Solution to Chapter 3 Analytical Exercises

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1 December 7, 003 Solutio to Chapter 3 Aalytical Exercises Hayashi Ecoometrics. If A is symmetric a iempotet, the A = A a AA = A. So x Ax = x AAx = x A Ax = z z 0 where z Ax.. (a) By assumptio, {x i, ε i } is joitly statioary a ergoic, so by ergoic theorem the first term of ( ) coverges almost surely to E(x i ε i ) which exists a is fiite by Assumptio 3.5. (b) z i x i ε i is the prouct of x i ε i a x i z i. By usig the Cauchy-Schwarts iequality, we obtai E( x i ε i x i z i ) E(x i ε i ) E(x i z i ). E(x i ε i ) exists a is fiite by Assumptio 3.5 a E(x i z i ) exists a is fiite by Assumptio 3.6. Therefore, E( x i z i x i ε i ) is fiite. Hece, E(x i z i x i ε i ) exists a is fiite. (c) By ergoic statioarity the sample average of z i x i ε i coverges i probability to some fiite umber. Because δ is cosistet for δ by Propositio 3., δ δ coverges to 0 i probability. Therefore, the seco term of ( ) coverges to zero i probability. () By ergoic statioarity a Assumptio 3.6 the sample average of zi x i coverges i probability to some fiite umber. As metioe i (c) δ δ coverges to 0 i probability. Therefore, the last term of ( ) vaishes. 3. (a) Q Σ xzs Σ xz Σ xzwσ xz (Σ xzwswσ xz ) Σ xzwσ xz = Σ xzc CΣ xz Σ xzwσ xz (Σ xzwc C WΣxz ) Σ xzwσ xz = H H Σ xzwσ xz (G G) Σ xzwσ xz = H H H G(G G) G H = H [I K G(G G) G ]H = H M G H. (b) First, we show that M G is symmetric a iempotet. M G = I K G(G(G G) ) = I K G((G G) G ) = I K G(G G) G = M G. M G M G = I K I K G(G G) G I K I K G(G G) G + G(G G) G G(G G) G = I K G(G G) G = M G. Thus, M G is symmetric a iempotet. For ay L-imesioal vector x, x Qx = x H M G Hx Therefore, Q is positive semiefiite. = z M G z (where z Hx) 0 (sice M G is positive semiefiite).

2 4. (the aswer o p. 54 of the book simplifie) If W is as efie i the hit, the WSW = W a Σ xzwσ xz = Σ zz A Σ zz. So (3.5.) reuces to the asymptotic variace of the OLS estimator. By (3.5.), it is o smaller tha (Σ xz S Σ xz ), which is the asymptotic variace of the efficiet GMM estimator. 5. (a) From the expressio for δ(ŝ ) (give i (3.5.)) a the expressio for g ( δ) (give i (3.4.)), it is easy to show that g ( δ(ŝ )) = Bs xy. But Bs xy = Bg because Bs xy = (I K S xz (S xz Ŝ S xz ) S xz Ŝ )s xy = (I K S xz (S xz Ŝ S xz ) S xz Ŝ )(S xz δ + g) (sice y i = z iδ + ε i ) = (S xz S xz (S xz Ŝ S xz ) S xz Ŝ S xz )δ + (I K S xz (S xz Ŝ S xz ) S xz Ŝ )g = (S xz S xz )δ + Bg = Bg. (b) Sice Ŝ = C C, we obtai B Ŝ B = B C C B = (C B) (C B). But C B = C(I K S xz (S xz Ŝ S xz ) S xz Ŝ ) = C CS xz (S xz C CS xz ) S xz C C = C A(A A) A C (where A CS xz ) = [I K A(A A) A ]C MC. So B Ŝ B = (MC) (MC) = C M MC. It shoul be routie to show that M is symmetric a iempotet. Thus B Ŝ B = C MC. The rak of M equals its trace, which is trace(m) = trace(i K A(A A) A ) = trace(i K ) trace(a(a A) A ) = trace(i K ) trace(a A(A A) ) = K trace(i L ) = K L. (c) As efie i (b), C C = Ŝ. Let D be such that D D = S. The choice of C a D is ot uique, but it woul be possible to choose C so that plim C = D. Now, v (Cg) = C( g). By usig the Ergoic Statioary Martigale Differeces CLT, we obtai g N(0, S). So v = C( g) N(0, Avar(v)) where Avar(v) = DSD = D(D D) D = DD D D = I K.

