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For a Galton-Watson process with one ancestor, the branching extinction probability is the probability that some generation is empty:The equality uses the absorbing state at zero. The complementary branching survival probability $1-q$ is the probability that every generation is nonempty. The Galton-Watson extinction fixed point computes $q$ from the offspring distribution.

For $p_0=p_1=1/4$, $p_2=1/2$, one ancestor has $q=1/2$. Starting instead from $m$ ancestors with independent descendant families givesAll $m$ families must die out. With three ancestors, eventual extinction has probability $1/8$ and survival has probability $7/8$; by generation two, extinction has probability $(11/32{)}^3=1331/32768\approx 0.040619$. Extra ancestors change the extinction probability without changing the offspring distribution. ^[ref]

For finite offspring mean $\mu$, the extinction classification is:

The critical exception is $p_1=1$: every individual has exactly one child, $F(s)=s$, and $Z_n=1$ forever from one ancestor. Thus $\mu =1$ but $q=0$; a blanket claim that every critical Galton-Watson process becomes extinct would be false. These cases concern the Galton-Watson process with its common offspring distribution and independence assumptions. ^[ref]

This probability answers whether any lineage persists, which the offspring mean alone cannot quantify in the supercritical case. A branching process conditioned on extinction uses the same fixed point to reweight reproduction and describe families known eventually to die out.

A Galton-Watson process is a branching process that counts individuals by generation. Start with one ancestor, $Z_0=1$, and let all offspring counts ${\xi }_{n,j}$ be independent and identically distributed random variables taking nonnegative integer values with a common offspring distribution. ThenThe empty sum is zero: zero is an absorbing state, so once a generation is empty every later generation is empty. The same offspring distribution applies to every individual in every generation. ^[ref]

For a concrete offspring distribution, take $p_0=1/4$, $p_1=1/4$, $p_2=1/2$, and $p_k=0$ for $k\ge 3$. An individual has no children with probability $1/4$, one with probability $1/4$, and two with probability $1/2$. The offspring expected value is $\mu =5/4$. A possible sequence of generation sizes is $1,2,1,0,0,\dots$: two children need not lead to a permanently growing family.

The probability generating function $F(s)=1/4+s/4+s^2/2$ gives the exact generation probability distributions through branching-process generating-function iteration. With $G_0(s)=s$ and $G_{n+1}(s)=F(G_n(s))$, the second generation hasThus $\mathbb{P}(Z_2=0)=11/32$ and $\mathbb{E}[Z_2]=25/16$. The figure shows full generation probability mass functions, computed by polynomial function composition, through generation six; no simulated frequencies are used.

The contrast is striking: the expected generation size in a Galton-Watson process grows as $(5/4{)}^n$, while the branching extinction probability is $1/2$. By generation six, $\mathbb{P}(Z_6=0)\approx 0.462186$ although $\mathbb{E}[Z_6]\approx 3.814697$. Larger surviving families carry the rising expected value; that mean is an average over possible families, not a forecast of one family's path. The Galton-Watson extinction fixed point turns an unbounded number of future generations into a single equation for this probability.

The historical question was the disappearance of surnames passed through male descendants. In Watson and Galton, On the Probability of the Extinction of Families, pages 139–140, Francis Galton formulates a constant-law model counting sons who reach adulthood, and Watson computes later generation probability distributions by composing generating functions. Those are assumptions of their model, rather than a claim that actual human reproduction follows this offspring distribution.

Start the Galton-Watson process with one ancestor, $Z_0=1$, and let every individual independently have a number of offspring distributed as $Z$. With independent copies $Z_{t,j}$,where an empty sum is zero. Extinction is the event that some generation is empty; zero is then absorbing. Write $q$ for its probability and $1-q$ for the survival probability.

Set $q_t=P(Z_t=0)$, with $q_0=0$. Conditional on the ancestor's number of children, extinction by generation $t+1$ requires independent extinction by generation $t$ in every child subtree. ThusThe events $\{Z_t=0\}$ increase, so $q_t\uparrow q$. The probability generating function is continuous on $[0,1]$, by domination of $s^Z$ by $1$, giving $q=f_Z(q)$. If $r\in [0,1]$ is any fixed point, then $q_0=0\le r$ and monotonicity of $f_Z$ inductively gives $q_t\le r$. Taking limits proves the Galton-Watson extinction fixed point characterization:In particular it is the smallest nonnegative fixed point wherever the generating series is defined, since it already lies in $[0,1]$.

