Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Wald distribution ▷ Skewness

Theorem: Let $X$ be a random variable following a Wald distribution:

$\label{eq:wald} X \sim \mathrm{Wald}(\gamma,\alpha) \; .$

Then the skewness of $X$ is

$\label{eq:wald-skew} \mathrm{Skew}(X) = \frac{3}{\sqrt{\alpha\gamma}} \; .$

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. Since $X$ follows an Wald distribution, the mean of $X$ is given by

$\label{eq:wald-mean} \mu = \mathrm{E}(X) = \frac{\alpha}{\gamma}$

and the standard deviation of $X$ is given by

$\label{eq:wald-var} \sigma = \sqrt{\mathrm{Var}(X)} = \sqrt{\frac{\alpha}{\gamma^3}}\; .$

Substituting \eqref{eq:wald-mean} and \eqref{eq:wald-var} into \eqref{eq:skew-mean} gives:

$\label{eq:skew-mean-alt} \begin{split} \mathrm{Skew}(X) &= \frac{\mathrm{E}(X^3)-3\mu\sigma^2-\mu^3}{\sigma^3}\\ &= \frac{\mathrm{E}(X^3) - 3\left(\frac{\alpha}{\gamma}\right)\left(\frac{\alpha}{\gamma^3}\right)-\left(\frac{\alpha}{\gamma}\right)^3}{\left(\sqrt{\frac{\alpha}{\gamma^3}}\right)^3}\\ &= \frac{\gamma^{9/2}}{\alpha^{3/2}}\left[\mathrm{E}(X^3) - \frac{3\alpha^2}{\gamma^4}-\frac{\alpha^3}{\gamma^3}\right] \; . \end{split}$

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

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

based on the relationship between raw moment and moment-generating function. First, we differentiate the moment-generating function

$\label{eq:wald-mgf} M_X(t) = \exp\left[\alpha \gamma - \sqrt{\alpha^2(\gamma^2-2t)}\right]$

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

$\label{eq:wald-skew-s1} \begin{split} M_X'(t) &= \exp\left[\alpha \gamma - \sqrt{\alpha^2(\gamma^2-2t)}\right] \cdot -\frac{1}{2}\left(\alpha^2(\gamma^2-2t)\right)^{-1/2}\cdot -2\alpha^2 \\ &= \exp\left[\alpha \gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right] \cdot \frac{\alpha^2}{\sqrt{\alpha^2(\gamma^2-2t)}} \\ &= \alpha \cdot \exp\left[\alpha \gamma -\sqrt{\alpha^2(\gamma^2-2t)}\right] \cdot (\gamma^2-2t)^{-1/2} \; . \end{split}$

Now we use the product rule to obtain the second derivative:

$\label{eq:wald-skew-s2} \begin{split} M_X''(t) &= \alpha \cdot \exp\left[\alpha \gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right]\cdot (\gamma^2-2t)^{-1/2}\cdot -\frac{1}{2}\left(\alpha^2(\gamma^2-2t)\right)^{-1/2}\cdot -2\alpha^2 \\ &+ \alpha \cdot \exp\left[\alpha \gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right]\cdot -\frac{1}{2}(\gamma^2-2t)^{-3/2}\cdot -2 \\ &= \alpha^2\cdot \exp\left[\alpha \gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right]\cdot (\gamma^2-2t)^{-1} \\ &+ \alpha\cdot \exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right]\cdot (\gamma^2-2t)^{-3/2} \\ & = \frac{\alpha^2}{\gamma^2-2t}\exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right] + \frac{\alpha}{(\gamma^2-2t)^{3/2}}\exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right] \; . \end{split}$

Finally, one more application of the chain rule will give us the third derivative. To start, we will decompose the second derivative obtained in \eqref{eq:wald-skew-s2} as

$\label{eq:wald-skew-s3} M''(t) = f(t) + g(t)$

where

$\label{eq:wald-skew-split1} f(t) = \frac{\alpha^2}{\gamma^2-2t}\exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-t2)}\right]$

and

$\label{eq:wald-skew-split2} g(t) = \frac{\alpha}{(\gamma^2-2t)^{3/2}}\exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right] \; .$

With this decomposition, $M_X’’‘(t) = f’(t) + g’(t)$. Applying the product rule to $f$ gives:

$\label{eq:wald-skew-f} \begin{split} f'(t) &= 2\alpha^2(\gamma^2-2t)^{-2}\exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right] \\ &+ \alpha^2(\gamma^2-2t)^{-1}\exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right]\cdot -\frac{1}{2}\left[\alpha^2(\gamma^2-2t)\right]^{-1/2}\cdot -2\alpha^2\\ &= \frac{2\alpha^2}{(\gamma^2-2t)^2}\exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right] \\ &+ \frac{\alpha^3}{(\gamma^2-2t)^{3/2}}\exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right] \; . \end{split}$

Similarly, applying the product rule to $g$ gives:

$\label{eq:wald-skew-g} \begin{split} g'(t) &= -\frac{3}{2}\alpha(\gamma^2-2t)^{-5/2}(-2)\exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right] \\ &+ \alpha(\gamma^2-2t)^{-3/2}\exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right]\cdot -\frac{1}{2}\left[\alpha^2(\gamma^2-2t)\right]^{-1/2}\cdot -2\alpha^2\\ &= \frac{3\alpha}{(\gamma^2-2t)^{5/2}}\exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right] \\ &+ \frac{\alpha^2}{(\gamma^2-2t)^2}\exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right] \; . \end{split}$

Applying \eqref{eq:wald-moment}, together with \eqref{eq:wald-skew-f} and \eqref{eq:wald-skew-g}, yields

$\label{eq:wald-skew-s4} \begin{split} \mathrm{E}(X^3) &= M_X'''(0)\\ &= f'(0) + g'(0)\\ &= \left[\frac{2\alpha^2}{\gamma^4}+\frac{\alpha^3}{\gamma^3}\right] + \left[\frac{3\alpha}{\gamma^5}+\frac{\alpha^2}{\gamma^4}\right]\\ &= \frac{3\alpha^2}{\gamma^4} + \frac{\alpha^3}{\gamma^3} + \frac{3\alpha}{\gamma^5} \; . \end{split}$

We now substitute \eqref{eq:wald-skew-s4} into \eqref{eq:skew-mean-alt}, giving

$\label{eq:wald-skew-s5} \begin{split} \mathrm{Skew}(X) &= \frac{\gamma^{9/2}}{\alpha^{3/2}}\left[\mathrm{E}(X^3) - \frac{3\alpha^2}{\gamma^4}-\frac{\alpha^3}{\gamma^3}\right] \\ &= \frac{\gamma^{9/2}}{\alpha^{3/2}}\left[\frac{3\alpha^2}{\gamma^4} + \frac{\alpha^3}{\gamma^3} + \frac{3\alpha}{\gamma^5} - \frac{3\alpha^2}{\gamma^4}-\frac{\alpha^3}{\gamma^3}\right] \\ &= \frac{\gamma^{9/2}}{\alpha^{3/2}} \cdot \frac{3\alpha}{\gamma^5}\\ &= \frac{3}{\alpha^{1/2}\cdot \gamma^{1/2}}\\ &= \frac{3}{\sqrt{\alpha\gamma}} \; . \end{split}$

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

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Metadata: ID: P421 | shortcut: wald-skew | author: tomfaulkenberry | date: 2023-10-24, 12:00.