TITLE: Proof that Markov Chains converges to the stationary distribution
QUESTION [5 upvotes]: Let $P$ is a transition matrix of a Markov Chain, which is irreducible, aperiodic and lets assume $\pi$ is its stationary distribution: $\pi = \pi P$. Does anyone knows the proof for the following fact (given $\pi_0$ is valid probability distribution):
$$\forall \pi_0 \lim_{n \to \infty}\pi_0 P^n = \pi$$
I know the proof when if $P=Q\Sigma Q^{-1}$is the eigenvector decomposition and $Q^{-1} = Q^{T}$. But how do I prove it in the general case, when the eigenvectors are not orthogonal to each other?

REPLY [3 votes]: An elegant probabilistic proof uses the notion of coupling. See Theorem 8.6 of Billingsley's Probability & Measure. For another proof see Chapter 4 of Markov Chains and Mixing Times here http://pages.uoregon.edu/dlevin/MARKOV/