TITLE: Extinction probability in a population with competition
QUESTION [12 upvotes]: 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
$$\lim_{N\to\infty}P_N^{1/N}=\space ?$$

REPLY [2 votes]: 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 $$\mathbb{P}(Y_n=0)\leq1-\frac{1}{2^n}<1$$