EE5620 Stochastic Processes
1
EE5620  Homework #2
Solutions
Problem assignments:
1. Problem 4.19 in textbook.
This problem again reﬂects the importance of Gaussian density, where with only mean and
variance of a random variable, we can model the random variable as Gaussian in the sense
that the Gaussian density maximizes the entropy of the random variable.
In this problem, the only things we have are:
Z
∞
∞
p
(
x
)
dx
= 1
(1)
Z
∞
∞
xp
(
x
)
dx
=
μ
(2)
Z
∞
∞
x
2
p
(
x
)
dx
=
μ
2
+
σ
2
.
(3)
We want to ﬁnd a
p
(
x
) that maximizes the entropy
H
[
X
]
,

Z
∞
∞
p
(
x
)ln
p
(
x
)
dx,
while satisfying the above three conditions. This is a typical constrained optimization problem,
and we can resort to Lagrange Multiplier technique to ﬁnd a solution.
The cost function for the constrained optimization problem here is
Q
(
p
(
x
)) =
Z
∞
∞
£

p
(
x
)ln
p
(
x
) +
λ
1
p
(
x
) +
λ
2
xp
(
x
) +
λ
3
x
2
p
(
x
)
/
dx,
where
λ
1
, λ
2
,
and
λ
3
are the Lagrange multipliers. Taking the derivative with respect to
p
(
x
)
and letting the result equal to zero give
ln
p
(
x
) =

1 +
λ
1
+
λ
2
x
+
λ
3
x
2
.
So, we have
p
(
x
) =
K
·
e
λ
2
x
+
λ
3
x
2
,
(4)
where
K
=
e

1+
λ
1
.
We will solve for these 3 Lagrange multipliers using (1), (2), and (3).
Actually, we don’t need to solve for
λ
1
,
λ
2
and
λ
3
explicitly. Based on (1) and (4), we know
K
·
Z
∞
∞
e
λ
2
x
+
λ
3
x
2
dx
= 1
,
where the exponent is quadratic with respect to
x
. Merely from this observation, we can
conclude that
p
(
x
) is a Gaussian density with proper coeﬃcient
K
. Then, from (2) and (3),
we know
p
(
x
) is Gaussian with mean
μ
and variance
σ
2
.
¥
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2
2. This problem is not graded.
3. Problem 5.9 in textbook.
(a) The matrix is not positive semideﬁnite. It has negative eigenvalues.
(b) The diagonal terms of a covariance matrix (i.e. the variances of associated random vari
ables) should be nonnegative.
(c) Since we are considering real random vector, all entries of the covariance matrix should
be real.
(d) Not symmetric.
4. Problem 5.16 in textbook.
From the problem, we can ﬁnd the inverse of the covariance matrix of
x
= [
X
1
,X
2
]
T
is
K
x

1
=
•
1
3
/
4
3
/
4
1
‚
.
That is the exponent inside the parenthesis can be represented by
X
2
1
+
3
2
X
1
X
2
+
X
2
2
=
x
T
K
x

1
x
.
If we can decompose
K
x

1
into
K
x

1
=
EΛE
T
, then the above expression becomes
x
T
EΛE
T
x
.
We can say the transformation
E
T
x
results in two independent random variables since
Λ
is
diagonal. So, we can ﬁnd a transformation
A
=
E
T
such that is
Ax
independent. One such decomposition is to take
E
T
=
•
1
1
1

1
‚
,
and
Λ
=
•
7
/
8
0
0
1
/
8
‚
.
Therefore the exponent can be expressed by
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 Winter '10
 GFung
 Probability theory, Covariance matrix, Kxy Ky

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