毛刺分布¶
有两个形状参数 \(c,d > 0\) 支持的是 \(x \in [0,\infty)\) 。
\Begin{Eqnarray*}\tExtrm{设}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\f25 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{，否则}\quad 0\\
m_{n}&=&\Left(2^{1/d}-1\f25 Right)^{-1/c}\end{eqnarray*}
实施： scipy.stats.burr