econometrics_solutions to analy - fumio hayashi

44
Nov. 22, 2003, revised Dec. 27, 2003 Hayashi Econometrics Solution to Chapter 1 Analytical Exercises 1. (Reproducing the answer on p. 84 of the book) (y-X β) (y - X β) = [(y - Xb)+ X(b - β)] [(y - Xb)+ X(b - β)] (by the add-and-subtract strategy) = [(y - Xb) +(b - β) X ][(y - Xb)+ X(b - β)] =(y - Xb) (y - Xb)+(b - β) X (y - Xb) +(y - Xb) X(b - β)+(b - β) X X(b - β) =(y - Xb) (y - Xb) + 2(b - β) X (y - Xb)+(b - β) X X(b - β) (since (b - β) X (y - Xb)=(y - Xb) X(b - β)) =(y - Xb) (y - Xb)+(b - β) X X(b - β) (since X (y - Xb)= 0 by the normal equations) (y - Xb) (y - Xb) (since (b - β) X X(b - β)= z z = n i=1 z 2 i 0 where z X(b - β)). 2. (a), (b). If X is an n × K matrix of full column rank, then X X is symmetric and invertible. It is very straightforward to show (and indeed you’ve been asked to show in the text) that M X I n - X(X X) -1 X is symmetric and idempotent and that M X X = 0. In this question, set X = 1 (vector of ones). (c) M 1 y =[I n - 1(1 1) -1 1 ]y = y - 1 n 11 y (since 1 1 = n) = y - 1 n 1 n i=1 y i = y - 1· y (d) Replace “y” by “X” in (c). 3. Special case of the solution to the next exercise. 4. From the normal equations (1.2.3) of the text, we obtain (a) X 1 X 2 [X 1 . . . X 2 ] b 1 b 2 = X 1 X 2 y. Using the rules of multiplication of partitioned matrices, it is straightforward to derive (*) and (**) from the above. 1

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Page 1: Econometrics_solutions to Analy - Fumio Hayashi

Nov. 22, 2003, revised Dec. 27, 2003 Hayashi Econometrics

Solution to Chapter 1 Analytical Exercises

1. (Reproducing the answer on p. 84 of the book)

(y−Xβ)′(y −Xβ) = [(y −Xb) + X(b− β)]′[(y −Xb) + X(b− β)](by the add-and-subtract strategy)

= [(y −Xb)′ + (b− β)′X′][(y −Xb) + X(b− β)]

= (y −Xb)′(y −Xb) + (b− β)′X′(y −Xb)

+ (y −Xb)′X(b− β) + (b− β)′X′X(b− β)

= (y −Xb)′(y −Xb) + 2(b− β)′X′(y −Xb) + (b− β)′X′X(b− β)

(since (b− β)′X′(y −Xb) = (y −Xb)′X(b− β))

= (y −Xb)′(y −Xb) + (b− β)′X′X(b− β)(since X′(y −Xb) = 0 by the normal equations)

≥ (y −Xb)′(y −Xb)

(since (b− β)′X′X(b− β) = z′z =n∑i=1

z2i ≥ 0 where z ≡ X(b− β)).

2. (a), (b). If X is an n×K matrix of full column rank, then X′X is symmetric and invertible.It is very straightforward to show (and indeed you’ve been asked to show in the text) thatMX ≡ In−X(X′X)−1X′ is symmetric and idempotent and that MXX = 0. In this question,set X = 1 (vector of ones).

(c)

M1y = [In − 1(1′1)−11′]y

= y − 1n

11′y (since 1′1 = n)

= y − 1n

1n∑i=1

yi = y − 1· y

(d) Replace “y” by “X” in (c).

3. Special case of the solution to the next exercise.

4. From the normal equations (1.2.3) of the text, we obtain

(a) [X′1X′2

][X1

... X2][

b1

b2

]=[

X′1X′2

]y.

Using the rules of multiplication of partitioned matrices, it is straightforward to derive (∗)and (∗∗) from the above.

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Page 2: Econometrics_solutions to Analy - Fumio Hayashi

(b) By premultiplying both sides of (∗) in the question by X1(X′1X1)−1, we obtain

X1(X′1X1)−1X′1X1b1 = −X1(X′1X1)−1X′1X2b2 + X1(X′1X1)−1X′1y

⇔ X1b1 = −P1X2b2 + P1y

Substitution of this into (∗∗) yields

X′2(−P1X2b2 + P1y) + X′2X2b2 = X′2y

⇔ X′2(I−P1)X2b2 = X′2(I−P1)y⇔ X′2M1X2b2 = X′2M1y

⇔ X′2M′1M1X2b2 = X′2M

′1M1y (since M1 is symmetric & idempotent)

⇔ X′2X2b2 = X′2y.

Therefore,

b2 = (X′2X2)−1X′2y

(The matrix X′2X2 is invertible because X2 is of full column rank. To see that X2 is of fullcolumn rank, suppose not. Then there exists a non-zero vector c such that X2c = 0. But

X2c = X2c−X1d where d ≡ (X′1X1)−1X′1X2c. That is, Xπ = 0 for π ≡[−dc

]. This is

a contradiction because X = [X1

... X2] is of full column rank and π 6= 0.)(c) By premultiplying both sides of y = X1b1 + X2b2 + e by M1, we obtain

M1y = M1X1b1 + M1X2b2 + M1e.

Since M1X1 = 0 and y ≡M1y, the above equation can be rewritten as

y = M1X2b2 + M1e

= X2b2 + M1e.

M1e = e because

M1e = (I−P1)e= e−P1e

= e−X1(X′1X1)−1X′1e

= e (since X′1e = 0 by normal equations).

(d) From (b), we have

b2 = (X′2X2)−1X′2y

= (X′2X2)−1X′2M′1M1y

= (X′2X2)−1X′2y.

Therefore, b2 is the OLS coefficient estimator for the regression y on X2. The residualvector from the regression is

y − X2b2 = (y − y) + (y − X2b2)

= (y −M1y) + (y − X2b2)= (y −M1y) + e (by (c))= P1y + e.

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Page 3: Econometrics_solutions to Analy - Fumio Hayashi

This does not equal e because P1y is not necessarily zero. The SSR from the regressionof y on X2 can be written as

(y − X2b2)′(y − X2b2) = (P1y + e)′(P1y + e)

= (P1y)′(P1y) + e′e (since P1e = X1(X′1X1)−1X′1e = 0).

This does not equal e′e if P1y is not zero.

(e) From (c), y = X2b2 + e. So

y′y = (X2b2 + e)′(X2b2 + e)

= b′2X′2X2b2 + e′e (since X2e = 0).

Since b2 = (X′2X2)−1X′2y, we have b′2X′2X2b2 = y′X2(X′2M1X2)−1X2y.

(f) (i) Let b1 be the OLS coefficient estimator for the regression of y on X1. Then

b1 = (X′1X1)−1X′1y

= (X′1X1)−1X′1M1y

= (X′1X1)−1(M1X1)′y= 0 (since M1X1 = 0).

So SSR1 = (y −X1b1)′(y −X1b1) = y′y.(ii) Since the residual vector from the regression of y on X2 equals e by (c), SSR2 = e′e.

(iii) From the Frisch-Waugh Theorem, the residuals from the regression of y on X1 andX2 equal those from the regression of M1y (= y) on M1X2 (= X2). So SSR3 = e′e.

5. (a) The hint is as good as the answer.

(b) Let ε ≡ y−Xβ, the residuals from the restricted regression. By using the add-and-subtractstrategy, we obtain

ε ≡ y −Xβ = (y −Xb) + X(b− β).

So

SSRR = [(y −Xb) + X(b− β)]′[(y −Xb) + X(b− β)]

= (y −Xb)′(y −Xb) + (b− β)′X′X(b− β) (since X′(y −Xb) = 0).

But SSRU = (y −Xb)′(y −Xb), so

SSRR − SSRU = (b− β)′X′X(b− β)

= (Rb− r)′[R(X′X)−1R′]−1(Rb− r) (using the expresion for β from (a))

= λ′R(X′X)−1R′λ (using the expresion for λ from (a))

= ε′X(X′X)−1X′ε (by the first order conditions that X′(y −Xβ) = R′λ)

= ε′Pε.

(c) The F -ratio is defined as

F ≡ (Rb− r)′[R(X′X)−1R′]−1(Rb− r)/rs2

(where r = #r) (1.4.9)

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Page 4: Econometrics_solutions to Analy - Fumio Hayashi

Since (Rb− r)′[R(X′X)−1R′]−1(Rb− r) = SSRR − SSRU as shown above, the F -ratiocan be rewritten as

F =(SSRR − SSRU )/r

s2

=(SSRR − SSRU )/r

e′e/(n−K)

=(SSRR − SSRU )/rSSRU/(n−K)

Therefore, (1.4.9)=(1.4.11).

6. (a) Unrestricted model: y = Xβ + ε, where

y(N×1)

=

y1

...yn

, X(N×K)

=

1 x12 . . . x1K

......

. . ....

1 xn2 . . . xnK

, β(K×1)

=

β1

...βn

.Restricted model: y = Xβ + ε, Rβ = r, where

R((K−1)×K)

=

0 1 0 . . . 00 0 1 . . . 0...

.... . .

0 0 1

, r((K−1)×1)

=

0...0

.Obviously, the restricted OLS estimator of β is

β(K×1)

=

y0...0

. So Xβ =

yy...y

= 1· y.

(You can use the formula for the unrestricted OLS derived in the previous exercise, β =b − (X′X)−1R′[R(X′X)−1R′]−1(Rb − r), to verify this.) If SSRU and SSRR are theminimized sums of squared residuals from the unrestricted and restricted models, they arecalculated as

SSRR = (y −Xβ)′(y −Xβ) =n∑i=1

(yi − y)2

SSRU = (y −Xb)′(y −Xb) = e′e =n∑i=1

e2i

Therefore,

SSRR − SSRU =n∑i=1

(yi − y)2 −n∑i=1

e2i . (A)

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Page 5: Econometrics_solutions to Analy - Fumio Hayashi

On the other hand,

(b− β)′(X′X)(b− β) = (Xb−Xβ)′(Xb−Xβ)

=n∑i=1

(yi − y)2.

Since SSRR − SSRU = (b− β)′(X′X)(b− β) (as shown in Exercise 5(b)),

n∑i=1

(yi − y)2 −n∑i=1

e2i =

n∑i=1

(yi − y)2. (B)

(b)

F =(SSRR − SSRU )/(K − 1)∑n

i=1 e2i /(n−K)

(by Exercise 5(c))

=(∑ni=1(yi − y)2 −

∑ni=1 e

2i )/(K − 1)∑n

i=1 e2i /(n−K)

(by equation (A) above)

=∑ni=1(yi − y)2/(K − 1)∑n

i=1 e2i /(n−K)

(by equation (B) above)

=

Pni=1(byi−y)2/(K−1)Pn

i=1(yi−y)2Pni=1 e

2i/(n−K)Pn

i=1(yi−y)2

(by dividing both numerator & denominator byn∑i=1

(yi − y)2)

=R2/(K − 1)

(1−R2)/(n−K)(by the definition or R2).

7. (Reproducing the answer on pp. 84-85 of the book)

(a) βGLS − β = Aε where A ≡ (X′V−1X)−1X′V−1 and b − βGLS = Bε where B ≡(X′X)−1X′ − (X′V−1X)−1X′V−1. So

Cov(βGLS − β,b− βGLS)= Cov(Aε,Bε)= A Var(ε)B′

= σ2AVB′.

It is straightforward to show that AVB′ = 0.

(b) For the choice of H indicated in the hint,

Var(β)−Var(βGLS) = −CV−1q C′.

If C 6= 0, then there exists a nonzero vector z such that C′z ≡ v 6= 0. For such z,

z′[Var(β)−Var(βGLS)]z = −v′V−1q v < 0 (since Vq is positive definite),

which is a contradiction because βGLS is efficient.

5

Page 6: Econometrics_solutions to Analy - Fumio Hayashi

Nov. 25, 2003, Revised February 23, 2010 Hayashi Econometrics

Solution to Chapter 2 Analytical Exercises

1. For any ε > 0,

Prob(|zn| > ε) =1n→ 0 as n →∞.

So, plim zn = 0. On the other hand,

E(zn) =n− 1

n· 0 +

1n·n2 = n,

which means that limn→∞ E(zn) = ∞.

2. As shown in the hint,

(zn − µ)2 = (zn − E(zn))2 + 2(zn − E(zn))(E(zn)− µ) + (E(zn)− µ)2.

Take the expectation of both sides to obtain

E[(zn − µ)2] = E[(zn − E(zn))2] + 2 E[zn − E(zn)](E(zn)− µ) + (E(zn)− µ)2

= Var(zn) + (E(zn)− µ)2 (because E[zn − E(zn)] = E(zn)− E(zn) = 0).

Take the limit as n →∞ of both sides to obtain

limn→∞

E[(zn − µ)2] = limn→∞

Var(zn) + limn→∞

(E(zn)− µ)2

= 0 (because limn→∞

E(zn) = µ, limn→∞

Var(zn) = 0).

Therefore, zn →m.s. µ. By Lemma 2.2(a), this implies zn →p µ.

