For a finite reversible irreducible Markov chain with transition matrix $P$, why is $P(x,y)>0$ equivalent to $P(y,x)>0$?

Question

Let $P$ be the transition matrix for a reversible irreducible finite Markov chain with respect to the stationary measure $\pi$.

How can I prove $P(x,y)>0$ iff $P(y,x)>0$?

What I know is that by defintion of reversibility,

$$\pi(x)P(x,y)=\pi(y)P(y,x)$$.

I tried to show the equivalence $P(x,y)=0$ iff $P(y,x)=0$, which makes it sufficient to show $P(x,y)=0$ with $\pi(y)=0$ is impossible.

Any help is appreciated.

Answer 1

A hint, in the form of a possible strategy for how you might approach this:

1. If no state $x$ has $\pi(x) = 0$, then the result is obvious.
2. If one state $x$ has $\pi(x) = 0$, show that this is infectious; that is, any state $y$ with $p(y,x) > 0$ must have $\pi(y) = 0$ as well. Use this, and irreducibility, to argue that there is some pair of states $a$ and $b$ such that $\pi(a) = 0$, $\pi(b) \neq 0$, and $\pi(b, a) > 0$, which is a contradiction.