Math 6010
Solutions to homework 1
2, p. 15. (a) We may notice that X1 − 2X2 + X3 = a0 X, where a := (1 , −2 , 1)0 .
Therefore, Var (X1 − 2X2 + X3 ) = a0 Var(X)a, which is
 
 

1
4
5 2 3
(1 , −2 , 1) 2 3 0 −2 = (1 , −2 , 1) −4 = 18.
1
6
3 0 3
(b) We may write Y = AX, where
1
A :=
1
1
1
0
.
1
Therefore, Var(Y ) = AVar(X)A0 , which is



5 2 3
1 1
12
1 1 0 
2 3 0 1 1 =
15
1 1 1
3 0 3
0 1
15
.
21
Pn
− x̄)2 = x0 Ax for every x, where


− n1
1 − n1 − n1 · · · − n1

..
..
..  .
..
A :=  ...
.
.
.
. 
1
1
1
−n
− n · · · − n 1 − n1
Pn
Therefore, E(Q1 ) = tr(AΣ) + µ0 Aµ. Because µ0 Aµ = i=1 (µi − µ̄)2 , it
follows that µ0 Aµ = 0 in this particular case, because the µi ’s are all the
same.
P P
Next, we recall that tr(AΣ) = i j Ai,j Σi,j . Because Σi,j = 0 when
j > i + 1, and since Σ and A are both symmetric,
3, p. 15. Recall that
i=1 (xi
E(Q1 ) =
XX
i
j
=
Ai,j Σi,j =
1−
n
X
Ai,i Σi,i + 2
i=1
1
n
X
n
i=1
n
Σi,i −
i=1
1
n
X
2X
Σi,i+1 .
n i=1
Ai,i+1 Σi,i+1
This does the job for Q1 . As regards Q2 , we appeal to Example 1.8 (p.
10) to see that
Q2 = 2
n
X
Xi2 − X12 − Xn2 − 2
i=1
=2
n
X
n−1
X
Xi Xi+1 + (X1 − Xn )2
i=1
Xi2 − 2
n−1
X
i=1
Xi Xi+1 − 2X1 Xn .
i=1
Because E(X1 Xn ) = E(X1 )E(Xn ) = µ2 [I am assuming that n > 3;
otherwise the result is silly], it follows that
E(Q2 ) = 2
n
X
E(Xi2 )
i=1
−2
n−1
X
E(Xi Xi+1 ) − 2µ2 .
i=1
Now, E(Xi2 ) = Var(Xi )+(EXi )2 = Σi,i +µ2 , and E(Xi Xi+1 ) = Σi,i+1 +µ2
similarly. Therefore,
E(Q2 ) = 2
n
X
Σi,i +2nµ2 −2
i=1
n−1
X
Σi,i+1 −2(n−1)µ2 −2µ2 = 2
i=1
n
X
Σi,i −2
i=1
n−1
X
Σi,i+1 .
i=1
Therefore,
# " n
#
n
n
n−1
X
X
2X
1 X
Σi,i −
Σi,i+1 − 2
Σi,i − 2
Σi,i+1
E(3Q1 − Q2 ) = 3 1 −
n i=1
n i=1
i=1
i=1
n
n−1
6 X
3 X
Σi,i + 2 −
Σi,i+1
= 1−
n i=1
n i=1
n=1
n
n−3 X
n−3X
Σi,i+1 .
Σi,i + 2
=
n i=1
n
i=1
"
Divide both sides by n(n − 3) in order to see that
n
n−1
3Q1 − Q2
1 X
2 X
E
= 2
Σi,i + 2
Σi,i+1 .
n(n − 3)
n i=1
n i=1
Finally, recall that
Var(X1 + · · · + Xn ) =
n X
n
X
Cov(Xi , Xj ) =
i=1 j=1
n
X
i=1
Σi,i + 2
n−1
X
Σi,i+1 .
i=1
[The sum is the sum of the diagonal entries plus twice the sum of the
entires that lie above the diagonal, by symmetry.] Therefore,
Var(X̄) =
n
n−1
1
1 X
2 X
Var(X
+
·
·
·
+
X
)
=
Σ
+
Σi,i+1 ,
1
n
i,i
n2
n2 i=1
n2 i=1
which as we have seen is equal to the expectation of (3Q1 − Q2 )/n(n − 3).
2
4, p. 16. The matrix of the quadratic form (x1 − x2 )2 + (x2 − x3 )2 + (x3 − x1 )2 is


