题型 5. 关于分位点

1.25

设 $F\sim F\left(m,n\right)$,证明: $F_{1-\alpha }\left(m,n\right)=\frac{1}{F_{\alpha }\left(n,m\right)}$.

证

由分位点定义:

$$\begin{array}{rl}1-\alpha & =P\left\{F>F_{1-\alpha }(m,n)\right\}\\ & =P\left\{\frac{1}{F}<\frac{1}{F_{1-\alpha }(m,n)}\right\}\\ & =1-P\left\{\frac{1}{F}>\frac{1}{F_{1-\alpha }(m,n)}\right\}.\end{array}$$

则 $P\left\{\frac{1}{F}>\frac{1}{F_{1-\alpha }\left(m,n\right)}\right\}=\alpha$,

由 $F$ 分布性质可知 $\frac{1}{F}\sim F\left(n,m\right)$, 故 $\frac{1}{F_{1-\alpha }\left(m,n\right)}=F_{\alpha }\left(n,m\right)$,

即 $F_{1-\alpha }\left(m,n\right)=\frac{1}{F_{\alpha }\left(n,m\right)}$.