It follows that
p
X
(
k
)=
(
(111
−
k
)
3
−
(110
−
k
)
3
10
3
,
if
k
= 101
,...
110,
0
,
otherwise.
(An alternative solution is based on the notion of a CDF, which will be introduced in
Chapter 3.)
(b) Since
X
i
is uniformly distributed over the integers in the range [101
,
110], we have
E
[
X
i
] = (101 + 110)
/
2 = 105
.
5. The expected value of
X
is
E
[
X
]=
∞
X
k
=
−∞
k
·
p
X
(
k
110
X
k
=101
k
·
p
x
(
k
110
X
k
=101
k
·
(111
−
k
)
3
−
(110
−
k
)
3
10
3
.
The above expression can be evaluated to be equal to 103.025. The expected improve
ment is therefore 105.5  103.025 = 2.475.
Solution to Problem 2.31.
The marginal PMF
p
Y
is given by the binomial formula
p
Y
(
y
±
4
y
²
³
1
6
´
y
³
5
6
´
4
−
y
,y
=0
,
1
,...,
4
.
To compute the conditional PMF
p
X

Y
, note that given that
Y
=
y
,
X
is the number
of 1’s in the remaining 4
−
y
rolls, each of which can take the 5 values 1
,
3
,
4
,
5
,
6w
ith
equal probability 1/5. Thus, the conditional PMF
p
X

Y
is binomial with parameters
4
−
y
and
p
=1
/
5:
p
X

Y
(
x

y
±
4
−
y
x
²
³
1
5
´
x
³
4
5
´
4
−
y
−
x
,
for all nonnegative integers
x
and
y
such that 0
≤
x
+
y
≤
4. The joint PMF is now
given by
p
X,Y
(
x, y
p
Y
(
y
)
p
X

Y
(
x

y
)
=
±
4
y
²
³
1
6
´
y
³
5
6
´
4
−
y
±
4
−
y
x
²
³
1
5
´
x
³
4
5
´
4
−
y
−
x
,
for all nonnegative integers
x
and
y
such that 0
≤
x
+
y
≤
4. For other values of
x
and
y
,wehave
p
X,Y
(
x, y
)=0
.
Solution to Problem 2.32.
Let
X
i
be the random variable taking the value 1 or 0
depending on whether the ±rst partner of the
i
th couple has survived or not. Let
Y
i
be the corresponding random variable for the second partner of the
i
th couple. Then,
we have
S
=
∑
m
i
=1
X
i
Y
i
, and by using the total expectation theorem,
E
[
S

A
=
a
m
X
i
=1
E
[
X
i
Y
i

A
=
a
]
=
m
E
[
X
1
Y
1

A
=
a
]
=
m
E
[
Y
1

X
1
,A
=
a
]
P
(
X
1

A
=
a
)
=
m
P
(
Y
1

X
1
=
a
)
P
(
X
1

A
=
a
)
.
31