out 413.12. The deductible on the fourth loss is then 86.88 for a payment of 204.29 – 86.88 = 117.51. The total paid by the insurer is 20.18 + 86.75 + 117.51 = 224.44.

Question #5 Key: A The density function is 2 /( ) xf x x e θθ − −= and the likelihood function is

2 /186 2 /91 2 / 66 / 60 7

3 0.148184

1

( ) (186 ) (91 ) (66 ) ( )

( ) ln ( ) 3ln( ) 0.148184( ) 3 0.148184 0

3/ 0.148184 20.25.

L e e e ee

l Ll

θ θ θ θ

θ

θ θ θ θ

θθ θ θ θ

θ θθ

− − − − − − −

−

−

=

∝= = −

′ = − == =

The mode is / 2 20.25 / 2 10.125θ = = . Question #6 Key: D We have ( ) 4 and 4E( ) 4(600) 2400μ θ θ μ θ= = = = . The average loss for Years 1 and 2 is 1650 and so 1800 (1650) (1 )(2400)Z Z= + − which gives Z = 0.8. Because there were two years,

0.8 2 /(2 )Z k= = + which gives k = 0.5. For three years, the revised value is 3 /(3 0.5) 6 / 7Z = + = and the revised credibility estimate (using the new sample mean of 2021), (6 / 7)(2021) (1/ 7)(2400) 2075.14+ = . Question #7 Key: B The uncensored observations are 4 and 8 (values beyond 11 are not needed). The two r values are 10 and 5 and the two s values are 2 and 1. The Kaplan-Meier estimate is

ˆ(11) (8 /10)(4 / 5) 0.64S = = and Greenwood’s estimate is 2 2 1(0.64) 0.0307210(8) 5(4)⎛ ⎞

+ =⎜ ⎟⎝ ⎠

.

Question #8 Key: C There are two ways to approach this problem. One is LaGrange’s formula:

(3 4)(3 5) (3 2)(3 5) (3 2)(3 4)(3) 25 20 30 18.33.(2 4)(2 5) (4 2)(4 5) (5 2)(5 4)

f − − − − − −= + + =

− − − − − −

Or, if the equation is 2( )f x a bx cx= + + then three equations must be satisfied: 25 2 420 4 1630 5 25

a b ca b ca b c

= + += + += + +

The solutions is a = 63.3333, b = -27.5, and c = 4.1667. The answer is 63.3333 27.5(3) 4.1667(9) 18.33− + = . Question #9 Key: E

| ~ ( , )mS Q bin m Q and ~ (1,99)Q beta . Then 1( ) [ ( | )] ( ) 0.01

1 99m mE S E E S Q E mQ m m= = = =+

. For the mean to be at least 50, m must be at

least 5,000. Question #10 Key: D The posterior distribution is

4 5090 7 2 2 3 17 150( | ) ( ) ( ) ( ) ( ) edata e e e e e

λλ λ λ λ λλπ λ λ λ λ λ

λ

−− − − − −∝ = which is a gamma distribution

with parameters 18 and 1/150. For one risk, the estimated value is the mean, 18/150. For 100 risks it is 100(18)/150 = 12. Alternatively, The prior distribution is gamma with α = 4 and β = 50. The posterior will be continue to be gamma, with α/ = α + no. of claims = 4 + 14 = 18 and β/ = β + no. of exposures = 50 + 100 = 150. Mean of the posterior = α / β = 18/150 = 0.12. Expected number of claims for the portfolio = 0.12 (100) = 12.

Question #11 Key: E

2 2

0.95 Pr(0.95 1.05 )~ ( , / 1.44 / )

0.95 1.050.95 Pr1.2 / 1.2 /

0.95 Pr( 0.05 /1.2 0.05 /1.2)

0.05 /1.2 1.962212.76.

XX N n n

Zn n

n Z n

nn

μ μ

μ σ μ

μ μ μ μμ μ

= < <

=

⎛ ⎞− −= < <⎜ ⎟

⎝ ⎠

= − < <

==

Question #12 Key: B

5000 5000

2 5000 5000 15000

1 1

2 2( ) (1 ) (1 ) 2 (1 )

0 1

( ) 5000ln(2) 5000ln( ) 15000ln(1 )( ) 5000 15000(1 ) 0

ˆ 0.25(0.25) 5000ln(2) 5000ln(0.25) 15000ln(0.75) 7780.97.

L q q q q q q

l q q ql q q qql

− −

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − − = −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦= + + −

′ = − − ==

= + + = −

Question #13 Key: C The estimate of the overall mean, μ, is the sample mean, per vehicle, which is 7/10 = 0.7. With the Poisson assumption, this is also the estimate of v = EPV. The means for the two insureds are 2/5 = 0.4 and 5/5 = 1.0. The estimate of a is the usual non-parametric estimate,

VHM = 2 25(0.4 0.7) 5(1.0 0.7) (2 1)(0.7)ˆ 0.04110 (25 25)

10

a − + − − −= =

− +

(The above formula: Loss Models page 596, Herzog page 116, Dean page 25) Then, k = 0.7/0.04 = 17.5 and so Z = 5/(5+17.5) = 2/9. The estimate for insured A is (2/9)(0.4) + (7/9)(0.7) = 0.6333.

Question #14 Key: A Item (i) indicates that X must one of the four given values. Item (ii) indicates that X cannot be 200 Item (iii) indicates that X cannot be 400. First assume X = 100. Then the values of r are 5, 3, 2, and 1 and the values of s are 2, 1, 1, and

1. Then 2 1 1 1ˆ (410) 2.235 3 2 1

H = + + + = and thus the answer is 100. As a check, if X = 300, the r

values are 5, 4, 3, and 1 and the s values are 1, 1, 2, and 1. Then, 1 1 2 1ˆ (410) 2.125 4 3 1

H = + + + = .

Question #15 Key: B The estimator of the Poisson parameter is the sample mean. Then,

ˆ( ) ( )ˆ( ) ( ) /

. . / / 1/

E E X

Var Var X n

c v n n

λ λ

λ λ

λ λ λ

= =

= =

= =

It is estimated by 1/ 1/ 39 0.1601.nλ = = Question #16 Key: E

( ) 11

1 15(0.125)

5(0.125) 5(0.5)

Pr( 5 | 8) Pr( 8)Pr 8 | 5Pr( 5 | 8) Pr( 8) Pr( 5 | 2) Pr( 2)

0.125 (0.8) 0.867035.0.125 (0.8) 0.5 (0.2)

XXX X

ee e

θ θθθ θ θ θ

−

− −

= = == = =

= = = + = = =

= =+

Then, 2 1 1( | 5) ( | 5) 0.867035(8) 0.132965(2) 7.202.E X X E Xθ= = = = + =

Question #17 Key: D

We have ( ) (1) (2)1 (1 )(1 )Tq q q′ ′= − − − and so ( ) ( ) ( )

(1)(2)

1 1 0.051 11 1 0.05 0.95

T T Tq q qqq

− − −′ = − = − =′− −

. Then,

(1) (1)20 0 20 200.05/ 0.95 0.05263, 0.132 / 0.95 0.1389q q′ ′= = = = , and

(1)40 0 0.9474(0.8611) 0.8158p′ = = . Out of 1000 at age 0, 816 are expected to survive to age 40.

