命题 6.3.5 (正态分布的顺序统计量)

设总体 $X$ 的分布函数为 $F_X$, 分布密度函数为 $f_X$, 则对 $k=1,2,\cdots ,n$

{f}_{{X}_{\left( k\right) }}\left( x\right) = \frac{n!}{\left( {n - k}\right) !\left( {k - 1}\right) !}{\left\lbrack {F}_{X}\left( x\right) \right\rbrack }^{k - 1}{\left\lbrack 1 - {F}_{X}\left( x\right) \right\rbrack }^{n - k}{f}_{X}\left( x\right). \tag{6.3.17} $$特别地,

{f}{{X}{\left( 1\right) }}\left( x\right) = n{\left\lbrack 1 - {F}{X}\left( x\right) \right\rbrack }^{n - 1}{f}{X}\left( x\right), \tag{6.3.18}

{f}{{X}{\left( n\right) }}\left( x\right) = n{\left\lbrack {F}{X}\left( x\right) \right\rbrack }^{n - 1}{f}{X}\left( x\right). \tag{6.3.19}