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Solutions to the Fall 2015

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 2015 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 2015 by Howard C. Mahler.

Study Aid 2015-ST-5BHoward [email protected]/Teaching

1. For two Poisson processes, N1 and N2, you are given:

�

• N1 has intensity function λ1(t) =

�

2t for 0 < t ≤ 1t3 for t > 1

⎧ ⎨ ⎩

�

• N2 is a homogenous Poisson process.

�

• Var[N1(3)] = 4 Var[N2(3)] Calculate the intensity of N2 at t = 3. A. Less than 1 B. At least 1, but less than 3 C. At least 3, but less than 5 D. At least 5, but less than 7 E. At least 7

1. B. m1(3) =

�

2t dt0

1

∫ +

�

t3 dt1

3

∫ = 1 + (34 - 14)/4 = 21.

Thus, N1(3) is Poisson with mean 21, and thus variance 21.

N2(3) is Poisson with mean 3λ, and thus variance 3λ.We are given that Var[N1(3)] = 4 Var[N2(3)].

�

⇒ 21 = (4)(3λ).

�

⇒ λ = 21/12 = 1.75.Comment: Since N2 is homogeneous, its intensity is the same at t = 3 as at any other time.

HCMSA-2015-ST-5B Solutions Fall 2015 CAS Exam ST, 11/12/15, Page 1

2. For a health insurance policy, the annual number of claims follows the Poisson process with the mean, λ, equal to 5. Calculate the probability that the third claim occurs after one year, that is Pr(T3 > 1). A. Less than 0.10 B. At least 0.10, but less than 0.11 C. At least 0.11, but less than 0.12 D. At least 0.12, but less than 0.13 E. At least 0.13

2. D. The number of claims by time 1 is Poisson with mean 5.Prob[third claim occurs after one year] = Prob[at most 2 claims by time 1] =f(0) + f(1) + f(2) = e-5 (1 + 5 + 52/2) = 12.47%.


3. The number of customers who use an ATM for withdrawals follows the Poisson process with the rate equal to 100 per day. The amounts withdrawn are distributed as follows:

Amount Withdrawn Probability $20 0.25 $40 0.50$50 0.10 $100 0.15

Calculate the standard deviation of the total daily withdrawals. A. Less than $500 B. At least $500, but less than $510 C. At least $510, but less than $520 D. At least $520, but less than $530 E. At least $530

3. C. The second moment of the amount distribution is:(0.25)(202) + (0.50)(402) + (0.10)(502) + (0.15)(1002) = 2650.Variance of this Compound Poisson is: λ E[X2] = (100)(2650) = 265,000.

Standard deviation of the total daily withdrawals is:

�

265,000 = 514.8.


4. You are given the following:

�

• X is a uniformly distributed random variable with the probability density function: fX(x) = 1/θ 0 ≤ x ≤ θ.

�

•

�

θ̂ = c Xmax where Xmax is the largest Xi from a sample of size n.

Determine the value of c that makes

�

θ̂ an unbiased estimator of θ when n = 100.

A. Less than 1.00 B. At least 1.00, but less than 1.02 C. At least 1.02, but less than 1.04 D. At least 1.04, but less than 1.06 E. At least 1.06

4. B. The expected value of the ith order statistic of a sample of size n from a uniform distribution from 0 to θ is: θ i/(n+1).

�

⇒ E[Xmax] = θ 100/101.

We want θ = E[

�

θ̂ ] = c E[Xmax] = c θ 100/101.

�

⇒ c = 101/100 = 1.01.


5. You are given:

�

• An insurance product with a per loss limit of 200 covers losses from an exponential distribution with parameter θ.

�

• Based on the following table, the maximum likelihood estimate of θ is 168. Size of Loss Number of Claims Sum of Losses Less than 200 1,114 142,752 At least 200 N 200 x N

Calculate N for the table above. A. Less than 250 B. At least 250, but less than 550 C. At least 550, but less than 850 D. At least 850, but less than 1,150 E. At least 1,150

5. A. For the Exponential Distribution

�

θ̂ =

�

sum of the paymentsnumber of uncensored values .

Thus, 168 = (142,752 + 200N) / 1114.

�

⇒ N = 222. Comment: The data has been censored from above at 200.


