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Burr Distribution

There are two shape parameters \(c,d > 0\) and the support is \(x \in [0,\infty)\).

\begin{eqnarray*} \textrm{Let }k & = & \Gamma\left(d\right)\Gamma\left(1-\frac{2}{c}\right)\Gamma\left(\frac{2}{c}+d\right)-\Gamma^{2}\left(1-\frac{1}{c}\right)\Gamma^{2}\left(\frac{1}{c}+d\right)\\ f\left(x;c,d\right) & = & \frac{cd}{x^{c+1}\left(1+x^{-c}\right)^{d+1}} \\ F\left(x;c,d\right) & = & \left(1+x^{-c}\right)^{-d}\\ G\left(q;c,d\right) & = & \left(q^{-1/d}-1\right)^{-1/c}\\ \mu & = & \frac{\Gamma\left(1-\frac{1}{c}\right)\Gamma\left(\frac{1}{c}+d\right)}{\Gamma\left(d\right)}\\ \mu_{2} & = & \frac{k}{\Gamma^{2}\left(d\right)}\\ \gamma_{1} & = & \frac{1}{\sqrt{k^{3}}}\left[2\Gamma^{3}\left(1-\frac{1}{c}\right)\Gamma^{3}\left(\frac{1}{c}+d\right)+\Gamma^{2}\left(d\right)\Gamma\left(1-\frac{3}{c}\right)\Gamma\left(\frac{3}{c}+d\right)\right.\\ & & \left.-3\Gamma\left(d\right)\Gamma\left(1-\frac{2}{c}\right)\Gamma\left(1-\frac{1}{c}\right)\Gamma\left(\frac{1}{c}+d\right)\Gamma\left(\frac{2}{c}+d\right)\right]\\ \gamma_{2} & = & -3+\frac{1}{k^{2}}\left[6\Gamma\left(d\right)\Gamma\left(1-\frac{2}{c}\right)\Gamma^{2}\left(1-\frac{1}{c}\right)\Gamma^{2}\left(\frac{1}{c}+d\right)\Gamma\left(\frac{2}{c}+d\right)\right.\\ & & -3\Gamma^{4}\left(1-\frac{1}{c}\right)\Gamma^{4}\left(\frac{1}{c}+d\right)+\Gamma^{3}\left(d\right)\Gamma\left(1-\frac{4}{c}\right)\Gamma\left(\frac{4}{c}+d\right)\\ & & \left.-4\Gamma^{2}\left(d\right)\Gamma\left(1-\frac{3}{c}\right)\Gamma\left(1-\frac{1}{c}\right)\Gamma\left(\frac{1}{c}+d\right)\Gamma\left(\frac{3}{c}+d\right)\right]\\ m_{d} & = & \left(\frac{cd-1}{c+1}\right)^{1/c}\,\text{if }\quad cd>1 \text{, otherwise }\quad 0\\ m_{n} & = & \left(2^{1/d}-1\right)^{-1/c}\end{eqnarray*}

Implementation: scipy.stats.burr