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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 τ 2 / (1 + 2 2 ) = P Z ≤ - τ ( σ 2 /n )( τ 2 + σ 2 /n ) ¯ x = P Z τ ( σ 2 /n )( τ 2 + σ 2 /n ) ¯ x . b. Using the fact that if θ = 0, ¯ X n(0 , σ 2 /n ), the p-value is P ( ¯ X ¯ x ) = P Z ¯ x - 0 σ/ n = P Z 1 σ/ n ¯ x c. For σ 2 = τ 2 = 1, P ( θ 0 | x ) = P Z 1 (1 /n )(1 + 1 /n ) ¯ x and P ( ¯ X ¯ x ) = P Z 1 1 /n ¯ x . Because 1 (1 /n )(1 + 1 /n ) < 1 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 ( σ 2 /n )( τ 2 + σ 2 /n ) = 1 ( σ 2 /n )(1+ σ 2 / ( τ 2 n )) 1 σ/ n , the constant in the p-value. So the indicated equality is true. 8.55 The formulas for the risk functions are obtained from (8.3.14) using the power function β ( θ ) = Φ( - z α + θ 0 - θ ), where Φ is the standard normal cdf.
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