STAT 333 Assignment 3 SOLUTIONS
Type I Problems
1. #10, page 265 of the text, 9
th
edition.
Gary’s mood is given by an MC with S = {C, S, G} and transition matrix
=
5
.
0
3
.
0
2
.
0
3
.
0
4
.
0
3
.
0
1
.
0
4
.
0
5
.
0
P
.
If we want to examine the probability he is not in a glum mood (state G) in *any* of the next
three days, we can modify the matrix to make him stay in that state if he ever enters it. (In other
words, make Glum an absorbing state.)
=
1
0
0
3
.
0
4
.
0
3
.
0
1
.
0
4
.
0
5
.
0
'
P
.
Then, if we multiply this matrix by itself 3 times, we will have the 3-step probabilities for this
modification.
=
1
0
0
561
.
0
220
.
0
219
.
0
415
.
0
292
.
0
293
.
0
)
'
(
3
P
. The third element in the first row (P
(3)
C,G
= 0.415) represent the
probability that, given he was Cheerful, he moves to the Glum state any time in the next three
days. So we want 1 – 0.415 = 0.585.
This can also be done by listing all possible paths of Cheerful and So-So, and summing up.
2. #11, page 265 of the text, 9
th
edition.
Because the process is a Markov Chain, knowing Gary hasn’t been cheerful in a week is
irrelevant if we know the state he was in 4 days ago. (We can ignore the history bringing him to
that point.)
We again modify the matrix so that Cheerful is an absorbing state.
=
5
.
0
3
.
0
2
.
0
3
.
0
4
.
0
3
.
0
0
0
1
'
'
P
.
Taking the 4
th
power of the matrix,
=
1885
.
0
1593
.
0
6522
.
0
1593
.
0
1354
.
0
7053
.
0
0
0
1
)
'
'
(
4
P
. We want the probability he goes from Glum to Glum in 4
steps (P
(4)
G,G
= 0.1885), but conditional on the fact that he did not every go to the Cheerful state
in those 4 steps (1 - P
(4)
G,C
= 1 - 0.6522 = 0.3478). So the probability is 0.1885/0.3478 = 0.542
This can also be done (similarly to #10) by brute force, though there are more cases.
3. #25, page 267 of the text, 9
th
edition.
There are two ways to approach this problem. The state space is always {0, 1, …, k}. You can
define the Markov Chain such that X
n
= the number of pairs of shoes at
a fixed door
(say the