Wooldridge_5e_AppE_IM - APPENDIX E SOLUTIONS TO PROBLEMS...

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271 APPENDIX E SOLUTIONS TO PROBLEMS E.1 This follows directly from partitioned matrix multiplication in Appendix D. Write X = 1 2 n x x x , X = ( 1 x 2 x n x ), and y = 1 2 n y y y Therefore, X X = 1 n t t t = x x and X y = 1 n t t t = x y . An equivalent expression for ˆ β is ˆ β = 1 1 1 n t t t n = x x 1 1 n t t t n y = x which, when we plug in y t = x t β + u t for each t and do some algebra, can be written as ˆ β = β + 1 1 1 n t t t n = x x 1 1 n t t t n u = x . As shown in Section E.4, this expression is the basis for the asymptotic analysis of OLS using matrices. E.2 (i) Following the hint, we have SSR( b ) = ( y Xb ) ( y Xb ) = [ ˆ u + X ( ˆ β b )] [ ˆ u + X ( ˆ β b )] = ˆ u ˆ u + ˆ u X ( ˆ β b ) + ( ˆ β b ) X ˆ u + ( ˆ β b ) X X ( ˆ β b ). But by the first order conditions for OLS, X ˆ u = 0 , and so ( X ˆ u ) = ˆ u X = 0 . But then SSR( b ) = ˆ u ˆ u + ( ˆ β b ) X X ( ˆ β b ), which is what we wanted to show. (ii) If X has a rank k then X X is positive definite, which implies that ( ˆ β b ) X X ( ˆ β b ) > 0 for all b ˆ β . The term ˆ u ˆ u does not depend on b , and so SSR( b ) – SSR( ˆ β ) = ( ˆ β b )
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