A Constant-Rate Chain With Geometric Stationary Probabilities

The Constant-Rate Birth-Death Model

Consider a population whose size changes by one at each transition. When the population is zero, births can still occur but deaths cannot; at every positive population size, births and deaths are both possible. If the birth rate is smaller than the death rate, the balance between upward and downward movement produces a stationary distribution that decreases geometrically with population size.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((X_t)_{t\ge0}\) be a continuous-time Markov chain with state space \(S=\{0,1,2,\ldots\}\). Fix rates \(\lambda,\mu\) with \(0\le\lambda<\mu\) and \(\mu>0\). From state \(0\), the chain moves to state \(1\) at rate \(\lambda\). From each state \(i\ge1\), it moves to \(i+1\) at rate \(\lambda\) and to \(i-1\) at rate \(\mu\). The generator \(Q=(q_{ij})\) therefore has \(q_{01}=\lambda\), \(q_{00}=-\lambda\), and, for \(i\ge1\), \(q_{i,i+1}=\lambda\), \(q_{i,i-1}=\mu\), and \(q_{ii}=-(\lambda+\mu)\); all other entries are zero.

The row sums check the rate convention: at state \(0\), \(-\lambda+\lambda=0\), while at each \(i\ge1\), \(\mu-(\lambda+\mu)+\lambda=0\). The total rate out of any state is at most \(\lambda+\mu\). In particular, the chain has bounded rates, so it can be constructed using a Poisson clock of rate \(\lambda+\mu\), with some clock rings producing no change of state. It cannot make infinitely many transitions in a finite time interval.

From Balance Weights to a Probability Distribution

For neighboring states, the rate of flow from \(i\) to \(i+1\) under weights \(\pi_i\) is \(\pi_i\lambda\), while the reverse flow is \(\pi_{i+1}\mu\). Equating these gives \(\pi_{i+1}=(\lambda/\mu)\pi_i\). Define \(\rho=\lambda/\mu\). Since \(0\le\lambda<\mu\), \(0\le\rho<1\), so the weights are proportional to \(\rho^i\).

Proof. For any integer \(m\ge0\), multiplying the finite sum by \(1-r\) gives \((1-r)\sum_{i=0}^{m}r^i=1-r^{m+1}\). Because \(0\le r<1\), \(r^{m+1}\) tends to zero as \(m\) tends to infinity. Taking the limit gives \((1-r)\sum_{i=0}^{\infty}r^i=1\), and division by \(1-r>0\) gives the stated sum. Multiplication by \(1-r\) makes the weights sum to one. At \(r=0\), the same formula gives probability \(1\) at zero, since \(0^0\) here denotes the zeroth term of the series, which is \(1\), and every positive power is zero. \(\square\)

This calculation suggests the candidate distribution \(\pi_i=(1-\rho)\rho^i\). Normalization alone is not enough to establish stationarity for a continuous-time chain. Here it can be verified directly by using a discrete transition kernel for the Poisson-clock construction.

Proof. Set \(c=\lambda+\mu>0\), and define a discrete transition kernel \(R\) by \(R=I+Q/c\), where \(I\) is the identity matrix. Its nonzero entries at state \(0\) are \(R_{00}=\mu/c\) and \(R_{01}=\lambda/c\); at each \(i\ge1\), they are \(R_{i,i-1}=\mu/c\) and \(R_{i,i+1}=\lambda/c\). The row at zero sums to \((\mu+\lambda)/c=1\), and every positive-state row also sums to \((\mu+\lambda)/c=1\). Thus \(R\) is a transition kernel. In generator terms, the row at zero sums to \(-\lambda+\lambda=0\), and each positive-state row sums to \(\mu-(\lambda+\mu)+\lambda=0\).

By the geometric normalization lemma, \(\pi\) is a probability distribution. Its neighboring probabilities satisfy \(\pi_{i+1}\mu=\pi_i\lambda\) for every \(i\ge0\). For state \(0\), the incoming mass under \(R\) is \(\pi_0R_{00}+\pi_1R_{10}=(\pi_0+\pi_1)\mu/c\). Since \(\pi_1=\rho\pi_0\), this equals \(\pi_0(1+\lambda/\mu)\mu/c=\pi_0\). For each \(j\ge1\), the incoming mass is \[ \pi_{j-1}\frac{\lambda}{c}+\pi_{j+1}\frac{\mu}{c} =\pi_j\frac{\mu}{c}+\pi_j\frac{\lambda}{c} =\pi_j. \] These identities also hold when \(\lambda=0\), since then \(\pi_1,\pi_2,\ldots\) are all zero. Therefore \(\pi R=\pi\), and induction gives \(\pi R^k=\pi\) for every integer \(k\ge0\).

Because \(Q=c(R-I)\), the Poisson-clock construction has transition semigroup \[ P(t)=e^{tQ}=e^{-ct}\sum_{k=0}^{\infty}\frac{(ct)^k}{k!}R^k. \] Indeed, the number of clock rings by time \(t\) is Poisson with mean \(ct\), and conditional on \(k\) rings the state is updated \(k\) times using \(R\); a self-transition represents a ring with no actual change. Hence \[ \pi P(t)=e^{-ct}\sum_{k=0}^{\infty}\frac{(ct)^k}{k!}\pi R^k =\pi e^{-ct}\sum_{k=0}^{\infty}\frac{(ct)^k}{k!} =\pi. \] The last sum equals \(e^{ct}\). Thus the distribution is unchanged at every \(t\ge0\), proving stationarity. \(\square\)

Worked Distributions and Their Checks

What the Geometric Formula Does—and Does Not—Say

The condition \(\lambda<\mu\) is exactly what makes the proposed weights summable: it makes \(\rho<1\). If \(\lambda=\mu>0\), the adjacent-state balance weights would be constant, and their infinite sum would diverge. If \(\lambda>\mu\), the weights proportional to \((\lambda/\mu)^i\) grow rather than decay, so they also cannot be normalized. These observations explain why this geometric construction requires \(\lambda<\mu\); they do not by themselves establish a full classification of the chain in those other regimes.

Common mistake: reversing the rate ratio. With \(\lambda=2\) and \(\mu=5\), using \(\mu/\lambda=5/2\) would give weights proportional to \((5/2)^i\). The sum of these weights diverges because each term grows, so no finite normalizing constant can turn them into probabilities. The correct flow equation is \(\pi_i\lambda=\pi_{i+1}\mu\), which gives \(\pi_{i+1}/\pi_i=\lambda/\mu=2/5\).

A second boundary error is to assign death rate \(\mu\) at state zero. There is no transition from zero to \(-1\), so the correct generator row there has \(q_{00}=-\lambda\) and \(q_{01}=\lambda\). Setting \(q_{00}=-(\lambda+\mu)\) while keeping \(q_{01}=\lambda\) makes the row sum \(-\mu\), not zero, and is not a generator for this model.

Finally, stationarity is a statement about the distribution when the initial state is sampled from \(\pi\): at every later time that distribution remains \(\pi\). It is not a claim that the chain started at a fixed state already has distribution \(\pi\), nor does the verification above prove a limiting-distribution result for arbitrary initial conditions. The central calculation here is more specific: normalize the neighboring balance weights, then verify invariance using a valid transition kernel.