But clearly the long run fraction of time the chain

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Unformatted text preview: also the number of cycles completed up to time n, that is, Vnj  = maxfk : Sk  ng: To ease the notation, let Vn denote Vn j . Notice that SVn  n SVn +1; Stochastic Processes J. Chang, March 30, 1999 1.9. A SLLN FOR MARKOV CHAINS and divide by Vn to obtain Page 1-31 SVn  n Vn Vn SVn +1 : Vn Since j is recurrent, Vn ! 1 with probability one as n ! 1. Thus, by the ordinary Strong Law of Large Numbers for iid random variables, we have both SVn ! E T  jj Vn and SVn+1 = SVn+1 Vn Vn + 1 V n+1 Vn ! E j Tj   1 = E j Tj  with probability one. Note that the last two displays hold whether E j Tj  is nite or in nite. Thus, n=Vn ! E j Tj  with probability one, so that Vn ! 1 n E j Tj with probability one, which is what we wanted to show. Next, to treat the general case where i may be di erent from j , note that Pi fTj 1g = 1 by Theorem 1.35. Thus, with probability one, a path starting from i behaves as follows. It starts by going from i to j in some nite number Tj of steps, and then proceeds on from state j in such a way that the long run fraction of time that Xt = j for t  Tj approaches 1=E j Tj . But clearly the long run fraction of time the chain is at j is not a ected by the behavior of the chain on the nite segment X0 ; : : : ; XTj ,1 . So with probability one, the long run fraction of time that Xn = j for n  0 must approach 1=E j Tj . The following result follows directly from Theorem 1.57 by the Bounded Convergence Theorem from the Appendix. That is, we are using the following fact: if Zn ! c with probability one as n ! 1 and the random variables Zn all take values in the same bounded interval, then we also have E Zn  ! c. To apply this in our situation, note that we have 1 Zn := n n X t=1 I fXt = j g ! E 1T jj with probability one as n ! 1, and also each Zn lies in the interval 0,1 . Finally, use the fact that the expectation of an indicator random variable is just the probability of the corresponding event. 1.58 Corollary. For an irreducible Markov chain, we have n 1X 1 lim n P t i; j  = E T  n!1 jj t=1 for all states i and j . Stochastic Processes J. Chang, March 30, 1999 1. MARKOV CHAINS Page 1-32 There's something suggestive here. Consider for the moment an irreducible, aperiodic Markov chain having a stationary distribution . From the Basic Limit Theorem, we know that, P n i; j  ! j  as n ! 1. However, it is simple fact that if a sequence of numbers converges to a limit, then the sequenceP Cesaro averages" converges to the same limit; of that is, if at ! a as t ! 1, then 1=n n=1 at ! a as n ! 1. Thus, the Cesaro averages t of P ni; j  must converge to j . However, the previous Corollary shows that the Cesaro averages converge to 1=E j Tj . Thus, it follows that j  = E 1T  : jj It turns out that the aperiodicity assumption is not needed for this last conclusion; we'll see this in the next result. Incidentally, we could have proved this result much earlier; for example we don't need the Basic Limit Theorem in the development. 1.59 Theorem. An irreducible, positive recurr...
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