# Generalized Extreme Value Distribution#

Extreme value distributions with one shape parameter $$c$$.

If $$c>0$$, the support is $$-\infty<x\leq1/c.$$ If $$c<0$$, the support is $$\frac{1}{c}\leq x<\infty.$$

\begin{eqnarray*} f\left(x;c\right) & = & \exp\left(-\left(1-cx\right)^{1/c}\right)\left(1-cx\right)^{1/c-1}\\ F\left(x;c\right) & = & \exp\left(-\left(1-cx\right)^{1/c}\right)\\ G\left(q;c\right) & = & \frac{1}{c}\left(1-\left(-\log q\right)^{c}\right)\end{eqnarray*}
$\mu_{n}^{\prime}=\frac{1}{c^{n}} \sum_{k=0}^{n} \binom{n}{k} \left(-1\right)^{k}\Gamma\left(ck+1\right)\quad\text{if } cn>-1$

So,

\begin{eqnarray*} \mu_{1}^{\prime} & = & \frac{1}{c}\left(1-\Gamma\left(1+c\right)\right)\quad c>-1\\ \mu_{2}^{\prime} & = & \frac{1}{c^{2}}\left(1-2\Gamma\left(1+c\right)+\Gamma\left(1+2c\right)\right)\quad c>-\frac{1}{2}\\ \mu_{3}^{\prime} & = & \frac{1}{c^{3}}\left(1-3\Gamma\left(1+c\right)+3\Gamma\left(1+2c\right)-\Gamma\left(1+3c\right)\right)\quad c>-\frac{1}{3}\\ \mu_{4}^{\prime} & = & \frac{1}{c^{4}}\left(1-4\Gamma\left(1+c\right)+6\Gamma\left(1+2c\right)-4\Gamma\left(1+3c\right)+\Gamma\left(1+4c\right)\right)\quad c>-\frac{1}{4}\end{eqnarray*}

For $$c=0$$ the distribution is the same as the (left-skewed) Gumbel distribution, and the support is $$\mathbb{R}$$.

\begin{eqnarray*} f\left(x;0\right) & = & \exp\left(-e^{-x}\right)e^{-x}\\ F\left(x;0\right) & = & \exp\left(-e^{-x}\right)\\ G\left(q;0\right) & = & -\log\left(-\log q\right)\end{eqnarray*}
\begin{eqnarray*} \mu & = & \gamma=-\psi_{0}\left(1\right)\\ \mu_{2} & = & \frac{\pi^{2}}{6}\\ \gamma_{1} & = & \frac{12\sqrt{6}}{\pi^{3}}\zeta\left(3\right)\\ \gamma_{2} & = & \frac{12}{5}\end{eqnarray*}

Implementation: scipy.stats.genextreme