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De nition 352 Let i be a recurrent state The period of state i is said to be d if (n) d is the greatest common divisor of the indices n 1 for which pii > 0 A state i with period d = 1 is said to be aperiodic Lemma 355 (a) Let C be an irreducible set consisting of recurrent states Then all states in C have the same period (n) (b) If state i is aperiodic, then there is an integer n0 such that pii > 0 for all n n0 Proof (a) Denote by d(k) the period of state k C Choose i, j C with j = i By Lemma 352 we have i j and j i Hence there are integers v, w 1 (v) (w) (n) such that pij > 0 and pji > 0 Let n be any positive integer with pjj > 0 Then

(n+v+w) the rst inequality in (351) implies that pii > 0 and so n+v +w is divisible (n) by d(i) Thus we nd that n is divisible by d(i) whenever pjj > 0 This implies that d(i) d(j ) For reasons of symmetry, d(j ) d(i) Hence d(i) = d(j ) which veri es part (a) (n) (b) Let A = {n 1 | pii > 0} The index set A is closed in the sense that (n+m) (n) (m) pii pii Since n + m A when n A and m A This follows from pii

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state i is aperiodic, there are integers a A and b A whose greatest common divisor is equal to 1 An elementary result in number theory states that there exist integers r and s such that gcd (a, b) = ar + bs The integers r and s are not necessarily non-negative Let p and q be any positive integers such that both p and q are larger than a max(|r|, |s|) Take m = pa +qb Since m+a = (p +1)a +qb, part (b) of the lemma follows by proving that m + k A for k = 0, , a 1 (n) We then have pii > 0 for all n m Noting that ar + bs = 1, it follows that m + k = pa + qb + k(ar + bs) = (p + kr)a + (q + ks)b The integers p + kr and q + ks are positive Hence, by the closedness of A, the integers (p + kr)a and (q + ks)b belong to A and so the integer m + k A for any k = 0, , a 1 Finite state space There are a number of basic results that hold for nite-state Markov chains but not for Markov chains with in nitely many states In an in nite-state Markov chain it may happen that there is no recurrent state, as is demonstrated by the Markov chain example with state space I = {1, 2, } and one-step transition probabilities with pi,i+1 = 1 for all i 1 In this example all states are transient The next lemma shows that a nite-state Markov chain always has recurrent states Lemma 356 Each nite closed set of states has at least one recurrent state Proof Let C be a closed set of states Then, for any i C,.

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n = 1, 2,

(352)

Assume now that all states j C are transient In Lemma 323 it was shown that (n) limn pij = 0 for all i I if state j is transient Let n in (352) By the niteness of C, it is permissible to interchange the order of limit and summation Hence we obtain the contradiction 0 = 1 when all states in C are transient This ends the proof In most applications the Markov chain has no two disjoint closed sets (usually there is a state that is accessible from any other state) The next theorem summarizes a number of useful results for nite-state Markov chains having no two disjoint closed sets Theorem 357 Let {Xn } be a nite-state Markov chain Suppose that the Markov chain has no two disjoint closed sets Denote by R the set of recurrent states Then (a) fij = 1 for all i I and j R (b) ij < for all i I and j R, where the mean rst-passage times ij are (n) de ned by ij = nfij n=1

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