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**akbar** Let $\displaystyle N_n$, $\displaystyle n\geq 1$ be the size of a population of bacteria at time step n. At each step each bacteria produces a number of offspring and dies. The number of offspring is independent for each bacteria and is distributed according to the Poisson law with parameter $\displaystyle \lambda = 2$.

Assuming that $\displaystyle N_1=a>0$, the problem is to find the probability that the population will eventually die, e.g. $\displaystyle \mathbb{P}(N_n=0$ for some $\displaystyle n\geq 1)$.

Note there is a hint: find $\displaystyle c$ such that $\displaystyle \exp(-cN_n)$ is a martingale.

Here's a way to justify the answer: $\displaystyle M_n=e^{-c N_n}$ is a martingale, and it is positive, hence it converges almost-surely to a random variable $\displaystyle M_\infty\geq 0$. We deduce, taking logarithm, that either $\displaystyle N_n$ converges to a finite limit (if $\displaystyle M_\infty>0$) or to $\displaystyle +\infty$ (if $\displaystyle M_\infty=0$). Of course, since $\displaystyle N_n$ is an integer, the first case implies that this sequence is constant eventually. Obviously this can only happen if the population dies out. I let you devise your own argument for that (for instance, show $\displaystyle P(N_n=N_{n+1}=\cdots=N_{n+k}=m)\to_k 0$ for all $\displaystyle n,m>0$ and conclude, or just recall a result on Markov chains).

As a consequence, $\displaystyle M_\infty$ is either equal to 0 or 1, corresponding to a growth to infinity and to extinction respectively, and thus

$\displaystyle e^{-ca}=E[M_1]=E[M_\infty]=P(\text{extinction})$.

(The martingale is bounded hence the middle equality is just a consequence of the bounded convergence theorem $\displaystyle E[M_1]=E[M_n]\to E[M_\infty]$; no optional sampling here or whatever)