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Massachusetts
Institute
of
Technology
Department
of
Electrical
Engineering
&
Computer
Science
6.041/6.431:
Probabilistic
Systems
Analysis
(Fall
2010)
Problem
Set
5:
Solutions
1.
(a)
Because of the required normalization property of any joint PDF,
�
2
2
�
2
�
2
3
1
3
2
1 =
ax dy
dx
=
ax
(2
−
x
)
dx
=
a
2
2
−
1
2
−
+
=
a
3
3
3
x
=1
y
=
x
x
=1
so
a
= 3
/
2.
(b)
For
1
≤
y
≤
2,
y
a
3
f
Y
(
y
) =
ax
dx
=
(
y
2
−
1) =
(
y
2
−
1)
,
2
4
1
and
f
Y
(
y
) = 0
otherwise.
(c)
First
notice that
for
1
≤
x
≤
3
/
2,
f
X,Y
(
x,
3
/
2)
(3
/
2)
x
8
x
f
X

Y
(
x

3
/
2) =
=
�
� �
2
�
=
.
f
Y
(3
/
2)
3
3
−
1
2
5
4
2
Therefore,
�
3
/
2
1 8
x
E
[1
/X

Y
= 3
/
2]
=
dx
= 4
/
5
.
x
5
1
2. (a)
By
definition
f
X,Y
(
x, y
) =
f
X
(
x
)
f
Y

X
(
y

x
).
f
X
(
x
) =
ax
as shown in the
graph.
We
have
that
�
40
1 =
ax dx
=
800
a.
0
So
f
X
(
x
) =
x/
800. From
the problem
statement
f
Y

X
(
y

x
) =
2
1
x
for
y
∈
[0
,
2
x
].
Therefore,
1
/
1600
,
if 0
≤
x
≤
4 and 0
< y
<
2
x,
f
X,Y
(
x, y
) =
0
,
otherwise
.
(b)
Paul makes a
positive profit
if
Y > X
.
This occurs with probability
� �
40
2
x
1
1
P
(
Y
> X
) =
f
X,Y
(
x, y
)
dy dx
=
dy dx
=
.
1600
2
y>x
0
x
We could have also
arrived at
this answer by realizing that for
each possible
value
of
X
,
there
is a 1
/
2
probability that
Y
> X
.
(c)
The joint density function satisfies
f
X,Z
(
x, z
) =
f
X
(
x
)
f
Z

X
(
z

x
).
Since
Z
is conditionally
uniformly distributed given
X
,
f
Z

X
(
z

x
) =
2
1
x
for
−
x
≤
z
≤
x
.
Therefore,
f
X,Z
(
x, z
) =
1
/
1600
for
0
≤
x
≤
40
and
−
x
≤
z
≤
x
.
The
marginal
density
f
z
(
z
) is calculated as
40
1
40
−
z

,
if

z

<
40
,
f
Z
(
z
) =
f
X,Z
(
x, z
)
dx
=
dx
=
1600
x
x
=

z

1600
0
,
otherwise.
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�
Massachusetts
Institute
of
Technology
Department
of
Electrical
Engineering
&
Computer
Science
6.041/6.431:
Probabilistic
Systems
Analysis
(Fall
2010)
3. (a)
In order for
X
and
Y
to be independent,
any observation of
X
should not give
any information
on
Y
. If
X
is observed to
be equal
to 0,
then
Y
must be
0.
In other
words,
f
Y
{
X
=0
}
(
y

0)
negationslash
f
Y
(
y
).
Therefore,
X
and
Y
are
not
independent.
=
1.5
1
0
,
otherwise.
0.5
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
x
2
1.8
1.6
1.4
x/
2
,
if
0
≤
x
≤
1
,
(b)
f
X
(
x
) =
−
3
x/
2 + 3
,
if
1
< x
≤
2
,
f
X

Y
(
x

0
.
5)
f
Y

X
(
y

0
.
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 Fall '10
 Prof.DimitriBertsekas
 Electrical Engineering, Conditional Probability, Probability theory, Massachusetts Institute of Technology, Probabilistic Systems Analysis, Department of Electrical Engineering & Computer Science

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