Past exam of the graph Theory and Probabilistic Combinatorics course of the University of Oxford / 2008 / B / 6 / c / Solution

Let $Z_r$ be the size of generation $r$, with $Z_0=1$. Its offspring mean is $2p$, so conditioning on the previous generation gives $\mathbb{E}{Z}_r=(2p{)}^r$. At $p=1/2$,The interchange uses Tonelli's theorem for nonnegative variables. Critical extinction is nevertheless almost sure: its probability is the smallest root in $[0,1]$ of $s=(1+s^2)/2$, whose only root is one. Hence $K$ is finite almost surely. If all its finite atoms had the exponential bound, then $\mathbb{E}K\le {\sum }_{k\ge 1}k{e}^{-ck}<\infty$, a contradiction.

For $p>1/2$ the extinction probability is $(1-p)/p<1$. There is positive mass at $K=\infty$, so the mean is infinite even if every finite atom decays exponentially. This is precisely why the critical contradiction does not extend to the supercritical case.