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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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 Spring '10
 DURRETT
 The Land

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