The Finite Birth-Death Stationary Formula

From Rates to Stationary Probabilities

A finite birth-death process moves between neighboring states: births increase the state by one, and deaths decrease it by one. When every neighboring pair communicates in both directions, the process has a stationary distribution. Its probabilities can be found directly from the rates: each successive weight is the preceding weight multiplied by a birth rate and divided by the death rate across that same edge. The remaining task is to normalize the weights so that they sum to one.

Let \((X_t)_{t\ge0}\) be a continuous-time Markov chain on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with finite state space \(S=\{0,1,\ldots,N\}\), where \(N\ge1\). From state \(i\), the chain jumps to \(i+1\) at rate \(\lambda_i\) and to \(i-1\) at rate \(\mu_i\). At the boundaries set \(\mu_0=0\) and \(\lambda_N=0\). Assume throughout that \(\lambda_i>0\) for \(0\le i<N\) and \(\mu_i>0\) for \(1\le i\le N\). These strict inequalities make the finite chain irreducible: every state can be reached from every other by a sequence of neighboring jumps.

The generator \(Q=(q_{ij})\) has \(q_{i,i+1}=\lambda_i\), \(q_{i,i-1}=\mu_i\), and \(q_{ii}=-(\lambda_i+\mu_i)\), with entries outside these transitions equal to zero. Thus every row sums to zero. A probability row vector \(\pi\) is stationary if, when it is the initial distribution, the distribution of \(X_t\) remains \(\pi\) for every \(t\ge0\). For a finite-state chain, the transition matrix at time \(t\) is \(P(t)=e^{tQ}\).

Why the Formula Gives the Stationary Distribution

For a finite chain, stationarity is equivalent to the generator equation \(\pi Q=0\). Here is a proof of that equivalence. If \(\pi\) is stationary, then \(\pi e^{tQ}=\pi\) for all \(t\ge0\). Differentiating at \(t=0\), using the power series for the matrix exponential, gives \(\pi Q=0\). Conversely, if \(\pi Q=0\), then \(\pi Q^m=0\) for every integer \(m\ge1\), by repeated multiplication. Therefore the exponential series gives \(\pi e^{tQ}=\pi\), including its \(m=0\) term \(\pi\). Hence \(\pi\) is stationary.

Proof. Define the flux across the edge from \(i\) to \(i+1\) by \(J_i=\pi_i\lambda_i-\pi_{i+1}\mu_{i+1}\), for \(0\le i<N\). The coordinate \(0\) of \(\pi Q=0\) says \(-\pi_0\lambda_0+\pi_1\mu_1=0\), so \(J_0=0\). For \(1\le j<N\), coordinate \(j\) says \(\pi_{j-1}\lambda_{j-1}-\pi_j(\lambda_j+\mu_j)+\pi_{j+1}\mu_{j+1}=0\). Rearranging gives \(J_{j-1}=J_j\). Since \(J_0=0\), induction yields \(J_i=0\) for every edge. Thus the generator equation is equivalent to the adjacent-state equalities \(\pi_i\lambda_i=\pi_{i+1}\mu_{i+1}\). As \(\mu_{i+1}>0\), these equalities give \(\pi_{i+1}=\pi_i\lambda_i/\mu_{i+1}\). Iterating from \(\pi_0\) gives \(\pi_k=\pi_0w_k\), with the product in the theorem. The weights are positive and their finite sum is positive, so normalization forces \(\pi_0=1/\sum_{\ell=0}^Nw_\ell\). Conversely, these normalized weights satisfy every adjacent-state equality, hence \(\pi Q=0\), and therefore are stationary. Any stationary probability vector must satisfy the same recurrence and normalization, so it is this vector. \(\square\)

The product is indexed by edges, not just by states: the step from \(r-1\) to \(r\) uses the birth rate \(\lambda_{r-1}\) and the reverse death rate \(\mu_r\). No limiting-distribution claim is needed to use the formula. It constructs the stationary distribution directly, and the finite sum makes normalization possible.

A Rate Pattern with a Binomial Distribution

The product formula can reveal familiar probability distributions. Consider a process on \(\{0,1,2,3\}\) with birth rates \(\lambda_i=3-i\) for \(i=0,1,2\), \(\lambda_3=0\), and death rates \(\mu_i=2i\) for \(i=1,2,3\), \(\mu_0=0\). These rates model births that become less frequent as the state approaches capacity and deaths that become more frequent as the state increases.

Boundary Conditions and a Common Indexing Error

The assumptions on the rates matter. If an interior birth or death rate is zero, the chain may fail to communicate in both directions, and the positive product formula need not describe a unique stationary distribution on the whole state space. For example, if \(\lambda_0=0\), state \(0\) cannot move upward; a stationary distribution concentrated at \(0\) may coexist with stationary distributions on a separate closed class. The irreducibility assumption avoids this ambiguity. At the boundaries, \(\mu_0=0\) and \(\lambda_N=0\) are not extra factors in the product: there is no edge below \(0\) or above \(N\).

Common mistake: Using \(\lambda_i/\mu_i\) instead of \(\lambda_i/\mu_{i+1}\) misaligns the rates. In the three-state example, this error would incorrectly set \(w_2=(1/2)(\lambda_1/\mu_1)=(1/2)(3/4)=3/8\), rather than the correct \(w_2=(1/2)(\lambda_1/\mu_2)=(1/2)(3/1)=3/2\). Normalizing the incorrect weights \(1,1/2,3/8\) gives \((8/15,4/15,3/15)\). But the flux across the edge from \(1\) to \(2\) would then be \((4/15)(3)=12/15\), while the reverse flux would be \((3/15)(1)=3/15\). The mismatch shows that the incorrect vector does not satisfy \(\pi Q=0\). Matching the birth and death rates across each same edge prevents this error.

In applications, \(\pi_i\) describes the long-run stationary fraction of time assigned to state \(i\) when the process is run in stationarity. The finite formula is useful because it avoids solving a full system of generator equations: compute neighboring weight ratios, form the products, and normalize. The generator and flux checks remain valuable, particularly when rates have state-dependent indices or boundary conventions.