f分布k阶矩推导

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已于 2023-05-20 13:36:18 修改

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文章标签：概率论

于 2023-05-20 13:17:22 首次发布

本文链接：https://blog.youkuaiyun.com/weixin_51362945/article/details/130777613

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$Y=\frac{X_{1}/n_{1}}{X_{2}/n_{2}}$

其中， $X\sim \chi^{2}(n)\sim Ga(\frac{n}{2},\frac{1}{2})$

$f(x)=\frac{\lambda^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x},x> 0$

转换的雅可比式： $\left | \frac{\varphi (x_{1},x_{2})}{\varphi (y,x_{2})} \right |=\begin{vmatrix} \frac{n_{1}x_{2}}{n_{2}} &\,\,\frac{n_{1}y}{n_{2}} \\ 0 & 1 \end{vmatrix}=\left |\frac{n_{1}x_{2}}{n_{2}} \right |$

$\begin{aligned} f_{Y}(y)=&\int_{0}^{\infty} \frac{n_{1}x_{2}}{n_{2}}f_{X_{1}}(yx_{2}\frac{n_{1}}{n_{2}})f_{X_{2}}(x_{2})dx_{2}\\ =&\int_{0}^{\infty} \frac{n_{1}x_{2}}{n_{2}}\frac{1}{\Gamma(\frac{n_{1}}{2})\Gamma(\frac{n_{2}}{2})2^{\frac{n_{1}+n_{2}}{2}}}\left ( \frac{n_{1}}{n_{2}} \right )^{\frac{n_{1}}{2}}y^{\frac{n_{1}}{2}-1}\\&\,\,\,\,\,\,\,x_{2}^{\frac{n_{1}+n_{2}}{2}-1}e^{-\frac{x_{2}}{2}(1+\frac{n_{1}}{n_{2}}y)}dx_{2}\\ &t=\frac{x_{2}}{2}(1+\frac{n_{1}}{n_{2}}y) \\=&\int_{0}^{\infty} \frac{1}{\Gamma(\frac{n_{1}}{2})\Gamma(\frac{n_{2}}{2})2^{\frac{n_{1}+n_{2}}{2}}}\left ( \frac{n_{1}}{n_{2}} \right )^{\frac{n_{1}}{2}}y^{\frac{n_{1}}{2}-1}\\&\,\,\,\,\,\,\,t^{\frac{n_{1}+n_{2}}{2}-1}e^{-t}\left ( \frac{1}{2}\left(1+\frac{n_{1}}{n_{2}}y\right) \right )^{-\frac{n_{1}+n_{2}}{2}}dt\\ =&\frac{\Gamma(\frac{n_{1}+n_{2}}{2})}{\Gamma(\frac{n_{1}}{2})\Gamma(\frac{n_{2}}{2})}\left ( \frac{n_{1}}{n_{2}} \right )^{\frac{n_{1}}{2}}y^{\frac{n_{1}}{2}-1}\left ( 1+\frac{n_{1}}{n_{2}}y \right )^{-\frac{n_{1}+n_{2}}{2}} \end{aligned}$

k阶矩 $E(Y^{k})$ 为：

$\int_{0}^{\infty}y^{k}\frac{\Gamma(\frac{n_{1}+n_{2}}{2})}{\Gamma(\frac{n_{1}}{2})\Gamma(\frac{n_{2}}{2})}\left ( \frac{n_{1}}{n_{2}} \right )^{\frac{n_{1}}{2}}y^{\frac{n_{1}}{2}-1}\left ( 1+\frac{n_{1}}{n_{2}}y \right )^{-\frac{n_{1}+n_{2}}{2}}dy$

上式中积分变换难以直接看出怎么解，先利用f分布和其他分布关系来求：

$E(Y^{k})=\left(\frac{n_{2}}{n_{1}} \right )^{k}EX_{1}^{k}E\left(\frac{1}{X_{2}} \right )^{k}$

