X y 1 x px xn y tx n n0 denes a stationary

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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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