Spring 2005 - Exam C SOA Exam - MATH · Exam C: Spring 2005 - 7 - GO ON TO NEXT PAGE 7. Loss data for 925 policies with deductibles of 300 and 500 and policy limits of 5,000 and 10,000

Exam C: Spring 2005 - 1 - GO ON TO NEXT PAGE

**BEGINNING OF EXAMINATION**

1. You are given:

(i) A random sample of five observations from a population is:

0.2 0.7 0.9 1.1 1.3

(ii) You use the Kolmogorov-Smirnov test for testing the null hypothesis, 0H , that the probability density function for the population is:

( )( )5

4 , 01

f x xx

= >+

(iii) Critical values for the Kolmogorov-Smirnov test are:

Level of Significance 0.10 0.05 0.025 0.01 Critical Value

n22.1

n36.1

n48.1

n63.1

Determine the result of the test. (A) Do not reject 0H at the 0.10 significance level. (B) Reject 0H at the 0.10 significance level, but not at the 0.05 significance level. (C) Reject 0H at the 0.05 significance level, but not at the 0.025 significance level. (D) Reject 0H at the 0.025 significance level, but not at the 0.01 significance level. (E) Reject 0H at the 0.01 significance level.


2. You are given: (i) The number of claims follows a negative binomial distribution with parameters r and

3=β . (ii) Claim severity has the following distribution:

Claim Size Probability 1 0.4 10 0.4 100 0.2

(iii) The number of claims is independent of the severity of claims. Determine the expected number of claims needed for aggregate losses to be within 10% of expected aggregate losses with 95% probability. (A) Less than 1200 (B) At least 1200, but less than 1600 (C) At least 1600, but less than 2000 (D) At least 2000, but less than 2400 (E) At least 2400


3. You are given: (i) A mortality study covers n lives. (ii) None were censored and no two deaths occurred at the same time. (iii) =kt time of the thk death

(iv) A Nelson-Aalen estimate of the cumulative hazard rate function is 38039)(ˆ

2 =tH .

Determine the Kaplan-Meier product-limit estimate of the survival function at time 9t . (A) Less than 0.56 (B) At least 0.56, but less than 0.58 (C) At least 0.58, but less than 0.60 (D) At least 0.60, but less than 0.62 (E) At least 0.62


4. Three observed values of the random variable X are:

1 1 4 You estimate the third central moment of X using the estimator:

( ) ( )31 2 3

1, ,3 ig X X X X X= −∑

Determine the bootstrap estimate of the mean-squared error of g. (A) Less than 3.0 (B) At least 3.0, but less than 3.5 (C) At least 3.5, but less than 4.0 (D) At least 4.0, but less than 4.5 (E) At least 4.5


5. You are given the following p-p plot:

0.0

0.2

0.4

0.6

0.8

1.0

0.0 0.2 0.4 0.6 0.8 1.0

Fn(x)

F(x)

The plot is based on the sample:

1 2 3 15 30 50 51 99 100 Determine the fitted model underlying the p-p plot. (A) F(x) = 1 – 0.25x− , x ≥ 1

(B) F(x) = x / (1 + x), x ≥ 0 (C) Uniform on [1, 100]

(D) Exponential with mean 10

(E) Normal with mean 40 and standard deviation 40


6. You are given: (i) Claims are conditionally independent and identically Poisson distributed with

mean Θ . (ii) The prior distribution function of Θ is:

2.61( ) 11

F θθ

⎛ ⎞= − ⎜ ⎟+⎝ ⎠, 0θ >

Five claims are observed. Determine the Bühlmann credibility factor. (A) Less than 0.6 (B) At least 0.6, but less than 0.7 (C) At least 0.7, but less than 0.8 (D) At least 0.8, but less than 0.9 (E) At least 0.9


7. 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(300 , 500] 50 − 50(500 , 1,000] 50 75 125(1,000 , 5,000) 150 150 300(5,000 , 10,000) 100 200 300At 5,000 40 80 120At 10,000 10 20 30

Total 400 525 925 Using the Kaplan-Meier approximation for large data sets, with 1α = and 0β = , estimate F(5000). (A) 0.25

(B) 0.32

(C) 0.40

(D) 0.51 (E) 0.55


8. You are given the following knots:

(0,0) (2,0) and derivative values: ( )0 2f ′ = −

( )2 2f ′ = . Determine the value of the squared norm measure of curvature for the cubic spline that satisfies these conditions. (A) 16/15 (B) 8/3 (C) 4 (D) 8

(E) 16


9-10. Use the following information for questions 9 and 10. The time to an accident follows an exponential distribution. A random sample of size two has a mean time of 6. Let Y denote the mean of a new sample of size two.

9. Determine the maximum likelihood estimate of ( )Pr 10Y > . (A) 0.04 (B) 0.07 (C) 0.11 (D) 0.15 (E) 0.19


9-10. (Repeated for convenience) Use the following information for questions 9 and 10. The time to an accident follows an exponential distribution. A random sample of size two has a mean time of 6. Let Y denote the mean of a new sample of size two.

