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!
 A: 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".
A: 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!
A: Using user abkds and Did 's idea here and modifying, organizing a bit . This simple approach starts directly from definition of stationary distribution and Birth Death chain , but may not be as intuitive as Lopsy's answer .
OP assumed the existence of stationary distribution $\pi$ , by definition
$$
\pi(x) = \sum_{k\in \mathbb{Z}} \pi(k)p(k,x) 
$$
by $|x-y|>1 \implies p(x,y)=0 $ ,
$$
= \pi(x-1) p(x-1,x) + \pi(x+1) p(x+1,x) + \pi(x)(1-p(x,x-1)-p(x,x+1))
$$
rearranging yields
$$
\color{red}{ \pi(x)p(x,x-1) } + \pi(x)p(x,x+1)= \color{red}{ \pi(x-1) p(x-1,x) } + \pi(x+1) p(x+1,x) , \; \forall x\in  \mathbb{Z} \tag{*}
$$
$*$ is obtained by pairwise addition of the detailed balance equations , but they are not equivalent , as proven here . We need an additional condition : $\exists n \in \mathbb{Z},\; \pi(n-1)p(n-1,n)= \pi(n)p(n,n-1) $
along with $*$ to imply the detailed balance equations .
Let state space becomes $\mathbb{Z}_{N} = \{-N,..,0,..,N\} \subset \mathbb{Z} $ , we have $ \pi(N-1)p(N-1,N)= \pi(N)p(N,N-1)$ and a version of $*$ with $\mathbb{Z}$ replaced by $\mathbb{Z}_{N}\setminus\{-N,N\} $ . Now let
inetger $N\to \infty$ , we have $\mathbb{Z}_{N} \to \mathbb{Z} $ , so $*$ implies the detailed balance equations now . (I'm unsure if this's rigorous enough) $\qquad \square$
So we only used 3 conditions :

*

*stationary distribution exists

*$|x-y|>1 \implies p(x,y)=0 $ (or as OP says : "simply connected")

*there exists a pair of state such that the detailed balance equation holds (I think even trivially holds : $0=0$ would work)

