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Dr. Hackney STA Solutions pg 152

# Dr. Hackney STA Solutions pg 152 - 9-10 Solutions Manual...

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9-10 Solutions Manual for Statistical Inference The log of this is - λ/ 2 t - λ + λ 1 - t/ 2 λ = t 2 λ - t 2 + 2 λ = t 2 - ( t 2 / λ ) + 2 t 2 / 2 as λ → ∞ , so the mgf converges to e t 2 / 2 , the mgf of a standard normal. b. Since P ( χ 2 2 Y χ 2 2 Y,α ) = α for all λ , χ 2 2 Y,α - 2 λ 8 λ z α as λ → ∞ . In standardizing (9.2.22), the upper bound is nb nb +1 χ 2 2( Y + a ) ,α/ 2 - 2 λ 8 λ = 8( λ + a ) 8 λ nb nb +1 [ χ 2 2( Y + a ) ,α/ 2 - 2( λ + a )] 8( λ + a ) + nb nb +1 2( λ + a ) - 2 λ 8( λ + a ) . While the first quantity in square brackets z α/ 2 , the second one has limit lim λ →∞ - 2 1 nb +1 λ + a nb nb +1 8( λ + a ) → -∞ , so the coverage probability goes to zero. 9.33 a. Since 0 C a ( x ) for every x , P (0 C a ( X ) | μ = 0) = 1. If μ > 0, P ( μ C a ( X )) = P ( μ max { 0 , X + a } ) = P ( μ X + a ) (since μ > 0) = P ( Z ≥ - a ) ( Z n(0 , 1)) = . 95 ( a = 1 . 645 . ) A similar calculation holds for μ < 0. b. The credible probability is max(0 ,x + a ) min(0 ,x - a ) 1 2 π e - 1 2 ( μ - x ) 2 = max( - x,a ) min( - x, - a ) 1 2 π e - 1 2 t 2 dt = P (min( - x, - a ) Z max( - x, a )) . To evaluate this probability we have two cases: (i) | x | ≤ a credible probability = P ( | Z | ≤ a ) (ii) | x | > a
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