Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Wald distribution ▷ Variance

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

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

Then, the variance of $X$ is

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

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

\[\label{eq:var-mean} \mathrm{Var}(X) = \mathrm{E}(X^2)-\mathrm{E}(X)^2.\]

We then use the moment-generating function of the Wald distribution to calculate

\[\label{eq:wald-moment} \mathrm{E}(X^2) = M_X''(0) \; .\]

First we differentiate

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

with respect to $t$. Using the chain rule gives

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

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

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

Applying \eqref{eq:wald-moment} yields

\[\label{eq:wald-var-s3} \begin{split} \mathrm{E}(X^2) &= M_X''(0) \\ &= \alpha \cdot \exp\left[\alpha \gamma-\sqrt{\alpha^2(\gamma^2-2(0))}\right]\left[\frac{\alpha}{\gamma^2-2(0)}+\frac{1}{\sqrt{(\gamma^2-2(0))^3}}\right] \\ &= \alpha \cdot \exp\left[\alpha \gamma - \alpha \gamma\right] \cdot \left[\frac{\alpha}{\gamma^2} + \frac{1}{\gamma^3}\right] \\ &= \frac{\alpha^2}{\gamma^2} + \frac{\alpha}{\gamma^3} \; . \end{split}\]

Since the mean of a Wald distribution is given by $\mathrm{E}(X)=\alpha/\gamma$, we can apply \eqref{eq:var-mean} to show

\[\label{eq:wald-var-s4} \begin{split} \mathrm{Var}(X) &= \mathrm{E}(X^2) - \mathrm{E}(X)^2 \\ &= \frac{\alpha^2}{\gamma^2} + \frac{\alpha}{\gamma^3} - \left(\frac{\alpha}{\gamma}\right)^2 \\ &= \frac{\alpha}{\gamma^3} \end{split}\]

which completes the proof of \eqref{eq:wald-var}.

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Metadata: ID: P170 | shortcut: wald-var | author: tomfaulkenberry | date: 2020-09-13, 12:00.