样本极值在经济学、环境科学、气候学、水文学、精算学和拍卖理论等领域有着重要的应用. 令$ X_{1:n} $和$ X_{n:n} $表示随机变量$ X_{1}, X_{2}, \cdots , X_{n} $的最小和最大次序统计量. 例如, 在经济学中, 最小次序统计量$ X_{1:n} $和最大次序统计量$ X_{n:n} $分别定义了第一次价格采购拍卖的获胜者价格和第一次价格密封投标拍卖的中标价格^{[7, 11]}. 在过去的三十年里, 很多学者的研究集中在服从不同分布的随机变量的样本极值, 如指数分布、威布尔分布、伽玛分布和比例失效率分布, 相关结果参见文献[3-5, 9, 10, 13, 16-18]. 关于独立随机变量的综合研究结论见文献[2]. 最近, Archimedean copula刻画的相依随机变量的一般研究成果可参见文献[6, 8, 12].

随机变量$ X_{1}, \cdots , X_{n} $被称为服从尺度比例失效率模型(scale proportional hazards model, 简称SPH), 如果$ X_{i} $的生存函数$ \overline{F_{i}}(x)=\overline{F}^{\alpha_{i}}(\lambda_{i}x), $$ \alpha_{i}>0, \lambda_{i}>0, i\in I_{n}=\{1, 2, \cdots , n\}, $这里$ \overline{F} $是基本生存函数, $ \boldsymbol{\alpha}=(\alpha_{1}, \cdots , \alpha_{n}) $是脆弱参数向量, $ \boldsymbol{\lambda}=(\lambda_{1}, \cdots , \lambda_{n}) $是尺度参数向量. 记号$ \boldsymbol{X}=(X_{1}, \cdots , X_{n})\thicksim SPH(\overline{F}, \boldsymbol{\alpha}, \boldsymbol{\lambda}, \phi) $表明随机变量的相依是具有生成子$ \phi $的Archimedean copula和$ X_{i}\thicksim \overline{F}^{\alpha_{i}}(\lambda_{i}x) $, 对任意的$ i\in I_{n} $. 很显然比例失效率模型是尺度比例失效率模型的一个特殊情形. 文献[8]给出了Archimedean copula刻画的SPH模型的样本极值的普通随机序和分散序的随机比较. 沿着这条研究路线, 我们研究两组来自于不同Archimedean copula刻画的SPH模型的最小次序统计量的随机比较. 在某种数学意义下, 一个由尺度比例失效率分布的不同脆弱参数和尺度参数构成的矩阵变化到另一个矩阵时, 该文研究了在一定的条件下, 来自于第一个尺度比例失效率分布的最小次序统计量在普通随机序下小于变化到的参数矩阵对应的尺度比例失效率分布的最小次序统计量.

本文的组织结构如下: 第2节简要给出相关的预备知识, 如普通随机序, 多元链式优化序, Archimedean copula的定义. 第3节, 在多元链式优化序下, 给出了两组来自于不同相依尺度比例失效率分布的最小次序统计量的普通随机序比较的一些充分条件. 同时也给出了一些数值例子来说明得到的结果的正确性. 最后, 第4节给出了本文的总结.

定义2.1   随机变量$ Y $在普通随机序下小于随机变量$ X $, 记为$ X\geq_{st} Y $, 若$ \bar{F}(x)=P(X>x) \geq P(Y>x)=\bar{G}(x) $对所有的$ x $成立.

一个方阵$ \Pi $被称为是置换矩阵, 如果矩阵的每行和每列只有一个元素是1, 其他的元素都是0. 显然, 通过交换$ n\times n $单位阵的行(或列)可以得到$ n! $个置换矩阵. 一个$ n\times n $矩阵$ P=(p_{ij}) $被称为双随机的, 如果元素$ p_{ij}\geq 0 $, $ i, j=1, \cdots n $, 且每行和每列的所有元素之和是1. 一个$ T $-transform矩阵具有如下的表达式$ T_{w}=wI_{n}+(1-w)\Pi, $这里$ 0 \leq w \leq 1 $和$ \Pi $是一个仅交换两个分量得到的置换矩阵. 令$ T_{w_{1}}=w_{1}I_{n}+(1-w_{1})\Pi_{1} $和$ T_{w_{2}}=w_{2}I_{n}+(1-w_{2})\Pi_{2} $是两个$ T $-transform矩阵, 这里$ \Pi_{1} $和$ \Pi_{2} $是两个仅交换两个分量得到的置换矩阵. 当$ \Pi_{1}=\Pi_{2}, $则$ T_{w_{1}} $和$ T_{w_{2}} $具有相同的结构. 当$ \Pi_{1} $和$ \Pi_{2} $不相同, 则$ T_{w_{1}} $和$ T_{w_{2}} $是具有不同结构. 有限个具有相同结构的$ T $-transform矩阵的乘积仍然是一个$ T $-transform矩阵, 且结构保持不变. 然而, 此结论对于有限个具有不同结构的$ T $-transform矩阵的乘积不成立.

接下来, 我们给出向量优化序和多元链式优化序的定义.

定义2.2  令$ \boldsymbol{\lambda}=(\lambda_{1}, \cdots , \lambda_{n}) $和$ \boldsymbol{\lambda^{\ast}}= (\lambda_{1}^{\ast}, \cdots , \lambda_{n}^{\ast}) $是两个实向量, $ \lambda_{(1)}\leq \dots\leq \lambda_{(n)} $和$ \lambda_{(1)}^{\ast}\leq \dots\leq \lambda_{(n)}^{\ast} $表示相应的有序元素. 则$ \boldsymbol{\lambda^{\ast}} $说成被$ \boldsymbol{\lambda} $优化, 记为$ \boldsymbol{\lambda}\succeq_{m}\boldsymbol{\lambda^{\ast}} $, 如果

$ \sum\limits_{i=1}^{k}\lambda_{(i)}\leq \sum\limits_{i=1}^{k}\lambda_{(i)}^{\ast}, $

$ k=1, 2, \cdots , n-1 $, 和$ \sum\limits_{i=1}^{n}\lambda_{i}= \sum\limits_{i=1}^{n}\lambda_{i}^{\ast} $.

