Springer Texts in Business and Economics

Simultaneous Equations Model

11.1 The Inconsistency of OLS. The OLS estimator from Eq. (11.14) yields 8ols =

T T

E Ptqt/ E Pt2 where pt = Pt — l3 and qt = Qt — Q. Substituting qt = 8pt C

t=i t=i

TT

(u2t — U2) from (11.14) we get 8ois = 8 C E Pt(u2t — N/Y, P2. Using (11.18),

t=i t=i

T

we get plim £ Pt(U2t — U2)/T = (012 — 022)/(8 — ") where Oij = cov. Uit, Ujt)

t=1

for i, j = 1,2 and t = 1,2,.., T. Using (11.20) we get

Plim 8ois = 8 C [(CT12 — 022)/(8 — ")]/[(011 C CT22 — 2ст12)/(8 — ")2] = 8 C (o12 — 022)(8 — ")/(o11 C 022 — 2°12).

11.2 When Is the IVEstimator Consistent?

a. ForEq.(11.30)y1 = a^y2 C '1зУз C "11X1 C "12X2 C U1.Whenweregress

T

У2 on X1, X2 and X3 to get У2 = у2 C V2, the residuals satisfy E y2tV2t = 0

t=1

TTT

and 22 V2tXu = 22 V2tX2t = 22 V2tX3t = 0.

t=1 t=1 t=1

Similarly, when we regress y3 on X1, X2 and X4 to get y3 = y3 C

T T T

V3, the residuals satisfy E y3t83t = 0 and E V3tXu = E ^3tX2t =

t=1 t=1 t=1

T

E v3tX4t = 0. In the second stage regression y1 = a^y2 C '13y3 C

t=1

"11X1 C "12X2 C £1 where Є1 = '12 (y2 — 382) C '13(y3 — 383) C u =

T

'12882 C'13883Cu1. For this to yield consistent estimates, we needE 2t©1t =

t=1

Ey^m E y3t©1t = E y3^; £ X1t£1t = E X1tu1t and E X2t£1t =

t=1 t=1 t=1 t=1 t=1 t=1

E X2tu1t. But E y 2t©1t = '12 E 882ty2t C '13 E 88 3t:8 2t C E y2tu1t with

t=1 t=1 t=1 t=1 t=1

T

E V2ty2t = 0 because 882 are the residuals and y2 are the Predicted val-

t=1

T

ues from the same regression. However, 883ty2t is not necessarily zero

t=1

B. H. Baltagi, Solutions Manual for Econometrics, SPringer Texts in Business and Economics, DOI 10.1007/978-3-642-54548-1—11, © SPringer-Verlag Berlin Heidelberg 2015

because V3 is only orthogonal to Xi, X2 and X4, while У2 is a perfect linear

T

combination of Xi, X2 and X3. Hence, Y $3tX3t is not necessarily zero,

t=i

T

which makes Y v3ty2t not necessarily zero.

t=i

T T T T

Similarly, y3t©it = ai2 £ V2ty3t C 'nY v3ty3t C Y y3tuit with

t=i t=i t=i t=i

TT

Y v3ty3t = 0 and Y V2ty^3t not necessarily zero. This is because V2t is

t=i t=i

only orthogonal to Xi, X2 and X3, while y3 is a perfect linear combina-

T

tion of Xi, X2 and X4. Hence, Y v2tX4t is not necessarily zero, which

t=i

T

makes Y V2ty3t not necessarily zero. Since both Xi and X2 are included

t=i

TT

in the first stage regressions, we have J]Xit£it = Y XitUit using the

t=i t=i

T T T T

fact that XitV2t = Y XitV3t = 0. Also, X2teit = Y X2tuit since

t=i t=i t=i t=i

TT

Y X2tV2t = Y X2tV3t = 0.

t=i t=i

b. The first stage regressions regress y2 and y3 on X2, X3 and X4 to get

T

y2 = y2 C V2 and y3 = y3 C V3 with the residuals satisfying Y V2ty2t =

t=i

Y V2tX2t = J2 V2tX3t = J2 V2tX4t = 0 and Y V3ty3t = J2 V3tX2t =

t=i t=i t=i t=i t=i

TT

v3tX3t = V v3tX4t = 0. In the second stage regression yi = ai2y2 C

t=i t=i

'i3y3 C PiiXi C "i2X2 C £i where ei = ап(у2 - У2) C ai3.y3 - У3) C ui = ai2v2 C ai3v3 C ui. For this to yield consistent estimates, we need p y2t©it = p y2tuit; p y3teit = p y3tuit; p X^it = p Xituit and

t=i t=i t=i t=i t=i t=i

P X2t©it = P X2tuit.

t=i t=i

The first two equalities are satisfied because

T T T T

y2tv2t = y2tv3t = 0 and y3tv2t = y3tv3t = 0

t=i t=i t=i t=i

since the same set of X’s are included in both first stage regressions.

T T T T

Y X2t©it = Y X2tuit because X2tV2t = Y X2tV3t = 0 since X2 is

t=i t=i t=i t=i

T T

included in both first-stage regressions. However, X1te1t ф X1tu1t

t=i t=i

TT

since y^X1tv2t ф 0 and E XitV3t ф 0

t=i t=i

because Xi was not included in the first stage regressions.

11.3 Just-Identification and Simple IV. If Eq. (11.34) is just-identified, then X2 is of the same dimension as Yi, i. e., both are Txgi. Hence, Zi is of the same dimension as X, both of dimension Tx (g1 + k1). Therefore, X0Z1 is a square non-singular matrix of dimension (g1 + k1). Hence, (ZjX) 1 exists. But from (11.36), we know that

•i,2SLS = (ZjPxZi)-1 ZjPxyi = [Z1X(X, X)-1X, Z1]“1 Z1X(X, X)-1X, y1

= (X, Z1)-1(X, X) (z;x)-1 (z;x) (x0x)-1x0y1 = (x0ZO-1x0y1

as required. This is exactly the IV estimator with W = X given in (11.41). It is important to emphasize that this is only feasible if X0Z1 is square and non-singular.

11.4 2SLS Can Be Obtained as GLS. Premultiplying (11.34) by X0 we get X0y1 = X, Z181 + X0u1 with X0u1 having mean zero and var(X0u1) = o11X0X since var(u1) = o11 IT. Performing GLS on this transformed equation yields

•i, gls = [Z01X(anX0X)-1X0Z1]-1Z01X(an X0X)-1X0y1 = (Z^Zi)-^0^

as required.

11.5 The Equivalence of 3SLS and 2SLS.

a. From (11.46), (i) if X is diagonal then X-1 is diagonal and X-1 <S> PX is block-diagonal with the i-th block consisting of PX/(tii. Also, Z is block - diagonal, therefore, fZ0[X-1 <S> PX]Z}-1 is block-diagonal with the i-th block consisting of (jii (Z0PXZ^ 1. In other words,

y0 0 .. Zoy

/ZiPXZi/CT11 0 .. 0

0 Z2 Px Z2 /f^22 .. 0

у 0 0 .. Z0PxZo/6ooy

VZ0Pxyo/ООО у

Hence, from (11.46) 83SLS = {Z0[C 1 <8> Px]Z} 1{Z0[^ 1 (g> Px]y}

(ii) If every equation in the system is just-identified, then ZiX is square and non-singular for i = 1,2, ..,G. In problem 11.3, we saw that for say the first equation, both Z1 and X have the same dimension Tx(g1 + k1). Hence, Z1X is square and of dimension (g1 + k1). This holds for every

equation of the just-identified system. In fact, problem 11.3 proved that 8il2SLS = (X, Zi)_1 X0yifori = 1,2,.., G. Also, from (11.44), we get

83SLS = {diag [Z0X] [t-1 ® (X0X)-1] diag [X0Zi]}-1

{diag [Z0X] [£-1 ® (X0X)-1] (Ig ® X0)y}.

