Solution to Chapter 2 Analytical Exercises

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1 Nov. 25, 23, Revised Dec. 27, 23 Hayashi Ecoometrics Solutio to Chapter 2 Aalytical Exercises. For ay ε >, So, plim z =. O the other had, which meas that lim E(z =. 2. As show i the hit, Prob( z > ε = as. E(z = + 2 =, (z µ 2 = (z E(z 2 + 2(z E(z (E(z µ + (E(z µ 2. Take the expectatio of both sides to obtai E(z µ 2 = E(z E(z Ez E(z (E(z µ + (E(z µ 2 = Var(z + (E(z µ 2 (because Ez E(z = E(z E(z =. Take the limit as of both sides to obtai lim E(z µ 2 = lim Var(z + lim (E(z µ 2 = (because lim E(z = µ, Therefore, z m.s. µ. By Lemma 2.2(a, this implies z p µ. lim Var(z =. 3. (a Sice a i.i.d. process is ergodic statioary, Assumptio 2.2 is implied by Assumptio 2.2. Assumptios 2. ad 2.2 imply that g i x i ε i is i.i.d. Sice a i.i.d. process with mea zero is mds (martigale differeces, Assumptio 2.5 is implied by Assumptios 2.2 ad 2.5. (b Rewrite the OLS estimator as b β = (X X X ε = S xx g. (A Sice by Assumptio 2.2 {x i } is i.i.d., {x i x i } is i.i.d. So by Kolmogorov s Secod Strog LLN, we obtai S xx p Σ xx The covergece is actually almost surely, but almost sure covergece implies covergece i probability. Sice Σ xx is ivertible by Assumptio 2.4, by Lemma 2.3(a we get S xx p Σ xx.

2 Similarly, uder Assumptio 2. ad 2.2 {g i } is i.i.d. By Kolmogorov s Secod Strog LLN, we obtai g p E(g i, which is zero by Assumptio 2.3. So by Lemma 2.3(a, S xx g p Σ xx =. Therefore, plim (b β = which implies that the OLS estimator b is cosistet. Next, we prove that the OLS estimator b is asymptotically ormal. Rewrite equatio(a above as (b β = S xx g. As already observed, {g i } is i.i.d. with E(g i =. The variace of g i equals E(g i g i = S sice E(g i = by Assumptio 2.3. So by the Lideberg-Levy CLT, g d N(, S. Furthermore, as already oted, S xx p Σ xx. Thus by Lemma 2.4(c, (b β d N(, Σ xx S Σ xx. 4. The hit is as good as the aswer. 5. As show i the solutio to Chapter Aalytical Exercise 5, SSR R SSR U ca be writte as SSR R SSR U = (Rb r R(X X R (Rb r. Usig the restrictios of the ull hypothesis, Rb r = R(b β Also R(X X R = RS xxr. So Thus = R(X X X ε (sice b β = (X X X ε = RS xxg (where g x i ε i.. SSR R SSR U = ( g S xx R (R S xx R R S xx( g. i= SSR R SSR U s 2 = ( g S xx R (s 2 R S xx R R S xx( g = z A z, where z R S xx( g, A s 2 R S xx R. 2

3 By Assumptio 2.2, plim S xx = Σ xx. By Assumptio 2.5, g d N(, S. So by Lemma 2.4(c, we have: z d N(, RΣ xx SΣ xx R. But, as show i (2.6.4, S = σ 2 Σ xx uder coditioal homoekedasticity (Assumptio 2.7. So the expressio for the variace of the limitig distributio above becomes Thus we have show: RΣ xx SΣ xx R = σ 2 RΣ xxr A. z d z, z N(, A. As already observed, S xx p Σ xx. By Assumptio 2.7, σ 2 = E(ε 2 i. So by Propositio 2.2, s 2 p σ 2. Thus by Lemma 2.3(a (the Cotiuous Mappig Theorem, A p A. Therefore, by Lemma 2.4(d, z A z d z A z. But sice Var(z = A, the distributio of z A z is chi-squared with #z degrees of freedom. 6. For simplicity, we assumed i Sectio 2.8 that {y i, x i } is i.i.d. Collectig all the assumptios made i Sectio 2.8, (i (liearity y i = x i β + ε i. (ii (radom sample {y i, x i } is i.i.d. (iii (rak coditio E(x i x i is o-sigular. (iv E(ε 2 i x ix i is o-sigular. (v (stroger versio of orthogoality E(ε i x i = (see ( (vi (parameterized coditioal heteroskedasticity E(ε 2 i x i = z i α. These coditios together are stroger tha Assumptios (a We wish to verify Assumptios for the regressio equatio ( Clearly, Assumptio 2. about the regressio equatio (2.8.8 is satisfied by (i about the origial regressio. Assumptio 2.2 about (2.8.8 (that {ε 2 i, x i} is ergodic statioary is satisfied by (i ad (ii. To see that Assumptio 2.3 about (2.8.8 (that E(z i η i = is satisfied, ote first that E(η i x i = by costructio. Sice z i is a fuctio of x i, we have E(η i z i = by the Law of Iterated Expectatio. Therefore, Assumptio 2.3 is satisfied. The additioal assumptio eeded for (2.8.8 is Assumptio 2.4 that E(z i z i be osigular. With Assumptios satisfied for (2.8.8, the OLS estimator α is cosistet by Propositio 2.(a applied to ( (b Note that α α = ( α α ( α α ad use the hit. (c Regardig the first term of (, by Kolmogorov s LLN, the sample mea i that term coverges i probability to E(x i ε i z i provided this populatio mea exists. But E(x i ε i z i = Ez i x i E(ε i z i. By (v (that E(ε i x i = ad the Law of Iterated Expectatios, E(ε i z i =. Thus E(x i ε i z i =. Furthermore, plim(b β = sice b is cosistet whe Assumptios 3

