Solutions to the Spring 2016 CAS Exam SThowardmahler.com/Teaching/ST_files/SolsS16CASST.pdfComment:...

Solutions to the Spring 2016

CAS Exam ST

There were 25 questions in total, of equal value, on this 2.5 hour exam. There was a 10 minute reading period in addition to the 2.5 hours.

The Exam ST is copyright 2016 by the Casualty Actuarial Society. The exam is available from the CAS. The solutions and comments are solely the responsibility of the author.

CAS Exam STprepared by

Howard C. Mahler, FCASCopyright 2016 by Howard C. Mahler.

Howard [email protected]/Teaching

1. You are given that N(t) follows the Poisson process with rate λ = 2. Calculate Pr[N(2) = 3 I N(5) = 7]. A. Less than 0.25 B. At least 0.25, but less than 0.35 C. At least 0.35, but less than 0.45 D. At least 0.45, but less than 0.55 E. At least 0.55

1. B. Given we have 7 claims by time 5, their times are independently, identically distributed uniform distributions on (0, 5). Thus there is a 2/5 = 40% chance that each one is before time 2.Thus the number of claims from time 0 to 2 is Binomial with m = 7 and q = 40%.Pr[N(2) = 3 I N(5) = 7] is the density at 3 of this Binomial Distribution:

€

73⎛

⎝ ⎜ ⎞

⎠ ⎟ (0.43) (0.64) = (35) (0.43) (0.64) = 29.0%.

Alternately, Pr[N(2) = 3 I N(5) = 7] =

€

Pr[N(2) = 3 and N(5) = 7]Pr[N(5) = 7] =

€

Pr[N(2) = 3] Pr[N(5) - N(2) = 4]Pr[N(5) = 7] =

€

{(3λ)4 e- 3λ / 4!} {(5λ)6 e- 5λ / 6!}(8λ)10 e- 8 λ / 10! =

€

(43) (64)107

€

7!3! 4! =

29.0%. Comment: It turns out that the answer does not depend on the value of λ.

Solutions Spring 2016 CAS Exam ST, HCM 5/24/16, Page 1

2. For a health insurance policy the annual number of claims follows the Poisson process with mean λ = 5. Calculate the probability that the time interval between the second and third claims exceeds 0.6 years. A. Less than 0.01 B. At least 0.01, but less than 0.04 C. At least 0.04, but less than 0.07 D. At least 0.07, but less than 0.10 E. At least 0.10

2. C. The time between successive claims is Exponential with mean: 1/5 = 0.2.S(0.6) = exp[-0.6/0.2] = e-3 = 4.98%.


3. For a block of one-year insurance policies you are given:

€

• Aggregate losses follow a compound Poisson process.

€

• Claim frequency follows a non-homogenous Poisson process with the following monthly rates: Month 1 2 3 4 5 6 7 8 9 10 11 12 Rate 5 5 5 2 2 2 2 2 2 3 3 3

€

• Claim severities follow an exponential distribution with θ = 500. Calculate the standard deviation of the annual aggregate losses. A. Less than 4,500 B. At least 4,500 but less than 5,000 C. At least 5,000 but less than 5,500 D. At least 5,500 but less than 6,000 E. More than 6,000

3. A. The number of claims is Poisson with mean:5 + 5 + 5 + 2 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 3 = 36. The second moment of severity is: 2θ2 = (2)(5002) = 500,000.Thus the variance of the aggregate annual loss is: (36)(500,000) = 18,000,000.The standard deviation is:

€

18,000,000 = 4243.


4. You are given:

€

• Loss amounts follow the probability density function

f(x) =

€

α 3αxα + 1, x > 3

• Observed loss amounts of 20, 30, 60, and 90. Calculate the maximum likelihood estimate of α. A. Less than 0.25 B. At least 0.25, but less than 0.35 C. At least 0.35, but less than 0.45 D. At least 0.45, but less than 0.55 E. At least 0.55

4. C. ln f(x) = ln(α) + α ln[3] - (α+1)ln[x].Loglikelihood is: 4ln(α) + 4α ln[3] - (α+1) (ln[20] + ln[30] + ln[60] + ln[90]).Setting the partial derivative of the loglikelihood with respect to alpha equal to zero:4/α + 4 ln[3] - (ln[20] + ln[30] + ln[60] + ln[90]) = 0.