Question #18 Key: D

2

2

5 5

1 1 1 4( 4 )( )

( 4)4

( ) 2 ln( 4 ) 5ln( 4)2 5( ) 04 4

2 50 (29)25 25

15.

ppL

l p

lp

lp

p

ωωω ω ω ωωωω

ωω ω ω

ωω ω

− −⎛ ⎞⎜ ⎟ − −⎝ ⎠= =

−−⎛ ⎞⎜ ⎟⎝ ⎠

= − − − −

′ = − =− − −

′= = −−

=

The denominator in the likelihood function is S(4) to the power of five to reflect the fact that it is known that each observation is greater than 4. Question #19 Key: B

VHM = 2 2

( ) ( )( ) 0.1 (1 1/ 2) 0.088623

( ) (0.1) (1 2 / 2) 0.088623 0.002146500 0.92371.

500 0.088623/ 0.002146

vv E

a Var

Z

μ λ λ λμ λ

λ

= == = = Γ + =

= = Γ + − =

= =+

The estimate for one insured for one month is 0.92371(35 / 500) 0.07629(0.088623) 0.07142.+ = For 300 insureds for 12 months it is (300)(12)(0.07142) = 257.11. Question #20 Key: D With no censoring the r values are 12, 9, 8, 7, 6, 4, and 3 and the s values are 3, 1, 1, 1, 2, 1, 1 (the two values at 7500 are not needed). Then,

13 1 1 1 2 1 1ˆ (7000) 1.5456.

12 9 8 7 6 4 3H = + + + + + + =

With censoring, there are only five uncensored values with r values of 9, 8, 7, 4, and 3 and all five s values are 1. Then,

21 1 1 1 1ˆ (7000) 0.9623.9 8 7 4 3

H = + + + + = The absolute difference is 0.5833.

Question #21 Key: A The simulated paid loss is 1exp[0.494 ( ) 13.294]u−Φ + where 1( )u−Φ is the inverse of the standard normal distribution function. The four simulated paid losses are 450,161, 330,041, 939,798, and 688,451 for an average of 602,113. The multiplier for unpaid losses is

0.851 0.747(2006 2005)0.801(2006 2005) 0.3795e− −− = and the answer is 0.3795(602,113) = 228,502 Question #22 Key: A The deduction to get the SBC is ( / 2) ln( ) ( / 2) ln(260) 2.78r n r r= = where r is the number of parameters. The SBC values are then -416.78, -417.56, -419.34, -420.12, and -425.68. The largest value is the first one, so model I is to be selected. Question #23 Key: E

( ) 11

1 11

1 3

Pr( | 1) Pr( 1)Pr 1|Pr( | 1) Pr( 1) Pr( | 3) Pr( 3)

(0.75) 0.2759! .3 0.2759 0.1245(3 )(0.75) (0.25)

! !

r r

X rX rX r X r

er

e er r

λ λλλ λ λ λ

−

− −

= = == = =

= = = + = = =

= =++

Then, 0.2759 0.1245(3 )2.98 (1) (3)

0.2759 0.1245(3 ) 0.2759 0.1245(3 )0.2759 0.3735(3 ) .0.2759 0.1245(3 )

r

r r

r

r

= ++ +

+=

+

Rearrange to obtain 0.82218 0.037103(3 ) 0.2759 0.03735(3 )0.54628 0.00025(3 )

7.

r r

r

r

+ = +

==

Because the risks are Poisson, (μ = EPV, a = VHM): ( ) 0.75(1) 0.25(3) 1.5

( ) 0.75(1) 0.25(9) 2.25 0.751 1/ 3

1 1.5 / 0.75

v Ea Var

Z

μ λλ

= = = + == = + − =

= =+

and the estimate is (1/3)(7) + (2/3)(1.5) = 3.33.

Question #24 Key: E The uniform kernel spreads the probability of 0.1 to 10 units each side of an observation. So the observation at 25 contributes a density of 0.005 from 15 to 35, contributing nothing to survival past age 40. The same applies to the point at 30. For the next 7 points: 35 contributes probability from 25 to 45 for 5(0.005) = 0.025 above age 40. 35 contributes probability from 25 to 45 for 5(0.005) = 0.025 above age 40. 37 contributes probability from 27 to 47 for 7(0.005) = 0.035 above age 40. 39 contributes probability from 29 to 49 for 9(0.005) = 0.045 above age 40. 45 contributes probability from 35 to 55 for 15(0.005) = 0.075 above age 40. 47 contributes probability from 37 to 57 for 17(0.005) = 0.085 above age 40. 49 contributes probability from 39 to 59 for 19(0.005) = 0.095 above age 40. The observation at 55 contributes all 0.1 of probability. The total is 0.485. Question #25 Key: D

(3) 2 (3/ 2)(4) (1/ 4)(8) 6(3) (3 / 2)(2)(2) (1/ 4)(3)(4) 3(4) (3) (4 3) (3) 6 1(3) 9.

fff f f

= + − =′ = − =

′= + − = + =

Question #26 Key: A

( )

1

2 2 2

~ )

~ ( , )

~ ( , / )

( / ) ( )( 1) ( 1) / .

n

ii

X Exp

X n

X n n

E X n n n n n

θ

θ

θ

θ θ

=

(

Γ

Γ

= + = +

∑

The second line follows because an exponential distribution is a gamma distribution with α = 1 and the sum of independent gamma random variables is gamma with the “α” parameters added. The third line follows because the gamma distribution is a scale distribution. Multiplying by 1/n retains the gamma distribution with the “θ” parameter multiplied by 1/n.

Question #27 Key: C The sample means are 3, 5, and 4 and the overall mean is 4. Then,

2 2 2

1 0 0 1 0 0 1 1 1 1 1 1 8ˆ3(4 1) 9

(3 4) (5 4) (4 4) 8 / 9 7ˆ 0.78.3 1 4 9

v

a

+ + + + + + + + + + += =

−

− + − + −= − = =

−

Question #28 Key: C The ordered values are: 22t, 25t, 27h, 28t, 31h, 33t, 35h, 39t, 42h, and 45h where t is a traditional car and h is a hybrid car. The s values are all 1 because there are no duplicate values. The c values are 1 for traditional cars and 1e− for hybrid cars. Then

0 1 1 1 1 1

1 1 1 1 1ˆ (32) 1.0358.5 5 4 5 3 5 3 4 2 4

He e e e e− − − − −= + + + + =

+ + + + +

Question #29 Key: C

2 2 3 3( | 2) 6 (1 ) 6 (1 ) (1 )q q q q q q qπ = − − ∝ − The mode can be determined by setting the derivative equal to zero.

2 3 3 2( | 2) 3 (1 ) 3 (1 ) 0(1 ) 0

0.5.

q q q q qq q

q

π ′ ∝ − − − =− − ==

Question #30 Key: B For the severity distribution the mean is 5,000 and the variance is 10,0002/12. For credibility based on accuracy with regard to the number of claims,

222000 , 1.8

0.03z z⎛ ⎞= =⎜ ⎟

⎝ ⎠

where z is the appropriate value from the standard normal distribution. For credibility based on accuracy with regard to the total cost of claims, the number of claims needed is

2 2

2 2

10000 /121 960.0.05 5000

z ⎛ ⎞+ =⎜ ⎟

⎝ ⎠

Question #31 Key: C

ˆ (10)ˆ(10) 0.5751 3 5 7ˆ (10) ln(0.575) 0.5534 .50 49 12

HS e

Hk

−= =

= − = = + + +

The solution is k = 36. Question #32 Key: A The annual dental charges are simulated from

/100011000ln(1 ).

xu ex u

−= −= − −

The four simulated values are 356.67, 2525.73, 1203.97, and 83.38. The reimbursements are 205.34 (80% of 256.67), 1000 (the maximum), 883.18 (80% of 1103.97), and 0. The total is 2088.52 and the average is 522.13. Question #33 Key: B

9 6 59 11( ) 1 (10 )

10 10 25 25( ) 9ln(10 ) 11ln( )

9 11( ) 010

11(10 ) 9110 20

110 / 20 5.5.

