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Exponential Power Distribution

One positive shape parameter \(b\). The support is \(x\geq0.\)

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

Implementation: scipy.stats.exponpow