Probability Generating Function of Poisson Distribution

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Theorem

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


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

$\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$


Proof

From the definition of p.g.f:

$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$

From the definition of the Poisson distribution:

$\ds \forall k \in \N, k \ge 0: \map {p_X} k = \frac {e^{-\lambda} \lambda^k} {k!}$

So:

\(\ds \map {\Pi_X} s\) \(=\) \(\ds \sum_{k \mathop \ge 0} \frac {e^{-\lambda} \lambda^k} {k!} s^k\)
\(\ds \) \(=\) \(\ds e^{-\lambda} \sum_{k \mathop \ge 0} \frac {\paren {\lambda s}^k} {k!}\)
\(\ds \) \(=\) \(\ds e^{-\lambda} e^{\lambda s}\) Taylor Series Expansion for Exponential Function
\(\ds \) \(=\) \(\ds e^{-\lambda + \lambda s}\)

Hence the result.

$\blacksquare$


Sources