4 (which are implied by Assumptios (i-(vi above are satisfied for the origial regressio. Therefore, the first term of ( coverges i probability to zero. Regardig the secod term of (, the sample mea i that term coverges i probability to E(x 2 i z i provided this populatio mea exists. The the secod term coverges i probability to zero because plim(b β =. (d Multiplyig both sides of ( by, ( ( α α = z i z i z i v i i= i= ( = z i z i 2 (b β x i ε i z i + (b β (b β i= i= x 2 i z i. Uder Assumptios for the origial regressio (which are implied by Assumptios (i-(vi above, (b β coverges i distributio to a radom variable. As show i (c, i= x iε i z i p. So by Lemma 2.4(b the first term i the brackets vaishes (coverges to zero i probability. As show i (c, (b β i= x2 i z i vaishes provided E(x 2 i z i exists ad is fiite. So by Lemma 2.4(b the secod term, too, vaishes. Therefore, ( α α vaishes, provided that E(zi z i is o-sigular. 7. This exercise is about the model i Sectio 2.8, so we cotiue to maitai Assumptios (i- (vi listed i the solutio to the previous exercise. Give the hit, the oly thig to show is that the LHS of ( equals Σ xx S Σ xx, or more specifically, that plim X VX = S. Write S as S = E(ε 2 i x i x i = EE(ε 2 i x i x i x i = E(z iα x i x i (sice E(ε 2 i x i = z iα by (vi. Sice x i is i.i.d. by (ii ad sice z i is a fuctio of x i, z i αx ix i is i.i.d. So its sample mea coverges i probability to its populatio mea E(z i α x ix i, which equals S. The sample mea ca be writte as z iαx i x i i= = v i x i x i (by the defiitio of v i, where v i is the i-th diagoal elemet of V i= = X VX. 8. See the hit. 9. (a E(g t g t, g t 2,..., g 2 = EE(g t ε t, ε t 2,..., ε g t, g t 2,..., g 2 (by the Law of Iterated Expectatios = EE(ε t ε t ε t, ε t 2,..., ε g t, g t 2,..., g 2 = Eε t E(ε t ε t, ε t 2,..., ε g t, g t 2,..., g 2 (by the liearity of coditioal expectatios = (sice E(ε t ε t, ε t 2,..., ε =. i= 4

5 (b E(g 2 t = E(ε 2 t ε 2 t But = EE(ε 2 t ε 2 t ε t, ε t 2,..., ε (by the Law of Total Expectatios = EE(ε 2 t ε t, ε t 2,..., ε ε 2 t (by the liearity of coditioal expectatios = E(σ 2 ε 2 t (sice E(ε 2 t ε t, ε t 2,..., ε = σ 2 = σ 2 E(ε 2 t. E(ε 2 t = EE(ε 2 t ε t, ε t 2,..., ε = E(σ 2 = σ 2. (c If {ε t } is ergodic statioary, the {ε t ε t } is ergodic statioary (see, e.g., Remark 5.3 o p. 488 of S. Karli ad H. Taylor, A First Course i Stochastic Processes, 2d. ed., Academic Press, 975, which states that For ay fuctio φ, the sequece Y = φ(x, X +,... geerates a ergodic statioary process wheever {X } is ergodic statioary. Thus the Billigsley CLT (see p. 6 of the text is applicable to γ = t=j+ g t. (d Sice ε 2 t is ergodic statioary, γ coverges i probability to E(ε 2 t = σ 2. As show i (c, γ d N(, σ 4. So by Lemma 2.4(c bγ bγ d N(,.. (a Clearly, E(y t = for all t =, 2,.... ( + θ 2 + θ2σ 2 ε 2 for j = (θ + θ θ 2 σε 2 for j =, Cov(y t, y t j = θ 2 σε 2 for j = 2, for j > 2, (b So either E(y t or Cov(y t, y t j depeds o t. E(y t y t j, y t j,..., y, y = E(y t ε t j, ε t j,..., ε, ε (as oted i the hit = E(ε t + θ ε t + θ 2 ε t 2 ε t j, ε t j,..., ε, ε ε t + θ ε t + θ 2 ε t 2 for j =, θ ε t + θ 2 ε t 2 for j =, = θ 2 ε t 2 for j = 2, for j > 2, which gives the desired result. 5

