Math 6010
Solutions to homework 2
3, p. 15. Recall that
Pn
− x̄)2 = x0 Ax for every x, where


− n1
1 − n1 − n1 · · · − n1

..
..
..  .
..
A :=  ...
.
.
. 
.
1
1
1
−n
− n · · · − n 1 − n1
i=1 (xi
Pn
2
Therefore, E(Q1 ) = tr(AΣ) + µ0 Aµ. Because µ0 Aµ =
i=1 (µi − µ̄)
0
and the µi ’s are allPthe
same
it
follows
that
µ
Aµ
=
0
and
therefore
P
E(Q1 ) = tr(AΣ) = i j Ai,j Σi,j . Because Σi,j = 0 when j > i + 1, and
since Σ and A are both symmetric,
E(Q1 ) =
XX
i
j
=
Ai,j Σi,j =
1−
n
X
Ai,i Σi,i + 2
i=1
1
n
X
n
n
X
Ai,i+1 Σi,i+1
i=1
n
Σi,i −
i=1
2X
Σi,i+1 .
n 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 ) − 2
i=1
E(Xi2 )
n−1
X
E(Xi Xi+1 ) − 2µ2 .
i=1
Now,
= Var(Xi )+(EXi ) = Σi,i +µ2 , and E(Xi Xi+1 ) = Σi,i+1 +µ2
similarly. Therefore,
E(Q2 ) = 2
n
X
i=1
2
Σi,i +2nµ2 −2
n−1
X
Σi,i+1 −2(n−1)µ2 −2µ2 = 2
i=1
n
X
i=1
1
Σi,i −2
n−1
X
i=1
Σi,i+1 .
Therefore,
"
# " n
#
n
n
n−1
X
X
1 X
2X
E(3Q1 − Q2 ) = 3 1 −
Σi,i+1 − 2
Σi,i −
Σi,i − 2
Σi,i+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 + 2
=
Σi,i+1 .
n i=1
n
i=1
Divide both sides by n(n − 3) in order to see that
E
3Q1 − Q2
n(n − 3)
=
n
n−1
1 X
2 X
Σ
+
Σi,i+1 .
i,i
n2 i=1
n2 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
2 X
1 X
Σ
+
Σi,i+1 ,
Var(X
+
·
·
·
+
X
)
=
i,i
1
n
n2
n2 i=1
n2 i=1
which as we have seen is equal to the expectation of (3Q1 − Q2 )/n(n − 3).
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
2
1 2
1
x dx = ,
2
3
and
µ4 = E(Xi4 ) =
Z
1
−1
1 4
1
x dx = .
2
5
24
Therefore, (µ4 − 3µ22 )a0 a = − 15
= − 58 . Since A is symmetric,
tr(A2 ) =
3 X
3
X
Ai,j Ai,j =
3 X
3
X
i=1 j=1
A2i,j = 18.
i=1 j=1
Therefore, the answer to this question is − 85 + 4 =
12
5 .
5, p. 16. Observe that X0 AX and X0 BX are 1-dimensional random variables. Consequently,
0
0
X AX · X BX =
=
n X
n
X
Xi Ai,j Xj
i=1 j=1
n X
n X
n X
n
X
·
n X
n
X
Xk Bk,l Xl
k=1 l=1
Ai,j Bk,l Xi Xj Xk Xl .
i=1 j=1 k=1 l=1
Take expectations to see that
E [X0 AX , X0 BX] = E (X0 AX · X0 BX)
n
n X
n X
n X
X
Ai,j Bk,l E (Xi Xj Xk Xl ) .
=
i=1 j=1 k=1 l=1
Because the Xi ’s are independent and have mean and third moment zero,
all of these terms are zero except in the following cases: (i) When i = j =
k = l; (ii) i = j 6= k = l; (iii) i = k 6= j = l; (iv) i = l 6= j = k. In
case (i), E(Xi Xj Xk Xl ) = E(X14 ) = 3σ 4 . In the remaining cases (i)–(iv),
E(Xi Xj Xk Xl ) = |E(X12 )|2 = σ 4 . Therefore,
E [X0 AX , X0 BX] = 3σ 4
n
X
Ai,i Bi,i + σ 4
n X
n
X
Ai,i Bk,k
i=1 k=1
k6=i
n X
n
n X
n
X
X
σ4
Ai,j Bi,j + σ 4
Ai,j Bj,i ,
i=1 j=1
i=1 j=1
j6=i
j6=i
i=1
+
considering each case separately in order from (i)–(iv). Because A and
B are symmetric, the last two quantities are equal. This reduces the
computation to the following:
E [X0 AX , X0 BX] = 3σ 4
n
X
Ai,i Bi,i + σ 4
i=1
+ 2σ 4
n X
n
X
i=1 j=1
j6=i
3
n X
n
X
i=1 k=1
k6=i
Ai,j Bi,j .
Ai,i Bk,k
Now,
n X
n
X
i=1 j=1
j6=i
Ai,j Bi,j =
n X
n
X
Ai,j Bi,j −
i=1 j=1
= tr(AB) −
n
X
Ai,i Bi,i
i=1
n
X
Ai,i Bi,i .
i=1
Consequently,
E [X0 AX , X0 BX] = σ 4
n
X
Ai,i Bi,i + σ 4
i=1
n X
n
X
Ai,i Bk,k
i=1 k=1
k6=i
+ 2σ 4 tr(AB)
n X
n
X
= σ4
Ai,i Bk,k + 2σ 4 tr(AB)
i=1 k=1
4
= σ tr(A)tr(B) + 2σ 4 tr(AB)
= E (X0 AX) E (X0 BX) + 2σ 4 tr(AB),
thanks to Theorem 1.5. This shows readily that
Cov [X0 AX , X0 BX] = 2σ 4 tr(AB).
4