But each Zi0X is square and non-singular, hence

83SLS = [diag (X'Zi)-1 (£ ® (X0X)) diag (Z0X)-1

{diag (Z0X) [e-1 ® (X0X)-1] (Ig ® X0)y}.

which upon cancellation yields

O3SLS = diag(X0Zi)-1 (Ig ® X0)y

/(X, Z1)-1 0 .. 0 1 0 X'yA

b. Premultiplying (11.43) by (IG <g> PX) yields y* = Z*8 + u* where y* = (IG <8>PX )y, Z* = (IG <S>PX )Zandu* = (IG <g>PX)u. OLS on this transformed model yields

Sols = (Z*0Z*)-1Z*0y* = [Z0(Ig ® Px)Z]-1 [Z0(Ig ® PX)y]

which is the 2SLS estimator of each equation in (11.43). Note that u* has mean zero and var(u*) = £ <S> PX since var(u) = £ <S> IT. The generalized inverse of var(u*) is £-1 <S> PX. Hence, GLS on this transformed model yields

8gls = [Z*0(£-1 ® Px)Z*]-1Z*0(£-1 ® Px)y*

= [Z0(Ig <8> Px)(£ 1 <S> Px)(Ig <S> Px)Z] 1Z0(Ig <8> Px)(£ 1 <S> Px) x (Ig <S> Px)y

= [Z0(£-1 ® Px)Z]-1Z0[£-1 ® Px]y.

Using the Milliken and Albohali condition for the equivalence of OLS and GLS given in Eq. (9.7) of Chap. 9, we can obtain a necessary and sufficient condition for 2SLS to be equivalent to 3SLS. After all, the last two esti­mators are respectively OLS and GLS on the transformed * system. This condition gives Z*0(£-1 <S> PX)PZ* = 0 since Z* is the matrix of regres­sors and £ <S> PX is the corresponding variance-covariance matrix. Note that Z* = diag[PXZ0] = diag[Z0] or a block-diagonal matrix with the i-th block being the matrix of regressors of the second-stage regression of 2SLS. Also, PZ* = diag[PZ. ] andPZ* = diag[Pz]. Hence, the above condition reduces to оijZ0PZj = 0 for i ф j and i, j = 1,2,.., G.

If £ is diagonal, this automatically satisfies this condition since cry = оij=0 for i ф j. Also, if each equation in the system is just-identified, then Z0X and X0Zj are square and non-singular. Hence,

I0Z (z0j 1 Zj = Z0X(X0X)-1X0Zj ZjX(X0X)-1X0Zj 1

Z0X(X0X)-1X0 = Z0X(X0X)-1X0 = Z0PX = Z0

after some cancellations. Hence, Z0P = Z0 — Z0P = Z0 — Z0 = 0, under

0 Z 0 0 Z 0 0

just-identification of each equation in the system.

11.6 a. Writing this system of two equations in the matrix form given by (A.1) we

(?)

b. There are two zero restrictions on Г. These restrictions state that Wt and Lt do not appear in the demand equation. Therefore, the first equation is over­identified by the order condition of identification. There are two excluded exogenous variables and only one right hand side included endogenous vari­able. However, the supply equation is not identified by the order condition of identification. There are no excluded exogenous variables, but there is one right hand side included endogenous variable.

c. The transformed matrix FB should satisfy the following normalization restrictions: fll + fl2 = 1 and f2i + f22 = 1.

These are the same normalization restrictions given in (A.6). Also, Fr must satisfy the following zero restrictions:

f110 - f12e = 0 and f110 - f12f = 0.

If either e or f are different from zero, so that at least Wt or Lt appear in the supply equation, then f12 = 0. This means that f11 + 0 = 1 or f11 = 1, from the first normalization equation. Hence, the first row of F is indeed the first row of an identity matrix and the demand equation is identified. However, there are no zero restrictions on the second equation and in the absence of any additional restrictions f21 ф 0 and the second row of F is not necessarily the second row of an identity matrix. Hence, the supply equation is not identified.

11.7 a. In this case,

b. There are four zero restrictions on Г. These restrictions state that Yt and At are absent from the supply equation whereas Wt and Lt are absent from the demand equation. Therefore, both equations are over-identified. For each equation, there is one right hand sided included endogenous variable and two excluded exogenous variables.

c. The transformed matrix FB should satisfy the following normalization restrictions:

fll + fl2 = 1 and f2i + f22 = 1.

Also, Fr must satisfy the following zero restrictions:

—f21 c C f220 = 0 - f21 d + f220 = 0.

and

fii0 — f12g = 0 fii0 — f12h = 0.

If either c or d are different from zero, so that at least Yt or At appear in the demand equation, then f2i = 0 and from the second normalization equation we deduce that f22 = i. Hence, the second row of F is indeed the second row of an identity matrix and the supply equation is identified. Similarly, if either g or h are different from zero, so that at least Wt or Lt appear in the supply equation, then fi2 = 0 and from the first normalization equation we deduce that fii = i. Hence, the first row of F is indeed the first row of an identity matrix and the demand equation is identified.

11.8 a. From example (A.1) and Eq. (A.5), we get

i b —a —c 0

i —e —d 0 —f

Therefore, ® for the first equation consists of only one restriction, namely that Wt is not in that equation, or yi3 = 0. This makes ф0 = (0,0,0,0, i), since

'ф = 0 gives уїз = 0. Therefore,

**=0=(.”,).

This is of rank one as long as f ф 0. Similarly, for * the second equation consists of only one restriction, namely that Yt is not in that equation, or У22 = 0. This makes * = (0,0,0,1,0/, since al* = 0 gives у22 = 0. 'У 12 I

/(e + b).

This can be easily verified by solving the two equations in (A.3) and (A.4) for Pt and Qt in terms of the constant, Yt and Wt.

From part (a) and Eq. (A.17), we can show that the parameters of the first structural equation can be obtained from al [W, *] = 0 where '1 = ("11, "12, "її, у12, уїз) represents the parameters of the first structural equation. W0 = [П, I3] and *0 = (0,0,0,0,1) as seen in part (a).

"11^11 + "12^21 + У11 = 0 "11^12 + "12^22 + У12 = 0 "11^13 + "12^23 + У13 = 0 у1з = 0

If we normalize by setting "11 = 1, we get "12 = —л13/л23 also

Л13 л22л13 — л12л23

У12 = —12 H---------------- Л22 = ---------------------------------- and

Л23 Л23

Л13 ТС21Л13 — ли Л23

У11 = — л11 H---------------- л21 = ------------------------------------ .

л23 л23

One can easily verify from the reduced form parameters that "12 = - Л13/Н23 = - bf/ — f = b.

Similarly, "12 = (л 22 л 13 - л 12 л 23)/ Л23 = (cbf + ecf)/ - f(e + b) = - c. Finally, "її = (л 2i л із - лїї л23)/ л 23 = (abf + aef)/ - f(e + b) = - a. Similarly, for the second structural equation, we get,

If we normalize by setting "21 = 1, we get "22 = — л 12/ л 22 also

л 12 л 23л 12 - л 13л 22 ,

"23 = - л13 Н---------------- л 23 = ---------------------------------- and

22 22

л 12 л12 л 21 - л 11 л 22

"21 = — л11 Н---------------- л 21 = ----------------------------------- .

