To treat the cases 1 we will use a one step

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Unformatted text preview: 22 1 ⇡ (x) = Ex T x If Ex Tx = 1, then this gives ⇡ (x) = 0. This observation motivates: Theorem 1.29. For an irreducible chain the following are equivalent: (i) Some state is positive recurrent. (ii) There is a stationary distribution ⇡ . (iii) All states are positive recurrent. 56 CHAPTER 1. MARKOV CHAINS Proof. The stationary measure constructed in Theorem 1.20 has total mass X µ(y ) = y = 1 XX Px (Xn = y, Tx > n) n=0 y 1 X Px (Tx > n) = Ex Tx n=0 so (i) implies (ii). Noting that irreducibility implies ⇡ (y ) > 0 for all y and then using ⇡ (y ) = 1/Ey Ty shows that (ii) implies (iii). It is trivial that (iii) implies (i). Our next example may at first seem to be quite di↵erent. In a branching process 0 is an absorbing state, so by Theorem 1.5 all the other states are transient. However, as the story unfolds we will see that branching processes have the same trichotomy as random walks do. Example 1.52. Branching Processes. Consider a population in which each individual in the nth generation gives birth to an independent and identically distributed number of children. The number of individuals at time n, Xn is a Markov chain with transition probability given in Example 1.8. As announced t...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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