# Probability Generating Function of Poisson Distribution

## Theorem

Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.

Then the p.g.f. of $X$ is:

$\Pi_X \left({s}\right) = e^{-\lambda \left({1-s}\right)}$

## Proof

From the definition of p.g.f:

$\displaystyle \Pi_X \left({s}\right) = \sum_{x \mathop \ge 0} p_X \left({x}\right) s^x$

From the definition of the Poisson distribution:

$\displaystyle \forall k \in \N, k \ge 0: p_X \left({k}\right) = \frac {e^{-\lambda} \lambda^k} {k!}$

So:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \Pi_X \left({s}\right)$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sum_{k \mathop \ge 0} \frac {e^{-\lambda} \lambda^k} {k!} s^k$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle e^{-\lambda} \sum_{k \mathop \ge 0} \frac {\left({\lambda s}\right)^k} {k!}$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle e^{-\lambda} e^{\lambda s}$$ $$\displaystyle$$ $$\displaystyle$$ by Taylor Series Expansion for Exponential Function $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle e^{-\lambda + \lambda s}$$ $$\displaystyle$$ $$\displaystyle$$

Hence the result.

$\blacksquare$