定义2.3  令$ A=\{a_{ij}\} $和$ B=\{b_{ij}\} $是两个$ m\times n $矩阵, $ a_{1}^{R}, \cdots , a_{m}^{R} $和$ b_{1}^{R}, \cdots , b_{m}^{R} $分别是$ A $和$ B $的行向量. 则

(ⅰ) $ A $多元链式优化$ B $ (记为$ A \gg B $), 如果存在有限个$ n\times n $$ T $-transform矩阵$ T_{w_{1}}, \cdots , T_{w_{k}} $使得$ B = AT_{w_{1}}T_{w_{2}}\cdots T_{w_{k}} $;

(ⅱ) $ A $行优化$ B $ (记为$ A >^{row} B $), 如果$ a_{i}^{R}\succeq_{m} b_{i}^{R} $, $ i = 1, \cdots , m. $

由上面的定义可得$ A \gg B \Rightarrow A >^{row} B. $关于多元链式优化序更多的详细结论, 有兴趣的读者可见文献[14].

令

$ \begin{eqnarray*} S_{n}=\biggl\{(\mathbf{x}, \mathbf{y})=\left( \begin{array}{c} x_{1}\cdots x_{n} \\ y_{1}\cdots y_{n} \\ \end{array} \right) : x_{i}> 0, y_{j}> 0, (x_{i}-x_{j})(y_{i}-y_{j})\geq 0, i, j = 1, \cdots , n \biggr\}. \end{eqnarray*} $

定理2.4   一个可微函数$ \phi: (0, \infty)^{4}\rightarrow (0, \infty) $满足

$ \phi(A)\geq(\leq) \phi(B) \; \boldsymbol{对所有}\; A, B \; \boldsymbol{使得}\; A\in S_{2}\; \boldsymbol{和} \; A\gg B $

当且仅当

ⅰ) $ \phi(A)=\phi(A\Pi) $对所有的置换矩阵$ \Pi $;

ⅱ) $ \sum\limits_{i=1}^2 (a_{ik}-a_{ij})[\phi_{ik}(A)-\phi_{ij}(A)]\geq (\leq)0 $对所有$ j, k=1, 2, $这里$ \phi_{ij}(A)=\partial\phi(A)/\partial a_{ij}. $

注2.5  定理2.4的证明类似于文献[1]的定理2的证明, 为了简洁, 这里忽略其证明过程.

Archimedean copula是一种常见的刻画相依性的工具. 设$ \phi: $$ [0, +\infty) \rightarrow [0, 1] $是一个非增加的连续函数, 且满足$ \phi(0)=1 $和$ \phi(+\infty)=0 $, 令$ \psi=\phi^{-1} $是对应的逆函数. 则

$ C_{\phi}(u_{1}, \cdots , u_{n})=\phi(\psi(u_{1})+\cdots+\psi(u_{n})), \; \; \boldsymbol{对于 }\; \; (u_{1}, \cdots , u_{n})\in [0, 1]^{n}, $

被称为是一个具有生成子$ \phi $的Archimedean copula, 其中$ (-1)^{k}\phi^{(k)}(x)\geq 0 $, $ k=0, \cdots , n-2 $和$ (-1)^{n-2}\phi^{(n-2)}(x) $是非增加的凸函数. 生成子$ \phi $被称为是严格的, 若$ \phi (+\infty)=0 $. Archimedean copula包含很多著名的copulas, 比如Clayton copula和Ali-Mikhail-Haq copula. 关于更多的Archimedean copula信息及其应用, 读者可以见文献[6, 12, 15].

引理2.6^[12]  对于两个$ n $维Archimedean copula $ C_{\phi_{1}}(u) $和$ C_{\phi_{2}}(u), $如果$ \psi_{2}\circ \phi_{1} $具有超可加性, 则$ C_{\phi_{1}}(u)\leq C_{\phi_{2}}(u) $对所有$ u\in [0, 1]^{n}. $

在本节, 我们考虑来自于不同相依结构的尺度比例失效率分布的两组样本最小次序统计量的普通随机序. 下面的定理表明, 若一个尺度比例失效率分布的参数构成的矩阵链式优化另一个比例失效率分布的参数构成的矩阵, 则基于某些充分条件, 来自于第一个分布的样本最小次序统计量在普通随机序下小于来自于第二个分布的样本最小次序统计量.

定理3.1   令$ (X_{1}, X_{2})\thicksim SPH(\overline{F}, \boldsymbol{ \alpha}, \boldsymbol{\lambda }, \phi_{1}) $和$ (X_{1}^{*}, X_{2}^{*})\thicksim SPH(\overline{F}, \boldsymbol{\alpha^{*} }, \boldsymbol{\lambda^{*}}, \phi_{2}) $. 假定下列条件成立:

(ⅰ) $ \overline{F} $是对数凹;

(ⅱ) $ \psi_{2}\circ \phi_{1} $具有超可加性;

(ⅲ) $ \phi_{2} $是对数凸.

则, 对于$ \left(\begin{array}{ccc}\alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\end{array}\right)\in S_{2}, $我们有$ \left(\begin{array}{ccc}\alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\end{array}\right)\gg \left(\begin{array}{ccc}\alpha_{1}^{*} & \alpha_{2}^{*} \\ \lambda_{1}^{*} &\lambda_{2}^{*}\end{array}\right)\Rightarrow X_{1:2}\leq_{st}X^{*}_{1:2} $.