3. (a) Since an i.i.d. process is ergodic stationary, Assumption 2.2 is implied by Assumption 2.2′.Assumptions 2.1 and 2.2′ imply that gi ≡ xi· εi is i.i.d. Since an i.i.d. process with meanzero is mds (martingale differences), Assumption 2.5 is implied by Assumptions 2.2′ and2.5′.

(b) Rewrite the OLS estimator as

b− β = (X′X)−1X′ε = S−1xx g. (A)

Since by Assumption 2.2′ xi is i.i.d., xix′i is i.i.d. So by Kolmogorov’s Second StrongLLN, we obtain

Sxx →p

Σxx

The convergence is actually almost surely, but almost sure convergence implies convergencein probability. Since Σxx is invertible by Assumption 2.4, by Lemma 2.3(a) we get

S−1xx →

pΣ−1

xx .

1

Page 7: Econometrics_solutions to Analy - Fumio Hayashi

Similarly, under Assumption 2.1 and 2.2′ gi is i.i.d. By Kolmogorov’s Second StrongLLN, we obtain

g →p

E(gi),

which is zero by Assumption 2.3. So by Lemma 2.3(a),

S−1xx g →

pΣ−1

xx ·0 = 0.

Therefore, plimn→∞(b− β) = 0 which implies that the OLS estimator b is consistent.Next, we prove that the OLS estimator b is asymptotically normal. Rewrite equation(A)

above as√

n(b− β) = S−1xx

√ng.

As already observed, gi is i.i.d. with E(gi) = 0. The variance of gi equals E(gig′i) = S

since E(gi) = 0 by Assumption 2.3. So by the Lindeberg-Levy CLT,√

ng →d

N(0,S).

Furthermore, as already noted, S−1xx →p Σ−1

xx . Thus by Lemma 2.4(c),√

n(b− β) →d

N(0,Σ−1xx SΣ−1

xx ).

4. The hint is as good as the answer.

5. As shown in the solution to Chapter 1 Analytical Exercise 5, SSRR − SSRU can be written as

SSRR − SSRU = (Rb− r)′[R(X′X)−1R′]−1(Rb− r).

Using the restrictions of the null hypothesis,

Rb− r = R(b− β)

= R(X′X)−1X′ε (since b− β = (X′X)−1X′ε)

= RS−1xxg (where g ≡ 1

n

n∑

i=1

xi· εi.).

Also [R(X′X)−1R]−1 = n· [RS−1xxR]−1. So

SSRR − SSRU = (√

ng)′S−1xx R′(RS−1

xx R′)−1RS−1xx (

√ng).

Thus

SSRR − SSRU

s2= (

√ng)′S−1

xx R′(s2 RS−1xx R′)−1RS−1

xx (√

ng)

= z′n A−1n zn,

wherezn ≡ RS−1

xx (√

ng), An ≡ s2 RS−1xx R′.

By Assumption 2.2, plimSxx = Σxx. By Assumption 2.5,√

ng →d N(0,S). So byLemma 2.4(c), we have:

zn →d

N(0,RΣ−1xx SΣ−1

xx R′).

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Page 8: Econometrics_solutions to Analy - Fumio Hayashi

But, as shown in (2.6.4), S = σ2Σxx under conditional homoekedasticity (Assumption 2.7).So the expression for the variance of the limiting distribution above becomes

RΣ−1xx SΣ−1

xx R′ = σ2 RΣ−1xxR′ ≡ A.

Thus we have shown:zn →

dz, z ∼ N(0,A).

As already observed, Sxx →p Σxx. By Assumption 2.7, σ2 = E(ε2i ). So by Proposition

2.2, s2 →p σ2. Thus by Lemma 2.3(a) (the “Continuous Mapping Theorem”), An →p A.Therefore, by Lemma 2.4(d),

z′n A−1n zn →

dz′A−1z.

But since Var(z) = A, the distribution of z′A−1z is chi-squared with #z degrees of freedom.

6. For simplicity, we assumed in Section 2.8 that yi,xi is i.i.d. Collecting all the assumptionsmade in Section 2.8,

(i) (linearity) yi = x′iβ + εi.

(ii) (random sample) yi,xi is i.i.d.

(iii) (rank condition) E(xix′i) is non-singular.

(iv) E(ε2i xix′i) is non-singular.

(v) (stronger version of orthogonality) E(εi|xi) = 0 (see (2.8.5)).

(vi) (parameterized conditional heteroskedasticity) E(ε2i |xi) = z′iα.

These conditions together are stronger than Assumptions 2.1-2.5.

(a) We wish to verify Assumptions 2.1-2.3 for the regression equation (2.8.8). Clearly, As-sumption 2.1 about the regression equation (2.8.8) is satisfied by (i) about the originalregression. Assumption 2.2 about (2.8.8) (that ε2

i ,xi is ergodic stationary) is satisfiedby (i) and (ii). To see that Assumption 2.3 about (2.8.8) (that E(zi ηi) = 0) is satis-fied, note first that E(ηi|xi) = 0 by construction. Since zi is a function of xi, we haveE(ηi|zi) = 0 by the Law of Iterated Expectation. Therefore, Assumption 2.3 is satisfied.

The additional assumption needed for (2.8.8) is Assumption 2.4 that E(ziz′i) be non-singular. With Assumptions 2.1-2.4 satisfied for (2.8.8), the OLS estimator α is consistentby Proposition 2.1(a) applied to (2.8.8).

(b) Note that α− α = (α−α)− (α−α) and use the hint.

(c) Regarding the first term of (∗∗), by Kolmogorov’s LLN, the sample mean in that termconverges in probability to E(xiεizi) provided this population mean exists. But

E(xiεizi) = E[zi·xi· E(εi|zi)].

By (v) (that E(εi|xi) = 0) and the Law of Iterated Expectations, E(εi|zi) = 0. ThusE(xiεizi) = 0. Furthermore, plim(b − β) = 0 since b is consistent when Assumptions2.1-2.4 (which are implied by Assumptions (i)-(vi) above) are satisfied for the originalregression. Therefore, the first term of (∗∗) converges in probability to zero.

Regarding the second term of (∗∗), the sample mean in that term converges in prob-ability to E(x2

i zi) provided this population mean exists. Then the second term convergesin probability to zero because plim(b− β) = 0.

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Page 9: Econometrics_solutions to Analy - Fumio Hayashi

(d) Multiplying both sides of (∗) by√

n,

√n(α− α) =

( 1n

n∑

i=1

ziz′i)−1 1√

n

n∑

i=1

zi· vi

=( 1

n

n∑

i=1

ziz′i)−1

[−2√

n(b− β)1n

n∑

i=1

xiεizi +√

n(b− β)· (b− β)1n

n∑

i=1

x2i zi

].

Under Assumptions 2.1-2.5 for the original regression (which are implied by Assumptions(i)-(vi) above),

√n(b − β) converges in distribution to a random variable. As shown in

(c), 1n

∑ni=1 xiεizi →p 0. So by Lemma 2.4(b) the first term in the brackets vanishes

(converges to zero in probability). As shown in (c), (b− β) 1n

∑ni=1 x2

i zi vanishes providedE(x2

i zi) exists and is finite. So by Lemma 2.4(b) the second term, too, vanishes. Therefore,√n(α− α) vanishes, provided that E(ziz′i) is non-singular.

7. This exercise is about the model in Section 2.8, so we continue to maintain Assumptions (i)-(vi) listed in the solution to the previous exercise. Given the hint, the only thing to show isthat the LHS of (∗∗) equals Σ−1

xx SΣ−1xx , or more specifically, that plim 1

nX′VX = S. Write Sas

S = E(ε2i xix′i)

= E[E(ε2i |xi)xix′i]

= E(z′iαxix′i) (since E(ε2i |xi) = z′iα by (vi)).

Since xi is i.i.d. by (ii) and since zi is a function of xi, z′iαxix′i is i.i.d. So its sample meanconverges in probability to its population mean E(z′iαxix′i), which equals S. The samplemean can be written as

1n

n∑

i=1

z′iαxix′i

=1n

n∑

i=1

vixix′i (by the definition of vi, where vi is the i-th diagonal element of V)

=1nX′VX.

8. See the hint.

9. (a)

E(gt|gt−1, gt−2, . . . , g2)= E[E(gt|εt−1, εt−2, . . . , ε1)|gt−1, gt−2, . . . , g2] (by the Law of Iterated Expectations)= E[E(εt· εt−1|εt−1, εt−2, . . . , ε1)|gt−1, gt−2, . . . , g2]= E[εt−1 E(εt|εt−1, εt−2, . . . , ε1)|gt−1, gt−2, . . . , g2] (by the linearity of conditional expectations)= 0 (since E(εt|εt−1, εt−2, . . . , ε1) = 0).

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Page 10: Econometrics_solutions to Analy - Fumio Hayashi

(b)

E(g2t ) = E(ε2

t · ε2t−1)

= E[E(ε2t · ε2

t−1|εt−1, εt−2, . . . , ε1)] (by the Law of Total Expectations)

= E[E(ε2t |εt−1, εt−2, . . . , ε1)ε2

t−1] (by the linearity of conditional expectations)

= E(σ2ε2t−1) (since E(ε2

t |εt−1, εt−2, . . . , ε1) = σ2)

= σ2 E(ε2t−1).

But

E(ε2t−1) = E[E(ε2

t−1|εt−2, εt−3, . . . , ε1)] = E(σ2) = σ2.

(c) If εt is ergodic stationary, then εt· εt−1 is ergodic stationary (see, e.g., Remark 5.3 on p.488 of S. Karlin and H. Taylor, A First Course in Stochastic Processes, 2nd. ed., AcademicPress, 1975, which states that “For any function φ, the sequence Yn = φ(Xn, Xn+1, . . . )generates an ergodic stationary process whenever Xn is ergodic stationary”.) Thus theBillingsley CLT (see p. 106 of the text) is applicable to

√nγ1 =

√n 1

n

∑nt=j+1 gt.

(d) Since ε2t is ergodic stationary, γ0 converges in probability to E(ε2

t ) = σ2. As shown in (c),√nγ1 →d N(0, σ4). So by Lemma 2.4(c)

√n γ1

γ0→d N(0, 1).

10. (a) Clearly, E(yt) = 0 for all t = 1, 2, . . . .

Cov(yt, yt−j) =

(1 + θ21 + θ2

2)σ2ε for j = 0

(θ1 + θ1θ2)σ2ε for j = 1,

θ2σ2ε for j = 2,

0 for j > 2,

So neither E(yt) nor Cov(yt, yt−j) depends on t.

(b)

E(yt|yt−j , yt−j−1, . . . , y0, y−1)= E(yt|εt−j , εt−j−1, . . . , ε0, ε−1) (as noted in the hint)= E(εt + θ1εt−1 + θ2εt−2|εt−j , εt−j−1, . . . , ε0, ε−1)

=

εt + θ1εt−1 + θ2εt−2 for j = 0,θ1εt−1 + θ2εt−2 for j = 1,θ2εt−2 for j = 2,0 for j > 2,

which gives the desired result.

5

Page 11: Econometrics_solutions to Analy - Fumio Hayashi

(c)

Var(√

n y) =1n

[Cov(y1, y1 + · · ·+ yn) + · · ·+ Cov(yn, y1 + · · ·+ yn)]

=1n

[(γ0 + γ1 + · · ·+ γn−2 + γn−1) + (γ1 + γ0 + γ1 + · · ·+ γn−2)

+ · · ·+ (γn−1 + γn−2 + · · ·+ γ1 + γ0)]

=1n

[nγ0 + 2(n− 1)γ1 + · · ·+ 2(n− j)γj + · · ·+ 2γn−1]

= γ0 + 2n−1∑

j=1

(1− j

n

)γj .

(This is just reproducing (6.5.2) of the book.) Since γj = 0 for j > 2, one obtains thedesired result.

(d) To use Lemma 2.1, one sets zn =√

ny. However, Lemma 2.1, as stated in the book, inad-vertently misses the required condition that there exist an M > 0 such that E(|zn|s+δ) < Mfor all n for some δ > 0. Provided this technical condition is satisfied, the variance of thelimiting distribution of

√ny is the limit of Var(

√ny), which is γ0 + 2(γ1 + γ2).

11. (a) In the auxiliary regression, the vector of the dependent variable is e and the matrix of

regressors is [X... E]. Using the OLS formula,

α = B−1

[1nX′e1nE′e

].

X′e = 0 by the normal equations for the original regression. The j-th element of 1nE′e is

1n

(ej+1e1 + · · ·+ enen−j) =1n

n∑

t=j+1

etet−j .

which equals γj defined in (2.10.9).(b) The j-th column of 1

nX′E is 1n

∑nt=j+1 xt· et−j (which, incidentally, equals µj defined on

p. 147 of the book). Rewrite it as follows.

1n

n∑

t=j+1

xt· et−j

=1n

n∑

t=j+1

xt(εt−j − x′t−j(b− β))

=1n

n∑

t=j+1

xt· εt−j − 1

n

n∑

t=j+1

xtx′t−j

(b− β)

The last term vanishes because b is consistent for β. Thus 1n

∑nt=j+1 xt· et−j converges in

probability to E(xt· εt−j).The (i, j) element of the symmetric matrix 1

nE′E is, for i ≥ j,

1n

(e1+i−je1 + · · ·+ en−jen−i) =1n

n−j∑

t=1+i−j

etet−(i−j).

