Detailed Balance For Birth-Death Chains

Balancing Flow Across Neighboring States

In a birth-death chain, movement is restricted to neighboring states: a birth takes the process from \(n\) to \(n+1\), and a death takes it from \(n\) to \(n-1\). A stationary distribution must balance probability flow between each pair of neighbors. This local balance is especially useful because it turns a system of stationary equations into a simple recursion.

Let \((X_t)_{t\ge0}\) be a continuous-time Markov chain on the state space \(S=\{0,1,2,\ldots\}\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \(\lambda_n\) be its birth rate from \(n\) to \(n+1\), and let \(\mu_n\) be its death rate from \(n\) to \(n-1\). At the lower boundary, \(\mu_0=0\). The generator \(Q=(q_{ij})\) has entries \(q_{n,n+1}=\lambda_n\), \(q_{n,n-1}=\mu_n\) for \(n\ge1\), \(q_{n,n}=-(\lambda_n+\mu_n)\), and all other entries zero. Each row sums to zero. Unless stated otherwise, assume the rates across all neighboring edges are positive. For the stationarity result below, we also assume the total rates are bounded: \(\sup_n(\lambda_n+\mu_n)<\infty\).

The equation for an edge says that, under \(\pi\), the rate of probability flow from \(n\) to \(n+1\) equals the flow in the reverse direction. To see why stationarity forces these equations, first write the stationary balance equation at each state. For \(n\ge1\), the flow into \(n\) from its two neighbors must equal the flow out:

At state \(0\), the corresponding equation is \(0=-\pi_0\lambda_0+\pi_1\mu_1\). Define the net upward flow across the edge from \(n\) to \(n+1\) by \(J_n=\pi_n\lambda_n-\pi_{n+1}\mu_{n+1}\). The equation at \(0\) gives \(J_0=0\). For \(n\ge1\), the displayed stationary equation is \(J_{n-1}-J_n=0\), so \(J_n=J_{n-1}\). Induction gives \(J_n=0\) on every edge, which is exactly detailed balance.

Proof. Bounded rates make the generator a bounded operator, so the transition matrices are differentiable at time \(0\), with derivative \(Q\). If \(\pi\) is stationary, then \(\pi P(t)=\pi\), and differentiating at \(t=0\) gives \(\pi Q=0\). Its coordinate equations are the stationary balance equations above. The equation at \(0\) gives \(J_0=0\), and each equation for \(n\ge1\) gives \(J_n=J_{n-1}\). Thus every \(J_n=0\), proving necessity.

For the converse, choose a finite number \(\nu\) such that \(\nu\ge\sup_n(\lambda_n+\mu_n)\), and define \(R=I+Q/\nu\). Every row of \(R\) sums to \(1\), and its entries are nonnegative because \(R_{nn}=1-(\lambda_n+\mu_n)/\nu\ge0\). The adjacent-state equations imply \(\pi_iR_{ij}=\pi_jR_{ji}\) on every edge; this identity also holds for diagonal entries, and off-edge entries are both zero. For each \(j\), summing over \(i\) gives \(\sum_i\pi_iR_{ij}=\pi_j\sum_iR_{ji}=\pi_j\), because row \(j\) of \(R\) sums to \(1\). Hence \(\pi R=\pi\), and repeated multiplication gives \(\pi R^k=\pi\) for every integer \(k\ge0\). Bounded-rate uniformization expresses the transition matrix as \(P(t)=e^{-\nu t}\sum_{k\ge0}(\nu t)^kR^k/k!\). Therefore \(\pi P(t)=e^{-\nu t}\sum_{k\ge0}(\nu t)^k\pi/k!=\pi\), since the scalar coefficients sum to \(1\). This proves stationarity. \(\square\)

Recursive Weights and Normalization

When the rates across edges are positive, detailed balance determines every probability relative to \(\pi_0\). Set \(w_0=1\). The equation on the edge from \(n\) to \(n+1\) gives the recursion \(w_{n+1}=w_n\lambda_n/\mu_{n+1}\), so

If \(Z=\sum_{n\ge0}w_n\) is finite, then \(\pi_n=w_n/Z\) is a probability distribution and satisfies detailed balance. Under the bounded-rate assumption, the theorem shows that it is stationary. Conversely, any stationary distribution must obey the same recursion. Its probabilities are therefore proportional to these weights, and normalization fixes the proportionality constant. If \(Z\) is infinite, these weights cannot be normalized to a probability distribution.

What the Conditions Do—and Do Not—Give

Summability is the step that turns a positive sequence of balanced weights into a probability distribution. Bounded total rates are the condition used here to pass from those local equations to stationarity of the continuous-time process: uniformization then represents the process using a stochastic matrix and a Poisson clock. If rates are unbounded, this proof does not apply as written. In that setting, additional care about the continuous-time chain and the validity of generator balance equations is needed; summability by itself is not a substitute for checking those conditions.

Common mistake: reversing the rate ratio. In the three-state example, using \(\mu_1/\lambda_0\) instead of \(\lambda_0/\mu_1\) would give the incorrect weight \(w_1=1/2\). But the edge balance would then require \(\pi_0\lambda_0=\pi_1\mu_1\); with weights \(w_0=1,w_1=1/2\), the two sides are \(2\) and \(1/2\), not equal. The correct recursion is \(w_{n+1}=w_n\lambda_n/\mu_{n+1}\), because it is obtained by solving \(\pi_n\lambda_n=\pi_{n+1}\mu_{n+1}\) for \(\pi_{n+1}\).

Detailed balance is useful beyond giving a recipe for stationary probabilities. It provides an edge-by-edge check on a proposed answer and explains why the stationary flow has no net direction across any neighboring pair. For a birth-death chain on a line with a reflecting lower boundary, the boundary equation forces the flow across the first edge to be zero; the balance equations then force it to be zero across every edge. This is the key reason the full stationary calculation reduces to a one-step recursion.