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

Theorem: Let $y = \left\lbrace y_1, \ldots, y_n \right\rbrace$ be a set of observed data independent and identically distributed according to an ex-Gaussian distribution with parameters $\mu$, $\sigma$, and $\lambda$:

$\label{eq:exq} y_i \sim \mathrm{ex-Gaussian}(\mu,\sigma,\lambda), \quad i = 1, \ldots, n \; .$

Then, the method-of-moments estimates for the parameters $\mu$, $\sigma$, and $\lambda$ are given by

$\label{eq:exg-MoM} \begin{split} \hat{\mu} &= \bar{y} - \sqrt[3]{\frac{\bar{s}\cdot \bar{v}^{3/2}}{2}}\\ \hat{\sigma} &= \sqrt{\bar{v}\cdot\left(1 - \sqrt[3]{\frac{\bar{s}^2}{4}}\right)}\\ \hat{\lambda} &= \sqrt[3]{\frac{2}{\bar{s}\cdot \bar{v}^{3/2}}} \; , \end{split}$

where $\bar{y}$ is the sample mean, $\bar{v}$ is the sample variance and $\bar{s}$ is the sample skewness

$\label{eq:y-mean-var-skew} \begin{split} \bar{y} &= \frac{1}{n} \sum_{i=1}^n y_i \\ \bar{v} &= \frac{1}{n-1} \sum_{i=1}^n (y_i - \bar{y})^2 \\ \bar{s} &= \frac{\frac{1}{n}\sum_{i=1}^n (y_i-\bar{y})^3}{\left[\frac{1}{n}\sum_{i=1}^n(y_i-\bar{y})^2\right]^{3/2}} \; . \end{split}$

Proof: The mean, variance, and skewness of the ex-Gaussian distribution in terms of the parameters $\mu$, $\sigma$, and $\lambda$ are given by

$\label{eq:exg-E-Var-Skew} \begin{split} \mathrm{E}(X) &= \mu + \frac{1}{\lambda} \\ \mathrm{Var}(X) &= \sigma^2 + \frac{1}{\lambda^2}\\ \mathrm{Skew}(X) &= \frac{2}{\lambda^3\left(\sigma^2+\frac{1}{\lambda^2}\right)^{3/2}} \; . \end{split}$

Thus, matching the moments requires us to solve the following system of equations for $\mu$, $\sigma$, and $\lambda$:

$\label{eq:exg-mean-var-skew} \begin{split} \bar{y} &= \mu + \frac{1}{\lambda} \\ \bar{v} &= \sigma^2 + \frac{1}{\lambda^2}\\ \bar{s} &= \frac{2}{\lambda^3\left(\sigma^2+\frac{1}{\lambda^2}\right)^{3/2}} \; . \end{split}$

To this end, our first step is to substitute the second equation of \eqref{eq:exg-mean-var-skew} into the third equation:

$\label{eq:lambda-s1} \begin{split} \bar{s} &= \frac{2}{\lambda^3\left(\sigma^2+\frac{1}{\lambda^2}\right)^{3/2}} \\ &= \frac{2}{\lambda^3\cdot \bar{v}^{3/2}} \; . \end{split}$

Re-expressing \eqref{eq:lambda-s1} in terms of $\lambda^3$ and taking the cube root gives:

$\label{eq:lambda-s2} \lambda = \sqrt[3]{\frac{2}{\bar{s}\cdot \bar{v}^{3/2}}} \; .$

Next, we solve the first equation of \eqref{eq:exg-mean-var-skew} for $\mu$ and substitute \eqref{eq:lambda-s2}:

$\label{eq:mu-s1} \begin{split} \mu &= \bar{y} - \frac{1}{\lambda}\\ &= \bar{y} - \sqrt[3]{\frac{\bar{s}\cdot\bar{v}^{3/2}}{2}} \; . \end{split}$

Finally, we solve the second equation of \eqref{eq:exg-mean-var-skew} for $\sigma$:

$\label{eq:sigma-s1} \sigma^2 = \bar{v} - \frac{1}{\lambda^2} \; .$

Taking the square root gives and substituting \eqref{eq:lambda-s2} gives:

$\label{eq:sigma-s2} \begin{split} \sigma &= \sqrt{\bar{v}-\frac{1}{\lambda^2}} \\ &= \sqrt{\bar{v} - \left(\sqrt[3]{\frac{\bar{s}\cdot \bar{v}^{3/2}}{2}}\right)^2} \\ &= \sqrt{\bar{v} - \bar{v}\cdot\sqrt[3]{\frac{\bar{s}^2}{4}}} \\ &= \sqrt{\bar{v}\cdot \left(1- \sqrt[3]{\frac{\bar{s}^2}{4}}\right)} \; . \end{split}$

Together, \eqref{eq:mu-s1}, \eqref{eq:sigma-s2}, and \eqref{eq:lambda-s2} constitute the method-of-moment estimates of $\mu$, $\sigma$, and $\lambda$.

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Metadata: ID: P424 | shortcut: exg-mome | author: tomfaulkenberry | date: 2023-10-30, 12:00.