Problem 15.8 Medium Free 1 variation

Extinction of a Bacterial Colony

A single bacterium sits on a plate. Every minute, each living bacterium independently dies, stays as one cell, splits into two, or splits into three, each with probability $1/4$ , and then the process repeats for all resulting bacteria. Find the probability that the colony eventually dies out, given that it has the form $\sqrt{a}-b$ for integers $a,b$ , and report $a+b$ .

Work it out, then compare with the answer.

SolutionShow the worked solution

Let $q$ be the extinction probability starting from one cell. The offspring generating function is

G(s)=\tfrac14\left(1+s+s^2+s^3\right).

Key equation. Condition on what the first cell does. Its $k$ offspring start independent copies of the process. The colony dies out exactly when all $k$ lineages die out, which has probability $q^k$ . Hence $q=G(q)$ .

Which root. Let $q_n=\mathbb{P}(\text{extinct by generation } n)$ . Then $q_0=0$ and $q_n=G(q_{n-1})$ , and $q_n\uparrow q$ . Suppose $r\ge 0$ is any root of $s=G(s)$ . Since $G$ is increasing on $[0,\infty)$ , induction gives $q_n=G(q_{n-1})\le G(r)=r$ . So $q$ is the smallest non-negative root.

Solving. The equation $4s=1+s+s^2+s^3$ becomes

s^3+s^2-3s+1=0 \iff (s-1)(s^2+2s-1)=0.

The roots are $1$ and $-1\pm\sqrt2$ . The smallest non-negative root is $\sqrt2-1\approx 0.414$ . This is consistent with the mean number of offspring, $\tfrac14(0+1+2+3)=\tfrac32>1$ , which forces $q<1$ .

Thus $q=\sqrt2-1$ , which has the form $\sqrt a-b$ with $a=2$ and $b=1$ , so $a+b=3$ .

\boxed{\mathbb{P}(\text{extinction})=\sqrt2-1,\qquad a+b=3}

Key idea. Extinction from one ancestor is the smallest fixed point of the offspring generating function in $[0,1]$ .

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 15.8a

Suppose a colony starts from a single cell, and each generation every living cell independently either dies with no offspring or produces some random positive number of offspring, according to a fixed offspring distribution. Given that a colony starting from one cell goes extinct with probability $1/3$ , find the probability that a colony starting from three independent cells eventually goes extinct.

Work it out, then compare with the answer.

Solution

Each of the three initial cells starts its own lineage, and the three lineages evolve independently. Each is a copy of the one-cell process. The colony dies out exactly when all three lineages die out. So the probability is the product of three single-lineage extinction probabilities, each equal to $\tfrac13$ :

\boxed{\left(\tfrac13\right)^3=\tfrac1{27}}

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