= Galton-Watson extinction fixed point
{c}
{title2=$q=\min\{s\in[0,1]:F(s)=s\}$}
For a <Galton-Watson process> with one ancestor and offspring <probability generating function> $F$, the <branching extinction probability> $q$ is the smallest <fixed point> of $F$ in $[0,1]$. A direct way to find it is to compute extinction by generation $n$:
$$
q_0=0,\qquad q_n=\mathbb P(Z_n=0),\qquad q_{n+1}=F(q_n),\qquad q_n\uparrow q.
$$
Zero is an <absorbing state>, so extinction by generation $n$ is exactly the <event> $Z_n=0$. https://ocw.mit.edu/courses/18-445-introduction-to-stochastic-processes-spring-2015/27aaeb9eb9a9cecce93d9ad74413ddf5_MIT18_445S15_lecture19.pdf[Hao Wu, MIT 18.445, Lecture 19, slide 4]{ref}
For $F(s)=1/4+s/4+s^2/2$, the equation is
$$
F(s)-s=\frac12\left(s-\frac12\right)(s-1)=0.
$$
Both $1/2$ and $1$ are <fixed points>, but iteration from zero selects $q=1/2$. The first values are $q_1=1/4$, $q_2=11/32$, $q_3=809/2048\approx0.395020$, and $q_6\approx0.462186$. Iteration started at $1$ would stay at $1$ forever and would give the wrong extinction answer for this <offspring distribution>.
\Image[/branching-extinction-fixed-point.png]
{title=The smallest extinction fixed point selected by iteration from zero}
The vertical steps in the figure apply the offspring <probability generating function>; horizontal steps return to the diagonal to start the next iteration. This gives a useful numerical calculation even when solving $F(s)=s$ explicitly is difficult. For the displayed law, every finite iteration is a lower bound for the eventual <branching extinction probability>; at generation six, about $0.037814$ additional extinction <probability> remains beyond the current empty families.
Why the smallest <fixed point>? The ancestor's children's descendant families are <independent>; given $k$ children, all are empty $n$ generations later with <probability> $q_n^k$. Averaging gives $q_{n+1}=\sum_kp_kq_n^k=F(q_n)$. The increasing extinction <events> give the <limit> $q$, and <continuity> of the <probability generating function> yields $F(q)=q$. For any other <fixed point> $r\in[0,1]$, $q_0\leq r$ and <monotonicity> of $F$ imply $q_n\leq r$ at every step; hence $q\leq r$. https://www.uni-muenster.de/Stochastik/lehre/WS1011/SpezielleStochastischeProzesse/Ch_1.pdf[University of Münster branching-process notes, Section 1.4, Theorem 1.7]{ref}
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