Merge pull request #91 from tomfaulkenberry/master

add [wald-mgf, wald-mean, wald-var]

Author: JoramSoch

P/wald-mean.md

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+---
+layout: proof
+mathjax: true
+
+author: "Thomas J. Faulkenberry"
+affiliation: "Tarleton State University"
+e_mail: "faulkenberry@tarleton.edu"
+date: 2020-09-13 12:00:00
+
+title: "Mean of the Wald distribution"
+chapter: "Probability Distributions"
+section: "Univariate continuous distributions"
+topic: "Wald distribution"
+theorem: "Mean"
+
+sources:
+
+proof_id: "P169"
+shortcut: "wald-mean"
+username: "tomfaulkenberry"
+---
+
+**Theorem:** Let $X$ be a positive [random variable](/D/rvar) following a [Wald distribution](/D/wald):
+
+$$ \label{eq:wald}
+X \sim \mathrm{Wald}(\gamma, \alpha) \; .
+$$
+
+
+Then, the [mean or expected value](/D/mean) 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](/D/mom) of $X$, so we can use the [moment-generating function of the Wald distribution](/P/wald-mgf) to calculate
+
+$$ \label{eq:wald-moment}
+\mathrm{E}(X) = M_X'(0) \; .
+$$
+
+First we differentiate $M_X(t) = \exp\left[\alpha \gamma - \sqrt{\alpha^2(\gamma^2-2t)}\right]$ 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{wald-mean-s1} at $t=0$ gives the desired result:
+
+$$ \label{eq:wald-mean-s2}
+\begin{split}
+  M_X'(0) &= \exp\left[\alpha \gamma-\sqrt{\alpha^2(\gamma^2-2(0))}\right] \cdot \frac{\alpha^2}{\sqrt{\alpha^2(\gamma^2-2(0))}}\\
+          &= \exp\left[\alpha \gamma - \sqrt{\alpha^2 \cdot \gamma^2}\right]\cdot \frac{\alpha^2}{\sqrt{\alpha^2\cdot \gamma^2}}\\
+          &= \exp[0] \cdot \frac{\alpha^2}{\alpha \gamma}\\
+  &= \frac{\alpha}{\gamma}\\ \; .
+\end{split}
+$$

P/wald-mgf.md

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+---
+layout: proof
+mathjax: true
+
+author: "Thomas J. Faulkenberry"
+affiliation: "Tarleton State University"
+e_mail: "faulkenberry@tarleton.edu"
+date: 2020-09-13 12:00:00
+
+title: "Moment-generating function of the Wald distribution"
+chapter: "Probability Distributions"
+section: "Univariate continuous distributions"
+topic: "Wald distribution"
+theorem: "Moment-generating function"
+
+sources:
+  - authors: "Siegrist, K."
+    year: 2020
+    title: "The Wald Distribution"
+    in: "Random (formerly Virtual Laboratories in Probability and Statistics)"
+    pages: "retrieved on 2020-09-13"
+    url: "https://www.randomservices.org/random/special/Wald.html"
+
+  - authors: "National Institute of Standards and Technology"
+    year: 2020
+    title: "NIST Digital Library of Mathematical Functions"
+    pages: "retrieved on 2020-09-13"
+    url: "https://dlmf.nist.gov"
+
+proof_id: "P168"
+shortcut: "wald-mgf"
+username: "tomfaulkenberry"
+---
+
+
+**Theorem:** Let $X$ be a positive [random variable](/D/rvar) following a [Wald distribution](/D/wald):
+
+$$ \label{eq:wald}
+X \sim \mathrm{Wald}(\gamma, \alpha) \; .
+$$
+
+
+Then, the [moment-generating function](/D/mgf) of $X$ is
+
+
+$$ \label{eq:wald-mgf}
+M_X(t) = \exp \left[ \alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right].
+$$
+
+
+**Proof:** The [probability density function of the Wald distribution](/P/wald-pdf) is
+
+$$ \label{eq:wald-pdf}
+f_X(x) = \frac{\alpha}{\sqrt{2\pi x^3}}\exp\left(-\frac{(\alpha-\gamma x)^2}{2x}\right)
+$$
+
+and the [moment-generating function](/D/mgf) 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](/D/mean), 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](https://dlmf.nist.gov/10.25), 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](https://dlmf.nist.gov/10.39.2) 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](https://dlmf.nist.gov/10.32.10) 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}.

P/wald-var.md

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+---
+layout: proof
+mathjax: true
+
+author: "Thomas J. Faulkenberry"
+affiliation: "Tarleton State University"
+e_mail: "faulkenberry@tarleton.edu"
+date: 2020-09-13 12:00:00
+
+title: "Variance of the Wald distribution"
+chapter: "Probability Distributions"
+section: "Univariate continuous distributions"
+topic: "Wald distribution"
+theorem: "Variance"
+
+sources:
+
+proof_id: "P170"
+shortcut: "wald-var"
+username: "tomfaulkenberry"
+---
+
+**Theorem:** Let $X$ be a positive [random variable](/D/rvar) following a [Wald distribution](/D/wald):
+
+$$ \label{eq:wald}
+X \sim \mathrm{Wald}(\gamma, \alpha) \; .
+$$
+
+
+Then, the [variance](/D/var) 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](/P/var-mean):
+
+$$ \label{eq:var-mean}
+\mathrm{Var}(X) = \mathrm{E}(X^2)-\mathrm{E}(X)^2.
+$$
+
+We then use the [moment-generating function of the Wald distribution](/P/wald-mgf) to calculate
+
+$$ \label{eq:wald-moment}
+\mathrm{E}(X^2) = M_X''(0) \; .
+$$
+
+First we differentiate $M_X(t) = \exp\left[\alpha \gamma - \sqrt{\alpha^2(\gamma^2-2t)}\right]$ 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](/P/wald-mean) 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}.