Moment Generating Function of Poisson Distribution

Theorem
Let $X \sim \Poisson \lambda$ for some $\lambda \in \R_{> 0}$.

Then the moment generating function $M_X$ of $X$ is given by:


 * $\map {M_X} t = e^{\lambda \paren {e^t - 1} }$

Proof
From the definition of the Poisson distribution, $X$ has probability mass function:


 * $\map \Pr {X = n} = \dfrac {\lambda^n e^{-\lambda} } {n!}$

From the definition of a moment generating function:


 * $\displaystyle \map {M_X} t = \expect {e^{t X} } = \sum_{n \mathop = 0}^\infty \map \Pr {X = n} e^{t n}$

So: