Homework #6 Solutions
1. Note: All page numbers in this problem refer to the class notes for Chapter 5.
(a) Let K = G2 G1 . Then AKA = AG2 A AG1 A = A A = 0. Thus G2 = G1 + K
G1 AKAG1 .
(b) Note that, ( ) is a function of and u,
Homework #5 Solutions
1. (a) It follows from Problem 3 of Homework #3 that the vector of residuals is
XG11 X )y XG12 b, that E ( ) = 0, and that I XG11 X is symmetric and idempotent.
Thus / N (0, I XG11 X ). Then by Problem
Homework #4 Solutions
1. If x has a multivariate normal distribution N (, ), then its m.g.f. is
E [exp(t x)] = exp t +
for t Rn .
Thus, for every n 1 vector of constants a, we have that the m.g.f. of a x is
E [exp(sa x
Homework #3 Solutions
1. (a) For the unconstrained model, the regression line is E (y ) = 0 + 2 x for x t,
and is E (y ) = (0 + 1 ) + (2 + 3 )x for x > t. This regression has a jump of
1 + 3 t at x = t, which may be undesirable.
Homework #2 Solutions
1. (a) Yes! If 1 and 2 are both estimable, then there exist vectors a1 and a2 such
that E (a1 y ) = 1 and E (a2 y ) = 2 , in which case E (a1 y + a2 y ) = 1 + 2 .
(b) No! Suppose that is non-estimable. Then,
Homework #1 Solutions
1. (a) Part (i) of Corollary 1 to Lemma 1.4 follows from Lemma 1.4 and the corollary
to Lemma 1.1. Parts (ii) and (iii) are obtained by taking transposition on both
sides of the identities in part (i) and in