ECE 3100
Fall 2009
Homework #6 Solutions
1. (a)
Notice that there are
21
pairs of integers
(
x, y
)
that are in the set:
Ω
=
{
(
x, y
) :

2
≤
x
≤
4
,

1
≤
y

x
≤
1
}
.
Consequently since we are considering a uniform distribution we deduce that the joint PMF of
X
and
Y
is
P
X,Y
(
x, y
) =
1
21
for
(
x, y
)
∈
Ω
and
0
otherwise.
Note that for each value of
x
∈
{

2
, ...,
4
}
, the random variable
Y
can take three values, namely
x

1
,
x
or
x
+ 1
.
Consequently,
P
X
(
x
) =
P
X,Y
(
x, x

1)+
P
X,Y
(
x, x
)+
P
X,Y
(
x, x
+1) =
3
21
for all
x
∈
{

2
, ...,
4
}
and
P
X
(
x
) = 0
otherwise. The mean of
X
is thus equal to
E
[
X
] =
∑
4
x
=

2
x
P
X
(
x
) =
3
21
∑
4
x
=

2
x
= 1
.
Similarly for each value of
y
∈
Ω
, the random variable
X
can take at most three values, namely
y

1
,
y
and
y
+ 1
,
since
x
has to be also in
{

2
, ...,
4
}
we deduce that for each value of
y
∈
Ω
the random variable
X
∈
Ω
y
=
{
y

1
, y, y
+ 1
}
∩
{

2
, ...,
4
}
. Consequently,
P
Y
(
y
) =
∑
x
∈
Ω
y
P
X,Y
(
x, y
) =
1
21

Ω
y

. We obtain therefore the
following values for the PMF of
Y
,
P
Y
(
y
) =
1
21
for
y
∈
{

3
,
5
}
,
P
Y
(
y
) =
2
21
for
y
∈
{

2
,
4
}
,
P
Y
(
y
) =
3
21
for
y
∈
{

1
, ...,
3
}
and
P
Y
(
y
) = 0
otherwise. The mean of
Y
is thus equal to
E
[
Y
] =
∑
5
y
=

3
y
P
Y
(
y
) = 1
.
1. (b)
The gain is
G
= 100
X
+ 200
Y
, i.e.,
E
[
G
] = 100
E
[
X
] + 200
E
[
Y
] = 300
.
2.
Let
X
i
be the Bernoulli random variable taking the value 1 if the first member of the
i
th
couple is alive, and 0 otherwise.