22 22

One can easily verify from the reduced form parameters that "22 = - л 12/ л 22 = - ec/c = - e. Similarly,

"23 = (л 23л 13 - л 13л 22)/ л 22 = (-ecf - bdf)/c(e + b) = - f. Finally, "21 = (л 12 л 21 - лпл 22)/ л 22 = (-dec - bdc)/c(e + b) = - d.

11.9 For the simultaneous model in problem 11.6, the reduced form parameters

are given

This can be easily verified by solving for Pt and Qt in terms of the constant, Wt and Lt. From the solution to 11.6, we get

A = [B, Г] =

For the first structural equation,

This can be rewritten as

"іітс її + Р12Л 21 + "її = 0 "11 те 12 + Р12Л 22 + "12 = 0 "її те 13 + Р12Л 23 + "13 = 0 "12 = 0 "13 = 0

If we normalize by setting "11 = 1,weget"12 = — тт13/тт23; "12 = — т 12/ тт22.

Also, "11 = — ти + (^12/^22)^21 =

(т 13/ тт23)тт21 = —21—13-------------------- 11—23 one can easily verify from the reduced form

23

parameters that

"12 = - т 13/T23 = - bf/ — f = b. Also "12 = — т 12/т22 = —be/ — e = b. Similarly,

Also,

n 21n 13 — n 11n 23 abf — bcf + adf + bcf a(bf + df)

Y11 n 23 —f(b + d) —(bf + df) a

For the second structural equation there are no zero restrictions. Therefore, (A.15) reduces to

(0,0,0,0,0)

This can be rewritten as

"21n 11 + "22n 21 + "21 = 0 "21n 12 + "22 n 22 + "22 = 0

"21 n 13 + "22 n 23 + "23 = 0

Three equations in five unknowns. When we normalize by setting "21 = 1, we still get three equations in four unknowns for which there is no unique solution.

11.10 Just-Identified Model.

a. The generalized instrumental variable estimator for 81 based on W is given below (11.41)

8Uv = (Z^wZ!)-1 ZiPWy!

Z1W (w0w) 1 W0 Z1

= (W0Z1)—!(W0W) (ZiW 1 (ZiW (W0W)—!W—1y1 = (W0 Z1)—1W0y1

since W0Z1 is square and non-singular under just-identification. This is exactly the expression for the simple instrumental variable estimator for 81 given in (11.38).

b. Let the minimized value of the criterion function be

Q = (yi — Zi8i, iv),Pw(Yi — Z18 i, iv).

Substituting the expression for 81jv and expanding terms, we get

Q = yiPWy1 — yiW (ZiW)-1 ZiPWy1 — yiPWZ1(W, Zi)-1W, yi + y1W (Z1W)-1 Z1 PwZ 1 ( W0Z1)-1 W0y 1 Q = y,1PWy1 — y1W (Z1W)-1 Z,1W(W, W)-1 W0y1 — y'^^W/^W^ (W,1Z1)-1 W0y1 + y1W (Z1W)-1 Z,1W(W/W)-1W0Z1(W, Z1)-1W, y1 Q = y1Pwy1 — y1Pwy1 — y1Pwy1 + y1Pwy1 = 0

as required.

c. Let Z1 = PWZ1 be the set of second stage regressors obtained from regress­ing each variable in Z1 on the set of instrumental variables W. For the just-identified case

PZ1 = PpwZ1 = Z1 (Z1Z1)- Z1 = PwZ1 (Z1PWZ1)-1 Z1Pw

= W(W0W)-1W0 Z1 [Z,1W(W, W)-1 W0 Z1]-1 Z-1W(W0W)-1W0 = W(W0W)-1W0Z1(W0Z1)-1(W0W) (Z,1W)-1 Z,1W(W0W)-1W0 = W(W0W)-1W0 = PW.

Hence, the residual sum of squares of the second stage regression for the just-identified model, yields y1P£ y1 = y1PWy1 since P% = PW. This is exactly the residual sum of squares from regression y1 on the matrix of instruments W.

11.11 The More Valid Instruments the Better. If the set of instruments W1 is spanned by the space of the set of instruments W2, then PW2Wi = W1. The corresponding two instrumental variables estimators are

with asymptotic covariance matrices cm plim (Z,1PwiZ1/T)_1 fori = 1,2.

The difference between the asymptotic covariance matrix of 81,w1 and that of 8i, w2 is positive semi-definite if

is positive semi-definite. This holds, if Z,1PW2Z1 — Z,1PW1Z1 is positive semi­definite. To show this, we prove that PW2 — PW1 is idempotent.

(Pw2 — Pw1 )(Pw2 — Pw1 ) = Pw2 — Pw2Pw1 — Pw1Pw2 + Pw1

= PW2 — PW1 — PW1 + PW1 = PW2 — PW1.

This uses the fact that PW2PW1 = PW2W1 (W'W1) 1 W' = W1 (W'W1) 1 W1 = Pw1, since PW2W1 = W1.

11.12 Testing for Over-Identification. The main point here is that W spans the same space as [Z 1,W*] and that both are of full rank '. In fact, W* is a subset of instruments W, of dimension (' — k1 — g1), that are linearly independent of Z1 = PwZ1.

a. Using instruments W, the second stage regression on y1 = Z181+W*y+u1 regresses y1 on PW[Z1,W*] = [Z 1,W*] since Z 1 = PWZ1 and W* = PWW*. But the matrix of instruments W is of the same dimension as the matrix of regressors [Z 1,W*]. Hence, this equation is just-identified, and from problem 11.10, we deduce that the residual sum of squares of this second stage regression is exactly the residual sum of squares obtained by regressing y1 on the matrix of instruments W, i. e.,

URSS* = yjPwy1 = yiy1 — yiPwy1.

b. Using instruments W, the second stage regression on y1 = Z181 + u1

regresses y1 on Z 1 = PWZ1. Hence, the RRSS* = y,1P21 y1 = y1y1 — y,1Pz 1 y1 where P^ = I — 1 and P^ = Z1 (^Z1) Z1 =

PWZ1 (Z,1PWZ1) 1 Z'jPw. Therefore, using the results in part (a) we get RRSS* - URSS* = y,1PWy1 - ylPZiy1 as given in (11.49).

c. Hausman (1983) proposed regressing the 2SLS residuals (y1 — Z1812SLS) on the set of instruments W and obtaining nR^ where R is the uncentered R2 of this regression. Note that the regression sum of squares yields

(y1 — Z1Q1,2sls )0Pw(y1 — Z1§1,2SLs)

= y1PWy1 — y,1PWZ1(Z,1PWZ1)_1Z,1PWy1 — y,1PWZ1(Z,1PWZ1)_1Z,1PWy1 + y,1PWZ1(Z1PWZ1)-1Z,1PWZ1 (z;PwZ1)“1 Z1PWy1 = y1Pwy1 — y1PwZ1 (Z^PwZ1) 1 Z1Pwy1 = y,1PWy1 — yiPZl y1 = RRSS* — URSS*

as given in part (b). The total sum of squares of this regression, uncentered, is the 2SLS residuals sum of squares given by

(y1 — Z1Q1,2sls )0(y1 — Z1Q1,2SLs) = T5n

where an is given in (11.50). Hence, the test statistic

RRSS* — URSS* Regression Sum of Squares

&П Uncentered Total Sum of Squares / T

= T(uncentered R2) = T R2 as required.

