威布尔最大极值分布¶
定义为 \(x<0\) 和 \(c>0\) 。
\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* }
平均值是上面给出的右偏Frechet分布的负值，其他统计参数可以从
\[\mu_{n}^{\prime}=\left(-1\right)^{n}\Gamma\left(1+\frac{n}{c}\right).\]
\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{案例}
-\Left(\frac{c}{c}\right)^{\frac{1}{c}}&\text{if}\；c>1\\
0&\text{if}\；c<=1
\结束{案例}\\
m_{n}&=&-\ln\Left(2\Right)^{\frac{1}{c}}
\end{eqnarray*}
\[H\Left [X\right] =-\frac{\gamma}{c}-\log\left(c\right)+\gamma+1\]
哪里 \(\gamma\) 欧拉常数是否等于
\[\γ\约0.57721566490153286061。\]