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