# Inverse Normal (Inverse Gaussian) Distribution¶

The standard form involves the shape parameter $$\mu$$ (in most definitions, $$L=0.0$$ is used). (In terms of the regress documentation $$\mu=A/B$$ ) and $$B=S$$ and $$L$$ is not a parameter in that distribution. A standard form is $$x>0$$

\begin{eqnarray*} f\left(x;\mu\right) & = & \frac{1}{\sqrt{2\pi x^{3}}}\exp\left(-\frac{\left(x-\mu\right)^{2}}{2x\mu^{2}}\right).\\ F\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)\\ G\left(q;\mu\right) & = & F^{-1}\left(q;\mu\right)\end{eqnarray*}
\begin{eqnarray*} \mu & = & \mu\\ \mu_{2} & = & \mu^{3}\\ \gamma_{1} & = & 3\sqrt{\mu}\\ \gamma_{2} & = & 15\mu\\ m_{d} & = & \frac{\mu}{2}\left(\sqrt{9\mu^{2}+4}-3\mu\right)\end{eqnarray*}

This is related to the canonical form or JKB “two-parameter “inverse Gaussian when written in it’s full form with scale parameter $$S$$ and location parameter $$L$$ by taking $$L=0$$ and $$S\equiv\lambda,$$ then $$\mu S$$ is equal to $$\mu_{2}$$ where $$\mu_{2}$$ is the parameter used by JKB. We prefer this form because of it’s consistent use of the scale parameter. Notice that in JKB the skew $$\left(\sqrt{\beta_{1}}\right)$$ and the kurtosis ( $$\beta_{2}-3$$ ) are both functions only of $$\mu_{2}/\lambda=\mu S/S=\mu$$ as shown here, while the variance and mean of the standard form here are transformed appropriately.

Implementation: scipy.stats.invgauss

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