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)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
.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.
- Spring '10
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