6. You are given two independent samples from an exponential distribution with mean, θ. The samples have three and ten observations, respectively. The relative efficiency of the first sample to the second sample is the ratio of the lower bound of the variance of the estimator of the mean, θ, for the second sample to the lower bound of the variance of

the estimator of the mean, θ, for first sample. Calculate the relative efficiency of the first sample to the second one using the Rao-Cramer Lower Bound. A. Less than 20% B. At least 20%, but less than 40% C. At least 40%, but less than 60% D. At least 60%, but less than 80% E. At least 80% Note: I have rewritten this past exam question.

6. B. The Rao-Cramer lower bound is inversely proportional to the sample size.Thus the relative efficiency of the first sample to the second sample is: (1/10) / (1/3) = 0.3 Comment: The original exam question read: “The relative efficiency is the ratio of the lower bound of the variance of the estimator the mean, θ, for the first sample to the lower bound of the variance of

the estimator the mean, θ , for the second sample.” This is the reciprocal of the definition of the relative efficiency of the first sample to the second sample as in Section 9.2 of Wackerly, Mendenhall, and Scheaffer. For two unbiased estimators of the same quantity, define the relative

efficiency of estimator B compared to estimator A as

�

Variance[A]Variance[B]. This would be typically applied

to compare two different estimators applied to the same data, rather than as in this exam question to compare a single estimator applied to two data sets of different sizes.The Rao-Cramer lower bound for an unbiased estimator of the mean of the Exponential is: θ2/n,

which is in fact the variance of the maximum likelihood estimator of θ.

The maximum Iikelihood estimator of θ, which is the same as the method of moments estimator of θ, is more efficient when applied to the second larger sample than when applied to the first smaller sample; smaller variance corresponds to greater efficiency. Therefore, the relative efficiency of the estimator when applied to the first sample to the estimator when applied the second sample is less than 1.


7. You are given:

�

• Assume losses follow a Single-parameter Pareto distribution.

�

• θ = 150

�

• Losses given are 200, 300, 350, 400. Find the Maximum Likelihood Estimate of α.A. Less than 0.8 B. At least 0.8, but less than 1.2 C. At least 1.2, but less than 1.6 D. At least 1.6, but less than 2.0 E. At least 2.0

7. C. For a Single Parameter Pareto, with θ = 150, f(x) = α 150α / xα+1.

Σ ln f(xi) = Σ ln(α) + α ln(150) - (α+1)ln(xi).

�

∂ ln[f(xi)]∑∂α

= Σ {1/α + ln(150) - ln(xi)} = N/α - Σln(xi/150).

Set equal to zero the partial derivative with respect to α of the loglikelihood: N/α - Σln(xi/150) = 0.

⇒ α = N / Σln(xi / 150) = 4/ {ln(2/1.5) + ln(2) + ln(3.5/1.5) + ln(4/1.5)} = 1.424.

Comment: For a Single Parameter Pareto Distribution, the maximum likelihood fit: α =

�

Nln[xi / θ]∑

.


8. Let Y1, Y2, ..., Yn be a random sample of size n from a distribution with the probability density function.

f(y | θ) =

�

2yθ

exp[-y2 / θ], 0 ≤ y < ∞0, elsewhere

⎧ ⎨ ⎪

⎩ ⎪

Determine which of the following formulas is a sufficient statistic for θ using the factorization criterion.

A.

�

Yii=1

n

∑ B.

�

Yii=1

n

∑ 2 C.

�

Yii=1

n

∑ 4 D.

�

exp[Yi]i=1

n

∑E. The distribution has no sufficient statistic for θ.

8. B. The likelihood is:

�

f(Yi)i=1

n

∏ =

�

2Yiθ

exp[-Yi2 / θ]i=1

n

∏ = {exp[-

�

Yii=1

n

∑ 2 / θ] / θn} {2n

�

Yii=1

n

∏ }.

We have factored the likelihood into 2 terms, the second of which is not a function of the parameter θ.

The first term is a function of θ and

�

Yii=1

n

∑ 2 .

Thus

�

Yii=1

n

∑ 2 is sufficient statistic for θ.

Comment: The given density is a Weibull Distribution with τ = 2, parameterized somewhat differently than in Appendix A attached to the exam.Rewriting the given density as f(x) = 2x exp[-x2 / β] / β, this is a Weibull with τ = 2 and β = θ2.

