Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ ex-Gaussian distribution ▷ Moment-generating function

Theorem: Let $X$ be a random variable following an ex-Gaussian distribution:

\[\label{eq:exg} X \sim \text{ex-Gaussian}(\mu, \sigma, \lambda) \; .\]

Then, the moment generating function of $X$ is

\[\label{eq:exg-mgf} M_X(t) = \left( \frac{\lambda}{\lambda-t} \right) \exp \left[ \mu t + \frac{1}{2}\sigma^2t^2 \right] \; .\]

Proof: Suppose $X$ follows an ex-Gaussian distribution. Then, $X=A+B$ where $A$ and $B$ are independent, $A$ is normally distributed with mean $\mu$ and variance $\sigma^2$, and $B$ is exponentially distributed with rate $\lambda$. Then the moment generating function for $A$ is given by

\[\label{eq:norm-mgf} M_A(t) = \exp \left[ \mu t + \frac{1}{2}\sigma^2t^2 \right]\]

and the moment generating function for $B$ is given by

\[\label{eq:exp-mgf} M_B(t) = \frac{\lambda}{\lambda - t} \; .\]

By definition, $X$ is a linear combination of independent random variables $A$ and $B$, so the moment generating function of $X$ is the product of $M_A(t)$ and $M_B(t)$. That is,

\[\label{eq:exg-mgf-s1} \begin{split} M_X(t) &= M_A(t)\cdot M_B(t)\\ &= \exp\left[ \mu t + \frac{1}{2}\sigma^2t^2 \right] \cdot \left( \frac{\lambda}{\lambda-t} \right)\\ &= \left( \frac{\lambda}{\lambda-t} \right) \exp\left[ \mu t + \frac{1}{2}\sigma^2t^2 \right] \; . \end{split}\]

This finishes the proof of \eqref{eq:exg-mgf}.

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Metadata: ID: P404 | shortcut: exg-mgf | author: tomfaulkenberry | date: 2023-04-19, 12:00.