Evolution by death and immigration of Poisson distributed population 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.

 A: 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:
$$
     \mathcal{P}_{n}(z) = \exp\left( \mu (z-1) \frac{1-(1-p)^{n+1}}{p}\right)
$$
Thus $X_n$ is Poisson distributed with mean
$$ 
  \mathbb{E}(X_n) = \mu \frac{1-(1-p)^{n+1}}{p}
$$
A: $X_0$ 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 $\mbox{Poisson}\left(\mu\frac{1 -  (1-p)^{N+1}}{p}\right)$
