Stochastic

# 1 2 m n n x 1 m with 0 1 and n n xn

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Unformatted text preview: in the denominator, (ii) the indices decrease by 1 each time, (iii) the answer will not depend on q` (which is 0) or p`+i . For a concrete example to illustrate the use of this formula consider Example 1.31. Ehrenfest chain. For concreteness, suppose there are three balls. In this case the transition probability is 0 1 2 3 0 1 2 3 0 3 /3 0 0 1/3 0 2 /3 0 0 2/3 0 1/3 0 0 3 /3 0 Setting ⇡ (0) = c and using (1.12) we have ⇡ (1) = 3c, ⇡ (2) = ⇡ (1) = 3c ⇡ (3) = ⇡ (2)/3 = c. 33 1.6. SPECIAL EXAMPLES The sum of the ⇡ ’s is 8c, so we pick c = 1/8 to get ⇡ (0) = 1/8, ⇡ (1) = 3/8, ⇡ (2) = 3/8, ⇡ (3) = 1/8 Knowing the answer, one can look at the last equation and see that ⇡ represents the distribution of the number of Heads when we ﬂip three coins, then guess and verify that in general that the binomial distribution with p = 1/2 is the stationary distribution: ⇡ (x) = 2 Here m! = 1 · 2 · · · (m n ✓◆ n x 1) · m, with 0! = 1, and ✓◆ n n! = x!(n x)! x is the binomial coe cient which gives the number of ways of choosing x objects out of a set of n. To check that our...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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