€

⇒ α = 0.3775.Comment: This is Single Parameter Pareto Distribution, with θ = 3.

€

α̂ =

€

Nln[xi / θ]∑

=

€

4ln[20 / 3] + ln[30 / 3] + ln[60 / 3] + ln[90 / 3] = 0.3775.


5. You are given:

€

• A random variable X has an exponential distribution with mean θ.

€

• An estimator of θ has a mean of 2 and a variance of 0.5. Calculate the mean square error of this estimator for θ = 2.1. A. Less than 0.3 B. At least 0.3, but less than 0.4 C. At least 0.4, but less than 0.5 D. At least 0.5, but less than 0.6 E. At least 0.6

5. D. Bias = the expected value of the estimator minus the true value = 2 - 2.1 = -0.1.MSE = Variance + Bias2 = 0.5 + (-0.1)2 = 0.51.Comment: The first bullet in the question is not used.


6. A random variable, Y, with parameter θ is a member of the exponential family of distributions if its density function has the form:

f(y; θ) = exp[a(y)b(θ) + c(θ) + d(y)]. It is known that, for the exponential family of distributions,

E[a(Y)] = - c'(θ) / b'(θ).For one member of the exponential family, you are given, for y = 0, 1, 2, ... , n

€

• a(y) = y

€

• b(θ) = logθ

€

• c(θ) = -θ

€

• d(y) = -logy!

€

•

€

θ̂ is an unbiased estimator of θ

Find the smallest possible value of the variance of

€

θ̂ .

A. θ B. θ/n C. θ/

€

n D. θ2/n E. 0

6. B. ln f(y) = a(y)b(θ) + c(θ) + d(y).

€

∂ lnf(y)∂θ

= a(y) bʼ(θ) + cʼ(θ).

€

∂2 lnf(y)∂θ2 = a(y) bʼʼ(θ) + cʼʼ(θ).

E[

€

∂2 lnf(y)∂θ2 ] = E[a(y)] bʼʼ(θ) + cʼʼ(θ) = {- c'(θ) / b'(θ)} bʼʼ(θ) + cʼʼ(θ).

Here b(θ) = logθ. bʼ(θ) = 1/θ. bʼʼ(θ) = -1/θ2. c(θ) = -θ . cʼ(θ) = -1. cʼʼ(θ) = 0.

€

⇒ E[

€

∂2 lnf(y)∂θ2 ] = {-(-1) / (1/θ)} (-1/θ2) + 0 = -1/θ.

€

⇒ Rao-Cramer lower bound = -1 / {n (-1/θ)} = θ/n. Alternately, f(y; θ) = exp[a(y)b(θ) + c(θ) + d(y)] = exp[y logθ - θ - logy!] = e−θ θy / y!.This is a Poisson Distribution with mean θ.The method of moments is equal to maximum likelihood.

€

θ̂ =

€

X .

€

⇒ Var[

€

θ̂ ] = Var[

€

X ] = Var[X] / n = θ/n.Comment: The Rao-Cramer lower bound goes down as 1/n, eliminating choices A, C, and E.


7. You are given the following information:

€

• Loss severity follows an exponential distribution with mean θ.

€

• Small claims are handled directly by the insured, and no information about losses less than $5,000 is available to you.

€

•You observe three losses with the following claim payments: $6,000 $7,500 $10,000

Calculate the maximum likelihood estimate of θ. A. Less than 2,500 B. At least 2,500, but less than 3,000 C. At least 3,000, but less than 3,500 D. At least 3,500, but less than 4,000 E. At least 4,000

7. B. If we were to subtract the truncation point of 5000 from each of the given sizes of loss,we would have the payments if there had been a 5000 deductible: 1000, 2500, 5000.Then for the Exponential Distribution:

€

θ̂= (sum of the payments) / (uncensored values) = (1000 + 2500 + 5000) / 3 = 2833.

Alternately, f(x) = e-x/θ / θ. S(x) = e--x/θ.

