Taking x 1 2 p1 p1 p 1 q2 q2

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Unformatted text preview: ution with detailed balance since ⇡ (1)p(1, 3) = 0 but p(1, 3) > 0 so we would have to have ⇡ (3) = 0 and using ⇡ (3)p(3, i) = ⇡ (i)p(i, 3) we conclude all the ⇡ (i) = 0. This chain is doubly stochastic so (1/3, 1/3, 1/3) is a stationary distribution. Example 1.30. Birth and death chains are defined by the property that the state space is some sequence of integers `, ` + 1, . . . r 1, r and it is impossible to jump by more than one: p(x, y ) = 0 when |x y| > 1 Suppose that the transition probability has p(x, x + 1) = px p(x, x 1) = qx p(x, x) = 1 px qx for x < r for x > ` for ` x r while the other p(x, y ) = 0. If x < r detailed balance between x and x + 1 implies ⇡ (x)px = ⇡ (x + 1)qx+1 , so ⇡ (x + 1) = px · ⇡ (x) qx+1 (1.12) Using this with x = ` gives ⇡ (` + 1) = ⇡ (`)p` /q`+1 . Taking x = ` + 1 ⇡ (` + 2) = p`+1 p`+1 · p` · ⇡ (` + 1) = · ⇡ (`) q`+2 q`+2 · q`+1 Extrapolating from the first two results we see that in general ⇡ (` + i) = ⇡ (`) · p`+i 1 · p`+i 2 · · · p`+1 · p` q`+i · q`+i 1 · · · q`+2 · q`+1 To keep the indexing straight note that: (i) there are i terms in the numerator and...
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