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Condition on the current generation size. By independence of offspring families, $\mathbb{E}[s^{Z_{n+1}}\mid Z_n=k]=G(s{)}^k$. Taking the expected value gives the probability generating function recursionInduction therefore proves the branching-process generating-function iterationwhere the right side means $n$-fold composition, not an ordinary power.

Let $m=G^{\prime }(1)$ and ${\sigma }^2=G^{\prime \prime }(1)+m-m^2$ be the offspring mean and variance, assuming a finite second moment. Differentiating the generating-function recursion at one gives $\mathbb{E}{Z}_n=m^n$ and, with $a_n=G_n^{\prime \prime }(1)$,Since $\mathrm{Var}(Z_n)=a_n+m^n-m^{2n}$, this impliesIteration and summation of the geometric progression give the mean and variance of a Galton-Watson generation:For $m=0$ the offspring count is identically zero, so every later generation has zero variance. The zeroth generation always has zero variance. Finiteness of the needed moments is an assumption for this finite-valued variance formula, not a consequence of merely having a probability generating function. With finite positive mean but infinite offspring variance, all positive-generation variances are infinite.

For the total count, make the generation convention explicit. If the first $n$ generations include the ancestor, put $T_n=Z_0+\cdots +Z_{n-1}$ and $H_n(s)=\mathbb{E}[s^{T_n}]$, with $T_0=0$. Given $k$ children of the ancestor, each starts an independent depth-$n$ descendant tree with generating function $H_n$. The ancestor contributes an additional factor $s$, soThis is the finite-horizon total progeny generating function. If instead “first $n$ generations” means descendants $Z_1+\cdots +Z_n$, excluding $Z_0$, write ${\hat{H}}_n$ for that convention. Conditioning on the children now gives ${\hat{H}}_{n+1}(s)=G(s{\hat{H}}_n(s))$, with ${\hat{H}}_0=1$ and ${\hat{H}}_1=G$. The two conventions encode the same tree but different counts.