6

Page 12: Econometrics_solutions to Analy - Fumio Hayashi

Using the relation et = εt − x′t(b− β), this can be rewritten as

1n

n−j∑

t=1+i−j

εtεt−(i−j) −1n

n−j∑

t=1+i−j

(xtεt−(i−j) + xt−(i−j)εt)′(b− β)

− (b− β)′( 1

n

n−j∑

t=1+i−j

xtx′t−(i−j)

)(b− β).

The type of argument that is by now routine (similar to the one used on p. 145 for(2.10.10)) shows that this expression converges in probability to γi−j , which is σ2 for i = jand zero for i 6= j.

(c) As shown in (b), plim B = B. Since Σxx is non-singular, B is non-singular. So B−1

converges in probability to B−1. Also, using an argument similar to the one used in (b)for showing that plim 1

nE′E = Ip, we can show that plim γ = 0. Thus the formula in (a)shows that α converges in probability to zero.

(d) (The hint should have been: “ 1nE′e = γ. Show that SSR

n = 1ne′e− α′

[0γ

].” The SSR from

the auxiliary regression can be written as

1n

SSR =1n

(e− [X... E]α)′(e− [X

... E]α)

=1n

(e− [X... E]α)′e (by the normal equation for the auxiliary regression)

=1ne′e− 1

nα′[X

... E]′e

=1ne′e− α′

[1nX′e1nE′e

]

=1ne′e− α′

[0γ

](since X′e = 0 and

1nE′e = γ).

As shown in (c), plim α = 0 and plim γ = 0. By Proposition 2.2, we have plim 1ne′e = σ2.

Hence SSR/n (and therefore SSR/(n−K − p)) converges to σ2 in probability.

(e) Let

R ≡[

0(p×K)

... Ip

], V ≡ [X

... E].

The F -ratio is for the hypothesis that Rα = 0. The F -ratio can be written as

F =(Rα)′

[R(V′V)−1R′]−1 (Rα)/p

SSR/(n−K − p). (∗)

7

Page 13: Econometrics_solutions to Analy - Fumio Hayashi

Using the expression for α in (a) above, Rα can be written as

Rα =[

0(p×K)

... Ip

]B−1

0(K×1)

γ(p×1)

=[

0(p×K)

... Ip

]

B11

(K×K)B12

(K×p)

B21

(p×K)B22

(p×p)

0(K×1)

γ(p×1)

= B22 γ. (∗∗)Also, R(V′V)−1R′ in the expression for F can be written as

R(V′V)−1R′ =1nRB−1 R′ (since

1nV′V = B)

=1n

[0

(p×K)

... Ip

]

B11

(K×K)B12

(K×p)

B21

(p×K)B22

(p×p)

[0

(K×p)

Ip

]

=1nB22. (∗ ∗ ∗)

Substitution of (∗ ∗ ∗) and (∗∗) into (∗) produces the desired result.(f) Just apply the formula for partitioned inverses.

(g) Since√

nρ−√nγ/σ2 →p 0 and Φ →p Φ, it should be clear that the modified Box-PierceQ (= n· ρ′(Ip− Φ)−1ρ) is asymptotically equivalent to nγ′(Ip−Φ)−1γ/σ4. Regarding thepF statistic given in (e) above, consider the expression for B22 given in (f) above. Sincethe j-th element of 1

nX′E is µj defined right below (2.10.19) on p. 147, we have

s2Φ =( 1

nE′X

)S−1

xx

( 1nX′E

),

soB22 =

[ 1nE′E− s2Φ

]−1

.

As shown in (b), 1nE′E →p σ2Ip. Therefore, B22 →p

1σ2 (Ip −Φ)−1, and pF is asymptoti-

cally equivalent to nγ′(Ip −Φ)−1γ/σ4.

12. The hints are almost as good as the answer. Here, we give solutions to (b) and (c) only.

(b) We only prove the first convergence result.

1n

r∑t=1

xtx′t =r

n

(1r

r∑t=1

xtx′t

)= λ

(1r

r∑t=1

xtx′t

).

The term in parentheses converges in probability to Σxx as n (and hence r) goes to infinity.(c) We only prove the first convergence result.

1√n

r∑t=1

xt · εt =√

r

n

(1√r

r∑t=1

xt · εt

)=√

λ

(1√r

r∑t=1

xt · εt

).

The term in parentheses converges in distribution to N(0, σ2Σxx) as n (and hence r) goesto infinity. So the whole expression converges in distribution to N(0, λ σ2Σxx).

8

Page 14: Econometrics_solutions to Analy - Fumio Hayashi

December 27, 2003 Hayashi Econometrics

Solution to Chapter 3 Analytical Exercises

1. If A is symmetric and idempotent, then A′ = A and AA = A. So x′Ax = x′AAx =x′A′Ax = z′z ≥ 0 where z ≡ Ax.

2. (a) By assumption, xi, εi is jointly stationary and ergodic, so by ergodic theorem the firstterm of (∗) converges almost surely to E(x2

i ε2i ) which exists and is finite by Assumption

3.5.(b) zix

2i εi is the product of xiεi and xizi. By using the Cauchy-Schwarts inequality, we obtain

E(|xiεi · xizi|) ≤√

E(x2i ε

2i ) E(x2

i z2i ).

E(x2i ε

2i ) exists and is finite by Assumption 3.5 and E(x2

i z2i ) exists and is finite by Assump-

tion 3.6. Therefore, E(|xizi · xiεi|) is finite. Hence, E(xizi · xiεi) exists and is finite.(c) By ergodic stationarity the sample average of zix2

i εi converges in probability to some finitenumber. Because δ is consistent for δ by Proposition 3.1, δ−δ converges to 0 in probability.Therefore, the second term of (∗) converges to zero in probability.

(d) By ergodic stationarity and Assumption 3.6 the sample average of z2i x

2i converges in prob-

ability to some finite number. As mentioned in (c) δ − δ converges to 0 in probability.Therefore, the last term of (∗) vanishes.

3. (a)

Q ≡ Σ′xzS−1Σxz −Σ′xzWΣxz(Σ′xzWSWΣxz)−1Σ′xzWΣxz

= Σ′xzC′CΣxz −Σ′xzWΣxz(Σ′xzWC−1C′−1WΣxz)−1Σ′xzWΣxz

= H′H−Σ′xzWΣxz(G′G)−1Σ′xzWΣxz

= H′H−H′G(G′G)−1G′H

= H′[IK −G(G′G)−1G′]H= H′MGH.

(b) First, we show that MG is symmetric and idempotent.

MG′ = IK −G(G(G′G)−1)′

= IK −G((G′G)−1′G′)= IK −G(G′G)−1G′

= MG.

MGMG = IKIK −G(G′G)−1G′IK − IK G(G′G)−1G′ + G(G′G)−1G′G(G′G)−1G′

= IK −G(G′G)−1G′

= MG.

Thus, MG is symmetric and idempotent. For any L-dimensional vector x,

x′Qx = x′H′MGHx

= z′MGz (where z ≡ Hx)≥ 0 (since MG is positive semidefinite).

Therefore, Q is positive semidefinite.

1

Page 15: Econometrics_solutions to Analy - Fumio Hayashi

4. (the answer on p. 254 of the book simplified) If W is as defined in the hint, then

WSW = W and Σ′xzWΣxz = ΣzzA−1Σzz.

So (3.5.1) reduces to the asymptotic variance of the OLS estimator. By (3.5.11), it is no smallerthan (Σ′xz S−1Σxz)−1, which is the asymptotic variance of the efficient GMM estimator.

5. (a) From the expression for δ(S−1) (given in (3.5.12)) and the expression for gn(δ) (given in(3.4.2)), it is easy to show that gn(δ(S−1)) = Bsxy. But Bsxy = Bg because

Bsxy = (IK − Sxz(S′xz S−1Sxz)−1S′xz S−1)sxy

= (IK − Sxz(S′xz S−1Sxz)−1S′xz S−1)(Sxzδ + g) (since yi = z′iδ + εi)

= (Sxz − Sxz(S′xz S−1Sxz)−1S′xz S−1Sxz)δ + (IK − Sxz(S′xz S−1Sxz)−1S′xz S−1)g

= (Sxz − Sxz)δ + Bg

= Bg.

(b) Since S−1 = C′C, we obtain B′S−1B = B′C′CB = (CB)′(CB). But

CB = C(IK − Sxz(S′xz S−1Sxz)−1S′xz S−1)= C−CSxz(S′xz C′CSxz)−1S′xz C′C

= C−A(A′A)−1A′C (where A ≡ CSxz)= [IK −A(A′A)−1A′]C≡ MC.

So B′S−1B = (MC)′(MC) = C′M′MC. It should be routine to show that M is symmet-ric and idempotent. Thus B′S−1B = C′MC.

The rank of M equals its trace, which is

trace(M) = trace(IK −A(A′A)−1A′)= trace(IK)− trace(A(A′A)−1A′)= trace(IK)− trace(A′A(A′A)−1)= K − trace(IL)= K − L.

(c) As defined in (b), C′C = S−1. Let D be such that D′D = S−1. The choice of C and Dis not unique, but it would be possible to choose C so that plim C = D. Now,

v ≡√n(Cg) = C(

√ng).

By using the Ergodic Stationary Martingale Differences CLT, we obtain√ng→d N(0,S).

So

v = C(√ng)→

dN(0,Avar(v))

where

Avar(v) = DSD′

= D(D′D)−1D′

= DD−1D−1′D′

= IK .

2

Page 16: Econometrics_solutions to Analy - Fumio Hayashi

(d)

J(δ(S−1), S−1) = n · gn(δ(S−1)) S−1gn(δ(S−1))

= n · (Bg)′S−1(Bg) (by (a))

= n · g′B′S−1Bg

= n · g′C′MCg (by (b))= v′Mv (since v ≡

√nCg).

Since v →d N(0, IK) and M is idempotent, v′Mv is asymptotically chi-squared withdegrees of freedom equaling the rank of M = K − L.

6. From Exercise 5, J = n·g′B′S−1Bg. Also from Exercise 5, Bg = Bsxy.

7. For the most parts, the hints are nearly the answer. Here, we provide answers to (d), (f), (g),(i), and (j).

(d) As shown in (c), J1 = v′1M1v1. It suffices to prove that v1 = C1F′C−1v.

v1 ≡√nC1g1

=√nC1F′g

=√nC1F′C−1Cg

= C1F′C−1√nCg

= C1F′C−1v (since v ≡√nCg).

(f) Use the hint to show that A′D = 0 if A′1M1 = 0. It should be easy to show that A′1M1 = 0from the definition of M1.

(g) By the definition of M in Exercise 5, MD = D − A(A′A)−1A′D. So MD = D sinceA′D = 0 as shown in the previous part. Since both M and D are symmetric, DM =D′M′ = (MD)′ = D′ = D. As shown in part (e), D is idempotent. Also, M is idempotentas shown in Exercise 5. So (M−D)2 = M2 −DM−MD + D2 = M−D. As shown inExercise 5, the trace of M is K − L. As shown in (e), the trace of D is K1 − L. So thetrace of M−D is K −K1. The rank of a symmetric and idempotent matrix is its trace.

(i) It has been shown in Exercise 6 that g′C′MCg = s′xyC′MCsxy since C′MC = B′S−1B.Here, we show that g′C′DCg = s′xyC′DCsxy.

g′C′DCg = g′FC′1M1C1F′g (C′DC = FC′1M1C1F′ by the definition of D in (d))

= g′FB′1(S11)−1 B1F′g (since C′1M1C1 = B′1(S11)−1 B1 from (a))

= g′1B′1(S11)−1 B1g1 (since g1 = F′g).

From the definition of B1 and the fact that sx1y = Sx1zδ + g1, it follows that B1g1 =B1sx1y. So

g′1B′1(S11)−1 B1g1 = s′x1yB′1(S11)−1 B1sx1y

= s′xyFB′1(S11)−1 B1F′sxy (since sx1y = F′sxy)

= s′xyFC′1M1C1F′sxy (since B′1(S11)−1 B1 = C′1M1C1 from (a))

= s′xyC′DCsxy.

3

Page 17: Econometrics_solutions to Analy - Fumio Hayashi

(j) M−D is positive semi-definite because it is symmetric and idempotent.

8. (a) Solve the first-order conditions in the hint for δ to obtain

δ = δ(W)− 12n

(S′xzWSxz)−1R′λ.

Substitute this into the constraint Rδ = r to obtain the expression for λ in the question.Then substitute this expression for λ into the above equation to obtain the expression forδ in the question.

(b) The hint is almost the answer.

(c) What needs to be shown is that n·(δ(W) − δ)′(S′xzWSxz)(δ(W) − δ) equals the Waldstatistic. But this is immediate from substitution of the expression for δ in (a).

9. (a) By applying (3.4.11), we obtain[√n(δ1 − δ)√n(δ1 − δ)

]=

[(S′xzW1Sxz)−1S′xzW1

(S′xzW2Sxz)−1S′xzW2

]√ng.

By using Billingsley CLT, we have√ng→

dN(0,S).

Also, we have [(S′xzW1Sxz)−1S′xzW1

(S′xzW2Sxz)−1S′xzW2

]→p

[Q−1

1 Σ′xzW1

Q−12 Σ′xzW2

].

