Poisson Offspring and Its Generating Function

Poisson Offspring in a Galton–Watson Process

In a Galton–Watson process, each individual produces a random number of offspring, independently of the other individuals and of reproduction in other generations. When the offspring count is Poisson, its probability generating function has a particularly simple form. That function gives the offspring moments directly, and independence lets us calculate the distribution of the next generation for any fixed initial population.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let the state space be \(S=\{0,1,2,\ldots\}\), and let time be the discrete generation index \(n\ge 0\). The random variable \(Z_n\) is the number of individuals in generation \(n\). Fix an initial population \(Z_0=z\), where \(z\in S\). For each generation \(n\) and individual \(r\), let \(\xi_{n,r}\) be that individual’s number of offspring. Assume the variables \(\xi_{n,r}\) are independent and all have the Poisson distribution with parameter \(\lambda\ge 0\). Thus, for \(k\in S\),

When \(\lambda=0\), this means \(\xi_{n,r}=0\) with probability one. The population is constructed by counting the offspring of the individuals in the current generation:

where a sum with no terms is defined to be zero. Let \(\mathcal{F}_n\) contain the initial population and the reproduction variables from generations \(0,\ldots,n-1\). Then \(Z_n\) is known given \(\mathcal{F}_n\), while the offspring variables used to form \(Z_{n+1}\) are independent of \(\mathcal{F}_n\).

The Offspring Generating Function and Moments

For a nonnegative integer-valued random variable \(\xi\), its probability generating function is \(G(s)=\mathbb{E}[s^\xi]\), for \(0\le s\le 1\). For Poisson offspring, substitute the probability mass function into this definition:

Proof. The exponential power series converges for every real \(s\), so it can be differentiated term by term. Differentiating the defining series gives \(G'(s)=\sum_{k\ge1} k\mathbb{P}(\xi=k)s^{k-1}\), and \(G''(s)=\sum_{k\ge2}k(k-1)\mathbb{P}(\xi=k)s^{k-2}\). Evaluating at \(s=1\) therefore gives \(G'(1)=\mathbb{E}[\xi]\) and \(G''(1)=\mathbb{E}[\xi(\xi-1)]\). From the explicit generating function, \(G'(s)=\lambda e^{\lambda(s-1)}\) and \(G''(s)=\lambda^2e^{\lambda(s-1)}\). Hence \(G'(1)=\lambda\) and \(G''(1)=\lambda^2\). Since \(\xi^2=\xi(\xi-1)+\xi\), it follows that \(\mathbb{E}[\xi^2]=\lambda^2+\lambda\). By the definition of variance, \(\operatorname{Var}(\xi)=\mathbb{E}[\xi^2]-(\mathbb{E}[\xi])^2=\lambda^2+\lambda-\lambda^2=\lambda\). These calculations also hold at \(\lambda=0\), where \(\xi=0\) almost surely. \(\square\)

The second derivative gives the factorial second moment, not the ordinary second moment: \(G''(1)=\mathbb{E}[\xi(\xi-1)]\). To obtain \(\mathbb{E}[\xi^2]\), one must add \(\mathbb{E}[\xi]\). This distinction matters whenever a variance is computed from a generating function.

The Next Generation from a Fixed Population

The generating function also provides a direct way to find the distribution of \(Z_1\). With \(Z_0=z\) fixed, \(Z_1\) is the sum of \(z\) independent Poisson offspring counts. The probability generating function of a sum of independent nonnegative integer-valued variables is the product of their generating functions: independence factors the expectation of the product \(s^{\xi_{0,1}}\cdots s^{\xi_{0,z}}\). This gives the following result.

Proof. If \(z=0\), the sum defining \(Z_1\) has no terms, so \(Z_1=0\) with probability one. If \(\lambda=0\), every individual has zero offspring with probability one, so again \(Z_1=0\). These cases satisfy the stated result. Now suppose \(z\lambda>0\). For \(0\le s\le1\), independence of the \(z\) offspring counts gives

Expanding the final exponential as a power series gives \(e^{z\lambda(s-1)}=e^{-z\lambda}\sum_{k=0}^{\infty}(z\lambda)^k s^k/k!\). Its coefficient of \(s^k\) is \(e^{-z\lambda}(z\lambda)^k/k!\), which is the probability \(\mathbb{P}(Z_1=k)\). Thus \(Z_1\) is Poisson with parameter \(z\lambda\). Applying the Poisson moment formulas already derived, with parameter \(z\lambda\), gives \(\mathbb{E}[Z_1]=z\lambda\) and \(\operatorname{Var}(Z_1)=z\lambda\). For the zero-parameter cases, both quantities are zero, as stated. \(\square\)

This result concerns the population after one reproduction step from a fixed initial size. It accounts for all \(z\) individuals: the parameter of the total offspring distribution is \(z\lambda\), not just \(\lambda\).

A Common Mistake

A common error is to use the offspring distribution for one individual as though it were the distribution of the next generation, even when the initial population contains several individuals. For example, if \(Z_0=2\) and \(\lambda=\tfrac12\), one individual has probability \(e^{-1/2}\) of having no offspring. It would be wrong to conclude that \(\mathbb{P}(Z_1=0)=e^{-1/2}\). For the whole next generation to be empty, both initial individuals must have no offspring. Independence gives

The same answer follows from the total parameter \(z\lambda=2\cdot\tfrac12=1\), since a Poisson random variable with parameter \(1\) is zero with probability \(e^{-1}\). The individual’s generating function must be raised to the power \(z\) to account for all \(z\) initial individuals.

The calculations here depend on independence and a fixed initial population. If the number of initial individuals is itself random, the distribution of the next generation must also account for that randomness; it is not obtained by substituting a non-integer average population into the fixed-\(z\) formula. For a fixed \(z\), the practical sequence is simple: compute \(z\lambda\), use the Poisson probability formula for the requested count, and check the parameter and the boundary case \(z\lambda=0\).