Positive Survival Probability in the Supercritical Case

Why a Supercritical Process Can Still Become Extinct

A Galton–Watson process can have an expected population that grows from generation to generation and still have a positive chance of dying out. The mean alone does not say whether a particular family line survives. The offspring generating function gives a sharper answer: when the offspring mean is finite and greater than one, its fixed points determine the extinction probability, and that probability is strictly less than one.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(Z_n)_{n\ge0}$ be a discrete-time Galton–Watson process with state space $S=\{0,1,2,\ldots\}$ and $Z_0=1$. Each individual independently produces $\xi$ children, where $\mathbb{P}(\xi=k)=p_k$, $k\ge0$. Write $G(s)=\mathbb{E}[s^\xi]=\sum_{k\ge0}p_ks^k$ for $0\le s\le1$, and $m=\mathbb{E}[\xi]=\sum_{k\ge0}kp_k$. Assume throughout the main result that $m<\infty$ and $m>1$.

Definition: Extinction by generation $n$ is the event $\{Z_n=0\}$, and the extinction probability is $q=\mathbb{P}(Z_n=0\text{ for some }n\ge0)$. Since zero is absorbing, extinction is equivalent to $Z_n=0$ at some finite generation. Survival is the complementary event, so its probability is $1-q$.

The Extinction Recursion

Set $x_n=\mathbb{P}(Z_n=0)$, with $x_0=0$ because $Z_0=1$. The recursion for $x_n$ comes from the branching property, not just from averaging a function of the current population. Given that the first individual has $k$ children, each of their $k$ descendant processes is an independent Galton–Watson process. Each is extinct by $n$ generations after its own start with probability $x_n$. Therefore, the probability all $k$ descendant processes are extinct by then is $x_n^k$. For $k=0$, this probability is $1$, using the empty-product convention $x_n^0=1$, including when $x_n=0$.

Proposition: For every $n\ge0$, $x_{n+1}=G(x_n)$. Moreover, $x_n$ increases to the extinction probability $q$, and $q$ is the least fixed point of $G$ in $[0,1]$.

Proof. Condition on the number $\xi$ of children of the initial individual. For each possible value $k$, the branching property and independence give probability $x_n^k$ that all $k$ descendant processes have zero population at the corresponding generation. Summing over $k$ yields $$ x_{n+1}=\sum_{k=0}^{\infty}p_kx_n^k=G(x_n). $$ The events $\{Z_n=0\}$ are increasing in $n$, because once the population is zero it remains zero. Their union is the extinction event, so continuity of probability for increasing events gives $x_n\uparrow q$. Since $G$ is continuous on $[0,1]$, taking limits in $x_{n+1}=G(x_n)$ gives $q=G(q)$. Continuity of $G$ follows, for example, by dominated convergence applied to $s^\xi$, which is bounded by $1$ for $s\in[0,1]$. Finally, let $r\in[0,1]$ be any fixed point of $G$. Since $x_0=0\le r$ and $G$ is nondecreasing on $[0,1]$, induction gives $x_n\le r$ for every $n$: if $x_n\le r$, then $x_{n+1}=G(x_n)\le G(r)=r$. Taking limits gives $q\le r$. Thus $q$ is the least fixed point. $\square$

Why There Is a Fixed Point Below One

One fixed point is always $1$, because $G(1)=\sum_kp_k=1$. To obtain a fixed point strictly below $1$, compare $G(s)$ with $s$ just to the left of $1$. Finite mean is important here: it determines the limiting slope of the generating function at $1$ from below.

Theorem: If $m=\mathbb{E}[\xi]$ is finite and $m>1$, then $G$ has a fixed point $q<1$, and the extinction probability equals this least fixed point. In fact, $G$ has exactly two fixed points in $[0,1]$: $q$ and $1$. Consequently, the survival probability $1-q$ is strictly positive.

Proof. For $0\le s<1$, expand each finite geometric sum to obtain $$ \frac{1-G(s)}{1-s} =\sum_{k=0}^{\infty}p_k\frac{1-s^k}{1-s} =\sum_{k=0}^{\infty}p_k\sum_{\ell=0}^{k-1}s^\ell. $$ For $k=0$, the inner sum is empty and equals $0$. As $s\uparrow1$, each inner sum increases to $k$. The monotone convergence theorem therefore gives $$ \lim_{s\uparrow1}\frac{1-G(s)}{1-s} =\sum_{k=0}^{\infty}kp_k=m>1. $$ It follows that some $s_0\in[0,1)$, chosen sufficiently close to $1$, satisfies $(1-G(s_0))/(1-s_0)>1$. Rearranging this strict inequality gives $G(s_0)<s_0$.

Define $f(s)=G(s)-s$. Then $f(1)=0$ and $f(s_0)<0$. Also $f(0)=p_0\ge0$. If $p_0>0$, the intermediate value theorem gives a root $r\in(0,s_0)$ of $f$; if $p_0=0$, then $r=0$ is already a root. In either case there is a fixed point $r<1$. By the proposition, the extinction probability $q$ is a fixed point and $q\le r<1$. Hence the survival probability $1-q$ is positive.

It remains to justify the claim that there are exactly two fixed points. Since $m>1$, at least one $p_k$ with $k\ge2$ is positive; otherwise $\xi\le1$ almost surely and $m\le1$. For every integer $k\ge2$, the function $s\mapsto s^k$ is strictly convex on $[0,1]$, and each $s\mapsto s^j$, $j\ge0$, is convex there. Thus $G$, which includes a strictly positive multiple of at least one strictly convex power and otherwise nonnegative multiples of convex powers, is strictly convex. Subtracting the linear function $s$ preserves strict convexity, so $f$ is strictly convex. A strictly convex function cannot have three distinct zeros: if $a<b<c$ were zeros, strict convexity at $b$, expressed as a strict convex combination of $a$ and $c$, would give $f(b)<0$, contradicting $f(b)=0$. Since $q<1$ and $1$ are two distinct zeros, these are exactly the two fixed points. $\square$

Worked Examples

Worked Example: A Positive but Nonzero Extinction Probability

Suppose each individual has either $0$ or $2$ children, with $p_0=\tfrac14$ and $p_2=\tfrac34$. The probabilities sum to $\tfrac14+\tfrac34=1$, and the mean is $m=0\cdot\tfrac14+2\cdot\tfrac34=\tfrac32>1$. Find the extinction probability.

