TITLE: Prove that a constant multiplied by a Poisson random variable is not Poisson
QUESTION [5 upvotes]: 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. $a = 1$?

REPLY [4 votes]: 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 $a=1$.