TITLE: Simple proof that stationary birth-death chains are reversible
QUESTION [6 upvotes]: A Markov chain with state space $\mathbb{Z}$ is a birth-death chain if the transition probabilities satisfy $p(x,y) = 0$ for $|x-y| > 1$.  That is, the only possible transitions are to move one state to the left or right or to stay still.
Suppose such a chain has a stationary distribution $\pi$.  It is then a "well-known" fact that $\pi$ satisfies the detailed balance equation
$$\pi(x) p(x, y) = \pi(y) p(y,x), \quad x,y \in \mathbb{Z}.$$
That is, the chain is reversible.
I am looking for a simple proof of this fact (for a class I am teaching).  The only proofs I've found go via Kolmogorov's criterion for reversibility, which seems like a lot of work.  If possible, the proof should give some insight as to why the particular structure of a birth-death chain causes reversibility to hold.
Thanks!

REPLY [5 votes]: Lopsy's answer has the right idea, I think.  Here's how I worked out the details.
If $x$ and $y$ are not adjacent then the detailed balance equation is trivially true.  If they are, say $y = x+1$, let $L = \{z : z \le x\}$ be the set of states on the left side.  The key, as Lopsy says, is that the only route in or out of $L$ is via the transition between $x$ and $y$.
Consider $P_\pi(X_1 \in L)$.  On the one hand, since $\pi$ is stationary, $P_\pi(X_1 \in L) = P_\pi(X_0 \in L)$.  On the other hand, we can break up the event $\{X_1 \in L\}$ and write
$$\begin{align} P_\pi(X_1 \in L) &= P_\pi(X_1 \in L, X_0 \in L) + P_\pi(X_1 \in L, X_0 \notin L) \\\\ &= P_\pi(X_0 \in L) - P_\pi(X_1 \notin L, X_0 \in L) + P_\pi(X_1 \in L, X_0 \notin L).\end{align}$$
Thus $P_\pi(X_1 \notin L, X_0 \in L) = P_\pi(X_1 \in L, X_0 \notin L)$ which is Lopsy's "total flow" statement.  However, $P_\pi(X_1 \notin L, X_0 \in L) = P_\pi(X_1 = y, X_0 = x) = \pi(x) p(x,y)$, and likewise the other side is $\pi(y) p(y,x)$.
This also makes clear that the essential feature of a birth-death chain is that it is "simply connected".