Extinction Probability of a Branching Process
A branching process starts with a single cell at generation 0. Each cell independently produces a random number of offspring in the next generation, where each of $0, 1, 2, 3$ children is equally likely (probability
(a) Write down the probability generating function $f(s) = E[s^{Z_1}]$ and compute the mean offspring count $m = f'(1)$.
(b) Show that the extinction probability $q = P(\exists\, n : Z_n = 0)$ is the smallest root in $[0,1]$ of the fixed-point equation $s = f(s)$. Solve this equation explicitly and give a numerical value for $q$.
(c) Classify the process as subcritical, critical, or supercritical, and interpret what this means for the extinction probability.
Hints
- Write out the PGF as a polynomial and compute $f'(1)$ to get the mean. The mean tells you immediately whether extinction is certain or not.
- For the extinction probability, use the recursive argument: extinction from one ancestor requires extinction from all children independently, giving $q = f(q)$. Factor out the known root $s = 1$.
- After factoring $(s-1)$ from $s^3 + s^2 - 3s + 1 = 0$, you get a quadratic $s^2 + 2s - 1 = 0$. The smallest non-negative root is the extinction probability.
Worked Solution
How to Think About It: This is a classic branching process question. The key tool is the probability generating function (PGF), which converts the recursive structure of the process into a functional equation. The extinction probability satisfies a fixed-point equation because extinction from one ancestor means extinction from all of its children -- and that event factors nicely through the PGF. Before doing any algebra, note that the mean number of offspring is $(0+1+2+3)/4 = 1.5 > 1$, so the process is supercritical. That immediately tells you the extinction probability is strictly less than 1 -- the population has a positive chance of surviving forever.
Quick Estimate: With mean offspring $m = 1.5$, the population grows on average by 50% each generation. A rough heuristic for branching processes: the survival probability is approximately
Part (a): Probability Generating Function
Each cell produces $k$ offspring with probability
$f(s) = E[s^{Z_1}] = \frac{1}{4}(1 + s + s^2 + s^3)$
This is a geometric series, so equivalently $f(s) = \frac{1-s^4}{4(1-s)}$ for $s \neq 1$.
The mean offspring count:
$m = f'(1) = \frac{1}{4}(0 + 1 + 2 + 3) = \frac{3}{2}$
Part (b): Extinction Probability
Let $q_n = P(Z_n = 0)$ be the probability of extinction by generation $n$. Starting from one cell, extinction by generation $n$ requires that every child's subtree goes extinct by generation $n-1$. Since children produce independent copies of the process, conditioning on the number of children $k$ gives:
$q_n = \sum_{k=0}^{3} P(k \text{ children}) \cdot q_{n-1}^k = f(q_{n-1})$
Starting from $q_0 = 0$ (cannot be extinct at generation 0 since we start with one cell), the sequence $q_0, q_1, q_2, \ldots$ is increasing and bounded by 1, so it converges to a limit $q = \lim_{n \to \infty} q_n$. Taking limits in the recursion gives:
$q = f(q)$
Since $q_0 = 0$ and $f$ is increasing and convex on $[0,1]$, the sequence converges to the smallest fixed point of $f$ in $[0,1]$.
Now solve $s = f(s)$:
$s = \frac{1}{4}(1 + s + s^2 + s^3)$
$4s = 1 + s + s^2 + s^3$
$s^3 + s^2 - 3s + 1 = 0$
We know $s = 1$ is always a root (every branching process has $s = 1$ as a fixed point of its PGF). Factor it out:
$s^3 + s^2 - 3s + 1 = (s - 1)(s^2 + 2s - 1) = 0$
The quadratic $s^2 + 2s - 1 = 0$ gives:
$s = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 \pm \sqrt{2}$
The three roots are $s = 1$, $s = \sqrt{2} - 1 \approx 0.4142$, and $s = -1 - \sqrt{2} \approx -2.414$. The last root is outside $[0,1]$, so the smallest root in $[0,1]$ is:
$q = \sqrt{2} - 1 \approx 0.4142$
Part (c): Classification
Since $m = 3/2 > 1$, the process is supercritical. The standard result for branching processes says:
- Subcritical ($m < 1$): extinction is certain ($q = 1$)
- Critical ($m = 1$): extinction is certain ($q = 1$), but slower
- Supercritical ($m > 1$): extinction probability $q < 1$ is the unique root of $s = f(s)$ in $[0,1)$
Here the survival probability is
Answer: The PGF is $f(s) = \frac{1}{4}(1 + s + s^2 + s^3)$ with mean $m = 3/2$. The process is supercritical, and the extinction probability is $q = \sqrt{2} - 1 \approx 0.4142$.
Intuition
The PGF is the natural tool for branching processes because independence of offspring turns multiplication of survival events into composition of generating functions. The extinction probability satisfies $q = f(q)$ because for the whole population to die out, every child of the original ancestor must independently start a subtree that dies out -- and the probability of that event for $k$ children is $q^k$, which is exactly what the PGF computes when you plug in $q$.
The supercritical case ($m > 1$) is the interesting one: the process has enough reproductive capacity that there is a positive chance of indefinite survival, but the extinction probability is still nonzero because there is always a chance of an unlucky early run of zeros. The value $q = \sqrt{2} - 1$ is elegant and exact -- a rare case where the fixed-point equation factors cleanly. In practice, branching process models underlie everything from default contagion in credit portfolios to epidemic modeling. The key lesson is that even when average growth is positive, stochastic extinction remains a real possibility, and its probability is determined entirely by the offspring distribution through the PGF fixed-point equation.