Each observation contributes to the Iikelihood: f(x) / S(5000) = e-(x-5000)/θ / θ.ln[f(x) / S(5000)] = -(x-5000)/θ - ln[θ].Thus the loglikelihood is: -(6000 - 5000)/θ - ln[θ] - (7500 - 5000)/θ - ln[θ] - (10,000 - 5000)/θ - ln[θ] = -8500/θ - 3 ln[θ]. Set the partial derivative of the loglikelihood with respect to theta equal to zero:0 = 8500/θ2 - 3/θ.

€

⇒ θ = 8500/3 = 2833.


8. You are given the following random sample: 0.15 0.25 0.55 0.60 1.10

The probability density function given below is selected to be fit to the random sample:

f(x) =

€

1/ 2 for 0 ≤ x ≤ a1 for a ≤ x ≤ 1 + a / 2⎧ ⎨ ⎩

where 0 ≤ a ≤ 2.

Select the range of the maximum likelihood estimate of a. A. Less than 0.15 B. At least 0.15, but less than 0.20 C. At least 0.20, but less than 0.25 D. At least 0.25, but less than 0.30 E. At least 0.30

8. C. In order to get the biggest Iikelihood, we want as many of the observations to be in the second interval. Therefore we want “a” to be as small as possible.However, since we observe 1.10, we must have: 1.10 ≤ 1 + a/2.

€

⇒ a ≥ 0.2.If 0.2. ≤ a < 0.25, then the likelihood is: (1/2)(14) = 1/2, which is the maximum Iikelihood.Comment: Since the parameter is part of the support of the density, one has to be careful.If a = 0.25, then the likelihood is: (1/22)(13) = 1/4.If a < 0.2, then the likelihood is 0.



€

• A random variable X is uniformly distributed on the interval (0, θ).

€

• θ is unknown.

€

• For a random sample of size n, an estimate of θ is Yn = max[X1, X2, ... , Xn].

€

• E(Yn) = [n/(n+1)] θDetermine which one of the following is an unbiased estimator of θ.

A. Yn B.

€

1n- 1 Yn C.

€

n2 Yn D.

€

n+ 1n Yn E.

€

1n+ 1 Yn

9. D. E[

€

n+ 1n Yn] =

€

n+ 1n {n/(n+1)} θ = θ.

Thus

€

n+ 1n Yn is an unbiased estimator of θ.

Comment: With the final bullet, this question should be very easy.


10. Some people are right-hand dominant; others are left-hand dominant. The same is true for eyes. You are given the following table of observed values:

Dominant Hand Dominant Eye Right Hand Left Hand Total Right Eye 8402 207 8609 Left Eye 1308 104 1412 Total 9710 311 10,021

€

• H0: Eye and hand dominance are independent

€

• HA: Eye and hand dominance are not independent Calculate the chi-squared statistic. A. Less than 60 B. At least 60, but less than 70 C. At least 70, but less than 80 D. At least 80, but less than 90 E. At least 90

10. E. Here is the calculation of the contributions to the Chi-Square Statistic of 99.28:Right Hand Left Hand Sum

Right Eye 8402 207 8609Left Eye 1308 104 1412

Sum 9710 311 10021

Expected Number Expected Number SumRight Eye 8341.82 267.18 8609.00Left Eye 1368.18 43.82 1412.00

Sum 9710.00 311.00 10021.00

Contribution to Chi-Square Contribution to Chi-Square SumRight Eye 0.43 13.55 13.99Left Eye 2.65 82.64 85.29

Sum 3.08 96.20 99.28For example, in the first row and first column, the expected number is: (9710)(8609)/10,021 = 8341.82.The contribution from the first row and first column is: (8402 - 8341.82)2 / 8341.82 = 0.43.Comment: There is: (2 - 1)(2 - 1) = 1 degree of freedom.Using a computer, the p-value is less than 10-10.



€

• A random variable follows the normal distribution with mean µ and standard deviation 3.

€

• H0: µ = 8.

€

• H1: µ = 7.

€

• It is required that errors of Type I and II have probabilities of 0.05 and 0.01, respectively. Calculate that necessary sample size based on the Neyman-Pearson Lemma. A. Less than 50 B. At least 50, but less than 100 C. At least 100, but less than 150 D. At least 150, but less than 200 E. At least 200

11. C. With a known variance, the best critical region from the Neyman-Pearson Lemma is the usual critical region for a one-sided Normal test.