3 () J( δ(ŝ ), Ŝ ) = g ( δ(ŝ )) Ŝ g ( δ(ŝ )) = ( Bg) Ŝ ( Bg) (by (a)) = g B Ŝ Bg = g C MCg (by (b)) = v Mv (sice v Cg). Sice v N(0, I K ) a M is iempotet, v Mv is asymptotically chi-square with egrees of freeom equalig the rak of M = K L. 6. From Exercise 5, J = g B Ŝ Bg. Also from Exercise 5, Bg = Bsxy. 7. For the most parts, the hits are early the aswer. Here, we provie aswers to (), (f), (g), (i), a (j). () As show i (c), J = v M v. It suffices to prove that v = C F C v. v C g = C F g = C F C Cg = C F C Cg = C F C v (sice v Cg). (f) Use the hit to show that A D = 0 if A M = 0. It shoul be easy to show that A M = 0 from the efiitio of M. (g) By the efiitio of M i Exercise 5, MD = D A(A A) A D. So MD = D sice A D = 0 as show i the previous part. Sice both M a D are symmetric, DM = D M = (MD) = D = D. As show i part (e), D is iempotet. Also, M is iempotet as show i Exercise 5. So (M D) = M DM MD + D = M D. As show i Exercise 5, the trace of M is K L. As show i (e), the trace of D is K L. So the trace of M D is K K. The rak of a symmetric a iempotet matrix is its trace. (i) It has bee show i Exercise 6 that g C MCg = s xyc MCs xy sice C MC = B Ŝ B. Here, we show that g C DCg = s xyc DCs xy. g C DCg = g FC M C F g (C DC = FC M C F by the efiitio of D i ()) = g F B (Ŝ) B F g (sice C M C = B (Ŝ) B from (a)) = g B (Ŝ) B g (sice g = F g). From the efiitio of B a the fact that s x y = S x zδ + g, it follows that B g = B s x y. So g B (Ŝ) B g = s B x y (Ŝ) B s xy = s xyf B (Ŝ) B F s xy (sice s x y = F s xy ) = s xyfc M C F s xy (sice B (Ŝ) B = C M C from (a)) = s xyc DCs xy. 3

4 (j) M D is positive semi-efiite because it is symmetric a iempotet. 8. (a) Solve the first-orer coitios i the hit for δ to obtai δ = δ(ŵ) (S xzŵs xz) R λ. Substitute this ito the costrait Rδ = r to obtai the expressio for λ i the questio. The substitute this expressio for λ ito the above equatio to obtai the expressio for δ i the questio. (b) The hit is almost the aswer. (c) What ees to be show is that ( δ(ŵ) δ) (S xzŵsxz)( δ(ŵ) δ) equals the Wal statistic. But this is immeiate from substitutio of the expressio for δ i (a). 9. (a) By applyig (3.4.), we obtai [ ] [ ] ( δ δ) (S xz Ŵ S xz ) S xzŵ g. = ( δ δ) (S xzŵs xz ) S xzŵ By usig Billigsley CLT, we have g N(0, S). Also, we have [ ] (S xz Ŵ S xz ) S xzŵ (S xzŵs xz ) S xzŵ p [ ] Q Σ xzw Q. Σ xzw Therefore, by Lemma.4(c), [ ] ( δ δ) N ( δ δ) = N ( 0, ( 0, [ ] Q Σ xzw Q Σ xzw [ ]) A A. A A S (W Σ xz Q ).. W Σ xz Q ) (b) q ca be rewritte as q = ( δ δ ) = ( δ δ) ( δ δ) = [ ] [ ] ( δ δ). ( δ δ) Therefore, we obtai q N(0, Avar(q)). where Avar(q) = [ ] [ ] [ ] A A = A A A + A A A. 4

5 0. (a) (c) Sice W = S, Q, A, A, a A ca be rewritte as follows: Q = Σ xzw Σ xz = Σ xzs Σ xz, A = Q Σ xzw S S Σ xz Q = Q (Σ xzw Σ xz )Q = Q Q Q = Q, A = Q Σ xzs SW Σ xz Q = Q, A = (Σ xzs Σ xz ) Σ xzs SS Σ xz (Σ xzs Σ xz ) = (Σ xzs Σ xz ) = Q. Substitutio of these ito the expressio for Avar(q) i (b), we obtai (b) From the efiitio of δ, Avar(q) = A Q = A (Σ xzs Σ xz ) σ xz E(x i z i ) = E(x i (x i β + v i )) δ δ = = Avar( δ(ŵ)) Avar( δ(ŝ )). = β E(x i ) + E(x i v i ) = βσ x 0 (by assumptios (), (3), a (4)). ( ) x i z i x i ε i = s xz x i ε i. We have x i z i = x i (x i β + v i ) = x i β + x iv i, which, beig a fuctio of (x i, η i ), is ergoic statioary by assumptio (). So by the Ergoic theorem, s xz p σ xz. Sice σ xz 0 by (a), we have s xz p σxz. By assumptio (), E(x i ε i ) = 0. So by assumptio (), we have x iε i p 0. Thus δ δ p 0. (c) s xz x i z i = = (x i β + x i v i ) x i + p 0 E(x i ) + E(x i v i ) = 0 x i v i (sice β = ) 5

6 () sxz = x i + x i v i. By assumptio () a the Ergoic Theorem, the first term of RHS coverges i probability to E(x i ) = σ x > 0. Assumptio () a the Martigale Differeces CLT imply that x i v i a N(0, s ). Therefore, by Lemma.4(a), we obtai sxz σ x + a. (e) δ δ ca be rewritte as δ δ = ( s xz ) g. From assumptio () a the Martigale Differeces CLT, we obtai g b N(0, s ). where s is the (, ) elemet of S. By usig the result of () a Lemma.3(b), δ δ (σ x + a) b. (a, b) are joitly ormal because the joit istributio is the limitig istributio of [ ] g g = ( x. iv i ) (f) Because δ δ coverges i istributio to (σ x + a) b which is ot zero, the aswer is No. 6