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5.3.7

hard二级题目发布者: ai-batch

题干

设总体 XXX 的分布函数 F(x)F(x)F(x) 是连续的,x(1),x(2),⋯ ,x(n)x_{(1)}, x_{(2)}, \cdots, x_{(n)}x(1)​,x(2)​,⋯,x(n)​ 为取自此总体的次序统计量,设 ηi=F(x(i))\eta_i = F(x_{(i)})ηi​=F(x(i)​),试证:

(1) η1⩽η2⩽⋯⩽ηn\eta_1 \leqslant \eta_2 \leqslant \cdots \leqslant \eta_nη1​⩽η2​⩽⋯⩽ηn​,且 ηi\eta_iηi​ 是来自均匀分布 U(0,1)U(0, 1)U(0,1) 总体的次序统计量;

(2) E(ηi)=in+1E(\eta_i) = \dfrac{i}{n+1}E(ηi​)=n+1i​,Var(ηi)=i(n+1−i)(n+1)2(n+2)\mathrm{Var}(\eta_i) = \dfrac{i(n+1-i)}{(n+1)^2(n+2)}Var(ηi​)=(n+1)2(n+2)i(n+1−i)​,1⩽i⩽n1 \leqslant i \leqslant n1⩽i⩽n;

(3) ηi\eta_iηi​ 和 ηj\eta_jηj​ 的协方差矩阵为

(a1(1−a1)n+2a1(1−a2)n+2a1(1−a2)n+2a2(1−a2)n+2),\begin{pmatrix} \dfrac{a_1(1-a_1)}{n+2} & \dfrac{a_1(1-a_2)}{n+2} \\ \dfrac{a_1(1-a_2)}{n+2} & \dfrac{a_2(1-a_2)}{n+2} \end{pmatrix},​n+2a1​(1−a1​)​n+2a1​(1−a2​)​​n+2a1​(1−a2​)​n+2a2​(1−a2​)​​​,

其中 a1=in+1a_1 = \dfrac{i}{n+1}a1​=n+1i​,a2=jn+1a_2 = \dfrac{j}{n+1}a2​=n+1j​。

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