Common Errors in Generation Moments and Extinction Calculations

Keep the Random Variables Straight

In a Galton–Watson process, offspring are sampled for each individual, while generation sizes are random sums of those offspring. Confusing the variance of one individual's offspring with the variance of a whole generation, or replacing a conditional mean with an unconditional one, can produce answers that look plausible but are wrong. Extinction calculations have a related trap: the fixed-point equation can have more than one solution, and the extinction probability is not selected by solving that equation alone.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \(Z_n\) be the number of individuals in generation \(n\), with \(Z_0=1\). Each individual independently produces a nonnegative integer number of offspring with distribution \(\mathbb{P}(\xi=r)=p_r\), where \(\sum_{r=0}^{\infty}p_r=1\). The offspring generating function is \(G(s)=\mathbb{E}[s^\xi]=\sum_{r=0}^{\infty}p_rs^r\) for \(0\le s\le1\). Write \(m=\mathbb{E}\xi\) and, when finite, \(\sigma^2=\operatorname{Var}(\xi)\). Let \(\mathcal{F}_n\) record the reproduction through generation \(n-1\), so \(Z_n\) is known given \(\mathcal{F}_n\). Reproduction by the \(Z_n\) individuals is independent of \(\mathcal{F}_n\).

Proof. Given \(\mathcal{F}_n\), let \(\xi_1,\ldots,\xi_{Z_n}\) denote the offspring counts of the individuals in generation \(n\). Then \(Z_{n+1}=\sum_{r=1}^{Z_n}\xi_r\). If \(Z_n=k\), this is a sum of \(k\) independent copies of \(\xi\), so its conditional mean is \(km\), and its conditional variance is \(k\sigma^2\). This also covers \(k=0\): the sum is empty and equals zero, so both conditional moments are zero. Since \(Z_n\) is \(\mathcal{F}_n\)-measurable, the stated identities follow. \(\square\)

Taking expectations in the first identity gives \(\mathbb{E}Z_{n+1}=m\mathbb{E}Z_n\), hence \(\mathbb{E}Z_n=m^n\) when \(Z_0=1\). But the conditional variance identity cannot be used as if it were the unconditional variance. The law of total variance, when the second moments are finite, says

Substituting the conditional identities gives the generation-variance recursion. Its two terms have distinct sources: the first is the average variation in offspring counts within a generation, and the second is variation already present in the size of the current generation.

Proof. The law of total variance and the conditional identities give \(\operatorname{Var}(Z_{n+1})=\mathbb{E}[\sigma^2Z_n]+\operatorname{Var}(mZ_n)=\sigma^2\mathbb{E}Z_n+m^2v_n\). Since \(\mathbb{E}Z_n=m^n\), the recursion follows, with \(v_0=0\). If \(m=1\), repeated substitution adds \(\sigma^2\) at each step, so \(v_n=n\sigma^2\). If \(m>0\) and \(m\ne1\), repeated substitution gives \(v_n=\sigma^2\sum_{j=0}^{n-1}m^{2(n-1-j)}m^j=\sigma^2\sum_{j=0}^{n-1}m^{2n-2-j}\). This finite geometric sum equals \(\sigma^2m^{n-1}(m^n-1)/(m-1)\). Finally, if \(m=0\), nonnegativity of \(\xi\) and \(\mathbb{E}\xi=0\) imply \(\mathbb{P}(\xi=0)=1\); every generation after generation zero is empty. \(\square\)

This calculation illustrates why every conditioning step should keep the number of summands explicit. Given \(Z_1=2\), there are two offspring counts to add, not four; each count is zero or two. The total-variance calculation is usually shorter, but direct computation is a useful independent check.

Fixed Points Need a Root-Selection Rule

Let \(q=\mathbb{P}(Z_n=0\text{ for some }n\ge0)\) when \(Z_0=1\). Since zero is absorbing, this is the probability of eventual extinction. Conditioning on the first individual's offspring gives the fixed-point equation \(q=G(q)\): if that individual has \(r\) children, extinction requires extinction of each of the \(r\) independent descendant processes, with probability \(q^r\).

Proof. Define \(q_n=\mathbb{P}(Z_n=0)\), with \(q_0=0\) because \(Z_0=1\). Given that the first individual has \(r\) children, all \(r\) descendant populations must be zero by generation \(n\) for \(Z_{n+1}=0\). Independence gives \(q_{n+1}=\sum_{r=0}^{\infty}p_rq_n^r=G(q_n)\), where the \(r=0\) term is \(p_0\). The events \(\{Z_n=0\}\) increase with \(n\), so \(q_n\) increases to \(q\), the probability of eventual extinction. The generating function is continuous on \([0,1]\), so taking limits in \(q_{n+1}=G(q_n)\) gives \(q=G(q)\). Now let \(s\in[0,1]\) be any other fixed point. Since \(q_0=0\le s\), and \(G\) is nondecreasing on \([0,1]\), induction gives \(q_n\le s\) for every \(n\): if \(q_n\le s\), then \(q_{n+1}=G(q_n)\le G(s)=s\). Taking limits yields \(q\le s\). Thus \(q\) is the least fixed point. \(\square\)

The theorem supplies the root-selection rule even when the fixed-point equation has several roots: choose the least one in \([0,1]\), not an arbitrary algebraic solution. Iteration from \(q_0=0\) also explains the rule computationally. A numerical solver that returns a different fixed point has not, by that fact alone, found the extinction probability.

The Deterministic One-Child Exception

A common shortcut is to assert that mean offspring number \(m=1\) implies certain extinction. That statement needs a nondegeneracy condition: the deterministic law \(\xi=1\) is an exception. Here \(G(s)=s\), so every \(s\in[0,1]\) is a fixed point. The least-fixed-point theorem still applies and gives \(q=0\); the population remains exactly one forever. In particular, the fixed-point equation alone does not imply \(q=1\), and a rule that selects the root \(1\) whenever \(m=1\) fails in this case.

Common Errors and a Reliable Check

Three errors account for many incorrect calculations. First, confusing \(\operatorname{Var}(\xi)\) with \(\operatorname{Var}(Z_n)\) ignores the random number of individuals contributing offspring. In the one-or-three offspring example, \(\operatorname{Var}(\xi)=1\), but \(\operatorname{Var}(Z_2)=6\). Second, using only \(\mathbb{E}[\operatorname{Var}(Z_{n+1}\mid\mathcal{F}_n)]\) drops the variation in the conditional mean. In that same example, the two total-variance terms at generation two are \(2\) and \(4\); keeping only the first gives \(2\), not \(6\). Third, solving \(s=G(s)\) without identifying the least root can confuse a fixed point with the extinction probability.

A practical calculation should make the conditioning visible: write \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=mZ_n\) and \(\operatorname{Var}(Z_{n+1}\mid\mathcal{F}_n)=\sigma^2Z_n\) before taking unconditional moments. For extinction, list the fixed points in \([0,1]\) and then apply the least-fixed-point characterization. Finally, check whether the offspring law is deterministic with exactly one child; that process has mean one but never becomes extinct.