Total progeny of a branching process ($T={\sum }_{j\ge 0}{Z}_j$)

The total progeny counts all individuals, including the initial ancestor, in a Galton-Watson process. It is infinite on survival. If the offspring mean is $m<1$, the expected generation size in a Galton-Watson process and the monotone convergence theorem give $\mathbb{E}T=1/(1-m)$.

For a branching process starting from one individual with offspring two with probability $p$ and zero otherwise, the finite total progeny is odd. Exploring individuals gives an active queue beginning at one, with increments $+1$ or $-1$. A first hit of zero at time $2j+1$ requires $j$ positive and $j+1$ negative increments and no earlier empty queue. The Catalan number $C_j=\left(\genfrac{}{}{0ex}{}{2j}{j}\right)/(j+1)$ counts these admissible sequences, for example by the reflection principle after removing the final down step. Independence then gives the displayed mass formula.

If $p\ne 1/2$, $b=4p(1-p)<1$ and $C_j\le 4^j$, so $\mathbb{P}(K=2j+1)\le (1-p)b^j$. Choosing $c=\min \{-\log (1-p),-\frac{1}{2}\log b\}>0$ bounds every finite mass by $e^{-c(2j+1)}$. This does not exclude an atom at infinity. In the supercritical case that atom has mass $(2p-1)/p$. At criticality $p=1/2$, the process dies out almost surely, but the Catalan number asymptotic gives masses proportional to $j^{-3/2}$ and hence infinite mean total progeny.

If a Galton-Watson process starts from one ancestor and has bounded nonnegative integer offspring $Z$ with $\mathbb{E}Z<1$, its total progeny $T$ obeys $\mathbb{P}(T>n)\le e^{-cn}$ for some $c>0$ and all integers $n\ge 1$. Explore one individual at a time. The unexplored queue has size $1+{\sum }_{j=1}^t(Z_j-1)$ until its first hit of zero, whose hitting time is $T$. Hence $T>n$ implies ${\sum }_{j=1}^n(Z_j-1)\ge 0$. Boundedness makes $M(\theta )=\mathbb{E}{e}^{\theta (Z-1)}$ finite near zero, with $M(0)=1$ and $M^{\prime }(0)=\mathbb{E}Z-1<0$. Choose $\theta >0$ with $M(\theta )<1$. The exponential Markov inequality gives the desired bound with $c=-\log M(\theta )$. The same proof works whenever the offspring law has an exponential moment near zero, even without boundedness.

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, $F(z)=e^{m(z-1)}$.

Consider a Galton-Watson process started by one individual, with offspring count $2$ with probability $p<1/2$ and $0$ with probability $q=1-p$. Let $N$ count individuals producing two offspring. At the first generation, either $N=0$, or $N=1+N_1+N_2$ for independent copies $N_1,N_2$. Its probability generating function therefore satisfies $G=q+ps{G}^2$ andThe branch is fixed by $G(0)=q$. Differentiation at $s=1$ gives $\mathbb{E}[N]=p/(1-2p)$. Alternatively, generation $t$ has expected size $(2p{)}^t$, so its expected splitting count is $p(2p{)}^t$ and summing gives the same result. The total progeny is $2N+1$, since every split creates two new individuals.

Count the ancestor and the next $n-1$ generations of a Galton-Watson process. Conditioning on the ancestor's offspring, their independent descendant trees each have the preceding finite-horizon probability generating function, proving the displayed recursion. If the count instead excludes the ancestor, the recursion is ${\hat{H}}_{n+1}(s)=G(s{\hat{H}}_n(s))$ with ${\hat{H}}_0=1$. The indexing convention must be specified.