STAT 510
Homework 1
Due Date: 11:00 A.M., Wednesday, January 20
1. The very last expression on slide 4 of slide set 1 shows how to compute a matrix product as a sum of
matrices. Write down the matrices summed together to give the matrix product AB, where


1
5
3
3
−2
4
A =  4 −1  and B =
.
5 −1
2 8
0
1
2. Explain why the relationship between t and F distributions described on slide 24 of slide set 1 is true.
3. Imagine extending a string from (0, 0), the origin in IR2 , to a random point (x, y) in IR2 , where
x ∼ N (2, 1) independent of y ∼ N (0, 1). Use R to find the probability that the string will need to be
longer than 3 units to reach from (0, 0) to (x, y).
iid
4. Suppose z1 , z2 ∼ N (0, 1). Find the distribution of the following random variables.
(a) (z1 − z2 )2 /2
(b) (z1 + z2 )/|z1 − z2 |
iid
5. Suppose y1 , . . . , yn ∼ N (µ, σ 2 ). Let y = [y1 , . . . , yn ]0 , and let ȳ· =
(a) Show that s2 =
(b) Prove that (n −
1 Pn
2
0
i=1 (yi − ȳ· ) can be written as y By
n−1
1)s2 /σ 2 ∼ χ2n−1 using the result on slide 18
1
n
Pn
i=1 yi .
for some matrix B.
of slide set 1.
6. The main purpose of this question is to prove several results about orthogonal project matrices stated
on slide 5 of slide set 2 (A Review of Some Key Linear Model Results).
(a) Prove that a matrix A is 0 if and only if A0 A = 0. (Hint: What are the diagonal elements of
A0 A?)
(b) Prove that X 0 XA = X 0 XB if and only if XA = XB. (Note that the “if” part of the proof,
i.e.,
XA = XB =⇒ X 0 XA = X 0 XB,
holds trivially. Thus, proving the converse, i.e.,
X 0 XA = X 0 XB =⇒ XA = XB,
is the challenging part. One proof starts like this
X 0 XA = X 0 XB
=⇒
X 0 XA − X 0 XB = 0
=⇒
X 0 X(A − B) = 0
Now if you multiply on the left by the appropriate matrix, you can use the result of part (a) to
help complete the proof.)
(c) Use the definition of generalized inverse and the result of part (b) to prove that
X(X 0 X)− X 0 X = X
for any (X 0 X)− a generalized inverse of X 0 X.
(d) Prove that if A is any symmetric matrix and G is any generalized inverse of A, then it must be
true that G0 is also a generalized inverse of A.
(e) Use the results of problems (c) and (d) to prove that
X 0 X(X 0 X)− X 0 = X 0
for any (X 0 X)− a generalized inverse of X 0 X.
(f) Show that idempotency of P X (i.e., P X P X = P X ) follows from the result of part (c) and,
alternatively, from the result of part (e).
(g) Use parts (c) and (e) to prove that
XG1 X 0 = XG2 X 0
for any two generalized inverses of X 0 X denoted by G1 and G2 . (This says that P X is the
same matrix no matter which generalized inverse of X 0 X is used to compute it.)
(h) Use (d) and (g) to prove that P X is symmetric (i.e., P 0X = P X ).
7. Suppose X is an n × p matrix and y is an n × 1 vector. Suppose z ∈ C (X) and z 6= P X y. Prove
that ky − zk > ky − P X yk. Hint: Note that for any vector a and any vector b 6= 0 such that a0 b = 0
ka + bk2 = (a + b)0 (a + b) = (a0 + b0 )(a + b)
= a0 a + a0 b + b0 a + b0 b = a0 a + 2a0 b + b0 b
= kak2 + kbk2 + 2a0 b
= kak2 + kbk2 (because a0 b = 0).
> kak2 (because b 6= 0).
Now note that
ky − zk2 = ky − P X y + P X y − zk2 = . . .
Page 2