# Direct proof of unique invariant distribution for ergodic, positive-recurrent Markov chain

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!