Let $m=\mathbb{E}Z=f_Z^{\prime }(1-)$, allowing $m=\infty$. The probability generating function is increasing and convex, with $f_Z(1)=1$. If $m<1$, its tangent inequality gives $f_Z(s)\ge 1+m(s-1)>s$ for $s<1$, so $q=1$. If $m=1$ and $P(Z=1)\ne 1$, there is positive mass at both zero and some integer at least two; strict convexity gives $f_Z(s)>s$ for $s<1$, again $q=1$. The exceptional case $Z=1$ almost surely has one individual forever, so $q=0$ despite $m=1$.

If $m>1$, the left derivative at $1$ gives $f_Z(s)-s<0$ for $s<1$ sufficiently close to $1$. Since $f_Z(0)\ge 0$, continuity supplies a fixed point below $1$, possibly $0$. Thus survival has positive probability precisely when $m>1$, apart from deterministic one-child reproduction.

For the random bipartite graph, early exploration of a component alternates between the two vertex classes. Before many vertices are exposed, each discovered vertex has approximately $\mathrm{Poisson}(c)$ new neighbors in the other class, since $\mathrm{Bin}(n-O(1),c/n)$ tends to that distribution. The two-type mean offspring matrix iswith spectral radius $c$; equivalently, the mean growth over two generations is $c^2$. The branching-process comparison therefore predicts a giant component exactly whenFor $c\le 1$, the local process dies out almost surely, so no linear-sized component is expected; $c=1$ is the critical case, not a positive-survival exception. For $c>1$, the symmetric survival probability is $1-q>0$, where $q=e^{c(q-1)}$, predicting approximately $2n(1-q)$ vertices in the giant. This last passage from local survival to a unique giant is the usual random-graph exploration argument; the branching-process reasoning here supplies the requested threshold explanation rather than a proof of that global theorem.

An uneventful day leaves one aphid, a one-offspring day leaves the parent plus its offspring, and a two-offspring day leaves three aphids. Death leaves none. Thus the daily descendant counts are $0,1,2,3$ with probabilities $(1-q)t$, $q$, $(1-q)r$, $(1-q)s$, respectively. The mean and probability generating function areIndependence across individuals and days makes the daily population a Galton-Watson process, with each surviving parent counted as one of the next day's descendants.

The standard Galton-Watson extinction fixed point result states that the eventual extinction probability of a branching process $e$ from one ancestor is the smallest root in $[0,1]$ of $G(e)=e$. Moreover a nondegenerate offspring law of mean at most one has extinction probability of a branching process one. Here, since $q<1$, the fixed-point equation reduces towhich contains no $q$. This proves laziness invariance of branching extinction: uneventful days change the timescale, not eventual extinction. The excluded case $q=1$ would keep one aphid forever.

If $2r+3s\le 1$, the offspring mean is at most one. In this parameter range $t>0$: if $t=0$, then $r+s=1$ would imply $2r+3s\ge 2$. Thus the offspring law is not deterministically one, and the extinction criterion applies. Extinction is certain.

For the specified probabilities, the fixed-point equation factors asIts roots in $[0,1]$ are $1/2$ and $1$, and the smaller root is the extinction probability of a branching process:

The Galton-Watson extinction fixed point equation isIts smallest root in $[0,1]$ is $q=(3\sqrt{3}-5)/4$, not the other allowed fixed point $1$. Consequently the large-depth escape probability isFor finite $N$, the exact escape probability is $1-q_N$ and is slightly larger. Indeed ${\varphi }^{\prime }(q)=2-\sqrt{3}<1$, and the mean value theorem on $[q_N,q]$ gives $0\le q-q_N\le q(2-\sqrt{3}{)}^N$. This also quantifies how quickly the finite-depth answer approaches the displayed limit.

