Course4Solutions_1101[1].PDF - Course 4 Solutions November...

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Course 4 Solutions November 2001 Exams
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November 2001 Course 4 Solutions -1- November, 2001 Society of Actuaries Question #1 Answer is B From the Yule-Walker equations: 11 12 2 1 . rf r r ff =+ Substituting the given quantities yields: 0.5 3 0.53 0.2 2 0.5 3. - The solution is f 1 090 = . and 2 0.70 f =- . The next Yule-Walker equation is: 3 1 2 21 0.90 ( 0.22 ) 0.70(0.53) 0.57. r f r fr +- Question #2 Answer is E For an interval running from c to d , the uniform density function is ( ) / [ ( )] f x g n dc where g is the number of observations in the interval and n is the sample size. The contribution to the second raw moment for this interval is: 3 33 2 () . 3 3 d d c c g g x g x dx n d c n d c n - == - -- For this problem, the second raw moment is: 3 3 3 3 3 3 1 30(2 5 0 ) 32(5 0 25 ) 20(10 0 50 ) 8(20 0 100) 3958.33. 9 0 3(2 5 0 ) 3(5 0 25 ) 3(10 0 50 ) 3(20 0 100)  - +++=  - --- 
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November 2001 Course 4 Solutions -2- Question #3 Answer is B Because the Bayes and Bühlmann results must be identical, this problem can be solved either way. For the Bühlmann approach, ( ) () v m l ll == . Then, noting that the prior distribution is a gamma distribution with parameters 50 and 1/500, we have: 2 50/50 0 0.1 ( ) 0.1 0 0.0002 / 500 1500/(150 0 500 ) 0.75 7 5 210 0.19. 60 0 900 E vE a Var k va Z X ml l l = = = += + + The credibility estimate is 0.75(0.19 ) 0.25(0.1 ) 0.1675. For 1100 policies, the expected number of claims is 1100(0.1675) = 184.25. For the Bayes approach, the posterior density is proportional to (because in a given year the number of claims has a Poisson distribution with parameter l times the number of policies) 60 0 7 5 90 0 21 0 5 0 500 33 5 2000 (60 0 ) (90 0 ) (50 0) 75 ! 210 ! (50) e ee e l ll l l l l - -- - Γ which is a gamma density with parameters 335 and 1/2000. The expected number of claims per policy is 335/2000 = 0.1675 and the expected number of claims in the next year is 184.25. Question #4 Answer is B All but B can be seen as true from various items on pages 91-96 of Survival Analysis . B is false because if the last observed time is a death time, then the number of deaths is equal to the number at risk (that is, d = Y ). Thus the survival function will be multiplied by zero and will become zero.
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November 2001 Course 4 Solutions -3- Question #5 Answer is B s N i 2 2 2 2394 6 399 = - = = $ e s s x i $ . . b 2 2 2 399 162 2463 = = = s $ . b = 1569 .95 1.943 t = The confidence interval is ( 29 ( 29 35.6 9 1.943 15.6 9 66.2 , 5.2 - ± = -- Question #6 Answer is E The q - q plot takes the ordered values and plots the j th point at j /( n+ 1) on the horizontal axis and at ( ;) j Fx q on the vertical axis. For small values, the model assigns more probability to being below that value than occurred in the sample. This indicates that the model has a heavier left tail than the data. For large values, the model again assigns more probability to being below that value (and so less probability to being above that value). This indicates that the model has a lighter right tail than the data. Of the five answer choices, only E is consistent with these observations. In addition, note that as you go from 0.4 to 0.6 on the horizontal axis (thus looking at the middle 20% of the data), the q - q plot increases from about 0.3 to 0.4 indicating that the model puts only about 10% of the probability in this range, thus confirming answer E.
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Course4Solutions_1101[1].PDF - Course 4 Solutions November...

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