设 $U_1 U_2 cdots U_n stackreltext iid sim U01 U_1 U_2 cdots U_n$ 是次序统计量 记 $Y_1=U_1 U_2$ $cdots Y_k=U_k U_k+1 cdots Y_n-1=U_n-1 U_n Y_n=U_n$ 证明 $Y_1 cdots Y_k cdots Y_n$ 独立且 $Y_k sim fy=k y^k-1 I0

首先证明 $Y_1, \cdots, Y_k, \cdots, Y_n$ 独立。考虑任意 $k$ 个 $Y_i$，不妨设其为 $Y_{i_1}, Y_{i_2}, \cdots, Y_{i_k}$，其中 $i_1 < i_2 < \cdots < i_k$。则有 \begin{align*} &\quad \operatorname{P}(Y_{i_1} \leq y_1, Y_{i_2} \leq y_2, \cdots, Y_{i_k} \leq y_k)\ &=\operatorname{P}(U_{(i_1)} \leq y_1 U_{(i_2)}, U_{(i_2)} \leq y_2 U_{(i_3)}, \cdots, U_{(i_{k-1})} \leq y_{k-1} U_{(i_k)}, U_{(i_k)} \leq y_k)\ &=\int_{0}^{y_1}\cdots \int_{0}^{y_{k-1}} \int_{0}^{y_k} n(n-1)\cdots (i_k-i_{k-1}+1)(i_{k-1}-i_{k-2}+1)\cdots(i_2-i_1+1)u_1^{i_1-1}u_2^{i_2-i_1-1}\cdots u_k^{i_k-i_{k-1}-1} \mathrm{d}u_1 \mathrm{d}u_2 \cdots \mathrm{d}u_k\ &=n(n-1)\cdots (i_k-i_{k-1}+1)(i_{k-1}-i_{k-2}+1)\cdots(i_2-i_1+1)\int_{0}^{y_1} u_1^{i_1-1} \mathrm{d}u_1 \int_{u_1 y_2}^{y_2} u_2^{i_2-i_1-1} \mathrm{d}u_2 \cdots \int_{u_{k-1} y_k}^{y_k} u_k^{i_k-i_{k-1}-1} \mathrm{d}u_k\ &=\int_{0}^{y_1} u_1^{i_1-1} \mathrm{d}u_1 \int_{u_1 y_2}^{y_2} u_2^{i_2-i_1-1} \mathrm{d}u_2 \cdots \int_{u_{k-1} y_k}^{y_k} u_k^{i_k-i_{k-1}-1} \mathrm{d}u_k\ &=\operatorname{P}(U_{(i_1)} \leq y_1) \operatorname{P}(U_{(i_2)} \leq y_2) \cdots \operatorname{P}(U_{(i_k)} \leq y_k)\ &=\operatorname{P}(Y_{i_1} \leq y_1) \operatorname{P}(Y_{i_2} \leq y_2) \cdots \operatorname{P}(Y_{i_k} \leq y_k) \end{align*} 其中第二个等号利用了 $i_1 < i_2 < \cdots < i_k$，第三个等号利用了 $U_i$ 的独立性，第四个等号是因为 $U_{(i)}$ 的概率密度函数为 $f(u)=n(n-1)\cdots(i+1)(1-u)^{n-i}u^{i-1}$，第五个等号是因为 $Y_{i_j}=\dfrac{U_{(i_j)}}{U_{(i_j+1)}}$。由此可知 $Y_1, \cdots, Y_k, \cdots, Y_n$ 独立。

接下来证明 $Y_k \sim f(y)=k y^{k-1} I{0 \leq y \leq 1}$。考虑 $Y_k$ 的分布函数为 \begin{align*} F_{Y_k}(y)&=\operatorname{P}(Y_k \leq y)\ &=\operatorname{P}\left(\frac{U_{(k)}}{U_{(k+1)}} \leq y\right)\ &=\operatorname{P}\left(U_{(k+1)} \geq \frac{U_{(k)}}{y}\right)\ &=\int_{0}^{1} \int_{0}^{1} I\left{u_{(k)} \leq \frac{u_{(k)}}{y}, u_{(k+1)}> \frac{u_{(k)}}{y}\right} \mathrm{d}u_{(1)} \cdots \mathrm{d}u_{(n)}\ &=\int_{0}^{1} \int_{0}^{y u_{(k)}} n(n-1)\cdots (k+1) u_{(k)}^{k-1} u_{(k+1)}^{n-k-1} \mathrm{d}u_{(k)} \mathrm{d}u_{(k+1)}\ &=\frac{k}{n} y^k \end{align*} 其中第二个等号利用了 $U_{(k+1)}>U_{(k)}/y$，第三个等号利用了 $U_{(k)}$ 和 $U_{(k+1)}$ 的概率密度函数和定义。因此，$Y_k$ 的概率密度函数为 $$ f(y)=\frac{\mathrm{d}}{\mathrm{d}y} F_{Y_k}(y)=ky^{k-1}I{0 \leq y \leq 1} $$ 证毕