10. Use the delta method to approximate the variance of the maximum likelihood estimator of (10)YF .

(A) 0.08 (B) 0.12 (C) 0.16 (D) 0.19 (E) 0.22


11. You are given: (i) The number of claims in a year for a selected risk follows a Poisson distribution with

mean λ . (ii) The severity of claims for the selected risk follows an exponential distribution with

mean .θ (iii) The number of claims is independent of the severity of claims. (iv) The prior distribution of λ is exponential with mean 1. (v) The prior distribution of θ is Poisson with mean 1. (vi) A priori, λ and θ are independent. Using Bühlmann’s credibility for aggregate losses, determine k. (A) 1 (B) 4/3 (C) 2 (D) 3 (E) 4


12. A company insures 100 people age 65. The annual probability of death for each person is 0.03. The deaths are independent. Use the inversion method to simulate the number of deaths in a year. Do this three times using:

1

2

3

0.200.030.09

uuu

===

Calculate the average of the simulated values. (A) 1

3

(B) 1

(C) 53

(D) 7

3

(E) 3


13. You are given claim count data for which the sample mean is roughly equal to the sample variance. Thus you would like to use a claim count model that has its mean equal to its variance. An obvious choice is the Poisson distribution. Determine which of the following models may also be appropriate. (A) A mixture of two binomial distributions with different means (B) A mixture of two Poisson distributions with different means (C) A mixture of two negative binomial distributions with different means (D) None of (A), (B) or (C) (E) All of (A), (B) and (C)


14. You are given: (i) Annual claim frequencies follow a Poisson distribution with mean λ. (ii) The prior distribution of λ has probability density function:

( ) / 6 /121 1(0.4) (0.6) , 06 12

e eλ λπ λ λ− −= + >

Ten claims are observed for an insured in Year 1. Determine the Bayesian expected number of claims for the insured in Year 2. (A) 9.6 (B) 9.7 (C) 9.8 (D) 9.9 (E) 10.0


15. Twelve policyholders were monitored from the starting date of the policy to the time of first claim. The observed data are as follows:

Time of First Claim 1 2 3 4 5 6 7 Number of Claims 2 1 2 2 1 2 2

Using the Nelson-Aalen estimator, calculate the 95% linear confidence interval for the cumulative hazard rate function H(4.5). (A) (0.189, 1.361) (B) (0.206, 1.545) (C) (0.248, 1.402) (D) (0.283, 1.266) (E) (0.314, 1.437)


16. For the random variable X, you are given: (i) [ ]E X θ= , 0θ >

(ii) ( )2

Var25

X θ=

(iii) ˆ ,1

k Xk

θ =+

0k >

(iv) ( ) ( )2

ˆ ˆMSE 2 biasθ θθ θ⎡ ⎤= ⎣ ⎦

Determine k. (A) 0.2 (B) 0.5 (C) 2 (D) 5 (E) 25


17. You are given: (i) The annual number of claims on a given policy has a geometric distribution with

parameter β. (ii) The prior distribution of β has the Pareto density function

( 1)( ) , 0 ( 1) α

απ β ββ += < < ∞+

,

where α is a known constant greater than 2. A randomly selected policy had x claims in Year 1. Determine the Bühlmann credibility estimate of the number of claims for the selected policy in Year 2.

(A) 11α −

(B) ( 1) 1( 1)

xαα α α−

+−

(C) x

(D) 1xα+

(E) 11

xα+−


18. You are given: (i) The Cox proportional hazards model is used for the following data:

Gender (Z1)

Age (Z2)

Time to First Accident

(X) 0 20 3 0 30 > 6 1 30 7 1 40 > 8

(ii) The baseline hazard rate function is a constant equal to 1/θ. (iii) The regression coefficients for gender and age are β1 and β2, respectively. Determine the loglikelihood function when θ = 18, β1 = 0.1, and β2 = 0.01. (A) Less than −10 (B) At least −10, but less than −7.5 (C) At least −7.5, but less than −5 (D) At least −5, but less than −2.5 (E) At least −2.5


19. 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 the chi-square goodness-of-fit test becomes larger with increased sample size.

(E) None of (A), (B), (C) or (D) is true.


20. For a particular policy, the conditional probability of the annual number of claims given θΘ = , and the probability distribution of Θ are as follows:

Number of claims 0 1 2 Probability 2θ θ 1 3θ−

θ 0.05 0.30 Probability 0.80 0.20

Two claims are observed in Year 1. Calculate the Bühlmann credibility estimate of the number of claims in Year 2. (A) Less than 1.68

(B) At least 1.68, but less than 1.70

(C) At least 1.70, but less than 1.72

(D) At least 1.72, but less than 1.74

(E) At least 1.74


21. You are given: (i) The annual number of claims for a policyholder follows a Poisson distribution with

mean Λ . (ii) The prior distribution of Λ is gamma with probability density function:

f (λ ) = 5 2(2 )

24e λλλ

−

, 0λ >

An insured is selected at random and observed to have 1 5x = claims during Year 1 and

2 3x = claims during Year 2. Determine ( )1 2E 5, 3x xΛ = = . (A) 3.00 (B) 3.25 (C) 3.50 (D) 3.75 (E) 4.00


22. You are given the kernel:

( )

2

otherwise

2 1 ( ) , 1 1

0,y

x y y x y

k xπ

⎧ − − − ≤ ≤ +⎪⎪

= ⎨⎪⎪⎩

You are also given the following random sample:

1 3 3 5

Determine which of the following graphs shows the shape of the kernel density estimator.

