Dr. Hackney STA Solutions pg 158

Dr. Hackney STA Solutions pg 158 - /I ( ). 10 . 4 a. Write...

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Chapter 10 Asymptotic Evaluations 10.1 First calculate some moments for this distribution. E X = θ/ 3 , E X 2 = 1 / 3 , Var X = 1 3 - θ 2 9 . So 3 ¯ X n is an unbiased estimator of θ with variance Var(3 ¯ X n ) = 9(Var X ) /n = (3 - θ 2 ) /n 0 as n → ∞ . So by Theorem 10.1.3, 3 ¯ X n is a consistent estimator of θ . 10 . 3 a. The log likelihood is - n 2 log (2 πθ ) - 1 2 X ( x i - θ ) /θ. Differentiate and set equal to zero, and a little algebra will show that the MLE is the root of θ 2 + θ - W = 0. The roots of this equation are ( - 1 ± 1 + 4 W ) / 2, and the MLE is the root with the plus sign, as it has to be nonnegative. b. The second derivative of the log likelihood is ( - 2 x 2 i + ) / (2 θ 3 ), yielding an expected Fisher information of I ( θ ) = - E θ - 2 X 2 i + 2 θ 3 = 2 + n 2 θ 2 , and by Theorem 10.1.12 the variance of the MLE is 1
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Unformatted text preview: /I ( ). 10 . 4 a. Write X i Y i X 2 i = X i ( X i + i ) X 2 i = 1 + X i i X 2 i . From normality and independence E X i i = 0 , Var X i i = 2 ( 2 + 2 ) , E X 2 i = 2 + 2 , Var X 2 i = 2 2 (2 2 + 2 ) , and Cov( X i ,X i i ) = 0. Applying the formulas of Example 5.5.27, the asymptotic mean and variance are E X i Y i X 2 i 1 and Var X i Y i X 2 i n 2 ( 2 + 2 ) [ n ( 2 + 2 )] 2 = 2 n ( 2 + 2 ) b. Y i X i = + i X i with approximate mean and variance 2 / ( n 2 )....
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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