The maximum Iikelihood fit is:

�

θ̂ =

�

xiτ / n∑( )1 / τ =

�

xi2 / n∑( )1/ 2.

�

⇒

�

β^ =

�

xi2 / n∑ .

As it has to be, the maximum likelihood fit of beta is a function of its sufficient statistic.


9. Let X1, X2, ..., Xn, be a random sample of independent and identically distributed observations from a population with probability density function

f(x | θ) =

�

x3

6θ4 exp[-x / θ], 0 ≤ x < ∞0, elsewhere

⎧ ⎨ ⎪

⎩ ⎪

Determine the maximum likelihood estimator of θ.

A.

�

x B.

�

x /4 C.

�

x 2 D. 3

�

x 2 E.

�

n

3 / xii=1

n

∑

9. B. This is a Gamma Distribution with α = 4.Maximum Likelihood is equal to the Method of Moments.4 θ =

�

X .

�

⇒ θ =

�

X /4.

Alternately, ln f(xi) = 3 ln(xi) - ln(6) - 4 ln(θ) - xi/θ.

0 =

�

∂ ln[f(xi)]∑∂θ

= -4n / θ + Σxi / θ2.

�

⇒ θ = (Σxi/n) / 4 =

�

X /4.


10. You are given:

�

• A random sample of size n is observed from a Pareto distribution with parameters θ = 20 and

unknown α.

�

• H0: α = 4.

�

• H1: α = 6. Using the Neyman-Pearson Lemma, determine the form of the best critical region.

A.

�

ln(xi)∑ ≥ c B.

�

ln(xi)∑ ≤ c C.

�

x ≥ c

D.

�

ln(xi + 20)∑ ≥ c E.

�

ln(xi + 20)∑ ≤ c

10. E. f(x) = α 20α / (x + 20)α+1. ln f(x) = ln(α) + α ln(20) - (α+1)ln(x+20).

The loglikelihood is: n ln(α) + n α ln(20) - (α+1)

�

ln(xi + 20)∑ .

The best critical region is of the form:(loglikelihood for H1) - (loglikelihood for H0) ≥ k, for some constant k.

n ln(6) + n 6 ln(20) - 7

�

ln(xi + 20)∑ - {n ln(4) + n 4 ln(20) - 5

�

ln(xi + 20)∑ } ≥ k.

�

⇔

-2

�

ln(xi + 20)∑ ≥ k + n ln(4) - n ln(6) - 2 n ln(20) = b.

�

⇔

�

ln(xi + 20)∑ ≤ b/(-2) = c.

The best critical region is of the form:

�

ln(xi + 20)∑ ≤ c.

Comment: Note that the direction of the inequality reverses when one divides both sides by a negative number.The critical region, where we reject H0 in favor of H1, is when the loglikelihood for H1: α = 6 is big

compared to the loglikelihood for H0: α = 4. This occurs when

�

ln(xi + 20)∑ is smaller.

If instead H1: α = 2, then the best critical region is of the form:

�

ln(xi + 20)∑ ≥ c.


11. You are given:

�

• A sample of size n is observed from a normal distribution with variance 64.

�

• H0: µ = 2.

�

• H1: µ = 6.

�

• H0 is rejected when

�

X ≥ 5.25. Calculate the smallest sample size, n, needed so that the probability of a Type II error is less than 0.10. A. Less than 150 B. At least 150, but less than 200 C. At least 200, but less than 250 D. At least 250, but less than 300 E. At least 300

11. B. 10% = Prob[Type II Error] = Prob[Fail to reject when µ = 6] =

= 1 - Prob[

�

X ≥ 5.25 when µ = 6] = Φ[

�

5.25 - 664 / n

].

�

⇒

�

5.25 - 664 / n

= -1.282.

�

⇒ n = 187.


12. You are given:

�

• 100 decks of 52 cards are combined and shuffled to put the cards in a random order.

�

• Each deck contains only red and black cards.

�

• The first 4 cards are revealed to be all red cards.

�

• H0: 50% of the cards are red.

