Assignment 1 Solutions
IEOR173, Spring 2019
February 7, 2019
Problem 1
(a) By conditioning,
μ
=
p
0
E
[
X

X
= 1] + (1

p
0
)
E
[
X

X
6
= 1]
=
p
0
+ (1

p
0
)
E
[
X

X
6
= 1]
.
Therefore,
E
[
X

X
6
= 1] =
μ

p
0
1

p
0
.
(b) Again, by conditioning,
E
[
X
2
] =
σ
2
+
μ
2
=
p
0
E
[
X
2

X
= 1] + (1

p
0
)
E
[
X
2

X
6
= 1]
=
p
0
+ (1

p
0
)
E
[
X
2

X
6
= 1]
.
Hence,
E
[
X
2

X
6
= 1] =
σ
2
+
μ
2

p
0
1

p
0
,
and
var(
X

X
6
= 1) =
E
[
X
2

X
6
= 1]

(
E
[
X

X
6
= 1])
2
=
σ
2
+
μ
2

p
0
1

p
0

μ

p
0
1

p
0
2
.
Problem 2
(a) By taking expectation on the both sides of the recursive equation, we have
E
[
X
n
+1
] =
c
+
ρE
[
X
n
]
.
1
Because

ρ

<
1,
{
E
[
X
n
]
}
converges. Let
α
be the limit. Then, by taking the limit on both
side,
α
satisfies
α
=
c
+
ρα
⇐⇒
α
=
c
1

ρ
.
(b) Similar to the previous problem, by taking the variance of both sides, we have
var(
X
n
+1
) =
ρ
2
var(
X
n
) +
σ
2
.
{
var(
X
n
)
}
converges because

ρ

<
1. Let
β
be the limit. Then,
β
satisfies
β
=
ρ
2
β
+
σ
2
⇐⇒
β
=
σ
2
1

ρ
2
.
Problem 3
In this problem, the answer is not unique. An answer gets full credit if it is convincing
enough.
The followings are examples of possible answers.
(a) No. It seems the future development depends on the trend, which means the transition property
varies on the past few periods. To make it Markov, we need to keep track of the several past
periods.
(b) Yes. When an agent comes to the auction, she only observes the current bid and decides the
next bid. Thus, the bid process does not depend on the history and only on the current state.
Problem 4
(a) No.
The intuition for this question is that knowing just the summation
X
n
+
Z
n
does not
offer enough information, as there may be many possible combinations.
Combining (i) the