ACTSC 433 Midterm 1 Soln S-2012-W

ACTSC 433 Midterm 1 Soln S-2012-W - 35 months is S (35) =...

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SOLUTIONS TO TEST # 1 – ACTSC 433/833, WINTER 2012 1. (a) The distribution function of X ( n ) is F X ( n ) ( x ) = 0 , x < 0 , x 2 n θ 2 n , 0 x < θ, 1 , x θ. Thus, E ( ˆ θ n ) = R 0 (1 - F X ( n ) ( x )) dx = θ ± 1 - 1 2 n +1 ² θ as n → ∞ . Hence, ˆ θ n is asymptotically unbiased for θ . (b) The probability density function of T n = X ( n ) θ is f T ( n ) ( t ) = ( 2 nt 2 n - 1 , 0 < t < 1 , 0 , otherwise . (c) Let λ 1 = λ and λ 2 = 1 so that Pr { λ 1 < T n < λ 2 } = α , which implies that Pr { X ( n ) < θ < X ( n ) (1 - α ) - 1 2 n } = α . 2. We have ˆ S ( y 9 ) ˆ S ( y 6 ) = 0 . 85 = ± 1 - 1 n - 6 ²± 1 - 1 n - 7 ²± 1 - 1 n - 8 ² . Thus, n = 26, ˆ S ( y 5 ) = 21 26 , and d V ar ( ˆ S ( y 5 )) = 0 . 005974. 3. (a) Note that F n ( x ) = 1 n n i =1 I ( X i x ). Thus, E ( F n ( x )) = nF ( x ) n = F ( x ) and V ar ( F n ( x )) = nF ( x )(1 - F ( x )) n 2 0 as n → ∞ . Hence, F n ( x ) is a consistent estimator of F ( x ). (b) We see that ˆ θ n = 1 n n i =1 I ( X i a, or X i > b ) = F n ( a ) + 1 - F n ( b ) is a consistent empirical estimator for θ . That is because E ( ˆ θ n ) = nE ( I ( X i a, or X i > b )) n = n = θ and V ar ( ˆ θ n ) = nV ar ( I ( X i a, or X i > b )) n 2 = θ (1 - θ ) n 0 as n → ∞ . 4. We are given ˆ H (32) = 3 15 + 2 12 + k 10 + 2 10 - k = 49 60 . Thus, k = 2, ˆ H (35) = 49 60 + 1 6 = 59 60 , and the Nelson-Aalen estimate for the probability that one cancer patient will survive at least
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Unformatted text preview: 35 months is S (35) = e-59 / 60 = 0 . 37406. 5. (a) The values of y j ,s j and r j for all possible values of j are given in the following table: y j 0.7 2.8 3.4 3.8 4.7 5.1 s j 1 1 1 1 2 1 r j 8 10 9 7 6 4 (b) The Brown-Hollander-Kowar estimate for the probability that a member in the group will survive at least 8 years is S (8) = ( S (5 . 1)) 8 / 6 = 0 . 20083 . 6. We have H ( y 5 )- H ( y 4 ) = s 5 r 5 = 2 r 5 and H ( y 6 )- H ( y 5 ) = s 6 r 6 = 1 r 6 . Hence, r 5 = 10, r 6 = 5, and the number of censored observations between death times y 5 and y 6 is 10+0-2-5 =3....
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This note was uploaded on 02/28/2012 for the course ACTSC 433 taught by Professor Johnny during the Winter '09 term at Waterloo.

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