This is asymptotically distributed as x2 with ' — (g1 + k1) degrees of freedom. Large values of this test statistic reject the null hypothesis.

d. The GNR for the unrestricted model given in (11.47) computes the resid­uals from the restricted model under Ho; " = 0. These are the 2SLS residuals (y1 — Z1q12SLS) based on the set of instruments W. Next, one differentiates the model with respect to its parameters and evaluates these derivatives at the restricted estimates. For instrumental variables W one has to premultiply the right hand side of the GNR by PW, this yields (yi - Zi81,2sls) = PWZ1b1 + PWW*b2 + residuals.

But Z1 = PWZ1 and PWW* = W* since W* is a subset of W. Hence, the GNR becomes (y1 — Z181,2SLS) = Z 1b1 + W*b2 + residuals. How­ever, [Z1, W*] spans the same space as W, see part (a). Hence, this GNR regresses 2SLS residuals on the matrix of instruments W and computes TR where Ru is the uncentered R2 of this regression. This is Hausman’s (1983) test statistic derived in part (c).

11.15 a. The artificial regression in (11.55) is given by

y 1 = Z181 + (Y1 — Y1)^ + residuals

This can be rewritten as y1 = Z181 + IіWY1q + residuals where we made use of the fact that Y1 = PWY1 with Y1 — Y1 = Y1 — PWY1 = IіWY1. Note that to residual out the matrix of regressors PWY1, we need

P pwy1 = I — PwY1(Y1PwY1)-1Y1Pw

so that, by the FWL Theorem, the OLS estimate of 81 can also be obtained from the following regression

P PwY1 y1 = P PwY1 Z181 + residuals.

Note that

Ppwy1 Z1 = Z1 — PwY1(Y1PwY1)-1Y1 PwZ1

with PWZ1 = [PWY1,0] and PWX1 = 0 since X1 is part of the instruments in W. Hence, PpwY1 Z1 = [Y1,X1] — [PwY1,0] = [PWY1,X1] = [Y1 ,X1] = Pw[Y1,X1] = PwZi = Z1 In this case,

8l, ols = (Z1PpwY1Z1)-1 Z1PPWY1 y1 = (Z1Z1)- ^ 1y1

= (Z1PwZ1) 1 Z1Pwy1 = 8 1,iv as required.

b. The FWL Theorem also states that the residuals from the regressions in part (a) are the same. Hence, their residuals sum of squares are the same. The last regression computes an estimate of the var(81,ojs) as s2 (Z^PwZ^ 1 where s2 is the residual sum of squares divided by the degrees of freedom of the regression. In (11.55), this is the MSE of that regression, since it has the same residuals sum of squares. Note that when r ф 0 in (11.55), IV estimation is necessary and s11 underestimates o11 and will have to be replaced by

(y1 — Z181,Iy),(y1 — Z1§1,iv)/T.

11.16 Recursive Systems.

a. The order condition of identification yields one excluded exogenous vari­able (x3) from the first equation and no right hand side endogenous variable i. e., k2 = 1 andg1 = 0. Therefore, the first equation is over-identified with degree of over-identification equal to one. The second equation has two excluded exogenous variables (x1 and x2) and one right hand side endoge­nous variable (y1) so that k2 = 2 and g1 = 1. Hence, the second equation is over-identified of degree one. The rank condition of identification can be based on

"12 0

0 "23_

(x1t, x2t, x3t). For the first equation, the set of

where ' is the first row of A. For the second equation, the set of zero restrictions yield

where a2 is the second row of A. Hence, for the first equation

which is of rank 1 in general. Hence, the first equation is identified by the rank condition of identification. Similarly, for the second equation

"11 "12 0 0

which is of rank 1 as long as either "11 or "12 are different from zero. This ensures that either x1 or x2 appear in the first equation to identify the second equation.

b. The first equation is already in reduced form format y1t — —"lix1t - "I2x2t + u1t

y1t is a function of x1t, x2t and only u1t, not u2t. For the second equation, one can substitute for y1t to get

y2t — -"23x3t — "21 (—"11 x1t — "I2x2t + u1t) + u2t

— "21"11x1t + "21" 12x2t - "23x3t + u2t - "21 u1t

so that y2t is a function of x1t, x2t, x3t and the error is a linear combination of u1t and u2t.

c. OLS on the first structural equation is also OLS on the first reduced form equation with only exogenous variables on the right hand side. Since x1 and x2 are not correlated with u1t, this yields consistent estimates for "11
and y12. OLS on the second equation regresses y2t on y1t and x3t. Since x3t is not correlated with u2t, the only possible correlation is from y1t and u2t. However, from the first reduced form equation y1t is only a function of exogenous variables and u1t. Since u1t and u2t are not correlated because £ is diagonal for a recursive system, OLS estimates of "21 and y23 are consistent.

d. Assuming that ut ~ N(0, £) where u( = (u1t, u2t), this exercise shows that OLS is MLE for this recursive system. Conditional on the set of x’s, the likelihood function is given by

L(B, Г, £) = (2 Tt)-T/2|B|T|£|-T/2 exp(-2 £u0£-1utj so that

1 T

logL = -(T/2) log2 к + Tlog |B| - (T/2) log |£| - ^ ^u0£ 1ut.

2 t=1

Since B is triangular, |B| = 1 so that log |B| = 0. Therefore, the only place

T

in logL where B and Г parameters appear is in u0£-1ut. Maximizing

t=1

T

logL with respect to B and Г is equivalent to minimizing ^ u0£-1ut with

t=1

respect to B and Г. But £ is diagonal, so £-1 is diagonal and

T T T

£u0£-4 = J2 u2t/CT11 + £ u2‘/CT22.

t=1 t=1 t=1

The partial derivatives of logL with respect to the coefficients of the first

T

structural equation are simply the partial derivatives of J^u2t/on with

t=1

respect to those coefficients. Setting these partial derivatives equal to zero yields the OLS estimates of the first structural equation. Similarly, the partial derivatives of logL with respects to the coefficients of the second

T

structural equation are simply the derivatives of u22t/o22 with respect to

t=1

those coefficients. Setting these partial derivatives equal to zero yields the OLS estimates of the second structural equation. Hence, MLE is equivalent to OLS for a recursive system.

11.17 Hausman s Specification Test: 2SLS Versus 3SLS. This is based on Baltagi (1989).

a. The two equations model considered is

yi = Z181 + ui and y2 = Z282 + u2 (1)

where yi and y2 are endogenous; Z1 = [y2,x1,x2], Z2 = [y1,x3]; and x1, x2 and x3 are exogenous (the y’s and x’s are Tx1 vectors). As noted in the problem, 8 = 2SLS, 8 = 3SLS, and the corresponding residuals are denoted by u and u. = (a, "1, "2) and 82 = (у, "3) with ay ф 1,so that

the system is complete and we can solve for the reduced form.