The eventual extinction probability is the smallest solution $\boxed{q\in [0,1]\text{ of }{F}_1(q)=q}$. This is the Galton-Watson extinction fixed point result for one initial individual. Equivalently,since having an empty generation is an absorbing event and the extinction-by-generation events increase. In particular extinction is certain when the smallest fixed point is one. With $p_0>0$, an offspring mean at most one implies this, whereas a mean greater than one gives a fixed point below one. The fixed-point statement itself requires no finite-mean assumption.

Summing the geometric series for the offspring probability generating function givesThe fixed-point equation $s=1/(2-s)$ is $(s-1{)}^2=0$, whose only root in $[0,1]$ is one. The Galton-Watson extinction fixed point result therefore proves extinction is certain.

For the explicit iterates, the proposed formula at $n=1$ is $1/(2-s)$, so the base case holds. If it holds at $n$, part (a) givesThis is exactly the same formula with $n$ replaced by $n+1$. By induction,Evaluating the probability generating function at zero selects the mass of an empty generation:These probabilities increase to one, also confirming the extinction conclusion directly.

For the binomial branching process, the offspring probability generating function is $f(s)=(1-p+ps{)}^n$. The Galton-Watson extinction fixed point gives $1-\rho =f(1-\rho )=(1-p\rho {)}^n$.

The printed finite-binomial upper bound is false without an additional asymptotic qualification. For $n=2$ and $p=(1+\epsilon )/2$, solving the fixed-point equation exactly givesFor instance $\epsilon =1/20$ gives $\rho =80/441>1/10=2\epsilon$. More generally, at every fixed $n>1$ the Taylor expansion gives $\rho =2n\epsilon /(n-1)+O_n({\epsilon }^2)$, again contradicting that upper bound for sufficiently small positive $\epsilon$.

Here is the corrected binomial branching survival correction. The Poisson branching process of mean $1+\epsilon$ has branching survival probability $r$ satisfying $-\log (1-r)=(1+\epsilon )r$. Expanding the logarithm givesThus $2\epsilon /(1+2\epsilon )\le r\le 2\epsilon$, which implies the requested lower estimate $r\ge 2\epsilon -4{\epsilon }^2$. Since $(1-p+ps{)}^n\le e^{np(s-1)}$ on $[0,1]$, iteration of the two offspring probability generating functions gives $\rho \ge r$.

For the actual binomial branching process, expansion of its exact fixed-point equation givesIf $(1+\epsilon {)}^2<n$, all coefficients are nonnegative. Keeping the first term provesIn particular, for $n\to \infty$ and $\epsilon =o(1)$ this gives $\rho =2\epsilon +O({\epsilon }^2+\epsilon /n)$. The printed bounds also hold for the binomial branching process in an explicit large-$n$ regime: $0<\epsilon \le 1/4$ and $n\epsilon \ge 2$. Indeed these conditions give $n{p}^2\le \epsilon$ and $n{p}^3\le 1/4$. In the preceding nonnegative series, evaluation at $x=2\epsilon$ givesThe series is increasing, and its value at the actual branching survival probability $\rho$ is $\epsilon$, so $\rho \le 2\epsilon$. Together with the lower bound already proved, this recovers $2\epsilon -4{\epsilon }^2\le \rho \le 2\epsilon$ in that regime. In particular it applies eventually to part (ii). The counterexample shows why a regime condition is needed for a literal finite-$n$ statement.

Let $Z_n$ be the size of generation $n$. This is a Galton-Watson process with offspring meanThe expected generation size in a Galton-Watson process is thereforeIts probability generating function isBy the Galton-Watson extinction fixed point, the extinction probability $q$ is the smallest solution in $[0,1]$ of $G(q)=q$. Factoring givesso $q=(\sqrt{29}-5)/2$. The probability of production continuing forever is

Let $d_n$ be the probability of extinction by generation $n$. Conditioning on the initial individual's offspring givesThe events increase to eventual extinction, so $d_n\uparrow d$ and continuity gives $d=G_X(d)$. If $r\ge 0$ is any other fixed point, induction from $d_0=0\le r$ and monotonicity of $G_X$ gives $d_n\le r$ for every $n$. Hence $d\le r$, proving the Galton-Watson extinction fixed point characterization.