Math 6010
Solutions to homework 2
4, p. 31. We can write X = µ+AZ and Y = ν +BW , where Z and W are vectors
of i.i.d. standard normal random variables. Note that Z and W must
be independent. This is intuitively clear. But if you want a completely
rigorous proof, then here it is: Indeed,
0
0
0
0
0
0
E et AZ+s BW = e−t µ−s ν E et X+s Y
0 0 0
0
= e−t µ−s ν E et X · E es Y ,
because X and Y are independent. It follows then that
0
0 0
0
E et AZ+s BW = E et AZ E es BW ,
for every s and t. Because u := t0 A and v := s0 B are arbitrary vectors
[since t and s are], it follows that the MGF of (Z1 , · · · , Zn , W1 , . . . , Wm )0
is the product of the respective MGFs of Z and W . Therefore, W and
Z are independent by the independence theorem.
Let m := aµ + bν to see that
aX + bY = m + AZ + BW := m + CY ,
where
C :=
A
0
0
B
and
Y :=
X
W
.
We are done, because Y is a vector of i.i.d. standard normals.
7, p. 32. It is best to solve this problem directly.
First of all, let me emphasize that
n
1X
Yi .
Ȳ :=
n i=1
Note that the (j , k)th element of vv 0 —for any p-vector v—is vj vk . Apply
this with A := Yi − Ȳ to see that the (j , k)th element of (Yi − Ȳ )(Yi − Ȳ )0
is
(Yi,j − Ȳ·,j )(Yi,k − Ȳ·,k ) = Yi,j Yi,k − Ȳ·,j Yi,k − Yi,j Ȳ·,k + Ȳ·,j Ȳ·,k ,
1
where
n
Ȳ·,m
1X
:=
Yν,m .
n ν=1
Now,
E(Yi,j Yi,k ) = Σj,k ,
E(Yi,j Yν,k ) = 0
for i 6= ν.
Therefore,
n
1X
1
E(Yi,j Yν,k ) = Σj,k ,
n ν=1
n
E(Yi,j Ȳ·,k ) =
and
E(Ȳ·,j Ȳ·,k ) =
n
n
1
1 XX
E(Yτ,j Yν,k ) = Σj,k .
2
n ν=1 τ =1
n
We can put everything together to find that
E
h
0
(Yi − Ȳ )(Yi − Ȳ )
i
j,k
1
= Σj,k − Σj,k =
n
n−1
n
Σj,k .
We can write this, in equivalent form, using linear algebra:
n−1
E (Yi − Ȳ )(Yi − Ȳ )0 =
Σ.
n
We add from i = 1, . . . , n and then divide by n − 1 to see that
"
#
n
1 X
0
E
(Yi − Ȳ )(Yi − Ȳ ) = Σ,
n − 1 i=1
as desired.
15., p. 32. (a) Let us observe that Y1 , Y2 , Y3 , Y4 are i.i.d. standard normals. Therefore, Y1 Y2 and Y3 Y4 are i.i.d. In particular, Q := Y1 Y2 − Y3 Y4 is a
symmetric random variable, so that P{Q < 0} = 1/2 6= 0. Therefore, Q cannot have a χ2 distribution of any kind [since the latter are
distributions on nonnegative numbers].
(b) We may note that if

0
1/2
A := 
0
0
1/2
0
0
0

0
0
0
0 

0
−1/2
−1/2
0
then
y 0 Ay = y1 y2 − y3 y4
for all four-vectors y = (y1 , y2 , y3 , y4 )0 .
[Consider separately the quadratic forms for y1 y2 and y3 y4 , and then
subtract the matrices.] In other words, this A is the matrix for the
2
quadratic form Q. The eigenvalues of A are ±1/2 with multiplicity
of two each. Therefore, we can apply (2.10) on page 27 in order to
see that
1
1
1
1
Q = Y 0 AY = Z12 + Z22 − Z32 − Z42 ,
2
2
2
2
where Z1 , . . . , Z4 are i.i.d. standard normals. Consequently,
2 2 2
2
MQ (t) = E etZ1 /2 E etZ2 /2 E e−tZ3 /2 E e−tZ4 /2 = E12 E22 ,
2
2
where E1 := E(etZ1 /2 ) and E2 := E(e−tZ1 /2 ). Because Z12 ∼ χ21 ,
h
i2 h
i2 1 2 1 2
MQ (t) = Mχ21 (t/2) · Mχ21 (−t/2) = √
· √
,
1−t
1+t
if −1 < t < 1 and other wise infinity. In other words, we find once
again that MQ (t) = 1/(1 − t2 ) when |t| < 1 and ∞ when |t| ≥ 1.
3