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Dr. Hackney STA Solutions pg 183

# Dr. Hackney STA Solutions pg 183 - y i(ˆ α ˆ βx i 2 σ...

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Second Edition 11-11 A solution can be found with Lagrange multipliers, but verifying that it is a minimum is excruciating. So instead we note that X i a i = 1 a i = 1 n + k ( b i - ¯ b ) , for some constants k,b 1 ,b 2 ,...,b n , and X i a i x i = 0 k = - ¯ x i ( b i - ¯ b )( x i - ¯ x ) and a i = 1 n - ¯ x ( b i - ¯ b ) i ( b i - ¯ b )( x i - ¯ x ) . Now X i a 2 i = X i ± 1 n - ¯ x ( b i - ¯ b ) i ( b i - ¯ b )( x i - ¯ x ) ² 2 = 1 n + ¯ x 2 i ( b i - ¯ b ) 2 [ i ( b i - ¯ b )( x i - ¯ x )] 2 , since the cross term is zero. So we need to minimize the last term. From Cauchy-Schwarz we know that i ( b i - ¯ b ) 2 [ i ( b i - ¯ b )( x i - ¯ x )] 2 1 i ( x i - ¯ x )] 2 , and the minimum is attained at b i = x i . Substituting back we get that the minimizing a i is 1 n - ¯ x ( x i - ¯ x ) i ( x i - ¯ x ) 2 , which results in i a i Y i = ¯ Y - ˆ β ¯ x , the least squares estimator. 11.28 To calculate max σ 2 L ( σ 2 | y, ˆ α ˆ β ) = max σ 2 ³ 1 2 πσ 2 ´ n/ 2 e - 1 2 Σ i [ y i - α + ˆ βx i )] 2 2 take logs and diﬀerentiate with respect to σ 2 to get d 2 log L ( σ 2 | y, ˆ α, ˆ β ) = - n 2 σ 2 + 1 2 i [
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Unformatted text preview: y i-(ˆ α + ˆ βx i )] 2 ( σ 2 ) 2 . Set this equal to zero and solve for σ 2 . The solution is ˆ σ 2 . 11.29 a. Eˆ ± i = E( Y i-ˆ α-ˆ βx i ) = ( α + βx i )-α-βx i = 0 . b. Varˆ ± i = E[ Y i-ˆ α-ˆ βx i ] 2 = E[( Y i-α-βx i )-(ˆ α-α )-x i ( ˆ β-β )] 2 = Var Y i + Varˆ α + x 2 i Var ˆ β-2Cov( Y i , ˆ α )-2 x i Cov( Y i , ˆ β ) + 2 x i Cov(ˆ α, ˆ β ) . 11.30 a. Straightforward algebra shows ˆ α = ¯ y-ˆ β ¯ x = X 1 n y i-¯ x ∑ ( x i-¯ x ) y i ∑ ( x i-¯ x ) 2 = X ± 1 n-¯ x ( x i-¯ x ) ∑ ( x i-¯ x ) 2 ² y i ....
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