# 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:

• $\operatorname{Im} \left({X}\right) = \N$
• $\Pr \left({X = k}\right) = \dfrac 1 {k!} \lambda^k e^{-\lambda}$

Then:

 $\displaystyle \Pr \left({\Omega}\right)$ $=$ $\displaystyle \sum_{k \mathop \in \N} \frac 1 {k!} \lambda^k e^{-\lambda}$ $\displaystyle$ $=$ $\displaystyle e^{-\lambda} \sum_{k \mathop \in \N} \frac 1 {k!} \lambda^k$ $\displaystyle$ $=$ $\displaystyle e^{-\lambda} e^{\lambda}$ Taylor Series Expansion for Exponential Function $\displaystyle$ $=$ $\displaystyle 1$

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

$\blacksquare$