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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.

Michael Hardy
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denny
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  • Possible answer on http://stats.stackexchange.com/questions/35042/what-is-the-distribution-of-an-average-of-poisson-random-variables – Clement C. Oct 03 '15 at 13:44
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    https://proofwiki.org/wiki/Sum_of_Independent_Poisson_Random_Variables_is_Poisson – Jack D'Aurizio Oct 03 '15 at 13:46
  • @JackD'Aurizio Dividing by $n$, though, makes is non-Poisson. – Clement C. Oct 03 '15 at 13:47
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    Have you tried something -besides asking here? Can you work out the sum of two Poisson variables? – leonbloy Oct 03 '15 at 13:47
  • @ClementC.: that does not change the fact that $X_1+\ldots+X_n$ is Poisson distributed, the multiplication by $\frac{1}{n}$ is not that relevant. – Jack D'Aurizio Oct 03 '15 at 13:51
  • Indeed -- it's going to be a scaled Poisson. Simply, I don't think the resulting distribution itself has a name (besides "it's a scaled Poisson"). – Clement C. Oct 03 '15 at 13:52
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    If the random variables are independent, then $\sum_{i} X_i \sim Poisson\left(\sum_{i} \lambda_i \right)$. –  Oct 03 '15 at 14:04

2 Answers2

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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 $\left\{0, \dfrac 1 n, \dfrac 2 n, \dfrac 3 n, \ldots \right\}\vphantom{\dfrac{\displaystyle\sum}\sum}$.

Michael Hardy
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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.

Math1000
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