Merge pull request #88 from tomfaulkenberry/master

add wald and wald-pdf

Author: JoramSoch

D/wald.md

@@ -0,0 +1,43 @@
+---
+layout: definition
+mathjax: true
+
+author: "Thomas J. Faulkenberry"
+affiliation: "Tarleton State University"
+e_mail: "faulkenberry@tarleton.edu"
+date: 2020-09-04 12:00:00
+
+title: "Wald distribution"
+chapter: "Probability Distributions"
+section: "Univariate continuous distributions"
+topic: "Wald distribution"
+definition: "Definition"
+
+sources:
+  - authors: "Anders, R., Alario, F. -X., and van Maanen, L."
+    year: 2016
+    title: "The Shifted Wald Distribution for Response Time Data Analysis"
+    in: "Psychological Methods"
+    pages: "vol. 21, no. 3, pp. 309-327"
+    url: "https://dx.doi.org/10.1037/met0000066"
+    doi: "10.1037/met0000066"
+
+def_id: "D95"
+shortcut: "wald"
+username: "tomfaulkenberry"
+---
+
+
+**Definition**: Let $X$ be a [random variable](/D/rvar). Then, $X$ is said to follow a Wald distribution with drift rate $\gamma$ and threshold $\alpha$
+
+$$ \label{eq:wald}
+X \sim \mathrm{Wald}(\gamma, \alpha) \; ,
+$$
+
+if and only if its [probability density function](/D/pdf) is given by
+
+$$ \label{eq:wald-pdf}
+\mathrm{Wald}(x; \gamma, \alpha) = \frac{\alpha}{\sqrt{2\pi x^3}}\exp\Bigl(-\frac{(\alpha-\gamma x)^2}{2x}\Bigr)
+$$
+
+where $\gamma > 0$, $\alpha > 0$, and the density is zero if $x \leq 0$.

P/bf-ep.md

@@ -64,7 +64,7 @@ $$
 Now, both the constrained model $m_1$ and the [encompassing model](/D/encm) $m_e$ contain the same parameter vector $\theta$. Choose a specific value of $\theta$, say $\theta'$, that exists in the support of both models $m_1$ and $m_e$ (we can do this, because $m_1$ is nested within $m_e$). Then, for this parameter value $\theta'$, we have $p(y \mid \theta',m_1)=p(y \mid \theta',m_e)$, so the expression for the Bayes factor in equation \eqref{eq:bayesfactor} reduces to an expression involving only the priors and posteriors for $\theta'$ under $m_1$ and $m_e$:
 
 $$ \label{eq:bayesfactor2}
-  B_{1e} = \frac{p(\theta' \mid m_1) / p(\theta' \mid y,m_1)}{p(\theta' \mid m_e) / p(\theta' \mid y,m_e)}.
+  \text{BF}_{1e} = \frac{p(\theta' \mid m_1) / p(\theta' \mid y,m_1)}{p(\theta' \mid m_e) / p(\theta' \mid y,m_e)}.
 $$
 
 Because $m_1$ is nested within $m_e$ via an inequality constraint, the prior $p(\theta' \mid m_1)$ is simply a truncation of the encompassing prior $p(\theta' \mid m_e)$. Thus, we can express $p(\theta' \mid m_1)$ in terms of the encompassing prior $p(\theta' \mid m_e)$ by multiplying the encompassing prior by an indicator function over $m_1$ and then normalizing the resulting product.  That is,
@@ -91,7 +91,7 @@ $$
 Plugging \eqref{eq:prior} and \eqref{eq:posterior} into \eqref{eq:bayesfactor2}, this gives us
 
 $$ \label{eq:bayesfactor3}
-  B_{1e} = \frac{c \cdot p(\theta' \mid m_e) / d \cdot p(\theta' \mid y,m_e)}{p(\theta' \mid m_e) / p(\theta' \mid y,m_e)} = \frac{c}{d} = \frac{1/d}{1/c},
+  \text{BF}_{1e} = \frac{c \cdot p(\theta' \mid m_e) / d \cdot p(\theta' \mid y,m_e)}{p(\theta' \mid m_e) / p(\theta' \mid y,m_e)} = \frac{c}{d} = \frac{1/d}{1/c},
 $$
 
 which completes the proof. Note that by definition, $1/d$ represents the proportion of the posterior distribution for $\theta$ under the [encompassing model](/D/encm) $m_e$ that agrees with the constraints imposed by $m_1$.  Similarly, $1/c$ represents the proportion of the prior distribution for $\theta$ under the [encompassing model](/D/encm) $m_e$ that agrees with the constraints imposed by $m_1$.

P/wald-pdf.md

@@ -0,0 +1,37 @@
+---
+layout: proof
+mathjax: true
+
+author: "Thomas J. Faulkenberry"
+affiliation: "Tarleton State University"
+e_mail: "faulkenberry@tarleton.edu"
+date: 2020-09-04 12:00:00
+
+title: "Probability density function of the Wald distribution"
+chapter: "Probability Distributions"
+section: "Univariate continuous distributions"
+topic: "Wald distribution"
+theorem: "Probability density function"
+
+sources:
+
+proof_id: "P162"
+shortcut: "wald"
+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 [probability density function](/D/pdf) of $X$ is
+
+$$ \label{eq:wald-pdf}
+f_X(x) = \frac{\alpha}{\sqrt{2\pi x^3}}\exp\Bigl(-\frac{(\alpha-\gamma x)^2}{2x}\Bigr) \; .
+$$
+
+
+**Proof:** This follows directly from the [definition of the Wald distribution](/D/wald).