Therefore, by Lemma 2.4(c),[√n(δ1 − δ)√n(δ1 − δ)

]→d N

(0,

[Q−1

1 Σ′xzW1

Q−12 Σ′xzW2

]S (W1ΣxzQ−1

1

... W2ΣxzQ−12 ))

= N

(0,

[A11 A12

A21 A22

]).

(b)√nq can be rewritten as

√nq =

√n(δ1 − δ2) =

√n(δ1 − δ)−

√n(δ2 − δ) =

[1 −1

] [√n(δ1 − δ)√n(δ2 − δ)

].

Therefore, we obtain√nq→

dN(0,Avar(q)).

where

Avar(q) =[1 −1

] [A11 A12

A21 A22

] [1−1

]= A11 + A22 −A12 −A21.

4

Page 18: Econometrics_solutions to Analy - Fumio Hayashi

(c) Since W2 = S−1, Q2, A12, A21, and A22 can be rewritten as follows:

Q2 = Σ′xzW2Σxz

= Σ′xzS−1Σxz,

A12 = Q−11 Σ′xzW1S S−1ΣxzQ−1

2

= Q−11 (Σ′xzW1Σxz)Q−1

2

= Q−11 Q1Q−1

2

= Q−12 ,

A21 = Q−12 Σ

xzS−1SW1ΣxzQ−1

1

= Q−12 ,

A22 = (Σ′xzS−1Σxz)−1Σ′xzS

−1SS−1Σxz(Σ′xzS−1Σxz)−1

= (Σ′xzS−1Σxz)−1

= Q−12 .

Substitution of these into the expression for Avar(q) in (b), we obtain

Avar(q) = A11 −Q−12

= A11 − (Σ′xzS−1Σxz)−1

= Avar(δ(W1))−Avar(δ(S−1)).

10. (a)

σxz ≡ E(xizi) = E(xi(xiβ + vi))= β E(x2

i ) + E(xivi)= βσ2

x 6= 0 (by assumptions (2), (3), and (4)).

(b) From the definition of δ,

δ − δ =

(1n

n∑i=1

xizi

)−11n

n∑i=1

xiεi = s−1xz

1n

n∑i=1

xiεi.

We have xizi = xi(xiβ + vi) = x2iβ + xivi, which, being a function of (xi,ηi), is ergodic

stationary by assumption (1). So by the Ergodic theorem, sxz →p σxz. Since σxz 6= 0 by(a), we have s−1

xz →p σ−1xz . By assumption (2), E(xiεi) = 0. So by assumption (1), we have

1n

∑ni=1 xiεi →p 0. Thus δ − δ →p 0.

(c)

sxz ≡ 1n

n∑i=1

xizi

=1n

n∑i=1

(x2iβ + xivi)

=1√n

1n

n∑i=1

x2i +

1n

n∑i=1

xivi (since β =1√n

)

→p 0 · E(x2i ) + E(xivi)

= 0

5

Page 19: Econometrics_solutions to Analy - Fumio Hayashi

(d)

√nsxz =

1n

n∑i=1

x2i +

1√n

n∑i=1

xivi.

By assumption (1) and the Ergodic Theorem, the first term of RHS converges in probabilityto E(x2

i ) = σ2x > 0. Assumption (2) and the Martingale Differences CLT imply that

1√n

n∑i=1

xivi →da ∼ N(0, s22).

Therefore, by Lemma 2.4(a), we obtain√nsxz →

dσ2x + a.

(e) δ − δ can be rewritten as

δ − δ = (√nsxz)−1

√ng1.

From assumption (2) and the Martingale Differences CLT, we obtain√ng1 →

db ∼ N(0, s11).

where s11 is the (1, 1) element of S. By using the result of (d) and Lemma 2.3(b),

δ − δ →d

(σ2x + a)−1b.

(a, b) are jointly normal because the joint distribution is the limiting distribution of

√ng =

[ √ng1√

n( 1n

∑ni=1 xivi)

].

(f) Because δ− δ converges in distribution to (σ2x + a)−1b which is not zero, the answer is No.

6

Page 20: Econometrics_solutions to Analy - Fumio Hayashi

January 8, 2004,answer to 3(c)(i) simplified, February 23, 2004 Hayashi Econometrics

Solution to Chapter 4 Analytical Exercises

1. It should be easy to show that Amh = 1nZ′mPZh and that cmh = 1

nZ′mPyh. Going backto the formula (4.5.12) on p. 278 of the book, the first matrix on the RHS (the matrix tobe inverted) is a partitioned matrix whose (m,h) block is Amh. It should be easy to see

that it equals 1n [Z′(Σ

−1⊗ P)Z]. Similarly, the second matrix on the RHS of (4.5.12) equals

1nZ′(Σ

−1⊗P)y.

2. The sprinkled hints are as good as the answer.

3. (b) (amplification of the answer given on p. 320) In this part only, for notational brevity, letzi be a

∑m Lm × 1 stacked vector collecting (zi1, . . . , ziM ).

E(εim | Z)= E(εim | z1, z2, . . . , zn) (since Z collects zi’s)= E(εim | zi) (since (εim, zi) is independent of zj (j 6= i))= 0 (by the strengthened orthogonality conditions).

The (i, j) element of the n× n matrix E(εmε′h | Z) is E(εim εjh | Z).

E(εim εjh | Z) = E(εim εjh | z1, z2, . . . , zn)= E(εim εjh | zi, zj) (since (εim, zi, εjh, zj) is independent of zk (k 6= i, j)).

For j 6= i, this becomes

E(εim εjh | zi, zj)= E [E(εim εjh | zi, zj , εjh) | zi, zj ] (by the Law of Iterated Expectations)= E [εjh E(εim | zi, zj , εjh) | zi, zj ] (by linearity of conditional expectations)= E [εjh E(εim | zi) | zi, zj ] (since (εim, zi) is independent of (εjh, zj))= 0 (since E(εim | zi) = 0).

For j = i,

E(εim εjh | Z) = E(εim εih | Z) = E(εim εih | zi).

Since xim = xi and xi is the union of (zi1, . . . , ziM ) in the SUR model, the conditionalhomoskedasticity assumption, Assumption 4.7, states that E(εim εih | zi) = E(εim εih |xi) = σmh.

(c) (i) We need to show that Assumptions 4.1-4.5, 4.7 and (4.5.18) together imply As-sumptions 1.1-1.3 and (1.6.1). Assumption 1.1 (linearity) is obviously satisfied. As-sumption 1.2 (strict exogeneity) and (1.6.1) have been verified in 3(b). That leavesAssumption 1.3 (the rank condition that Z (defined in Analytical Exercise 1) be offull column rank). Since Z is block diagonal, it suffices to show that Zm is of fullcolumn rank for m = 1, 2, . . . ,M . The proof goes as follows. By Assumption 4.5,

1

Page 21: Econometrics_solutions to Analy - Fumio Hayashi

S is non-singular. By Assumption 4.7 and the condition (implied by (4.5.18)) thatthe set of instruments be common across equations, we have S = Σ ⊗ E(xix′i) (asin (4.5.9)). So the square matrix E(xix′i) is non-singular. Since 1

nX′X (where X isthe n×K data matrix, as defined in Analytical Exercise 1) converges almost surelyto E(xix′i), the n ×K data matrix X is of full column rank for sufficiently large n.Since Zm consists of columns selected from the columns of X, Zm is of full columnrank as well.

(ii) The hint is the answer.(iii) The unbiasedness of δSUR follows from (i), (ii), and Proposition 1.7(a).(iv) Avar(δSUR) is (4.5.15) where Amh is given by (4.5.16′) on p. 280. The hint shows

that it equals the plim of n ·Var(δSUR | Z).(d) For the most part, the answer is a straightforward modification of the answer to (c). The

only part that is not so straightforward is to show in part (i) that the Mn × L matrixZ is of full column rank. Let Dm be the Dm matrix introduced in the answer to (c), sozim = D′mxi and Zm = XDm. Since the dimension of xi is K and that of zim is L, thematrix Dm is K × L. The

∑Mm=1Km × L matrix Σxz in Assumption 4.4′ can be written

as

Σxz(KM×L)

= [IM ⊗ E(xix′i)]D where D(KM×L)

D1

...DM

.Since Σxz is of full column rank by Assumption 4.4′ and since E(xix′i) is non-singular, Dis of full column rank. So Z = (IM ⊗ X)D is of full column rank if X is of full columnrank. X is of full column rank for sufficiently large n if E(xix′i) is non-singular.

4. (a) Assumptions 4.1-4.5 imply that the Avar of the efficient multiple-equation GMM esti-mator is (Σ′xzS

−1Σxz)−1. Assumption 4.2 implies that the plim of Sxz is Σxz. UnderAssumptions 4.1, 4.2, and 4.6, the plim of S is S.

(b) The claim to be shown is just a restatement of Propositions 3.4 and 3.5.

(c) Use (A9) and (A6) of the book’s Appendix A. Sxz and W are block diagonal, so WSxz(S′xzWSxz)−1

is block diagonal.(d) If the same residuals are used in both the efficient equation-by-equation GMM and the

efficient multiple-equation GMM, then the S in (∗∗) and the S in (S′xzS−1Sxz)−1 are

numerically the same. The rest follows from the inequality in the question and the hint.(e) Yes.(f) The hint is the answer.

5. (a) For the LW69 equation, the instruments (1,MED) are 2 in number while the number ofthe regressors is 3. So the order condition is not satisfied for the equation.

(b) (reproducing the answer on pp. 320-321)1 E(S69) E(IQ)1 E(S80) E(IQ)

E(MED) E(S69 ·MED) E(IQ ·MED)E(MED) E(S80 ·MED) E(IQ ·MED)

β0

β1

β2

=

E(LW69)E(LW80)

E(LW69 ·MED)E(LW80 ·MED)

.The condition for the system to be identified is that the 4 × 3 coefficient matrix is of fullcolumn rank.

2

Page 22: Econometrics_solutions to Analy - Fumio Hayashi

(c) (reproducing the answer on p. 321) If IQ and MED are uncorrelated, then E(IQ ·MED) =E(IQ) · E(MED) and the third column of the coefficient matrix is E(IQ) times the firstcolumn. So the matrix cannot be of full column rank.

6. (reproducing the answer on p. 321) εim = yim − z′imδm = εim − z′im(δm − δm). So

1n

n∑i=1

[εim − z′im(δm − δm)][εih − z′ih(δh − δh)] = (1) + (2) + (3) + (4),

where

(1) =1n

n∑i=1

εim εih,

(2) = −(δm − δm)′1n

n∑i=1

zim · εih,

(3) = −(δh − δh)′1n

n∑i=1

zih · εim,

(4) = (δm − δm)′( 1n

n∑i=1

zimz′ih)

(δh − δh).

As usual, under Assumption 4.1 and 4.2, (1)→p σmh (≡ E(εim εih)).For (4), by Assumption 4.2 and the assumption that E(zimz′ih) is finite, 1

n

∑i ·

zimz′ih converges in probability to a (finite) matrix. So (4)→p 0.Regarding (2), by Cauchy-Schwartz,

E(|zimj · εih|) ≤√

E(z2imj) · E(ε2

ih),

where zimj is the j-th element of zim. So E(zim ·εih) is finite and (2)→p 0 because δm−δm →p

0. Similarly, (3)→p 0.

7. (a) Let B, Sxz, and W be as defined in the hint. Also let

sxy(MK×1)

=

1n

∑ni=1 xi · yi1

...1n

∑ni=1 xi · yiM

.Then

δ3SLS =(S′xzWSxz

)−1

S′xzWsxy

=[(I⊗B′)(Σ

−1⊗ S−1

xx )(I⊗B)]−1

(I⊗B′)(Σ−1⊗ S−1

xx )sxy

=(Σ−1⊗B′S−1

xxB)−1 (

Σ−1⊗B′S−1

xx

)sxy

=(Σ⊗ (B′S−1

xxB)−1)(

Σ−1⊗B′S−1

xx

)sxy

=(IM ⊗ (B′S−1

xxB)−1B′S−1xx

)sxy

=

(B′S−1xxB)−1B′S−1

xx1n

∑ni=1 xi · yi1

...(B′S−1

xxB)−1B′S−1xx

1n

∑ni=1 xi · yiM

,

3

Page 23: Econometrics_solutions to Analy - Fumio Hayashi

which is a stacked vector of 2SLS estimators.

(b) The hint is the answer.

8. (a) The efficient multiple-equation GMM estimator is(S′xz S−1Sxz

)−1

S′xz S−1sxy,

where Sxz and sxy are as defined in (4.2.2) on p. 266 and S−1 is a consistent estimator ofS. Since xim = zim here, Sxz is square. So the above formula becomes

S−1xz S S′xz

−1S′xzS−1sxy = S−1

xz sxy,

which is a stacked vector of OLS estimators.

(b) The SUR is efficient multiple-equation GMM under conditional homoskedasticity when theset of orthogonality conditions is E(zim · εih) = 0 for all m,h. The OLS estimator derivedabove is (trivially) efficient multiple-equation GMM under conditional homoskedasticitywhen the set of orthogonality conditions is E(zim · εim) = 0 for all m. Since the sets oforthogonality conditions differ, the efficient GMM estimators differ.

9. The hint is the answer (to derive the formula in (b) of the hint, use the SUR formula youderived in Analytical Exercise 2(b)).

10. (a) Avar(δ1,2SLS) = σ11A−111 .

(b) Avar(δ1,3SLS) equals G−1. The hint shows that G = 1σ11

A11.

