Reciprocal Inverse Gaussian Distribution#

The pdf is found from the inverse gaussian (IG), $f_{RIG}\left(x;\mu\right)=\frac{1}{x^{2}}f_{IG}\left(\frac{1}{x};\mu\right)$ defined for $x\geq0$ as

\begin{array}{rcl} f_{I G} (x; μ) & = & \frac{1}{\sqrt{2 π x^{3}}} \exp (- \frac{{(x - μ)}^{2}}{2 x μ^{2}}) . \\ F_{I G} (x; μ) & = & Φ (\frac{1}{\sqrt{x}} \frac{x - μ}{μ}) + \exp (\frac{2}{μ}) Φ (- \frac{1}{\sqrt{x}} \frac{x + μ}{μ}) \end{array}

$\begin{eqnarray*} f_{IG}\left(x;\mu\right) & = & \frac{1}{\sqrt{2\pi x^{3}}}\exp\left(-\frac{\left(x-\mu\right)^{2}}{2x\mu^{2}}\right).\\ F_{IG}\left(x;\mu\right) & = & \Phi\left(\frac{1}{\sqrt{x}}\frac{x-\mu}{\mu}\right)+\exp\left(\frac{2}{\mu}\right)\Phi\left(-\frac{1}{\sqrt{x}}\frac{x+\mu}{\mu}\right)\end{eqnarray*}$

\begin{array}{rcl} f_{R I G} (x; μ) & = & \frac{1}{\sqrt{2 π x}} \exp (- \frac{{(1 - μ x)}^{2}}{2 x μ^{2}}) \\ F_{R I G} (x; μ) & = & 1 - F_{I G} (\frac{1}{x}, μ) \\ = & 1 - Φ (\frac{1}{\sqrt{x}} \frac{1 - μ x}{μ}) - \exp (\frac{2}{μ}) Φ (- \frac{1}{\sqrt{x}} \frac{1 + μ x}{μ}) \end{array}

$\begin{eqnarray*} f_{RIG}\left(x;\mu\right) & = & \frac{1}{\sqrt{2\pi x}}\exp\left(-\frac{\left(1-\mu x\right)^{2}}{2x\mu^{2}}\right)\\ F_{RIG}\left(x;\mu\right) & = & 1-F_{IG}\left(\frac{1}{x},\mu\right)\\ & = & 1-\Phi\left(\frac{1}{\sqrt{x}}\frac{1-\mu x}{\mu}\right)-\exp\left(\frac{2}{\mu}\right)\Phi\left(-\frac{1}{\sqrt{x}}\frac{1+\mu x}{\mu}\right)\end{eqnarray*}$

Implementation: scipy.stats.recipinvgauss