Extinction probability in a population with competition

Question

Suppose we have a colony of bacteria. At the end of each day, each bacterium produces an exact copy of itself with probability $p$ and then dies with probability $q$. However, $q$ is not constant, but a function of $N$, the total number of bacteria: $$q=p\bigg(1-\frac{1}{N}\bigg)$$ So in larger populations of bacteria, each bacterium is more likely to die (because of competition, say).

To clarify, $N$ counts the number of bacteria before new ones were born. For instance, if there are $2$ bacteria on one day and they both reproduce to form $4$ bacteria, both of them still have exactly $p/2$ chance of dying (not $3p/4$). And the babies that have just been born cannot die immediately.

Let $P_N$ be the probability that a bacteria colony consisting of $N$ bacteria initially eventually goes extinct. Can we find an asymptotic formula for $P_N$? I suspect that we will have $$P_N\sim \alpha^N$$ for some $\alpha$, but I don’t know how to calculate this constant.

I did manage to figure out that if we keep $q$ constant, then the probability of eventual extinction starting with $N$ bacteria is exactly equal to $$\bigg(1-\frac{p-q}{p(1-q)}\bigg)^N$$ for $p>q$, and equal to $1$ for $p\le q$. But that problem was much easier because “newborn” bacteria were independent from their parents, whereas in this problem the chance of each bacterium’s survival is dependent on the overall population size.

So, really my question is: what is the value of $$\lim_{N\to\infty}P_N^{1/N}=\space ?$$

Answer 1

(query on the exact terms of the problem, too long for a comment)

One bacterium, at time $t$ , with an extant population $n(t)$, has
- $p$ , constant, probability to generate an additional bacterium, so that it contributes $+1$ to $n(t+1)$; - $q=p(1-1/n(t))$, depending on $n(t)$, probability to die and contributing $-1$ to $n(t+1)$;
- and consequently $r=1-p-q$ probability of just surviving and contribute $0$ to $n(t+1)$.

This is a classical [birth-death process][1] , which fundamentally is continuous in time, referring to live organism as bacteria.
The standard approach is to assume that in a small interval $\Delta t$ the probability of having more than one birth/death is negligible (higher order infinitesimal wrt $\Delta t$).
In the post, the example of two bacteria that replicate in one day hints that the adopted discretization in time is not obeying to the above hypothesis.
Although it would be possible to reduce the time unit from a day to an hour, or even less, so as to achieve that the hypothesis could be realistic, it seems that the OP is considering the one day lapse as a sort of "juvenile quisciency" of no fertility and no mortality.

Is this the correct interpretation of the scheme being adopted ?

Answer 2

The extinction probability is $1$ for any initial size population and any $p \in (0,1)$ (for $p=0,1$, the extinction probability is obviously $0$). Fix $p \in (0,1)$. Assume for the sake of contradiction that there is some $N_0 \ge 1$ such that the probability of no extinction with initial size $N_0$ is greater than $0$.

Our probability space is with respect to initial population sizes of $N_0$, probability $p$ of reproduction, and probability $p(1-\frac{1}{N})$ probability of death when population size is $N$. Let $s: \mathbb{Z}^{\ge 0} \to \mathbb{Z}^{\ge 0}$ represent population size ($s(t)$ is size of population on day $t$, $s(0) := N_0$).

.

Claim 1: $P(\{\omega \in \Omega : \lim_{t \to \infty} s(t) = \infty | \text{no extinction}\}) = 1$.

Proof: For $M \ge 1$, let $E_M := \{\omega \in \Omega : \liminf_{t \to \infty} s(t) = M \text{and no extinction}\}$. Then, since $M$ is a natural number, $P(E_M) \le P(\{\omega \in \Omega : s(t) = M \text{infinitely often}\})$. Note that the probability that the population goes directly from size $M$ to size $0$ is just some constant $c_M \in (0,1)$ depending on $M$. So, if we let $A_n$ denote the event that the population size does not go directly to $0$ after being at $M$ for the $n^{th}$ time, we see that $P(E_M) \le P(\cap_{n \ge 1} A_n) = \lim_{N \to \infty} P(A_1\cap\dots\cap A_N) = \lim_{N \to \infty} P(A_1)P(A_2 | A_1)P(A_3 | A_1,A_2)\dots P(A_N | A_1,\dots, A_{n-1}) = \lim_{N \to \infty} c_M^N = 0$, since [not going directly to $0$ after being at size $M$ for the $n^{th}$ time] does not affect the probability of [directly going to $0$ after being at size $M$ for the $(n+1)^{st}$ time].

.

Lemma 1: Let $(x_n)_{n \ge 1}$ be a sequence of positive integers with $\lim_{n \to \infty} x_n = \infty$. Then, there is a sequence $N_1 < N_2 < \dots$ of positive integers so that, for each $i \ge 1$, there is some $n \ge 1$ with $x_n = N_i$ and $x_m \ge N_i$ for each $m \ge n$.

Proof: Elementary.

.

Claim 2: For all $N \ge 1$, $\lim_{M \to \infty} P(\text{population size reaches at least $M$ if initial size was $N$ and there are constant probabilities $p$ and $q$ of reproduction and death}) = 1-P(\text{extinction with constant $p,q$ and initial size $N$})$.

