Closed-Form Variances for Generation Sizes

From the Recursion to a Closed Form

Knowing the expected size of each generation does not describe how much that size varies from one realization of a branching process to another. The variance answers that question. Once the offspring distribution has finite variance, the generation-size variances satisfy a simple recursion. Solving it requires care because the forcing term depends on the generation mean, and the answer changes form at the critical mean \(m=1\).

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). In a Galton–Watson process, each individual independently produces a number of children with the same nonnegative integer-valued distribution \(\xi\). Write \(m=\mathbb{E}[\xi]\) and \(\sigma^2=\operatorname{Var}(\xi)\), and assume \(\mathbb{E}[\xi^2]<\infty\). Let \(Z_n\) be the number of individuals in generation \(n\), starting with one ancestor, so \(Z_0=1\). Let \(\mathcal{F}_n\) contain the branching information through generation \(n\). Conditional on \(\mathcal{F}_n\), the offspring counts of the \(Z_n\) individuals in generation \(n\) are independent copies of \(\xi\), and their sum is \(Z_{n+1}\).

Here is the conditional-moment argument behind the recursion. Given \(\mathcal{F}_n\), \(Z_{n+1}\) is a sum of \(Z_n\) independent offspring counts. If \(Z_n=0\), that is an empty sum, with mean and variance both zero. In all cases, therefore, \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=mZ_n\) and \(\operatorname{Var}(Z_{n+1}\mid\mathcal{F}_n)=\sigma^2Z_n\). Taking expectations gives \(\mathbb{E}[Z_{n+1}]=m\mathbb{E}[Z_n]\), and induction from \(Z_0=1\) gives \(m^n\).

For the variance, decompose \(Z_{n+1}-\mathbb{E}[Z_{n+1}]\) into \(Z_{n+1}-\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]\) and \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]-\mathbb{E}[Z_{n+1}]\). The conditional expectation of the first term given \(\mathcal{F}_n\) is zero, so the cross term has expectation zero. Taking the expected squares gives the conditional-variance decomposition \(\operatorname{Var}(Z_{n+1})=\mathbb{E}[\operatorname{Var}(Z_{n+1}\mid\mathcal{F}_n)]+\operatorname{Var}(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n])\). Substituting the conditional moments and \(\mathbb{E}[Z_n]=m^n\) yields the stated recursion.

Proof. First suppose \(m>0\) and \(m\ne1\). Repeated substitution in \(v_{n+1}=m^2v_n+\sigma^2m^n\), starting from \(v_0=0\), gives for \(n\ge1\) \(v_n=\sigma^2\sum_{k=0}^{n-1}m^{2(n-1-k)}m^k\). The term indexed by \(k\) has exponent \(2n-2-k\). As \(k\) runs from \(0\) to \(n-1\), those exponents run from \(2n-2\) down to \(n-1\). Thus the sum equals \(\sigma^2m^{n-1}\sum_{r=0}^{n-1}m^r\). Multiplying the finite geometric sum by \(m-1\) gives \((m-1)\sum_{r=0}^{n-1}m^r=m^n-1\), so the claimed formula follows. At \(n=0\), the variance is \(v_0=0\) by definition. Next suppose \(m=1\). The recursion becomes \(v_{n+1}=v_n+\sigma^2\); induction from \(v_0=0\) gives \(v_n=n\sigma^2\). Finally, if \(m=0\), then \(\xi\ge0\) and \(\mathbb{E}[\xi]=0\) imply \(\xi=0\) almost surely: any positive probability of \(\xi\ge1\) would make its expectation positive. There are no individuals after generation zero, so \(Z_n=0\) almost surely and \(v_n=0\) for \(n\ge1\). \(\square\)

The separate critical expression is not a technical exception to be filled in by blindly substituting \(m=1\) into a fraction with denominator \(m-1\). It comes from summing \(n\) equal contributions in the recursion. In the noncritical case, each generation’s offspring noise is multiplied by a different power of \(m\). That weighting is what produces the geometric sum.

Worked Variance Calculations

Proof. The definition of variance gives \(\operatorname{Var}(Z_n)=\mathbb{E}[Z_n^2]-(\mathbb{E}[Z_n])^2\). Rearranging and using \(\mathbb{E}[Z_n]=m^n\) yields \(\mathbb{E}[Z_n^2]=v_n+m^{2n}\) when \(m>0\), and the theorem supplies \(v_n\) in both the critical and noncritical cases. When \(m=0\), the process starts with \(Z_0=1\) and has \(Z_n=0\) almost surely for \(n\ge1\), as established in the theorem’s proof. Squaring these values gives the stated second moments. \(\square\)

Interpreting and Checking the Formulas

The variance measures spread around the mean, not the mean itself. Two offspring laws can have the same \(m\), and hence the same expected generation sizes \(m^n\), but different \(\sigma^2\), and therefore different generation-size variances. In every formula here, \(\sigma^2\) is the variance of one individual’s offspring count; it is not the variance of \(Z_n\). The recursion carries this one-generation randomness forward through the factor \(m^2\), while the new offspring noise contributes \(\sigma^2m^n\).

The boundary cases are useful checks. At \(n=0\), \(Z_0=1\) deterministically, so every valid variance formula must give zero. At \(m=1\), the recursion adds \(\sigma^2\) at each generation and gives \(n\sigma^2\). If \(\sigma^2=0\), offspring counts are deterministic, so all generation sizes are deterministic and the formulas give zero variance. When \(m=0\), the process ends after its initial ancestor; treating that case separately avoids ambiguous powers such as \(0^0\).

Common mistake: using the wrong geometric ratio. For the Poisson offspring law with parameter \(2\), a tempting but incorrect calculation is \(v_3=2(1+4+16)=42\), treating the contributions as successive powers of \(m^2=4\). That does not follow from the recursion: the added term at step \(n+1\) is \(\sigma^2m^n\), not a constant. The actual expansion is \(v_3=\sigma^2(m^4+m^3+m^2)=2(16+8+4)=56\). Substituting into the final recursion also verifies the correction: \(m^2v_2+\sigma^2m^2=4\cdot12+2\cdot4=56\), whereas \(42\) would not satisfy it.

For practical calculations, first compute \(m\) and \(\sigma^2\) from the offspring law, then choose the matching case \(m=0\), \(m=1\), or \(m>0\) with \(m\ne1\). Check the generation index: the formula is for \(Z_n\) starting from \(Z_0=1\), not for a process with a different initial population. Finally, substitute the result into \(v_{n+1}=m^2v_n+\sigma^2m^n\) at a generation where both sides can be computed.