贝塔质数分布¶
有两个形状参数 \(a,b > 0\) 支持的是 \(x \in [0,\infty)\) 。注意CDF评估使用公式。控球3.194.1分。Gradshteyn&Ryzhik的313版(第六版)。
\Begin{eqnarray*}f\Left(x；\Alpha，\Beta\Right)&=&\frac{\Gamma\left(\alpha+\beta\right)}{\Gamma\left(\alpha\right)\Gamma\left(\beta\right)}x^{\alpha-1}\left(1+x\right)^{-\alpha-\beta}\\
F\Left(x；\Alpha，\Beta\Right)&=&\frac{\Gamma\left(\alpha+\beta\right)}{\alpha\Gamma\left(\alpha\right)\Gamma\left(\beta\right)}x^{\alpha}\，_{2}F_{1}\Left(\Alpha+\Beta，\Alpha；1+\Alpha；-x\Right)\\
G\Left(q；\alpha，\beta\right)&=&F^{-1}\Left(x；\alpha，\beta\right)\end{eqnarray*}
\[\begin{split}\MU_{n}^{\Prime}=\Left\{
\BEGIN{array}{ccc}
\frac{\Gamma\left(n+\alpha\right)\Gamma\left(\beta-n\right)}{\Gamma\left(\alpha\right)\Gamma\left(\beta\right)}=\frac{\left(\alpha\right)_{n}}{\left(\beta-n\right)_{n}}&&\测试版>n\\
\infty&&\mathm{否则}
\end{array}\右。\end{split}\]
所以呢，
\Begin{eqnarray*}\mu&=&\frac{\alpha}{\beta-1}\quad\texpm{for}\beta>1\\
\MU_{2}&=&\frac{\alpha\left(\alpha+1\right)}{\left(\beta-2\right)\left(\beta-1\right)}-\frac{\alpha^{2}}{\left(\beta-1\right)^{2}}\quad\textrm{for}\测试版>2\\
\Gamma_{1}&=&\frac{\frac{\alpha\left(\alpha+1\right)\left(\alpha+2\right)}{\left(\beta-3\right)\left(\beta-2\right)\left(\beta-1\right)}-3\mu\mu_{2}-\mu^{3}}{\mu_{2}^{3/2}}\quad\textrm{for}\Beta>3\\
\Gamma_{2}&=&\frac{\MU_{4}}{\MU_{2}^{2}}-3\\
\MU_{4}&=&\frac{\alpha\left(\alpha+1\right)\left(\alpha+2\right)\left(\alpha+3\right)}{\left(\beta-4\right)\left(\beta-3\right)\left(\beta-2\right)\left(\beta-1\right)}-4\mu\mu_{3}-6\mu^{2}\mu_{2}-\mu^{4}\quad\textrm{for}\BETA>4\end{等式*}