�

• HA: More than 50% of the cards are red. Calculate the smallest significance level for which H0 can be rejected. A. Less than 0.005 B. At least 0.005, but less than 0.010 C. At least 0.010, but less than 0.050 D. At least 0.050, but less than 0.100 E. At least 0.100

12. D. For H0 true, the probability of the observation (or more extreme) is approximately:

(1/2)4 = 1/16 = 0.625.Comment: For H0 true, given there are a total of 5200 cards, the chance of the observation is:(1/2) (2599/5199) (2598/5198) (2597/5197) = 0.06243.


13. You are given:

�

• The following table of values for samples A and B:

Number of Observations

�

1n x∑

�

1n (x - x )2∑

Sample A 11 30 5 Sample B 17 30 6

�

• H0: σA2 = σB2

�

• HA: σA2 > σB2

�

• Samples A and B are Normally distributed Calculate the smallest p-value for which H0 can be rejected. A. Less than 0.010 B. At least 0.010, but less than 0.025 C. At least 0.025, but less than 0.050 D. At least 0.050, but less than 0.100 E. At least 0.100

13. E. The sample variance of A is: (5)(11/10) = 5.5.The sample variance of B is: (6)(17/16) = 6.375.The F-Statistic is: 5.5/6.375 = 0.8627, with 10 and 16 degrees of freedom.From the table, the 10% critical value is 2.208.Since 0.8627 < 2.028, we do not reject at 10%.Comment: “Calculate the smallest p-value for which H0 can be rejected” is not correct language.It should have read either “Calculate the smallest significance level for which H0 can be rejected” or“Calculate the p-value.”Since 0.8627 < 1.580, we do not reject at 20%.Using a computer, the p-value is 0.582.I suspect that this exam question was somehow not presented as intended.


14. You observe the following distribution for a random sample of 100 claims. Layer Number of Claims Less than 5,000 31 Between 5,000 and 10,000 19 Between 10,000 and 15,000 26 Greater than 15,000 24

You are testing the null hypothesis that claim severity follows a Weibull distribution with θ = 10,000

and τ = 1.1, against the alternative hypothesis that claim severity does not follow the distribution specified in the null hypothesis. Using the chi-square goodness-of-fit test to evaluate the null hypothesis, what is the conclusion? A. Do not reject the null hypothesis at the 5.0% significance level B. Reject the null hypothesis at the 5.0% significance level, but not at the 2.5% significance level C. Reject the null hypothesis at the 2.5% significance level, but not at the 1.0% significance level D. Reject the null hypothesis at the 1.0% significance level, but not at the 0.5% significance level E. Reject the null hypothesis at the 0.5% significance level

14. C. For the Weibull, S(x) = exp[-(x/10,000)1.1].S(5000) = exp[-(5000/10,000)1.1] = 0.6272.S(10,000) = exp[-(10,000/10,000)1.1] = 0.3679.S(15,000) = exp[-(15,000/10,000)1.1] = 0.2097.Thus the expected numbers of claims are: (100)(1 - 0.6272) = 37.28. (100)(0.6272 - 0.3679) = 25.93. (100)(0.3679 - 0.2097) = 15.82. (100)(0.2097) = 20.97.Chi-Square Statistic is:

�

(31 - 37.28)237.28 +

�

(19 - 25.93)225.93 +

�

(26 - 15.82)215.82 +

�

(24 - 20.97)220.97 = 9.605.

There are 4 - 1 = 3 degrees of freedom.Since 9.35 < 9.605 < 11.34, reject at 2.5% but not at 1%.Comment: Using a computer, the p-value is 2.2%.


15. You are given that a random variable follows one of two discrete probability distribution functions where each categorical value from I to XII is assigned a probability. The two probability distribution functions are displayed in the table below:

I II III IV V VI VII VIII IX X XI XII D0 0.025 0.025 0.050 0.075 0.125 0.200 0.200 0.125 0.075 0.050 0.025 0.025 D1 0.200 0.125 0.075 0.050 0.025 0.025 0.025 0.025 0.050 0.075 0.125 0.200

�

• H0: D0 is the probability distribution function

�

• H1: D1 is the probability distribution function

�

• The significance level is set at 10%. Determine the combination of values that defines the critical region which provides the most powerful test. A. I + II + III B. I + II + X C. I + II + XI + XII D. III + X E. X + XI + XII

15. C. Using the Likelihood Ratio Test, we reject when the ratio Prob[D1] / Prob[D0] is large.This ratio is largest, 8, for I and XII.The significance level of a critical region consisting of I and XII is: 2.5% + 2.5 % = 5% < 10%.The next largest ratio, 5, is for II and XI.The significance level of a critical region consisting of I, II, XI, and XII is: (4)(2.5%) = 10%.Comment: The significance level is the probability of the observation assuming H0 is true,so we add up the probabilities for D0 for the possibilities in the critical region.