0.05 = Prob[Type I error] = Prob[reject H0 | H0] = Prob[

€

X < c | µ = 8] = Φ[

€

c - 83 / n

].

€

⇒ -1.645 =

€

c - 83 / n

.

0.01 = Prob[Type II error] = Prob[fail to reject H0 | H1] = Prob[

€

X < c | µ = 7] = 1 - Φ[

€

c - 73 / n

].

€

⇒ 2.326 =

€

c - 73 / n

.

Dividing the two equations: (c-8) / (c-7) = -1.645/2.326.

€

⇒ c = 7.5857.

€

⇒

€

n = (3)(2.326)/(7.5857 - 7) = 11.914.

€

⇒ n = 141.9.

€

⇒ n is at least 142.



€

• Your company has developed a new training program for claims adjusters to increase their productivity.

€

• Assume that the number of claims closed follows a normal distribution, with equal variance before and after the training program.

€

• H0: the number of claims closed is the same before and after training.

€

• H1: the number of claims closed is not the same before and after training. The table below shows the paired results (before and after training) by adjuster.

Adjuster ID # Claims Closed Before Training # Claims Closed After Training 1 7 5 2 6 3 3 7 7 4 8 12 5 4 10 6 7 13 7 5 10 8 7 14 9 8 6 10 2 6 TOTAL 61 86

Calculate the minimum significance level at which you reject H0. A. Less than 0.1% B. At least 0.1%, but less than 1.0% C. At least 1.0%, but less than 2.0% D. At least 2.0%, but less than 5.0% E. At least 5.0%


12. E. Take the differences, after training minus before training: -2, -3, 0, 4, 6, 6, 5, 7, -2, 4.The mean of these differences is (86 - 61)/10 = 2.5, with sample variance of 14.772.

t =

€

2.514.772 /10

= 2.060 with 10 - 1 = 9 degrees of freedom.

1.833 < 2.060 < 2.262. For a two-sided test: reject at 10%, but not at 5%.Comment: The given choices such 0.1% do not make sense for the t-table attached to the exam.If one took the differences in the opposite order from what I did, one gets minus my t-statistic, but would arrive at the same conclusion for the t-test.While the language of the setup might lead one to believe we would perform a one-sided test, the given form of H1 corresponds to a two-sided test.A one-sided test would have had insteadH1: the number of claims closed is greater after training than before training. The language for the null and alternative hypothesis would have been better to have been:

€

• H0: the expected number of claims closed is the same before and after training.

€

• H1: the expected number of claims closed is not the same before and after training.


13. You are given the following information about a process that follows the normal distribution:

€

• The variance is known and σ2 = 25.

€

• H0: µ = 35.

€

• H1: µ = 30.

€

• The sample size, n, is equal to 16. Determine the minimum possible Type I error such that the probability of Type II error is no more than 2.5%. A. Less than 2% B. At least 2%, but less than 3% C. At least 3%, but less than 4% D. At least 4%, but less than 5% E. At least 5%

13. B. 2.5% = Prob[Type II error] = Prob[Fail to reject | H1] = 1 - Φ[

€

c - 3025 /16

] .

€

⇒

€

c - 305 / 4 = 1.960.

€

⇒ c = 32.45.

Prob[Type I error] = Prob[reject | H0] = Φ[

€

32.45 - 355 / 4 ] = Φ[-2.04] = 2.07%.


14. You are given the following information about two loss severity distributions fit to a sample of 275 closed claims:

€

• For the Exponential distribution, the natural logarithm of the likelihood function evaluated at the maximum likelihood estimate is -828.37.

€

• For the Weibull distribution, the natural logarithm of the likelihood function evaluated at the maximum likelihood estimate is -826.23.

€

• The Exponential distribution is a subset of the Weibull distribution.

€

• The null hypothesis is that the exponential distribution provides a better fit than the Weibull distribution.

Calculate the significance level at which one would reject the null hypothesis. A. Less than 0.5% B. At least 0.5%, but less than 1.0% C. At least 1.0%, but less than 2.5% D. At least 2.5%, but less than 5.0% E. At least 5.0%

14. D. Likelihood Ratio Test Statistic is: (2) {-826.23 - (-828.37)} = 4.28.Since there is difference of one parameter between the Weibull and Exponential, we compare to the Chi-Square Distribution with one degree of freedom.3.84 < 4.28 < 5.02. Thus we reject at 5% and not at 2.5%.



