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Generalized Pareto Distribution¶
Shape parameter \(c\neq0\) and defined for \(x\geq0\) for all \(c\) and \(x<\frac{1}{\left|c\right|}\) if \(c\) is negative.
\[ \begin{eqnarray*} f\left(x;c\right) & = & \left(1+cx\right)^{-1-\frac{1}{c}}\\ F\left(x;c\right) & = & 1-\frac{1}{\left(1+cx\right)^{1/c}}\\ G\left(q;c\right) & = & \frac{1}{c}\left[\left(\frac{1}{1-q}\right)^{c}-1\right]\end{eqnarray*}\]
\[\begin{split}M\left(t\right)=\left\{ \begin{array}{cc} \left(-\frac{t}{c}\right)^{\frac{1}{c}}e^{-\frac{t}{c}}\left[\Gamma\left(1-\frac{1}{c}\right)+\Gamma\left(-\frac{1}{c},-\frac{t}{c}\right)-\pi\csc\left(\frac{\pi}{c}\right)/\Gamma\left(\frac{1}{c}\right)\right] & c>0\\ \left(\frac{\left|c\right|}{t}\right)^{1/\left|c\right|}\Gamma\left[\frac{1}{\left|c\right|},\frac{t}{\left|c\right|}\right] & c<0\end{array}\right.\end{split}\]
\[\begin{split}\mu_{n}^{\prime}=\frac{\left(-1\right)^{n}}{c^{n}}\sum_{k=0}^{n}\left(\begin{array}{c} n\\ k\end{array}\right)\frac{\left(-1\right)^{k}}{1-ck}\quad cn<1\end{split}\]
\[ \begin{eqnarray*} \mu_{1}^{\prime} & = & \frac{1}{1-c}\quad c<1\\ \mu_{2}^{\prime} & = & \frac{2}{\left(1-2c\right)\left(1-c\right)}\quad c<\frac{1}{2}\\ \mu_{3}^{\prime} & = & \frac{6}{\left(1-c\right)\left(1-2c\right)\left(1-3c\right)}\quad c<\frac{1}{3}\\ \mu_{4}^{\prime} & = & \frac{24}{\left(1-c\right)\left(1-2c\right)\left(1-3c\right)\left(1-4c\right)}\quad c<\frac{1}{4}\end{eqnarray*}\]
Thus,
\[ \begin{eqnarray*} \mu & = & \mu_{1}^{\prime}\\ \mu_{2} & = & \mu_{2}^{\prime}-\mu^{2}\\ \gamma_{1} & = & \frac{\mu_{3}^{\prime}-3\mu\mu_{2}-\mu^{3}}{\mu_{2}^{3/2}}\\ \gamma_{2} & = & \frac{\mu_{4}^{\prime}-4\mu\mu_{3}-6\mu^{2}\mu_{2}-\mu^{4}}{\mu_{2}^{2}}-3\end{eqnarray*}\]
\[\begin{split}h\left[X\right]=1+c\quad c>0\end{split}\]
Implementation: scipy.stats.genpareto