Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Wald distribution ▷ Moment-generating function

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

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

Then, the moment-generating function of $X$ is

$\label{eq:wald-mgf} M_X(t) = \exp \left[ \alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right] \; .$ $\label{eq:wald-pdf} f_X(x) = \frac{\alpha}{\sqrt{2\pi x^3}}\exp\left(-\frac{(\alpha-\gamma x)^2}{2x}\right)$

and the moment-generating function is defined as

$\label{eq:mgf-var} M_X(t) = \mathrm{E} \left[ e^{tX} \right] \; .$

Using the definition of expected value for continuous random variables, the moment-generating function of $X$ therefore is

$\label{eq:wald-mgf-s1} \begin{split} M_X(t) &= \int_0^{\infty} e^{tx} \cdot \frac{\alpha}{\sqrt{2\pi x^3}}\cdot \exp\left[-\frac{(\alpha-\gamma x)^2}{2x}\right]dx \\ &= \frac{\alpha}{\sqrt{2\pi}}\int_0^{\infty} x^{-3/2}\cdot \exp\left[tx - \frac{(\alpha-\gamma x)^2}{2x}\right]dx \; . \end{split}$

To evaluate this integral, we will need two identities about modified Bessel functions of the second kind, denoted $K_{p}$. The function $K_{p}$ (for $p\in \mathbb{R}$) is one of the two linearly independent solutions of the differential equation

$\label{eq:bessel-de} x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx}-(x^2+p^2)y=0 \; .$

The first of these identities gives an explicit solution for $K_{-1/2}$:

$\label{eq:bessel-fact1} K_{-1/2}(x) = \sqrt{\frac{\pi}{2x}} e^{-x} \; .$

The second of these identities gives an integral representation of $K_p$:

$\label{eq:bessel-fact2} K_p(\sqrt{ab}) = \frac{1}{2}\left(\frac{a}{b}\right)^{p/2} \int_0^{\infty}x^{p-1}\cdot \exp\left[-\frac{1}{2}\left(ax + \frac{b}{x}\right)\right]dx \; .$

Starting from \eqref{eq:wald-mgf-s1}, we can expand the binomial term and rearrange the moment generating function into the following form:

$\label{eq:wald-mgf-s2} \begin{split} M_X(t) &= \frac{\alpha}{\sqrt{2\pi}} \int_0^{\infty} x^{-3/2}\cdot \exp\left[ tx - \frac{\alpha^2}{2x} + \alpha\gamma - \frac{\gamma^2x}{2}\right]dx \\ &= \frac{\alpha}{\sqrt{2\pi}}\cdot e^{\alpha \gamma} \int_0^{\infty} x^{-3/2}\cdot \exp\left[\left(t-\frac{\gamma^2}{2}\right)x - \frac{\alpha^2}{2x}\right]dx \\ &= \frac{\alpha}{\sqrt{2\pi}}\cdot e^{\alpha \gamma} \int_0^{\infty} x^{-3/2}\cdot \exp \left[-\frac{1}{2}\left(\gamma^2-2t\right)x - \frac{1}{2}\cdot \frac{\alpha^2}{x}\right]dx \; . \end{split}$

The integral now has the form of the integral in \eqref{eq:bessel-fact2} with $p=-1/2$, $a=\gamma^2-2t$, and $b=\alpha^2$. This allows us to write the moment-generating function in terms of the modified Bessel function $K_{-1/2}$:

$\label{eq:wald-mgf-s3} M_X(t) = \frac{\alpha}{\sqrt{2\pi}}\cdot e^{\alpha \gamma}\cdot 2\left(\frac{\gamma^2-2t}{\alpha^2}\right)^{1/4}\cdot K_{-1/2}\left(\sqrt{\alpha^2(\gamma^2-2t)}\right).$

Combining with \eqref{eq:bessel-fact1} and simplifying gives

$\label{eq:wald-mgf-s4} \begin{split} M_X(t) &= \frac{\alpha}{\sqrt{2\pi}}\cdot e^{\alpha \gamma}\cdot 2\left(\frac{\gamma^2-2t}{\alpha^2}\right)^{1/4} \cdot \sqrt{\frac{\pi}{2\sqrt{\alpha^2(\gamma^2-2t)}}}\cdot \exp\left[-\sqrt{\alpha^2(\gamma^2-2t)}\right] \\ &= \frac{\alpha}{\sqrt{2}\cdot \sqrt{\pi}}\cdot e^{\alpha \gamma}\cdot 2 \cdot \frac{(\gamma^2-2t)^{1/4}}{\sqrt{\alpha}}\cdot \frac{\sqrt{\pi}}{\sqrt{2}\cdot \sqrt{\alpha}\cdot (\gamma^2-2t)^{1/4}}\cdot \exp\left[-\sqrt{\alpha^2(\gamma^2-2t)}\right] \\ &= e^{\alpha \gamma} \cdot \exp\left[-\sqrt{\alpha^2(\gamma^2-2t)}\right] \\ &= \exp\left[\alpha \gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right] \; . \end{split}$

This finishes the proof of \eqref{eq:wald-mgf}.

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Metadata: ID: P168 | shortcut: wald-mgf | author: tomfaulkenberry | date: 2020-09-13, 12:00.