Dr. Hackney STA Solutions pg 187

Dr. Hackney STA Solutions pg 187 - Second Edition 11-15...

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Second Edition 11-15 11.36 a. α = E( ¯ Y - ˆ β ¯ X ) = E h E( ¯ Y - ˆ β ¯ X | ¯ X ) i = E ± α + β ¯ X - β ¯ X ² = E α = α. E ˆ β = E[E( ˆ β | ¯ X )] = E β = β. b. Recall Var Y = Var[E( Y | X )] + E[Var( Y | X )] Cov( Y ,Z ) = Cov[E( Y | X ) , E( Z | X )] + E[Cov( Y,Z | X )] . Thus Varˆ α = E[Var(ˆ α | X )] = σ 2 E h X X 2 i . S XX i Var ˆ β = σ 2 E[1 /S XX ] Cov(ˆ α, ˆ β ) = E[Cov(ˆ α, ˆ β | ˆ X )] = - σ 2 E[ ¯ X/S XX ] . 11.37 This is almost the same problem as Exercise 11.35. The log likelihood is log L = - n 2 log(2 πσ 2 ) - 1 2 σ 2 X i ( y i - βx i ) 2 . The MLE is i x i y i / i x 2 i , with mean β and variance σ 2 / i x 2 i , the CRLB. 11.38 a. The model is y i = θx i + ± i , so the least squares estimate of θ is x i y i / x 2 i (regression through the origin). E ³∑ x i Y i x 2 i ´ = x i ( x i θ ) x 2 i = θ Var ³∑ x i Y i x 2 i ´ = x 2 i ( x i θ ) ( x 2 i ) 2 = θ x 3 i ( x 2 i ) 2 . The estimator is unbiased. b. The likelihood function is L ( θ | x ) = n Y i =1 e - θx i ( θx i ) y
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