The following ergodic theorem is well known.

Ergodic Theorem. Let $X$ be an ergodic (ie, irreducible and aperiodic) Markov chain on a countable state space $I$. Suppose that $X$ has an invariant distribution—$\pi$, say. Let $\mu$ be any distribution on $I$. Then,

$$ \mathbb P_i(X_t = j) \to \pi_j \quad\text{as}\quad t \to \infty \quad\text{for all}\quad i,j \in I. $$

In particular, since the limit exists, the invariant distribution is unique.

This is the way that I have seen uniqueness of the invariant distribution proved, eg via a coupling argument. It seems to me, though, that a direct, more algebraic, proof should exist.

Prove uniqueness of the invariant distirbution by direct, algebraic methods, not appealing to the probabilistic interpretation.

After all, it's just a system of linear equations! I haven't been able to find one, but below are some of my thoughts on the matter. They're not super insightful, though...

  1. Irreducibility is necessary, but aperiodicity isn't

    • $\pi P = P \iff \pi (I + P)/2 = \pi$, so can just make the chain lazy
    • this is great, since irreducibility implies that $-1$ is not an eigenvalue, but I'm not sure how to interpret aperiodicity algebraically
    • in fact, this lazification implies that we may assume that all eigenvalues of $P$ are non-negative—even strictly positive if we prefer, using $(2I + P)/3$
  2. $\pi P = P \iff \pi(I - P) = 0 \iff (I - P^T) \pi^T = 0$

    • so, we want to show that $I - P^T$ has a one-dimensional kernel
    • this is equivalent to showing that the multiplicity of eigenvalue $1$ of $P^T$ is $1$
  3. An invariant distribution can be constructed via the expected return times

    • this is just a sufficient condition for $\pi P = \pi$ to hold
    • I'm after a necessary condition
  4. Naturally, I've also searched a lot online, including SE, but have not been successful

If anyone can point me to a good reference online, or give a proof—or even an outline—that would be appreciated!



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