6 (c Var( y = Cov(y, y + + y + + Cov(y, y + + y = (γ + γ + + γ 2 + γ + (γ + γ + γ + + γ (γ + γ γ + γ = γ + 2( γ + + 2( jγ j + + 2γ ( = γ + 2 j γ j. j= (This is just reproducig (6.5.2 of the book. Sice γ j = for j > 2, oe obtais the desired result. (d To use Lemma 2., oe sets z = y. However, Lemma 2., as stated i the book, iadvertetly misses the required coditio that there exist a M > such that E( z s+δ < M for all for some δ >. Provided this techical coditio is satisfied, the variace of the limitig distributio of y is the limit of Var( y, which is γ + 2(γ + γ 2.. (a I the auxiliary regressio, the vector of the depedet variable is e ad the matrix of regressors is X.. E. Usig the OLS formula, α = B X e E e X e = by the ormal equatios for the origial regressio. The j-th elemet of E e is which equals γ j defied i (2..9. (e j+e + + e e j =. t=j+ e t e t j. (b The j-th colum of X E is t=j+ x t e t j (which, icidetally, equals µ j defied o p. 47 of the book. Rewrite it as follows. x t e t j t=j+ = = t=j+ t=j+ x t (ε t j x t j(b β x t ε t j t=j+ x t x t j (b β The last term vaishes because b is cosistet for β. Thus t=j+ x t e t j coverges i probability to E(x t ε t j. The (i, j elemet of the symmetric matrix E E is, for i j, (e +i je + + e j e i = j t=+i j e t e t (i j. 6

7 Usig the relatio e t = ε t x t(b β, this ca be rewritte as j t=+i j ε t ε t (i j (b β ( j t=+i j (x t ε t (i j + x t (i j ε t (b β j t=+i j x t x t (i j (b β. The type of argumet that is by ow routie (similar to the oe used o p. 45 for (2.. shows that this expressio coverges i probability to γ i j, which is σ 2 for i = j ad zero for i j. (c As show i (b, plim B = B. Sice Σ xx is o-sigular, B is o-sigular. So coverges i probability to B. Also, usig a argumet similar to the oe used i (b for showig that plim E E = I p, we ca show that plim γ =. Thus the formula i (a shows that α coverges i probability to zero. (d (The hit should have bee: E e = γ. Show that SSR = e e α. The SSR from γ the auxiliary regressio ca be writte as SSR = (e X.. E α (e X.. E α = (e X.. E α e (by the ormal equatio for the auxiliary regressio = e e α X.. E e = e e α X e = e e α γ E e (sice X e = ad E e = γ. B As show i (c, plim α = ad plim γ =. By Propositio 2.2, we have plim e e = σ 2. Hece SSR/ (ad therefore SSR/( K p coverges to σ 2 i probability. (e Let R (p K.. I p, V X. E. The F -ratio is for the hypothesis that Rα =. The F -ratio ca be writte as F = (R α R(V V R (R α/p. ( SSR/( K p 7

8 Usig the expressio for α i (a above, R α ca be writte as. R α =. I p B (K = (p K (p K. I p B (K K B 2 (p K γ (p B 2 (K p B 22 (p p (K γ (p = B 22 γ. ( Also, R(V V R i the expressio for F ca be writte as R(V V R = R B R = (p K (sice V V = B.. I p B (K K B 2 (p K B 2 (K p B 22 (p p (K p I p = B 22. ( Substitutio of ( ad ( ito ( produces the desired result. (f Just apply the formula for partitioed iverses. (g Sice ρ γ/σ 2 p ad Φ p Φ, it should be clear that the modified Box-Pierce Q (= ρ (I p Φ ρ is asymptotically equivalet to γ (I p Φ γ/σ4. Regardig the pf statistic give i (e above, cosider the expressio for B 22 give i (f above. Sice the j-th elemet of X E is µ j defied right below (2..9 o p. 47, we have ( s 2 Φ = E X Sxx( X E, so B 22 =. E E s 2 Φ As show i (b, E E p σ 2 I p. Therefore, B 22 p σ 2 (I p Φ, ad pf is asymptotically equivalet to γ (I p Φ γ/σ The hits are almost as good as the aswer. Here, we give solutios to (b ad (c oly. (b We oly prove the first covergece result. ( ( x t x t = r x t x t = λ r r t= t= x t x t. The term i paretheses coverges i probability to Σ xx as (ad hece r goes to ifiity. (c We oly prove the first covergece result. ( r x t ε t = x t ε t = ( λ x t ε t. r r t= t= The term i paretheses coverges i distributio to N(, σ 2 Σ xx as (ad hece r goes to ifiity. So the whole expressio coverges i distributio to N(, λ σ 2 Σ xx. t= t= 8

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