A deterministic three-cycle

Topic: Discrete-time Markov chains · more questions on this topic
‹ Two steps and the stationary law A doubly stochastic matrix ›

A chain on \{1,2,3\} moves from 1 to 2, from 2 to 3, and from 3 to 1, with probability 1. Find a stationary distribution, and describe \mathbb{P}(X_n=1\mid X_0=1) as n grows.

Worked answer (click to show)

The matrix is a permutation matrix, hence doubly stochastic, so the uniform row (1/3,1/3,1/3) is stationary. Started at 1, the chain is at 1 when n is a multiple of 3, and otherwise it is not. Thus \mathbb{P}(X_n=1\mid X_0=1) equals 1 when 3 divides n and equals 0 otherwise. The sequence has no limit. The chain has period 3.