Unformatted text preview: s are independent ⇡ (a, b) = ⇡ (a)⇡ (b) deﬁnes
¯
a stationary distribution for p, and Lemma 1.26 implies that all states are
¯
recurrent for p. Let (Xn , Yn ) denote the chain on S ⇥ S , and let T be the ﬁrst
¯
time that the two coordinates are equal, i.e., T = min{n 0 : Xn = Yn }. Let
V(x,x) = min{n
0 : Xn = Yn = x} be the time of the ﬁrst visit to (x, x).
Since p is irreducible and recurrent, V(x,x) < 1 with probability one. Since
¯
T V(x,x) for any x we must have
P (T < 1) = 1. (1.14) Step 3. By considering the time and place of the ﬁrst intersection and then
using the Markov property we have
P (Xn = y, T n) =
=
= n
XX m=1 x
n
XX
m=1 x
n
XX
m=1 x P (T = m, Xm = x, Xn = y )
P (T = m, Xm = x)P (Xn = y Xm = x)
P (T = m, Ym = x)P (Yn = y Ym = x) = P (Yn = y, T n)
Step 4. To ﬁnish up we observe that since the distributions of Xn and Yn
agree on {T n}
P (Xn = y ) P (Yn = y ) P (Xn = y, T > n) + P (Yn = y, T > n) and summing over y gives
X
P (Xn = y )
y P (Yn = y ) 2P (T > n) If we le...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
 Spring '10
 DURRETT
 The Land

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