Extinction probability in a population with competition
Solution 1:
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 $t$, $s(0) := N_0$).
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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].
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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.
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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 $N)$. $\square$
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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 $A(\omega) = (\overline{N_1},\overline{N_2},\dots)$.
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We now have all the machinery to finish the argument.
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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.
Solution 2:
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 $$\mathbb{P}(Y_n=0)\leq1-\frac{1}{2^n}<1$$