L

l

l

θ θ θ θθ θ θ

θ θ θ

θθ θ

θ θθ

θ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − ∝ −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= − +

′ = − + =−

− ==

= =

Question #34 Key: A The maximum likelihood estimate is ˆ 1000xθ = = . The quantity to be estimated is

( ) exp( 1500 / )S θ θ= − and 2( ) 1500 exp( 1500 / )S θ θ θ−′ = − . For the delta method, 2

2 2 2

ˆ ˆ ˆ[ ( )] [ ( )] ( )[1500(1000) exp( 1500 /1000)] (1000 / 6)0.01867.

Var S S Varθ θ θ−

′≅

= −=

This is based on 2ˆ( ) ( ) ( ) / /Var Var X Var X n nθ θ= = = . Question #35 Key: A Based on the information given

36 0.40.21

36 0.60.51

200 .

xn n

x yn n n

n x y

= +

= + +

= + +

Then, 0.21(200 ) 36 0.40.51(200 ) 36 0.6

x y xx y x y

+ + = ++ + = + +

and these linear equations can be solved for x = 119.37.

NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS

©S. Broverman, 2006 www.sambroverman.com

NOVEMBER 2006 SOA EXAM C/CAS 4 SOLUTIONS

1. The sample is of size 15. We assign smoothed percentiles to the sample points:"*& Þ Þ Þ #)! $&! Þ Þ Þ %&! %*! Þ Þ Þ " % & "! """' "' "' "' "' Þ Þ Þ Þ Þ Þ Þ Þ ÞÞ!'#& Þ Þ Þ Þ#&! Þ$! Þ$"#& Þ Þ Þ Þ'#& Þ'&! Þ')(& Þ Þ Þ ‚ "' % %Þ) & "! "!Þ% "" Since 4.8 is 80% of the way from 4 to 5, the smoothed empirical estimate of the 30th percentile is80% of the way from 280 to 350, which is 336. Since 10.5 is 40% of the way from 10 to 11, thesmoothed empirical estimate of the 65th percentile is 40% of the way from 450 to 490, which is466.

The cdf of the Burr distribution with is .α œ # JÐBÑ œ " Ò Ó""ÐBÎ Ñ) #

#

Applying the percentile matching method, we use the estimated percentiles in the cdf to getJÐ$$'Ñ œ " œ Þ$! JÐ%''Ñ œ " œ Þ'&Ò Ó Ò Ó" "

"Ð$$'Î Ñ "Ð%''Î Ñ) )# #

# # and .

After a little algebraic juggling, these equations becomeÐ$$'Î Ñ œ " œ Þ"*&##* Ð%''Î Ñ œ " œ Þ'*!$!*) )# #" "

Þ( Þ$&È È and .

Dividing the second equation by the first results in , and( )%''$$'

#œ $Þ&$&*

# œ œ $Þ)'68 $Þ&$&*

68 Ð Ñ%''$$'

. Answer: E

2. Denote by the number of claims in one year, and is the annual claim frequency. The priorR Adistribution of is and .A A A ATÐ œ !Þ#&Ñ œ Þ!& ß T Ð œ !Þ&!Ñ œ Þ#! ß T Ð œ "!!ÞÑ œ Þ(&The Bayesian estimate is

IÒR l R œ "Ó œ IÒR l œ !Þ#&Ó † T Ð œ !Þ#& l R œ "Ñ& 3 & 33œ" 3œ"

% %

A A

IÒR l œ !Þ&!Ó † T Ð œ !Þ&! l R œ "Ñ IÒR l œ "Þ!!Ó † T Ð œ "Þ!! l R œ "Ñ& 3 & 33œ" 3œ"

% %

A A A A

œ ÐÞ#&ÑT Ð œ Þ#& l R œ "Ñ ÐÞ#ÑT Ð œ Þ& l R œ "Ñ ÐÞ(&ÑT Ð œ " l R œ "ÑA A A3œ" 3œ" 3œ"

% % %

3 3 3 .

We find the posterior probabilities as follows, and note that if is Poisson with mean , thenR -

3œ"

%

3R %is Poisson with mean .-

Given: TÐ œ !Þ#&Ñ œ Þ!& T Ð œ !Þ&!Ñ œ Þ#! T Ð œ "!!ÞÑ œ Þ(&A A A

Given: TÐ R œ " l œ Þ#&Ñ T Ð R œ " l œ Þ&Ñ T Ð R œ " l œ "Ñ3œ" 3œ" 3œ"

% % %

3 3 3A A A

œ / œ #/ œ %/" # %

Ì Ì Ì

TÐ R œ " ∩ œ Þ#&Ñ T Ð R œ " ∩ œ Þ&Ñ3œ" 3œ"

% %

3 3A A

œ Þ!&/ œ Þ%/ œ $/" # %

Ì



TÐ R œ "Ñ œ TÐ R œ " ∩ œ Þ#&Ñ TÐ R œ " ∩ œ Þ&Ñ3œ" 3œ" 3œ"

% % %

3 3 3A A

TÐ R œ " ∩ œ "Ñ œ Þ!&/ Þ%/ $/3œ"

%

3" # %A .

Ì

TÐ œ Þ#& l R œ " Ñ œ œ Þ"%%#*& ßA3œ"

%

3Þ!&/

Þ!&/ Þ%/ $/

"

" # %

T Ð œ Þ& l R œ " Ñ œ œ Þ%#%''& ßA3œ"

%

3Þ%/

Þ!&/ Þ%/ $/

#

" # %

T Ð œ " l R œ " Ñ œ œ Þ%$"!%"A3œ"

%

3$/

Þ!&/ Þ%/ $/

%

" # % .

The Bayesian estimate is .ÐÞ#&ÑÐÞ"%%#*&Ñ ÐÞ&ÑÐÞ%#%''&Ñ Ð"ÑÐÞ%$"!%"Ñ œ Þ'(*Answer: D

3. The skewness coefficient is .IÒÐ\ Ñ ÓÐZ +<Ò\ÓÑ

. $

$Î#

The empirical estimate of is .. \ œ œ )!! #Ð%!!Ñ(Ð)!!Ñ"'!!

"!The empirical estimate of isZ +<Ò\Ó" ""! "!Ò Ð\ \Ñ Ó œ Ò#Ð%!! )!!Ñ (Ð)!! )!!Ñ Ð"'!! )!!Ñ Ó œ *'ß !!!

D 3

# # # # .The empirical estimate of isIÒÐ\ Ñ Ó. $

" ""! "!Ò Ð\ \Ñ Ó œ Ò Ð\ )!!Ñ Ó

D D3 3

$ $

œ Ò#Ð%!! )!!Ñ (Ð)!! )!!Ñ Ð"'!! )!!Ñ Ó œ $)ß %!!""!

$ $ $ .

