7.1 Invariance of the fitted values and residuals to non-singular transformations of the independent variables.
The regression model in (7.1) can be written as y = XCC-1" + u where Cisa non-singular matrix. LetX* = XC, theny = X*"* + u where "* = C-1".
a. PX* = X* (X*0X*)-1 X*0 = XC [C0X0XC]-1 C0X0 = XCC-1 (X0X)-1 c0-1 C0X0 = PX.
Hence, the regression of y on X* yields
y = X*" *ls = PX* y = PXy = X" ols which is the same fitted values as those from the regression of y on X. Since the dependent variable y is the same, the residuals from both regressions will be the same.
b. Multiplying each X by a constant is equivalent to post-multiplying the matrix X by a diagonal matrix C with a typical k-th element ck. Each Xk will be multiplied by the constant ck for k = 1,2,.., K... Read More
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
a. For the gamma distribution, E(ui) = 0 and var(ui) = 0. Hence, the disturbances of the simple regression have non-zero mean but constant variance. Adding and subtracting 0 on the right hand side of the regression we get Yi = (a + 0) + "Xi + (ui – 0) = a* + "Xi + ui* where a* = a + 0
and ui* = ui — 0 with E (ui*) = 0 and var (ui*) = 0. OLS yields the BLU estimators of a* and " since the ui* disturbances satisfy all the requirements of the Gauss-Markov Theorem, including the zero mean assumption. Hence, E(aols) = a* = a + 0 which is biased for a by the mean of the disturbances 0. ButE(s2) = var(u*) = var (ui) = 0. Therefore,
E Cols – s2) = E (aols) – E(s2) = a + 0 – 0 = a.
b. Similarly, for the x2 distribution, we have E(ui) = v and var(u^ = 2v... Read More
a. Using the MGF for the exponential distribution derived in problem 2.14e, we get
(1 — 0t).
Differentiating with respect to t yields
MX0(O) = O = E(X).
Differentiating MX(t) with respect to t yields 2O2 (1 – Ot) 2O2
(1 – Ot)4 (1 – Ot)3′
MXX (0) = 2O2 = E(X2).
var(X) = E(X2) – (E(X))2 = 2™2 – ™2 = ™2.
b. The likelihood function is given by
_ / 1 Y – P Xi/O
L (O)= 0e ‘d1
э logL (O) – n = 0
@O O C O2
solving for O one gets
Xi – nO D 0
Omle = X.
c. The method of moments equates E(X) = X. In this case, E(X) = O, hence O = X is the same as MLE.
d. E(X) = E(Xi)/n = n0/n = 0. Hence, X is unbiased for 0. Also,
var(X) = var(X)/n = 02/n which goes to zero as n! 1... Read More