11. Because there are as many orthogonality conditions as there are coefficients to be estimated,it is possible to choose δ so that gn(δ) defined in the hint is a zero vector. Solving

( 1n

n∑i=1

zi1·yi1 + · · ·+ 1n

n∑i=1

ziM ·yiM)−( 1n

n∑i=1

zi1z′i1 + · · ·+ 1n

n∑i=1

ziMz′iM)δ = 0

for δ, we obtain

δ =( 1n

n∑i=1

zi1z′i1 + · · ·+ 1n

n∑i=1

ziMz′iM)−1 ( 1

n

n∑i=1

zi1·yi1 + · · ·+ 1n

n∑i=1

ziM ·yiM),

which is none other than the pooled OLS estimator.

4

Page 24: Econometrics_solutions to Analy - Fumio Hayashi

January 9, 2004 Hayashi Econometrics

Solution to Chapter 5 Analytical Exercises

1. (a) Let (a′,b′)′ be the OLS estimate of (α′,β′)′. Define MD as in equation (4) of the hint.By the Frisch-Waugh theorem, b is the OLS coefficient estimate in the regression of MDyon MDF. The proof is complete if we can show the claim that

y = MDy and F = MDF,

where y and F are defined in (5.2.2) and (5.2.3). This is because the fixed-effects estimatorcan be written as (F′F)1 F′y (see (5.2.4)). But the above claim follows immediately if wecan show that MD = In ⊗Q, where Q ≡ IM − 1

M 1M1′M , the annihilator associated with1M .

MD = IMn − (In ⊗ 1M ) [(In ⊗ 1M )′(In ⊗ 1M )]−1 (In ⊗ 1M )′

= IMn − (In ⊗ 1M ) [(In ⊗ 1′M1M )]−1 (In ⊗ 1′M )

= IMn − (In ⊗ 1M ) [(In ⊗M)]−1 (In ⊗ 1′M )

= IMn − (In ⊗ 1M )(In ⊗1M

)(In ⊗ 1′M )

= IMn − (In ⊗1M

1M1′M )

= (In ⊗ IM )− (In ⊗1M

1M1′M )

= (In ⊗ (IM −1M

1M1′M ))

= In ⊗Q.

(b) As indicated in the hint to (a), we have a = (D′D)−1(D′y−D′Fb). It should be straight-forward to show that

D′D = M In, D′y =

1′My1

...1′Myn

, D′Fb =

1′MF1b...

1′MFnb

.Therefore,

a =

1M (1′My1 − 1′MF1b)

...1M (1′Myn − 1′MFnb)

.The desired result follows from this because b equals the fixed-effects estimator βFE and

1′Myi = (yi1 + · · ·+ yiM ) and 1′MFnb = 1′M

f ′i1...

f ′iM

b =

(M∑m=1

f ′im

)b.

1

Page 25: Econometrics_solutions to Analy - Fumio Hayashi

(c) What needs to be shown is that (3) and conditions (i)-(iv) listed in the question togetherimply Assumptions 1.1-1.4. Assumption 1.1 (linearity) is none other than (3). Assumption1.3 is a restatement of (iv). This leaves Assumptions 1.2 (strict exogeneity) and Assumption1.4 (spherical error term) to be verified. The following is an amplification of the answer to1.(c) on p. 363.

E(ηi |W) = E(ηi | F) (since D is a matrix of constants)= E(ηi | F1, . . . ,Fn)= E(ηi | Fi) (since (ηi,Fi) is indep. of Fj for j 6= i) by (i)= 0 (by (ii)).

Therefore, the regressors are strictly exogenous (Assumption 1.2). Also,

E(ηiη′i |W) = E(ηiη

′i | F)

= E(ηiη′i | Fi)

= σ2η IM (by the spherical error assumption (iii)).

For i 6= j,

E(ηiη′j |W) = E(ηiη

′j | F)

= E(ηiη′j | F1, . . . ,Fn)

= E(ηiη′j | Fi,Fj) (since (ηi,Fi,ηj ,Fj) is indep. of Fk for k 6= i, j by (i))

= E[E(ηiη′j | Fi,Fj ,ηi) | Fi,Fj ]

= E[ηi E(η′j | Fi,Fj ,ηi) | Fi,Fj ]= E[ηi E(η′j | Fj) | Fi,Fj ] (since (ηj ,Fj) is independent of (ηi,Fi) by (i))

= 0 (since E(η′j | Fj) by (ii)).

So E(ηη′ |W) = σ2η IMn (Assumption 1.4).

Since the assumptions of the classical regression model are satisfied, Propositions 1.1holds for the OLS estimator (a,b). The estimator is unbiased and the Gauss-Markovtheorem holds.

As shown in Analytical Exercise 4.(f) in Chapter 1, the residual vector from the originalregression (3) (which is to regress y on D and F) is numerically the same as the residualvector from the regression of y (= MDy) on F (= MDF)). So the two SSR’s are the same.

2. (a) It is evident that C′1M = 0 if C is what is referred to in the question as the matrix offirst differences. Next, to see that C′1M = 0 if C is an M × (M − 1) matrix created bydropping one column from Q, first note that by construction of Q, we have:

Q(M×M)

1M = 0(M×1)

,

which is a set of M equations. Drop one row from Q and call it C′ and drop the corre-sponding element from the 0 vector on the RHS. Then

C′((M−1)×M)

1M = 0((M−1)×1)

.

(b) By multiplying both sides of (5.1.1′′) on p. 329 by C′, we eliminate 1M ·biγ and 1M ·αi.

2

Page 26: Econometrics_solutions to Analy - Fumio Hayashi

(c) Below we verify the five conditions.

• The random sample condition is immediate from (5.1.2).• Regarding the orthogonality conditions, as mentioned in the hint, (5.1.8b) can be

written as E(ηi ⊗ xi) = 0. This implies the orthogonality conditions because

E(ηi ⊗ xi) = E[(C′ ⊗ IK)(ηi ⊗ xi)] = (C′ ⊗ IK) E(ηi ⊗ xi).

• As shown on pp. 363-364, the identification condition to be verified is equivalent to(5.1.15) (that E(QFi ⊗ xi) be of full column rank).• Since εi = 1M · αi + ηi, we have ηi ≡ C′ηi = C′εi. So ηiη

′i = C′εiε′iC and

E(ηiη′i | xi) = E(C′εiε′iC | xi) = C′ E(εiε′i | xi)C = C′ΣC.

(The last equality is by (5.1.5).)• By the definition of gi, we have: gig′i = ηiη

′i ⊗ xix′i. But as just shown above,

ηiη′i = C′εiε′iC. So

gig′i = C′εiε′iC⊗ xix′i = (C′ ⊗ IK)(εiε′i ⊗ xix′i)(C⊗ IK).

Thus

E(gig′i) = (C′ ⊗ IK) E[(εiε′i ⊗ xix′i)](C⊗ IK)= (C′ ⊗ IK) E(gig′i)(C⊗ IK) (since gi ≡ εi ⊗ xi).

Since E(gig′i) is non-singular by (5.1.6) and since C is of full column rank, E(gig′i) isnon-singular.

(d) Since Fi ≡ C′Fi, we can rewrite Sxz and sxy as

Sxz = (C′ ⊗ IK)( 1n

n∑i=1

Fi ⊗ xi), sxy = (C′ ⊗ IK)

( 1n

n∑i=1

yi ⊗ xi).

So

S′xzWSxz =( 1n

n∑i=1

F′i ⊗ x′i)

(C⊗ IK)

[(C′C)−1 ⊗

( 1n

n∑i=1

xix′i)−1

](C′ ⊗ IK)

( 1n

n∑i=1

Fi ⊗ xi)

=( 1n

n∑i=1

F′i ⊗ x′i)[

C(C′C)−1C′ ⊗( 1n

n∑i=1

xix′i)−1

]( 1n

n∑i=1

Fi ⊗ xi)

=( 1n

n∑i=1

F′i ⊗ x′i)[

Q⊗( 1n

n∑i=1

xix′i)−1

]( 1n

n∑i=1

Fi ⊗ xi)

(since C(C′C)−1C′ = Q, as mentioned in the hint).

Similarly,

S′xzWsxy =( 1n

n∑i=1

F′i ⊗ x′i)[

Q⊗( 1n

n∑i=1

xix′i)−1

]( 1n

n∑i=1

yi ⊗ xi).

3

Page 27: Econometrics_solutions to Analy - Fumio Hayashi

Noting that f ′im is the m-th row of Fi and writing out the Kronecker products in full, weobtain

S′xzWSxz =M∑m=1

M∑h=1

qmh

( 1n

n∑i=1

fimx′i)( 1

n

n∑i=1

xix′i)−1( 1

n

n∑i=1

xif ′ih),

S′xzWsxy =M∑m=1

M∑h=1

qmh

( 1n

n∑i=1

fimx′i)( 1

n

n∑i=1

xix′i)−1( 1

n

n∑i=1

xi · yih),

where qmh is the (m,h) element of Q. (This is just (4.6.6) with xim = xi, zim = fim,

W = Q⊗(

1n

∑ni=1 xix′i

)−1

.) Since xi includes all the elements of Fi, as noted in the hint,xi “dissappears”. So

S′xzWSxz =M∑m=1

M∑h=1

qmh

( 1n

n∑i=1

fimf ′ih)

=1n

n∑i=1

( M∑m=1

M∑h=1

qmhfimf ′ih),

S′xzWsxy =M∑m=1

M∑h=1

qmh1n

n∑i=1

fim · yih =1n

n∑i=1

( M∑m=1

M∑h=1

qmhfim · yih).

Using the “beautifying” formula (4.6.16b), this expression can be simplified as

S′xzWSxz =1n

n∑i=1

F′iQFi,

S′xzWsxy =1n

n∑i=1

F′iQyi.

So(S′xzWSxz

)−1

S′xzWsxy is the fixed-effects estimator.

(e) The previous part shows that the fixed-effects estimator is not efficient because the Win (10) does not satisfy the efficiency condition that plim W = S−1. Under conditionalhomoskedasticity, S = E(ηiη

′i) ⊗ E(xix′i). Thus, with Ψ being a consistent estimator of

E(ηiη′i), the efficient GMM estimator is given by setting

W = Ψ−1⊗( 1n

n∑i=1

xix′i)−1

.

This is none other than the random-effects estimator applied to the system of M −1 equa-tions (9). By setting Zi = Fi, Σ = Ψ, yi = yi in (4.6.8′) and (4.6.9′) on p. 293, weobtain (12) and (13) in the question. It is shown on pp. 292-293 that these “beautified”formulas are numerically equivalent versions of (4.6.8) and (4.6.9). By Proposition 4.7, therandom-effects estimator (4.6.8) is consistent and asymptotically normal and the asymp-totic variance is given by (4.6.9). As noted on p. 324, it should be routine to show thatthose conditions verified in (c) above are sufficient for the hypothesis of Proposition 4.7.In particular, the Σxz referred to in Assumption 4.4′ can be written as E(Fi⊗xi). In (c),we’ve verified that this matrix is of full column rank.

(f) Proposition 4.1, which is about the estimation of error cross moments for the multiple-equation model of Section 4.1, can easily be adapted to the common-coefficient model ofSection 4.6. Besides linearity, the required assumptions are (i) that the coefficient estimate

4

Page 28: Econometrics_solutions to Analy - Fumio Hayashi

(here βFE) used for calculating the residual vector be consistent and (ii) that the crossmoment between the vector of regressors from one equation (a row from Fi) and those fromanother (another row from Fi) exist and be finite. As seen in (d), the fixed-effects estimatorβFE is a GMM estimator. So it is consistent. As noted in (c), E(xix′i) is non-singular.Since xi contains all the elements of Fi, the cross moment assumption is satisfied.

(g) As noted in (e), the assumptions of Proposition 4.7 holds for the present model in question.It has been verified in (f) that Ψ defined in (14) is consistent. Therefore, Proposition 4.7(c)holds for the present model.

(h) Since ηi ≡ C′ηi, we have E(ηiη′i) = E(C′ηiη′iC) = σ2

ηC′C (the last equality is by

(15)). By setting Ψ = σ2ηC′C in the expression for W in the answer to (e) (thus setting

W = σ2ηC′C ⊗

(1n

∑ni=1 xix′i

)−1

), the estimator can be written as a GMM estimator

(S′xzWSxz)−1S′xzWsxy. Clearly, it is numerically equal to the GMM estimator with

W = C′C⊗(

1n

∑ni=1 xix′i

)−1

, which, as was verified in (d), is the fixed-effects estimator.

(i) Evidently, replacing C by B ≡ CA in (11) does not change Q. So the fixed-effectsestimator is invariant to the choice of C. To see that the numerical values of (12) and (13)are invariant to the choice of C, let Fi ≡ B′Fi and yi ≡ B′yi. That is, the original Mequations (5.1.1′′) are transformed into M − 1 equations by B = CA, not by C. ThenFi = A′Fi and yi = A′yi. If Ψ is the estimated error cross moment matrix when (14) isused with yi replacing yi and Fi replacing Fi, then we have: Ψ = A′ΨA. So

F′iΨ−1

Fi = F′iA(A′ΨA)−1A′Fi = F′iAA−1Ψ−1

(A′)−1A′Fi = F′iΨ−1

Fi.

Similarly, F′iΨ−1

yi = F′iΨ−1

yi.

