Unformatted text preview: he Binomial2m; 1=2
distribution by the Normal distribution N m; m=2, with the usual continuity correction:
PfBinomial2m; 1=2 = mg Pfm , 1=2 N m; m=2 m + 1=2g
= Pf,p=2 2=m p N 0; 1 1=2 2=mg
0 2=m = 1= 2 2=m = 1=pm:
Stochastic Processes J. Chang, March 30, 1999 1.6. IRREDUCIBILITY, PERIODICITY, AND RECURRENCE Page 1-21 Although this calculation does not follow as a direct consequence of the usual Central Limit
Theorem, it is an example of a local Central Limit Theorem." 1.41 Exercise The other 3-dimensional random walk . Consider a random walk on the 3-dimensional integer lattice; at each time the random walk moves with equal probability
to one of the 6 nearest neighbors, adding or subtracting 1 in just one of the three coordinates.
Show that this random walk is transient.
Hint: You want to show that some series converges. An upper bound on the terms will be
enough. How big is the largest probability in the Multinomialn; 1=3; 1=3; 1=3 distribution? Here are a few additional problems about a simple symmetric random walk fSn g in one
dimension starting from S0 = 0 at time 0.
1.42 Exercise. Let a and b be integers with a 0 b. De ning the hitting times
c = inf fn 0 : Sn = cg, show that the probability Pf b
a g is given by 0 , a=b , a.
Show that Pfg 1.43 Exercise. Let S0 ; S1 ; : : : be a simple, symmetric random walk in one dimension as
we have discussed, with S0 = 0. Show that fS1 6= 0; : : : ; S2n 6= 0g = PfS2n = 0g: P Now you can do a calculation that explains why the expected time to return to 0 is in nite. 1.44 Exercise. As in the previous exercise, consider a simple, symmetric random walk
started out at 0. Letting k 6= 0 be any xed state, show that the expected number of times the
random walk visits state k before returning to state 0 is 1. We'll end this section with a discussion of the relationship between recurrence and the
existence of a stationary distribution. The results will be useful in the next section.
Suppose a Markov chain has a stationary distribution . If the
state j is transient, then j = 0.
1.45 Proposition. Proof: Since is stationary, we have P n = for all n, so that 1.46
Stochastic Processes X
i iP n i; j = j for all n:
J. Chang, March 30, 1999 Page 1-22 1. MARKOV CHAINS However, since j is transient, Corollary 1.37 says that limn!1 P n i; j = 0 for all i. Thus,
the left side of 1.46 approaches 0 as n approaches 1, which implies that j must be 0.
The last bit of reasoning about equation 1.46 may look a little strange, but in fact
iP n i; j = 0 for all i and n. In light of what we now know, this is easy to see. Firstly,
if i is transient, then i = 0. Otherwise, if i is recurrent, then P n i; j = 0 for all n, since
if not, then j would be accessible from i, which would contradict the assumption that j is
1.47 Corollary. If an irreducible Markov chain has a stationary distribution, then the chain is recurrent. Proof: Being irreducible, the chain must be either recurrent or transient. However, if the chain were transient...
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