IEOR 161  Introduction to Stochastic Processes
Spring 2010
HW3 Solutions
** Note that the numbering is from Ross 9th Edition
3.42
Let X be the number of people who arrives before you. Because you are equally likely
to be the first, or second, ..., or eleventh arrival, we have
P
(
X
=
i
) =
1
11
, i
= 0
, ...,
10
Therefore,
E
[
X
] =
1
11
10
X
i
=1
i
= 5
and,
E
[
X
2
] =
1
11
10
X
i
=1
i
2
= 35
giving that,
V
(
X
) = 35

25 = 10
3.45
We are given that
E
[
X
n

X
n

1
] = 0
,
V
(
X
n

X
n

1
) =
βX
2
n

1
, X
0
=
x
0
a) It’s easy to see that,
E
[
X
n
] =
E
[
E
[
X
n

X
n

1
]] = 0
b) From a) we have that
V
(
X
n
) =
E
[
X
2
n
], where
E
[
X
2
n
]
=
E
[
E
[
X
2
n

X
n

1]]
=
E
[
V
(
X
n

X
n

1
)]
=
E
[
βX
2
n

1
]
=
β
E
[
X
2
n

1
]
=
β
n
X
2
0
=
β
n
x
2
0
3.50
Let’s start with the general case in b),
b)
P
(
N
=
n
)
=
1
3
10
n
(
.
3)
n
(
.
7)
10

n
+
10
n
(
.
5)
n
(
.
5)
10

n
+
10
n
(
.
7)
n
(
.
3)
10

n
1
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a)
P
(
N
= 0) = 0
.
00974.
c)
N
is clearly not binomial.
d) A fair game has an expected winnings of 0. For this game to be a fair game, the expected
number of heads needs to be 5 (win 5, lose 5). Now,
E
[
N
] = 3
1
3
+ 5
1
3
+ 7
1
3
= 5
therefore this is a fair game.
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 Spring '08
 Lim
 Operations Research, Probability theory, Xn, fair game

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