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Massachusetts Institute of Technology
Department of Electrical Engineering & Computer Science
6.041/6.431: Probabilistic Systems Analysis
(Spring 2006)
Problem Set 8 Solutions
1.
Let
A
t
(respectively,
B
t
)
be
a
Bernoulli random
variable
that
is equal to 1 if
and
only
if
the
t
th toss
resulted
in
1
(respectively, 2). We
have
E
[
A
t
B
t
] =
0 (since
A
t
= 0 implies
B
t
= 0)
and
1
1
E
[
A
t
B
s
] =
E
[
A
t
]
E
[
B
s
] =
k
·
k
for
s
�
=
t.
Thus,
E
[
X
1
X
2
]
=
E
[(
A
1
+
· · ·
+
A
n
)(
B
1
+
· · ·
+
B
n
)]
1
1
=
n
E
[
A
1
(
B
1
+
· · ·
+
B
n
)] =
n
(
n
−
1)
·
k
·
k
and
cov(
X
1
, X
2
)
=
E
[
X
1
X
2
]
−
E
[
X
1
]
E
[
X
2
]
n
(
n
−
1)
n
2
n
=
=
.
k
2
−
k
2
−
k
2
2.
(a) The
minimum mean
squared error estimator
g
(
Y
) is known
to be
g
(
Y
) =
E
[
X
Y
]. Let

us
first
find
f
X,Y
(
x, y
).
Since
Y
=
X
+
W
, we
can
write
1
x
) =
2
,
if
x
−
1
≤
y
≤
x
+
1;
f
Y

X
(
y

0
,
otherwise
and,
therefore,
1
if
x
−
1
≤
y
≤
x
+
1 and
5
≤
x
≤
10;
f
X,Y
(
x, y
) =
f
Y X
(
y

x
)
f
X
(
x
) =
10
,

·
0
,
otherwise
as
shown
in
the
plot
below.
o
o
x , y
x,y
f
(
y
o
5
10
) = 1/10
5
10
x
o
We now compute
E
[
X
Y
]
by first determining
f
X Y
(
x
y
).
This
can be
done
by



looking
at
the horizontal line
crossing the
compound PDF.
Since
f
X,Y
(
x, y
) is
uniformly
distributed in the defined region,
f
X Y
(
x
y
) is
uniformly distributed as
well.
Therefore,


5+(
y
+1)
,
if 4
≤
y <
6;
2
g
(
y
) =
E
[
X
Y
=
y
] =
y,
if 6
≤
y
≤
9;

10+(
y
−
1)
,
if 9
<
y
≤
11
.
2
The plot
of
g
(
y
) is
shown here.
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