Galton–Watson Branching Process: Will the Amoebas Die Out?

This classic problem in probability theory explores extinction in a branching process.

You begin with one amoeba. At each time step, each amoeba independently produces offspring based on the following rules:

All offspring behave identically in subsequent steps.

What is the probability that the population eventually goes extinct?

Step 1: Expected Offspring

Let \( X \) be the number of offspring from a single amoeba. Then:

\[ \mathbb{E}[X] = \frac{1}{4}(0 + 1 + 2 + 3) = \frac{6}{4} = 1.5 \]

Since the expected number of offspring is greater than 1, there’s a positive probability of survival.

Let \( q \) be the extinction probability. For a Galton–Watson process, \( q \) satisfies the fixed-point equation:

\[ q = f(q) \]

where \( f(s) \) is the generating function of the offspring distribution:

\[ f(s) = \frac{1}{4}(s^0 + s^1 + s^2 + s^3) = \frac{1}{4}(1 + s + s^2 + s^3) \]

So we solve:

\[ q = \frac{1}{4}(1 + q + q^2 + q^3) \]

Multiply both sides by 4:

\[ 4q = 1 + q + q^2 + q^3 \]

Rearranged:

\[ q^3 + q^2 - 3q + 1 = 0 \]

We find the smallest root in [0,1], which gives the extinction probability.

This cubic factors as:

\[ (q - 1)(q^2 + 2q - 1) = 0 \]

Solving \( q^2 + 2q - 1 = 0 \):

\[ q = -1 \pm \sqrt{2} \]

Only the solution \( q = \sqrt{2} - 1 \approx 0.4142 \) lies in \([0,1]\).

The probability that the amoeba population eventually dies out is:

\[ \boxed{\sqrt{2} - 1} \]