证   对$ x>0 $, $ X_{1:2} $的生存函数为

$ \overline{F}_{X_{1:2}}(x)=P(X_{1:2}>x)=P(X_{1}>x, X_{2}>x)=\phi_{1}\biggl(\sum\limits_{i=1}^{2}\psi_{1}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr). $

由于$ \psi_{2}\circ \phi_{1} $具有超可加性, 由引理2.6可得

$ \phi_{1}\biggl(\sum\limits_{i=1}^{2}\psi_{1}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr)\leq \phi_{2}\biggl(\sum\limits_{i=1}^{2}\psi_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr)\triangleq \overline{G}_{X_{1:2}}(x). $

于是, 为了证明$ X_{1:2}\leq_{st} X^{*}_{1:2}, $只需要证明对$ x>0 $, 下式成立

$ \overline{G}_{X_{1:2}}(x)=\phi_{2}\biggl(\sum\limits_{i=1}^{2}\psi_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr)\leq \phi_{2}\biggl(\sum\limits_{i=1}^{2}\psi_{2}(\overline{F}^{\alpha^{*}_{i}}(\lambda^{*}_{i}x))\biggr)=\overline{F}_{X^{*}_{1:2}}(x), $

当$ \left(\begin{array}{ccc}\alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\end{array}\right)\gg \left(\begin{array}{ccc}\alpha_{1}^{*} & \alpha_{2}^{*} \\ \lambda_{1}^{*} &\lambda_{2}^{*}\end{array}\right) $和$ \left(\begin{array}{ccc}\alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\end{array}\right)\in S_{2}. $

很容易看出$ \overline{G}_{X_{1:2}}(x) $是关于$ (\alpha_{i}, \lambda_{i}) $置换不变的, 于是定理2.4 (ⅰ)的条件是满足的. 接下来, 我们只需要证明定理2.4 (ⅱ)的条件也是满足的即可.

假设$ (\boldsymbol{\alpha}, \boldsymbol{\lambda})\in S_{2} $表明$ (\alpha_{1}-\alpha_{2})(\lambda_{1}-\lambda_{2})\geq 0. $这意味着$ \alpha_{1}\geq \alpha_{2} $和$ \lambda_{1}\geq \lambda_{2}, $或者$ \alpha_{1}\leq \alpha_{2} $和$ \lambda_{1}\leq \lambda_{2}. $我们只给出第一种情形$ \alpha_{1}\geq \alpha_{2} $和$ \lambda_{1}\geq \lambda_{2} $下定理2.4 (ⅱ)的条件成立的证明, 类似证明第二种情形.

$ \overline{G}_{X_{1:2}}(x) $关于$ \alpha_{i} $和$ \lambda_{i} $的偏导数分别为

$ \frac{\partial \overline{G}_{X_{1:2}}(x)}{\partial \alpha_{i}}=\phi_{2}'\biggl(\sum\limits_{i=1}^{2}\psi_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr) \psi'_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\overline{F}^{\alpha_{i}}(\lambda_{i}x)\ln\overline{F}(\lambda_{i}x), $

$ \frac{\partial \overline{G}_{X_{1:2}}(x)}{\partial \lambda_{i}}=\phi_{2}'\biggl(\sum\limits_{i=1}^{2}\psi_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr) \psi'_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x)) \overline{F}^{\alpha_{i}}(\lambda_{i}x) \biggl(-\alpha_{i}x\frac{f(\lambda_{i}x)}{\overline{F}(\lambda_{i}x)}\biggr). $

对于固定的$ x>0, $我们定义如下的$ g(\boldsymbol{\alpha}, \boldsymbol{\lambda}) $:

$ g({\bf{ \pmb{\mathit{ α}}}}, {\bf{ \pmb{\mathit{ λ}}}})=g_{1}({\bf{ \pmb{\mathit{ α}}}}, {\bf{ \pmb{\mathit{ λ}}}})+g_{2}({\bf{ \pmb{\mathit{ α}}}}, {\bf{ \pmb{\mathit{ λ}}}}), $

这里

(3.1) $ \begin{eqnarray} g_{1}({\bf{ \pmb{\mathit{ α}}}}, {\bf{ \pmb{\mathit{ λ}}}})&=&(\alpha_{1}-\alpha_{2})\biggl(\frac{\partial \overline{G}_{X_{1:2}}(x)}{\partial \alpha_{1}}-\frac{\partial \overline{G}_{X_{1:2}}(x)}{\partial \alpha_{2}}\biggr){}\\ &=&(\alpha_{1}-\alpha_{2})\phi_{2}'\biggl(\sum\limits_{i=1}^{2}\psi_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr)\biggl(\psi'_{2}(\overline{F}^{\alpha_{1}}(\lambda_{1}x)) \overline{F}^{\alpha_{1}}(\lambda_{1}x)\ln\overline{F}(\lambda_{1}x)\\ &&- \psi'_{2}(\overline{F}^{\alpha_{2}}(\lambda_{2}x)) \overline{F}^{\alpha_{2}}(\lambda_{2}x)\ln\overline{F}(\lambda_{2}x)\biggr), \end{eqnarray} $

(3.2) $ \begin{eqnarray} g_{2}({\bf{ \pmb{\mathit{ α}}}}, {\bf{ \pmb{\mathit{ λ}}}})&=&(\lambda_{1}-\lambda_{2})\biggl(\frac{\partial \overline{G}_{X_{1:2}}(x)}{\partial \lambda_{1}}-\frac{\partial \overline{G}_{X_{1:2}}(x)}{\partial \lambda_{2}}\biggr){}\\ &=&(\lambda_{1}-\lambda_{2})x \phi_{2}'\biggl(\sum\limits_{i=1}^{2}\psi_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr)\biggl( \frac{\alpha_{2}f(\lambda_{2}x)}{\overline{F}(\lambda_{2}x)}\psi'_{2}(\overline{F}^{\alpha_{2}}(\lambda_{2}x)) \overline{F}^{\alpha_{2}}(\lambda_{2}x)\\ &&- \frac{\alpha_{1}f(\lambda_{1}x)}{\overline{F}(\lambda_{1}x)}\psi'_{2}(\overline{F}^{\alpha_{1}}(\lambda_{1}x)) \overline{F}^{\alpha_{1}}(\lambda_{1}x)\biggr). \end{eqnarray} $

首先, 注意到$ \alpha_{1}\geq \alpha_{2}, \lambda_{1}\geq \lambda_{2}, $和由条件(ⅲ): $ t \psi'_{2}(t)=\frac{t}{\phi_{2}'(\phi_{2}^{-1}(t))}=\frac{\phi_{2}(\phi_{2}^{-1}(t))}{\phi_{2}'(\phi_{2}^{-1}(t))} $关于$ t>0 $是增加的, 我们可得

$ \overline{F}^{\alpha_{1}}(\lambda_{1}x)\leq \overline{F}^{\alpha_{1}}(\lambda_{2}x)\leq \overline{F}^{\alpha_{2}}(\lambda_{2}x), $