(A)

(C)

(E)

(B)

(D)


23. You are fitting a curve to a set of 1+n data points. Which of the following statements is true? (A) The collocation polynomial is recommended for extrapolation.

(B) Collocation polynomials pass through all data points, whereas cubic splines do not. (C) The curvature-adjusted cubic spline requires that 00 == nmm where ).( jj xfm ′′= (D) For a cubic runout spline, the second derivative is a linear function throughout the

intervals ],[ 20 xx and ],[ 2 nn xx − . (E) If )(xf is the natural cubic spline passing through the 1+n data points, and )(xh is

any function with continuous first and second derivatives that passes through the

same points, then 0 0

2 2[ ( )] [ ( )]n nx x

x xf x dx h x dx′′ ′′>∫ ∫ .


24. The following claim data were generated from a Pareto distribution:

130 20 350 218 1822 Using the method of moments to estimate the parameters of a Pareto distribution, calculate the limited expected value at 500. (A) Less than 250

(B) At least 250, but less than 280

(C) At least 280, but less than 310

(D) At least 310, but less than 340

(E) At least 340


25. You are given:

Group Year 1 Year 2 Year 3 Total Total Claims 1 10,000 15,000 25,000Number in Group 50 60 110Average 200 250 227.27Total Claims 2 16,000 18,000 34,000Number in Group 100 90 190Average 160 200 178.95Total Claims 59,000Number in Group 300Average 196.67

You are also given a = 651.03. Use the nonparametric empirical Bayes method to estimate the credibility factor for Group 1. (A) 0.48 (B) 0.50 (C) 0.52 (D) 0.54 (E) 0.56


26. You are given the following information regarding claim sizes for 100 claims:

Claim Size Number of Claims 0 - 1,000 16

1,000 - 3,000 22 3,000 - 5,000 25

5,000 - 10,000 18 10,000 - 25,000 10 25,000 - 50,000 5

50,000 - 100,000 3 over 100,000 1

Use the ogive to estimate the probability that a randomly chosen claim is between 2,000 and 6,000. (A) 0.36 (B) 0.40 (C) 0.45 (D) 0.47 (E) 0.50


27. You are given the following 20 bodily injury losses (before the deductible is applied):

Loss Number of Losses

Deductible Policy Limit

750 3 200 ∞ 200 3 0 10,000 300 4 0 20,000

>10,000 6 0 10,000 400 4 300 ∞

Past experience indicates that these losses follow a Pareto distribution with parameters α and 10,000θ = . Determine the maximum likelihood estimate of α . (A) Less than 2.0

(B) At least 2.0, but less than 3.0

(C) At least 3.0, but less than 4.0

(D) At least 4.0, but less than 5.0

(E) At least 5.0


28. You are given: (i) During a 2-year period, 100 policies had the following claims experience:

Total Claims in Years 1 and 2

Number of Policies

0 50 1 30 2 15 3 4 4 1

(ii) The number of claims per year follows a Poisson distribution. (iii) Each policyholder was insured for the entire 2-year period. A randomly selected policyholder had one claim over the 2-year period. Using semiparametric empirical Bayes estimation, determine the Bühlmann estimate for the number of claims in Year 3 for the same policyholder. (A) 0.380 (B) 0.387 (C) 0.393 (D) 0.403 (E) 0.443


29. For a study of losses on two classes of policies, you are given: (i) The Cox proportional hazards model was used with baseline hazard rate function:

02( ) , 0xh x xθ

= < < ∞

(ii) A single covariate Z was used with Z = 0 for policies in Class A and Z = 1 for policies

in Class B.

(iii) Two policies in Class A had losses of 1 and 3 while two policies in Class B had losses of 2 and 4.

