EE562A Homework 2 Solutions
Prepared by Hooman ShiraniMehr
Problem 1.
Let
X
= (
X
1
, X
2
, X
3
, X
4
)
be a Gaussian random vector where
E
[
X
i
] = 0
for
i
= 1
,
2
,
3
,
4
. Show that
E
[
X
1
X
2
X
3
X
4
] =
K
12
K
34
+
K
13
K
24
+
K
14
K
23
where
K
ij
is the
i, j
element of the covariance matrix
K
X
.
Solution.
Let
x
(
u
)
be an ndimensional vector of meanzero real Gaussian random variables. The expected value of the product of the
random variables in this vector can be computed by appropriate differentiation of the characteristic function of this Gaussian
random vector.
E
h
n
Y
t
=1
x
(
u, t
)
i
= (

i
)
n
∂
n
∂v
1
...∂v
n
E
h
exp
±
i
n
X
t
=1
v
t
x
(
u, t
)
²i³
³
³
v
=
0
The characteristic function can be substituted into this expression and the partial derivative calculated in an organized fashion.
E
h
n
Y
t
=1
x
(
u, t
)
i
= (

i
)
n
h
∂
n
∂v
1
...∂v
n
exp(

v
t
K
X
v
/
2)
i³
³
³
v
=
0
= (

i
)
n
h
∂
n
∂v
1
...∂v
n
∞
X
m
=0
(

1)
m
2
m
m
!
(

v
t
K
X
v
)
m
i³
³
³
v
=
0
= (

i
)
n
∞
X
m
=0
(

1)
m
2
m
m
!
h
∂
n
∂v
1
...∂v
n
±
n
X
j
1
=1
n
X
k
1
=1
v
j
1
v
k
1
K
x
(
j
1
, k
1
)
²
...
±
n
X
j
m
=1
n
X
k
m
=1
v
j
m
v
k
m
K
x
(
j
m
, k
m
)
²i³
³
³
v
=
0
(1)
Now the differentiation process reduces to differentiating products of the form
v
j
1
v
k
1
v
j
2
v
k
2
...v
j
m
v
k
m
=
n
Y
t
=1
v
n
t
t
where
n
t
simply counts the number of times that
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 Summer '07
 ToddBrun
 Normal Distribution, Variance, Covariance matrix, kx

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