If $X$, $Y$, and $Z$ are all independent Poisson random variables, each with parameter $\lambda$, can $\mathbb P(X+Y+Z=k)$ be simplified to $\mathbb P(3X=k)$, since $X$, $Y$, and $Z$ are the same? If so, would the distribution of $\mathbb P(X+Y+Z=k)=\mathbb P(3X=k)=\mathbb P(X=\frac{k}{3})$?
Distribution of sum of independent Poisson random variables
poisson distributionprobabilityrandom variables
Best Answer
No.
In that situation also $X+Y+Z$ will have Poisson-distribution, and this with parameter $3\lambda$.
$3X$ does not have Poisson-distribution. Observe for instance that $P(3X=1)=0$. This cannot be true for a Poisson-distribution.
addendum:
Let it be that $R,S$ are independent, that $R$ has Poisson distribution with parameter $\nu$ and $S$ has Poisson distribution with parameter $\mu$,
Then we have: $$\begin{aligned}P\left(R+S=n\right) & =\sum_{r+s=n}P\left(R=r,S=s\right)\\ & =\sum_{r+s=n}P\left(R=r\right)P\left(S=s\right)\\ & =\sum_{r+s=n}e^{-\nu}\frac{\nu^{r}}{r!}e^{-\mu}\frac{\mu^{s}}{s!}\\ & =e^{-\left(\nu+\mu\right)}\frac{1}{n!}\sum_{r+s=n}\binom{n}{r}\nu^{r}\mu^{s}\\ & =e^{-\left(\nu+\mu\right)}\frac{\left(\nu+\mu\right)^{n}}{n!} \end{aligned} $$ Proving that $R+S$ has Poisson-distribution with parameter $\nu+\mu$.