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Unformatted text preview: 5, 4, 6, and 7 are recurrent, completing our study of the Example
1.14 with the results we had claimed earlier.
In fact, the combination of Theorem 1.5 and 1.7 is su cient to classify the
states in any ﬁnite state Markov chain. An algorithm will be explained in the
proof of the following result.
Theorem 1.8. If the state space S is ﬁnite, then S can be written as a disjoint
union T [ R1 [ · · · [ Rk , where T is a set of transient states and the Ri , 1 i k ,
are closed irreducible sets of recurrent states.
Proof. Let T be the set of x for which there is a y so that x ! y but y 6! x.
The states in T are transient by Theorem 1.5. Our next step is to show that
all the remaining states, S T , are recurrent.
Pick an x 2 S T and let Cx = {y : x ! y }. Since x 62 T it has the
property if x ! y , then y ! x. To check that Cx is closed note that if y 2 Cx
and y ! z , then Lemma 1.4 implies x ! z so z 2 Cx . To check irreducibility,
note that if y, z 2 Cx , then by our ﬁrst observation y ! x a...
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 Spring '10
 DURRETT
 The Land

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