The estimated skewness coefficient is . Answer: B$)ß%!!ß!!!Ð*'ß!!!Ñ$Î#

œ "Þ#*"

4. To simulate the number of claims from uniform number , we find so that? 8J Ð8 "Ñ Ÿ ? J Ð8Ñ R ? œ ('&%R R . From the given cdf for , and , we haveJ Ð$Ñ œ Þ'&' Ÿ Þ('&% Þ(($ œ J Ð%Ñ R œ %R R . The simulated number of claims is .The cdf of the Weibull distribution isJ ÐBÑ œ " / œ " / œ " /\

ÐBÎ Ñ ÐBÎ#!!Ñ B Î%!ß!!!) 7 # # .Given uniform number , the simulated value of is the solution of the equationÐ!ß "Ñ ? B

? œ " / B œ Ò %!ß !!! 68Ð" ?ÑÓB Î%!ß!!! "Î## , or equivalently, .The simulated values of the loss amounts are:from , the simulated is , claim amount is 0 after deductible of 150 ;? œ Þ#($) B ""$Þ"$from , the simulated is , claim amount is 20.18 after deductible of 150;? œ Þ&"&# B "(!Þ")from , the simulated is , claim amount is 86.75 after deductible of 150;? œ Þ(&$( B #$'Þ(&from , the simulated is , claim amount is 117.52 after deductible of 86.87.? œ Þ'%)" B #!%Þ$*The deductible is 50 on the 4th claim to bring the total policyholder out of pocket to themaximum of 500 (113.13 out-of-pocket for first claim, 150 out-of-pocket for each of the next 2claims and 86.87 out-of-pocket for the 4th claim to bring the total out-of-pocket to 500).Insurer's aggregate payment is Answer: C#!Þ") )'Þ(& ""(Þ&# œ ##%Þ%& Þ



5. From the table of distributions, for the inverse exponential distribution with parameter , the)

mode is . The pdf of the inverse exponential is , and the log of the pdf is) )# B

/0ÐBÑ œ ÎB

#

)

68 0ÐBÑ œ 68 # 68 B ) ))B B" " , and the derivative of the log of the pdf is .

The log of the cdf is and the derivative of the log of the cdf is68 J ÐBÑ œ )B

. ". B) 68 J ÐBÑ œ .The likelihood function is , the loglikelihood isP œ 0Ð")'Ñ † 0Ð*"Ñ † 0Ð''Ñ † ÒJ Ð'!ÑÓ(

68P œ 68 0Ð")'Ñ 68 0Ð*"Ñ 68 0Ð''Ñ ( 68JÐ'!Ñ .The derivative of the loglikelihood is. " " " " " " ". ")' *" '' '!) ) ) )68P œ Ð Ñ Ð Ñ Ð Ñ (Ð Ñ .Setting and solving for results in the mle ..

.) 68P œ ! œ #!Þ#&s) )

The mle of the mode is . Answer: A#!Þ#&# œ "!Þ"

6. The annual loss will be . We are told that the conditional distribution of given is gamma\ \ )with and . The hypothetical mean is ,α ) ) α) )œ % IÒ\l Ó œ œ %and the process variance is .Z +<Ò\l Ó œ œ %) α) )# #

The expected hypothetical mean is . ) ) )œ IÒIÒ\l Ó Ó œ IÒ% Ó œ %IÒ Ó œ %Ð'!!Ñ œ #%!!(since we are told that the prior distribution of has mean 600).)

The Buhlmann credibility premium based on Years 1, 2 and 3 is ,^\ Ð" ^Ñ

.

where , and , and .\ œ œ #!#" œ #%!! ^ œ "%!!"*!!#('$ $

$ $. @+

@ œ œ IÒ% Ó œ %IÒ Óexpected process variance , and) )# #

+ œ œ Z +<Ò% Ó œ "'Z +<Ò Óvariance of hypothetical mean .) )

From the Buhlmann credibility premium based on Years 1 and 2, we have")!! œ ^\ Ð" ^Ñ \ œ œ "'&! œ #%!!

. . , where , (as above),"%!!"*!!

#

and . Therefore, .^ œ ")!! œ "'&!^ #%!!Ð" ^Ñ œ #%!! (&!^##@

+

Solving for results in , so that .^ ^ œ Þ) œ œ Þ&# @# +@

+

Then, going back to the premium based on Years 1, 2 and 3, we have ,^ œ œ œ$ $ '$ $Þ& (@

+

and the Buhlmann credibility premium is .Ð ÑÐ#!#"Ñ Ð" ÑÐ#%!!Ñ œ #!(&' '( (

Answer: D

7. The loss 11 that is nearest to 11 is 8, so . We assume that there is noŸ WÐ""Ñ œ WÐ)Ñs s

truncation since there is no indication that there is any truncation. The first payment amount isC œ % < œ "! = œ #" " ", and there are at risk and losses of that amount. The second paymentamount is , and there are at risk (only count losses above the last censoring point),C œ ) < œ &# #

and there is loss of that amount. The product-limit estimate is= œ "#

WÐ)Ñ œ Ð" ÑÐ" Ñ œ WÐ)Ñs s# " "'"! & #& . The Greenwood approximation to the variance of is

ÒWÐ)ÑÓ † Ò Ó œ Ð Ñ † Ò Ó œ Þ!$!(#s # #= =< Ð< = Ñ < Ð< = Ñ #& "!Ð"!#Ñ &Ð&"Ñ

"' # "" #

" " " # # # .

Answer: B



8. The interpolating polynomial can be found by the Lagrange method.With points , the 2nd degree interpolating polynomial isÐB ß C Ñ ß ÐB ß C Ñ ß ÐB ß C Ñ! ! " " # #

T ÐBÑ œ C † C † C †! " #ÐBB ÑÐBB Ñ ÐBB ÑÐBB Ñ ÐBB ÑÐBB ÑÐB B ÑÐB B Ñ ÐB B ÑÐB B Ñ ÐB B ÑÐB B Ñ

" # ! " ! "

! " ! # " ! " # # ! # " .

This is Ð#&Ñ † Ð#!Ñ † Ð$!Ñ † ÞÐB%ÑÐB&Ñ ÐB#ÑÐB&Ñ ÐB#ÑÐB%ÑÐ#%ÑÐ#&Ñ Ð%#ÑÐ%&Ñ Ð&#ÑÐ&%Ñ

Then .TÐ$Ñ œ Ð#&Ñ † Ð#!Ñ † Ð$!Ñ † œÐ$%ÑÐ$&Ñ Ð$#ÑÐ$&Ñ Ð$#ÑÐ$%ÑÐ#%ÑÐ#&Ñ Ð%#ÑÐ%&Ñ Ð&#ÑÐ&%Ñ

&&$

Answer: C

9. The Bernoulli distribution is a 0,1 distribution, with and .TÐ\ œ "Ñ œ ; TÐ\ œ !Ñ œ " ;W 7 ;7 has a binomial distribution with parameters and .Using the double expectation rule, we have .IÒW Ó œ IÒIÒW lUÓ Ó7 7

We know that , since has a binomial distribution with parameters andIÒW lU œ ;Ó œ 7; W 77 7

; IÒW Ó œ IÒ7UÓ œ 7IÒUÓ U. Then, . The pdf of is7>> >

Ð"**ÑÐ"Ñ Ð**Ñ !x *)x

**x† ; Ð" ;Ñ œ † Ð" ;Ñ œ **Ð" ;Ñ"" **" *) *) .The expected value of is;' '! !

" "*) *); † **Ð" ;Ñ .; œ ** Ò" Ð" ;ÑÓÐ" ;Ñ .;

œ **Ò Ð" ;Ñ .; Ð" ;Ñ .;Ó œ **Ò Ó œ' '! !

" "*) ** " " "** "!! "!! .

Then, .IÒW Ó œ 7IÒUÓ œ77"!!

In order for this to be at least 50, we must have . Answer: E7 &!!!

