 Solution

First take $0<p<1$ and $kp>1$. Part (a) gives an extinction probability $\eta \in (0,1)$. Write $p_j=\mathbb{P}(Z=j)$. Conditional on the ancestor having $j$ children, extinction requires extinction of all $j$ independent descendant trees and has probability ${\eta }^j$. Bayes' rule therefore gives the conditional offspring lawIts sum is $f(\eta )/\eta =1$. Conditional on extinction of all descendant trees, those trees remain independent, each with its own extinction-conditioned law. Thus the entire conditioned process is a Galton–Watson process with offspring generating functionFor binomial offspring, using $\eta =(1-p+p\eta {)}^k$ givesIts mean is $k\tilde{p}=f^{\prime }(\eta )<1$. Indeed, $f(s)-s$ is strictly convex and has its two zeros at $\eta$ and one, so its derivative at the first zero is strictly negative. This proves subcriticality directly from the generating function.

Let $T$ be total progeny, taking value infinity on survival. Apply part (b) to the conditioned process. For some $c>0$,Thus the unconditional tail tends to the positive survival probability $1-\eta$, with an exponentially small finite-extinction contribution. It does not decay to zero.

If $p=1$ and $kp>1$, extinction has probability zero and conditioning on it does not specify a probability law. In that deterministic case total progeny is infinite and $\mathbb{P}(T>n)=1$ for every finite $n$.

The nondegenerate law above is the binomial branching process conditioned on extinction.