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

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 skewness of $X$ is

$\label{eq:exg-skew} \mathrm{Skew}(X) = \frac{2}{\lambda^3\left( \sigma^2 + \frac{1}{\lambda^2}\right)^{\frac{3}{2}}} \; .$

Proof:

To compute the skewness of $X$, we partition the skewness into expected values:

$\label{eq:skew-mean} \mathrm{Skew}(X) = \frac{\mathrm{E}(X^3)-3\mu\sigma^2-\mu^3}{\sigma^3} \; ,$

where $\mu$ and $\sigma$ are the mean and standard deviation of $X$, respectively. To prevent confusion between the labels used for the ex-Gaussian parameters in \eqref{eq:exg} and the mean and standard deviation of $X$, we rewrite \eqref{eq:skew-mean} as

$\label{eq:skew-mean-alt} \mathrm{Skew}(X) = \frac{\mathrm{E}(X^3) - 3\cdot \mathrm{E}(X)\cdot \mathrm{Var}(X) - \mathrm{E}(X)^3}{\mathrm{Var}(X)^{\frac{3}{2}}} \; .$

Since $X$ follows an ex-Gaussian distribution, the mean of $X$ is given by

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

and the variance of $X$ is given by

$\label{eq:exg-var} \mathrm{Var}(X) = \sigma^2 + \frac{1}{\lambda^2} \; .$

Thus, the primary work is to compute $\mathrm{E}(X^3)$. To do this, we use the moment-generating function of the ex-Gaussian distribution to calculate

$\label{eq:exg-moment} \mathrm{E}(X^3) = M_X'''(0)$

First, we differentiate the moment-generating function of the ex-Gaussian distribution

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

with respect to $t$. Using the product rule and chain rule, we have:

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

We then use the product rule to obtain the second derivative:

$\label{eq:exg-skew-s2} \begin{split} M_X''(t) &= M_X'(t)\cdot \left[ \frac{1}{\lambda-t} + \mu + \sigma^2t\right] + M_X(t)\cdot \left[ \frac{1}{(\lambda-t)^2}+\sigma^2\right] \\ &= M_X(t)\cdot \left[ \frac{1}{\lambda-t} + \mu + \sigma^2t\right]^2 + M_X(t)\cdot \left[ \frac{1}{(\lambda-t)^2}+\sigma^2\right]\\ &= M_X(t) \cdot \left[ \left( \frac{1}{\lambda-t} + \mu + \sigma^2t\right)^2 + \frac{1}{(\lambda-t)^2} + \sigma^2\right] \; . \end{split}$

Finally, we use the product rule and chain rule to obtain the third derivative:

$\label{eq:exg-skew-s3} M_X'''(t) = M_X'(t)\left[\left(\frac{1}{\lambda-t}+\mu+\sigma^2t\right)^2 + \frac{1}{(\lambda-t)^2}+\sigma^2\right] + M_X(t)\left[2\left(\frac{1}{\lambda-t}+\mu+\sigma^2t\right)\left(\frac{1}{(\lambda-t)^2} + \sigma^2\right) + \frac{2}{(\lambda-t)^3}\right] \; .$

Applying \eqref{eq:exg-moment}, together with \eqref{eq:exg-skew-s3}, yields:

$\label{eq:exg-skew-s4} \begin{split} \mathrm{E}(X^3) &= M_X'''(0)\\ &= M_X'(0)\left[\left(\frac{1}{\lambda}+\mu\right)^2 + \frac{1}{\lambda^2} + \sigma^2\right] + M_X(0)\left[2\left(\frac{1}{\lambda}+\mu\right)\left(\frac{1}{\lambda^2}+\sigma^2\right)+\frac{2}{\lambda^3}\right]\\ &= \left(\mu + \frac{1}{\lambda}\right)\left(\frac{1}{\lambda^2}+\frac{2\mu}{\lambda} + \mu^2 + \frac{1}{\lambda^2}+\sigma^2\right) + \left(\frac{2}{\lambda^3}+\frac{2\sigma^2}{\lambda} + \frac{2\mu}{\lambda^2}+2\mu\sigma^2 + \frac{2}{\lambda^3}\right)\\ &= \left(\mu+\frac{1}{\lambda}\right)\left(\frac{2}{\lambda^2}+\frac{2\mu}{\lambda} + \mu^2+\sigma^2\right) + \left(\frac{4}{\lambda^3}+\frac{2\sigma^2}{\lambda} + \frac{2\mu}{\lambda^2} + 2\mu\sigma^2\right)\\ &= \frac{2\mu}{\lambda^2} + \frac{2\mu^2}{\lambda} + \mu^3 + \mu\sigma^2 + \frac{2}{\lambda^3} + \frac{2\mu}{\lambda^2} + \frac{\mu^2}{\lambda} + \frac{\sigma^2}{\lambda} + \frac{4}{\lambda^3} + \frac{2\sigma^2}{\lambda} + \frac{2\mu}{\lambda^2} + 2\mu\sigma^2\\ &= \frac{6\mu}{\lambda^2} + \frac{6}{\lambda^3} + \frac{3\mu^2+ 3\sigma^2}{\lambda} + 3\mu\sigma^2 + \mu^3 \; . \end{split}$

We now substitute \eqref{eq:exg-skew-s4}, \eqref{eq:exg-mean}, and \eqref{eq:exg-var} into the numerator of \eqref{eq:skew-mean-alt}, giving

$\label{eq:exg-skew-s5} \begin{split} \mathrm{E}(X^3) - 3\cdot \mathrm{E}(X)\cdot \mathrm{Var}(X) - \mathrm{E}(X)^3 &= \left(\frac{6\mu}{\lambda^2} + \frac{6}{\lambda^3} + \frac{3\mu^2+ 3\sigma^2}{\lambda} + 3\mu\sigma^2 + \mu^3\right) - 3\left(\mu+\frac{1}{\lambda}\right)\left(\sigma^2+\frac{1}{\lambda^2}\right)- \left(\mu+\frac{1}{\lambda}\right)^3\\ &= \frac{6\mu}{\lambda^2} + \frac{6}{\lambda^3} + \frac{3\mu^2+ 3\sigma^2}{\lambda} + 3\mu\sigma^2 + \mu^3 - 3\mu\sigma^2 - \frac{3\mu}{\lambda^2} - \frac{3\sigma^2}{\lambda} - \frac{3}{\lambda^3} - \mu^3 - \frac{3\mu^2}{\lambda} - \frac{3\mu}{\lambda^2}-\frac{1}{\lambda^3}\\ &= \frac{2}{\lambda^3} \; . \end{split}$

Thus, we have:

$\label{eq:exg-skew-s6} \begin{split} \mathrm{Skew}(X) &= \frac{\mathrm{E}(X^3) - 3\cdot \mathrm{E}(X)\cdot \mathrm{Var}(X) - \mathrm{E}(X)^3}{\mathrm{Var}(X)^{\frac{3}{2}}}\\ &= \frac{\frac{2}{\lambda^3}}{\left(\sigma^2 + \frac{1}{\lambda^2}\right)^{\frac{3}{2}}}\\ &= \frac{2}{\lambda^3\left(\sigma^2+\frac{1}{\lambda^2}\right)^{\frac{3}{2}}} \; . \end{split}$

This completes the proof of \eqref{eq:exg-skew}.

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Metadata: ID: P408 | shortcut: exg-skew | author: tomfaulkenberry | date: 2023-04-21, 12:00.