Consider a continuous time stochastic process where an individual is born at each tick of a poisson clock which has rate $n$. Each individual has a fixed life span of a unit of time. Let there be 0 individuals, initially. Does the process return to the 0 individuals state with probability 1? Is the mean time of returning to the 0 state finite?
PS. I need this result to establish the convergence of the average (over time) of a queue length in a (non-markovian) queue with abandonments (individuals abandon if not served after a unit of time)
If the birth of individuals is a Poisson process with rate $n$, then looking and the previous unit of time that ends at this instant, the number of births $b$ follows the following probability distribution: $$ P(b) = e^{-n} \frac{n^b}{b!} $$
The probability of having $0$ individuals at a given instant is given by $$ P(0) = e^{-n} $$ (since $n^0 = 1$ and $0! = 1$).
Assuming that $n$ is finite, eventually we will have $0$ individuals, on average after $1/P(0) = e^n$ units of time.
By the way, if you are thinking about taking care of the individuals, you should be able to do it at a rate higher than $n$ individuals per unit of time. For instance:
If, on average, we get $n=10$ births per unit of time and we can deal with, on average, $m=10$ individuals per unit of time, we will have individuals dying if the number of births $b > 10$, i.e., $b \geq 11$, that happens with a probability near $40\%$: $$ \begin{array}{cc} \hline j & P(b \geq j) \\ \hline 0 & 1 \\ 1 & 0.99995 \\ 2 & 0.9995 \\ 3 & 0.997 \\ 4 & 0.990 \\ 5 & 0.971 \\ 6 & 0.933 \\ 7 & 0.870 \\ 8 & 0.780 \\ 9 & 0.667 \\ 10 & 0.542 \\ 11 & 0.417 \\ 12 & 0.303 \\ 13 & 0.208 \\ 14 & 0.136 \\ 15 & 0.0835 \\ 16 & 0.0487 \\ 17 & 0.0270 \\ 18 & 0.0143 \\ 19 & 0.00719 \\ 20 & 0.00345 \\ \hline \\ \end{array} $$
Using the Normal approximation to the Poisson distribution, we should have around $97.73\%$ of the events up to $n + 2 \sqrt{n}$, in the above example it means that $b > 10 + 2 \times \sqrt{10} = 16.3$; and we should have around $99.87\%$ of the events up to $n + 3 \sqrt{n}$, meaning that $b > 19.5$