(3.3) $ \begin{equation} \ln\overline{F}(\lambda_{1}x)\leq \ln\overline{F}(\lambda_{2}x)\leq 0, \end{equation} $

(3.4) $ \begin{equation} \psi'_{2}(\overline{F}^{\alpha_{1}}(\lambda_{1}x)) \overline{F}^{\alpha_{1}}(\lambda_{1}x)\leq \psi'_{2}(\overline{F}^{\alpha_{2}}(\lambda_{2}x)) \overline{F}^{\alpha_{2}}(\lambda_{2}x)\leq 0. \end{equation} $

结合(3.1), (3.3)和(3.4)式, 可以得到$ g_{1}(\boldsymbol{\alpha}, \boldsymbol{\lambda})\leq 0. $

其次, $ \overline{F} $是对数凹的, 表明$ h(x)=\frac{f(x)}{\overline{F}(x)} $关于$ x>0 $是增加的. 于是, 注意到$ \alpha_{1}\geq \alpha_{2}, $$ \lambda_{1}\geq \lambda_{2} $和$ t \psi'_{2}(t) $关于$ t>0 $是增加的, 我们可得

(3.5) $ \begin{equation} 0\leq \frac{f(\lambda_{2}x)}{\overline{F}(\lambda_{2}x)}\leq \frac{f(\lambda_{1}x)}{\overline{F}(\lambda_{1}x)}, \end{equation} $

(3.6) $ \begin{equation} 0\leq -\psi'_{2}(\overline{F}^{\alpha_{2}}(\lambda_{2}x)) \overline{F}^{\alpha_{2}}(\lambda_{2}x)\leq -\psi'_{2}(\overline{F}^{\alpha_{1}}(\lambda_{1}x)) \overline{F}^{\alpha_{1}}(\lambda_{1}x). \end{equation} $

结合(3.2), (3.5)和(3.6)式, 可以得到$ g_{2}(\boldsymbol{\alpha}, \boldsymbol{\lambda})\leq 0. $

因此, $ g(\boldsymbol{\alpha}, \boldsymbol{\lambda})=g_{1}(\boldsymbol{\alpha}, \boldsymbol{\lambda})+g_{2}(\boldsymbol{\alpha}, \boldsymbol{\lambda})\leq 0, $这表明定理2.4(ii)的条件是满足的. 综上, 定理证明完成.

注3.2   (1) 定理3.1的两个条件"$ \psi_{2}\circ \phi_{1} $具有超可加性" 和"$ \phi_{2}(t) $是对数凸" 是很常见的, 对如下的三个著名copula很容易验证满足:

(ⅰ) Clayton copula, 其生成子$ \phi(t)=(\theta t+1)^{-\frac{1}{\theta}} $, $ \theta\geq0 $. 令$ \phi_{1}(t)=(\theta_{1} t+1)^{-\frac{1}{\theta_{1}}} $和$ \phi_{2}(t)=(\theta_{2} t+1)^{-\frac{1}{\theta_{2}}} $. 于是可得$ \psi_{2}\circ \phi_{1}(t)=\theta_{2}^{-1}[(\theta_{1} t+1)^{\frac{\theta_{2}}{\theta_{1}}}-1] $. 显然可见, 当$ \theta_{2}\geq \theta_{1} \geq 0, $$ [\psi_{2}\circ \phi_{1}(t)]^{''}=(\theta_{2}-\theta_{1})(\theta_{1} t+1)^{\frac{\theta_{2}}{\theta_{1}}-2}\geq 0 $, 这表明$ \psi_{2}\circ \phi_{1}(t) $具有超可加性, 而且$ \phi_{2}(t) $关于$ t>0 $是对数凸的.

(ⅱ) Gumbel-Hougaard copula, 其生成子$ \phi(t)={\rm e}^{-t^{\frac{1}{\theta}}} $, $ \theta\geq 1. $令$ \phi_{1}(t)={\rm e}^{-t^{\frac{1}{\theta_{1}}}} $和$ \phi_{2}(t)={\rm e}^{-t^{\frac{1}{\theta_{2}}}} $. 于是可得$ \psi_{2}\circ \phi_{1}(t)=t^{\frac{\theta_{2}}{\theta_{1}}} $. 显然可见, 当$ \theta_{2}\geq \theta_{1} \geq 1, $$ [\psi_{2}\circ \phi_{1}(t)]^{''}=\frac{\theta_{2}}{\theta_{1}}(\frac{\theta_{2}}{\theta_{1}}-1)t^{\frac{\theta_{2}}{\theta_{1}}-2}\geq 0 $, 这表明$ \psi_{2}\circ \phi_{1}(t) $具有超可加性, 而且$ \phi_{2}(t) $关于$ t>0 $是对数凸的.

(ⅲ) Ali-Mikhail-Haq copula, 其生成子$ \phi(t)=\frac{1-\theta}{{\rm e}^{t}-\theta} $, $ \theta\in[0, 1) $. 令$ \phi_{1}(t)=\frac{1-\theta_{1}}{{\rm e}^{t}-\theta_{1}} $和$ \phi_{2}(t)=\frac{1-\theta_{2}}{{\rm e}^{t}-\theta_{2}} $. 于是可得$ \psi_{2}\circ \phi_{1}(t)=\ln\biggl(\theta_{2}+\frac{1-\theta_{2}}{1-\theta_{1}}({\rm e}^{t}-\theta_{1})\biggr) $. 显然可见, 当$ 1>\theta_{2}\geq \theta_{1} \geq 0, $$ [\psi_{2}\circ \phi_{1}(t)]^{''}=\frac{\theta_{2}-\theta_{1}}{1-\theta_{2}}{\rm e}^{-t}(1+\frac{\theta_{2}-\theta_{1}}{1-\theta_{2}}{\rm e}^{-t})^{-2}\geq 0 $, 这表明$ \psi_{2}\circ \phi_{1}(t) $具有超可加性, 而且$ \phi_{2}(t) $关于$ t>0 $是对数凸的.

