Inverted Weibull Distribution#

There is one shape parameter $c>0$ and the support is $x\geq0$ . Then

\begin{array}{rcl} f (x; c) & = & c x^{- c - 1} \exp (- x^{- c}) \\ F (x; c) & = & \exp (- x^{- c}) \\ G (q; c) & = & {(- \log q)}^{- 1 / c} \end{array}

$\begin{eqnarray*} f\left(x;c\right) & = & cx^{-c-1}\exp\left(-x^{-c}\right)\\ F\left(x;c\right) & = & \exp\left(-x^{-c}\right)\\ G\left(q;c\right) & = & \left(-\log q\right)^{-1/c}\end{eqnarray*}$

h [X] = 1 + γ + \frac{γ}{c} - \log (c)

$h\left[X\right]=1+\gamma+\frac{\gamma}{c}-\log\left(c\right)$

where $\gamma$ is Euler’s constant.

Implementation: scipy.stats.invweibull