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The states are numbered or renumbered as i = 1, , N The algorithm uses the following four rules: (a) State i is absorbing if and only if bii = 1 and bij = 0 for j = i (b) If state j is absorbing and bij = 1, then state i is transient (c) If state j is transient and bij = 1, then state i is transient (d) If state i communicates with state j and state j communicates with state k, then state i communicates with state k The goal of the algorithm is to nd all recurrent subclasses and the set of transient states The algorithm rules (a), (b), (c) and (d) In particular, make repeated use of rule (d) is used to reduce the size of the Boolean matrix B whenever possible The algorithm works using the following steps: Step 1 Initialize the set T (i) := {i} for any state i Find all absorbing states by using rule (a) and classify T (i) = {i} as a recurrent subclass for each absorbing state i Classify any state i such that bij = 1 for some absorbing state j as a transient state Step 2 If all states are classi ed, then stop; otherwise, go to step 3 Step 3 Take an unclassi ed state i0 Since state i0 is not absorbing, there is another state i1 (say) that can be reached from state i0 in one step (ie bi0 i1 = 1) Continuing in this way, construct a chain of states i0 , i1 , until one of the following two exclusive possibilities occurs: A transient state is is found Then all states in T (i0 ) T (i1 ) T (is 1 ) are classi ed as transient according to rule (c) A state is is found that was already encountered during the development of the chain, ie is = ir for some r < s Go to step 4 Step 4 The circuit of communicating states ir , , is is replaced by a single aggregated state ir and the Boolean matrix B is adjusted accordingly This is done as follows: Replace column ir by the union of the columns ir , , is 1 and replace row ir by the union of the rows ir , , is 1 (the union of two Boolean vectors x and y to a Boolean vector z is de ned by zi = 0 if xi = yi = 0 and zi = 1 otherwise) Delete the row ik and the column ik for k = r + 1, , s 1 Let T (ir ) := T (ir ) T (ir+1 ) T (is 1 ) Having done this, there are two possibilities: State ir is absorbing for the new Boolean matrix B Then T (ir ) is classi ed as a recurrent subclass of states Classify any state that can reach the set T (ir ) in one step as a transient state (rule (b)) Go to step 2.

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State ir is not absorbing Then there exists a state j with bir j = 1 Go to step 3 and continue the chain i0 , , ir for the new Boolean matrix 352 Ergodic Theorems The theoretical analysis of Markov chains is much more subtle for the case of in nitely many states than for the case of nitely many states A nite-state Markov chain is always a regenerative process with a nite mean cycle length This is not true for in nite-state Markov chains Recall the example with I = {1, 2, } and pi,i+1 = 1 for all i I and recall the example of the symmetric random walk with I = {0, 1, 2, } and pi,i+1 = pi,i 1 = 1 for all i In the rst 2 example the Markov chain is not regenerative, while in the other example the Markov chain is regenerative but has an in nite mean cycle length In practical applications these pathological situations occur very rarely Typically there is a positive recurrent state that will ultimately be reached from any other state with probability one We therefore restrict our theoretical analysis to Markov chains which satisfy Assumption 331 Let R denote the set of recurrent states of the Markov chain {Xn } We rst prove the following lemma Lemma 358 Suppose that the Markov chain {Xn } satis es Assumption 331 Then the set R is not empty and is an irreducible set consisting of positive recurrent states For any j R, it holds that fij = 1 for all i I and jj < Proof The regeneration state r from Assumption 331 is recurrent and so R is not empty Since fir = 1 for all i I , the Markov chain {Xn } has no two disjoint closed sets Hence, by Theorem 354, the set R is an irreducible set of recurrent states Since rr < , it follows from part (b) of Theorem 353 that jj < for all j R In other words, each state j R is positive recurrent Also, by part (b) of Theorem 353, frj = 1 for all j R Together with the assumption fir = 1 for all i this implies fij = 1 for all i when j R This ends the proof De ne now the probabilities j by j = lim 1 n n.

.

n k=1 (k) pjj ,

(353)

In Theorem 331 it was shown that these limits exist Under Assumption 331, we have 1 n n lim and j = 1 > 0, jj j R (355)

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