16. You are given the following information:

�

• A random sample consisting of n observations is taken from a normal distribution with mean µ and valiance 1000.

�

• The sample mean is 3.5.

�

• H0: µ = 0.

�

• H1: µ ≠ 0.

�

• A 90% confidence interval calculated from this sample is (-2.109, 9.109) for the sample mean using the Central Limit Theorem.

Calculate the smallest significance level at which the null hypothesis would be rejected. A. Less than 10% B. At least 10%, but less than 18% C. At least 18%, but less than 26% D. At least 26%, but less than 34% E. At least 34%

16. D. The given 90% confidence interval is: 3.5 ± 5.609.Thus 5.609 = (1.645) (Standard Deviation of

�

X ).

�

⇒ Standard Deviation of

�

X = 3.4097.

p-value is: 2{1 - Φ[

�

3.5 - 03.4097 ]} = 2{1 - Φ[1.026]} = 30.5%.

Comment: 5.609 = 1.645

�

1000n .

�

⇒ n = 86.


17. You are given a random sample of 7 observations from a continuous distribution: 1.2 2.6 3.4 4.5 5.0 5.2 8.1

The following hypothesis test is performed using the Sign Test:

�

• H0: median = 6.0

�

• H1: median < 6.0 Calculate the p-value of this test using the exact calculation. A. Less than 5% B. At least 5%, but less than 7.5% C. At least 7.5%, but less than 10% D. At least 10%, but less than 12.5% E. At least 12.5%

17. B. 6 out of 7 values are less than 6.0.p-value is sum of the Binomial Densities for m = 7 and q = 1/2 at 6 and 7:7/27 + 1/27 = 0.0625.


18. You are given the following information:

�

• For a general liability policy loss amounts, Y, follow the Pareto distribution with probability density function:

f(y) = 3 θ3 (y + θ)-4, θ = 1000, 0 < y

�

• For reinsurance purposes we are interested in the distribution of the largest loss amount in a random sample of size 10, which is denoted by Y10. Calculate the 90th percentile of Y10. A. Less than 3,500 B. At least 3,500, but less than 3,550 C. At least 3,550, but less than 3,600 D. At least 3,600, but less than 3,650 E. At least 3,650

18. C. This is a Pareto Distribution with: F(x) = 1 -

�

10001000 + y⎛ ⎝ ⎜

⎞ ⎠ ⎟

3.

90% = Prob[Max ≤ y] = F(y)10.

�

⇒ F(y) = 0.98952.

�

⇒ 1 -

�

10001000 + y⎛ ⎝ ⎜

⎞ ⎠ ⎟

3 = 0.98952.

�

⇒ 0.01048 =

�

10001000 + y

⎛ ⎝ ⎜

⎞ ⎠ ⎟

3.

�

⇒ y = 3570.


19. An insurance company is examining the median age, m, of its policyholders. The company performs the following hypothesis test:

�

• H0: m = 40.

�

• H1: m ≠ 40. The following age values are observed from its portfolio of policies:

26 30 34 35.5 36 41 42 43 45 47 48 55 60 Calculate the signed-rank Wilcoxon statistic for this test. A. Less than 13.5 B. At least 13.5, but less than 14.5 C. At least 14.5, but less than 15.5 D. At least 15.5, but less than 16.5 E. At least 16.5

19. E. We subtract 40 from each of the age values, and rank the absolute differencesDiff.: -14 -10 -6 -4.5 -4 1 2 3 5 7 8 15 20Rank: 11 10 7 5 4 1 2 3 6 8 9 12 13

T+ = 1 + 2 + 3 + 6 + 8 + 9 + 12 + 13 = 54.T- = 11 + 10 + 7 + 5 + 4 = 37.The signed-rank Wilcoxon statistic is: T+ - T- = 54 - 37 = 17.Comment: This is the definition of the signed-rank Wilcoxon statistic for the one-sample test inSection 10.3 of Hogg, Craig, and McKean. In applying this test, they assume that the distribution is symmetric around its median; in that case the mean is equal to the median.This Wilcoxon statistic is not to be used directly with the Table attached to the exam, which is for the Wilcoxon Matched-Pairs, Signed-Rank Test, in other words the two sample test.