€

• f(x; θ) is the pdf of X, where θ represents one or more unknown parameters θ.

€

•

€

Ω is the set of all possible parameters for θ.

€

• H0: θ

€

∈ ω is a subset of

€

Ω.

€

• H1: θ

€

∈

€

Ω.

€

• λ. =

€

L(ω̂ )L(Ω̂)

, where the numerator is the maximum likelihood function with respect to θ

under the null hypothesis . Determine which of the following are true. I. λ ≤ 1 II. λ can be less than zeroIII. As λ increases the likelihood of rejecting the null hypothesis increases. A. I only B. II only C. I and II only D. I and IIl only E. II and III only

15. A. The maximum likelihood of the special case is less than or equal to that for the more general case. Since the likelihood of the special case is in the numerator, the ratio is ≤ 1. Statement 1 is true.Likelihoods are a product of densities and are thus positive (or zero). Thus Statement 2 is false.We reject H0 when the maximum likelihood of the more general case is large compared to the special case. This occurs when the given ratio is small. Thus Statement 3 is false.


16. Ten runners were ranked by judges on their speed (1 = fastest) prior to a race in which they then recorded the following times:

Runner Rank Time (min:sec)1 4 4:45 2 5 4:50 3 1 4:31 4 6 4:49 5 9 5:05 6 8 4:55 7 7 5:30 8 3 4:20 9 2 4:25 10 10 5:55

Calculate Spearman's Rank statistic, rs, for this data. A. Less than -0.75 B. At least -0.75, but less than -0.25 C. At least -0.25, but less than 0.25 D. At least 0.25, but less than 0.75 E. At least 0.75

16. E. Rank the finishing times:Runner Judges Rank Prior to Race Time Rank of Finishing Time1 4 4:45 42 5 4:50 63 1 4:31 34 6 4:49 55 9 5:05 86 8 4:55 77 7 5:30 98 3 4:20 19 2 4:25 210 10 5:55 10

Using the stat functions of the electronic calculator, determine the correlation between the ranks given by the judges prior to the race and the ranks of the finishing times: 0.903.


17. For a general liability policy, loss amounts, Y, follow the exponential distribution with probability density function:

f(y) =

€

1θ

e-y/θ θ = 1000, 0 < y.

For reinsurance purposes we are interested in the distribution of the median loss amount in a random sample of size 3, which is denoted by Y(2). Calculate the probability that Y(2) is less than 2000. A. At least 0.73, but less than 0.78 B. At least 0.78, but less than 0.83 C. At least 0.83, but less than 0.88 D. At least 0.88, but less than 0.93 E. At least 0.93

17. E. For the Exponential Distribution, F(2000) = 1 - e-2000/1000 = 0.8647.Prob[Y(2) < 2000] = Prob[exactly 2 values < 2000] + Prob[3 values < 2000] =

3F(2000)2S(2000) + F(2000)3 = (3)(0.86472)(1 - 0.8647) + 0.86473 = 95.00%.


18. You are given:

€

• A tutor claims that students will get more than 89.5 out of 100 questions right on a particular exam if the students use the tutor's service for 15 minutes a day.

€

• A random sample of 200 previous students was taken.

€

• No student scored exactly 89.5.

€

• H0: Students using the service have a mean score of 89.5 on the exam.

€

• H1: Students using the service have a mean score greater than 89.5 on the exam.

€

• H0 was rejected at the 5% significance level using the sign test.

€

• The normal approximation with continuity adjustment is used. Calculate the minimum number of students that scored more than 89.5 in the sample. A. 111 or fewer B. 112 C. 113 D. 114 E. 115 or more

18. C. Assuming the null hypothesis is true, the number of students with scores greater than 89.5 is Binomial with m = 200 and q = 1/2. Mean is 100 and variance is: (1/2)(1/2)(200) = 50.Assume that x students are observed to have scores greater than 89.5.For a one-sided test, using the normal approximation with continuity adjustment we want:

5% = 1 - Φ[

€

x - 1/ 2 - 10050

].

