Equal Offspring Means Can Give Different Extinction Risks

Why the Mean Is Not Enough

The mean number of children per individual describes average population growth, but it does not specify how offspring are distributed. A population in which every individual has one or two children behaves differently from one in which some individuals have no children and others have many—even if the average number of children is identical. Extinction depends on the distribution of offspring, not just its mean.

We study a Galton–Watson process \((Z_n)_{n\ge0}\) on the state space \(S=\{0,1,2,\ldots\}\), in discrete time, on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Start with one individual, \(Z_0=1\). Each individual independently has \(\xi\) children, where \(\xi\) is a nonnegative integer-valued random variable with probabilities \(p_k=\mathbb{P}(\xi=k)\). Thus \(Z_{n+1}\) is the sum of the offspring counts of the \(Z_n\) individuals in generation \(n\). We use the generation filtration \(\mathcal{F}_n\), containing the reproduction information through generation \(n-1\).

The mean is the derivative of \(G\) at \(1\) when it is finite: \(G'(1)=m\). This single value describes the slope at the endpoint \(s=1\); extinction probability, by contrast, is determined by the behavior of the whole function on \([0,1]\).

Extinction as a Fixed Point

We first establish the fixed-point fact used in the comparisons. Write \(q_n=\mathbb{P}(Z_n=0)\), the probability that extinction has occurred by generation \(n\). Conditioning on the number of children of the initial individual shows that \(q_n=G(q_{n-1})\) for \(n\ge1\), with \(q_0=0\). This gives a sequence of iterates of \(G\), starting at zero.

Proof. Let \(q_n=\mathbb{P}(Z_n=0)\). For \(n=0\), \(q_0=0\), since \(Z_0=1\). Given that the initial individual has \(k\) children, extinction by generation \(n\) occurs exactly when each of the \(k\) resulting family lines is extinct by generation \(n-1\). Those lines reproduce independently, so the conditional probability is \(q_{n-1}^k\). Averaging over \(k\) gives \(q_n=\sum_{k\ge0}p_kq_{n-1}^k=G(q_{n-1})\). The events \(\{Z_n=0\}\) increase with \(n\), because zero is absorbing, and their union is eventual extinction. Therefore \(q_n\) increases to \(q\). The function \(G\) is continuous on \([0,1]\): for any cutoff \(K\), its first \(K+1\) terms form a continuous polynomial, and the remaining sum is bounded in absolute value by \(\sum_{k>K}p_k\), which tends to zero as \(K\) increases. Taking limits in \(q_n=G(q_{n-1})\) gives \(q=G(q)\). Now let \(r\in[0,1]\) be any fixed point. The function \(G\) is nondecreasing on \([0,1]\), since each \(s^k\) is nondecreasing there. Starting with \(q_0=0\le r\), induction gives \(q_n=G(q_{n-1})\le G(r)=r\) for every \(n\). Taking limits yields \(q\le r\), so \(q\) is the smallest fixed point. \(\square\)

This result makes a generating-function comparison possible. It also explains why selecting a fixed point without checking which one is smallest can give the wrong extinction probability.

Proof. For \(\ell=1,2\), define \(q_{\ell,0}=0\) and \(q_{\ell,n+1}=G_\ell(q_{\ell,n})\). These iterates increase to \(q_\ell\), by the fixed-point argument above. We prove \(q_{1,n}\le q_{2,n}\) by induction. It holds at \(n=0\). If it holds at \(n\), monotonicity of \(G_1\), followed by the assumed pointwise order, gives \(q_{1,n+1}=G_1(q_{1,n})\le G_1(q_{2,n})\le G_2(q_{2,n})=q_{2,n+1}\). Thus the inequality holds for every \(n\). Taking limits gives \(q_1\le q_2\). \(\square\)

The comparison requires an order of the generating functions on the entire interval, not merely equal means or a comparison of their slopes at \(1\). The following examples show how the full functions determine the answer.

What the Comparison Does—and Does Not—Say

For the first and second laws, the generating functions can be ordered at every \(s\in[0,1]\); the comparison theorem then orders their extinction probabilities. In general, equal means alone do not supply such an order. Equal means say that the generating functions have the same derivative at \(1\), when the derivatives exist. They do not require the functions to coincide or to be ordered below \(1\). The three exact calculations make the consequence explicit: mean \(2\) is compatible with certain survival as well as with two distinct positive extinction probabilities.

Common mistake. A tempting but incorrect shortcut is to set \(q=1/m\). For the first law, \(m=2\), so that shortcut predicts \(q=\tfrac12\). But the exact calculation gives \(q=(\sqrt{3}-1)/2\), which is not \(\tfrac12\): equality would imply \(\sqrt{3}=2\), contradicting \(3\ne4\). For the third law the same shortcut predicts \(\tfrac12\), while the correct extinction probability is \(0\). The mean alone cannot replace the fixed-point calculation.

When comparing offspring laws, first compute each generating function and verify that the offspring probabilities sum to one and that the claimed means are equal. Then solve \(G(s)=s\), retain the smallest solution in \([0,1]\), and use a pointwise generating-function comparison when one is available. A comparison of means, or even the observation that two curves meet at \(s=1\), is not enough to establish an ordering of extinction risks.