10. The prior distribution of is gamma with and .- α )œ % œ Þ!#The model distribution is Poisson with a mean of .-There are a total of observations ( ) in Year 1, and the total number of8 œ "!! B ß B ß ÞÞÞß B" # "!!

claims is .DB œ (Ð"Ñ #Ð#Ñ "Ð$Ñ œ "%3

The gamma-prior/Poisson-model combination results in a posterior distribution which is alsogamma, with updated parameters, andα α Dw

3œ B œ % "% œ ")

)w œ œ œ))8 "

Þ!# Þ!#"!!ÐÞ!#Ñ" $ .

The Bayesian premium for one risk in Year 2 isIÒ\ lB ß B ß ÞÞÞß B Ó œ IÒ\ l Ó † Ð lB ß B ß ÞÞÞß B Ñ ."!" " # "!! "!" " # "!!!

∞' - 1 - -

œ † Ð lB ß B ß ÞÞÞß B Ñ . œ'!∞

" # "!!- 1 - - mean of posterior distributionœ œ Ð")ÑÐ Ñ œ Þ"#α )w w Þ!#

$ .The expected number of claims for the portfolio of 100 risks is ."!! ‚ Þ"# œ "#Alternatively, once we know and from the prior gamma distribution of , and and ,α ) - D8 B3

for the gamma/Poisson combination, the predictive distribution of is negative\ lB ß ÞÞÞß B8" " 8

binomial with and ,< œ B œα D "3))8 "

so that , as above. Answer: DIÒ\ lB ß ÞÞÞß B Ó œ < œ Ð B ÑÐ Ñ8" " 8 3" α D))8 "



11. We want .T Òl\ IÐ\Ñl Þ!&IÐ\ÑÓ Þ*&

According to the central limit theorem, has a distribution which is approximately normal.\

Also, and , where is the number of items in the sample.IÐ\Ñ œ IÐ\Ñ Z +<Ð\Ñ œ 8 Z +<Ð\Ñ

8

T Òl\ IÐ\Ñl Þ!&IÐ\ÑÓ can be "standardized" to be written in the form

T T Òl^l Ó ^Òl l Ó œ\IÐ\Ñ Þ!&IÐ\Ñ Þ!&IÐ\Ñ

Z +<Ð\Ñ Z +<Ð\Ñ

Z +<Ð\ÑÎ8È È È , where is standard normal.

In order for this probability to be at least .95, we must have .Þ!&IÐ\Ñ

Z +<Ð\ÑÎ8È "Þ*'

We are given that , so that the inequality becomes ,ÈZ +<Ð\Ñ œ "Þ#IÐ\Ñ "Þ*'Þ!& 8

"Þ#

Èwhich becomes . Answer: E8 ##"$

12. When is given, the maximum likelihood estimate of is found from .7 ; 7; œ \

There are 10,000 policies, so , where each has a binomial distribution\ œ \ \ â\

"!ß!!!" "!ß!!!

with . From the given data set, , so .7 œ # \ œ &!!! \ œ Þ&

D 3

Since , we get the mle for to be .7 œ # ; ; œ œ Þ#&s\

#The likelihood function for a discrete random variable is the product of the probability functionvalues at all the sample points. For the binomial with and , we7 œ # ; œ Þ#&have , 0Ð!Ñ œ TÐ\ œ !Ñ œ ÐÞ(&Ñ œ Þ&'#& 0Ð"Ñ œ TÐ\ œ "Ñ œ #ÐÞ#&ÑÐÞ(&Ñ œ Þ$(&#

and .0Ð#Ñ œ TÐ\ œ #Ñ œ ÐÞ#&Ñ œ Þ!'#&#

The likelihood function for the 10,000 data points is , since there are 5000ÐÞ&'#&Ñ ÐÞ$(&Ñ&!!! &!!!

'0's and 5000 '1's . The loglikelihood is&!!! 68ÐÞ&'#&Ñ &!!! 68ÐÞ$(&Ñ œ (ß ()" . Answer: B

13. There are 5 Type A vehicles observed with .7 œ ] ] ] ] ] œ #" # $ % &

The estimate for the number of claims for a Type A vehicle in Year 4 is ,^] Ð" ^Ñ

.

where , , and .] œ œ Þ% ^ œ œ IÐ\Ñ # &

& &@+

.

. is estimate as based on the entire data set of Type A and Type B, which is\

""#$##"$# œ Þ( .

From the data for Type A, we have hypothetical mean and for Type B weIÒ\lEÓ œ œ Þ%#&

have hypothetical Since is Poisson for each Type, we also haveIÒ\lFÓ œ œ "Þ \&&

Z +<Ò\lEÓ œ Þ% Z +<Ò\lFÓ œ " @ œ Þ( and , so that the expected process variance is , the sameas ..

We do not have individual data values for all of the vehicles, but we do have based on\ œ Þ%

E

7 œ & \ œ " 7 œ &

E F F observations and based on observations. We use the nonparametric

estimate of , . In this case insureds.+ + œ † Ò 7 Ð\ \Ñ @Ð< "Ñ Ó < œ #s s "

7 7"7

3œ"

<

3# 3œ"

<

3 3#

+ œ œ † Ò &ÐÞ% Þ(Ñ &Ð" Þ(Ñ Þ(Ð# "Ñ Ó œ Þ!%s"

"! Ð& & Ñ""!

# ## # .

Then, The semiparametric empirical Bayes estimate of the claim frequency^ œ œ Þ####&& Þ(

Þ!%

for a vehicle of Type A in Year 4 is . Answer: CÐÞ####ÑÐÞ%Ñ ÐÞ((()ÑÐÞ(Ñ œ Þ'$$



14. means that there are 4 distinct sample values but there are data points, so must be5 œ % & \"!!ß #!!ß $!! %!!or .LÐ%"!Ñ œ LÐ%!!Ñs s since there are no deaths after time 400.Since , there is only 1 at risk and 1 death at time 400, so .< œ " = œ "% %

Exactly one of and is 1 and the other is 2. By trial and error:= =" $

- if and , then and we have= œ " = œ # \ œ $!!" $

LÐ%!!Ñ œ œ #Þ"# #Þ"&s " " # "& % $ " ,so this is not correct.

Therefore, , and , s , and\ œ "!! = œ # œ "" $

LÐ%!!Ñ œ " œ #Þ#$s # " "& $ # . Answer: A

15. The mle of the Poisson parameter is . The mean of the estimator is\

IÐ\Ñ œ IÐ\Ñ œ Z +<Ð\Ñ œ œ

- , and the variance of the estimator is .Z +<Ð\Ñ8 "!

-

The coefficient of variation of the estimator is .È È ÈZ +<Ð\Ñ

IÐ\Ñ

Î"! "

"!œ œ

-

- -

The mle of is , so the estimated coefficient of variation is .- \ œ $Þ* œ Þ"' "

"!Ð$Þ*ÑÈAnswer: B

16. IÒ\ l\ œ &Ó œ IÒ\ l œ )Ó † T Ð œ )l\ œ &Ñ IÒ\ l œ #Ó † T Ð œ #l\ œ &Ñ# " # " # ") ) ) )œ Ð)Ñ † T Ð œ )l\ œ &Ñ Ð#Ñ † T Ð œ #l\ œ &Ñ) )" " .