(2) 由定义2.3, 当$ \lambda_{1}=\lambda_{2}=\lambda_{1}^{*}=\lambda_{2}^{*} $时, $ (\alpha_{1}, \alpha_{2})\succeq_{m}(\alpha_{1}^{*}, \alpha_{2}^{*}) $ (或者$ \alpha_{1}=\alpha_{2}=\alpha_{1}^{*}=\alpha_{2}^{*} $时, $ (\lambda_{1}, \lambda_{2})\succeq_{m}(\lambda_{1}^{*}, \lambda_{2}^{*}) $) 可由$ \left(\begin{array}{ccc}\alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\end{array}\right)\gg \left(\begin{array}{ccc}\alpha_{1}^{*} & \alpha_{2}^{*} \\ \lambda_{1}^{*} &\lambda_{2}^{*}\end{array}\right) $导出. 于是, 文献[8]中的定理3.1和3.3是本文定理3.1的特殊情形.

下面的反例表明, 当两个参数矩阵不属于$ S_{2} $时, 定理3.1的结论不一定成立.

例3.3   令基本生成函数$ \overline{F}(x)={\rm e}^{-x^{5}}, x>0. $于是显然可得$ \overline{F} $是对数凹的. 考虑分别具有生成子$ \phi_{1}(t)=(t+1)^{-1} $和$ \phi_{2}(t)=(2t+1)^{-\frac{1}{2}} $的Clayton copula. 于是, $ \psi_{2}\circ \phi_{1}(t)=\frac{1}{2}[(t+1)^{2}-1] $具有超可加性, 而且$ \phi_{2}(t) $关于$ t>0 $是对数凸的.

考虑如下的$ T $-transform矩阵: $ T_{0.6}=0.6\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)+0.4\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right). $

(ⅰ) 令$ \left( \begin{array}{cc} \alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\\ \end{array} \right)=\left( \begin{array}{cc} 50 & 2\\ 3 & 10 \\ \end{array} \right) $, $ \left( \begin{array}{cc} \alpha_{1}^{*} & \alpha_{2}^{*} \\ \lambda_{1}^{*} &\lambda_{2}^{*} \\ \end{array} \right)=\left( \begin{array}{cc} 30.8 & 21.2 \\ 5.8 & 7.2 \\ \end{array} \right) $. 我们发现这两个矩阵都不属于$ S_{2}. $而且$ \left( \begin{array}{cc} \alpha_{1}^{*} & \alpha_{2}^{*} \\ \lambda_{1}^{*} &\lambda_{2}^{*} \\ \end{array} \right)=\left( \begin{array}{cc} \alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\\ \end{array} \right)T_{0.6} $, 由定义2.3可得$ \left( \begin{array}{cc} \alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\\ \end{array} \right) \gg \left(\begin{array}{cc} \alpha_{1}^{*} & \alpha_{2}^{*} \\ \lambda_{1}^{*} &\lambda_{2}^{*} \\ \end{array} \right). $记

$ \begin{eqnarray*} h_{1}(x)&=&\overline{F}_{X_{1:2}}(x)-\overline{F}_{X_{1:2}^{*}}(x)\\ &=&\phi_{1}\biggl(\sum\limits_{i=1}^{2}\psi_{1}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr) -\phi_{2}\biggl(\sum\limits_{i=1}^{2}\psi_{2}(\overline{F}^{\alpha_{i}^{*}}(\lambda_{i}^{*}x))\biggr)\\ &=&({\rm e}^{50*3^{5}x^{5}}+{\rm e}^{2*10^{5}x^{5}}-1)^{-1}-({\rm e}^{61.6*5.8^{5}x^{5}}+{\rm e}^{42.4*7.2^{5}x^{5}}-1)^{-1/2}\\ &=&({\rm e}^{12150x^{5}}+{\rm e}^{200000x^{5}}-1)^{-1}-({\rm e}^{404315.8x^{5}}+{\rm e}^{820405.1x^{5}}-1)^{-1/2}. \end{eqnarray*} $

(ⅱ) 令$ \left( \begin{array}{cc} \alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\\ \end{array} \right)=\left( \begin{array}{cc} 0.5 & 0.2\\ 0.3 & 0.7 \\ \end{array} \right) $, $ \left( \begin{array}{cc} \alpha_{1}^{*} & \alpha_{2}^{*} \\ \lambda_{1}^{*} &\lambda_{2}^{*} \\ \end{array} \right)=\left( \begin{array}{cc} 0.38 & 0.32 \\ 0.46 & 0.54 \\ \end{array} \right) $. 我们发现这两个矩阵都不属于$ S_{2}. $而且$ \left( \begin{array}{cc} \alpha_{1}^{*} & \alpha_{2}^{*} \\ \lambda_{1}^{*} &\lambda_{2}^{*} \\ \end{array} \right)=\left( \begin{array}{cc} \alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\\ \end{array} \right)T_{0.6} $, 由定义2.3可得$ \left( \begin{array}{cc} \alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\\ \end{array} \right) \gg \left(\begin{array}{cc} \alpha_{1}^{*} & \alpha_{2}^{*} \\ \lambda_{1}^{*} &\lambda_{2}^{*} \\ \end{array} \right). $记

$ \begin{eqnarray*} h_{2}(x)&=&\phi_{1}\biggl(\sum\limits_{i=1}^{2}\psi_{1}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr) -\phi_{2}\biggl(\sum\limits_{i=1}^{2}\psi_{2}(\overline{F}^{\alpha_{i}^{*}}(\lambda_{i}^{*}x))\biggr)\\ &=&({\rm e}^{0.001215x^{5}}+{\rm e}^{0.033614x^{5}}-1)^{-1}-({\rm e}^{0.01565319x^{5}}+{\rm e}^{0.02938656x^{5}}-1)^{-1/2}. \end{eqnarray*} $

我们注意到$ h_{1}(0.1)=0.1165959>0 $和$ h_{2}(0.1)=-1.230912*10^{-7}<0 $. 因此, 通过取两种情形的两个矩阵都不属于$ S_{2}, $$ h_{1}(x) $和$ h_{2}(x) $的函数值在$ x=0.1 $处改变符号, 于是, $ X_{1:2}\nleq_{st}X^{*}_{1:2} $.

下面的例子说明定理3.1结论的正确性.

