Distribution of sum of independent Poisson random variables

Question

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 $\mathbb P(X+Y+Z=k)=\mathbb P(3X=k)=\mathbb P(X=\frac{k}{3})$?

Answer 1

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 $\nu+\mu$.

Answer 2

No. In fact, summing independent Poisson variables gets a new Poisson variable, with the parameters summed. Characteristic functions provide an easy proof, since $$\Bbb E\exp itX=e^{-\lambda}\cdot \sum_{k\ge 0}\frac{(\lambda e^{it})^k}{k!}=\exp \lambda (e^{it}-1).$$Or with less theory: $$P(X+Y=k)=\sum_{l=0}^k\frac{e^{-\lambda_X}\lambda_X^l}{l!}\frac{e^{-\lambda_Y}\lambda_Y^{k-l}}{(k-l)!}=\frac{e^{-(\lambda_X+\lambda_Y)}}{k!}\sum_l\binom{k}{l}\lambda_X^l\lambda_Y^{k-l}=\frac{e^{-(\lambda_X+\lambda_Y)}(\lambda_X+\lambda_Y)^k}{k!}.$$