Exam_C_Solutions_FALL_2006 - FALL 2006 EXAM C SOLUTIONS...

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FALL 2006 EXAM C SOLUTIONS Question #1 Key: E With n + 1 = 16, we need the 0.3(16) = 4.8 and 0.65(16) = 10.4 smallest observations. They are 0.2(280) + 0.8(350) = 336 and 0.6(450) + 0.4(490) = 466. The equations to solve are: 22 1/2 -1/2 0.3 1 and 0.65 1 336 466 (0.7) 1 (336/ ) and (0.35) 1 (466/ ) (0.7) 1 (336/ ) (0.35) 1 (466/ ) 0.282814 (336/ 466) ln(0.282814) ln(336/ 466) 3.8614. γγ γ θθ θ −− ⎛⎞ =− ⎜⎟ ++ ⎝⎠ =+ = = = = Question #2 Key: D Let E be the even of having 1 claim in the first four years. In four years, the total number of claims is Poisson(4 λ ). 1 2 4 Pr( | )Pr( ) (0.05) 0.01839 Pr( | ) 0.14427 Pr( ) Pr( ) Pr( ) (2)(0.2) 0.05413 Pr( | ) 0.42465 Pr( ) Pr( ) (4)(0.75) 0.05495 Pr( | ) 0.43108 Pr( ) Pr( ) : Pr( ) 0.01839 .05413 ET y p e I T y p e I e Type I E EE E e Type II E e Type III E Note E == = = = = .05495 0.12747 += The Bayesian estimate of the number of claims in Year 5 is: 0.14427(0.25) + 0.42465(0.5) + 0.43108(1) = 0.67947.
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Question #3 Key: B The sample mean is 0.2(400) + 0.7(800) + 0.1(1600) = 800. The sample variance is 22 2 0.2(400 800) 0.7(800 800) 0.1(1600 800) 96,000 −+ −= . The sample third central moment is 33 3 0.2(400 800) 0.7(800 800) 0.1(1600 800) 38,400,000 . The skewness coefficient is 1.5 38,400,000/96,000 1.29 = . Question #4 Key: C Because 0.656 < 0.7654 < 0.773, the simulated number of losses is 4. To simulate a loss by inversion, use 39 . 204 , 6481 . 0 75 . 236 , 7537 . 0 18 . 170 , 5152 . 0 12 . 113 , 2738 . 0 )) 1 ln( ( 200 )) 1 ln( ( ) / ( ) 1 ln( 1 1 ) ( 4 4 3 3 2 2 1 1 2 / 1 / 1 ) / ( ) / ( = = = = = = = = = = = = = = x u x u x u x u u u x x u e u u e x F x x τ θ With a deductible of 150, the first loss produces no payments and 113.12 toward the 500 limit. The second loss produces a payment of 20.18 and the insured is now out-of-pocket 263.12. The third loss produces a payment of 86.75 and the insured is out 413.12. The deductible on the fourth loss is then 86.88 for a payment of 204.29 – 86.88 = 117.51. The total paid by the insurer is 20.18 + 86.75 + 117.51 = 224.44.
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Question #5 Key: A The density function is 2/ () x f xx e θ −− = and the likelihood function is 2 /186 2 /91 2 /66 /60 7 3 0.148184 1 ( ) (186 ) (91 ) (66 ) ( ) ( ) ln ( ) 3ln( ) 0.148184 ( ) 3 0.148184 0 3/ 0.148184 20.25. Le e e e e lL l θθ = == =− = The mode is / 2 20.25/ 2 10.125 . Question #6 Key: D We have ( ) 4 and 4E( ) 4(600) 2400 μ = = . The average loss for Years 1 and 2 is 1650 and so 1800 (1650) (1 )(2400) ZZ =+ which gives Z = 0.8. Because there were two years, 0.8 2/(2 ) Z k + which gives k = 0.5. For three years, the revised value is 3/(3 0.5) 6/7 Z = += and the revised credibility estimate (using the new sample mean of 2021), (6/ 7)(2021) (1/ 7)(2400) 2075.14 + = . Question #7 Key: B The uncensored observations are 4 and 8 (values beyond 11 are not needed). The two r values are 10 and 5 and the two s values are 2 and 1. The Kaplan-Meier estimate is ˆ (11) (8/10)(4/5) 0.64 S and Greenwood’s estimate is 2 21 (0.64) 0.03072 10(8) 5(4) ⎛⎞ ⎜⎟ ⎝⎠ .
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Question #8 Key: C There are two ways to approach this problem. One is LaGrange’s formula: (3 4)(3 5) (3 2)(3 5) (3 2)(3 4) (3) 25 20 30 18.33.
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This note was uploaded on 11/14/2011 for the course MAT 2070 taught by Professor S.g. during the Spring '11 term at Université du Québec à Montréal.

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Exam_C_Solutions_FALL_2006 - FALL 2006 EXAM C SOLUTIONS...

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