# APPENDIX E

APPENDIX E

SOLUTIONS TO PROBLEMS

E.1 This follows directly from partitioned matrix multiplication in Appendix D. Write

x1 y1 xy2 2 X = , X = (x1x2 xn), and y = x y n n

Therefore, X X = x txt and X y = xtyt. An equivalent expression for β is

t 1

t 1

nn

1 1 = n x x nxyβtt tt

t 1 t 1

which, when we plug in yt = xt + ut for each t and do some algebra, can be written as

nn

1 1 = + n x x nxuβtt tt .

t 1 t 1

As shown in Section E.4, this expression is the basis for the asymptotic analysis of OLS using matrices.

– b)] [ u – + X(β + X(βE.2 (i) Following the hint, we have SSR(b) = (y – Xb) (y – Xb) = [u

– b) + (β – b) X u – b) X X(β – b). But by the first order conditions u + u X(β + (βb)] = u

1 1

nn

– b) X X(β – b), = 0, and so (X u ) = u X = 0. But then SSR(b) = u u + (βfor OLS, X u

which is what we wanted to show.

– b) X X(β – b) > (ii) If X has a rank k then X X is positive definite, which implies that (β

. The term u ) = (β – b) X X u does not depend on b, and so SSR(b) – SSR(β0 for all b β

– b) > 0 for b β . (β

= (Z Z)-1Z y = E.3 (i) We use the placeholder feature of the OLS formulas. By definition, β

. [(XA) (XA)]-1(XA) y = [A (X X)A]-1A X y = A-1(X X)-1(A )-1A X y = A-1(X X)-1X y = A-1β

and y . Plugging zt and β into the t = xtβ t = ztβ(ii) By definition of the fitted values, y

= y ) = xβ = (xtA)(A-1β . second equation gives y

t

tt

2(Z Z)-1 where 2 is (iii) The estimated variance matrix from the regression of y and Z is

the error variance estimate from this regression. From part (ii), the fitted values from the two

226

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