3.4 Markov Chains - Long Run Behaviour - Exercises Solutions.pdf - THE CHINESE UNIVERSITY OF HONG KONG Department of Statistics STAT3007 Introduction to

3.4 Markov Chains - Long Run Behaviour - Exercises Solutions.pdf

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THE CHINESE UNIVERSITY OF HONG KONG Department of Statistics STAT3007: Introduction to Stochastic Processes Markov Chains - Long Run Behaviour - Exercises Solutions 1. (Problem 4.1.2 in Pinsky and Karlin) We need to create an appropriate Markov Chain to help us answer the question. Since we are concerned with Urn A being empty, let the chain be X n = how many balls in Urn A after n turns. Then the state space is { 0 , 1 , 2 , 3 , 4 , 5 } . Say we’re in State 0. Then all the balls are in Urn B. One of these has to be chosen and moved to Urn A. Therefore p 01 = 1. Now say we’re in State 2. The single ball in Urn A is selected w.p. 1 / 5 and we would move to State 0. Otherwise, one of the four balls in Urn B is selected w.p. 4 / 5 and we move to State 2. Thus p 10 = 1 / 5 and p 12 = 4 / 5. Similarly thinking shows us p i,i - 1 = i/ 5 and p i,i +1 = (5 - i ) / 5 for i = 0 , 1 , . . . , 5. Thus the transition probability matrix is P = 0 1 0 0 0 0 1 / 5 0 4 / 5 0 0 0 0 2 / 5 0 3 / 5 0 0 0 0 3 / 5 0 2 / 5 0 0 0 0 4 / 5 0 1 / 5 0 0 0 0 1 0 . The long run fraction of time Urn A is empty is precisely the long run probability of the chain being in State 0, π 0 . Hence ee solve P T π = π, 5 i =1 π i = 1 to find π 0 = π 5 = 1 / 32, π 1 = π 4 = 5 / 32, π 2 = π 3 = 10 / 32. So in the long run the fraction of time urn A is empty is π 0 = 1 / 32. 2. (Problem 4.1.4 in Pinsky and Karlin) The limiting distribution of X n tells us the fraction of time spent in a state. We need to set-up an MC whose states are the possible transitions for the chain X n . Let Z n be such a chain, whose state space is all the possible transitions for X n . E.g. if X n has states 0 and 1, then Z n would have states 0
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