€

⇒

€

x - 1/ 2 - 10050

= 1.645.

€

⇒ x = 100.5 + 1.645

€

50 = 112.13.

€

⇒ The minimum number of students that scored more than 89.5 in the sample is 113.Comment: This is equivalent to the Sign Test in “Mahlerʼs Guide to Nonparametric Tests”; the hypotheses should have referred to the median rather than the mean.


19. A gun is fired using two different brands of bullets. • The results of the tests are provided

Bullet Brand A Bullet Brand B Muzzle Velocity Rank Muzzle Velocity Rank

1011 1 1014 4 1012 2 1015 5 1013 3 1016 6 1020 10 1017 7 1021 11 1018 8 1029 12 1019 9 1030 13 1050 19 1031 14 1051 20 1032 15 1052 21 1035 16 1055 22 1040 17 1045 18

€

• H0: MedianBrandA = MedianBrandB.

€

• H1: MedianBrandA ≠ MedianBrandB. • The normal approximation is used. Calculate the smallest value for significance level for which H0 can be rejected using the Mann-Whitney Test. A. Less than 1 % B. At least 1 %, but less than 5% C. At least 5%, but less than 10% D. At least 10%, but less than 15% E. At least 15%


19. E. Sum of the ranks for the first sample is 132.Mean for rank sum for the first sample (of size n1) = n1(n1+n2+1)/2 = (12)(10 + 12 + 1)/2 = 138. Variance of rank sum for the first sample (of size n1) = n1n2(n1+n2+1)/12 = (12)(10)(23)/12 = 230.

The probability value for the two sided test is: 2 Φ[

€

132 - 138230

] = 2 Φ[-0.396] = 69.2%.

Alternately, using the continuity correction, the p-value is:

2 Φ[

€

132.5 - 138230

] = 2 Φ[-0.363] = 71.7%.

Alternately, the sum of the ranks for the second sample is 121.Mean for rank sum for the second sample is: (10)(12 + 10 + 1)/2 = 115. Variance of rank sum for the second sample is: (10)(12)(23)/12 = 230.

The probability value for the two sided test is: 2 {1 - Φ[

€

121 - 115230

]} = 2 { 1 - Φ[0.396]} = 69.2%.

Comment: The choices in the question are all much smaller than the answer.


20. You are given:

€

• A multiple linear regression model was fit to 20 observations.

€

• There are five explanatory variables with fitted values for β1 through β5.

€

• The following summarizes the fitted coefficients excluding the intercept: Point Estimate Standard Error t-statistic

β1 0.020000 0.012000 1.661 β2 -0.004950 0.008750 -0.565 β3 0.216000 0.043200 5.000 β4 -0.034600 0.115000 -0.301 β5 -0.000294 0.000141 -2.090

Determine the number of coefficients in the table above for the five explanatory variables that are not statistically different from zero at a significance level of α = 10%, based on a two-tailed test. A. 1 B. 2 C. 3 D.4 E. 5

20. C. Including the intercept, we fit 6 coefficients. Thus the t-stats have 20 - 6 -14 degrees of freedom.From the table attached to the exam, the two-sided 10% critical vale is 1.761.We do not reject the null hypothesis when the absolute value of the t-statistic is less than 1.761.There are 3 such cases: β1, β2, β4.Comment: t-statistic = (point estimate) / (standard error).Thus, the first t-statistic should be: 0.020000 / 0.012000 = 1.667.


21. You are given the following linear regression model which is fitted to 11 observations: Y = β0 + β1X + ε

The coefficient of determination is R2 = 0.25. Calculate the F-statistic used to test for a linear relationship. A. Less than 1.5 B. At least 1.5, but less than 2.5 C. At least 2.5, but less than 3.5 D. At least 3.5, but less than 4.5 E. At least 4.5

21. C. F =

€

R2

1 - R2

€

N - k - 1k =

€

0.250.75

€

11 - 21 = 3, with 1 and 9 degrees of freedom.

Comment: Using a computer the p-value is 11.7%.


