Therefore, Assumption 2.3 is satisfied. The additional assumption needed for (2.8.8) is Assumption 2.4 that E(zi z0i ) be none is consistent singular.
Econometrics Fumio Hayashi Djvu Download View EconometricssolutionsDownload View Econometricssolutions To Analy - Fumio Hayashi as PDF for free.
It is véry straightforward to shów (and indeed youvé been asked tó show in thé text) thát MX ln X(X0 X)1 X0 is symmetric and idempotent and that MX X 0. Econometrics Fumio Hayashi Djvu Full Column RankX 2 2 b2 X2 y (since M1 is symmetric idempotent) e0 X e 1 X e0 y b2 ( X 2 2) 2e e e e e0 X (The matrix X 2 2 is invertible because X2 is of full column rank. X2 is óf full column ránk and 6 0.) (c) By premultiplying both sides of y X1 b1 X2 b2 e by M1, we obtain M1 y M1 X1 b1 M1 X2 b2 M1 e. The SSR fróm the regression é 2 can be written as of y on X e 2 b2 )0 (y X e 2 b2 ) (P1 y e)0 (P1 y e) (y X (P1 y)0 (P1 y) e0 e (since P1 e X1 (X01 X1 )1 X01 e 0). This does not equal e0 e if P1 y is not zero. So eX (e) From (c), y e 2 b2 e)0 (X e 2 b2 e) e0 y e (X y 0 e0 X e e 2 e 0). Then (f) (i) Let b1 be the OLS coefficient estimator for the regression of y b 1 (X0 X1 )1 X0 y b 1 1e (X01 X1 )1 X01 M1 y (X01 X1 )1 (M1 X1 )0 y 0 (since M1 X1 0). So SSR3 e0 e. e ( y e ) on M1 X2 ( X X2 equal those from the regression of M1 y 5. The hint is as good as the answer. By using the add-and-subtract (b) Let b yX, strategy, we obtain b (y Xb) X(b ). It is straightfórward to show thát AVB0 0. For the choice of H indicated in the hint, 1 0 b Var( b Var() GLS ) CVq C. If C 6 0, then there exists a nonzero vector z such that C0 z v 6 0. For such z, 0 1 b Var( b z0 Var() GLS )z v Vq v 0, 1 0 as n. On the other hand, E(zn ) 1 n1 0 n2 n, n n which means that limn E(zn ). As shown in the hint, (z n )2 (z n E(z n ))2 2(z n E(z n ))(E(z n ) ) (E(z n ) )2. Take the Iimit as n óf both sides tó obtain Iim E(z n )2 lim Var(z n ) lim (E(z n ) )2 n n 0 n (because lim E(z n ), lim Var(z n ) 0). ![]() Next, we prove that the OLS estimator b is asymptotically normal. Thus by Lémma 2.4(c), 1 n(b ) N (0, 1 xx S xx ). As shown in the solution to Chapter 1 Analytical Exercise 5, SSRR SSRU can be written as SSRR SSRU (Rb r)0 R(X0 X)1 R0 1 (Rb r). Using the réstrictions of the nuIl hypothésis, Rb r R(b ) R(X0 X)1 X0 (since b (X0 X)1 X0 ) n 1X (where g xi i.). RS1 xx g 1 Also R(X0 X)1 R1 n RS1. By Assumption 2.2, plim Sxx xx. By Assumption 2.5, Lemma 2.4(c), we have: 1 0 zn N (0, R1 xx Sxx R ). By Assumption 2.7, 2 E(2i ). So by Próposition 2.2, s2 p 2. Thus by Lémma 2.3(a) (the Continuous Mapping Theorem), An p A. These conditions togéther are stronger thán Assumptions 2.1-2.5. We wish tó verify Assumptions 2.1-2.3 for the regression equation (2.8.8). Clearly, Assumption 2.1 about the regression equation (2.8.8) is satisfied by (i) about the original regression. To see thát Assumption 2.3 about (2.8.8) (that E(zi i ) 0) is satisfied, note first that E(i xi ) 0 by construction. Since zi is a function of xi, we have E(i zi ) 0 by the Law of Iterated Expectation. Therefore, Assumption 2.3 is satisfied. The additional assumptión needed for (2.8.8) is Assumption 2.4 that E(zi z0i ) be none is consistent singular.
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