Determine the maximum likelihood estimate of the coefficient β. (A) − 0.7 (B) 0.5 (C) 0.7 (D) 1.0

(E) 1.6


30. A natural cubic spline is fit to ( ) 5h x x= at the knots 0 12, 0x x= − = and 2 2x = . Determine ( )0.5f ′′ − . (A) –1.0

(B) –0.5

(C) 0.0

(D) 0.5

(E) 1.0


31. Personal auto property damage claims in a certain region are known to follow the Weibull distribution:

( ) ( )0.2

1 , 0x

F x e xθ−= − >

A sample of four claims is:

130 240 300 540 The values of two additional claims are known to exceed 1000. Determine the maximum likelihood estimate of θ. (A) Less than 300 (B) At least 300, but less than 1200 (C) At least 1200, but less than 2100 (D) At least 2100, but less than 3000 (E) At least 3000


32. For five types of risks, you are given: (i) The expected number of claims in a year for these risks ranges from 1.0 to 4.0. (ii) The number of claims follows a Poisson distribution for each risk. During Year 1, n claims are observed for a randomly selected risk. For the same risk, both Bayes and Bühlmann credibility estimates of the number of claims in Year 2 are calculated for n = 0,1,2, ... ,9. Which graph represents these estimates?


(A)

(B)

(C)

(D)

(E)


33. You test the hypothesis that a given set of data comes from a known distribution with distribution function F(x). The following data were collected:

Interval ( )iF x Number of

Observations x < 2 0.035 5

2 ≤ x < 5 0.130 42 5 ≤ x < 7 0.630 137 7 ≤ x < 8 0.830 66

8 ≤ x 1.000 50 Total 300

where ix is the upper endpoint of each interval. You test the hypothesis using the chi-square goodness-of-fit test. Determine the result of the test. (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.


34. Unlimited claim severities for a warranty product follow the lognormal distribution with parameters 5.6µ = and 0.75σ = . You use simulation to generate severities. The following are six uniform (0, 1) random numbers:

0.6179 0.4602 0.9452 0.0808 0.7881 0.4207

Using these numbers and the inversion method, calculate the average payment per claim for a contract with a policy limit of 400. (A) Less than 300 (B) At least 300, but less than 320 (C) At least 320, but less than 340 (D) At least 340, but less than 360 (E) At least 360

Exam C: Spring 2005 - 36 - STOP

35. You are given: (i) The annual number of claims on a given policy has the geometric distribution with

parameter β. (ii) One-third of the policies have β = 2, and the remaining two-thirds have β = 5. A randomly selected policy had two claims in Year 1. Calculate the Bayesian expected number of claims for the selected policy in Year 2. (A) 3.4 (B) 3.6 (C) 3.8 (D) 4.0 (E) 4.2

**END OF EXAMINATION**

Exam C: Spring 2005 - 37 -

Exam C, Spring 2005

PRELIMINARY ANSWER KEY

Question # Answer Question # Answer

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

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

Exam C Solutions Spring 2005

Question #1 Key: D

The CDF is 4)1(11)(x

xF+

−=

Observation (x) F(x) compare to: Maximum difference 0.2 0.518 0, 0.2 0.518 0.7 0.880 0.2, 0.4 0.680 0.9 0.923 0.4, 0.6 0.523 1.1 0.949 0.6, 0.8 0.349 1.3 0.964 0.8, 1.0 0.164

The K-S statistic is the maximum from the last column, 0.680. Critical values are: 0.546, 0.608, 0.662, and 0.729 for the given levels of significance. The test statistic is between 0.662 (2.5%) and 0.729 (1.0%) and therefore the test is rejected at 0.025 and not at 0.01. Question #2 Key: E For claim severity,

2 2 2 2 2

1(0.4) 10(0.4) 100(0.2) 24.4,

1 (0.4) 10 (0.4) 100 (0.2) 24.4 1, 445.04.S

S

µ

σ

= + + =

= + + − =

For claim frequency, 23 , (1 ) 12 .F Fr r r rµ β σ β β= = = + =

For aggregate losses,

2 2 2 2 2

24.4(3 ) 73.2 ,

24.4 (12 ) 1, 445.04(3 ) 11, 479.44 .S F

S F S F

r r

r r r

µ µ µ

σ µ σ σ µ

= = =

= + = + =

For the given probability and tolerance, 20 (1.96 / 0.1) 384.16.λ = =

The number of observations needed is 2 2 2

0 / 384.16(11,479.44 ) /(73.2 ) 823.02 / .r r rλ σ µ = = The average observation produces 3r claims and so the required number of claims is (823.02 / )(3 ) 2,469.r r =

Question #3 Key: A

487.0,2003807993938039

)1(12

111)(ˆ 2

2 ===>=+−=>=−−

=−

+= nnnnnnn

nntH .

Discard the non-integer solution to have n = 20. The Kaplan-Meier Product-Limit Estimate is:

919 18 11 11ˆ( ) ... 0.55.20 19 12 20

S t = = =

Question #4 Key: E There are 27 possible bootstrap samples, which produce four different results. The results, their probabilities, and the values of g are: Bootstrap Sample Prob g 1, 1, 1 8/27 0 1, 1, 4 12/27 2 1, 4, 4 6/27 -2 4, 4, 4 1/27 0 The third central moment of the original sample is 2. Then,

MSE = ( ) ( ) ( ) ( )2 2 2 28 12 6 1 440 2 2 2 2 2 0 227 27 27 27 9

⎡ ⎤− + − + − − + − =⎢ ⎥⎣ ⎦.