We find the posterior probabilities and .TÐ œ )l\ œ &Ñ T Ð œ #l\ œ &Ñ) )" "

The pdf of the exponential distribution with mean is .) ") /

BÎ)

TÐ œ )Ñ œ Þ) T Ð œ #Ñ œ Þ#) )

TÐ\ œ &l œ )Ñ œ / TÐ\ œ &l œ #Ñ œ /" "&Î) &Î#) )" "

) # Ì Ì

TÐ\ œ & ∩ œ )Ñ œ Ð / ÑÐÞ)Ñ T Ð\ œ & ∩ œ #Ñ œ Ð / ÑÐÞ#Ñ" "&Î) &Î#) )" "

) # œ Þ"/ œ Þ"/Þ'#& #Þ&

Ì TÐ\ œ &Ñ œ TÐ\ œ & ∩ œ )Ñ TÐ\ œ & ∩ œ #Ñ" " ") ) œ Þ"/ Þ"/Þ'#& #Þ&

Ì

TÐ œ )l\ œ &Ñ œ œ œ Þ)'() "T Ð œ)∩\ œ&Ñ

T Ð\ œ&Ñ Þ"/ Þ"/Þ"/) "

"

Þ'#&

Þ'#& #Þ& ,

and .TÐ œ #l\ œ &Ñ œ œ œ Þ"$$) "T Ð œ#∩\ œ&Ñ

T Ð\ œ&Ñ Þ"/ Þ"/Þ"/) "

"

#Þ&

Þ'#& #Þ&

IÒ\ l\ œ &Ó œ Ð)ÑÐÞ)'(Ñ Ð#ÑÐÞ"$$Ñ œ (Þ## " .

Note that we have used the notation , but since is exponential, it is actually aTÐ\ œ #Ñ \"

density and not a probability. Answer: E



17. is the probability that a person at age departs due to some decrement by time .; - -4ÐX Ñ

4 4"

For a two decrement model .; œ " Ð" ; ÑÐ" ; Ñ4 4 4ÐX Ñ wÐ"Ñ wÐ#Ñ

Using these equations, we getÞ" œ ; œ " Ð" ; ÑÐ" ; Ñ œ " Ð" ; ÑÐ" Þ!&Ñ p " ; œ Þ*%($')ß! ! ! ! !

ÐX Ñ wÐ"Ñ wÐ#Ñ wÐ"Ñ wÐ"Ñ

Þ")# œ ; œ " Ð" ; ÑÐ" Þ!&Ñ p " ; œ Þ)'"!&$" " "ÐX Ñ wÐ"Ñ wÐ"Ñ .

Since Group A is affected only by decrement 1, the survival probability to age 40 for Group A isÐ" ; ÑÐ" ; Ñ œ Þ)"&(! "

wÐ"Ñ wÐ"Ñ , and the expected number of survivors to age 40 from 1000 GroupA individuals observed at 0 is . Answer: D"!!!ÐÞ)"&(Ñ œ )"'

18. To formulate the likelihood function we must first note that since observation of the bulbsbegins at 4 hours, all data is conditional given that . There are 3 known burnout times, soX %these would be included as conditional density in the likelihood function, and the 2 bulbs thatsurvive more hours would be included as conditional survival probabilities. The likelihood:function is .P œ 0Ð&lX %Ñ † 0Ð*lX %Ñ † 0Ð"$lX %Ñ † ÒT ÐX % :lX %ÑÓ#

For the uniform distribution on , , ,Ð!ß Ñ 0Ð>Ñ œ TÐX %Ñ œ= " %= =

=

so , and .0Ð>lX %Ñ œ œ TÐX % :lX %Ñ œ"Î

Ð %ÑÎ % %" %:=

= = = ==

The likelihood function is P œ Ð Ñ Ð Ñ œ †"% % Ð %Ñ

%: Ð %:Ñ= = =

= =$ ##

&

Then, , and .68 P œ #68Ð % :Ñ &68Ð %Ñ 68P œ = = . # &. %: %= = =

The mle of occurs where , so substituting results in= =..= 68 P œ ! œ #*

# &#*%: #*% œ ! : œ "& , from which we get . Answer: D

19. This Buhlmann-Straub situation can be considered an ordinary Buhlmann model with8 œ &!! ^\ Ð" ^Ñ

observations. We will find , which will be the Buhlmann credibility.estimate of the expected number of claims for one insured for one month. The expected numberof claims for 300 insureds for 12 months will be times as large.$!! ‚ "#

\ œ œ Þ!( "!"""%

&!! for the given data.\ is the number of claims for one insured for one month.The hypothetical mean is and the process variance is .IÒ\l Ó œ Z +<Ò\l Ó œ- - - -The first and second moments of the Weibull distribution areIÒ Ó œ Ð" Ñ œ ÐÞ"Ñ Ð"Þ&Ñ œ Þ!))'#$- ) > >"

7

and .IÒ Ó œ Ð" Ñ œ ÐÞ"Ñ Ð#Ñ œ Þ!"- ) > ># # ##7

Then, expected hypothetical mean. - -œ œ IÒIÒ\l Ó Ó œ IÒ Ó œ Þ!))'#$ ß@ œ œ IÒ Z +<Ò\l Ó Ó œ IÒ Ó œ Þ!))'#$expected process variance , and- -+ œ œ Z +<ÒIÒ\l Ó Óvariance of hypothetical mean -œ Z +<Ò Ó œ IÒ Ó ÐIÒ ÓÑ œ Þ!" ÐÞ!))'#$Ñ œ Þ!!#"%&*'%- - -# # # .

Then, ,^ œ œ œ Þ*#$(!'&!! &!!&!! &!!

@+

Þ!))'#$Þ!!#"%&'*%

and the credibility estimate for the number of claims for one month for the insured isÐÞ*#$(!'ÑÐÞ!(Ñ ÐÞ!('#*%ÑÐÞ!))'#$Ñ œ Þ!("%#" .The estimate for the expected number of claims for the next year for 300 insureds is$!! ‚ "# ‚ ÐÞ!("%#"Ñ œ #&( . Answer: B



20. .L Ð(!!!Ñ œ œ "Þ&%&'$&s"$ " " " # " ""# * ) ( ' % $

L Ð(!!!Ñ œ œ Þ*&#$!#s#" " " " "* ) ( % $ .

Note that since the three observations at 2500 and the two at 6000 become censored, we lose $"#

and from , so we lose a total of .# $ #' "# ' L Ð(!!!Ñ œ Þ&)$s"

The censoring at 7500 is irrelevant since the estimate at 7000 is based on deaths up to 7000 butnot at 7500. Answer: D

21. The cdf of the lognormal distribution is JÐ>Ñ œ ÞFÐ Ñ68 >"$Þ#*%!Þ%*%

The four simulated losses are:? œ Þ#)(( œ Ð Þ&'Ñ œ p Þ&' œ p > œ %&!ß "'"F FÐ Ñ68 >"$Þ#*% 68 >"$Þ#*%

!Þ%*% !Þ%*% ,? œ Þ"#"! œ Ð "Þ"(Ñ œ p "Þ"( œ p > œ $$$ß !%"F FÐ Ñ68 >"$Þ#*% 68 >"$Þ#*%

!Þ%*% !Þ%*% ,? œ !Þ)#$) œ ÐÞ*$Ñ œ p Þ*$ œ p > œ *$*ß (*)F FÐ Ñ68 >"$Þ#*% 68 >"$Þ#*%

!Þ%*% !Þ%*% ,? œ !Þ'"(* œ ÐÞ$Ñ œ p Þ$ œ p > œ '))ß %&"F FÐ Ñ68 >"$Þ#*% 68 >"$Þ#*%

