STA3007
Applied probability (0809)
Solution to Assignment 4
Chapter 3
Problem 4.1
Method I
Let 1
,
2
,
3
,
4
,
5
,
6
,
7
,
8 denote the different patterns of three tosses, HHH, HHT, HTH,
HTT, THH, THT, TTH, TTT, respectively. Then, we can get the following transition
matrix without stoping rule
P
1 =
HHH
1
/
2
1
/
2
0
0
0
0
0
0
HHT
0
0
1
/
2
1
/
2
0
0
0
0
HTH
0
0
0
0
1
/
2
1
/
2
0
0
HTT
0
0
0
0
0
0
1
/
2
1
/
2
THH
1
/
2
1
/
2
0
0
0
0
0
0
THT
0
0
1
/
2
1
/
2
0
0
0
0
TTH
0
0
0
0
1
/
2
1
/
2
0
0
TTT
0
0
0
0
0
0
1
/
2
1
/
2
.
From the above transition matrix and our intuition, we can see that all states are equally
distributed on the state space. If that HHT or HTH appears is taken as stoping rule, the
transition matrix is changed, and given as follows
P
=
HHH
1
/
2
1
/
2
0
0
0
0
0
0
HHT
0
1
0
0
0
0
0
0
HTH
0
0
1
0
0
0
0
0
HTT
0
0
0
0
0
0
1
/
2
1
/
2
THH
1
/
2
1
/
2
0
0
0
0
0
0
THT
0
0
1
/
2
1
/
2
0
0
0
0
TTH
0
0
0
0
1
/
2
1
/
2
0
0
TTT
0
0
0
0
0
0
1
/
2
1
/
2
.
Obviously, there are 2 state, HHH and THH, which can go to state HHT with probability
1
/
2, but only 1 state, THT, to state HTH with probability 1
/
2. Therefore, HHT may
appear more frequently than HTH on average.
Method II
Let 1
,
2
,
3
,
0 denote the states, H, HH, HHT and others different from the first three,
respectively, then we can get the following transition matrix
P
1 =
1
/
2
1
/
2
0
0
1
/
2
0
1
/
2
0
0
0
1
/
2
1
/
2
0
0
0
1
.
1
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where state 3 is an absorption state. Let
ν
i
= E[
T

X
0
=
i
] for
i
= 0
,
1
,
2. Then by first
step analysis, the following system equation holds
ν
0
=
1 +
1
2
ν
0
+
1
2
ν
1
ν
1
=
1 +
1
2
ν
0
+
1
2
ν
2
ν
2
=
1 +
1
2
ν
2
Solving the above system equation, we have
ν
0
= 8.
Let 1
,
2
,
3
,
0 denote the states, H, HT, HTH and others different from the first three,
respectively, then we can get the following transition matrix
P
1 =
1
/
2
1
/
2
0
0
0
1
/
2
1
/
2
0
1
/
2
0
0
1
/
2
0
0
0
1
.
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 Spring '11
 afdsa
 Probability theory, Markov chain, ﬁrst step analysis, νi

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