Dr. Hackney STA Solutions pg 142

Dr. Hackney STA Solutions pg 142 - Second Edition 8-21 8.54...

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Second Edition 8-21 8.54 a. From Exercise 7.22, the posterior distribution of θ | x is normal with mean [ τ 2 / ( τ 2 + σ 2 /n )]¯ x and variance τ 2 / (1 + 2 2 ). So P ( θ 0 | x ) = P ± Z 0 - [ τ 2 / ( τ 2 + σ 2 /n )]¯ x p τ 2 / (1 + 2 2 ) ! = P ± Z ≤ - τ p ( σ 2 /n )( τ 2 + σ 2 /n ) ¯ x ! = P ± Z τ p ( σ 2 /n )( τ 2 + σ 2 /n ) ¯ x ! . b. Using the fact that if θ = 0, ¯ X n(0 2 /n ), the p-value is P ( ¯ X ¯ x ) = P ² Z ¯ x - 0 σ/ n ³ = P ² Z 1 σ/ n ¯ x ³ c. For σ 2 = τ 2 = 1, P ( θ 0 | x ) = P ± Z 1 p (1 /n )(1 + 1 /n ) ¯ x ! and P ( ¯ X ¯ x ) = P ± Z 1 p 1 /n ¯ x ! . Because 1 p (1 /n )(1 + 1 /n ) < 1 p 1 /n , the Bayes probability is larger than the p-value if ¯ x 0. (Note: The inequality is in the opposite direction for ¯ x < 0, but the primary interest would be in large values of ¯ x .) d. As τ 2 → ∞ , the constant in the Bayes probability, τ p ( σ 2 /n )( τ 2 + σ 2 /n ) = 1 p ( σ 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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