Deriving the Variance Recursion

Why Condition on the Current Generation?

In a Galton–Watson process, each individual produces a random number of children, and the next generation is the sum of those offspring counts. The size \(Z_n\) of the current generation therefore controls both the expected size and the variability of the next one. Conditioning on the current generation makes that relationship precise: the conditional variance accumulates across individuals, while variation already present in \(Z_n\) is carried forward through the offspring mean.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space. For this tutorial, take \(Z_0=z_0\) to be a fixed finite nonnegative integer. Let \(\xi_{n,r}\), for \(n\ge 0\) and \(r\ge 1\), be independent, identically distributed, nonnegative integer-valued random variables. The variable \(\xi_{n,r}\) is the number of children of the \(r\)-th individual in generation \(n\). Assume that the offspring distribution has finite second moment, and write

The generation sizes are defined recursively by \(Z_{n+1}=\sum_{r=1}^{Z_n}\xi_{n,r}\), with an empty sum equal to zero. Let \(\mathcal{F}_n\) be the information from the process through generation \(n\), before the offspring of generation \(n\) are used. In particular, \(Z_n\) is \(\mathcal{F}_n\)-measurable, and the offspring variables \(\xi_{n,r}\) are independent of \(\mathcal{F}_n\).

The Conditional Moment Calculation

For any fixed integer \(k\ge 0\), the sum of \(k\) independent offspring counts has mean \(km\) and variance \(k\sigma^2\). Its second moment is the sum of its variance and the square of its mean. This also holds when \(k=0\), since the sum is then zero.

Proof. Conditional on \(\mathcal{F}_n\), the value \(Z_n=k\) is fixed, and the offspring variables \(\xi_{n,1},\ldots,\xi_{n,k}\) are independent, each with mean \(m\) and variance \(\sigma^2\). Thus the conditional mean of their sum is \(km\), and its conditional variance is \(k\sigma^2\). Its conditional second moment is \(k\sigma^2+(km)^2\). These expressions apply at \(k=0\) as well: each is zero. Replacing \(k\) by the \(\mathcal{F}_n\)-measurable variable \(Z_n\) gives the three stated conditional identities. \(\square\)

The conditional second-moment identity is important for the order of the proof. The law of total variance is used for square-integrable random variables. We first use the conditional second moment to show that \(Z_{n+1}\) is square-integrable; only then do we apply total variance.

The Variance Recursion

Proof. We first establish that \(\mathbb{E}[Z_n^2]<\infty\) for every \(n\). It holds at \(n=0\), because \(Z_0=z_0\) is fixed. Suppose it holds at some \(n\). Since \(Z_n\) is nonnegative and \(Z_n\le 1+Z_n^2\), \(\mathbb{E}[Z_n]\) is finite. Taking expectations of the conditional second-moment identity gives \(\mathbb{E}[Z_{n+1}^2]=\sigma^2\mathbb{E}[Z_n]+m^2\mathbb{E}[Z_n^2]<\infty\). This proves the required square-integrability by induction.

Now \(Z_n\) and \(Z_{n+1}\) are square-integrable, so total variance applies. Using the conditional variance and conditional mean from the proposition, \(\operatorname{Var}(Z_{n+1})=\mathbb{E}[\operatorname{Var}(Z_{n+1}\mid\mathcal{F}_n)]+\operatorname{Var}(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n])=\sigma^2\mathbb{E}[Z_n]+\operatorname{Var}(mZ_n)\). Since \(m\) is a constant, \(\operatorname{Var}(mZ_n)=m^2\operatorname{Var}(Z_n)\). Substitution proves the recursion. \(\square\)

The same conditional-mean calculation gives \(\mathbb{E}[Z_{n+1}]=m\mathbb{E}[Z_n]\). The variance recursion uses this expected generation size as an input, but it does not require solving the recursion across generations.

Why Independent Offspring Matter

Conditional on \(Z_n=k\), the next generation is a sum of \(k\) offspring counts. If those counts are independent, the conditional variance is \(k\sigma^2\). If they are dependent, the variance of the sum also contains covariance terms. The formula for a Galton–Watson process cannot be transferred unchanged to a model with shared or correlated reproduction.

Common pitfall. Applying total variance before establishing that \(Z_{n+1}\) is square-integrable leaves a gap: the induction hypothesis may give \(\mathbb{E}[Z_n^2]<\infty\), but that alone does not yet state that \(\mathbb{E}[Z_{n+1}^2]<\infty\). The repair is to first take expectations in \(\mathbb{E}[Z_{n+1}^2\mid\mathcal{F}_n]=\sigma^2Z_n+m^2Z_n^2\). The resulting finite value proves square-integrability of \(Z_{n+1}\); total variance can then be applied without circularity.

The recursion also makes the roles of the assumptions visible. Finite offspring variance ensures that the conditional second moment is finite when \(Z_n\) has finite second moment. Independent reproduction gives the term \(\sigma^2Z_n\), rather than a sum containing covariance contributions. If \(Z_n=0\), both terms on the right-hand side of the variance recursion are zero, consistent with \(Z_{n+1}=0\) with certainty.