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Dr. Hackney STA Solutions pg 56

# Dr. Hackney STA Solutions pg 56 - 4-12 Solutions Manual for...

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4-12 Solutions Manual for Statistical Inference = nαβ ( α + β + 1) ( α + β 2 )( α + β + 1) - nαβ ( α + β 2 )( α + β + 1) + n 2 Var P = n α α + β β α + β - n Var P + n 2 Var P = n E P (1 - E P ) + n ( n - 1)Var P. b. Var( Y ) = E(Var( Y | Λ)) + Var(E( Y | Λ)) = EΛ + Var(Λ) = μ + 1 α μ 2 since EΛ = μ = αβ and Var(Λ) = αβ 2 = ( αβ ) 2 α = μ 2 α . The “extra-Poisson” variation is 1 α μ 2 . 4.37 a. Let Y = X i . P ( Y = k ) = P ( Y = k, 1 2 < c = 1 2 (1 + p ) < 1) = 1 0 ( Y = k | c = 1 2 (1 + p )) P ( P = p ) dp = 1 0 n k [ 1 2 (1 + p )] k [1 - 1 2 (1 + p )] n - k Γ( a + b ) Γ( a )Γ( b ) p a - 1 (1 - p ) b - 1 dp = 1 0 n k (1 + p ) k 2 k (1 - p ) n - k 2 n - k Γ( a + b ) Γ( a )Γ( b ) p a - 1 (1 - p ) b - 1 dp = n k Γ( a + b ) 2 n Γ( a )Γ( b ) k j =0 1 0 p k + a - 1 (1 - p ) n - k + b - 1 dp = n k Γ( a + b ) 2 n Γ( a )Γ( b ) k j =0 k j Γ( k + a )Γ( n - k + b ) Γ( n + a + b ) = k j =0 ( k j ) 2 n n k Γ( a + b ) Γ( a )Γ( b ) Γ( k + a )Γ( n - k + b ) Γ( n + a + b ) . A mixture of beta-binomial. b. E Y = E(E( Y | c )) = E[ nc ] = E n 1 2 (1 + p ) = n 2 1 + a a + b . Using the results in Exercise 4.35(a),
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