20. Let Y1 < Y2 < ... < Y15 be the order statistics of a random sample from a uniform distribution on [0, 1]. Calculate the probability that 0.75 < Y15 < 0.85. A. Less than 5% B. At least 5%, but less than 6% C. At least 6%, but less than 7% D. At least 7%, but less than 8% E. At least 8%.

20. D. Prob[Y15 ≤ y] = F(y)15 = y15.

Prob[Y15 ≤ 0.75] = 0.7515 = 0.01336.

Prob[Y15 ≤ 0.85] = 0.8515 = 0.08735.Prob[0.75 < Y15 < 0.85] = 0.08735 - 0.01336 = 0.07399.


21. You wish to explain Y using the following multiple regression model and 32 observations: Y = β0 + β1X1 + β2X2 + β3X3

A linear regression package generates the following table of summary statistics: Estimated Coefficient Standard Error

Intercept 44.200 5.960 β1 -0.295 0.118

β1 9.110 6.860

β2 -8.700 1.200

For the Intercept and each of the betas, you decide to reject the null hypothesis which is that the Estimated Coefficient is zero at α = 10% significance. Which variables have coefficients significantly different from zero? A. Intercept B. Intercept, X1 C. Intercept, X2 D. Intercept, X1, X3 E. Intercept, X2, X3

21. D. In each case, we compare to a t-distribution with 32 - 4 = 28 degrees of freedom.The 10% critical value is 1.701 for a two-sided test.In each case, the t-statistic is: (Estimated Coefficient) / (Standard Error).

Estimated Coefficient Standard Error t-statIntercept 44.200 5.960 7.416 reject at 10%β1 -0.295 0.118 -2.5 reject at 10%

β1 9.110 6.860 1.328

β2 -8.700 1.200 -7.25 reject at 10%

Coefficients significantly different from zero: Intercept, X1 , X3 .

Comment: The p-values are: 4.5 x 10-8, 1.9%, 19.5%, 6.8 x 10-8.


22. In an experiment with three blocks, three treatments, and a total of nine observations you are given the following summary of results:

Sum of Squares Mean Square Between Treatments 146/9 73/9 Between Blocks 50/9 25/9 Residual 10/9 5/18 Total 206/9

�

• H0: The treatment means are all equal.

�

• H1: The treatment means are not all equal.

�

• A block is defined to be a group of three observations.

�

• The results are Normally distributed. Calculate the smallest p-value for which H0 can be rejected. (A) Less than 1% (B) At least 1%, but less than 2.5% (C) At least 2.5%, but less than 5% (D) At least 5%, but less than 10% (E) At least 10%


22. A. In order to test whether the treatment means are all equal, the F-Statistic is:

�

Treatment Mean SquareError Mean Square =

�

Treatment Mean SquareResidual Mean Square = (73/9) / (5/18) = 29.2.

The number of degrees of freedom of the numerator is: (number of treatments) - 1 = 3 - 1 = 2.The number of degrees of freedom of the numerator is: (same size) - (number of blocks) - (number of treatments) + 1 = 9 - 3 - 3 + 1 = 4.The 1% critical value for 2 and 4 degrees of freedom is 18.00.Since 29.2 > 18.00, the p-value is less than 1%.Comment: An example of two-way analysis of variance. See Section 13.9 of Mathematical Statistics with Applications by Wackerly, Mendenhall, and Scheaffer. Using a computer, the p-value is 0.41%.One can infer that the degrees of freedom for the numerator: (146/9) / (73/9) = 2.One can infer that the degrees of freedom for the denominator: (10/9) / (5/18) = 4.In order to test instead whether the block means are all equal, the F-Statistic is:

�

Block Mean SquareError Mean Square =

�

Block Mean SquareResidual Mean Square = (25/9) / (5/18) = 10.

