A branching population can restart through outside arrivals. Derive its count recursion, a Poisson equilibrium, and why that equilibrium is a special case.
Branching process with immigration
A branching process with immigration is a population model in which each existing individual produces a random number of descendants and an independent stream of outsiders joins each generation. It extends the ordinary branching process: even when every existing family eventually dies out, repeated outside arrivals can sustain a nonzero long-run population.
For a quant building a count model, this separates two mechanisms: activity carried forward from existing units and fresh external demand. The distinction matters when forecasting whether a quiet period ends the process or merely pauses it.
Two sources of the next count
Let Xn be the nonnegative integer population in generation n. Write
Xn+1=i=1∑Xnξn+1,i+In+1.
The offspring counts ξn,i are independent and identically distributed across individuals and generations; the immigration counts In are also i.i.d. and independent of all offspring and the initial population. New immigrants enter the next generation and begin reproducing at the following transition. An empty sum equals zero. These timing and independence assumptions define the discrete-generation model considered here. Singh, §5.2 develops this immigration extension.
With finite offspring mean m=E[ξ] and immigration mean λ=E[I],
E[Xn+1∣Xn]=mXn+λ,E[Xn+1]=mE[Xn]+λ.
When m<1, this recursion gives
E[Xn]=mnE[X0]+λ1−m1−mn⟶1−mλ.
This is a balance between fading descendants and replenishment. Under these finite-mean, subcritical assumptions a stationary law exists; the limiting mean alone does not identify its shape. The single-type result is covered by the more general construction and moment analysis in Barczy, Nedényi and Pap, §§2 and 4.
Half persist; two arrive on average
Consider a fictional demand population. Each current unit independently contributes either one unit to the next generation, with probability 0.5, or none. Independently, a Poisson-distributed number of fresh units arrives, with mean 2 per generation:
ξ∼Bernoulli(0.5),I∼Poisson(2),X0=0.
This is thinning: no unit creates more than one descendant. The first expected counts are
0⟶2⟶3⟶3.5,E[Xn]=4(1−2−n)⟶4.
The expectation can be fractional although every realised count is an integer. Nor does a limiting mean of four mean the population eventually stays at four.
Exact fictional Bernoulli 0.5 offspring and Poisson 2 immigration model. From an initial count of zero, expected counts are 0, 2, 3 and 3.5, tending to 4. The stationary Poisson 4 distribution has variance 4 and probability zero 1.83 percent; from zero the next count is positive with probability 86.47 percent.
Exact calculations for a fictional count model. The upper panel shows expectations, not a realised path. The lower panel shows the stationary Poisson(4) probability mass, grouping the entire 11-or-more tail.
Here we can identify the full stationary distribution. Let Gn(s)=EsXn] be the population’s [probability generating function, with offspring PGF f and immigration PGF h. Independence gives Gn+1(s)=Gn(f(s))h(s). In this example, f(s)=0.5+0.5s and h(s)=exp(2(s−1)). Substituting a Poisson(4) PGF yields
exp(4((0.5+0.5s)−1))exp(2(s−1))=exp(4(s−1)).
The law is preserved by one transition. More strongly, starting from zero, the same calculation inductively gives Xn∼Poisson(4(1−2−n)), so the distributions converge to Poisson(4).
Empty does not mean finished
With no immigration, zero is absorbing. With P(I>0)>0, an empty population can restart: Xn+1=In+1 whenever Xn=0. In the example,
P(Xn+1>0∣Xn=0)=1−e−2≈86.47%.
Yet a zero count remains possible: in equilibrium its probability is e−4≈1.83%. Immigration removes permanent absorption at zero; it does not guarantee a positive count every generation. Extinction of a particular family and emptiness of the entire replenished population are different events.
The Poisson equilibrium is also special. If offspring instead follow Poisson(0.5), keeping the same immigration, the stationary mean remains four but the stationary variance becomes 16/3≈5.33, rather than four. To see why, for finite offspring variance σ2 and immigration variance τ2, conditional variance gives
Var(Xn+1)=m2Var(Xn)+σ2E[Xn]+τ2.
At stationarity this gives (σ2λ/(1−m)+τ2)/(1−m2). Bernoulli(0.5) has variance 1/4; Poisson(0.5) has variance 1/2. Equal reproduction means therefore need not imply equal count risk.
The demand interpretation is illustrative, not an estimated trading model. Common shocks, changing arrival rates, finite capacity or dependence between existing demand and new arrivals can invalidate these assumptions. A model with multiple descendants also describes propagation differently from simple persistence.
