# How can a Markov chain have more than one but a finite amount of stationary distributions?

Here's my understanding of it: Assume we have an $$n\times n$$ stochastic matrix $$P$$ that represents our Markov chain such that $$x$$ and $$y$$ are stationary distributions for $$P$$. Then

$$P(x) = x$$

$$P(y) = y$$

$$P(ax+by) = P(ax) + P(by) = aP(x) + bP(y)$$ where $$ax+by$$ is a convex linear combination

$$= ax + by$$

meaning that $$ax+by$$ is a stationary distribution, so there is an infinite amount of stationary distributions of P if there are at least 2.

Does this mean a Markov chain either has one or infinitely many stationary distributions?

• For a perhaps unenlightening example, let $$P = \begin{pmatrix}1&0\\0&1\end{pmatrix}.$$ Then $\pi P = \pi$ for any row vector $\pi$, so any distribution is a stationary distribution for $P$. – Math1000 Feb 17 at 5:27

Yes. Let $$\mu$$ and $$\nu$$ be two distinct stationary distributions. Now choose randomly between $$\mu$$ and $$\nu$$ with probabilities $$p$$ and $$1-p$$, and whatever distribution is chosen, choose according to it an initial state. Then the distribution of the state at time $$0$$ is $$p\mu+(1-p)\nu$$. If we chose $$\mu$$, then our distribution at time $$1$$ is still $$\mu$$ (because $$\mu$$ is stationary), and if we chose $$\nu$$ our distribution at time $$1$$ is $$\nu$$. Therefore our distribution at time $$1$$ is still $$p\mu+(1-p)\nu$$, so $$p\mu+(1-p)\nu$$ is stationary. This is just a probabilistic proof of what you proved algebraically.