ECE 3100
Fall 2009
Homework #8: Solutions
1. (a) The area of the half circle is
πr
2
2
, since we consider a uniform probability over the semicircle, the joint PDF of
X
and
Y
will be
f
X,Y
(
x, y
) =
2
πr
2
for
(
x, y
)
∈
SC
=
{
(
x, y
)

y
≥
0
, x
2
+
y
2
≤
r
2
}
(i.e., in the semicircle) and
f
X,Y
(
x, y
) = 0
otherwise.
1. (b) The density of
Y
can be found using the following formula
f
Y
(
y
) =
Z
x
∈
A
y
f
X,Y
(
x, y
)
dx,
where
A
y
represents the range of possible values of
X
when
Y
=
y
. For any fixed
y
the range of possible values of
X
is
A
y
= [

p
r
2

y
2
,
p
r
2

y
2
]
, consequently we deduce that
f
Y
(
y
) =
Z
√
r
2

y
2

√
r
2

y
2
2
πr
2
dx
=
4
p
r
2

y
2
πr
2
,
for
0
≤
y
≤
r,
and
f
Y
(
y
) = 0
for
y > r
. The expectation of
Y
is computed as follows
E
[
Y
] =
4
πr
2
Z
r
0
y
p
r
2

y
2
dy.
Using the change of variable
t
=
r
2

y
2
, we have
dt
=

2
ydy
, the last integral becomes
E
[
Y
] =
2
πr
2
Z
r
2
0
t
1
2
dt
=
2
πr
2
[
2
3
t
3
2
]
r
2
0
=
4
r
3
π
.
1. (c) The expectation of
Y
can be computed using the joint pdf
f
X,Y
(
x, y
)
as follows
E
[
Y
] =
Z Z
(
x,y
)
∈
SC
yf
X,Y
(
x, y
)
dxdy
=
2
πr
2
Z Z
(
x,y
)
∈
SC
ydxdy.
In order to compute the last double integral we need to make a change of variables from the cartesian coordinates to
the polar coordinates, let
x
=
ρ
cos(
θ
)
and
y
=
ρ
sin(
θ
)
, with
ρ
∈
[0
.r
]
and
θ
∈
[0
, π
]
, then since
dxdy
=
ρdρdθ
, we
obtain
E
[
Y
] =
2
πr
2
Z
π
0
Z
r
0
ρ
sin(
θ
)
ρdρdθ
=
2
πr
2
Z
π
0
sin(
θ
)
dθ
Z
r
0
ρ
2
dρ
=
4
r
3
π
.
2. (a) Yes
X
is a continuous random variable, it has the following PDF
f
X
(
x
) =
1
if
0
≤
x
≤
1
,
0
otherwise
.
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 '05
 HAAS
 Probability theory, CDF, joint PDF, continuous random variable

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