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Consider a branching process with immediate offspring distribution $\xi \sim \operatorname{Bin}(m, p)$, where $m$ is a constant. Let $\phi(s)$ be the generating function of $\xi$, i.e. $\phi(s) = (1 - p + s p)^m$, and let $\phi_n(s) := \underbrace{(\phi \circ \phi \circ \ldots \circ \phi)}_{n\text{ times}}(s)$.

Let $p$ be such that we know that the branching process will go extinct (i.e. $\mathbb{E}\xi = m p \leq 1$). Let $T$ denote the time (generation) at which the branching process will go extinct.

I am trying to find what is $\mathbb{E}T$. What I know is that $\mathbb{P}(T = t) = \phi_t(0) - \phi_{t - 1}(0)$, and so $$ \mathbb{E}T = \sum\limits_{t = 1}^{+\infty} t \mathbb{P}(T = t) = \sum\limits_{t = 1}^{+\infty} t (\phi_t(0) - \phi_{t - 1}(0))\text{.} $$

However, I cannot get to a closed form. Any ideas? Could there perhaps be a simpler way?

d125q
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  • I am assuming the offspring distribution is binomial $(m,p$ for some fixed $m$ such that we know there will be extinction because $mp \leq 0$ (or maybe we need $mp<0$. The way the problem is stated there is some confusion about whether the $n$ in Bin$(n,p)$ is the same as the generation $n$ used later in the statement of the problem. (in the latter case, for non-zero $p$, we cannot be certain in probability that the process will eventually go extinct.) – Mark Fischler May 13 '15 at 20:40
  • @MarkFischler: Indeed, the $n$ (or $m$ in your case) is a constant. I will edit my post to clarify this. – d125q May 13 '15 at 20:41
  • You can simplify a little: Treating $\phi_0(0)$ as zero, $\Bbb{E}T = \sum_0^\infty (1-\phi_t(0))$. But that still requires figuring out $\phi_t(0)$ which for $m>1$ is highly non-trivial unless one finds some nice trick. – Mark Fischler May 13 '15 at 23:30

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