Stochastic Processes
nctuee07f
Homework 1
Solutions
1.
f
X
(
x
)
=
d
dx
P
[
X
≤
x
]
=
d
dx
X
n
P
[
X
≤
x

Y
=
n
]
P
[
Y
=
n
]
(by total probability theorem)
=
X
n
d
dx
P
[
X
≤
x

Y
=
n
]
¶
P
[
Y
=
n
]
=
X
n
f
X

Y
(
x

Y
=
n
)
P
[
Y
=
n
]
.
2. The probability that
S
=

1, given that we have observed
Y
=
y
is
P
[
S
=

1

Y
=
y
], which,
according to the formula on page 112 of Topic 1, can be calculated by
P
[
S
=

1

Y
=
y
]
=
P
[
S
=

1]
f
Y

S
(
y

S
=

1)
P
[
S
= 1]
f
Y

S
(
y

S
= 1) +
P
[
S
=

1]
f
Y

S
(
y

S
=

1)
.
With the given received signal model, the conditional density of
Y
given
S
=

1 is just
Gaussian with mean

1 and variance
σ
2
.
Thus, we have
f
Y

S
(
y

S
=

1) =
1
√
2
πσ
2
e

(
y
+1)
2
2
σ
2
.
And, similarly,
f
Y

S
(
y

S
= 1) =
1
√
2
πσ
2
e

(
y

1)
2
2
σ
2
.
After a little algebraic effort, it gives
P
[
S
=

1

Y
=
y
] =
e

y
σ
2
e
y
σ
2
+
e

y
σ
2
.
3.
(a) This identity holds true for both discrete and continuous random variables.
Consider
discrete
X
and
Y
.
E
[
E
[
X

Y
]]
=
X
y
E
[
X

Y
=
y
]
p
Y
(
y
)
=
X
y
ˆ
X
x
xp
X

Y
(
x

y
)
!
p
Y
(
y
)
=
X
x
x
ˆ
X
y
p
X

Y
(
x

y
)
p
Y
(
y
)
!
=
X
x
x
ˆ
X
y
p
X,Y
(
x, y
)
!
=
X
x
xp
X
(
x
)
=
E
[
X
]
.
(b) The result of this problem is intuitively obvious. The intuition is, given that
Y
=
y
, there
is nothing random about
h
(
Y
)

Y
=
y
.
Thus, it serves as a deterministic (nonrandom) value
and can be pulled out of the expectation.
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But we still need to justify it mathematically. We consider the discrete random variables
case here. Continuous random variables case can also be shown in a similar fashion.
Let
X
and
Y
be two discrete random variables.
From the rule of expected value, the
conditional expectation is
E
h
g
(
X
)
·
h
(
Y
)
fl
fl
fl
Y
=
y
i
=
X
x,y
0
g
(
x
)
h
(
y
0
)
p
X,Y

Y
(
x, y
0

y
)
=
X
x,y
0
g
(
x
)
h
(
y
0
)
P
[
X
=
x, Y
=
y
0
, Y
=
y
]
P
[
Y
=
y
]
,
where
P
[
X
=
x, Y
=
y
0
, Y
=
y
] =
‰
P
[
X
=
x, Y
=
y
]
if
y
0
=
y
,
0
otherwise.
Thus, the conditional expectation is evaluated only when
y
0
=
y
, which gives
E
h
g
(
X
)
·
h
(
Y
)
fl
fl
fl
Y
=
y
i
=
X
x,y
0
g
(
x
)
h
(
y
0
)
P
[
X
=
x, Y
=
y
0
, Y
=
y
]
P
[
Y
=
y
]
=
X
x
g
(
x
)
h
(
y
)
P
[
X
=
x, Y
=
y
]
P
[
Y
=
y
]
(a result from
y
0
=
y
)
=
h
(
y
)
X
x
g
(
x
)
P
[
X
=
x, Y
=
y
]
P
[
Y
=
y
]
=
h
(
y
)
ˆ
X
x
g
(
x
)
p
X

Y
(
x
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 Winter '10
 GFung
 Linear Algebra, Probability theory, Invertible matrix, Y, matrix inversion lemma

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