Proof: Let $F_M$ be the event that the population size reaches at least $M$ if the initial size was $N$ and there are constant probabilities $p$ and $q$ of reproduction and death. By the same proof in claim $1$, $P($ no extinction with constant $p,q$ and initial size $N)\le P(F_M)$ for each $M$. And of course, $\lim_{M \to \infty} P(F_M) = P(\limsup_M F_M) \le P($no extinction with constant $p,q$ and initial size $N)$. $\square$

.

Now we can define a map $A : \{\text{no extinction}\} \subseteq \Omega \to \{\text{sequences of increasing positive integers}\}$ as follows. Fix an $\omega$ that doesn't lead to extinction. By claim $1$ and Lemma $1$, there are $N_1 < N_2 < \dots$ so that for each $i$, there is some $t$ so that $s(t) = N_i$ and $s(t') \ge N_i$ for all $t' \ge t$. By claim $2$, given any $(\epsilon_i)_i, \epsilon_i > 0$, we may take a subsequence $(\overline{N_1},\overline{N_2},\dots)$ of the sequence $(N_1,N_2,\dots)$ so that $P($ population size reaches at least $\overline{N_{i+1}}$ if initial size was $\overline{N_i}$ and there are constant probabilities $p$ and $q := p(1-\frac{1}{\overline{N_i}})$ of reproduction and death$) \le 1-(1-\frac{p-q}{p(1-q)})^{\overline{N_i}}+\epsilon_i$. Define $A(\omega) = (\overline{N_1},\overline{N_2},\dots)$.

.

We now have all the machinery to finish the argument.

.

Fix $\omega \in \Omega$ that doesn't lead to extinction, and let $(N_1,N_2,\dots) = A(\omega)$. Let $\overline{E}_{N_i,N_{i+1}}$ denote the event of going from a localmin population size of $N_i$ to a localmin population size of $N_{i+1}$ (we say an ordered pair $(N,t)$ is a 'localmin' if (1) $s(t) = N$ and (2) $s(t') \ge N$ for all $t' \ge t$). For each $i$, $P(\overline{E}_{N_i,N_{i+1}}) \le P(\text{population size goes from $N_i$ to $N_{i+1}$}) \le P(\text{population size goes from $N_i$ to $N_{i+1}$ if death probability is constant at $q := p(1-\frac{1}{N_i})$}) \le 1-(1-\frac{p-q}{p(1-q)})^{N_i}+\epsilon_i$, which is at most a constant $c_p < 1$ for all $i \ge 1$, if the $\epsilon_i$'s are small enough. Therefore, $P(\cap_{i \ge 1} \overline{E}_{N_i,N_{i+1}}) = \lim_{J \to \infty} P(\cap_{i=1}^J \overline{E}_{N_i,N_{i+1}}) = \lim_{J \to \infty} P(\overline{E}_{N_1,N_2})P(\overline{E}_{N_2,N_3} | \overline{E}_{N_1,N_2})\dots P(\overline{E}_{N_J,N_{J+1}} | \overline{E}_{N_1,N_2},\dots,\overline{E}_{N_{J-1},N_J}) \le \lim_{J \to \infty} c_p^J = 0$, where we used the fact that the probability of going from a localmin of $N_i$ to $N_{i+1}$ once the size is $N_i$ of course does not depend on the history.

Answer 3

Notice: this is a draft, it lays a context for those who would like to research.

Let $X_N$ the number of bacteria the next day, considering $N$ bacteria the current day. Since replication and death are independent, the expected value of $X_N$ (with $N>0$, else $X_0=0$) is: $$\mathbb{E}(X_N)=N+pN-qN=N+pN-p(N-1)=N+p$$

Now let's define a markovian process $(Y_n)$ with $Y_0=N$ and $Y_{n+1}=X_{Y_n}$. Then $$P_N=\mathbb{P}(\exists n,Y_n=0)=\mathbb{P}\left(\bigcup_{n=0}^\infty\{Y_n=0\}\right)$$ Note that $P_N\leq\mathbb{P}(Y_n\to0)$ and these quantities have different meanings.

First, $\{Y_n=0\}$ is an increasing sequence of events, because $X_0=0$. Therefore $$P_N=\lim_{n\to\infty}\mathbb{P}(Y_n=0)$$

Second, $$\mathbb{E}(Y_{n+1})=\mathbb{E}(X_{Y_n})=\mathbb{E}(\mathbb{E}(X_{Y_n}|Y_n))$$ $$\begin{align} &=\mathbb{E}\left(\sum_{k=0}^\infty X_k\mathbb{P}(Y_n=k)\right)\\ &=\sum_{k=0}^\infty\mathbb{E}(X_k)\mathbb{P}(Y_n=k)\\ &=\sum_{k=1}^\infty(k+p)\mathbb{P}(Y_n=k)\\ &=\mathbb{E}(Y_n)+p(1-\mathbb{P}(Y_n=0)) \end{align}$$ with Tonelli's theorem (positive random variables). Therefore, by induction, $$\mathbb{E}(Y_n)=N+pn-p\sum_{i=0}^{n-1}\mathbb{P}(Y_i=0)$$ and especially $\mathbb{E}(Y_n)\geq N$. Also $0\leq Y_n\leq N2^n$ hence $$\mathbb{P}(Y_n=0)\leq1-\frac{1}{2^n}<1$$