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
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