IE 300 / GE 331 Spring 2009
Homework #5 Solutions
Problem 1.
(a) For 1
<

v

<
2, the crosssection of the joint pdf surface along the line
v
is a
rectangle extending from
u
=

1 to
u
= 1. Therefore the conditional pdf of
X
given
Y
is Uniform [

1
,
1]. Similarly for

v
 ≤
1, the crosssection extends from
u
=

1 to
u
= 2, so the conditional pdf of
X
given
Y
is Uniform [

1
,
2].
(b) Using the uniformity of the conditional pdfs, we have:
E
[
X

Y
=
v
] =
(
0
,
1
<

v

<
2
1
2
,

v
 ≤
1
.
= rect(
v/
2)
,
var[
X

Y
=
v
] =
(
1
/
3
,
1
<

v

<
2
3
/
4
,

v
 ≤
1
.
(c) It is easy to see that
f
X
(
u
) =
0
.
2
,
1
< u <
2
0
.
4
,

u
 ≤
1
0
,
else
.
Hence,
E
[
X
] =
Z
∞
∞
u
·
f
X
(
u
) d
u
=
Z
1

1
0
.
4
·
u
d
u
+
Z
2
1
0
.
2
·
u
d
u
= 0 + 0
.
2
·
u
2
2
2
1
=
3
10
.
1
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IE 300 / GE 331: Homework #5 Solutions
2
Alternative Solution
It is easy to see that
f
Y
(
v
) =
0
.
2
,
1
<

v

<
2
0
.
3
,

v
 ≤
1
0
,
else
.
Hence,
E
[
X
] =
Z
∞
∞
E
[
X

Y
=
v
]
·
f
Y
(
v
) d
v
=
Z
∞
∞
rect(
v/
2)
·
f
Y
(
v
) d
v
=
Z
1

1
1
2
·
0
.
3 d
v
=
3
10
.
Problem 2.
(a)
f
X
,
Y
(
u, v
) =
f
X
(
u
)
f
Y
(
v
) =
1
2
π
exp[

u
2
+
v
2
2
]
.
(b) The region over which the joint pdf must be integrated in order to find
P
{X
2
+
Y
2
>
2
α
2
}
is shown in the lefthand figure below.
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 Spring '09
 NegarKayavash
 var, x,y, X&Y

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