Let X = [x1,x2,x3] and PX = X(X0X)_1X0, the 3SLS equations for model (1) are given by

Z0 (Ё_1 <S> Px) ZQ = Z0 (Ё_1 <S> Px) y (2)

where Z = diag[Zi], y0 = (y1,y2), and Ё_1 = [(8ij] is the inverse of the estimated variance-covariance matrix obtained from 2SLS residuals. Equation (2) can also be written as

Z (Ё_1 <S> Px) u = 0, (3)

or more explicitly as

511Z,1Px^ 1 +1812Z11Pxu2 = 0 (4)

512Z2Pxu 1 C 522Z2Pxu2 = 0 (5)

Using the fact that Z1X is a square non-singular matrix, one can premulti­ply (4) by (X0X) (Z01X)“1 to get

511X0ti 1 C cr12X0ti2 = 0. (6)

b. Writing Eq. (5) and (6) in terms of Q1 and Q2, one gets

512 (Z2PxZ1) 81 C 522 (Z2PxZ2) 82 = 512 (Z2PxyO C (822 (Z2Pxy2) (7)
and

511X, ZiQi C &12X0Z28l2 = 511X, yi C 812X0y2. (8)

Premultiply (8) by 512Z'2X(X'X)~1 and subtract it from 511 times (7). This eliminates 81, and solves for 82:

82 = (Z2PxZ2)_1 Z2PXy2 = 82. (9)

Therefore, the 3SLS estimator of the over-identified equation is equal to its 2SLS counterpart.

Substituting (9) into (8), and rearranging terms, one gets 511X, Z1Q1 = 511X0y1 +512X0u2.

Premultiplying by Z,1X(X, X)_1/511, and solving for 81, one gets Q1 = (Z'^xZ!)-1 Z1Pxy1 C (512/511) (ZjPxZi)"1 ZjPxu. (10)

Using the fact that 512 = — 512/|£|, and 511 = 522/|£|, (10) becomes 81 = 81 — (512 /522) (Z'iPxZi)"1 Z'iPxu2. 01

Therefore, the 3SLS estimator of the just-identified equation differs from its 2SLS (or indirect least squares) counterpart by a linear combination of the 2SLS (or 3SLS) residuals of the over-identified equation; see Theil (1971).

c. A Hausman-type test based on and is given by

m = (Qi —81)0 [v (81) — v (81)] 1 (8i — Qi)

where V(8') = 5n (Z'PxZi) 1

and V(8i) = (1/5'') (Z'PxZi)-1 C (^/o^) (Z'PxZi)"1

(Z'iPXZ2) (Z2PxZ2)-1 (Z2PxZi) (Z'PxZi)"1 .

The latter term is obtained by using the partitioned inverse of Z0(£"' <S>

PX)Z. Using (11), the Hausman test statistic becomes m = (5i22/5222) (u2PxZi) [(5ii — 1/5n) (Z'iPxZi)

— (5i22/522) (Z'iPxZ2) (Z2PxZ2)-1 (Z2PxZi)]-1 (Z'iPxu2) .

However, (on — 1/(7n) = (@12І®22); this enables us to write

where Zi = PXZi is the matrix of second stage regressors of 2SLS, for i = 1,2. This statistic is asymptotically distributed as /J, and can be given the following interpretation:

Claim: m = TR2, where R2 is the R-squared of the regression of u2 on the set of second stage regressors of both equations, Z1 and Z2.

Proof. This result follows immediately using the fact that cr22 =

u2u2/T = total sums of squares/T, and the fact that the regression sum of squares is given by

where the last equality follows from partitioned inverse and the fact that

Z 2u 2 = o.

11.18 a. Using the order condition of identification, the first equation has two excluded exogenous variables x2 and x3 and only one right hand sided included endogenous variable y2. Hence,

k2 = 2 > g1 = 1

with yt = (y1t, y2t) and xt = (x1t, x2t, x3t). The first structural equation has the following zero restrictions:

where 'і is the first row of A. Similarly, the second structural equation has the following zero restriction:

where a2 is the second row of A. Hence, A¥1 = which has

"22 "2з

rank one provided either "22 or "23 is different from zero. Similarly, A¥2 = /"11

0

identified by the rank condition of identification. b. For the first equation, the OLS normal equations are given by

(ЗДІ “12 = Z1y1

V"" 1V ols

where Z1 = [y2,x1]. Hence,

Z1Z1

obtained from the matrices of cross-products provided by problem 11.18.

Hence, the OLS normal equations are

8 1^/p 12 10 20j "и

solving for the OLS estimators, we get

OLS on the second structural equation can be deduced in a similar fashion.

c. For the first equation, the 2SLS normal equations are given by

where Z' = [y2,x1] andX = [x1,x2,x3]. Hence, Z'X =

with

Therefore, the 2SLS normal equations are

2SLS on the second structural equation can be deduced in a similar fashion.

d. The first equation is over-identified and the reduced form parameter esti­mates will give more than one solution to the structural form parameters.

However, the second equation is just-identified. Hence, 2SLS reduces to indirect least squares which in this case solves uniquely for the structural parameters from the reduced form parameters. ILS on the second equation yields

" 21

"22 = (X, Zi)-1X, y2

V23 ILS

where Z1 = [y1,x2,x3] andX = [x1,x2,x3]. Therefore,

and

Note that X0Z1 for the first equation is of dimension 3 x 2 and is not even square.

11.19 Supply and Demand Equations for Traded Money. This is based on Laffer (1970).

b. OLS Estimation

SYSLIN Procedure
Ordinary Least Squares Estimation

Model: SUPPLY Dependent variable: LNTM_P

Root MSE 0.03068 R-Square 0.9442

DepMean 5.48051 Adj R-SQ 0.9380

C. V. 0.55982

OLS ESTIMATION OF MONEY DEMAND FUNCTION
SYSLIN Procedure
Ordinary Least Squares Estimation

Model: DEMAND Dependent variable: LNTM. P

c. 2SLS Estimation

SYSLIN Procedure
Two-Stage Least Squares Estimation

Model: SUPPLY Dependent variable: LNTM. P

2SLS ESTIMATION OF MONEY DEMAND FUNCTION
SYSLIN Procedure
Two-Stage Least Squares Estimation

Model: DEMAND

Dependent variable: LNTM. P

Analysis of Variance

Root MSE 0.02201 R-Square 0.9748

Dep Mean 5.48051 Adj R-SQ 0.9685

C. V. 0.40165

The total number of instruments equals the number of parameters in the equation. The equation is just identified, and the test for over identification is not computed.

d. 3 SLS Estimation

SYSLIN Procedure

Three-Stage Least Squares Estimation
Cross Model Covariance

Sigma SUPPLY DEMAND

SUPPLY 0.0009879044 -0.000199017

DEMAND -0.000199017 0.000484546

Cross Model Covariance

Corr SUPPLY DEMAND

SUPPLY 1 -0.287650003

DEMAND -0.287650003 1

System Weighted MSE: 0.53916 with 34 degrees of freedom. System Weighted R-Square: 0.9858 Model: SUPPLY Dependent variable: LNTM. P

3SLS ESTIMATION OF MODEY SUPPLY AND DEMAND FUNCTION

SYSLIN Procedure

Three-Stage Least Squares Estimation
Parameter Estimates

Model: DEMAND
Dependent variable: LNTM. P

Parameter Estimates

e. HAUSMANTEST

Hausman Test

2SLS vs OLS for SUPPLY Equation

HAUSMAN

6.7586205

2SLS vs OLS for DEMAND Equation

HAUSMAN

2.3088302

SAS PROGRAM

Data Laffer1; Input TM RM Y S2 i S1 P; Cards;

Data Laffer; Set Laffer1;

LNTM_P=LOG(TM/P);

LNRM_P=LOG(RM/P);

LNi=LOG(i);

LNY_P=LOG(Y/P);

LNS1=LOG(S1);