例3.4   令基本生成函数$ \overline{F}(x)={\rm e}^{-x^{4}}, x>0. $于是显然可得$ \overline{F} $是对数凹的. 考虑分别具有生成子$ \phi_{1}(t)={\rm e}^{-t^{\frac{1}{2}}} $和$ \phi_{2}(t)={\rm e}^{-t^{\frac{1}{4}}} $的Gumbel-Hougaard copula. 于是$ \psi_{2}\circ \phi_{1}(t)=t^{2} $具有超可加性, 而且$ \phi_{2}(t) $关于$ t>0 $是对数凸的.

令$ \left( \begin{array}{cc} \alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\\ \end{array} \right)=\left( \begin{array}{cc} 0.1 & 1\\ 2 & 4 \\ \end{array} \right)\in S_{2} $, $ \left( \begin{array}{cc} \alpha_{1}^{*} & \alpha_{2}^{*} \\ \lambda_{1}^{*} &\lambda_{2}^{*} \\ \end{array} \right)=\left( \begin{array}{cc} 0.82 & 0.28 \\ 3.6 & 2.4 \\ \end{array} \right) $. 考虑$ T $-transform矩阵: $ T_{0.2}=0.2\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)+0.8\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right). $于是, $ \left( \begin{array}{cc} \alpha_{1}^{*} & \alpha_{2}^{*} \\ \lambda_{1}^{*} &\lambda_{2}^{*} \\ \end{array} \right)=\left( \begin{array}{cc} \alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\\ \end{array} \right)T_{0.2} $, 由定义2.3可得$ \left( \begin{array}{cc} \alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\\ \end{array} \right) \gg \left(\begin{array}{cc} \alpha_{1}^{*} & \alpha_{2}^{*} \\ \lambda_{1}^{*} &\lambda_{2}^{*} \\ \end{array} \right). $从而, 由定理3.1可得$ X_{1:2}\leq_{st}X^{*}_{1:2} $, 也即: 对所有的$ x>0 $,

$ \begin{eqnarray*} \overline{F}_{X_{1:2}}(x)=\phi_{1}\biggl(\sum\limits_{i=1}^{2}\psi_{1}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr)\leq \phi_{2}\biggl(\sum\limits_{i=1}^{2}\psi_{2}(\overline{F}^{\alpha_{i}^{*}}(\lambda_{i}^{*}x))\biggr)=\overline{F}_{X_{1:2}^{*}}(x). \end{eqnarray*} $

图 1中给出了$ X_{1:2} $和$ X_{1:2}^{*} $的生存函数曲线, 与定理3.1的结论是一致的.

下面的推论, 是定理3.1的特殊情形, 给出了一组来自于具有Archimedean copula相依结构的尺度比例失效率分布的样本最小次序统计量的分布函数下界.

推论3.5   令$ \overline{\alpha}=\frac{1}{2}(\alpha_{1}+\alpha_{2}) $和$ \overline{\lambda}=\frac{1}{2}(\lambda_{1}+\lambda_{2}). $于是, 在定理3.1的条件下, 可得

$ F_{X_{1:2}}(x)\geq 1-\phi_{2}\Bigl(2\psi_{2}(\overline{F}^{\overline{\alpha}}(\overline{\lambda}x))\Bigr). $

证   显然, $ \left(\begin{array}{ccc}\overline{\alpha} & \overline{\alpha} \\ \overline{\lambda} & \overline{\lambda}\end{array}\right)=\left(\begin{array}{ccc}\alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\end{array}\right)T_{0.5}. $于是可得$ \left(\begin{array}{ccc}\alpha_{1} & \alpha_{2} \\ \lambda_{1} &\lambda_{2}\end{array}\right)\gg \left(\begin{array}{ccc}\overline{\alpha} & \overline{\alpha} \\ \overline{\lambda} & \overline{\lambda}\end{array}\right). $由定理3.1, 我们可得到推论结论成立.

最后, 我们将定理3.1的结论延拓到$ n>2 $的情形.

定理3.6   令$ (X_{1}, \cdots , X_{n})\thicksim SPH(\overline{F}, \boldsymbol{\alpha}, \boldsymbol{\lambda}, \phi_{1}) $和$ (X_{1}^{*}, \cdots , X_{n}^{*})\thicksim SPH(\overline{F}, \boldsymbol{\alpha^{*}}, \boldsymbol{\lambda^{*}}, \phi_{2}) $. 假定下列条件成立:

(ⅰ) $ \overline{F} $是对数凹;

(ⅱ) $ \psi_{2}\circ \phi_{1} $具有超可加性;

(ⅲ) $ \phi_{2} $是对数凸.

假设$ \left( \begin{array}{c} \alpha_{1}\cdots \alpha_{n} \\ \lambda_{1}\cdots \lambda_{n} \\ \end{array} \right)\in S_{n} $和$ \left( \begin{array}{c} \alpha_{1}^{*}\cdots \alpha_{n}^{*} \\ \lambda_{1}^{*}\cdots \lambda_{n}^{*} \\ \end{array} \right)=\left( \begin{array}{c} \alpha_{1}\cdots \alpha_{n} \\ \lambda_{1}\cdots \lambda_{n} \\ \end{array} \right)T_{\omega}. $于是可得$ X_{1:n}\leq_{st}X^{*}_{1:n}. $

证   对$ x>0 $, $ X_{1:n} $的生存函数为

$ \overline{F}_{X_{1:n}}(x)=\phi_{1}\biggl(\sum\limits_{i=1}^{n}\psi_{1}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr). $

由于$ \psi_{2}\circ \phi_{1} $具有超可加性, 由引理2.6可得

$ \phi_{1}\biggl(\sum\limits_{i=1}^{n}\psi_{1}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr)\leq \phi_{2}\biggl(\sum\limits_{i=1}^{n}\psi_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr)\triangleq \overline{G}_{X_{1:n}}(x). $

于是, 为了证明$ X_{1:n}\leq_{st}X^{*}_{1:n}, $只需证明对$ x>0 $, 下式成立:

$ \overline{G}_{X_{1:n}}(x)=\phi_{2}\biggl(\sum\limits_{i=1}^{n}\psi_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr)\leq \phi_{2}\biggl(\sum\limits_{i=1}^{n}\psi_{2}(\overline{F}^{\alpha^{*}_{i}}(\lambda^{*}_{i}x))\biggr)=\overline{F}_{X^{*}_{1:n}}(x), $

