# Simple proof that stationary birth-death chains are reversible

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!

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The parallel with flows, intensities, voltages and the like, is explained and pushed much further in a small book one can only recommend: P.G. Doyle and J.L. Snell, Random walks and electric networks, now freely available. – Did Feb 10 '12 at 6:23

Draw a permeable barrier between the states $x$ and $y$, dividing the number line into two parts. Since your Markov chain is a birth-death chain, the only passage through this barrier is the transition connecting $x$ and $y$. The LHS of your equality is the probability of a particle, at any given time, to be moving from $x$ to $y$; that is, the probability of it going through the barrier in one direction. The RHS of your equality is the probability of a particle, at any given time, to be moving from $y$ to $x$; that is, the probability of it going through the barrier in the opposite direction.

Since $\pi$ is stationary, the total flow through the barrier must be $0$ - that is, the LHS and the RHS must be the same, as desired!

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Thanks, this is very helpful. I'm adding another answer to expand on it. – Nate Eldredge Feb 10 '12 at 3:37

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".

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