Statistics 611
Homework #6 Solutions
Fall 2010
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,
Statistics 611
Homework #5 Solutions
Fall 2010
= (I
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
Statistics 611
Homework #4 Solutions
Fall 2010
1. If x has a multivariate normal distribution N (, ), then its m.g.f. is
E [exp(t x)] = exp t +
1
t t
2
for t Rn .
Thus, for every n 1 vector of constants a, we have that the m.g.f. of a x is
1
2
E [exp(sa x
Statistics 611
Homework #3 Solutions
Fall 2010
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.
F
Statistics 611
Homework #2 Solutions
Fall 2010
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,
Statistics 611
Homework #1 Solutions
Fall 2010
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