{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ACTSC 433 Midterm 1 Soln S-2012-W

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

This preview shows page 1. Sign up to view the full content.

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 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 n t 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 1 - 1 n - 7 1 - 1 n - 8 . 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 35 months is ˆ 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 ˆ S (8) = ( ˆ S (5 . 1)) 8 / 6 = 0 . 20083 . 6. We have ˆ H ( y 5 )-ˆ H ( y 4 ) = s 5 r 5 = 2 r 5 and ˆ H ( y 6 )-ˆ 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....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online