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Dr. Hackney STA Solutions pg 158

# Dr. Hackney STA Solutions pg 158 - /I θ 10 4 a Write ∑ X...

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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 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 ( θ ). 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 = σ 2 ( μ 2 + τ 2 ) , E X 2 i = μ 2 + τ 2 , Var X 2 i = 2 τ 2 (2 μ 2 + τ 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 ² ≈ 1 and Var ±∑ X i Y i ∑ X 2 i ² ≈ nσ 2 ( μ 2 + τ 2 ) [ n ( μ 2 + τ 2 )] 2 = σ 2 n ( μ 2 + τ 2 ) b. ∑ Y i ∑ X i = β + ∑ ± i ∑ X i with approximate mean β and variance σ 2 / ( nμ 2 )....
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