!Þ%*% !Þ%*% .For losses based on contract year 2005, the ratio is .C Þ)!"/ œ Þ$(*&Þ(%(

The estimate of the average unpaid losses for 2005 isÐÞ$(*&ÑÐ Ñ œ ##)ß ()'

%&!ß"'"$$$ß!%"*$*ß(*)'))ß%&"% . Answer: A

22. According to the Schwarz Bayesian criterion, we compare for each estimatedj 68 8<#

model, where is the maximized loglikelihood, is the number of parameters estimated and isj < 8the number of data points. The model favored is the one with the largest value of .j 68 8<

#We get the following valuesModel I: , Model II: , %"% 68 #'! œ %"'Þ() %"# 68 #'! œ %"(Þ&'" #

# #

Model III: , Model IV: , %"" 68 #'! œ %"*Þ$% %!* 68 #'! œ %#!Þ"#$ %# #

Model V: . %!* 68 #'! œ %#&Þ')'#

Model I has the largest value. Answer: A

23. Year 1 gives us one observed value of (annual number of claims) of .\ <The Bayesian estimate for the second year isIÒ\ l\ œ <Ó œ IÒ\ l œ "Ó † T Ð œ "l\ œ <Ñ IÒ\ l œ $Ó † T Ð œ $l\ œ <Ñ# " # " # "- - - -œ Ð"Ñ † T Ð œ "l\ œ <Ñ Ð$Ñ † T Ð œ $l\ œ <Ñ œ #Þ*)- -" " .

If we denote , then , and thenTÐ œ "l\ œ <Ñ œ - TÐ œ $l\ œ <Ñ œ " -- -" "

- $Ð" -Ñ œ #Þ*) - œ TÐ œ "l\ œ <Ñ œ Þ!" gives us .- "

We can formulate asTÐ\ œ <Ñ"

T Ð\ œ <Ñ œ TÐ\ œ <l œ "Ñ † T Ð œ "Ñ TÐ\ œ <l œ $Ñ † T Ð œ $Ñ" " "- - - -

œ † ÐÞ(&Ñ † ÐÞ#&ÑÐ Ñ Ð Ñ/ / †$<x <x

" $ <

.ThenTÐ œ "l\ œ <Ñ œ œ- "

T Ð œ"∩\ œ<Ñ T Ð\ œ<l œ"Ñ†T Ð œ"ÑT Ð\ œ<Ñ T Ð\ œ<Ñ- - -" "

" "

œ œ œ Þ!"Ð/ Î<xÑÐÞ(&Ñ

Ð Ñ†ÐÞ(&ÑÐ Ñ†ÐÞ#&Ñ

"

/ / $" $† <

<x <x

$$/ †$# < .

It follows that , and .$ œ #"*%Þ&& < œ (<



The Buhlmann credibility estimate is .^\ Ð" ^Ñ

.

\ œ < œ ( 8 œ " ^ œ and observed value, ."

"@+

Since is Poisson with mean , we have that the hypothetical mean is and the\l IÒ\l Ó œ- - - -process variance is . Then,Z +<Ò\l Ó œ- -. -œ œ IÒ Ó œ Ð"ÑÐÞ(&Ñ Ð$ÑÐÞ#&Ñ œ "Þ&expected hypothetical mean ,@ œ œ IÒ Ó œ Ð"ÑÐÞ(&Ñ Ð$ÑÐÞ#&Ñ œ "Þ&expected process variance , and-+ œ œ Ð$ "Ñ ÐÞ(&ÑÐÞ#&Ñ œ Þ(&variance of hypothetical mean .#

Then, , and the Buhlmann credibility estimate is .^ œ œ Ð ÑÐ(Ñ Ð ÑÐ"Þ&Ñ œ $Þ$$" " " #" $ $ $"Þ&

Þ(&

Answer: E

24. There are 9 distinct observed values, The empirical probabilities are for all -:ÐCÑ œ Þ" Cvalues except . We wish to estimate .:Ð$&Ñ œ Þ# T ÐX %!Ñ œ " JÐ%!ÑWith bandwidth , the interval for the uniform kernel at observed value is from to, œ "! C C "!

C "! JÐ%!Ñ J Ð%!Ñ œ :ÐC Ñ † O Ð%!Ñs . The kernel density estimate of is .3œ"

*

3 C3

For any that is , the interval around that is totally to the left of , soC Ÿ $! C B œ %!3 3

J Ð%!Ñ œ " C C œ #& $! C &!C 3 33 for that ; this applies to and . For any , the interval around thatC B œ %! J Ð%!Ñ œ ! C C œ &&3 C 3 is totally to the right of , so for that ; this applies to .3

For any for which , C C "! Ÿ %! Ÿ C "! J Ð%!Ñ œ Ò%! ÐC "!ÑÓÐÞ!&Ñ3 3 3 C 33

(this is the are of the rectangle whose base is from to 40 with height C "! œ œ Þ!&ÑÞ3" "#, #!

The kernel density estimate of isJÐ%!Ñ

J Ð%!Ñ œ ÐÞ"ÑÐ"Ñ ÐÞ"ÑÐ"Ñ ÐÞ#ÑÒ%! Ð$& "!ÑÓÐÞ!&Ñs

ÐÞ"ÑÒ%! Ð$( "!Ñ %! Ð$* "!Ñ %! Ð%& "!Ñ . %! Ð%( "!Ñ %! Ð%* "!ÑÓÐÞ!&Ñ ! œ Þ&"&The estimate of is . Answer: EWÐ%!Ñ " Þ&"& œ Þ%)&

25. The method of extrapolation for a natural cubic spline uses a straight line extrapolation fromthe right endpoint of the data set. The line has the same slope as the spline at the rightendpoint. for , so .0 ÐBÑ œ $ÐB "Ñ Ð ÑÐB "Ñ " Ÿ B Ÿ $ 0 Ð$Ñ œ $w # w$

%

0Ð$Ñ œ ' B œ $ , so the equation of the extrapolation line starting at is' $ÐB $Ñ œ $B $ 0Ð%Ñ $Ð%Ñ $ œ * . The extrapolated value of is . Answer: D

26. has an exponential distribution with mean , so .\ Z +<Ò\Ó œ) )#

For a sample of size and any distribution, it is always true that ,8 Z +<Ò\Ó œ Z +<Ò\Ó

8

and . Therefore, in this case, and .IÒ\Ó œ IÒ\Ó Z +<Ò\Ó œ IÒ\Ó œ )#

8 )

From the definition of variance, we know thatZ +<Ò\Ó œ IÒ\ Ó ÐIÒ\ÓÑ IÒ\ Ó œ Z +<Ò\Ó ÐIÒ\ÓÑ

# ## # , so that .It follows that for the exponential distribution. Answer: AIÒ\ Ó œ œ

# # #)#

8 88") )Ð Ñ



27. There are policyholders, and observations for each policyholder.< œ $ 8 œ %We use the equal sample size version of nonparametric empirical Bayes estimation.\ œ $ ß \ œ & ß \ œ % ß \ œ %

" # $ and .

@ œ Ð\ \ Ñ œ ÒÐ# $Ñ Ð$ $Ñ Ð$ $Ñ Ð% $Ñ Ó œs

" "4 "4œ"

%# # # # #" " #

%" $ $ .

@ œ @ œ @ œ œs s s# $# % s s s )$ $ $ *

@ @ @ , and , so that ." # $

+ œ Ð\ \Ñ œ ÒÐ$ %Ñ Ð& %Ñ Ð% %Ñ Ó œ Þ(()s " @ "

$" % # %s )Î*

3œ"

$

3# # # # .

