# Taking x 1 2 p1 p1 p 1 q2 q2

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ution with detailed balance since ⇡ (1)p(1, 3) = 0 but p(1, 3) &gt; 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| &gt; 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 &lt; r for x &gt; ` for ` x r while the other p(x, y ) = 0. If x &lt; 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...
View Full Document

Ask a homework question - tutors are online