Does the following constitute a proof that the multiplication of a Poisson random variable $K$ with an integer constant $a$ is not itself Poisson? That is,

$f_K(k) = \frac{\lambda^k}{k!} e^{-\lambda}$

$L = aK$

Does not imply $L$ is a Poisson random variable.

Intuitively, multiplication of an integer-valued distribution leaves "gaps" on the number line, which would mean $aK$ cannot be poisson.

Here is my attempted proof:

$E[K] = \lambda$

$E[L] = a\lambda$

$Var(L) = a^2 \lambda$

If $L$ is Poisson, then $E[L] = Var(L)$

Which is a contradiction. Additionally, does that mean that $L$ is poisson iff. $a = 1$?


2 Answers 2


Your observation about the gaps is right. One can look in particular to $P(L=1)$ which must be $e^{-\lambda}>0$ (for some $\lambda)$ for a Poisson. But $L$ cannot be 1, unless $a=1$. So, you are right, $L$ is Poisson iff $a=1$.


Short proof:

for a Poisson variable $N$, you have $$ \forall k\in\Bbb N \ \ \ P(N=k) = e^{-\lambda} \frac {\lambda^k}{k!} \neq 0 $$

Tis is not the case of the variable $aN$ if $a\neq 1$.


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