Question #5 Key: A Pick one of the points, say the fifth one. The vertical coordinate is F(30) from the model and should be slightly less than 0.6. Inserting 30 into the five answers produces 0.573, 0.096, 0.293, 0.950, and something less than 0.5. Only the model in answer A is close. Question #6 Key: C, E The distribution of Θ is Pareto with parameters 1 and 2.6. Then,

21 2( ) 0.625, ( ) 0.625 1.6927,2.6 1 1.6(0.6)

5/ 0.625 /1.6927 0.3692, 0.9312.5 0.3692

v EPPV E a VHM Var

k v a Z

= = Θ = = = = Θ = − =−

= = = = =+

Question #7 Key: E At 300, there are 400 policies available, of which 350 survive to 500. At 500 the risk set increases to 875, of which 750 survive to 1,000. Of the 750 at 1,000, 450 survive to 5,000. The

probability of surviving to 5,000 is 350 750 450 0.45.400 875 750

= The distribution function is 1 – 0.45 =

0.55. Alternatively, the formulas in Loss Models could be applied as follows:

j Interval 1( , ]j jc c + dj uj xj Pj rj qj ˆ ( )jF c 0 (300, 500] 400 0 50 0 400 0.125 0 1 (500, 1,000] 525 0 125 350 875 0.142857 0.125 2 (1,000, 5,000] 0 120 300 750 750 0.4 0.25 3 (5,000, 10,000] 0 30 300 330 330 0.90909 0.55

where dj = number of observations with a lower truncation point of cj, uj = number of observations censored from above at cj+1, xj = number of uncensored observations in the interval 1( , ]j jc c + . Then, 1j j j j jP P d u x+ = + − − , j j jr P d= + (using 1α = and 0β = ), and /j j jq x r= .

Finally, ( )1

0

ˆ ( ) 1 1j

j ii

F c q−

=

= − −∏ .

Question #8 Key: D Let the function be 2 3( ) .f x a bx cx dx= + + + The four conditions imply the following:

(0) 0 0,(2) 0 2 4 8 0,'(0) 2 2,'(2) 2 4 12 2.

f af b c df bf b c d

= ⇒ == ⇒ + + == − ⇒ = −= ⇒ + + =

The solution is c = 1 and d = 0 and thus the function is 2( ) 2f x x x= − . The curvature measure is 2 22 2

0 0[ "( )] 2 8.f x dx dx= =∫ ∫

Question #9 Key: D For an exponential distribution, the maximum likelihood estimate of θ is the sample mean, 6. Let 1 2Y X X= + where each X has an exponential distribution with mean 6. The sample mean is Y/2 and Y has a gamma distribution with parameters 2 and 6. Then

/ 6

20

/ 6 20/ 6/ 6 20 / 6

20

Pr( / 2 10) Pr( 20)36

20 0.1546.6 6

x

xx

xeY Y dx

xe ee e

−∞

∞− −− −

> = > =

= − − = + =

∫

Question #10 Key: A

From Question 9, 10/

10 / 10 / 110(10) 1 1 (1 10 ) ( ).eF e e gθ

θ θ θ θθ

−− − −= − − = − + =

20 /20 / 1 20/

2 2 3

20 20 400'( ) (1 20 ) .eg e eθ

θ θθ θθ θ θ

−− − −= − + + = −

At the maximum likelihood estimate of 6, '(6) 0.066063.g = − The maximum likelihood estimator is the sample mean. Its variance is the variance of one observation divided by the sample size. For the exponential distribution the variance is the square of the mean, so the estimated variance of the sample mean is 36/2 = 18. The answer is

2( 0.066063) (18) 0.079.− = Question #11 Key B

2

2

2 2 2

( , ) ( | , ) ,( , ) ( | , ) 2 ,

( 2 ) 1(2)(1 1) 4,( ) ( ) ( ) [ ( ) ( )] 2(2) 1 3,

/ 4 / 3.

E Sv Var Sv EVPV Ea VHM Var E E E Ek v a

µ λ θ λ θ λθ

λ θ λ θ λ θ

λ θ

λθ λ θ λ θ

= =

= =

= = = + =

= = = − = − == =

Question #12 Key: B The distribution is binomial with m = 100 and q = 0.03. The first three probabilities are:

100 990 1

98 22

0.97 0.04755, 100(0.97) (0.03) 0.14707,100(99) (0.97) (0.03) 0.22515.

2

p p

p

= = = =

= =

Values between 0 and 0.04755 simulate a 0, between 0.04755 and 0.19462 simulate a 1, and between 0.19462 and 0.41977 simulate a 2. The three simulated values are 2, 0, and 1. The mean is 1. Question #13 Key: A A mixture of two Poissons or negative binomials will always have a variance greater than its mean. A mixture of two binomials could have a variance equal to its mean, because a single binomial has a variance less than its mean. Question #14 Key: D The posterior distribution can be found from

( )10

/ 6 /12 10 7 / 6 13 /120.4 0.6( |10) 0.8 0.6 .10! 6 12

e e e e eλ

λ λ λ λλπ λ λ−

− − − −⎛ ⎞∝ + ∝ +⎜ ⎟⎝ ⎠

The required constant is found from

( )10 7 / 6 13 /12 11 11

00.8 0.6 0.8(10!)(6 / 7) 0.6(10!)(12 /13) 0.395536(10!).e e dλ λλ λ

∞ − −+ = + =∫

The posterior mean is

( )11 7 / 6 13 /12

0

12 12

1( |10) 0.8 0.60.395536(10!)

