nctuee07f
Stochastic Processes
Midterm 2
Solutions
1.
Circularly Symmetric Complex Gaussian
(5+10+10=25 points)
(a)
X
is circularly symmetric if
e
jθ
X
has the same distribution as
X
for any
θ
.
(b) (
⇒
)
We know
X
is complex Gaussian, its real part and imaginary part are jointly
Gaussian. To show they both are identically distributed, we need to show their
means and variances are the same.
X
is circularly symmetric, then
E
[
X
] = 0 and
E
[
XX
T
] = 0
.
Since
E
[
X
] =
E
[
X
r
] +
jE
[
X
i
]
,
E
[
X
] = 0 implies the real and imaginary part have the same mean; i.e.
E
[
X
r
] =
E
[
X
i
] = 0
.
Next, carrying out the pseudocovariance yields
E
[
XX
T
] =
E
[(
X
r
+
jX
i
)
2
]
=
E
[
X
2
r

X
2
i
] + 2
jE
[
X
r
X
i
]
,
and
E
[
XX
T
] = 0 suggests that the real part and imaginary part of
E
[
XX
T
] are
both zero. That is
E
[
X
2
r

X
2
i
] = 0 and
E
[
X
r
X
i
] = 0
.
From the above, we have reached the conclusion that
X
r
and
X
i
have the same
variance
E
[
X
2
r
] =
E
[
X
2
i
]. And,
E
[
X
r
X
i
] = 0 indicates that
X
r
and
X
i
are
statistically independent.
Therefore,
X
r
and
X
i
are zero mean i.i.d. Gaussian random variables.
(
⇐
)
For this part, we need to prove
e
jθ
X
follows complex Gaussian density with the
same mean, covariance, and pseudocovariance as
X
.
— Justify
e
jθ
X
is complex Gaussian
Let
Y
,
e
jθ
X
. Carrying out
Y
gives
Y
=
X
r
cos
θ

X
i
sin
θ

{z
}
,
Y
r
+
j
(
X
r
sin
θ
+
X
i
cos
θ

{z
}
,
Y
i
)
,
for which the real part
Y
r
and the imaginary part
Y
i
are linear combination
of
X
r
and
X
i
. Thus,
Y
is indeed complex Gaussian (
i.e.
Y
r
and
Y
i
are jointly
Gaussian).
— Justify
e
jθ
X
has the same mean, covariance, and pseudocovariance as
X
i. Since
X
r
and
X
i
both have zero mean. It is clear that
Y
r
and
Y
i
also
both have zero mean.
1
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E
[
Y Y
H
] =
E
[
e
jθ
X
(
e
jθ
X
)
H
] =
E
[
XX
H
].
iii. The pseudocovariance of
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 Winter '10
 GFung
 Normal Distribution, Variance, Xr Xi, Circularly Symmetric Complex Gaussian, Yr Yi

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