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sol3 - PSTAT 120C Assignment 3 Solutions May 2 2006 1 The...

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PSTAT 120C: Assignment # 3 Solutions May 2, 2006 1. The basic equality that we use over and over is that Cov( aX + bY, Z ) = a Cov( X, Z ) + b Cov( Y, Z ) . where a and b are constants and X, Y, and Z are random variables. This can be applied in light of the facts that Cov( y i , y i ) = Var( y i ) and for two independent observation Cov( y i , y j ) = 0. The covariance in question can therefore be written, Cov( e i , ˆ β 0 ) = Cov y i - ˆ β 0 - ˆ β 1 x i , ˆ β 0 = Cov( y i , ˆ β 0 ) - Cov( ˆ β 0 , ˆ β 0 ) - x i Cov( ˆ β 1 , ˆ β 0 ) . From the results in the book we can get Cov( ˆ β 0 , ˆ β 0 ) = Var( ˆ β 0 ) = σ 2 1 n i x 2 i n i =1 ( x i - ¯ x ) 2 Cov( ˆ β 0 , ˆ β 1 ) = - σ 2 ¯ x n i =1 ( x i - ¯ x ) 2 The tricky bit is the first term. Cov( y i , ˆ β 0 ) = Cov y i , ¯ y - ˆ β 1 ¯ x = Cov( y i , ¯ y ) - ¯ x Cov( y i , ˆ β 1 ) . (1) The first term in (1) is, as we’ve worked out before, Cov( y i , ¯ y ) = Cov y i , 1 n n j =1 y j = 1 n n j =1 Cov( y i , y j ) , but the y i ’s are independent so that Cov( y i , y j ) = 0 unless i = j in which case Cov( y i , y i ) = Var( y i ) = σ 2 . Thus, Cov( y i , ¯ y ) = 1 n n j =1 Cov( y i , y j ) = 1 n

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sol3 - PSTAT 120C Assignment 3 Solutions May 2 2006 1 The...

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