LNS2=LOG(S2);

T=21;

Proc syslin data=Lafferoutest=ols_s outcov;

SUPPLY: MODEL LNTM_P=LNRM_P LNi;

TITLE ‘OLS ESTIMATION OF MONEY SUPPLY FUNCTION’;

Proc syslin data=Lafferoutest=ols_s outcov;

DEMAND: MODEL LNTM_P=LNY_P LNi LNS1 LNS2;

TITLE ‘OLS ESTIMATION OF MONEY DEMAND FUNCTION’;

Proc syslin 2SLS data=Lafferoutest=TSLS_s outcov;

ENDO LNTMP LNi;

INST LNRM_P LNY_P LNS1 LNS2;

SUPPLY: MODEL LNTM_P=LNRM_P LNi/overid;

TITLE ‘2SLS ESTIMATION OF MONEY SUPPLY FUNCTION’;

Proc syslin 2SLS data=Lafferoutest=TSLS_d outcov;

ENDO LNTM_P LNi;

INST LNRM_P LNY_P LNS1 LNS2;

DEMAND: MODEL LNTM_P=LNY_P LNi LNS1 LNS2/overid; TITLE ‘2SLS ESTIMATION OF MONEY DEMAND FUNCTION’;

Proc SYSLIN 3SLS data=Lafferoutest=S_3SLS outcov; ENDO LNTMP LNi;

INST LNRM_P LNY_P LNS1 LNS2;

SUPPLY: MODEL LNTM_P=LNRM_P LNi;

DEMAND: MODEL LNTM_P=LNY_P LNi LNS1 LNS2; TITLE ‘3SLS ESTIMATION’;

RUN;

PROCIML;

TITLE ‘HAUSMAN TEST’;

use ols_s; read all var fintercep lnrm_p ini) into olsl;

olsbLs=ols1[1,]; ols_v_s=ols1[2:4,];

use ols_d; read all var f intercep lny_p lni lnsl lns2) into

ols2;

olsbLd=ols2[1,]; ols_v_d=ols2[2:6,];

use tsls_s; read all var f intercep lnrm_p lni} into tsls1;

tslsbts=tsls1[13,]; tsls_v_s=tsls1[14:16,];

use tsls_d; read all var f intercep lny_p lni lns1 lns2} into

tsls2;

tslsbtd=tsls2[13,]; tsls_v_d=tsls2[14:18,];

d=tslsbts ‘-olsbt_s‘; varcov=tsls_v_s-ols_v_s;

Hausman=d ‘*inv(varcov)*d;

print ‘Hauman Test’,, ‘2SLS vs OLS for SUPPLY Equation’,,

Hausman;

d=tslsbt_d ‘-olsbt_d‘; varcov=tsls_v_d-ols_v_d;

Hausman=d ‘*inv(varcov)*d;

print ‘2SLS vs OLS for DEMAND Equation’,, Hausman;

Similarly, the supply equation is just-identified by the order condition since k2 = gi = 1.

By the rank condition, only the supply equation need be checked for identification. In this case, ф0 = (0,1/ since y22 = 0. Therefore,

Аф = Р’гіф = ("22) = (?).

This is of rank one as long as a2 ф 0, i. e., as long as Xt is present in the demand equation. One can also premultiply this system of two equations by

F = fi1 fi2 f21 f22

the transformed matrix FB should satisfy the following normalization restrictions: f11 + f12 = 1 and f21 + f22 = 1.

Also, Fr must satisfy the following zero restriction:

—«2f21 + 0 f22 = 0.

If a2 ф 0, then f21 = 0 which also gives from the second normalization equation that f22 = 1. Hence, the second row of F is indeed the second row of an identity matrix and the supply equation is identified. However, for the demand equation, the only restriction is the normalization restriction f11 + f12 = 1 which is not enough for identification.

„ T T

b. The OLS estimator of the supply equation yields " 1ols = J2 ptqt/ J2 p2

t=1 t=1

where pt = Pt — l5 and qt = Qt — Q. Substituting qt = "1 pt + (u2t — u2/

TT

we get " 1,ols = "1 + J2 pt(u2t — 7)/ E pj2.

t=1 t=1

« "o '2

The reduced form equation for Pt yields Pt = ------------------------------ ——------------ Xt +

uit — u2t «1 + "1

T

Defining mxx = plim J2 x2/T, we get

b. Regress yi and y3 on all the X’s including the constant, i. e., [1,Xi, X2,X5]. Get y i andy 3 and regress y2 on a constant, y 1, y 3 andX2 in the second-step regression.

c. For the first equation, regressing y2 on X2 and X3 yields

y2 — ft21 C ft22X2 + O23X3 C 'V2 — y2 + 'O2 T T T

with £ v2t — J2 v2tX2t — J2 v2tX3t — 0 by the property of least squares.

t=i t=i t=i

Replacing y2 by y2 in Eq. (1) yields yi — ai C "2 (y2 C v^2) C "iXi C ui — ai C "2y2 C "iXi C ("2v2 C ui) .

T

In this case, y2tv2t — 0 because the predictors and residuals of the same

t=i

T

regression are uncorrelated. Also, y2tu1t — 0 since y2t is a linear com-

t=i

T

bination of X2 and X3 and the latter are exogenous. However, X1tv2t

t=i

is not necessarily zero since Xi was not included in the first stage regres­sion. Hence, this estimation method does not necessarily lead to consistent estimates.

d. The test for over-identification for Eq. (1) can be obtained from the F-test given in Eq. (11.48) with RRSS* obtained from the second stage regres­sion residual sum of squares of 2SLS run on Eq. (1), i. e., the regression of yi on [i,^2,Xi] where y2 is obtained from the regression of y2 on [i, X1, X2.., X5]. The URSS* is obtained from the residual sum of squares of the regression of yi on all the set of instruments, i. e., [i, X1,.., X5]. The URSS is the 2SLS residual sum of squares of Eq. (1) as reported from any 2SLS package. This is obtained as

2

Note that this differs from RRSS* in that y2t and not y2t is used in the computation of the residuals. ', the number of instruments is 6 and ki — 2 while gi — i. Hence, the numerator degrees of freedom of the F-test is
' — (gi + ki) = 3 whereas the denominator degrees of freedom of the F-test isT — ' = T — 6. This is equivalent to running 2SLS residuals on the matrix of all predetermined variables [1 ,X1,.., X5] and computing T times the uncentered R2 of this regression. This is asymptotically distributed as X2 with three degrees of freedom. Large values of this statistic reject the over-identification restrictions.

11.23 Identification and Estimation of a Simple Two-Equation Model. This solution is based upon Singh and Bhat (1988).

®12

®12 m2

The structural and reduced form parameters are related as follows:

or

' = Я11 — "^12 У = Я12 — Я11

We assume that the reduced-form parameters л11, я12, m2, m12 and m2 have been estimated. The identification problem is one of recovering a, ", ", oj2 and o12 and o| from the reduced-form parameters.

From the second equation, it is clear that у is identified. We now exam­ine the relation between £ and fi given by £ = BfiB0 where B = ^ — . Thus, we have

of = m1 — 2Pm12 + P2m2,

012 = - m2 + (1 + P)mi2 — Pm^ o-2 = m2 — 2m12 C m.

