Many chains do not have stationary distributions that

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Unformatted text preview: balance condition ⇡ is said to satisfy the detailed balance condition if ⇡ (x)p(x, y ) = ⇡ (y )p(y, x) (1.11) To see that this is a stronger condition than ⇡ p = ⇡ , we sum over x on each side to get X X ⇡ (x)p(x, y ) = ⇡ (y ) p(y, x) = ⇡ (y ) x x As in our earlier discussion of stationary distributions, we think of ⇡ (x) as giving the amount of sand at x, and one transition of the chain as sending a fraction p(x, y ) of the sand at x to y . In this case the detailed balance condition says that the amount of sand going from x to y in one step is exactly balanced by the amount going back from y to x. In contrast the condition ⇡ p = ⇡ says that after all the transfers are made, the amount of sand that ends up at each site is the same as the amount that starts there. Many chains do not have stationary distributions that satisfy the detailed balance condition. 32 CHAPTER 1. MARKOV CHAINS Example 1.29. Consider 1 .5 .3 .2 1 2 3 2 .5 .1 .4 3 0 .6 .4 There is no stationary distrib...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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