Poisson Distribution Gives Rise to Probability Mass Function

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

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

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

Proof
By definition:


 * $\Img X = \N$


 * $\map \Pr {X = k} = \dfrac 1 {k!} \lambda^k e^{-\lambda}$

Then:

So $X$ satisfies $\map \Pr \Omega = 1$, and hence the result.