Stochastic

Here we consider 5 to be adjacent to 0 so if we are

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Unformatted text preview: 2 . . . , L}. The chain goes to the right or left at each step with probability 1/2, subject to the rules that if it tries to go to the left from 0 or to the right from L it stays put. For example, when L = 4 the transition probability is 0 1 2 3 4 0 0.5 0.5 0 0 0 1 0.5 0 0.5 0 0 2 0 0.5 0 0.5 0 30 0 0.5 0 0.5 40 0 0 0.5 0.5 It is clear in the example L = 4 that each column adds up to 1. With a little thought one sees that this is true for any L, so the stationary distribution is uniform, ⇡ (i) = 1/(L + 1). 30 CHAPTER 1. MARKOV CHAINS Example 1.26. Tiny Board Game. Consider a circular board game with only six spaces {0, 1, 2, 3, 4, 5}. On each turn we roll a die with 1 on three sides, 2 on two sides, and 3 on one side to decide how far to move. Here we consider 5 to be adjacent to 0, so if we are there and we roll a 2 then the result is 5 + 2 mod 6 = 1, where i + k mod 6 is the remainder when i + k is divided by 6. In this case the transition probability is 0 1 2 3 4 5 0 0 0 0 1/6 1/3 1/2 1 2 3 4 5 1/3 1/3 1/6 0 0 0 1/2 1/3 1/6 0 0 0 1/2 1/3 1/6 0 0 0 1 / 2 1 /3 1/6 0 0 0 1 /2 1/3 1/6 0 0 0 It is clear...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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