Poisson Distribution Gives Rise to Probability Mass Function

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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.

$\blacksquare$


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