3. From (5.1.1′′), vi = C′(yi − Fiβ) = C′ηi. So E(viv′i) = E(C′ηiη′iC) = C′ E(ηiη′i)C =σ2ηC′C. By the hint,

plimSSRn

= trace[(C′C)−1 σ2

ηC′C]

= σ2η trace[IM−1] = σ2

η · (M − 1).

4. (a) bi is absent from the system of M equations (or bi is a zero vector).

yi =

yi1...yiM

, Fi =

yi0...

yi,M−1

.(b) Recursive substitution (starting with a substitution of the first equation of the system into

the second) yields the equation in the hint. Multiply both sides of the equation by ηih andtake expectations to obtain

E(yim · ηih) = E(ηim · ηih) + ρE(ηi,m−1 · ηih) + · · ·+ ρm−1 E(ηi1 · ηih)

+1− ρm

1− ρE(αi · ηih) + ρm E(yi0 · ηih)

= E(ηim · ηih) + ρE(ηi,m−1 · ηih) + · · ·+ ρm−1 E(ηi1 · ηih)(since E(αi · ηih) = 0 and E(yi0 · ηih) = 0)

=

ρm−h σ2

η if h = 1, 2, . . . ,m,0 if h = m+ 1,m+ 2, . . . .

5

Page 29: Econometrics_solutions to Analy - Fumio Hayashi

(c) That E(yim · ηih) = ρm−hσ2η for m ≥ h is shown in (b). Noting that Fi here is a vector,

not a matrix, we have:

E(F′iQηi) = E[trace(F′iQηi)]

= E[trace(ηiF′iQ)]

= trace[E(ηiF′i)Q]

= trace[E(ηiF′i)(IM −

1M

11′)]

= trace[E(ηiF′i)]−

1M

trace[E(ηiF′i)11′]

= trace[E(ηiF′i)]−

1M

1′ E(ηiF′i)1.

By the results shown in (b), E(ηiF′i) can be written as

E(ηiF′i) = σ2

η

0 1 ρ ρ2 · · · ρM−2

0 0 1 ρ · · · ρM−3

......

. . . . . . · · ·...

0 · · · · · · 0 1 ρ0 · · · · · · · · · 0 10 · · · · · · · · · · · · 0

.

So, in the above expression for E(F′iQηi), trace[E(ηiF′i)] = 0 and

1′ E(ηiF′i)1 = sum of the elements of E(ηiF

′i)

= sum of the first row + · · ·+ sum of the last row

= σ2η

[1− ρM−1

1− ρ+

1− ρM−2

1− ρ+ · · ·+ 1− ρ

1− ρ

]

= σ2η

M − 1−M ρ+ ρM

(1− ρ)2.

(d) (5.2.6) is violated because E(fim · ηih) = E(yi,m−1 · ηih) 6= 0 for h ≤ m− 1.

5. (a) The hint shows that

E(F′iFi) = E(QFi ⊗ xi)′(IM ⊗

[E(xix′i)

]−1)E(QFi ⊗ xi).

By (5.1.15), E(QFi ⊗ xi) is of full column rank. So the matrix product above is non-singular.

(b) By (5.1.5) and (5.1.6′), E(εiε′i) is non-singular.

(c) By the same sort of argument used in (a) and (b) and noting that Fi ≡ C′Fi, we have

E(F′i Ψ−1 Fi) = E(C′Fi ⊗ xi)′

(Ψ−1 ⊗

[E(xix′i)

]−1)E(C′Fi ⊗ xi).

We’ve verified in 2(c) that E(C′Fi ⊗ xi) is of full column rank.

6

Page 30: Econometrics_solutions to Analy - Fumio Hayashi

6. This question presumes that

xi =

fi1...

fiMbi

and fim = A′mxi.

(a) The m-th row of Fi is f ′im and f ′im = x′iAm.

(b) The rank condition (5.1.15) is that E(Fi ⊗ xi) be of full column rank (where Fi ≡ QFi).By the hint, E(Fi ⊗ xi) = [IM ⊗ E(xix′i)](Q ⊗ IK)A. Since E(xix′i) is non-singular,IM ⊗E(xix′i) is non-singular. Multiplication by a non-singular matrix does not alter rank.

7. The hint is the answer.

7

Page 31: Econometrics_solutions to Analy - Fumio Hayashi

September 10, 2004 Hayashi Econometrics

Solution to Chapter 6 Analytical Exercises

1. The hint is the answer.

2. (a) Let σn ≡∑n

j=0 ψ2j . Then

E[(yt,m − yt,n)2] = E

[( m∑j=n+1

ψjεt−j

)2]

= σ2m∑

j=n+1

ψ2j (since εt is white noise)

= σ2 |αm − αn|.

Since ψj is absolutely summable (and hence square summable), αn converges. So|αm−αn| → ∞ as m,n→∞. Therefore, E[(yt,m−yt,n)2] → 0 as m,n→∞, which meansyt,n converges in mean square in n by (i).

(b) Since yt,n →m.s. yt as shown in (a), E(yt) = limn→∞

E(yt,n) by (ii). But E(yt,n) = 0.

(c) Since yt,n − µ→m.s. yt − µ and yt−j,n − µ→m.s. yt−j − µ as n→∞,

E[(yt − µ)(yt−j − µ)] = limn→∞

E[(yt,,n − µ)(yt−j ,,n − µ)].

(d) (reproducing the answer on pp. 441-442 of the book) Since ψj is absolutely summable,ψj → 0 as j → ∞. So for any j, there exists an A > 0 such that |ψj+k| ≤ A for allj, k. So |ψj+k · ψk| ≤ A|ψk|. Since ψk (and hence Aψk) is absolutely summable, so isψj+k · ψk (k = 0, 1, 2, . . .) for any given j. Thus by (i),

|γj | = σ2

∣∣∣∣∣∞∑

k=0

ψj+kψk

∣∣∣∣∣ ≤ σ2∞∑

k=0

|ψj+kψk| = σ2∞∑

k=0

|ψj+k| |ψk| <∞.

Now set ajk in (ii) to |ψj+k| · |ψk|. Then

∞∑j=0

|ajk| =∞∑

j=0

|ψk| |ψj+k| ≤ |ψk|∞∑

j=0

|ψj | <∞.

Let

M ≡∞∑

j=0

|ψj | and sk ≡ |ψk|∞∑

j=0

|ψj+k|.

Then sk is summable because |sk| ≤ |ψk| ·M and ψk is absolutely summable. There-fore, by (ii),

∞∑j=0

( ∞∑k=0

|ψj+k| · |ψk|)<∞.

This and the first inequality above mean that γj is absolutely summable.

1

Page 32: Econometrics_solutions to Analy - Fumio Hayashi

3. (a)

γj = Cov(yt,n, yt−j,n)= Cov(h0xt + h1xt−1 + · · ·+ hnxt−n, h0xt−j + h1xt−j−1 + · · ·+ hnxt−j−n)

=n∑

k=0

n∑`=0

hkh` Cov(xt−k, xt−j−`)

=n∑

k=0

n∑`=0

hkh` γxj+`−k.

(b) Since hj is absolutely summable, we have yt,n →m.s. yt as n→∞ by Proposition 6.2(a).Then, using the facts (i) and (ii) displayed in Analytical Exercise 2, we can show:

n∑k=0

n∑`=0

hkh` γxj+`−k = Cov(yt,n, yt−j,n)

= E(yt,n yt−j,n)− E(yt,n) E(yt−j,n) → E(yt yt−j)− E(yt) E(yt−j) = Cov(yt, yt−j)

as n → ∞. That is,∑n

k=0

∑n`=0 hkh` γ

xj+`−k converges as n → ∞, which is the desired

result.

4. (a) (8) solves the difference equation yj − φ1yj−1 − φ2yj−2 = 0 because

yj − φ1yj−1 − φ2yj−2

= (c10λ−j1 + c20λ

−j2 )− φ1(c10λ

−j+11 + c20λ

−j+12 )− φ2(c10λ

−j+21 + c20λ

−j+22 )

= c10λ−j1 (1− φ1λ1 − φ2λ

21) + c20λ

−j1 (1− φ1λ2 − φ2λ

22)

= 0 (since λ1 and λ2 are the roots of 1− φ1z − φ2z2 = 0).

Writing down (8) for j = 0, 1 gives

y0 = c10 + c20, y1 = c10λ−11 + c20λ

−12 .

Solve this for (c10, c20) given (y0, y1, λ1, λ2).

(b) This should be easy.

(c) For j ≥ J , we have jnξj < bj . Define B as

B ≡ maxξ

b,2nξj

b2,3nξ3

b3, . . . ,

(J − 1)nξJ−1

bJ−1

.

Then, by construction,

B ≥ jnξj

bjor jnξj ≤ B bj

for j = 0, 1, .., J − 1. Choose A so that A > 1 and A > B. Then jnξj < bj < Abj forj ≥ J and jnξj ≤ B bj < Abj for all j = 0, 1, . . . , J − 1.

(d) The hint is the answer.

5. (a) Multiply both sides of (6.2.1′) by yt−j −µ and take the expectation of both sides to derivethe desired result.

(b) The result follows immediately from the MA representation yt−j − µ = εt−j + φ εt−j−1 +φ2 εt−j−2 + · · · .

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Page 33: Econometrics_solutions to Analy - Fumio Hayashi

(c) Immediate from (a) and (b).

(d) Set j = 1 in (10) to obtain γ1 − ργ0 = 0. Combine this with (9) to solve for (γ0, γ1):

γ0 =σ2

1− φ2, γ1 =

σ2

1− φ2φ.

Then use (10) as the first-order difference equation for j = 2, 3, . . . in γj with the initialcondition γ1 = σ2

1−φ2φ. This gives: γj = σ2

1−φ2φj , verifying (6.2.5).

6. (a) Should be obvious.

(b) By the definition of mean-square convergence, what needs to be shown is that E[(xt −xt,n)2] → 0 as n→∞.

E[(xt − xt,n)2] = E[(φnxt−n)2] (since xt = xt,n + φn xt−n)

= φ2n E(x2t−n)

→ 0 (since |φ| < 1 and E(x2t−n) <∞).

(c) Should be obvious.

7. (d) By the hint, what needs to be shown is that (F)nξt−n →m.s. 0. Let zn ≡ (F)nξt−n.Contrary to the suggestion of the hint, which is to show the mean-square convergence ofthe components of zn, here we show an equivalent claim (see Review Question 2 to Section2.1) that lim

n→∞E(z′nzn) = 0.

z′nzn = trace(z′nzn) = trace[ξ′t−n[(F)n]′[(F)n]ξt−n] = traceξt−nξ′t−n[(F)n]′[(F)n]

Since the trace and the expectations operator can be interchanged,

E(z′nzn) = traceE(ξt−nξ′t−n)[(F)n]′[(F)n].

Since ξt is covariance-stationary, we have E(ξt−nξ′t−n) = V (the autocovariance matrix).Since all the roots of the characteristic equation are less than one in absolute value, Fn =T(Λ)nT−1 converges to a zero matrix. We can therefore conclude that E(z′nzn) → 0.

(e) ψn is the (1,1) element of T(Λ)nT−1.

8. (a)

E(yt) =1− φt

1− φc+ φt E(y0) →

c

1− φ,

Var(yt) =1− φ2t

1− φ2σ2 + φ2t Var(y0) →

σ2

1− φ2,

Cov(yt, yt−j) = φj

[1− φ2(t−j)

1− φ2σ2 + φ2(t−j) Var(y0)

]→ φj σ2

1− φ2.

(b) This should be easy to verify given the above formulas.

9. (a) The hint is the answer.

(b) Since γj → 0, the result proved in (a) implies that 2n

∑nj=1 |γj | → 0. Also, γ0/n → 0. So

by the inequality for Var(y) shown in the question, Var(y) → 0.

3

Page 34: Econometrics_solutions to Analy - Fumio Hayashi

10. (a) By the hint,

n∑j=1

j aj ≤N∑

j=1

∣∣∣∣∣n∑

k=j

ak

∣∣∣∣∣ +n∑

j=N+1

∣∣∣∣∣n∑

k=j

ak

∣∣∣∣∣ < NM + (n−N)ε

2.

So1n

n∑j=1

j aj <NM

n+n−N

n

ε

2<NM

n+ε

2.

By taking n large enough, NM/n can be made less than ε/2.

(b) From (6.5.2),

Var(√n y) = γ0 + 2

n−1∑j=1

(1− j

n

)γj =

γ0 + 2n−1∑j=1

γj

− 2n

n−1∑j=1

j γj .

The term in brackets converges to∑∞

j=−∞ γj if γj is summable. (a) has shown that thelast term converges to zero if γj is summable.

4

Page 35: Econometrics_solutions to Analy - Fumio Hayashi

September 14, 2004 Hayashi Econometrics

Solution to Chapter 7 Analytical Exercises

1. (a) Since a(w) 6= 1 ⇔ f(y|x;θ) 6= f(y|x;θ0), we have Prob[a(w) 6= 1] = Prob[f(y|x;θ) 6=f(y|x;θ0)]. But Prob[f(y|x;θ) 6= f(y|x;θ0)] > 0 by hypothesis.

(b) Set c(x) = log(x) in Jensen’s inequality. a(w) is non-constant by (a).

(c) By the hint, E[a(w)|x] = 1. By the Law of Total Expectation, E[a(w)] = 1.

