Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ ex-Gaussian distribution ▷ Mean

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 mean or expected value of $X$ is

$\label{eq:exg-mean} \mathrm{E}(X) = \mu + \frac{1}{\lambda} \; .$

Proof: The mean or expected value $\mathrm{E}(X)$ is the first raw moment of $X$, so we can use the moment-generating function of the ex-Gaussian distribution to calculate

$\label{eq:mean-from-mgf} \mathrm{E}(X) = M_X'(0) \; .$

First, we differentiate

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

with respect to $t$. Using the product rule and chain rule gives:

$\label{eq:exg-mean-s1} \begin{split} M'_X(t) &= \frac{\lambda}{(\lambda-t)^2}\exp \left[ \mu t + \frac{1}{2}\sigma^2t^2\right] + \left(\frac{\lambda}{\lambda-t}\right)\exp\left[\mu t + \frac{1}{2}\sigma^2t^2\right] (\mu + \sigma^2t)\\ &= \left(\frac{\lambda}{\lambda-t}\right) \cdot \exp\left[\mu t + \frac{1}{2}\sigma^2t^2\right] \cdot \left[ \frac{1}{\lambda-t} +\mu +\sigma^2t \right] \; . \end{split}$

Evaluating \eqref{eq:exg-mean-s1} at $t=0$ gives the desired result:

$\label{eq:exg-mean-s2} \begin{split} M'_X(0) &= \left(\frac{\lambda}{\lambda-0}\right) \cdot \exp\left[\mu\cdot 0 + \frac{1}{2}\sigma^2\cdot 0^2\right] \cdot \left[ \frac{1}{\lambda-0} + \mu + \sigma^2\cdot 0 \right] \\ &= 1\cdot 1 \cdot \left[ \frac{1}{\lambda} + \mu \right]\\ &= \mu + \frac{1}{\lambda} \; . \end{split}$
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Metadata: ID: P405 | shortcut: exg-mean | author: tomfaulkenberry | date: 2023-04-19, 12:00.