Unformatted text preview: en in fact all of the following hold:
i Pi fTj 1g = 1; ii Pj fTi 1g = 1;
iii The state j is recurrent.
Proof: The proof will be given somewhat informally; it can be rigorized. Suppose i 6= j ,
since the result is trivial otherwise.
Firstly, let us observe that iii follows from i and ii: clearly if ii holds that is,
starting from j the chain is certain to visit i eventually and i holds so that starting from
i the chain is certain to visit j eventually , then iii must also hold since starting from j
the chain is certain to visit i, after which it will de nitely get back to j .
To prove i, let us imagine starting the chain in state i, so that X0 = i. With probability
one, the chain returns at some time Ti 1 to i. For the same reason, continuing the chain
after time Ti , the chain is sure to return to i for a second time. In fact, by continuing this
argument we see that, with probability one, the chain returns to i in nitely many times.
Thus, we may visualize the path followed by the Markov chain as a succession of in nitely
many cycles," where a cycle is a portion of the path between two successive visits to i.
That is, we'll say that the rst cycle is the segment X1 ; : : : ; XTi of the path, the second cycle
starts with XTi +1 and continues up to and including the second return to i, and so on. The
behaviors of the chain in successive cycles are independent and have identical probabilistic
characteristics. In particular, letting In = 1 if the chain visits j sometime during the nth
cycle and In = 0 otherwise, we see that I1 ; I2 ; : : : is an iid sequence of Bernoulli trials. Let
p denote the common success probability" p = Pfvisit j in a cycleg = Pi " Ti fXk = j g k=1 for these trials. Clearly if p were 0, then with probability one the chain would not visit j
in any cycle, which would contradict the assumption that j is accessible from i. Therefore,
Stochastic Processes J. Chang, March 30, 1999 1. MARKOV CHAINS Page 118 p 0. Now observe that in such a sequence of iid Bernoulli trials with a positive success
probability, with probability one we will eventually observe a success. In fact,
Pi fchain does not visit j in the rst n cyclesg = 1 , pn ! 0 as n ! 1. That is, with probability one, eventually there will be a cycle in which the
chain does visit j , so that i holds.
It is also easy to see that ii must hold. In fact, suppose to the contrary that Pj fTi =
1g 0. Combining this with the hypothesis that j is accessible from i, we see that it is
possible with positive probability for the chain to go from i to j in some nite amount of
time, and then, continuing from state j , never to return to i. But this contradicts the fact
that starting from i the chain must return to i in nitely many times with probability one.
Thus, ii holds, and we are done.
The cycle" idea used in the previous proof is powerful and important; we will be using
it again.
The next theorem gives a useful equivalent condition for recurrence. The statement
uses the notation Ni for the total number of visits of the Markov chain to the state i, tha...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
 Spring '10
 DURRETT
 Multiplication, Markov Chains

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