$\begin{aligned} E(X^{k})=&\int_{0}^{\infty}\frac{1}{\Gamma(\frac{n}{2})2^{\frac{n}{2}}}x^{\frac{n}{2}+k-1}e^{-\frac{1}{2}x}dx\\ &t=\frac{x}{2}\\ =&\int_{0}^{\infty}\frac{2^{k}}{\Gamma(\frac{n}{2})}t^{\frac{n}{2}+k-1}e^{-t}dt\\ =&\frac{2^{k}\Gamma(\frac{n}{2}+k)}{\Gamma(\frac{n}{2})}\\ =&2^{k}(\frac{n}{2}+k-1)(\frac{n}{2}+k-2)...\frac{n}{2}\\ =&(n+2k-2)(n+2k-4)...n\\ =&\frac{(n+2k-2)!!}{(n-2)!!} \end{aligned}$

则，

$\begin{aligned} E(Y^{k})=&\left(\frac{n_{2}}{n_{1}} \right )^{k}\frac{2^{k}\Gamma(\frac{n_{1}}{2}+k)}{\Gamma(\frac{n_{1}}{2})}\frac{2^{-k}\Gamma(\frac{n_{2}}{2}-k)}{\Gamma(\frac{n_{2}}{2})}\\ =&\left(\frac{n_{2}}{n_{1}} \right )^{k}\frac{\Gamma(\frac{n_{1}}{2}+k)\Gamma(\frac{n_{2}}{2}-k)}{\Gamma(\frac{n_{1}}{2})\Gamma(\frac{n_{2}}{2})} \end{aligned}$

对比上式与密度函数积分式的区别：

$\frac{\Gamma(\frac{n_{1}}{2}+k)\Gamma(\frac{n_{2}}{2}-k)}{\Gamma(\frac{n_{1}}{2})\Gamma(\frac{n_{2}}{2})}=\frac{\Gamma(\frac{n_{1}+n_{2}}{2})}{\Gamma(\frac{n_{1}}{2})\Gamma(\frac{n_{2}}{2})}\cdot B(\frac{n_{1}}{2}+k,\frac{n_{2}}{2}-k)$

所以证明下式等于上式右侧贝塔函数即可：

$\int_{0}^{\infty}y^{k}\left ( \frac{n_{1}}{n_{2}} \right )^{\frac{n_{1}}{2}+k}y^{\frac{n_{1}}{2}-1}\left ( 1+\frac{n_{1}}{n_{2}}y \right )^{-\frac{n_{1}+n_{2}}{2}}dy$

试图凑出与贝塔函数相似的元素，

设 $\frac{1}{1-t}=1+\frac{n_{1}}{n_{2}}y$ ，则 $t=\frac{\frac{n_{1}}{n_{2}}y}{1+\frac{n_{1}}{n_{2}}y}$ 范围是(0,1)

$\begin{aligned} &\int_{0}^{\infty}\left ( \frac{n_{1}}{n_{2}} \right )^{\frac{n_{1}}{2}+k}y^{\frac{n_{1}}{2}+k-1}\left ( 1+\frac{n_{1}}{n_{2}}y \right )^{-\frac{n_{1}+n_{2}}{2}}dy\\ &y=\frac{n_{2}}{n_{1}}\frac{t}{1-t},\left ( 1+\frac{n_{1}}{n_{2}}y \right )=\frac{1}{1-t}\\ =&\int_{0}^{1}\left ( \frac{n_{1}}{n_{2}} \right )^{-1}(\frac{t}{1-t})^{\frac{n_{1}}{2}+k-1}\left ( 1-t \right )^{\frac{n_{1}+n_{2}}{2}}\frac{n_{1}}{n_{2}}(1-t)^{-2}dt\\ =&\int_{0}^{1}t^{\frac{n_{1}}{2}+k-1}\left ( 1-t \right )^{\frac{n_{2}}{2}-k-1}dt\\ =&B(\frac{n_{1}}{2}+k,\frac{n_{2}}{2}-k) \end{aligned}$