当$ \left( \begin{array}{c} \alpha_{1}\cdots \alpha_{n} \\ \lambda_{1}\cdots \lambda_{n} \\ \end{array} \right)\in S_{n} $和$ \left( \begin{array}{c} \alpha_{1}^{*}\cdots \alpha_{n}^{*} \\ \lambda_{1}^{*}\cdots \lambda_{n}^{*} \\ \end{array} \right)=\left( \begin{array}{c} \alpha_{1}\cdots \alpha_{n} \\ \lambda_{1}\cdots \lambda_{n} \\ \end{array} \right)T_{\omega}. $

首先, 根据$ T $-transform矩阵的定义, $ \left( \begin{array}{c} \alpha_{1}^{*}\cdots \alpha_{n}^{*} \\ \lambda_{1}^{*}\cdots \lambda_{n}^{*} \\ \end{array} \right)=\left( \begin{array}{c} \alpha_{1}\cdots \alpha_{n} \\ \lambda_{1}\cdots \lambda_{n} \\ \end{array} \right)T_{\omega} $表明两个矩阵$ \left( \begin{array}{c} \alpha_{1}^{*}\cdots \alpha_{n}^{*} \\ \lambda_{1}^{*}\cdots \lambda_{n}^{*} \\ \end{array} \right) $和$ \left( \begin{array}{c} \alpha_{1}\cdots \alpha_{n} \\ \lambda_{1}\cdots \lambda_{n} \\ \end{array} \right) $有$ (n-2) $列相同, 第$ i $和$ j $列可能不同. 于是可得

$ \alpha_{i}^{*}=\omega\alpha_{i}+(1-\omega)\alpha_{j}, \alpha_{j}^{*}=(1-\omega)\alpha_{i}+\omega\alpha_{j}, $

$ \lambda_{i}^{*}=\omega\lambda_{i}+(1-\omega)\lambda_{j}, \lambda_{j}^{*}=(1-\omega)\lambda_{i}+\omega\lambda_{j}, $

(3.7) $ \begin{equation} \alpha_{s}^{*}=\alpha_{s}, \lambda_{s}^{*}=\lambda_{s}, s=\{1, 2, \cdots , n\}\setminus\{i, j\}. \end{equation} $

令

$ A_{0}=\left(\begin{array}{ccc}\alpha_{i} & \alpha_{j} \\ \lambda_{i} & \lambda_{j}\end{array}\right), B_{0}=\left(\begin{array}{ccc}\alpha_{i}^{*} & \alpha_{j}^{*} \\ \lambda_{i}^{*} & \lambda_{j}^{*}\end{array}\right), T_{0}=\left(\begin{array}{ccc}\omega & 1-\omega \\ 1-\omega & \omega\end{array}\right). $

由于$ \left( \begin{array}{c} \alpha_{1}\cdots \alpha_{n} \\ \lambda_{1}\cdots \lambda_{n} \\ \end{array} \right)\in S_{n}, $故$ A_{0}\in S_{2}. $进一步可得$ B_{0}=A_{0}T_{0}, $即$ A_{0}\gg B_{0}. $

结合以上观察, 以及定理3.1的证明, 我们可得

$ \phi_{2}\biggl(\psi_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))+\psi_{2}(\overline{F}^{\alpha_{j}}(\lambda_{j}x))\biggr)\leq \phi_{2}\biggl(\psi_{2}(\overline{F}^{{\alpha_{i}}^{*}}(\lambda_{i}^{*}x))+\psi_{2}(\overline{F}^{{\alpha_{j}}^{*}}(\lambda_{j}^{*}x))\biggr), $

也即

(3.8) $ \begin{equation} \psi_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))+\psi_{2}(\overline{F}^{\alpha_{j}}(\lambda_{j}x))\geq \psi_{2}(\overline{F}^{{\alpha_{i}}^{*}}(\lambda_{i}^{*}x))+\psi_{2}(\overline{F}^{{\alpha_{j}}^{*}}(\lambda_{j}^{*}x)). \end{equation} $

因此

$ \begin{eqnarray*} \overline{G}_{X_{1:n}}(x) &=&\phi_{2}\biggl(\sum\limits_{i=1}^{n}\psi_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))\biggr)\\ &=&\phi_{2}\biggl(\psi_{2}(\overline{F}^{\alpha_{i}}(\lambda_{i}x))+\psi_{2}(\overline{F}^{\alpha_{j}}(\lambda_{j}x))+\sum\limits_{s\neq i, j}\psi_{2}(\overline{F}^{\alpha_{s}}(\lambda_{s}x))\biggr)\\ &\leq& \phi_{2}\biggl(\psi_{2}(\overline{F}^{{\alpha_{i}}^{*}}(\lambda_{i}^{*}x))+\psi_{2}(\overline{F}^{{\alpha_{j}}^{*}}(\lambda_{j}^{*}x))+\sum\limits_{s\neq i, j}\psi_{2}(\overline{F}^{\alpha_{s}}(\lambda_{s}x))\biggr)\\ &=&\phi_{2}\biggl(\psi_{2}(\overline{F}^{{\alpha_{i}}^{*}}(\lambda_{i}^{*}x))+\psi_{2}(\overline{F}^{{\alpha_{j}}^{*}}(\lambda_{j}^{*}x))+\sum\limits_{s\neq i, j}\psi_{2}(\overline{F}^{{\alpha_{s}}^{*}}(\lambda_{s}^{*}x))\biggr)\\ &=&\phi_{2}\biggl(\sum\limits_{i=1}^{n}\psi_{2}(\overline{F}^{\alpha^{*}_{i}}(\lambda^{*}_{i}x))\biggr)=\overline{F}_{X^{*}_{1:n}}(x), \end{eqnarray*} $

其中, 第三步不等号是利用(3.8)式可得, 第四步等号是利用(3.7)式可得, 定理得证.

由于有限个具有相同结构的$ T $-transform矩阵的乘积仍然是$ T $-transform矩阵, 我们很容易得到如下的结论.

