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Let $X_1,X_2,..,X_N$ be independent identical samples from a Poission distribution with unknown parameter $\theta$

How would I find the sum of the distribution $X_1+X_2+X_3+..+X_N$ when n is finite. I looked up convolution but that only works for two random variables and I have multiple random variables and they are all poisson.

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  • $\begingroup$ Try proving by induction that $X_1 + \dots +X_N \sim \text{Poisson}(N\theta)$. The induction step will likely involve the total law of probability. $\endgroup$
    – user801306
    Commented Dec 10, 2020 at 20:04
  • $\begingroup$ There is also a way to do it with moment generating functions I think $\endgroup$
    – Fernando Martinez
    Commented Dec 10, 2020 at 20:11
  • $\begingroup$ You should definitely be able to do with with MGFs and CFs, too, but I overlooked the part when you said $\theta$ was unknown. $\endgroup$
    – user801306
    Commented Dec 10, 2020 at 20:13
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    $\begingroup$ Since $\theta$ is unknown, lets regard $\theta$ as a (continuous) random variable $\Theta \sim f_{\Theta}$. Set $$S_N=X_1+ \dots +X_N$$ Notice how $S_N|\Theta=\theta\sim\text{Poisson}(N\theta)$ so for any $k=0,1,2,.\dots$ we have that $$p_{S_{N}}(k)=P(S_N=k)=\int_{-\infty}^{\infty}\bigg[e^{-N\theta}\cdot \frac{(N\theta)^k}{k!}\cdot f_{\Theta}(\theta)\bigg]d\theta$$ This is of course assumes you can use the fact that $S_N|\Theta=\theta\sim\text{Poisson}(N\theta)$ $\endgroup$
    – user801306
    Commented Dec 10, 2020 at 20:14

1 Answer 1

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This what I did not sure if it is right. $W=X_1+X_2+..+X_N$ we have $M_w(t)=E(e^{X_1+X_2+..+X_N})=E(e^{x_1t}*e^{x_2*t}*..e^{x_n*t})$ And this equals

$M_{X_1}(T)*M_{X_2}(T)*..*M_{X_N}(t)=[M_X(t)]^n=[e^{\theta(e^t-1)}]^n$

So now you $W~Poisson(n*\theta))$ which is the same as pmf $(e^{-n*\theta}(n*\theta)^x)/x!$

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    $\begingroup$ $M_W(t)$ should only depend on $t$. Notice from independence we have $$M_{W}(t)=\mathbb{E}\Big(e^{X_1{t}}\times\dots\times e^{X_N{t}}\Big)=\Big[M_X(t)\Big]^N$$ where $X\sim \text{Poisson}(\theta)$. But you're also assuming $\theta$ isn't known, so this really is the moment generating function of $W|\Theta=\theta$ $\endgroup$
    – user801306
    Commented Dec 10, 2020 at 20:33
  • $\begingroup$ If you want to compute $M_{X|\Theta=\theta}(t)$ directly you can say $$M_{X|\Theta=\theta}(t)=\mathbb{E}\Big(e^{Xt}\Big|\Theta=\theta\Big)=\sum_{k=0}^{\infty}e^{kt}P(X=k|\Theta=\theta)$$ $\endgroup$
    – user801306
    Commented Dec 10, 2020 at 20:37
  • $\begingroup$ But if n is finite I can just the one in your first comment right. $\endgroup$
    – Fernando Martinez
    Commented Dec 10, 2020 at 20:39
  • $\begingroup$ With induction? $\endgroup$
    – user801306
    Commented Dec 10, 2020 at 20:40
  • $\begingroup$ No sorry I mean $M_w(t)=E(e^{X_1*t})*E(e^{x_2*t})*..E(e^{x_n*t})$ $\endgroup$
    – Fernando Martinez
    Commented Dec 10, 2020 at 20:43

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