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Unformatted text preview: 1 or ,1 with probability 1 2. That is,
the increments X1 ; X2 ; : : : would be iid taking the 2d values
1; 0; : : : ; 0; 0; 1; : : : ; 0; : : : ; 0; 0; : : : ; 1
with probability 1=2d each. This is indeed a popular model, and can be analyzed to reach
the conclusion recurrent in d 2 and transient in d 3" as well. But the concatenation of
d independent random walks" model we will consider is a bit simpler to analyze. Also, for all
you Brownian motion fans out there, our model is the random walk analog of ddimensional
Brownian motion, which is a concatenation of d independent onedimensional Brownian
motions.
We'll start with d = 1. It is obvious that S0 ; S1 ; : : : is an irreducible Markov chain.
Since recurrence is a class property, to show that every state is recurrent it su ces to show
that the state 0 is recurrent. Thus, by Theorem 1.36 we want to show that
Xn
1.39
E 0 N0 =
P 0; 0 = 1:
But P n 0; 0 = 0 n if n is odd, and for even n = 2m, say, P 2m 0; 0 is the probability that a
Binomial2m; 1=2 takes the value m, or P 2m 0; 0 = 2m 2,2m : m Stochastic Processes J. Chang, March 30, 1999 1. MARKOV CHAINS Page 120 This can be closely approximated in a convenient form by using Stirling's formula, which
says that
p
k! 2k k=ek ;
where the notation ak bk " means that ak =bk ! 1 as k ! 1. Applying Stirling's formula
gives
p
2m
2m 0; 0 = 2m! 2 2m 2m=e = p 1 :
P
m!2 22m
2mm=e2m 22m
m Pp Thus, from the fact that 1= m = 1 it follows that 1.39 holds, so that the random
walk is recurrent.
Now it's easy to see what happens in higher dimensions. In d = 2 dimensions, for
example, again we have an irreducible Markov chain, so we may determine the recurrence
or transience of chain by determining whether the sum
1
X 1.40 n=0 P0;0 fS2n = 0; 0g 12
is in nite or nite, where S2n is the vector S2n ; S2n , say. By the assumed independence
of the two components of the random walk, we have 1 1
1 = 0gP fS 2 = 0g p 1
p
=;
P
fS = 0; 0g = P fS
0;0 2m 0 0 2m 2m m m m so that 1.40 is in nite, and the random walk is again recurrent. However, in d = 3
dimensions, the analogous sum
1
X is nite, since n=0 P0;0;0 1
P0;0;0 fS2m = 0; 0; 0g = P0 fS2m fS2n = 0; 0; 0g 2
3
= 0gP0 fS2m = 0gP0 fS2m = 0g p1 3 m ; so that in three or more dimensions the random walk is transient.
The calculations are simple once we know that in one dimension P0 fS2m = 0g is of order
p
of magnitude 1= m. In a sense it is not very satisfactory to get this by using Stirling's formula and having huge exponentially large titans in the numerator and denominator ghting
p
it out and killing each other o , leaving just a humble m standing in the denominator
after the dust clears. In fact, it is easy to p without any unnecessary violence or calguess
culation that the order of magnitude is 1= mnote that the distribution of S2m , having
p
variance 2m, is spread out" over a p of order m, so that the probabilities of points
range
in that range should be of order 1= m. Another way to see the answer is to use a Normal approximation to the binomial distribution. We approximate t...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
 Spring '10
 DURRETT
 Multiplication, Markov Chains

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