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Let $q\in (0,1]$ be the branching extinction probability of a Galton-Watson process with offspring probability generating function $f$. Conditional on extinction, the probability of $j$ children is multiplied by $q^j/q$: every child's descendants must also become extinct. Hence the new probability generating function is $f_q$ and its mean is $f^{\prime }(q)$. If $f^{\prime }(q)<1$, the total progeny of a branching process has conditional expected value $1/(1-f^{\prime }(q))$, and unconditionally $\mathbb{E}[T;T<\infty ]=q/(1-f^{\prime }(q))$. The Markov inequality gives a finite-total-progeny tail bound without discarding the survival mass.

For a Galton-Watson process starting with one ancestor and finite offspring mean $\mu$, the expected value of the generation size is $\mathbb{E}[Z_n]={\mu }^n$. At $n=0$ it is one, including when $\mu =0$; zero offspring mean forces every later generation to be empty. With $m$ initial ancestors, the expected value is $m{\mu }^n$.

For $p_0=p_1=1/4$, $p_2=1/2$, the offspring mean is $5/4$. The exact calculations used in the generation probability mass function figure under Galton-Watson process give:

Nearly half the families are already empty at generation six, yet the expected value exceeds three. Since empty families contribute zero,whenever $q_n<1$. At generation six this conditional expectation is about $7.092969$. Conditioning on being nonempty now differs from conditioning on eventual survival: some presently nonempty families will still die out. The eventual branching extinction probability is $1/2$, even as the unconditional expected value grows without bound.

The expected value is useful for computing average population counts and, in the subcritical case, the expected value of the total progeny of a branching process. Its simple formula comes from one local calculation: linearity of expectation and the independent and identically distributed offspring counts giveThe tower property of conditional expectation gives the second equality; starting from $\mathbb{E}[Z_0]=1$ gives the formula. It describes an average over the generation probability distribution, not deterministic growth of an individual family. ^[ref]

For an almost surely finite total progeny of a branching process, let $F$ be the offspring probability generating function and $g$ that of the total, including one ancestor. Conditional on $k$ children, the total is one plus the sum of $k$ independent descendant totals. Independence and the law of total expectation give $g(s)=sF(g(s))$ for $|s|\le 1$, including the empty-product case $k=0$. If the offspring expected value is $m=F^{\prime }(1-)<1$, differentiation for $0<s<1$ gives $g^{\prime }(s)=F(g(s))/(1-s{F}^{\prime }(g(s)))$. The denominator is positive because $F^{\prime }(g(s))\le m$. Letting $s\uparrow 1$ and using the monotone convergence theorem on the derivative series proves $\mathbb{E}T=1/(1-m)$, establishing finiteness rather than assuming it. For Poisson branching process offspring, $F(z)=e^{m(z-1)}$.