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Generalized Pareto Distribution

There is one shape parameter \(c\neq0\). The support is \(x\geq0\) if \(c>0\), and \(0\leq x<\frac{1}{\left|c\right|}\) 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*}

\[\begin{split}M\left(t\right) = \left\{ \begin{array}{cc} \left(-\frac{t}{c}\right)^{\frac{1}{c}} e^{-\frac{t}{c}} \left[ \Gamma\left(1-\frac{1}{c}\right) + \left(\gamma\left(-\frac{1}{c},-\frac{t}{c}\right) / \Gamma\left(\frac{1}{-c}\right)\right) - \pi\csc\left(\frac{\pi}{c}\right)/\Gamma\left(\frac{1}{c}\right) \right] & c>0\\ \left( \frac{\left|c\right|}{t}\right)^{1/\left|c\right|} \Gamma\left(\frac{1}{\left|c\right|}, \frac{t}{\left|c\right|}\right) \frac{1}{\Gamma\left(\frac{1}{|c|}\right)} & c<0 \end{array} \right.\end{split}\]

\[\mu_{n}^{\prime}=\frac{\left(-1\right)^{n}}{c^{n}}\sum_{k=0}^{n}\binom{n}{k}\frac{\left(-1\right)^{k}}{1-ck}\quad \text{ if }cn<1\]

\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*}

\[h\left[X\right]=1+c\quad c>0\]

Implementation: scipy.stats.genpareto