spring05 Exam C Solutions feb06 (1)

spring05 Exam C Solutions feb06 (1) - Exam C Solutions...

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Exam C Solutions Spring 2005 Question #1 Key: D The CDF is 4 ) 1 ( 1 1 ) ( x x F + = 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, 22 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, 2 3, ( 1 ) 12. FF rr r r μβ σβ β == = += For aggregate losses, 2 24.4(3 ) 73.2 , 24.4 (12 ) 1,445.04(3 ) 11,479.44 . SF r μμμ σμ σσ = =+= + = For the given probability and tolerance, 2 0 (1.96/ 0.1) 384.16. λ The number of observations needed is 2 0 / 384.16(11,479.44 ) /(73.2 ) 823.02/ . r λσ The average observation produces 3 r claims and so the required number of claims is (823.02/ )(3 ) 2,469. =
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Question #3 Key: A 487 . 0 , 20 0 380 799 39 380 39 ) 1 ( 1 2 1 1 1 ) ( ˆ 2 2 = = => = + => = = + = n n n n n n n n n t H . Discard the non-integer solution to have n = 20. The Kaplan-Meier Product-Limit Estimate is: 9 19 18 11 11 ˆ ( ) ... 0.55. 20 19 12 20 St == = 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 = () 22 81 26 1 4 4 02 27 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.
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Question #6 Key: C or E This question was considered ambiguous, hence two solutions were accepted. Had the question begun with “Claim amounts are…” then the correct interpretation is that there are five observations (whose values are not stated), each from the Poisson distribution. In that case, the solution proceeds as follows: The distribution of Θ is Pareto with parameters 1 and 2.6. Then, 2 12 ( ) 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 EVPV E a VHM Var kva Z == Θ = = Θ = −= = = = + Alternatively, had the question begun with “The number of claims is …” then the correct interpretation is that there was a single observation whose value was 5. The only change is that now n = 1 rather than 5 and so 1 0.7304. 1 0.3692 Z + 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 (, ] jj cc + d j u j x j P j r j q j ˆ () j Fc 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 d j = number of observations with a lower truncation point of c j , u j = number of observations censored from above at c j+1 , x j = number of uncensored observations in the interval 1 + .
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spring05 Exam C Solutions feb06 (1) - Exam C Solutions...

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