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160a09_hw6sol

# 160a09_hw6sol - Winter 2009 PSTAT 160A Homework 6 solutions...

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Winter 2009 PSTAT 160A Homework # 6 solutions Problem 1. Pb 14 1) { 1 , 2 , 3 } recurrent. d = 1 2) { 1 , 2 , 3 , 4 } recurrent. d = 1 since can do 1-4-3-1 and 1-4-3-2-4-1: 3 and 5 steps. greatest divisor is 1. 3) { 2 } transient, { 1 , 3 } recurrent, { 4 , 5 } recurrent . all classes d = 1 since have positive P i i . 4) { 4 } transient, { 5 } transient, { 1 , 2 } recurrent. d = 1 5) with modified P 1 , the states are still all recurrent, the period is still d = 1 since it can do 1-2-1, 1-3-2-1 ie 2 or 3 steps. Problem 2. 1) P 5 has all entries positive. This means at the 5 th step, the chain can be in any state, starting from any step. This means all state communicates with each others. 2) ( P 5 ) 31 = . 2808 3) Need to make 1 absorbing chose Q as defined in the problem set. ( Q 5 ) 31 = . 8319 4) Add an extra absorbing state that can be reached only through 1. chose Q a as defined in the problem set. ( Q 5 a ) 31 = . 0720 5) Just 1-P(ever visited 1 in first 5 step)= 1 - . 8319 = . 1681. This also can be found by arguing that the chain with the extra state is in either 2 or 3, so 0 + . 1681 6) Oops, typo, same as 5)!
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