Weibull Maximum Extreme Value Distribution¶
Defined for \(x<0\) and \(c>0\) .
\begin{eqnarray*} f\left(x;c\right) & = & c\left(-x\right)^{c-1}\exp\left(-\left(-x\right)^{c}\right)\\ F\left(x;c\right) & = & \exp\left(-\left(-x\right)^{c}\right)\\ G\left(q;c\right) & = & -\left(-\log q\right)^{1/c}\end{eqnarray*}
The mean is the negative of the right-skewed Frechet distribution given above, and the other statistical parameters can be computed from
\[\mu_{n}^{\prime}=\left(-1\right)^{n}\Gamma\left(1+\frac{n}{c}\right).\]
\begin{eqnarray*}
\mu & = & -\Gamma\left(1+\frac{1}{c}\right) \\
\mu_{2} & = & \Gamma\left(1+\frac{2}{c}\right) -
\Gamma^{2}\left(1+\frac{1}{c}\right) \\
\gamma_{1} & = & -\frac{\Gamma\left(1+\frac{3}{c}\right) -
3\Gamma\left(1+\frac{2}{c}\right)\Gamma\left(1+\frac{1}{c}\right) +
2\Gamma^{3}\left(1+\frac{1}{c}\right)}
{\mu_{2}^{3/2}} \\
\gamma_{2} & = & \frac{\Gamma\left(1+\frac{4}{c}\right) -
4\Gamma\left(1+\frac{1}{c}\right)\Gamma\left(1+\frac{3}{c}\right) +
6\Gamma^{2}\left(1+\frac{1}{c}\right)\Gamma\left(1+\frac{2}{c}\right) -
3\Gamma^{4}\left(1+\frac{1}{c}\right)}
{\mu_{2}^{2}} - 3 \\
m_{d} & = & \begin{cases}
-\left(\frac{c-1}{c}\right)^{\frac{1}{c}} & \text{if}\; c > 1 \\
0 & \text{if}\; c <= 1
\end{cases} \\
m_{n} & = & -\ln\left(2\right)^{\frac{1}{c}}
\end{eqnarray*}
\[h\left[X\right]=-\frac{\gamma}{c}-\log\left(c\right)+\gamma+1\]
where \(\gamma\) is Euler’s constant and equal to
\[\gamma\approx0.57721566490153286061.\]
Implementation: scipy.stats.weibull_max