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Math 6010, Fall 2004: Homework
Homework 3
#2, page 23: Recall that Y ∼ Nn(μ, Σ) iff for all t ∈ R
n , t
′ Y ∼
N (t
′ μ, t
′ Σt). Choose t such that ti = 1 and tj = 0 for j 6 = i.
Then t
′ Y = Yi, t
′ μ = μi, and t
′ Σt = σi,i.
#3, page 23: Of course, Z = AY , where
A =
Therefore, Z ∼ N 2 (Aμ, AΣA
′ ). This is a bivariate normal;
Aμ =
AμA
′
#5, page 24: Each (Xi, Yi) is obtained from linear combination
of two i.i.d. standard normals. That is, Xi = ai, 1 Zi, 1 + ai, 2 Zi, 2
and Yi = bi, 1 Zi, 1 +bi, 2 Zi, 2 , where Z 1 , 1 , Z 1 , 2 , Z 2 , 1 , Z 2 , 2 ,... , Zn, 1 , Zn, 2
are i.i.d. standard normals, and ai,j ’s and bi,j ’s are constants.
Therefore,
X 1
Y 1
X 2
Y 2
Xn
Yn
A 1
A 2
An
Z 1 , 1
Z 1 , 2
Z 2 , 1
Z 2 , 2
Zn, 1
Zn, 1
where the empty parts of the matrix with A’s in it are are zero,
and
Aj =
aj, 1 aj, 2
bj, 1 bj, 2
1
This proves that (X 1 , Y 1 ,... , Xn, Yn)
′ is multivariate normal.
Therefore, so is
X
Y
X 1
Y 1
X 2
Y 2
Xn
Yn
It is easiest to compute the mean and variance matrix directly
though. Suppose EX 1 = μX , EY 1 = μY , VarX 1 = σ
2 X , VarY^1 =
σ
2 Y , and Cor(X^1 , Y^1 ) =^ ρ.^ Then,^ EX^ =^ EX^1 =^ μX^ ,^ EY^ =
EY 1 = μY , VarX = σ
2 X
/n, VarY = σ
2 Y
/n. Finally,
Cov(X, Y ) = Cov
n
n ∑
i=
Xi ,
n
n ∑
j=
Yj
n^2
n ∑
i=
n ∑
j=
Cov(Xi, Yj ) =
n^2
n ∑
i=
Cov(Xi, Yi)
ρσX σY
n
Therefore, Cor(X, Y ) = Cov(X, Y )/SD(X)SD(Y ) = ρ. Thus,
(X, Y ) ∼ N 2 (μ, Σ), where
μ =
μX
μY
σ
2 X /n^ ρ
ρ σ
2 Y /n
#6, page 24: Let μi = EY 2 and σ
2 i = VarYi.^ Also define^ ρ^ =
Cor(Y 1 , Y 2 ).
Define Z 1 = Y 1 + Y 2 and Z 2 = Y 1 − Y 2. Then we are told that
Z 1 and Z 2 are independent N (0, 1)’s. Note that
( Y 1
Y 2
= A
Z 1
Z 2
where A =
1 2
1 2 −
1 2
1 2
Therefore, (Y 1 , Y 2 )
′ is bivariate normal with
EY =
and VarY = AA
′
( (^1) n 1
′ n)
2 = n (^1) n 1
′ n. In particular,^ A
2 = (1 − ρ)
− 1
. In addition,
AVarX =
1 − ρ
AΣ
1 − ρA +
ρ √ 1 − ρ
A (^1) n 1
′ n =^
1 − ρA.
Therefore, AVarX is idempotent. The corollary on page 30
tells us then that ‖AX‖
2 ∼ χ
2 r where^ r^ = rank(AVarX). Note
that
‖AX‖
2
1 − ρ
∑^ n
i=
(Yi − Y )
2 .
Therefore, it suffices to prove that r = n − 1. That is, we wish
to prove that there is exactly one solution to AVar(X)x = 0.
This was proved in #5, page 32; simply set a = (^1) n there.
#11, page 32: One can check that Y = Aa, where A (n + 1
columns and n rows) as follows:
A =
φ 1 0 0 0 · · · 0 0
0 φ 1 0 0 · · · 0 0
0 0 φ 1 0 · · · 0 0
. . .
0 0 0 0 0 · · · φ 1
So Y ∼ Nn( 0 , σ
2 AA
′ ). To finish, we compute the n×n matrix,
AA
′
φ
2
φ φ
2
0 φ φ
2
.. .
0 0 0 0 · · · φ φ
2
0 0 0 0 · · · 0 φ φ
2
That is, (AA
′ )i,i = φ
2
′ )i,i+1 = (AA
′ )i,i− 1 = φ, and
for all j 6 ∈ {i, i ± 1 }, (AA
′ )i,j = 0.