22. The following two models were fit to 18 observations:

€

• Model 1: Y = β0 + β1X1 + β2X2 + ε

€

• Model 2: Y = β0 + β1X1 + β2X2 + β3X1X2 + β4X12 + β5X22 + ε The results of the regression are:

Model Number Error Sum of Squares Regression Sum of Squares 1 102 23 2 78 39

Calculate the value of the F-statistic used to test the hypothesis that β3 = β4 = β5 = 0. A. Less than 1.30 B. At least 1.30, but less than 1.40 C. At least 1.40, but less than 1.50 D. At least 1.50, but less than 1.60 E. At least 1.60

22. A. F =

€

(ESSR - ESSUR) / qESSUR / (N- k -1) =

€

(102 - 78) / 378 / (18 - 6) = 1.231, with 3 and 12 degrees of freedom.

Alternately, R12 = 23/(102 + 23) = 0.1840. R22 = 39/(78 + 39) = 0.3333.

F =

€

(RUR2 - RR2) / q(1 - RUR2)/ (N - k - 1) =

€

(0.3333 - 0.1840) / 3(1 - 0.3333) / (18 - 6) = 0.896, with 3 and 12 d.f.

Comment: Since the two models were fit to the same data, TSS = ESS + RSS should be the same. However, 102 + 23 = 125 ≠ 117 = 78 + 39. This is why the two different formulas for F produce different answers.For F = 1.231, using a computer the p-value is 34.1%.



€

• The number of claims per year for an individual follows a Poisson distribution with parameter λ.

€

• The parameter λ follows a gamma distribution with parameters α = 1 and θ = 5.

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• A randomly selected insured had 0 claims last year. Calculate the upper bound of the 95% equal-tailed Bayesian credible interval for the posterior distribution of λ. A. Less than 2.2 B. At least 2.2, but less than 2.9 C. At least 2.9, but less than 3.6 D. At least 3.6, but less than 4.3 E. At least 4.3

23. C. Posterior Gamma has parameters: αʼ = α + C = 1 + 0 = 1.1/θʼ = 1/θ + E = 1/5 + 1 = 6/5.

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⇒ θʼ = 5/6.Thus the posterior Distribution is Exponential with mean 5/6.For the upper bound, we want S(x) = 2.5%.

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⇒ 0.025 = exp[-x6/5].

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⇒ x = 3.074.Comment: The lower bound is where F(x) = 2.5%.

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⇒ 0.025 = 1 - exp[-x6/5].

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⇒ x = 0.0211.


24. The annual number of claims for a policy follows the Poisson distribution with mean λ, which varies by policy. The prior probability density function of λ is

h(λ) = λe−λ, λ > 0 In Year 1, three claims are reported for a policy. Calculate the coefficient of variation (the standard deviation divided by the mean) of the posterior distribution of λ for this policy. A. Less than 0.5 B. At least 0.5, but less than 1.0 C. At least 1.0, but less than 1.5 D. At least 1.5, but less than 2.0 E. More than 2.0

24. A. Prior distribution is Gamma with α = 2 and θ = 1.Posterior Gamma has parameters: αʼ = α + C = 2 + 3 = 5.1/θʼ = 1/θ + E = 1/1 + 1 = 2.

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⇒ θʼ = 1/2.Thus the posterior Distribution is Exponential with mean 5/6.

Coefficient of variation is:

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αθ2

αθ =

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1α

=

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15

= 0.447.


25. You are given:

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• The number of claims per year for a policyholder follows a Poisson distribution with mean λ.

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• The mean λ follows a gamma distribution with parameters α = 3 and θ = 10.

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• A policyholder had 1 claim last year. Calculate the mean of the posterior distribution of λ. A. Less than 2.0 B. At least 2.0, but less than 4.0 C. At least 4.0, but less than 6.0 D. At least 6.0, but less than 8.0 E. At least 8.0

25. B. Posterior Gamma has parameters: αʼ = α + C = 3 + 1 = 4.1/θʼ = 1/θ + E = 1/10 + 1 = 11/10.

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⇒ θʼ = 10/11.Posterior mean is: αθ = (4)(10/11) = 3.636.Alternately, using Buhlmann Credibility, K = 1/θ = 1/10.Z = 1/(1 + K) = 1/(1 + 1/10) = 10/11.Observed mean is 1. Prior mean is: (3)(10) = 30.Estimated future frequency is: (10/11)(1) + (1 - 10/11)(30) = 3.636.

END OF EXAMINATION


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