What is the distribution of an average of independent Poisson random variables? If I have random variables $X_1,X_2,\ldots,X_n$ that are Poisson distributed with parameters $λ_1,λ_2,\ldots,λ_n$, what is the distribution of $Y=1/n \sum X_i$? 
Who knows this can give an answer because I don't know how to do, please. 
 A: If $X_1,\ldots,X_n$ are independent (something not mentioned in the question) then $X_1+\cdots+X_n\sim\mathrm{Poisson}(\lambda_1+\cdots+\lambda_n)$.  And then the distribution of $(X_1+\cdots+X_n)/n$ is a rescaling of that Poisson distribution, supported on the set $\left\{0, \dfrac 1 n, \dfrac 2 n, \dfrac 3 n, \ldots \right\}\vphantom{\dfrac{\displaystyle\sum}\sum}$.
A: In general, if $X\sim\operatorname{Pois}(\lambda)$ and $Y\sim\operatorname{Pois}(\mu)$ are independent, then for $n\in\mathbb N\cup\{0\}$,
$$
\{X+Y=n\} = \bigcup_{k=0}^n \{X=k\}\cap\{Y=n-k\}
$$
and hence
\begin{align}
\mathbb P(X+Y=n) &= \sum_{k=0}^n \mathbb P(X=k)\mathbb P(Y=n-k)\\
&= \sum_{k=0}^n \left(\frac{e^{-\lambda}\lambda^k}{k!}\right)\left(\frac{e^{-\mu}\mu^{n-k}}{(n-k)!}\right)\\
&= \frac{e^{-\lambda+\mu}}{n!}\sum_{k=0}^n \binom nk \frac{}{}\lambda^k\mu^{n-k}\\
&= \frac{e^{-\lambda+\mu}(\lambda+\mu)^n}{n!},
\end{align}
so that $X+Y\sim\operatorname{Pois}(\lambda+\mu)$. Let $S_n = \sum_{i=1}^n X_i$ and $\lambda=\sum_{i=1}^n\lambda_i$. By induction it is clear that $S_n\sim\operatorname{Pois}(\lambda_{(n)})$. Hence if $Y_n = \frac1n S_n$, then $$\left\{Y_n = \frac mn\right\} = \{S_n = m\}$$ for $m\geqslant 0$. It follows that 
$$\mathbb P\left(Y_n = \frac mn\right) =\mathbb P(S_n=m) = \frac{e^{-\lambda_{(n)}}\lambda_{(n)}^m}{m!}, \; m=0,1,2,\ldots $$
An interesting aside - if $\lim_{n\to\infty}\lambda_{(n)} = \lambda<\infty$ then
\begin{align}
\lim_{n\to\infty} \mathbb P\left(Y_n = \frac mn\right) &= \lim_{n\to\infty}\frac{e^{-\lambda_{(n)}}\lambda_{(n)}^m}{m!}\\
&= \frac1{m!}\lim_{n\to\infty}e^{-\lambda_{(n)}}e^{m\log\lambda_{(n)}}\\
&= \frac1{m!}\lim_{n\to\infty}e^{- \lambda_{(n)}+m\log\lambda_{(n)}} \\
&= \frac1{m!}\exp\left(\lim_{n\to\infty} -\lambda_{(n)} + m\log\lambda_{(n)}\right)\\ 
&= \frac1{m!}\exp(-\lambda + m\log\lambda)\\
&= \frac{e^{-\lambda}\lambda^m}{m!}.
\end{align}
It follows that $\mathbb E[Y_n]\stackrel{n\to\infty}\longrightarrow 0$, so $Y_n$ converges in $L^1$ (and hence in probability and in distribution) to zero.
