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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.

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    Look at the very beginning example in http://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture26.pdf2012-11-01

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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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    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?.2018-11-14
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$X_0$ 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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    I assumed the o.p.'s con$f$usion w$a$s with the induction. That's what I spelled out.2012-11-01