5 05 0 0 1 5 5 1 0 2 0 45 5 3 3 0 0 4 7 rows 0 and

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Unformatted text preview: guess satisfies the detailed balance condition, we note that n! nx · x!(n x)! n n! x+1 =2 n · = ⇡ (x + 1)p(x + 1, x) (x + 1)!(n x 1)! n ⇡ (x)p(x, x + 1) = 2 n However the following proof in words is simpler. Create X0 by flipping coins, with heads = “in the left urn.” The transition from X0 to X1 is done by picking a coin at random and then flipping it over. It should be clear that all 2n outcomes of the coin tosses at time 1 are equally likely, so X1 has the binomial distribution. Example 1.32. Three machines, one repairman. Suppose that an o ce has three machines that each break with probability .1 each day, but when there is at least one broken, then with probability 0.5 the repairman can fix one of them for use the next day. If we ignore the possibility of two machines breaking on the same day, then the number of working machines can be modeled as a birth and death chain with the following transition matrix: 0 1 2 3 0 .5 .05 0 0 1 .5 .5 .1 0 2 0 .45 .5 .3 3 0 0 .4 .7 Rows 0 and 3 are easy to see. To expl...
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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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