ACTSC 433 Midterm 2 Soln S-2012-W

ACTSC 433 Midterm 2 Soln S-2012-W - SOLUTIONS TO MIDTERM...

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SOLUTIONS TO MIDTERM TEST # 2 – ACTSC 433/833, WINTER 2012 1. (a) The kernel density estimate for the probability that an individual aged 90 will survive at least 3 years using the uniform kernel with bandwidth b = 0 . 5 is ˆ S (3) = 1 8 (0 + 0 + 0 . 9 + 1 + 2 × 1 + 1 + 1) = 0 . 7375. (b) The uniform distribution function is F ( x ) = x 8 , 0 x 8. Thus, it is easy to determine that D = 1 . 8 8 < D 8 , 0 . 1 = 0 . 43. Hence, we accept the null hypothesis and the uniform distribution U (0 , 8) is appropriate for the time until death of an individual aged 90 at a 10% significance level. 2. (a) Note that L ( θ ) = θ 3 n Q n i =1 3 x 4 i , θ x (1) . Hence, the maximum likelihood estimator ˆ θ n = X (1) . (b) We have that E ( ˆ θ n ) = R 0 S X (1) ( x ) dx = 3 n 3 n - 1 θ θ as n → ∞ . Hence, ˆ θ n is an asymptotically unbiased estimator of θ . (c) We have E ( ˆ θ 2 n ) = R 0 2 xS X (1) ( x ) dx = 3 n 3 n - 2 θ 2 θ 2 . Thus, V ar ( ˆ θ n ) 0 as n → ∞ . Hence, ˆ θ n is a consistent estimator of θ . 3. (a) Note that Pr { T 65 1 } = 0 . 0951626 , Pr { 1 < T 65 < 2 } = 0 . 0861067, and Pr { T 65 2 } = 0 . 818731. Hence, it is easy to determine that Q = 5 . 34 < χ 2 2 , 0 . 05 = 5 . 99. Thus, we accept the null hypothesis and the exponential distribution function F ( t ) = 1 - e - 0 . 1 t , t > 0 is acceptable for the future lifetime T 65 . (b) Note that L ( α, β ) = α 10 (1 - α ) 15 β 15 (1
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