The Immigration-Death Process And Its Poisson Law

The Immigration-Death Model

Suppose individuals enter a population at a constant rate, and each individual dies independently at a constant rate. The total death rate then depends on the current population: with \(n\) individuals present, any one of the \(n\) can die. This simple difference between a per-individual rate and a total rate leads to a stationary law that is Poisson, rather than the geometric law associated with constant birth and death rates.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). The state space is \(S=\{0,1,2,\ldots\}\), and \((X_t)_{t\ge0}\) is a continuous-time Markov chain with its natural filtration \((\mathcal{F}_t)_{t\ge0}\). Immigration occurs at rate \(\nu>0\); each individual has an independent exponential lifetime with rate \(\mu>0\). Thus the process has a transition from \(n\) to \(n+1\) at rate \(\nu\), and from \(n\) to \(n-1\) at rate \(n\mu\) when \(n\ge1\). There are no other transitions.

This rate description defines a nonexplosive process. Starting from any state \(n<\infty\), there are only finitely many immigrants on every bounded time interval, almost surely, because immigration is a Poisson process. Each individual present initially or arriving during that interval can die at most once. Consequently, the total number of jumps on that interval is finite almost surely.

Detailed Balance Gives the Weights

For a birth-death process, detailed balance across neighboring states means that the stationary probability flow from \(n\) to \(n+1\) equals the flow back from \(n+1\) to \(n\). Write \(\pi_n\) for the proposed stationary probability of state \(n\). The equations are \(\pi_n q_{n,n+1}=\pi_{n+1}q_{n+1,n}\). Here they become \(\pi_n\nu=\pi_{n+1}(n+1)\mu\). The factor \(n+1\) is essential: each of the \(n+1\) individuals can die.

Proof. The detailed-balance equation at \(n\) gives \(\pi_{n+1}=\pi_n\frac{\nu}{(n+1)\mu}\). Starting from \(\pi_0\) and applying this recurrence successively yields \(\pi_n=\pi_0\frac{(\nu/\mu)^n}{n!}\) for every \(n\ge0\). This includes \(n=0\), and the recurrence determines every later term. Put \(a=\nu/\mu\). The exponential series gives \(\sum_{n=0}^{\infty}a^n/n!=e^a\), so normalization requires \(1=\sum_n\pi_n=\pi_0e^a\), or \(\pi_0=e^{-a}\). This proves the displayed formula and its normalization. It also proves uniqueness among normalized sequences satisfying these detailed-balance equations. To see that this law is stationary, use the individual-lifetime construction. If \(X_0\) has this Poisson law, each initial individual survives until time \(t\) with probability \(p=e^{-\mu t}\), independently. Thinning a Poisson count of mean \(a\) by independent survival probability \(p\) gives a Poisson count of mean \(ap\): its probability generating function is \(\exp(a((1-p+pz)-1))=\exp(ap(z-1))\). Immigrants arriving by time \(t\) survive to time \(t\) independently, with survival probability \(e^{-\mu(t-s)}\) for an arrival at time \(s\). Their surviving count is Poisson with mean \(\nu\int_0^t e^{-\mu(t-s)}\,ds=a(1-e^{-\mu t})\). For completeness, the Poisson marking calculation follows by conditioning on the number of arrivals: given \(m\) arrivals, their times are independent uniform points in \([0,t]\); averaging the product of their survival generating functions and then averaging over a Poisson number with mean \(\nu t\) gives the generating function \(\exp(\nu\int_0^t e^{-\mu(t-s)}(z-1)\,ds)\). Initial survivors and surviving immigrants are independent. Their sum therefore has generating function \(\exp(ap(z-1))\exp(a(1-p)(z-1))=\exp(a(z-1))\), the generating function of the original Poisson law. Thus the distribution of \(X_t\) is \(\pi\) whenever \(X_0\) has distribution \(\pi\), for every \(t\ge0\). \(\square\)

The proposition identifies an invariant law: when the process starts with this distribution, its one-time marginal distribution remains unchanged. This does not mean that the population stays fixed along each sample path; births and deaths continue to occur. The stationary mean and variance can be read from the Poisson probabilities, as shown below.

Mean and Variance

Let \(X\) have the stationary distribution just derived, so \(X\) is Poisson with parameter \(a=\nu/\mu\). For \(a>0\), its expectation follows by shifting the index in the defining series: \(\mathbb{E}[X]=e^{-a}\sum_{n=1}^{\infty}n a^n/n!=ae^{-a}\sum_{m=0}^{\infty}a^m/m!=a\). Likewise, \(n(n-1)/n!=1/(n-2)!\) for \(n\ge2\), so \(\mathbb{E}[X(X-1)]=a^2\). Since \(X^2=X(X-1)+X\), it follows that \(\mathbb{E}[X^2]=a^2+a\), and hence \(\operatorname{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X])^2=a\).

A Finite-Time Calculation

Stationarity concerns the distribution when the process starts in equilibrium. From a fixed initial population, the distribution at a later time is generally different. The same survival argument gives a direct calculation: each of the \(X_0\) initial individuals survives independently with probability \(e^{-\mu t}\), and surviving immigrants form an independent Poisson count with mean \((\nu/\mu)(1-e^{-\mu t})\).

Common Mistakes and Boundary Cases

A common mistake is to use \(\mu\), rather than \(n\mu\), as the total death rate from state \(n\). For a concrete check, take \(\nu=\mu=1\). The correct detailed-balance equation between states 1 and 2 is \(\pi_1\cdot1=\pi_2\cdot2\), so it requires \(\pi_2=\pi_1/2\). If one incorrectly uses a constant death rate of 1, one obtains \(\pi_2=\pi_1\), which fails the actual balance equation. The per-individual death rate is \(\mu\); the total rate is \(n\mu\).

The assumption \(\mu>0\) matters for the Poisson parameter \(\nu/\mu\). If \(\nu=0\) and \(\mu>0\), state 0 is absorbing, and the Poisson law with parameter zero is interpreted as unit mass at 0. If both rates vanish, every state is absorbing. The main derivation assumes \(\nu,\mu>0\), so it does not divide by zero. In all cases, verify the rates at the boundary state 0: there is no transition to state \(-1\).

Finally, a stationary distribution is a probability law preserved by the transition dynamics when used as the initial law. The Poisson calculation proves this preservation for the immigration-death process; it does not say that every initial condition already has that law, or that the population path is constant.