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⇡ (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
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
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
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
- Spring '10
- The Land