# 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

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Weibull Minimum Extreme Value Distribution