Poisson Distribution Gives Rise to Probability Mass Function

Theorem
Let $$X$$ be a discrete random variable on a probability space $$\left({\Omega, \Sigma, \Pr}\right)$$.

Let $$X$$ have the poisson distribution with parameter $\lambda$ (where $$\lambda > 0$$).

Then $$X$$ gives rise to a probability mass function.

Proof
By definition:


 * $$\operatorname{Im} \left({X}\right) = \N$$


 * $$\Pr \left({X = k}\right) = \frac 1 {k!} \lambda^k e^{-\lambda}$$

Then:

$$ $$ $$ $$

So $$X$$ satisfies $$\Pr \left({\Omega}\right) = 1$$, and hence the result.