0.8(11!)(6 / 7) 0.6(11!)(12 /13) 9.88.0.395536(10!)

E e e dλ λλ λ λ∞ − −= +

+= =

∫

Question #15 Key: A

2 1 2 2ˆ (4.5) 0.77460.12 10 9 7

H = + + + = The variance estimate is 2 2 2 2

2 1 2 2 0.089397.12 10 9 7

+ + + =

The confidence interval is 0.77460 1.96 0.089397 0.77460 0.58603.± = ± The interval is from 0.1886 to 1.3606.

Question #16 Key: D

2 2

2

2 2 22

2 2

22

2

ˆ( ) ,1 1

ˆ( ) ,1 25( 1)

ˆ( ) ,25( 1) ( 1)

22 .( 1)

kbias Ek k

k kVar Vark k

kMSE Var biask k

MSE biask

θθ θ θ θ

θ θθ

θ θθ

θ

= − = − = −+ +

⎛ ⎞= =⎜ ⎟+ +⎝ ⎠

= + = ++ +

= =+

Setting the last two equal and canceling the common terms gives 2

1 225k

+ = for k = 5.

Question #17 Key: D For the geometric distribution ( )µ β β= and ( ) (1 )v β β β= + . The prior density is Pareto with parameters α and 1. Then,

2 2

1( ) ,1

1 2[ (1 )] ,1 ( 1)( 2) ( 1)( 2)

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

1 1/ 1, .1

E

v EVPV E

a VHM Var

k v a Zk

µ βα

αβ βα α α α α

αβα α α α α

αα

= =−

= = + = + =− − − − −

= = = − =− − − − −

= = − = =+

The estimate is 1 1 1 11 .

1xx

α α α α+⎛ ⎞+ − =⎜ ⎟ −⎝ ⎠

Question #18 Key: C

The hazard rate function is 1 1 2 21( ) z zh x h eβ β

θ+= = , a constant. For the four observations, those

constants are

.1(0) .01(20) .1(0) .01(30)

.1(1) .01(30) .1(1) .01(40)

1 10.06786, 0.07449,18 181 10.08288, 0.09160.

18 18

e e

e e

+ +

+ +

= =

= =

Each observation is from an exponential distribution with density function hxhe− and survival function hxe− . The contributions to the loglikelihood function are ln( )h hx− and hx− , respectively. The answer is, ln(0.06786) 0.06786(3) 0.07499(6) ln(0.08288) 0.08288(7) 0.09160(8) 7.147.− − + − − = − Question #19 Key: E A is false. Using sample data gives a better than expected fit and therefore a test statistic that favors the null hypothesis, thus increasing the Type II error probability. The K-S test works only on individual data and so B is false. The A-D test emphasizes the tails, thus C is false. D is false because the critical value depends on the degrees of freedom which in turn depends on the number of cells, not the sample size. Question #20 Key: B

2 2 2

2 2 2 2 2

2

( ) 0.05(0.8) 0.3(0.2) 0.1,( ) 0.05 (0.8) 0.3 (0.2) 0.02,( ) 0(2 ) 1( ) 2(1 3 ) 2 5 ,( ) 0 (2 ) 1 ( ) 2 (1 3 ) (2 5 ) 9 25 ,

(2 5 ) 2 5(0.1) 1.5,(9 25 ) 9(0.1) 25(0.02) 0.4,

EE

vE

v EVPV Ea VHM V

θ

θµ θ θ θ θ θ

θ θ θ θ θ θ θµ θ

θ θ

= + =

= + == + + − = −

= + + − − − = −= − = − =

= = − = − =

= = 2(2 5 ) 25 ( ) 25(0.02 0.1 ) 0.25,1 5/ 0.4 / 0.25 1.6, ,

1 1.6 135 82 1.5 1.6923.

13 13

ar Var

k v a Z

P

θ θ− = = − =

= = = = =+

= + =

Question #21 Key: B

3 5 5 212 42( | 5,3) .

