Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

2 Answers 2

up vote 4 down vote accepted

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!

share|improve this answer
    
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".

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.