包络柯西分布¶
有一个形状参数 \(c\in\left(0,1\right)\) 在有支持的情况下 \(x\in\left[0,2\pi\right]\) 。
\BEGIN{eqnarray*}f\Left(x；c\right)&=&\frac{1-c^{2}}{2\pi\Left(1+c^{2}-2c\cos x\right)}\\
G_{c}\Left(x\Right)&=&\frac{1}{\pi}\arctan\left(\frac{1+c}{1-c}\tan\left(\frac{x}{2}\right)\right)\\
R_{c}\Left(Q\Right)&=&2\arctan\Left(\frac{1-c}{1+c}\tan\Left(\pi Q\Right)\Right)\\
f\Left(x；c\Right)&=&\Left\{
\BEGIN{array}{ccc}
G_{c}\Left(x\Right)&&0\leq x<\pi\\
1-g_{c}\Left(2\pi-x\right)&&\pi\leq x\leq2\pi
\end{数组}
\对。\\
g\Left(q；c\Right)&=&\Left\{
\BEGIN{array}{ccc}
R_{c}\Left(Q\Right)&&0\leq Q<\frac{1}{2}\\
2\pi-r_{c}\Left(1-q\right)&&\frac{1}{2}\leq q\leq1
\end{数组}
\对。\end{eqnarray*}
\[H\Left [X\right] =\log\Left(2\pi\Left(1-c^{2}\Right)\Right)。\]