The number of degrees of freedom of the numerator is: (number of blocks) - 1 = 3 - 1 = 2.The number of degrees of freedom of the numerator is: (same size) - (number of blocks) - (number of treatments) + 1 = 9 - 3 - 3 + 1 = 4.The 5% critical value for 2 and 4 degrees of freedom is 6.944.The 2% critical value for 2 and 4 degrees of freedom is 12.142.Since 6.944 < 10 < 12.142, the p-value is between 2% and 5%; using a computer it is 2.8%.With b blocks and k treatments, the ANOVA Table in general is:Source D.F. Sum of Squares Mean Square F-Statistic

Treatments k - 1 SST SST / (k-1)

�

SST / (k -1)SSE / (n - b- k +1)

Blocks b -1 SBB SBB / (b-1)

�

SBB / (b- 1)SSE / (n - b- k +1)

Error or Residual n - b - k + 1 SSE SSE / (n-b-k+1)Total n - 1 Total SSThe first F-Statistic is used to test whether all of the means of the treatments are equal, and has k-1 and n-b-k+1 degrees of freedom.The second F-Statistic is used to test whether all of the means of the blocks are equal, and has b-1 and n-b-k+1 degrees of freedom.


23. For a large number of health insurance policies, the annual number of claims per policy follows the binomial distribution with parameters m = 10 and q, which varies by policy. The prior probability density function of q is:

h(q) = 12q2(1- q), 0 < q < 1. In the first year, two claims are reported for a policy. Calculate the standard deviation of the posterior distribution of q for this policy. A. Less than 0.10 B. At least 0.10, but less than 0.20 C. At least 0.20, but less than 0.30 D. At least 0.30, but less than 0.40 E. At least 0.40

23. B. The prior distribution of q is Beta with a = 3 and b = 2.Thus the posterior distribution of q is Beta with: a = 3 + 2 = 5, and b = 2 + 10 - 2 = 10.The mean of the posterior distribution of q is: a/(a+b) = 5/15 = 1/3.The second moment of the posterior distribution of q is: (a)(a+1) / {(a+b)(a+b+1)} = (5)(6)/{(15)(16)} = 1/8. The variance of the posterior distribution of q is: 1/8 - 1/32 = 1/72.

The standard deviation of the posterior distribution is: 1/

�

72 = 0.118.


24. You are given:

�

• For a general liability policy, the log (natural logarithm) of paid claim amounts conditionally follows the normal distribution with mean µ, and variance σ2 = 1.

�

• The posterior distribution of µ follows the normal distribution with variance = 1/(n+1), where n denotes the number of observed claims paid.

The following are the five observed log of paid claim amounts: 3.22 4.34 5.98 7.32 2.78

Calculate the length of the symmetric 95% Bayesian credible interval for µ. 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

24. C. We are told that the posterior distribution of µ is Normal with variance: 1/ (5+1).

Thus the length of the symmetric 95% Bayesian credible interval for µ is: (2)(1.96)

�

1/ 6 = 1.60. Comment: The symmetric 95% Bayesian credible interval is: (mean of the posterior distribution) ± 1.96

�

1/ 6 .If the prior distribution of µ is also Normal with variance 1, then the posterior distribution of µ is

Normal with variance:

�

σ2 s2 C σ2 + s2 = 1/(n + 1).


25. You are given:

�

• Daily claim counts follow a Poisson distribution with mean λ.

�

• The prior distribution of λ has probability density function:

f(λ) =

�

15 e-λ/5

�

• Three claims were observed on the first day. Calculate the variance of the posterior distribution of λ. A. Less than 1.0 B. At least 1.0, but less than 2.0 C. At least 2.0, but less than 3.0 D. At least 3.0, but less than 4.0 E. At least 4.0

25. C. The prior distribution is Gamma with α = 1 and θ = 5.

The posterior distribution is Gamma with α = 1 + 3 = 4, and 1/θ = 1/5 + 1 = 6/5. θ = 5/6.

The variance of the posterior distribution is: α θ2 = (4)(5/6)2 = 2.778.

END OF EXAMINATION


solutions to the fall 2015 cas exam st - …howardmahler.com/teaching/st_files/solsf15casst.pdf3....

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