This is quite an interesting problem, but I'm not sure how to go about doing it. I know that by using some basic Poisson properties I can figure it out but I'm failing to see how. It goes like this:

A population comprises of $X_n$ individuals at time $n=1,2,3...$ Suppose that $X_0$ has Poisson ($\mu$) distribution. Between time $n$ and time $n+1$ each of the $X_n$ individuals dies with probability $p$, independently of the others. The population at time $n+1$ is formed from the survivors together with a random number of immigrants who arrive independently according to a Poisson ($\mu$) distribution. What is the distribution of $X_n$? I feel that I'm close but can't quite get it. It sounds like I could set up a series and then maybe prove it by induction.


Let's consider time $n$. We have $X_{n+1} = X_n - D_n + I_n$, where $\{I_n\}$ are iids with $\operatorname{Poi}(\mu)$ distribution, and $D_n$ is the number of deaths. For each individual, the probability of dying is $p$. The total mortality is the sum of $X_n$ independent $\operatorname{Bernoulli}(p)$ random variables, thus $D_n \sim \operatorname{Binom}(X_n, p)$. Now consider the probability generating function of $X_n$: $$ \mathbb{E}\left(z^{X_{n+1}}\right) = \mathbb{E}\left(z^{X_{n} - D_n}\right) \underbrace{ \mathbb{E}\left(z^{I_n}\right)}_{\exp(\mu (z-1))} = \exp(\mu (z-1)) \mathbb{E}\left(z^{X_n} \, \mathbb{E}\left(z^{-D_n}|X_n\right)\right) = \\ \exp(\mu (z-1)) \mathbb{E}\left((p+(1-p) z)^{X_n}\right) $$ That is $$ \mathcal{P}_{n+1}(z) = \mathrm{e}^{\mu(z-1)} \mathcal{P}_{n}\left(p+(1-p)z\right) \qquad \mathcal{P}_0(z) = \mathrm{e}^{\mu(z-1)} $$ Solving: $$ \mathcal{P}_{n}(z) = \exp\left( \mu (z-1) \frac{1-(1-p)^n}{p}\right) $$ Thus $X_n$ is Poisson distributed with mean $$ \mathbb{E}(X_n) = \mu \frac{1-(1-p)^n}{p} $$

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  • $\begingroup$ +1. An alternative is to write $X_{n+1}=Y_n+I_n$ with $Y_n$ Bin$(X_n,1-p)$ conditionally on $X_n$. The computations which follow are then somewhat easier (but only veeeeery slightly...). $\endgroup$ – Did Nov 1 '12 at 18:05
  • $\begingroup$ I do not understand how you went from $\mathbb{E}\left(z^{X_{n} - D_n}\right) \underbrace{ \mathbb{E}\left(z^{I_n}\right)}_{\exp(\mu (z-1))}$ to $\exp(\mu (z-1)) \mathbb{E}\left(z^{X_n} \, \mathbb{E}\left(z^{-D_n}|X_n\right)\right)$. Can you please explain?. $\endgroup$ – makansij Nov 14 '18 at 15:12

$X_0$ is $\mbox{Poisson}(\mu)$.

Mean number of $X_1$ is $\mu (1 - p)$ + $\mu$. My claim is that it is also Poisson.

If you write out a few more terms, you'll soon see that $X_n$ is $\mbox{Poisson}\left(\mu\frac{1 - (1-p)^{N+1}}{p}\right)$

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  • 3
    $\begingroup$ (pretty easy to prove)... Funny: this is what the exercise is about, no? $\endgroup$ – Did Nov 1 '12 at 15:21
  • $\begingroup$ You mean, on problems that are obviously homework problems, it is expected to spell out the answers for the o.p.? $\endgroup$ – John Nov 1 '12 at 17:53
  • $\begingroup$ You think this is what I mean? Sure. (And I eat children too.) $\endgroup$ – Did Nov 1 '12 at 17:56
  • $\begingroup$ I assumed the o.p.'s confusion was with the induction. That's what I spelled out. $\endgroup$ – John Nov 1 '12 at 18:08

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