Knowledge of mi, m12, and m2 is enough to recover o|. Hence, o| is iden­tified. Note, however, that o2 and o12 depend on ". Also, a is dependent on ". Hence, given any " = "*, we can solve for a, oj2 and o12. Thus, in principle, we have an infinity of solutions as the choice of " varies. Identifi­cation could have been studied using rank and order conditions. However, in a simple framework as this, it is easier to check identification directly. Identification can be reached by imposing an extra restriction f (", a, oj2, o12) = 0, which is independent of the three equations we have for of, o12 and a. For example, " = 0 gives a recursive structure. We now study the particular case where f (", a, o2, o12) = o12 = 0.

b. Identification with oj2 = 0: When o12 = 0, one can solve immediately for P from the o12 equation:

P = m12 — mf

m22 — m12

Therefore, P can be recovered from m^ m12, and m2. Given ", all the other parameters (of, a) can be recovered as discussed above.

c. OLS estimation of ft when aj2 = 0: The OLS estimator of " is

T T

л E (yt1— y 1 )(yt2— y2) E (ut1 — u1 )(yt1— y 1)

P ols = V = P C---------------- V

T

E(ut1 — u1)(ut2 — ЇЇ2)

t=1_____________________

T

E(yt2 — y2)2

t=1

The second term is non-zero in probability limit as yti and uti are correlated from the first structural equation. Hence "ols is not consistent.

d. From the second structural equation, we have yt2 — yt1 = у + ut2, which means that (yt2 — y2) — (yt1 — y 1) = ut2 — u2, or zt = ut2 — u2. When

an = 0, this zt is clearly uncorrelated with ut1 and by definition correlated withyt2. Therefore, zt = (yt2 —y2) — (yt1 —y1) maybe taken as instruments, and thus

T T

E zt(yt1 — yi) E(yt1 — yi)[(yt2 — y2) — (yt1 — yi)]

" t=i t=i

"IV = —-------------------------- = —-----------------------------------------------------------------

Ё zt(yt2 — y2) Ё(yt2 — y2)[(yt2 — y2) — (yt1 — yi)]

t=i t=i

obtained in part (d). The OLS estimators of the reduced-form parameters are consistent. Therefore, "ILS which is a continuous tranform of those estimators is also consistent. It is however, useful to show this consistency directly since it brings out the role of the identifying restriction.

From the structural equations, we have

Using the above relation and with some algebra, it can be shown that for

" Ф i:

T T

E(Ul - Ul)(Ut2 - Ui) + " E(^2 - П2)2

t=1 t=1

P ILS = T----------------------------------------- T since,

(utl - ui)(ut2 - u2) + (ut2 - u2)2

t=1 t=1

TT plimT £(ut! - TN1)(ut2 - 1N2) = 012, and plimT £(ut2 - u)2 = of.

t=1 t=1

Therefore, plim "ILS = (о12 + bof) / (o12 C of) . The restriction o12 = 0 implies plim pils = " which proves consistency. It is clear that"ILS can be interpreted as a method-of-moments estimator.

11.24 Errors in Measurement and the Wald (1940) Estimator. This is based on Farebrother (1987).

a. The simple IV estimator of " with instrument z is given by "IV =

n n n

E ziyi zixi. But E ziyi = sum of the yi’s from the second sam-

i=1 i=1 i=1

ple minus the sum of the yi’s from the first sample. If n is even, then

nn

Eziyi = |(y2 - y 1). Similarly, £ziXi = |(x2 - x 1), so that "iv =

i=1 i=1

(y2 - y1)/(x2 - X0 = "W.

b. Let p2 = of/ (^u C of) and x2 = of/ (o2 + of) then Wi = p2Xi* - x2u

has mean E(wi) = 0 since E (xi*) = E(ui) = 0. Also, var(wD = (P2)2°* + (£2)2ou2 = o*ou= (o* + ou).

Since wi is a linear combination of Normal random variables, it is also Normal.

E(xiwi) = p2E (xix*) - x2E(xiui) = p2of - x2o2 = 0.

c. Now £2xi C wi = X2 (xi* C ui) C p2xi* - £2ui = (x2 C p2)xi* = xi* since

X2 C p2 = 1.

Using this result and the fact that yi = "xi* C £i, we get

nn

PW = ^2 *yi/l>i

i=1 i=1

(n n n n

zixi + ^i/ ^ zixi

i=i i=i i=i i=i

so that E("W/x1, ..,xn) = "x2. This follows because wi and ei are independent of xi and therefore independent of zi which is a function of xi. Similarly, "ols = p xiyi p xi2 = "x2 + " p wixi p xi2 +

i=i i=i i=i i=i

nn

Exi£i xi2 so that E("ols/x1, ..,xn) = "x2 since wi and ei are inde-

i=i i=i

pendent of xi. Therefore, the bias of "ols = bias of "W = ("x2 — ") = _PCT,2 / (CT* + CT,2) for all choices of zi which are a function of xi.

11.25 Comparison oft-ratios. This is based onFarebrother(1991). LetZ1 = [y2,X], then by the FWL-Theorem, OLS on the first equation is equivalent to that on P Xyi = рхУ2а+р xui. This yields a ols = (у2РхУ2) 1 y^j5 Xyi and the residual sum of squares are equivalent

yiPzyi = yiPxyi — yiPxy2 (y2pxy^ i y2pxyi so that var(aols) = s^ (y213Xy2) i where s^ = yiPzyi/(T — K). The t-ratio for

a = 0 is

= (T — к)1/2(у2РхУі!) [(yiРхУі)(у2РхУ2) — (yipxy2)r|i/2 .

This expression is unchanged if the roles of yi and y2 are reversed so that the t-ratio for Hb у = 0 in the second equation is the same as that for Ho; a = 0 in the first equation.

Foryi andy2 jointly Normal with correlation coefficient p, Farebrother(1991) shows that the above two tests correspond to a test for p = 0 in the bivariate Normal model so that it makes sense that the two statistics are identical.

11.28 This gives the backup runs for Crime Data for 1987 for Tables 11.2 to 11.6

. ivreg lcrmrte (lprbarr lpolpc= ltaxpc lmix) lprbconv lprbpris lavgsen ldensity lwcon lwtuc lwtrd lwfir lwser lwmfg lwfed lwsta lwloc lpctymle lpctmin west central urban if

year==87

Instrumental variables (2SLS) regression

Number of obs= 90

F(20,69) = 17.35

Prob > F =0.0000 R-squared =0.8446 Adj R-squared =0.7996 Root MSE =.24568

Instrumented: lprbarr lpolpc

Instruments: lprbconv lprbpris lavgsen ldensity lwcon lwtuc lwtrd lwfir lwser lwmfg

lwfed lwsta lwloc lpctymle lpctmin west central urban ltaxpc lmix

. estimates store b2sls

. reg lcrmrte lprbarr lprbconv lprbpris lavgsen lpolpc ldensity lwcon lwtuc lwtrd lwfir lwser lwmfg lwfed lwsta lwloc lpctymle lpctmin west central urban if year==87

b = consistent under Ho and Ha; obtained from ivreg B = inconsistent under Ha, efficient under Ho; obtained from regress

Test: Ho: difference in coefficients not systematic chi2(20) = (b-B)'[(V_b-V_B)~ (-1)](b-B)

= 0.87

Prob > chi2 = 1.0000

. ivreg Icrmrte (Iprbarr lpolpc= Itaxpc Imix) Iprbconv Iprbpris lavgsen Idensity Iwcon Iwtuc Iwtrd Iwfir Iwser Iwmfg Iwfed Iwsta IwIoc IpctymIe Ipctmin west centraI urban if year==87, first

