8.1 Since H = PX is idempotent, it is positive semi-definite with b0H b > 0 for any arbitrary vector b. Specifically, for b0 = (1,0,.., 0/ we get hn > 0. Also, H2 = H. Hence,
hii =J2 hb2 > h2i > 0.
From this inequality, we deduce that hjy — h11 < 0 or that h11(h11 — 1/ < 0. But h11 > 0, hence 0 < h11 < 1. There is nothing particular about our choice of h11. The same proof holds for h22 or h33 or in general hii. Hence, 0 < hii < 1 for i = 1,2,.., n.
8.2 A Simple Regression With No Intercept. Consider yi = xi" + ui for i = 1,2,.., n
a. H = Px = x(x0x)_1x0 = xx0/x0x since x0x is a scalar. Therefore, hii =
x2/ x2 for i = 1,2,.., n. Note that the xi’s are not in deviation form as
in the case of a simple regression with an intercept. In this case tr(H/ =
tr(Px/ = tr(xx0//x0x ... Read More
9.1 GLS Is More Efficient than OLS.
a. Equation (7.5) of Chap. 7 gives "ois = " + (X’X)-1X’u so that E("ois) = " as long as X and u are uncorrelated and u has zero mean. Also,
var("ols) = E("ols – ")("ols – ")’ = E[(X, X)_1X, uu, X(X, X)_1]
= (X’X)-1X’ E(uu’)X(X’X)-1 = CT2(X, X)-1X’fiX(X’X)-1.
b. var("ols) – var("gls) = o2[(X’X)-1X’fiX(X’X)-1 – (X’fi-1X)-1]
= CT2[(X, X)-1X, fiX(X, X)-1 – (X’^-1X)-1X’^-1fifi-1 X(X’fi-1X)-1]
= ct2[(X’X)-1X’ – (X’fi-1X)-1X’fi-1]fi[X(X’X)-1 – fi-1X(X’fi-1X)-1]
= o2 AfiA’
where A = [(X’X)-1X’ – (X’fi-1X)-1X’fi-1]. The second equality post multiplies (X’fi-1X)-1 by (X’fi-1X)(X’fi-1X)-1 which is an identity of dimension K. The third equality follows since the cross-product terms give -2(X’fi-1X)-1... Read More
10.1 When Is OLS as Efficient as Zellner’s SUR?
a. From (10.2), OLS on this system gives
p /Р 1,ols I
p ols = "
"(X1X1) 1 0
"(x! xO 1 х1уГ
_(x2x^ 1 X2y2_
This is OLS on each equation taken separately. For (10.2), the estimated var("ols) is given by
where s2 = RSS/(2T — (K1 + K2)) and RSS denotes the residual sum of squares of this system. In fact, the RSS = e,1e1 + e2e2 = RSS1 + RSS2 where
ei = yi — Xi P i, ols fori = 1,2.
If OLS was applied on each equation separately, then
var (p 1,ol^ = s2 (X1X1)-1 with s2 = RSS1/(T — K1)
var (p2,ol^ = s2 (X2X2)-1 with s2 = RSS2/(T — K2).
Therefore, the estimates of the variance-covariance matrix of OLS from the system of two equations differs from OLS on each equation s... Read More
11.1 The Inconsistency of OLS. The OLS estimator from Eq. (11.14) yields 8ols =
E Ptqt/ E Pt2 where pt = Pt — l3 and qt = Qt — Q. Substituting qt = 8pt C
(u2t — U2) from (11.14) we get 8ois = 8 C E Pt(u2t — N/Y, P2. Using (11.18),
we get plim £ Pt(U2t — U2)/T = (012 — 022)/(8 — ") where Oij = cov. Uit, Ujt)
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
У2 on X1, X2 and X3 to get У2 = у2 C V2, the residuals satisfy E y2tV2t = 0
and 22 V2tXu = 22 V2tX2t = 22 V2tX3t = 0.
t=1 t=1 t=1
Simi... Read More
12.1 Fixed Effects and the Within Transformation.
a. Premultiplying (12.11) by Q one gets Qy = «Qint + QX" + QZpp + Qv
But PZp = Zp and QZp = 0. Also, PiNT = iNT and Qint = 0. Hence, this
transformed equation reduces to (12.12)
Qy = QX" + Qv
Now E(Qv) = QE(v) = 0 and var(Qv) = Q var(v)Q0 = o2Q, since var(v) = ov2Int
and Q is symmetric and idempotent.
b. For the general linear model y = X" + u with E(uu0) = Й, a necessary and sufficient condition for OLS to be equivalent to GLS is given by X0 fi_1PX where PX = I – PX and PX = X(X0X)_1 X0, see Eq.(9.7) of Chap.9. For Eq. (12.12), this condition can be written as
(X0Q)(Q/o2)P qx = 0
using the fact that Q is idempotent, the left hand side can be written as (X0Q)P qx/ov2
which is clearly 0, since PqX is the orthogonal projection of QX.
One ca... Read More
a. From (12.17) we get
Й = ct^In <8> Jt) + c^.In <8> It)
Replacing JT by TJT, and IT by (Et + JT) where ET is by definition (It — JT), one gets
Й = Tc^(In <8> Jt) + c^(In <8> Et) + c^.In <8> Jt)
collecting terms with the same matrices, we get
Й = (Tc^ C c2)(In <S> Jt) C cv2(In <S> Et) = стуР + cv2Q where Cj2 = Tc2 C c2.
b. p = z2(z;z2)“ z; = IN <S> JT is a projection matrix of Z2. Hence,
it is by definition symmetric and idempotent. Similarly, Q = INT — P is the orthogonal projection matrix of Z2. Hence, Q is also symmetric and idempotent. By definition, P + Q = INT. Also, PQ = P(Int—P) = P—P2 = P — P = 0.
c. From (12.18) and (12.19) one gets
П ^-1 = (ci2P C cv2Q) (%P C q) = P C Q = Int
Vc12 cv2 J
since P2 = P, Q2 = Q and PQ = 0 as verified in part (b)... Read More