Answer: C

28. The numbers at risk for each covariate class at each data point are:Ÿ $# ## #& #( #) $"D œ ! & % $ $ #: D œ " & & & % %: There is a single observation at each data point.The estimate of is .L Ð$#Ñ !

" " " " "&&/ %&/ $&/ $%/ #%/" " " " "

Substituting into this expression gives a value of 1.036 . Answer: C" œ "

29. The prior distribution is a beta distribution (with ), with and .) œ " + œ # , œ #The Bayesian combination of a beta prior distribution for with parameters and , and a; + ,binomial model distribution for with parameters and , and with observed values\ 7 ; 8B ß ÞÞÞß B ;" 8 results in a posterior distribution for which is also beta, with parameters+ œ + B , œ , 78 Bw w

3 3D D and The given beta prior has , , and the binomial model distribution has , and there+ œ " , œ " 7 œ %is (one year) observation with . The posterior distribution is beta with8 œ " B œ #+ œ # # œ % , œ # Ð%ÑÐ"Ñ # œ %w w and . The pdf of the posterior distribution is>> >

Ð%%ÑÐ%Ñ Ð%Ñ † ; Ð" ;Ñ ! ; "$ $ , . The mode of the posterior is point at which the posterior pdf

is maximized. This occurs where log pdf is maximized.Log pdf of posterior is , and the derivative is .68 "%! $68 ; 68Ð" ;Ñ $ $

; ";

Setting the derivative to 0 and solving for results in . This could have been anticipated; ; œ Þ&from the form of the posterior cdf which is symmetric on the interval . Answer: CÐ!ß "Ñ

30. (number of claims) is Poisson and we are told that the standard for full credibility for R Rbased on total number of claims (not total number of exposures of ) is 2000. This standard isR

Ð ÑDÞ!$ IÒRÓ

Z +<ÒRÓ: # † œ #!!! R Z +<ÒRÓ œ IÒRÓ. Since is Poisson, we have , so that

Ð ÑDÞ!$: #

: : :œ #!!! D T Ð D ^ D Ñ œ :. Note that is the standard normal value such that .Note also that in the first part of this problem we are applying limited fluctuation credulity to theclaim number random variable . In the second part of the problem, we are applying credibilityRto the compound aggregate claims distribution (with frequency and severity that isW R \uniform from 0 to 10,000). The standard applied to has "closeness" parameter 5% and the sameWprobability .:



The standard is based on expected number of claims (not expected exposures of ), and thatW

standard is . Since is Poisson, , and usingÐ ÑDÞ!& ÐIÒWÓÑ

Z +<ÒWÓ:#

# #† † IÒRÓ R Z +<ÒWÓ œ IÒRÓ † IÒ\ Ó

IÒWÓ œ IÒRÓ † IÒ\Ó , the standard becomesÐ Ñ Ð ÑD DÞ!& ÐIÒ\ÓÑ Þ!$

IÒ\ Ó: :#

## # #

:† œ #!!! D œ "Þ) . From we get , and from the uniform distribution for

\ IÒ\Ó œ &!!! IÒ\ Ó œ we get , . The expected number of claims needed for full# "!ß!!!$

#

credibility of under this standard is .W † œ † œ *'!Ð ÑDÞ!& ÐIÒ\ÓÑ ÐÞ!&Ñ Ð&!!!Ñ

IÒ\ Ó "!ß!!! Î$"Þ):# #

# # ##

Answer: B

31. The Nelson-Aalen estimate of is , where is the Nelson-Aalen estimateWÐ"!Ñ / LÐ"!ÑsLÐ"!Ñs

of . We are given that , so that .LÐ"!Ñ / œ Þ&(& LÐ"!Ñ œ Þ&&$%sLÐ"!Ñs

From the data set, we have , from which we get .LÐ"!Ñ œ œ Þ&&$% 5 œ $'s " $ & (&! %* 5 #"

Answer: C

32. The cdf of the exponential distribution with mean 1000 is , for .JÐBÑ œ " / B !Þ!!"B

The simulated dental charges, and related reimbursements are:? œ !Þ$! œ " / p B œ $&'Þ'( ‚ œÞ!!"B , 257.67 .8 206 is reimbursed ,? œ !Þ*# œ " / p B œ #&#&Þ($ ‚ œÞ!!"B , 2425.73 .8 1941, so 1000 is reimbursed ,? œ !Þ(! œ " / p B œ "#!$Þ*( ‚ œÞ!!"B , 1103.97 .8 883 is reimbursed ,? œ !Þ!) œ " / p B œ )$Þ$)Þ!!"B , 0 is reimbursed .Average annual reimbursement is . Answer: A#!'"!!!))$!

% œ &##

33. The likelihood function is the product of probabilities for each interval.The probability for the interval is .B Ÿ "! TÐ\ Ÿ "!Ñ œ JÐ"!Ñ œ " œ) )

"! "!"!

The probability for the interval is ."! B Ÿ #& JÐ#&Ñ JÐ"!Ñ œ œ Þ!') )"! #& )

The probability for the interval is .B #& " JÐ#&Ñ œ œ Þ!%)#& )

The likelihood function isP œ ÒJÐ"!ÑÓ † ÒJ Ð#&Ñ JÐ"!ÑÓ † Ò" JÐ#&ÑÓ œ Ð Ñ ÐÞ!' Ñ ÐÞ!% Ñ œ -Ð"! Ñ* ' & * ' & * """!

"!) ) ) ) ) ,

where .- œ ÐÞ"Ñ ÐÞ!'Ñ ÐÞ!%Ñ* ' &

The log of is , and the derivative isP 68P œ 68 - * 68Ð"! Ñ "" 68) ). * "" "". "! #) ) )68 P œ œ . Setting this equal to 0 and solving for results in the mle .) )

Answer: B



34. For the exponential distribution with mean , .) WÐ"&!!Ñ œ /"&!!Î)

If is the mle of , then the mle of is , and according to the delta method, the) )s WÐ"&!!Ñ /"&!!Îs)

variance pf this estimate is .Ð / Ñ † Z +<Ò Ós..)

"&!!Î #) )

For the exponential distribution, the mle of is , and the variance is) \ œ "!!!

Z +<Ò\Ó œ œ 8 œ ' œ "!!! s sZ +<Ò\Ó

8 8 '"!!!)# #

. Since and , the estimated variance of is .) ). "&!!.) )/ œ / †"&!!Î "&!!Î) ) Ð Ñ# . Applying the delta method, the variance of the estimate of

WÐ"&!!Ñ Ò/ † Ó † is , which is estimated to be"&!!Î #) Ð Ñ"&!!8))

#

#

Ò/ † Ó † œ Þ!")("&!!Î"!!! #Ð Ñ"&!! "!!!"!!! '#

#

. Answer: A

35. Using linearity (uniform distribution) within each interval, we haveJ Ð*!Ñ œ œ Þ#" $' Þ%B œ Þ#"88

$'Þ%B8 , so that .

Similarly, , so that .J Ð#"!Ñ œ œ Þ&" $' B Þ'C œ Þ&"88$'BÞ%C

8The total number of losses is .8 œ $' B C )% )! œ #!! B C œ 8The first two equations become $' Þ%B œ Þ#"Ð#!! B CÑ œ %# Þ#"B Þ#"Cand .$' B Þ'C œ Þ&"Ð#!! B CÑ œ "!# Þ&"B Þ&"CSolving these last two equations for and results in and . Answer: AB C B œ "#! C œ )!

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November 2006 - Exam C