First-stage regressions

90

3.11

0.0002

0.4742

0.3218

.33174

The test that the additional instruments are jointly significant in the first stage regression for Ipolpc is performed as follows:

. test lmix=ltaxpc=0

(1) lmix - ltaxpc = 0

(2) lmix = 0

F(2,69) = 10.56 Prob > F = 0.0001

. ivregress 2sls Icrmrte (Iprbarr lpolpc= Itaxpc Imix) Iprbconv Iprbpris lavgsen Idensity Iwcon Iwtuc Iwtrd Iwfir Iwser Iwmfg Iwfed Iwsta Iwloc Ipctymle Ipctmin west central urban if year==87

Instrumented: Iprbarr Ipolpc

Instruments: Iprbconv Iprbpris lavgsen Idensity Iwcon Iwtuc Iwtrd Iwfir Iwser Iwmfg

Iwfed Iwsta Iwloc Ipctymle Ipctmin west central urban Itaxpc Imix

. reg Iprbarr Imix Itaxpc Iprbconv Iprbpris lavgsen Idensity Iwcon Iwtuc Iwtrd Iwfir Iwser Iwmfg Iwfed Iwsta Iwloc Ipctymle Ipctmin west central urban if year==87

Then we can test that the additional instruments are jointly significant in the first stage regression for Iprbarr as follows:

. test lmix=ltaxpc=0

(1) lmix - ltaxpc = 0

(2) lmix = 0

F(2, 69) = 5.78
Prob > F = 0.0048

For lpolpc, the first stage regression is given by

. reg lpolpc lmix ltaxpc lprbconv lprbpris lavgsen ldensity lwcon lwtuc lwtrd lwfir lwser lwmfg lwfed lwsta lwloc lpctymle lpctmin west central urban

urban -.1139416 .1255201 -0.91 0.364 -.3599564 .1320732

.cons -1.159015 3.413247 -0.34 0.734 -7.848855 5.530825

Instrumented: Iprbarr Ipolpc

Instruments: lprbconv lprbpris lavgsen ldensity lwcon lwtuc lwtrd lwfir lwser lwmfg lwfed

lwsta lwloc lpctymle lpctmin west central urban ltaxpc lmix

. estat firststage

Shea's partial R-squared

Minimum eigenvalue statistic = 5.31166

Critical Values # of endogenous regressors: 2

Ho: Instruments are weak # of excluded instruments: 2

11.30 Growth and Inequality Reconsidered

a. The 3SLS results are replicated using Stata below:

reg3 (Growth: dly = yrt gov m2y inf swo dtot f_pcy d80 d90) (Inequality: gih = yrt m2y civ mlg mlgldc), exog(commod f. civ f_dem Ldtot f. flit f. gov Lint f_m2y f_swo

f. yrt pop urb lex lfr marea oil legor_fr legor_ge legor_mx legor_ sc legor_uk) endog (yrt gov m2y inf swo civ mlg mlgldc)

Three-stage least-squares regression

Endogenous variables: dly gih yrt gov m2y inf swo civ mlg mlgldc Exogenous variables: dtot f_pcy d80 d90 commod Lciv Ldem Ldtot f_flit Lgov Linf Lm2y Lswo Lyrt pop urb lex lfr marea oil legor_fr legor_ge legor_mx legor_sc legor_uk

b. The 3sls results of the respecified equations are replicated using Stata below:

. reg3 (Growth: dly = gih yrt gov m2y inf swo dtot f_pcy d80 d90) (Inequality: gih = dly yrt m2y civ mlg mlgldc), exog(f_civ Ldem Ldtot Lflit Lgov Linf f_m2y Lswo Lyrt pop urb lex lfr marea commod oil legor_fr legor_ge legor_mx legor_sc legon. uk) endog(yrt gov m2y inf swo civ mlg mlgldc)

Three-stage least-squares regression

Endogenous variables: dly gih yrt gov m2y inf swo civ mlg mlgldc Exogenous variables: dtot f_pcy d80 d90 f_civ Ldem Ldtot Lflit Lgov Linf Lm2y Lswo Lyrt pop urb lex lfr marea commod oil legorJr legor_ge legor_mx legor_sc legor_uk

11.31 Married Women Labor Supply

b. The following Stata output replicates Equation 2 of Table IV of Mroz (1987, p. 770) which runs 2SLS using the following instrumental variables:

. local B unem city motheduc fatheduc. local E exper expersq. local control age educ

. ivregress 2sls hours (lwage= ‘B’ ‘E’) nwifeinc kidslt6 kidsge6 ‘control’, vce(robust)

Instrumental variables (2SLS) regression Number of obs = 428

Wald chi2(6) = 18.67

Prob > chi2 = 0.0048

R-squared = .

Root MSE = 1141.3

Instrumented: Iwage

Instruments: nwifeinc kidslt6 kidsge6 age educ unem city motheduc fatheduc exper expersq

estat overid

Test of overidentifying restrictions: Score chi2(5) = 5.9522 (p = 0.3109)

estat firststage, forcenonrobust all
First-stage regression summary statistics

Shea’s partial R-squared

Minimum eigenvalue statistic = 3.40543

Critical Values # of endogenous regressors: 1

Ho: Instruments are weak # of excluded instruments: 6

Similarly, the following Stata output replicates Equation 3 of Table IV of Mroz(1987, p. 770) which runs 2SLS using the following instrumental variables: . local F2 age2 educ2 age_educ

. ivregress 2sls hours (lwage= ‘B’ ‘E’ ‘F2’) nwifeinc kidslt6 kidsge6 ‘control’, vce(robust)

Instrumented: lwage

Instruments: nwifeinc kidslt6 kidsge6 age educ unem city motheduc fatheduc exper expersq age2 educ2 age_educ

. estat overid

Test of overidentifying restrictions: Score chi2(8) = 19.8895 (p = 0.0108)

. estat firststage, forcenonrobust all First-stage regression summary statistics

Minimum eigenvalue statistic = 3.0035

Critical Values # of endogenous regressors: 1

Ho: Instruments are weak # of excluded instruments: 9

References

Baltagi, B. H. (1989), “A Hausman Specification Test in a Simultaneous Equations Model,”Econometric Theory, Solution 88.3.5, 5: 453-467.

Farebrother, R. W. (1987), “The Exact Bias of Wald’s Estimation,” Econometric Theory, Solution 85.3.1, 3: 162.

Farebrother, R. W. (1991), “Comparison of t-Ratios,” Econometric Theory, Solution 90.1.4,7: 145-146.

Hausman, J. A. (1983), “Specification and Estimation of Simultaneous Equation Models,” Chapter 7 in Griliches, Z. and Intriligator, M. D. (eds.) Handbook of Econometrics, Vol. I (North Holland: Amsterdam).

Laffer, A. B., (1970), “Trade Credit and the Money Market,” Journal of Political Economy, 78: 239-267.

Mroz, T. A. (1987), “The Sensitivity of an Empirical Model of Married Women’s Hours of Work to Economic and Statistical Assumptions,” Econometrica, 55: 765-799.

Singh, N. and A N. Bhat (1988), “Identification and Estimation of a Simple Two - Equation Model,” Econometric Theory, Solution 87.3.3, 4: 542-545.

Theil, H. (1971), Principles of Econometrics (Wiley: New York).

Wald, A. (1940), “Fitting of Straight Lines if Both Variables are Subject to Error,” Annals of Mathematical Statistics, 11: 284-300.

CHAPTER 12