Extinction probability
Section of Galton–Watson branching, part of Stochastic processes
Extinction probability
| Type | Section |
| Taught in | Galton–Watson branching |
| Part | Stochastic processes |
| Questions using it | 2 |
Let q_n=\mathbb{P}(Z_n=0)=f_n(0). Because 0 is absorbing, \{Z_n=0\}\subset\{Z_{n+1}=0\}, so the sequence q_n increases. It is bounded by 1, so it has a limit q=\mathbb{P}(Z_n=0\text{ for some }n), the probability of eventual extinction. The generating function is continuous on [0,1], and q_{n+1}=f(q_n), so the limit satisfies q=f(q). Starting the iteration at 0 shows that q is the smallest fixed point of f in [0,1].
Theorem 1 (Extinction criterion) For a non-degenerate Galton–Watson process with p_0>0, the extinction probability is q=1 when m\le 1, and q<1 when m>1. If p_0=0, extinction is impossible and q=0. The deterministic law \xi\equiv 1 has mean m=1 and extinction probability q=0.
The generating function of a non-negative integer law is increasing and convex on [0,1], with f(1)=1 and left derivative m at 1. Suppose m>1 and p_0>0. Then f(0)=p_0>0, so f(s)-s is positive at s=0. The slope at 1 is greater than 1, so just to the left of 1 the graph of f lies below the diagonal. A smaller root of f(s)=s therefore exists in (0,1), and the extinction probability is that root.
Suppose m\le 1, the law is not concentrated at a single point, and p_0>0. Convexity puts the graph of f strictly above the diagonal on [0,1): the slope coming into 1 is at most 1, and the graph is not the diagonal itself. There is then no fixed point below 1, so q=1. The deterministic one-child law is the excluded case. Its generating function is f(s)=s, every point is a fixed point, and the iteration that starts at 0 stays at 0.
The mean decides which side of the threshold the process is on. It does not, by itself, give the numerical value of q. Two supercritical offspring laws with the same mean can have very different extinction probabilities, because q is a fixed point of the whole function f.
Example 1 (Two offspring or none) Suppose \mathbb{P}(\xi=0)=1/4 and \mathbb{P}(\xi=2)=3/4.
Question. Find the mean, the offspring generating function, and the extinction probability.
Solution. The mean is m=0\cdot(1/4)+2\cdot(3/4)=3/2>1, so the process is supercritical and q<1. The generating function is
f(s)=\tfrac14+\tfrac34 s^2.
The fixed-point equation s=1/4+(3/4)s^2 rearranges to 3s^2-4s+1=0, or (3s-1)(s-1)=0. The roots are s=1/3 and s=1. The smallest root in [0,1] is q=1/3. The lineage survives with probability 2/3, while the expected population still grows as (3/2)^n.
Used in
- Taught in Galton–Watson branching
- Also used in Stochastic processes
- Questions that use it (2) — A critical offspring law · Several ancestors
- All 3 questions on Galton–Watson branching
See also
Extinction criterion · Offspring probability generating function · Galton–Watson process · Mean and variance by generation · Subcritical decay, and what the model leaves out · Other offspring laws · Total family size · Generations and dependence