Dr. Hackney STA Solutions pg 106

Dr. Hackney STA Solutions pg 106 - Second Edition 7-9 7.21...

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Second Edition 7-9 7.21 a. E 1 n i Y i x i = 1 n i E Y i x i = 1 n i βx i x i = β. b. Var 1 n i Y i x i = 1 n 2 i Var Y i x 2 i = σ 2 n 2 i 1 x 2 i . Using Example 4.7.8 with a i = 1 /x 2 i we obtain 1 n i 1 x 2 i n i x 2 i . Thus, Var ˆ β = σ 2 i x 2 i σ 2 n 2 i 1 x 2 i = Var 1 n i Y i x i . Because g ( u ) = 1 /u 2 is convex, using Jensen’s Inequality we have 1 ¯ x 2 1 n i 1 x 2 i . Thus, Var i Y i i x i = σ 2 n ¯ x 2 σ 2 n 2 i 1 x 2 i = Var 1 n i Y i x i . 7.22 a. f x, θ ) = f x | θ ) π ( θ ) = n 2 πσ e - n x - θ ) 2 / (2 σ 2 ) 1 2 πτ e - ( θ - μ ) 2 / 2 τ 2 . b. Factor the exponent in part (a) as - n 2 σ 2 x - θ ) 2 - 1 2 τ 2 ( θ - μ ) 2 = - 1 2 v 2 ( θ - δ ( x )) 2 - 1 τ 2 + σ 2 /n x - μ ) 2 , where δ ( x ) = ( τ 2 ¯ x +( σ 2 /n ) μ ) / ( τ 2 + σ 2 /n ) and v = ( σ 2 τ 2 /n ) ( τ + σ 2 /n ). Let n( a, b ) denote the pdf of a normal distribution with mean a and variance b . The above factorization shows that f ( x , θ ) = n( θ, σ 2 /n ) × n( μ, τ 2 ) = n( δ ( x ) , v 2 ) × n( μ, τ 2 + σ 2 /n ) , where the marginal distribution of ¯ X is n( μ, τ 2 + σ
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