# 33 random walks on graphs a graph is described by

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Unformatted text preview: ain row 1, we note that the state will only decrease by 1 if one machine breaks and the repairman fails to repair the one he is working on, an event of probability (.1)(.5), while the state can only increase by 1 if he succeeds and there is no new failure, an event of probability .5(.9). Similar reasoning shows p(2, 1) = (.2)(.5) and p(2, 3) = .5(.8). To ﬁnd the stationary distribution we use the recursive formula (1.12) to 34 CHAPTER 1. MARKOV CHAINS conclude that if ⇡ (0) = c then p0 0.5 =c· = 10c q1 0.05 p1 0.45 ⇡ (2) = ⇡ (1) · = 10c · = 45c q2 0.1 0.4 p2 = 45c · ⇡ (3) = ⇡ (2) · = 60c q3 0.3 ⇡ (1) = ⇡ (0) · The sum of the ⇡ ’s is 116c, so if we let c = 1/116 then we get ⇡ (3) = 60 , 116 ⇡ (2) = 45 , 116 ⇡ (1) = 10 , 116 ⇡ (0) = 1 116 There are many other Markov chains that are not birth and death chains but have stationary distributions that satisfy the detailed balance condition. A large number of possibilities are provided by Example 1.33. Random walks on graphs. A graph...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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