The generating function is $G(s)=\tfrac14+\tfrac34s^2$. Its fixed-point equation is $$ s=\tfrac14+\tfrac34s^2 \quad\Longleftrightarrow\quad 3s^2-4s+1=0 \quad\Longleftrightarrow\quad (3s-1)(s-1)=0. $$ The fixed points are $\tfrac13$ and $1$, so the least one, and hence the extinction probability, is $q=\tfrac13$. The survival probability is $1-q=\tfrac23$.

Check: $G(\tfrac13)=\tfrac14+\tfrac34\cdot\tfrac19=\tfrac14+\tfrac1{12}=\tfrac13$, and $G(1)=\tfrac14+\tfrac34=1$. The first iteration gives $x_1=G(0)=\tfrac14$; the next is $x_2=G(\tfrac14)=\tfrac14+\tfrac34\cdot\tfrac1{16}=\tfrac{19}{64}$, which is greater than $\tfrac14$ and less than $\tfrac13$, as expected for an increasing sequence converging to $q$.

Worked Example: No Chance of Extinction

Suppose $p_1=\tfrac12$ and $p_2=\tfrac12$, with all other offspring probabilities zero. These probabilities sum to $1$, and $m=1\cdot\tfrac12+2\cdot\tfrac12=\tfrac32>1$. Compute the extinction probability.

Here $G(s)=\tfrac12s+\tfrac12s^2$. Its fixed-point equation reduces to $$ s=\tfrac12s+\tfrac12s^2 \quad\Longleftrightarrow\quad 0=\tfrac12s(s-1). $$ The fixed points are $0$ and $1$, so the smaller one is $q=0$. Directly, every individual has at least one child. Starting from $Z_0=1$, induction gives $Z_n\ge1$ for every $n$, so extinction is impossible and survival has probability $1$.

Check: $G(0)=0$ and $G(1)=\tfrac12+\tfrac12=1$. Since $x_0=0$, the recursion gives $x_1=G(0)=0$, and then $x_n=0$ at every generation, agreeing with $q=0$.

Worked Example: An Irrational Extinction Probability

Suppose an individual has $0$ children with probability $\tfrac12$ and $3$ children with probability $\tfrac12$. The probabilities sum to $1$, and the mean is $m=0\cdot\tfrac12+3\cdot\tfrac12=\tfrac32>1$. Find the fixed points and identify the extinction probability.

The generating function is $G(s)=\tfrac12+\tfrac12s^3$. The fixed-point equation becomes $$ s=\tfrac12+\tfrac12s^3 \quad\Longleftrightarrow\quad s^3-2s+1=0 \quad\Longleftrightarrow\quad (s-1)(s^2+s-1)=0. $$ The quadratic roots are $\frac{-1+\sqrt5}{2}$ and $\frac{-1-\sqrt5}{2}$. The second is negative, while $q=\frac{\sqrt5-1}{2}$ lies between $0$ and $1$, since $1<\sqrt5<3$. The fixed points in $[0,1]$ are therefore $q$ and $1$, and $q$ is the extinction probability.

Check: The quadratic equation for $q$ gives $q^2+q=1$, hence $q^2=1-q$ and $q^3=q(1-q)=q-q^2=2q-1$. Substitution gives $G(q)=\tfrac12+\tfrac12(2q-1)=q$, verifying the fixed point.

A Common Mistake and Why the Conclusion Matters

A frequent sign error is to infer from $m>1$ that $G(s)>s$ for $s<1$ near $1$. The limiting-slope calculation gives the opposite comparison. For the first worked example, at $s=\tfrac12$, $$ G(\tfrac12)=\tfrac14+\tfrac34\cdot\tfrac14=\tfrac7{16}<\tfrac8{16}=\tfrac12. $$ So $G(s)<s$ at this value, not $G(s)>s$. This direction matters: it is what places a fixed point below $1$ and, through iteration from $x_0=0$, makes the extinction probability strictly less than one.

The conclusion distinguishes survival probability from expected population growth. A mean greater than one guarantees a positive chance of survival, but it does not guarantee survival with certainty: in the first example, the mean is $\tfrac32$ while extinction still has probability $\tfrac13$. Conversely, the second example has zero extinction probability because every individual has at least one child. In both cases the fixed-point iteration, rather than the mean alone, identifies the extinction probability.

Check Your Understanding

For each question, use the generating function and the least-fixed-point characterization; verify any proposed root by substitution.

For $p_0=\tfrac12$ and $p_2=\tfrac12$, check that the probabilities sum to one, compute the mean, and find the extinction probability.
Explain why the extinction events $\{Z_n=0\}$ increase with $n$, and how this gives $q=\lim_{n\to\infty}x_n$.
For a finite-mean offspring law with $m>1$, what is the limit of $(1-G(s))/(1-s)$ as $s\uparrow1$, and what comparison does it imply near $1$?
For the law $p_1=\tfrac12,\ p_2=\tfrac12$, compute $x_1$ and $x_2$ from $x_0=0$, and compare them with the extinction probability.
Why does showing that $G$ has a fixed point below $1$ imply positive survival only after also establishing that extinction probability is the least fixed point?