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

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

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

Then, the mean or expected value of $X$ is

\[\label{eq:wald-mean} \mathrm{E}(X) = \frac{\alpha}{\gamma} \; .\]

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

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

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

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

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

\[\label{eq:wald-mean-s2} \begin{split} M_X'(0) &= \exp\left[\alpha \gamma-\sqrt{\alpha^2(\gamma^2-2(0))}\right] \cdot \frac{\alpha^2}{\sqrt{\alpha^2(\gamma^2-2(0))}} \\ &= \exp\left[\alpha \gamma - \sqrt{\alpha^2 \cdot \gamma^2}\right]\cdot \frac{\alpha^2}{\sqrt{\alpha^2\cdot \gamma^2}} \\ &= \exp[0] \cdot \frac{\alpha^2}{\alpha \gamma} \\ &= \frac{\alpha}{\gamma} \; . \end{split}\]

Metadata: ID: P169 | shortcut: wald-mean | author: tomfaulkenberry | date: 2020-09-13, 12:00.