APPENDIX E
SOLUTIONS TO PROBLEMS
E.1 This follows directly from partitioned matrix multiplication in Appendix D. Write
?x1??y1?????xy2?2??????xn), and y = X = , X? = (x1x2 ???????x???y???n??n?
??Therefore, X?X = ?x?txt and X?y = ?xtyt. An equivalent expression for β is
t?1t?1nnn????1n??1??β = ?n?x?x nxytt???tt? ?t?1??t?1?which, when we plug in yt = xt? + ut for each t and do some algebra, can be written as
nn?????1?1?= ? + n??βxx nxu??tt???tt?. ?t?1??t?1?As shown in Section E.4, this expression is the basis for the asymptotic analysis of OLS using matrices.
? – b)]?[ u? – ? + X(β? + X(βE.2 (i) Following the hint, we have SSR(b) = (y – Xb)?(y – Xb) = [u? – b) + (β? – b)?X?u? – b)?X?X(β? – b). But by the first order conditions ??u? + u??X(β? + (βb)] = u?1?1? – b)?X?X(β? – b), ? = 0, and so (X?u?)? = u??X = 0. But then SSR(b) = u??u? + (βfor OLS, X?uwhich is what we wanted to show.
? – b) ?X?X(β? – b) > (ii) If X has a rank k then X?X is positive definite, which implies that (β?. The term u?) = (β?– b) ?X?X ??u? does not depend on b, and so SSR(b) – SSR(β0 for all b ? β?– b) > 0 for b ?β?. (β
E.3 (i) We use the placeholder feature of the OLS formulas. By definition, β = (Z?Z)-1Z?y =
?. [(XA)? (XA)]-1(XA)?y = [A?(X?X)A]-1A?X?y = A-1(X?X)-1(A?)-1A?X?y = A-1(X?X)-1X?y = A-1β
? and yt = zβ. Plugging zt and β into the ?t = xtβ(ii) By definition of the fitted values, yt?) = xβ? = y?. second equation gives y = (xtA)(A-1βttt
(iii) The estimated variance matrix from the regression of y and Z is ?2(Z?Z)-1 where ?2 is the error variance estimate from this regression. From part (ii), the fitted values from the two
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regressions are the same, which means the residuals must be the same for all t. (The dependent
?2. Further, as we showed in part (i), variable is the same in both regressions.) Therefore, ?2 = ??2A-1(X?X)-1(A-1)?, which is what we wanted to (Z?Z)-1 = A-1(X?X)-1(A?)-1, and so ?2(Z?Z)-1 = ?show.
(iv) The ?j are obtained from a regression of y on XA, where A is the k ? k diagonal matrix
?. But A-1 is easily seen to be the with 1, a2, , ak down the diagonal. From part (i), β = A-1β?1k ? k diagonal matrix with 1, a2, , ak?1 down its diagonal. Straightforward multiplication
? is ?? and the jth element is ??/aj, j = 2, , k. shows that the first element of A-1β1j
?2A-1(X?X)-1(A-1)?. But A-1 is a (v) From part (iii), the estimated variance matrix of β is ?symmetric, diagonal matrix, as described above. The estimated variance of ?j is the jth
th2?2A-1(X?X)-1A-1, which is easily seen to be = ??2cjj/a?diagonal element of ?, where c is the j jjj?cjj/|aj|, is se(?j), which is simply diagonal element of (X?X)-1. The square root of this, ?se(?j)/|aj|.
(vi) The t statistic for ?j is, as usual,
?/aj)/[se(??)/|aj|], ?j/se(?j) = (?jj
?|/|aj|)/[se(??)/|aj|] = |??|/se(??), which is just the absolute value and so the absolute value is (|?jjjj?. If aj > 0, the t statistics themselves are identical; if aj < 0, the t statistics of the t statistic for ?jare simply opposite in sign.
垐?|X)?Gβ?δ. |X)?E(Gβ|X)?GE(βE.4 (i) E(δ
垐?|X)]G??G[?2(X?X)?1]G???2G[(X?X)?1]G?. |X)?Var(Gβ|X)?G[Var(β (ii) Var(δ
(iii) The vector of regression coefficients from the regression y on XG-1 is
[(XG?1)?XG?1]?1(XG?1)?y?[(G?1)?X?XG?1]?1(G?1)?X?y
?G(X?X)?[(G?1)?]?1(G?)?1X?y?. ?G(X?X)?G?(G?)?1X?y?G(X?X)?X?y?δ
Further, as shown in Problem E.3, the residuals are the same as from the regression y on X, and
?2, is the same. Therefore, the estimated variance matrix is so the error variance estimate, ?
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伍德里奇计量经济学第六版答案Appendix-E
