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:

\(\ds \map \Pr \Omega\)

\(=\)

\(\ds \sum_{k \mathop \in \N} \frac 1 {k!} \lambda^k e^{-\lambda}\)

\(\ds \)

\(=\)

\(\ds e^{-\lambda} \sum_{k \mathop \in \N} \frac 1 {k!} \lambda^k\)

\(\ds \)

\(=\)

\(\ds e^{-\lambda} e^{\lambda}\)

Taylor Series Expansion for Exponential Function

\(\ds \)

\(=\)

\(\ds 1\)

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

$\blacksquare$

Sources

• 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 2.2$: Examples