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
] =
∞
k
=
−∞
k
·
p
X
(
k
) =
110
k
=101
k
·
p
x
(
k
) =
110
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
1
6
y
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
,
6 with
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
1
5
x
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
1
6
y
5
6
4
−
y
4
−
y
x
1
5
x
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
, we have
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 first 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
i
=1
E
[
X
i
Y
i

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

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

X
1
= 1
, A
=
a
]
P
(
X
1
= 1

A
=
a
)
=
m
P
(
Y
1
= 1

X
1
= 1
, A
=
a
)
P
(
X
1
= 1

A
=
a
)
.
31
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We have
P
(
Y
1
= 1

X
1
= 1
, A
=
a
) =
a
−
1
2
m
−
1
,
P
(
X
1
= 1

A
=
a
) =
a
2
m
.
Thus
E
[
S

A
=
a
] =
m
a
−
1
2
m
−
1
·
a
2
m
=
a
(
a
−
1)
2(2
m
−
1)
.
Note that
E
[
S

A
=
a
] does not depend on
p
.
Solution to Problem 2.38.
(a) Let
X
be the number of red lights that Alice
encounters. The PMF of
X
is binomial with
n
= 4 and
p
= 1
/
2. The mean and the
variance of
X
are
E
[
X
] =
np
= 2 and var(
X
) =
np
(1
−
p
) = 4
·
(1
/
2)
·
(1
/
2) = 1.
(b) The variance of Alice’s commuting time is the same as the variance of the time by
which Alice is delayed by the red lights. This is equal to the variance of 2
X
, which is
4var(
X
) = 4.
Solution to Problem 2.39.
Let
X
i
be the number of eggs Harry eats on day
i
.
Then, the
X
i
are independent random variables, uniformly distributed over the set
{
1
, . . . ,
6
}
. We have
X
=
∑
10
i
=1
X
i
, and
E
[
X
] =
E
10
i
=1
X
i
=
10
i
=1
E
[
X
i
] = 35
.
Similarly, we have
var(
X
) = var
10
i
=1
X
i
=
10
i
=1
var(
X
i
)
,
since the
X
i
are independent. Using the formula of Example 2.6, we have
var(
X
i
) =
(6
−
1)(6
−
1 + 2)
12
≈
2
.
9167
,
so that var(
X
)
≈
29
.
167.
Solution to Problem 2.40.
Associate a success with a paper that receives a grade
that has not been received before.
Let
X
i
be the number of papers between the
i
th
success and the (
i
+ 1)st success. Then we have
X
= 1 +
∑
5
i
=1
X
i
and hence
E
[
X
] = 1 +
5
i
=1
E
[
X
i
]
.
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 Spring '08
 staff
 Math, Integers, Normal Distribution, Variance, Probability theory, CDF, Xi Yi

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