5! 3! 24e e ef e

λ λ λλλ λ λλ λ

λ

− − −−∝ ∝ This is a gamma distribution with parameters 13

and 0.25. The expected value is 13(0.25) = 3.25. Alternatively, if the Poisson-gamma relationships are known, begin with the prior parameters

5α = and 2β = where 1/β θ= if the parameterization from Loss Models is considered. Then the posterior parameters are ' 5 5 3 13α = + + = where the second 5 and the 3 are the observations and ' 2 2 4β = + = where the second 2 is the number of observations. The posterior mean is then 13/4 = 3.25. Question #22 Key: D Because the kernel extends one unit each direction there is no overlap. The result will be three replications of the kernel. The middle one will be twice has high due to having two observations of 3 while only one observation at 1 and 5. Only graphs A and D fit this description. The kernel function is smooth, which rules out graph A. Question #23 Key: D A is false, the collocation polynomial is too wiggly. Both collocation polynomials and cubic splines pass through all points, so B is false. C is false because the curvature adjusted spline sets the values, but not necessarily to zero. The inequality in E should be the other direction. Answer D is true. Question #24 Key: C The two moment equations are

22508 , 701, 401.6 .1 ( 1)( 2)

θ θα α α

= =− − −

Dividing the square of the first equation into the

second equation gives 2

701,401.6 2( 1)2.7179366 .508 1

αα

−= =

− The solution is α = 4.785761. From

the first equation, θ = 1,923.167. The requested LEV is 3.785761

1,923.167 1,923.167( 500) 1 296.21.3.785761 1,923.167 500

E X⎡ ⎤⎛ ⎞∧ = − =⎢ ⎥⎜ ⎟+⎝ ⎠⎢ ⎥⎣ ⎦

Question #25 Key: B

2 2 2 250(200 227.27) 60(250 227.27) 100(160 178.95) 90(200 178.95)ˆ1 1

71,985.647,ˆ 71,985.647 / 651.03 110.57,

110ˆ 0.499.110 110.57

v EVPV

k

Z

− + − + − + −= =

+=

= =

= =+

Question #26 Key: B

100 100 100 100

100

100

(1,000) 0.16, (3,000) 0.38, (5,000) 0.63, (10,000) 0.81,(2,000) 0.5(0.16) 0.5(0.38) 0.27,(6,000) 0.8(0.63) 0.2(0.81) 0.666.

Pr(2,000 6,000) 0.666 0.27 0.396.

F F F FFF

X

= = = =

= + == + =

< < = − =

Question #27 Key: C

3 43 4 6

3 3 4 6 4

1 1 1 1

14

(750) (400)(200) (300) [1 (10,000)]1 (200) 1 (300)

10, 200 10,000 10,000 10,000 10,30010,750 10, 200 10,300 20,000 10, 400

10, 200

f fL f f FF F

α α α α α

α α α α α

α α α α

α

+ + + +

⎡ ⎤ ⎡ ⎤= −⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦= 3 13 4 3 3 6 4 4

14 13 3 6 4

10,000 10,300 10,750 20,000 10, 40010,000 10,750 20,000 10,400 ,

ln 14ln 13 ln(10,000) 3 ln(10,750) 6 ln(20,000) 4 ln(10,400)14ln 4.5327 .L

α α α α

α α α ααα α α α α

α α

− − − − − − −

− − −∝= + − − −

= −

The derivative is 14 / 4.5327α − and setting it equal to zero gives ˆ 3.089.α =

Question #28 Key: C

2 2 2 2 2

30 30 12 4ˆ 0.76.100

50(0 0.76) 30(1 0.76) 15(2 0.76) 4(3 0.76) 1(4 0.76)ˆ 0.7699

0.090909,0.76 1ˆ ˆ8.36, 0.10684,

0.090909 1 8.360.10684(1) 0.89316(0.76) 0.78564.

v EVPV x

a VHM

k Z

P

+ + += = = =

− + − + − + − + −= = −

=

= = = =+

= + =

The above analysis was based on the distribution of total claims for two years. Thus 0.78564 is the expected number of claims for the next two years. For the next one year the expected number is 0.78564/2 = 0.39282. Question #29 Key: A For members of Class A,

2 2/ /( ) 2 / , ( ) , ( ) (2 / )x xA Ah x x S x e f x x eθ θθ θ− −= = = and for members of

Class B, 2 2/ /( ) 2 / 2 / , ( ) , ( ) (2 / )x x

B Bh x xe x S x e f x x eβ γ θ γ θθ γ θ γ θ− −= = = = where eβγ = . The likelihood function is

1 1/ 1 9 / 1 4 / 1 16 /

4 2 (10 20 ) /

( , ) (1) (3) (2) (4)

2 6 4 8.

