Exam (elaborations) TEST BANK FOR Econometric Analysis 5th Edition By William H. Greene (Solution manual)

1. (a) Let . The normal equations are given by (3-12), , hence for each of the columns of X, x           = n x x X 1 . . 1 1 X′e = 0 = i i i k, we know that xk’e=0. This implies that Σ = 0and x e . i i e Σ 0 (b) Use Σ = 0 to conclude from the first normal equation that i i e a = y − bx . (c) Know that Σ = 0 and . It follows then that i i e Σ = 0 i i i x e Σ ( − ) = 0 i i i x x e . Further, the latter implies ( )( ) = 0 i Σ − bx i − − a i i x x y or (x − x)(y − y − b(x − x))= 0 i i i i Σ from which the result follows. 2. Suppose b is the least squares coefficient vector in the regression of y on X and c is any other Kx1 vector. Prove that the difference in the two sums of squared residuals is (y-Xc)′(y-Xc) - (y-Xb)′(y-Xb) = (c - b)′X′X(c - b). Prove that this difference is positive. Write c as b + (c - b). Then, the sum of squared residuals based on c is (y - Xc)′(y - Xc) = [y - X(b + (c - b))] ′[y - X(b + (c - b))] = [(y - Xb) + X(c - b)] ′[(y - Xb) + X(c - b)] = (y - Xb) ′(y - Xb) + (c - b) ′X′X(c - b) + 2(c - b) ′X′(y - Xb). But, the third term is zero, as 2(c - b) ′X′(y - Xb) = 2(c - b)X′e = 0. Therefore, (y - Xc) ′(y - Xc) = e′e + (c - b) ′X′X(c - b) or (y - Xc) ′(y - Xc) - e′e = (c - b) ′X′X(c - b). The right hand side can be written as d′d where d = X(c - b), so it is necessarily positive. This confirms what we knew at the outset, least squares is least squares. 3. Consider the least squares regression of y on K variables (with a constant), X. Consider an alternative set of regressors, Z = XP, where P is a nonsingular matrix. Thus, each column of Z is a mixture of some of the columns of X. Prove that the residual vectors in the regressions of y on X and y on Z are identical. What relevance does this have to the question of changing the fit of a regression by changing the units of measurement of the independent variables? The residual vector in the regression of y on X is MXy = [I - X(X′X)-1X′]y. The residual vector in the regression of y on Z is MZy = [I - Z(Z′Z)-1Z′]y = [I - XP((XP)′(XP))-1(XP)′)y = [I - XPP-1(X′X)-1(P′)-1P′X′)y = MXy Since the residual vectors are identical, the fits must be as well. Changing the units of measurement of the regressors is equivalent to postmultiplying by a diagonal P matrix whose kth diagonal element is the scale factor to be applied to the kth variable (1 if it is to be unchanged). It follows from the result above that this will not change the fit of the regression. 4. In the least squares regression of y on a constant and X, in order to compute the regression coefficients on X, we can first transform y to deviations from the mean, y , and, likewise, transform each column of X to deviations from the respective column means; second, regress the transformed y on the transformed X without a constant. Do we get the same result if we only transform y? What if we only transform X? 3