Solutions to Homework 6, ISyE 2027 Fall 2006
Problem 1
(a) geometric (b) Bernoulli (c) binomial (d) Poisson
Problem 2
(a) We know that 1 =
∑
4
k
=1
P
(
X
=
k
), so
1 =
4
k
=1
P
(
X
=
k
) =
c
4
k
=1
k
= 10
c
so
c
= 1
/
10.
(b) The expected value (or mean) of
X
is just
E
[
X
] =
4
k
=1
kP
(
X
=
k
) = (1
/
10)
4
k
=1
k
2
= 30
/
10 = 3
.
(c) To compute the probability mass function of
Y
, we first need to determine the
range of
Y
, or the set of possible values
Y
can take. It’s clear that the range of
Y
is
{
1
,
4
,
9
,
16
}
. Therefore, the probability mass function of
Y
is
P
(
Y
= 1) =
P
(
X
= 1) = 1
/
10
P
(
Y
= 4) =
P
(
X
= 2) = 2
/
10
P
(
Y
= 9) =
P
(
X
= 3) = 3
/
10
P
(
Y
= 16) =
P
(
X
= 4) = 4
/
10
(d) The expected value of
Y
=
X
2
is just
E
[
Y
] = (1)
P
(
Y
= 1) + (4)
P
(
Y
= 4) + (9)
P
(
Y
= 9) + (16)
P
(
Y
= 16)
= 1
/
10 + 8
/
10 + 27
/
10 + 64
/
10 = 100
/
10 = 10
.
(e) Using the “law of the unconscious statistician”, we see that
E
[
X
2
] =
4
k
=1
k
2
P
(
X
=
k
) =
4
k
=1
(1
/
10)
k
3
= 10
.
(f) Here
E
[
X
(
X

1)] =
E
[
X
2

X
] =
E
[
X
2
]

E
[
X
] = 10

3 = 7
.
(g) The variance of
X
is just
V ar
(
X
) =
E
[
X
2
]

(
E
[
X
])
2
= 10

9 = 1
.
(h) The standard deviation is just the square root of the variance, so in this case
it’s 1.
(i) If we condition on the event that
{
X
≤
2
}
, then
X
is either 1 or 2.
1
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2
P
(
X
= 1

X
≤
2) =
P
(
X
= 1
, X
≤
2)
P
(
X
≤
2)
=
P
(
X
= 1)
P
(
X
≤
2)
=
(1
/
10)
(3
/
10)
= 1
/
3
.
Similarly,
P
(
X
= 2

X
≤
2) =
P
(
X
= 2)
P
(
X
≤
2)
= 2
/
3
.
(j) The conditional expectation of
X
, given
{
X
≤
2
}
is
E
[
X

X
≤
2] = (1)
P
(
X
= 1

X
≤
2) + (2)
P
(
X
= 2

X
≤
2) = 1
/
3 + 4
/
3 = 5
/
3
.
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 Spring '08
 Zahrn
 Variance, Probability theory, 0 k, Probability mass function, k=1

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