ISyE 3232
Stochastic Manufacturing and Service Systems
Spring 2011
Professors Hayriye Ayhan and Jim Dai
Solutions to Homework 1
1. (a) Poisson (b) exponential (c) geometric (d) Bernoulli (e) binomial (f) normal
2. We are given that
E
[
X
] = 2 and
V ar
(
X
) = 16. Thus
a)
E
[6

2
X
] = 6

2
E
[
X
] = 2 and
V ar
(6

2
X
) = 4
V ar
(
X
) = 64.
b)
E
[(
X

3)
/
4] =
1
4
E
[
X
]

3
4
=

1
/
4 and
V ar
((
X

3)
/
4) =
1
16
V ar
(
X
) = 1.
3. We know that
a)
1 =
3
X
k
=1
P
(
X
= 2
k

1) =
3
X
k
=1
(2
k

1)
c
= 9
c
so it follows that c = 1/9.
b)
E
[
X
] =
∑
3
i
=1
(2
k

1)
P
(
X
= 2
k

1) = (1
/
9)(1 + 9 + 25) = 35
/
9
.
c)
E
[
X
2
] =
∑
3
i
=1
(2
k

1)
2
P
(
X
= 2
k

1) = (1
/
9)(1 + 27 + 125) = 17
.
d)
V ar
(
X
) =
E
[
X
2
]

(
E
[
X
])
2
= 17

(35
/
9)
2
= 1
.
8765
.
e) Note that
(
X

2)
+
=
0
with probability 1
/
9 + 3
/
9
2
with probability 5
/
9
Hence,
E
[(
X

2)
+
] = 2
*
5
/
9 = 10
/
9
.
4.
a)
P
(
X
=
k
) = 1
k
e

1
/k
!
b)
E
(
X
) = 1
c)
V ar
(
X
) = 1.
d) Suppose 0
≤
k
≤
1, k integer. Then
P
(
Y
=
k
) =
P
(
min
(
X,
2) =
k
) =
P
(
X
=
k
) = 1
k
e

1
/k
!
P
(
Y
= 2) =
P
(
min
(
X,
2) = 2) =
∞
X
k
=2
P
(
X
=
k
) = 1

1
X
k
=0
1
k
e

1
/k
!
.
e)
E
[
Y
] =
2
X
k
=0
kP
(
Y
=
k
) =
e

1
+ 2

2
1
X
k
=0
1
k
e

1
/k
!
.
5.
a) We know that
1 =
Z
∞
0
ce

2
s
ds
=
c/
2
so it follows that
c
= 2 (thus, Y is an exponential random variable with rate 2).
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 Billings
 Normal Distribution, Probability theory, Stochastic Manufacturing, Hayriye Ayhan

Click to edit the document details