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# sol4 - STA3007 Applied probability(08-09 Chapter 3 Solution...

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STA3007 Applied probability (08-09) 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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