A A B BL f f f f

e e e ee

θ θ γ θ γ θ

γ θ

γ θ

θ θ γθ γθ

θ γ

− − − − − − − −

− − +

=

=

∝

The logarithm and its derivatives are 1

1 1

1 2

( , ) 4 ln 2ln (10 20 )/ 2 20/ 4 (10 20 ) .

lll

γ θ θ γ γ θ

γ γ θ

θ θ γ θ

−

− −

− −

= − + − +

∂ ∂ = +

∂ ∂ = − + +

Setting each partial derivative equal to zero and solving the first equation for θ gives 10θ γ= . Substituting in the second equation yields

2 22

4 10 20 0, 400 100 200 , 0.5.10 100

γ γ γ γ γγ γ

+− + = = + = Then, ln( ) ln(0.5) 0.69.β γ= = = −

Question #30 Key: C Let the two spline functions be

2 30

2 31

( ) ( 2) ( 2) ( 2) ,

( ) .

f x a b x c x d x

f x e fx gx hx

= + + + + + +

= + + +

The eight spline conditions produce the following eight equations. 0 1

''0 0

' '1 0 1'' '' ''

0 1 1

( 2) 32 32, (0) 0 0,

( 2) 0 0, (0) 0 32 2 4 8 0,

(2) 32 2 4 8 32, (0) (0) 4 12 ,

(0) (0) 2 12 2 , (2) 0 2 12 0.

f a f e

f c f b c d

f f g h f f b c d f

f f c d g f g h

− = − ⇒ = − = ⇒ =

− = ⇒ = = ⇒ − + + + =

= ⇒ + + = = ⇒ + + =

= ⇒ + = = ⇒ + =

Eliminating one variable at a time leads to b = 16, d = 0, f = 16, g = 0, and h = 0. Then, 0 ( ) 32 16( 2) 16f x x x= − + + = and the second derivative is zero.

The above work could be avoided by noting that the three knots are (-2, -32), (0,0), and (2, 32). These points lie on a straight line and so the spline will be that straight line. Question #31 Key: E

The density function is 0.2

0.8( / )

0.2

0.2( ) xxf x e θ

θ

−−= . The likelihood function is

0.2 0.2 0.2 0.2 0.2 0.2

0.2 0.2

2

0.8 0.8 0.8 0.8(130/ ) (240/ ) (300/ ) (540 / ) (1000/ ) (1000 / )

0.2 0.2 0.2 0.2

0.8 (130

( ) (130) (240) (300) (540)[1 (1000)]0.2(130) 0.2(240) 0.2(300) 0.2(540)

L f f f f F

e e e e e e

e

θ θ θ θ θ θ

θ

θ

θ θ θ θθ

−

− − − −− − − − − −

− −

= −

=

∝0.2 0.2 0.2 0.2 0.2240 300 540 1000 1000 )

0.2 0.2 0.2 0.2 0.2 0.2 0.2

0.2

1 1.2

0.2

,( ) 0.8ln( ) (130 240 300 540 1000 1000 )

0.8ln( ) 20.2505 ,'( ) 0.8 0.2(20.2505) 0,

ˆ0.197526, 3,325.67.

l

l

θ θ θ

θ θ

θ θ θ

θ θ

+ + + + +

−

−

− −

−

= − − + + + + +

= − −

= − + =

= =

Question #32 Key: A Buhlmann estimates are on a straight line, which eliminates E. Bayes estimates are never outside the range of the prior distribution. Because graphs B and D include values outside the range 1 to 4, they cannot be correct. The Buhlmann estimates are the linear least squares approximation to the Bayes estimates. In graph C the Bayes estimates are consistently higher and so the Buhlmann estimates are not the best approximation. This leaves A as the only feasible choice.

Question #33 Key: C The expected counts are 300(0.035) = 10.5, 300(0.095) = 28.5, 300(0.5) = 150, 300(0.2) = 60, and 300(0.17) = 51 for the five groups. The test statistic is

2 2 2 2 2(5 10.5) (42 28.5) (137 150) (66 60) (50 51) 11.02.10.5 28.5 150 60 51− − − − −

+ + + + =

There are 5 – 1 = 4 degrees of freedom. From the table, the critical value for a 5% test is 9.488 and for a 2.5% test is 11.143. The hypothesis is rejected at 5%, but not at 2.5%. Question #34 Key: A To simulate a lognormal variate, first convert the uniform random number to a standard normal. This can be done be using the table of normal distribution values. For the six values given, the corresponding standard normal values are 0.3, -0.1, 1.6, -1.4, 0.8, and -0.2. Next, multiply each number by the standard deviation of 0.75 and add the mean of 5.6. This produces random observations from the normal 5.6, 0.752 distribution. These values are 5.825, 5,525, 6.8, 4.55, 6.2, and 5.45. To create lognormal observations, exponentiate these values. The results are 339, 251, 898, 95, 493, and 233. After imposing the policy limit of 400, the mean is (339 + 251 + 400 + 95 + 400 + 233)/6 = 286. Question #35 Key: C

For the geometric distribution, 2

1 3Pr( 2 | )(1 )

X βββ

= =+

and the expected value is β.

11

1 1

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

4 127 3 0.39024.4 1 25 2

27 3 216 3

XXX X

β βββ β β β

= = == = =

= = = + = = =

= =+

The expected value is then 0.39024(2) + 0.60976(5) = 3.83.

Documents

Spring 2005 - Exam C SOA Exam - MATH · Exam C: Spring 2005 - 7 - GO ON TO NEXT PAGE 7. Loss data for 925 policies with deductibles of 300 and 500 and policy limits of 5,000 and 10,000