EE 278
Handout #11
Statistical Signal Processing
Monday, July 27, 2009
Midterm Solution
1.
(a)
≥
.
P
(
A
) =
P
(
A

B
)
P
(
B
)+
P
(
A

B
c
)
P
(
B
c
)
≥
min
{
P
(
A

B
)
, P
(
A

B
c
)
}
(
P
(
B
)+
P
(
B
c
)) =
min
{
P
(
A

B
)
, P
(
A

B
c
)
}
.
(b)
≤
.
F
XY
(
x, y
) =
P
(
X
≤
x, Y
≤
y
)
1
≤
P
(
X
≤
x
) =
F
X
(
x
)
.
The inequality (1) is due to the fact that probability of intersection of two sets is
less than the probability of each of those sets.
(c) =. First, according to the symmetry of the problem we have,
E
X
2
1
X
2
1
+
X
2
2
=
E
X
2
2
X
2
1
+
X
2
2
.
Second,
E
X
2
1
X
2
1
+
X
2
2
+
E
X
2
2
X
2
1
+
X
2
2
=
E
X
2
1
+
X
2
2
X
2
1
+
X
2
2
= 1
.
Therefore each of the terms should be equal to
1
2
.
(d) =. We know that
E
(
V ar
(
X

Y
)) is the MSE of the best nonlinear or linear estima
tor of
X
from
Y
. Suppose that we call this estimator
g
(
Y
). Also,
E
(
V ar
(
X

Y
3
))
is the error of the best nonlinear estimator of
X
from
Y
3
and let’s call it
h
(
Y
3
).
Since
h
(
Y
3
) is the best estimator of
X
given
Y
3
we have,
E
(
X

h
(
Y
3
))
2
≤
E
(
X

g
(
3
√
Y
3
))
2
=
E
(
X

g
(
Y
))
2
.
On the other hand,
E
(
X

g
(
Y
))
2
≤
E
(
X

h
(
Y
3
))
2
.
From these two inequalities we see that
E
(
X

g
(
Y
))
2
=
E
(
X

h
(
Y
3
))
2
. Intu
itively speaking,
Y
3
is a onetoone mapping and it keeps all the information
Y
has about
X
.
(e)
≥
. First from the independence of
X
and
Y
we have,
E
X
4
Y
=
E
(
X
4
)
E
1
Y
.
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 Spring '09
 Eggers

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