(d) By combining (b) and (c), E[log(a(w))] < log(1) = 0. But log(a(w)) = log f(y|x;θ) −log f(y|x;θ0).

2. (a) (The answer on p. 505 is reproduced here.) Since f(y | x;θ) is a hypothetical density, itsintegral is unity: ∫

f(y | x;θ)dy = 1. (1)

This is an identity, valid for any θ ∈ Θ. Differentiating both sides of this identity withrespect to θ, we obtain

∂θ

∫f(y | x;θ)dy = 0

(p×1). (2)

If the order of differentiation and integration can be interchanged, then

∂θ

∫f(y | x;θ)dy =

∫∂

∂θf(y | x;θ)dy. (3)

But by the definition of the score, s(w;θ)f(y | x;θ) = ∂∂θ f(y | x;θ). Substituting this

into (3), we obtain ∫s(w;θ)f(y | x;θ)dy = 0

(p×1). (4)

This holds for any θ ∈ Θ, in particular, for θ0. Setting θ = θ0, we obtain∫s(w;θ0)f(y | x;θ0)dy = E[s(w;θ0) | x] = 0

(p×1). (5)

Then, by the Law of Total Expectations, we obtain the desired result.

(b) By the hint, ∫H(w;θ)f(y|x;θ)dy +

∫s(w;θ) s(w;θ)′f(y | x;θ)dy = 0

(p×p).

The desired result follows by setting θ = θ0.

3. (a) For the linear regression model with θ ≡ (β′, σ2)′, the objective function is the average loglikelihood:

Qn(θ) = −12

log(2π)− 12

log(σ2)− 12σ2

1n

n∑t=1

(yt − x′tβ)2.

1

Page 36: Econometrics_solutions to Analy - Fumio Hayashi

To obtain the concentrated average log likelihood, take the partial derivative with respectto σ2 and set it equal to 0, which yields

σ2 =1n

n∑t=1

(yt − x′tβ)2 ≡ 1n

SSR(β).

Substituting this into the average log likelihood, we obtain the concentrated average loglikelihood (concentrated with respect to σ2):

Qn(β,1n

SSR(β)) = −12

log(2π)− 12− 1

2log(

1n

SSR(β)).

The unconstrained ML estimator (β, σ2) of θ0 is obtained by maximizing this concentratedaverage log likelihood with respect to β, which yields β, and then setting σ2 = 1

nSSR(β).The constrained ML estimator, (β, σ2), is obtained from doing the same subject to theconstraint Rβ = c. But, as clear from the expression for the concentrated average loglikelihood shown above, maximizing the concentrated average log likelihood is equivalentto minimizing the sum of squared residuals SSR(β).

(b) Just substitute σ2 = 1nSSR(β) and σ2 = 1

nSSR(β) into the concentrated average loglikelihood above.

(c) As explained in the hint, both σ2 and σ2 are consistent for σ20 . Reproducing (part of)

(7.3.18) of Example 7.10,

−E[H(wt;θ0)

]=

1σ20

E(xtx′t) 0

0′ 12σ4

0

. (7.3.18)

Clearly, both Σ and Σ are consistent for −E[H(wt;θ0)

]because both σ2 and σ2 are

consistent for σ20 and 1

n

∑nt=1 xtx′t is consistent for E(xtx′t).

(d) The a(θ) and A(θ) in Table 7.2 for the present case are

a(θ) = Rβ − c, A(θ) =(

R(r×K)

... 0(r×1)

).

Also, observe that

∂Qn(θ)∂θ

=

[1

σ21n

∑nt=1 xt(yt − x′tβ)

− 12σ2 + 1

2σ41n

∑nt=1(yt − x′tβ)2

]=

1SSRR

[X′(y −Xβ)

0

]and

Qn(θ) = −12

log(2π)− 12− 1

2log(

1n

SSRU ), Qn(θ) = −12

log(2π)− 12− 1

2log(

1n

SSRR).

Substitute these expressions and the expression for Σ and Σ given in the question into theTable 7.2 formulas, and just do the matrix algebra.

(e) The hint is the answer.(f) Let x ≡ SSRR

SSRU. Then x ≥ 1 and W/n = x− 1, LR/n = log(x), and LM/n = 1− 1

x . Drawthe graph of these three functions of x with x in the horizontal axis. Observe that theirvalues at x = 1 are all 0 and the slopes at x = 1 are all one. Also observe that for x > 1,x− 1 > log(x) > 1− 1

x .

2

Page 37: Econometrics_solutions to Analy - Fumio Hayashi

September 22, 2004 Hayashi Econometrics

Solution to Chapter 8 Analytical Exercises

1. From the hint,n∑

t=1

(yt −Π′xt)(yt −Π′xt)′ =n∑

t=1

vtv′t + (Π−Π)′( n∑

t=1

xtx′t)(Π−Π).

But

(Π−Π)′( n∑

t=1

xtx′t)(Π−Π) =

n∑t=1

[Π−Π)′xtx′t(Π−Π)

]is positive semi-definite.

2. Since yt = Π′0xt + vt, we have yt −Π′xt = vt + (Π0 −Π)′xt. So

E[(yt −Π′xt)(yt −Π′xt)′] = E[(vt + (Π0 −Π)′xt)(vt + (Π0 −Π)′xt)′]

= E(vtv′t) + E[vtx′t(Π0 −Π)] + E[(Π0 −Π)′xtv′t] + (Π0 −Π)′ E(xtx′t)(Π0 −Π)

= E(vtv′t) + (Π0 −Π)′ E(xtx′t)(Π0 −Π) (since E(xtv′t) = 0).

SoΩ(Π) → Ω0 + (Π0 −Π)′ E(xtx′t)(Π0 −Π)

almost surely. By the matrix algebra result cited in the previous question,

|Ω0 + (Π0 −Π)′ E(xtx′t)(Π0 −Π)| ≥ |Ω0| > 0.

So for sufficiently large n, Ω(Π) is positive definite.

3. (a) Multiply both sides of z′tm =(y′tS

′m

... x′tC′m

)from left by xt to obtain

xtz′tm =[xty′tS

′m

... xtx′tC′m

]. (∗)

Do the same to the reduced form y′t = x′tΠ0 + v′t to obtain xty′t = xtx′tΠ0 + xtv′t.Substitute this into (∗) to obtain

xtz′tm =[xtx′tΠ0S′m

... xtx′tC′m

]+

[xtv′t

... 0]

= xtx′t[Π0S′m

... C′m

]+

[xtv′t

... 0].

Take the expected value of both sides and use the fact that E(xtv′t) = 0 to obtain thedesired result.

(b) Use the reduced form yt = Π′0xt + vt to derive

yt + Γ−1Bxt = vt + (Π′0 + Γ−1B)xt

as in the hint. So

(yt + Γ−1Bxt)(yt + Γ−1Bxt)′

= [vt + (Π′0 + Γ−1B)xt][vt + (Π′

0 + Γ−1B)xt]′

= vtv′t + (Π′0 + Γ−1B)xtv′t + vtx′t(Π

′0 + Γ−1B)′ + (Π′

0 + Γ−1B)xtx′t(Π′0 + Γ−1B)′.

1

Page 38: Econometrics_solutions to Analy - Fumio Hayashi

Taking the expected value and noting that E(xtv′t) = 0, we obtain

E[(yt + Γ−1Bxt)(yt + Γ−1Bxt)′] = E(vtv′t) + (Π′0 + Γ−1B) E(xtx′t)(Π

′0 + Γ−1B)′.

Since yt,xt is i.i.d., the probability limit of Ω(δ) is given by this expectation. In thisexpression, E(vtv′t) equals Γ−1

0 Σ0(Γ−10 )′ because by definition vt ≡ Γ−1

0 εt and Σ0 ≡E(εtε

′t).

(c) What needs to be proved is that plim |Ω(δ)| is minimized only if ΓΠ′0 + B = 0. Let A ≡

Γ−10 Σ0(Γ−1

0 )′ be the first term on the RHS of (7) and let D ≡ (Π′0 +Γ−1B) E(xtx′t)(Π

′0 +

Γ−1B)′ be the second term. Since Σ0 is positive definite and Γ−10 is non-singular, A is

positive definite. Since E(xtx′t) is positive definite, D is positive semi-definite. Then usethe following the matrix inequality (which is slightly different from the one mentioned inAnalytical Exercise 1 on p. 552):

(Theorem 22 on p. 21 of Matrix Differential Calculus with Applications in Statisticsand Econometrics by Jan R. Magnus and Heinz Neudecker, Wiley, 1988) Let A bepositive definite and B positive semi-definite. Then

|A + B| ≥ |A|

with equality if and only if B = 0.

Henceplim |Ω(δ)| = |A + D| ≥ |A| = |Γ−1

0 Σ0(Γ−10 )′|.

with equality “|A + D| = |A|” only if D = 0. Since E(xtx′t) is positive definite, D ≡(Π′

0 + Γ−1B) E(xtx′t)(Π′0 + Γ−1B)′ is a zero matrix only if Π′

0 + Γ−1B = 0, which holdsif and only if ΓΠ′

0 + B = 0 since Γ is non-singular (the parameter space is such that Γ isnon-singular).

(d) For m = 1, the LHS of (8) is

α′m =

[− γ11 1 −β11 −β12 0

].

The RHS is

e′m − δ′m(1×(Mm+Km))

Sm(Mm×M)

0(Mm×K)

0(Km×M)

Cm(Km×K)

=

[0 1 0 0 0

]−

[γ11 β11 β12

] 1 0 0 0 00 0 1 0 00 0 0 1 0

.

2

Page 39: Econometrics_solutions to Analy - Fumio Hayashi

(e) Since α′m is the m-th row of

... B], the m-th row of of the LHS of (9) equals

α′m

[Π′

0

IK

]=

e′m − δ′m(1×(Mm+Km))

Sm(Mm×M)

0(Mm×K)

0(Km×M)

Cm(Km×K)

Π′0

(M×K)

IK

(by (8))

= e′m

[Π′

0

IK

]− δ′m

[Sm 00 Cm

] [Π′

0

IK

]

= [[Π0

... IK ] em]′ − δ′m

[SmΠ′

0

Cm

]

= π′0m − δ′m

[SmΠ′

0

Cm

](by the definition of π0m).

(f) By definition (see (8.5.10)), Γ0Π′0 + B0 = 0. By the same argument given in (e) with δm

replaced by δ0m shows that δ0m is a solution to (10). Rewrite (10) by taking the transpose:

Ax = y with A ≡ [Π0S′m... C′

m],x ≡ δm,y ≡ π0m. (10′)

A necessary and sufficient condition that δ0m is the only solution to (10′) is that thecoefficient matrix in (10′), which is K × Lm (where Lm = Mm + Km), be of full columnrank (that is, the rank of the matrix be equal to the number of columns, which is Lm). Wehave shown in (a) that this condition is equivalent to the rank condition for identificationfor the m-th equation.

(g) The hint is the answer.

4. In this part, we let Fm stand for the K×Lm matrix [Π0S′m... C′

m]. Since xtK does not appearin the system, the last row of Π0 is a vector of zeros and the last row of C′

m is a vector ofzeros. So the last row of of Fm is a vector of zeros:

Fm =

Fm((K−1)×Lm)

0′(1×Lm)

.

Dropping xtK from the list of instruments means dropping the last row of Fm, which does notalter the full column rank condition. The asymptotic variance of the FIML estimator is givenin (4.5.15) with (4.5.16) on p. 278. Using (6) on (4.5.16), we obtain

Amh = F′m E(xtx′t)Fh =

[F′

m 0] [

E(xtx′t) E(xtK xt)E(xtK x′t) E(x2

tK)

] [Fh

0′

]= F′

m E(xtx′t)Fh.

This shows that the asymptotic variance is unchanged when xtK is dropped.

3

Page 40: Econometrics_solutions to Analy - Fumio Hayashi

September 16, 2004 Hayashi Econometrics

Solution to Chapter 9 Analytical Exercises

1. From the hint, we have

1T

T∑t=1

∆ξt · ξt−1 =12

(ξT√T

)2

− 12

(ξ0√T

)2

− 12T

T∑t=1

(∆ξt)2. (∗)

Consider the second term on the RHS of (∗). Since E(ξ0/√T ) → 0 and Var(ξ0/

√T ) → 0,

ξ0/√T converges in mean square (by Chevychev’s LLN), and hence in probability, to 0. So the

second term vanishes (converges in probability to zero) (this can actually be shown directlyfrom the definition of convergence in probability). Next, consider the expression ξT /

√T in the

first term on the RHS of (∗). It can be written as

ξT√T

=1√T

(ξ0 + ∆ξ1 + · · ·+ ∆ξT ) =ξ0√T

+√T

1T

T∑t=1

∆ξt.

As just seen, ξ0√T

vanishes. Since ∆ξt is I(0) satisfying (9.2.1)-(9.2.3), the hypothesis of Proposi-tion 6.9 is satisfied (in particular, the absolute summability in the hypothesis of the Propositionis satisfied because it is implied by the one-summability (9.2.3a)). So

√T

1T

T∑t=1

∆ξt →dλX, X ∼ N(0, 1).

where λ2 is the long-run variance of ∆ξt. Regarding the third term on the RHS of (∗), since∆ξt is ergodic stationary, 1

2T

∑Tt=1(∆ξt)

2 converges in probability to 12γ0. Finally, by Lemma

2.4(a) we conclude that the RHS of (∗) converges in distribution to λ2

2 X2 − 1

2γ0.

