Dr. Hackney STA Solutions pg 159

Dr. Hackney STA Solutions pg 159 - 10-2 Solutions Manual...

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10-2 Solutions Manual for Statistical Inference c. 1 n X Y i X i = β + 1 n X ± i X i with approximate mean β and variance σ 2 / ( 2 ). 10 . 5 a. The integral of E T 2 n is unbounded near zero. We have E T 2 n > r n 2 πσ 2 Z 1 0 1 x 2 e - ( x - μ ) 2 / 2 σ 2 dx > r n 2 πσ 2 K Z 1 0 1 x 2 dx = , where K = max 0 x 1 e - ( x - μ ) 2 / 2 σ 2 b. If we delete the interval ( - δ,δ ), then the integrand is bounded, that is, over the range of integration 1 /x 2 < 1 2 . c. Assume μ > 0. A similar argument works for μ < 0. Then P ( - δ < X < δ ) = P [ n ( - δ - μ ) < n ( X - μ ) < n ( δ - μ )] < P [ Z < n ( δ - μ )] , where Z n(0 , 1). For δ < μ , the probability goes to 0 as n → ∞ . 10 . 7 We need to assume that τ ( θ ) is differentiable at θ = θ 0 , the true value of the parameter. Then we apply Theorem 5.5.24 to Theorem 10.1.12. 10 . 9 We will do a more general problem that includes a ) and b ) as special cases. Suppose we want to estimate λ t e - λ /t ! = P ( X = t ). Let T = T ( X 1 ,...,X n ) = ± 1 if X 1 = t 0 if X 1 6 = t. Then E T = P ( T = 1) = P ( X 1 = t ), so T is an unbiased estimator. Since X i is a complete sufficient statistic for λ
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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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