= Solution
Let $q_n=\mathbb P(X_n=0)$ for a <Galton-Watson process> starting from one ancestor. Conditional on $X_1=k$, extinction by generation $n+1$ requires $k$ independent descendant processes to be extinct by generation $n$, so
$$
q_{n+1}=F(q_n),\qquad q_0=0.
$$
Because zero is absorbing, $q_n\uparrow q$, the eventual extinction probability, and continuity gives $q=F(q)$. If $r\geq0$ is any fixed point, monotonicity of $F$ and $q_0\leq r$ give $q_n\leq r$ inductively. Thus $q$ is the smallest nonnegative fixed point.
Here $F(s)=s/4+3s^3/4$. Every individual has at least one child, so \b[$q=0$]. The offspring mean is $F'(1)=5/2$, hence
$$
\boxed{\mathbb EX_n=(5/2)^n.}
$$
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