2 −1 −1
A := −1 2 −1 .
−1 −1 2
According to Theorem 1.6 (p. 10), the variance of (X1 − X2 )2 + (X2 −
X3 )2 + (X1 − X3 )2 = X 0 AX is
(µ4 − 3µ22 )a0 a + 2µ22 tr(A2 ),
because the means [i.e., the θ’s in that theorem] are all zero. Here, a :=
(2 , 2 , 2)0 is the diagonals-vector of A,
µ2 = E(Xi2 ) =
Z
1
−1
and
µ4 = E(Xi4 ) =
Z
1
−1
Therefore, (µ4 −
3µ22 )a0 a
tr(A2 ) =
=
24
− 15
3
3 X
X
1
1 2
x dx = ,
2
3
1
1 4
x dx = .
2
5
= − 58 . Since A is symmetric,
Ai,j Ai,j =
3
3 X
X
A2i,j = 18.
i=1 j=1
i=1 j=1
Therefore, the answer to this question is − 85 + 4 =
12
5 .
5, p. 16. This is a direct, but long, computation; the methods of Chapter 2 will
clarify things a great deal, but we do not need them [strictly speaking].
First recall that
E(X 0 AX) = σ 2 tr(A), E(X 0 BX) = σ 2 tr(B).
Therefore,
Cov(X 0 AX , X 0 BX) = E (X 0 AXX 0 BX) − σ 4 tr(A) tr(B).
Therefore, we need to compute the expectation of the random “quartic
form”
X 0 AXX 0 BX = (X 0 AX) (X 0 BX)
n X
n
n X
n
X
X
=
Xi Ai,j Xj
Xk Bk,` X`
=
i=1 j=1
n X
n X
n X
n
X
i=1 j=1 k=1 `=1
3
k=1 `=1
Ai,j Bk,` Xi Xj Xk X` .
That is, we want to find
E (X 0 AXX 0 BX) =
n X
n X
n X
n
X
Ai,j Bk,` E(Xi Xj Xk X` ).
i=1 j=1 k=1 `=1
Because the X’s are i.i.d. mean-zero random variables,


E(X14 )
if i = j = k = `,




2 2

if i = j 6= k = `,
 E(X1 )
E(Xi Xj Xk X` ) =
or if i = k 6= j = `,



or if i = ` 6= j = k,



0
otherwise.
Simple computations for normal distributions show that E(X14 ) = 3σ 4 and
E(X12 ) = σ 2 .1 Therefore,

4

3σ if i = j = k = `,



σ 4
if i = j 6= k = `,

E(Xi Xj Xk X` ) =
or if i = k 6= j = `,



or if i = ` 6= k = j,



0
otherwise.
Consequently,
E (X 0 AXX 0 BX) = 3σ 4
n
X
Ai,i Bi,i + σ 4
i=1
n X
n
X
Ai,i Bk,k + σ 4
i=1 k=1
k6=i
+ σ4
n X
n
X
n X
n
X
Ai,j Bi,j
i=1 j=1
j6=i
Ai,j Bi,j
i=1 j=1
j6=i
= σ4
n
X
i=1
Ai,i
n
X
k=1
Bk,k + 2σ 4
n X
n
X
Ai,j Bi,j
i=1 j=1
= σ 4 tr(A)tr(B) + 2σ 4 tr(AB).
1 Indeed, the second moment ought to be well known to you. And the fourth moment can
be computed as follows: Without loss of generality, σ = 1; else you consider Y1 := X1 /σ
instead of X1 . When σ = 1, we have
Z ∞
Z ∞
2
2
1
2
E(X14 ) = √
x4 e−x /2 dx = √
x4 e−x /2 dx
(by symmetry)
2π −∞
2π 0
Z ∞
2
dy
= √
(4y 2 )e−y √
(y := x2 /2)
2y
2π 0
Z ∞
4
4
5
4
3
3
4
3 1
1
= √
y 3/2 e−y dy = √ Γ
= √ · ·Γ
= √ · · ·Γ
=3
π 0
π
2
π 2
2
π 2 2
2
√
since Γ(1/2) = π. Alternatively, we can compute the fourth moment by taking four derivatives of the MGF of N(0 , σ 2 ).
4
This does the job.
5