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 3step 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 SoSo, 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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documentfront door) each day, or you can define Xn = the number of pairs of shoes at
the door they
happen to leave from
each day. Both are Markovian and both will get the correct answer. The
second case is slightly easier in terms of calculating the probabilities.
For 0 < i < k,
P
i,i
= P
i, i1
= P
i, ki
= P
i, ki+1
= ¼
The first probability is the situation where the runner returns to the same door she left from one
day, and then leaves from the same door the next day. The second is the situation where she
comes home to the opposite door she left from, but then leaves from the original door the next
day. Other cases are similarly defined and all equally likely.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Chisholm
 Probability, Probability theory, Markov chain

Click to edit the document details