定理3.7   令$ (X_{1}, \cdots , X_{n})\thicksim SPH(\overline{F}, \boldsymbol{\alpha}, \boldsymbol{\lambda}, \phi_{1}) $和$ (X_{1}^{*}, \cdots , X_{n}^{*})\thicksim SPH(\overline{F}, \boldsymbol{\alpha^{*}}, \boldsymbol{\lambda^{*}}, \phi_{2}) $. 假定下列条件成立:

(ⅰ) $ \overline{F} $是对数凹;

(ⅱ) $ \psi_{2}\circ \phi_{1} $具有超可加性;

(ⅲ) $ \phi_{2} $是对数凸.

假设$ \left( \begin{array}{c} \alpha_{1}\cdots \alpha_{n} \\ \lambda_{1}\cdots \lambda_{n} \\ \end{array} \right)\in S_{n} $和$ \left( \begin{array}{c} \alpha_{1}^{*}\cdots \alpha_{n}^{*} \\ \lambda_{1}^{*}\cdots \lambda_{n}^{*} \\ \end{array} \right)=\left( \begin{array}{c} \alpha_{1}\cdots \alpha_{n} \\ \lambda_{1}\cdots \lambda_{n} \\ \end{array} \right)T_{\omega_{1}}\cdots T_{\omega_{k}}, $这里$ T_{\omega_{i}}, i=1, \cdots, k, $具有相同的结构. 于是, 我们有$ X_{1:n}\leq_{st}X^{*}_{1:n}. $

若$ k $个$ T $-transform矩阵$ T_{\omega_{i}}, i=1, \cdots, k, $具有不同结构, 对应定理3.7, 我们给出下面的研究结论.

定理3.8   令$ (X_{1}, \cdots , X_{n})\thicksim SPH(\overline{F}, \boldsymbol{\alpha}, \boldsymbol{\lambda}, \phi_{1}) $和$ (X_{1}^{*}, \cdots , X_{n}^{*})\thicksim SPH(\overline{F}, \boldsymbol{\alpha^{*}}, \boldsymbol{\lambda^{*}}, \phi_{2}) $. 假定下列条件成立:

(ⅰ) $ \overline{F} $是对数凹;

(ⅱ) $ \psi_{2}\circ \phi_{1} $具有超可加性;

(ⅲ) $ \phi_{2} $是对数凸.

假设$ \left( \begin{array}{c} \alpha_{1}\cdots \alpha_{n} \\ \lambda_{1}\cdots \lambda_{n} \\ \end{array} \right)\in S_{n} $和$ \left( \begin{array}{c} \alpha_{1}\cdots \alpha_{n} \\ \lambda_{1}\cdots \lambda_{n} \\ \end{array} \right)T_{\omega_{1}}\cdots T_{\omega_{i}}\in S_{n} $, $ i=1, \cdots, k-1, $这里$ k\geq 2. $如果$ \left( \begin{array}{c} \alpha_{1}^{*}\cdots \alpha_{n}^{*} \\ \lambda_{1}^{*}\cdots \lambda_{n}^{*} \\ \end{array} \right)=\left( \begin{array}{c} \alpha_{1}\cdots \alpha_{n} \\ \lambda_{1}\cdots \lambda_{n} \\ \end{array} \right)T_{\omega_{1}}\cdots T_{\omega_{k}}, $则可得$ X_{1:n}\leq_{st}X^{*}_{1:n}. $

证   令

$ \left( \begin{array}{c} \alpha_{1}^{(j)}\cdots \alpha_{n}^{(j)} \\ \lambda_{1}^{(j)}\cdots \lambda_{n}^{(j)} \\ \end{array} \right)=\left( \begin{array}{c} \alpha_{1}\cdots \alpha_{n} \\ \lambda_{1}\cdots \lambda_{n} \\ \end{array} \right)T_{\omega_{1}}\cdots T_{\omega_{j}}, j=1, \cdots , k-1. $

令$ (Y_{1}^{(j)}, \cdots , Y_{n}^{(j)})\thicksim SPH(\overline{F}, \boldsymbol{\alpha^{(j)}}, \boldsymbol{\lambda^{(j)}}, \phi_{2}) , j=1, \cdots, k-1. $由定理的假设条件, 我们得到

$ \left( \begin{array}{c} \alpha_{1}^{(j)}\cdots \alpha_{n}^{(j)} \\ \lambda_{1}^{(j)}\cdots \lambda_{n}^{(j)} \\ \end{array} \right)\in S_{n}, $$ j=1, 2, \cdots, k-1. $

结合定理3.6的结论, 可得$ X_{1:n}\leq_{st}Y_{1:n}^{(1)}\leq_{st}Y_{1:n}^{(2)}\leq_{st} \cdots \leq_{st}Y_{1:n}^{(k-1)}\leq_{st}X_{1:n}^{*} $. 证毕.

在多元链式优化序下, 该文研究了两组来自于不同相依尺度比例失效率分布的最小次序统计量的随机比较. 作为对偶, 随机变量$ X_{1}, \cdots , X_{n} $服从尺度比例反失效率模型(scale proportional reverse hazards model, 简称$ SPRH $), 如果$ X_{i} $的分布函数$ F_{i}(x)=F^{\alpha_{i}}(\lambda_{i}x), $$ \alpha_{i}>0, \lambda_{i}>0, i\in I_{n}, $这里$ F $是基本分布函数. 记号$ \boldsymbol{X}=(X_{1}, \cdots , X_{n})\thicksim SPRH(F, \boldsymbol{\alpha}, \boldsymbol{\lambda}, \phi) $表明随机变量的相依是具有生成子$ \phi $的Archimedean copula和$ X_{i}\thicksim F^{\alpha_{i}}(\lambda_{i}x), i\in I_{n}. $于是, 多元链式优化序下, 两组来自于不同相依尺度比例反失效率分布的最大次序统计量的随机比较结果可以类似研究. 另一方面, 本文致力于研究两组来自于不同相依尺度比例失效率分布的最小次序统计量的具体普通随机序比较. 当然, 还有很多的随机序, 比如似然比序、平均剩余寿命序、凸序、增加凸序和星序. 我们正在致力于最小次序统计量的如前面提到的随机序的比较.