Moment Generating Function of Poisson Distribution

From ProofWiki
Jump to navigation Jump to search

Theorem

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

\(\displaystyle \map {M_X} t\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac {\lambda^n e^{-\lambda} } {n!} e^{t n}\)
\(\displaystyle \) \(=\) \(\displaystyle e^{-\lambda} \sum_{n \mathop = 0}^\infty \frac {\paren {\lambda e^t}^n} {n!}\)
\(\displaystyle \) \(=\) \(\displaystyle e^{-\lambda} e^{\lambda e^t}\) Power Series Expansion for Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle e^{\lambda \paren {e^t - 1} }\) Exponential of Sum

$\blacksquare$