# Eti n and hence t1 tn t eti n t there is a simple

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Unformatted text preview: ty p(0, j ) = fj +1 for j p(i, i 1) = 1 p(i, j ) = 0 for i 1 otherwise 0 In this chain 0 is always recurrent. If there are inﬁnitely many values of k with fk &gt; 0 then it is irreducible. If not and K is the largest value of k with fk &gt; 0 then {0, 1, . . . K 1} is a closed irreducible set. To deﬁne a stationary measure we will use the cycle trick, Theorem 1.20, with x = 0. Starting from 0 the chain will visit a site i at most omce before it 111 3.3. AGE AND RESIDUAL LIFE* returns to 0, and this will happen if and only if the ﬁrst jump is to a state i.e., t1 &gt; i. Thus the stationary measure is i, µ(i) = P (t1 &gt; i) Using (A.20) we see that 1 X µ(i) = Et1 i=0 so the chain is positive recurrent if and only if Et1 &lt; 1. In this case ⇡ (i) = P (t1 &gt; i)/Et1 (3.8) I0 J0 = {k : fk &gt; 0} so if the greatest common divisor of J0 is 1 then 0 is aperiodic. To argue the converse note that I0 consists of all ﬁnite sums of elements in J0 so g.c.d. I0 = g.c.d. J0 . Using the Markov chain convergenc...
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