2. (a) The hint is the answer.(b) From (a),

T · (ρµ − 1) =1T

∑Tt=1 ∆yt y

µt−1

1T 2

∑Tt=1(y

µt−1)2

.

Apply Proposition 9.2(d) to the numerator and Proposition 9.2(c) to the denominator.(c) Since yt is random walk, λ2 = γ0. Just set λ2 = γ0 in (4) of the question.(d) • First, a proof that α∗ →p 0. By the algebra of OLS,

α∗ =1T

T∑t=1

(yt − ρµyt−1)

=1T

T∑t=1

(∆yt − (ρµ − 1)yt−1)

=1T

T∑t=1

∆yt − (ρµ − 1)1T

T∑t=1

yt−1

=1T

T∑t=1

∆yt −1√T

(T · (ρµ − 1)

)( 1√T

1T

T∑t=1

yt−1

).

1

Page 41: Econometrics_solutions to Analy - Fumio Hayashi

The first term after the last equality, 1T

∑Tt=1 ∆yt, vanishes (converges to zero in proba-

bility) because ∆yt is ergodic stationary and E(∆yt) = 0. To show that the second termafter the last equality vanishes, we first note that 1√

T

(T · (ρµ − 1)

)vanishes because

T · (ρµ − 1) converges to a random variable by (b). By (6) in the hint, 1√T

1T

∑Tt=1 yt−1

converges to a random variable. Therefore, by Lemma 2.4(b), the whole second termvanishes.

• Now turn to s2. From the hint,

s2 =1

T − 1

T∑t=1

(∆yt − α∗)2 − 2T − 1

· [T · (ρµ − 1)] · 1T

T∑t=1

(∆yt − α∗) · yt−1

+1

T − 1· [T · (ρµ − 1)]2 · 1

T 2

T∑t=1

(yt−1)2. (∗)

Since α∗ →p 0, it should be easy to show that the first term on the RHS of (∗) convergesto γ0 in probability. Regarding the second term, rewrite it as

2T − 1

· [T · (ρµ− 1)] · 1T

T∑t=1

∆yt yt−1−2√T

T − 1· [T · (ρµ− 1)] · α∗ · 1√

T

1T

T∑t=1

yt−1. (∗∗)

By Proposition 9.2(b), 1T

∑Tt=1 ∆yt yt−1 converges to a random variable. So does T ·

(ρµ−1). Hence the first term of (∗∗) vanishes. Turning to the second term of (∗∗), (6)in the question means 1√

T1T

∑Tt=1 yt−1 converges to a random variable. It should now

be routine to show that the whole second term of (∗∗) vanishes. A similar argument,this time utilizing Proposition 9.2(a), shows that the third term of (∗) vanishes.

(e) By (7) in the hint and (3), a little algebra yields

tµ =ρµ − 1

s · 1√∑Tt=1(y

µt−1)

2

=1T

∑Tt=1 ∆yt y

µt−1

s ·√

1T 2

∑Tt=1(y

µt−1)2

.

Use Proposition 9.2(c) and (d) with λ2 = γ0 = σ2 and the fact that s is consistent for σto complete the proof.

3. (a) The hint is the answer.

(b) From (a), we have

T · (ρτ − 1) =1T

∑Tt=1 ∆yt yτt−1

1T 2

∑Tt=1(y

τt−1)2

.

Let ξt and ξτt be as defined in the hint. Then ∆yt = δ+∆ξt and yτt = ξτt . By construction,∑Tt=1 y

τt−1 = 0. So

T · (ρτ − 1) =1T

∑Tt=1 ∆ξt ξτt−1

1T 2

∑Tt=1(ξ

τt−1)2

.

Since ξt is driftless I(1), Proposition 9.2(e) and (f) can be used here.

(c) Just observe that λ2 = γ0 if yt is a random walk with or without drift.

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Page 42: Econometrics_solutions to Analy - Fumio Hayashi

4. From the hint,

1T

T∑t=1

yt−1 εt = ψ(1)1T

T∑t=1

wt−1 εt +1T

T∑t=1

ηt−1 εt + (y0 − η0)1T

T∑t=1

εt. (∗)

Consider first the second term on the RHS of (∗). Since ηt−1, which is a function of (εt−1, εt−2, . . . ),is independent of εt, we have: E(ηt−1εt) = E(ηt−1) E(εt) = 0. Then by the ergodic theoremthis second term vanishes. Regarding the third term of (∗), 1

T

∑Tt=1 εt →p 0. So the whole third

term vanishes. Lastly, consider the first term on the RHS of (∗). Since wt is random walk andεt = ∆wt, Proposition 9.2(b) with λ2 = γ0 = σ2 implies 1

T

∑Tt=1 wt−1 εt →d

(σ2

2

)[W (1)2− 1].

5. Comparing Proposition 9.6 and 9.7, the null is the same (that ∆yt is zero-mean stationaryAR(p), φ(L)∆yt = εt, whose MA representation is ∆yt = ψ(L)εt with ψ(L) ≡ φ(L)−1) butthe augmented autoregression in Proposition 9.7 has an intercept. The proof of Proposition9.7 (for p = 1) makes appropriate changes on the argument developed on pp. 587-590. Let band β be as defined in the hint. The AT and cT for the present case is

AT =

[1T 2

∑Tt=1(y

µt−1)

2 1√T

1T

∑Tt=1 y

µt−1 (∆yt−1)(µ)

1√T

1T

∑Tt=1(∆yt−1)(µ) yµt−1

1T

∑Tt=1[(∆yt−1)(µ)]2

],

cT =

[1T

∑Tt=1 y

µt−1 ε

µt

1√T

∑Tt=1(∆yt−1)(µ) εµt

]=

[1T

∑Tt=1 y

µt−1 εt

1√T

∑Tt=1(∆yt−1)(µ) εt

],

where εµt is the residual from the regression of εt on a constant for t = 1, 2, ..., T .

• (1,1) element of AT : Since yt is driftless I(1) under the null, Proposition 9.2(c) canbe used to claim that 1

T 2

∑Tt=1(y

µt−1)

2 →d λ2∫

(Wµ)2, where λ2 = σ2[ψ(1)]2 with σ2 ≡Var(εt).

• (2,2) element of AT : Since (∆yt−1)(µ) = ∆yt−1 − 1T

∑Tt=1 ∆yt−1, this element can be

written as1T

T∑t=1

[(∆yt−1)(µ)]2 =1T

T∑t=1

(∆yt−1)2 −

(1T

T∑t=1

∆yt−1

)2

.

Since E(∆yt−1) = 0 and E[(∆yt−1)2] = γ0 (the variance of ∆yt), this expression convergesin probability to γ0.

• Off diagonal elements of AT : it equals

1√T

1T

T∑t=1

(∆yt−1)(µ) yµt−1 =1√T

[1T

T∑t=1

(∆yt−1) yt−1

]−

(1√T

1T

T∑t=1

yt−1

)(1T

T∑t=1

∆yt−1

).

The term in the square bracket is (9.4.14), which is shown to converge to a random variable(Review Question 3 of Section 9.4). The next term, 1√

T1T

∑Tt=1 yt−1, converges to a ran-

dom variable by (6) assumed in Analytical Exercise 2(d). The last term, 1T

∑Tt=1 ∆yt−1,

converges to zero in probability. Therefore, the off-diagonal elements vanish.

Taken together, we have shown that AT is asymptotically diagonal:

AT →d

[λ2 ·

∫ 1

0[Wµ(r)]2 dr 0

0 γ0

],

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Page 43: Econometrics_solutions to Analy - Fumio Hayashi

so

(AT )−1 →d

[(λ2 ·

∫ 1

0[Wµ(r)]2 dr

)−1 00 γ−1

0

].

Now turn to cT .

• 1st element of cT : Recall that yµt−1 ≡ yt−1 − 1T

∑Tt=1 yt−1. Combine this with the BN

decomposition yt−1 = ψ(1)wt−1 + ηt−1 + (y0 − η0) with wt−1 ≡ ε1 + · · ·+ εt−1 to obtain

1T

T∑t=1

yµt−1 εt = ψ(1)1T

T∑t=1

wµt−1 εt +1T

T∑t=1

ηµt−1 εt,

where wµt−1 ≡ wt−1 − 1T

∑Tt=1 wt−1. η

µt−1 is defined similarly. Since ηt−1 is independent of

εt, the second term on the RHS vanishes. Noting that ∆wt = εt and applying Proposition9.2(d) to the random walk wt, we obtain

1T

T∑t=1

wµt−1 εt →d

(σ2

2

)[W (1)µ]2 − [W (0)µ]2 − 1

.

Therefore, the 1st element of cT converges in distribution to

c1 ≡ σ2 · ψ(1) · 12[W (1)µ]2 − [W (0)µ]2 − 1

.

• 2nd element of cT : Using the definition (∆yt−1)(µ) ≡ ∆yt−1− 1T

∑Tt=1 ∆yt−1, it should be

easy to show that it converges in distribution to

c2 ∼ N(0, γ0 · σ2).

Using the results derived so far, the modification to be made on (9.4.20) and (9.4.21) on p.590 for the present case where the augmented autoregression has an intercept is

T · (ρµ − 1) →d

σ2ψ(1)λ2

·12

[W (1)µ]2 − [W (0)µ]2 − 1

∫ 1

0[Wµ(r)]2 dr

orλ2

σ2ψ(1)· T · (ρµ − 1) →

dDFµρ ,

√T · (ζ1 − ζ1) →

dN(0,σ2

γ0

).

Repeating exactly the same argument that is given in the subsection entitled “Deriving TestStatistics” on p. 590, we can claim that λ2

σ2ψ(1) is consistently estimated by 1/(1 − ζ). Thiscompletes the proof of claim (9.4.34) of Proposition 9.7.

6. (a) The hint is the answer.(b) The proof should be straightforward.

7. The one-line proof displayed in the hint is (with i replaced by k to avoid confusion)

∞∑j=0

|αj | =∞∑j=0

∣∣∣∣∣∣−∞∑

k=j+1

ψk

∣∣∣∣∣∣ ≤∞∑j=0

∞∑k=j+1

|ψk| =∞∑k=0

k|ψk| <∞, (∗)

where ψk (k = 0, 1, 2, ...) is one-summable as assumed in (9.2.3a). We now justify each ofthe equalities and inequalities. For this purpose, we reproduce here the facts from calculusshown on pp. 429-430:

4

Page 44: Econometrics_solutions to Analy - Fumio Hayashi

(i) If ak is absolutely summable, then ak is summable (i.e., −∞ <∑∞k=0 ak <∞) and∣∣∣∣∣

∞∑k=0

ak

∣∣∣∣∣ ≤∞∑k=0

|ak|.

(ii) Consider a sequence with two subscripts, ajk (j, k = 0, 1, 2, . . .). Suppose∑∞j=0 |ajk| <

∞ for each k and let sk ≡∑∞j=0 |ajk|. Suppose sk is summable. Then∣∣∣∣∣

∞∑j=0

( ∞∑k=0

ajk

)∣∣∣∣∣ <∞ and∞∑j=0

( ∞∑k=0

ajk

)=

∞∑k=0

( ∞∑j=0

ajk

)<∞.

Since ψk is one-summable, it is absolutely summable. Let

ak =

ψk if k ≥ j + 1,0 otherwise.

Then ak is absolutely summable because ψk is absolutely summable. So by (i) above, wehave ∣∣∣∣∣−

∞∑k=j+1

ψk

∣∣∣∣∣ =∣∣∣∣∣∞∑

k=j+1

ψk

∣∣∣∣∣ =∣∣∣∣∣∞∑k=0

ak

∣∣∣∣∣ ≤∞∑k=0

|ak| =∞∑

k=j+1

|ψk|.

Summing over j = 0, 1, 2, ..., n, we obtain

n∑j=0

∣∣∣∣∣−∞∑

k=j+1

ψk

∣∣∣∣∣ ≤n∑j=0

∞∑k=j+1

|ψk|.

If the limit as n → ∞ of the RHS exists and is finite, then the limit of the LHS exists andis finite (this follows from the fact that if xn is non-decreasing in n and if xn ≤ A < ∞,then the limit of xn exists and is finite; set xn ≡

∑nj=0 | −

∑∞k=j+1 ψk|). Thus, provided that∑∞

j=0

∑∞k=j+1 |ψk| is well-defined, we have

∞∑j=0

∣∣∣∣∣−∞∑

k=j+1

ψk

∣∣∣∣∣ ≤∞∑j=0

∞∑k=j+1

|ψk|.

We now show that∑∞j=0

∑∞k=j+1 |ψk| is well-defined. In (ii), set ajk as

ajk =

|ψk| if k ≥ j + 1,0 otherwise.

Then∑∞j=0 |ajk| = k |ψk| <∞ for each k and sk = k |ψk|. By one-summability of ψk, sk

is summable. So the conditions in (ii) are satisfied for this choice of ajk. We therefore concludethat

∞∑j=0

∞∑k=j+1

|ψk| =∞∑j=0

( ∞∑k=0

ajk

)=

∞∑k=0

( ∞∑j=0

ajk

)=

∞∑k=0

k |ψk| <∞.

This completes the proof.

5