关于重尾分布的精细大偏差的研究一直以来是保险与金融行业中的研究重点,它有利于保险公司做出更好的决策及降低在经营过程中的风险.经典风险模型中精细大偏差主要关注渐近式

(1.1) $ \begin{align} P(S_{n}-n\mu>x)\sim n\overline{F}(x), \end{align} $

在$ n\rightarrow\infty $,对于$ x\in T_{n} $一致成立,即

$ \lim\limits_{n\rightarrow\infty}\sup\limits_{x\in T_{n}}\bigg|\frac{P(S_{n}-n\mu>x)}{n\overline{F}(x)}-1\bigg| = 0, $

其中$ S_{n} = \sum\limits_{i = 1}^{n}X_{i}(n\geq1) $是随机变量的部分和,经典结果参见文献[1-3].

在现实中,一场意外事故的发生可能导致多种理赔的同时发生.最经典的例子:一场交通事故,人会发生意外伤害事故,车会发生损坏,则可能导致车险和意外伤害保险同时发生.因此,在最近几年,一些研究者将精细大偏差推广到多维的情形,例如,文献[4]在多维风险模型下研究了随机和的精细大偏差,后来文献[5]在相依多维风险模型下得到部分和及随机和的精细大偏差.

本文考虑基于客户来到过程的二维风险模型的精细大偏差. $ \{N(t), t\geq0\} $为一计数过程,其中$ N(t) $表示到时刻$ t $为止保险公司出售的保单总数,每出售一张保单保险公司在一定时期内承担来自该保单的风险.设第$ i $个客户所拥有的保单潜在索赔额为$ \overrightarrow{X^{i}} = (X_{1}^{i}, X_{2}^{i})^{\top} $为一个二维随机向量,独立同分布于$ \overrightarrow{X} = (X_{1}, X_{2})^{\top}, E\overrightarrow{X} = \overrightarrow{\mu}, \overrightarrow{\mu} = (\mu_{1}, \mu_{2})^{\top} $, $ \overrightarrow{X} $有共同边缘分布$ F_{1}(x_{1}), F_{2}(x_{2}) $及联合分布$ F_{12}(x_{1}, x_{2}) $.用Bernonlli随机变量$ I_{i} $表示第$ i $个客户所拥有的保单是否发生索赔($ I_{i} = 1 $表示发生索赔), $ P(I_{i} = 1) = \theta_{i}, P(I_{i} = 0) = 1-\theta_{i} $,假设$ \{I_{i}, i\geq 1\} $相互独立.设每张保单的保费为$ (1+\delta)\overline{}\overrightarrow{\mu} $,其中常数$ \delta>0 $为安全负载系数.则保险公司来自第$ i $份保单的净索赔额为$ I_{i}\overrightarrow{X^{i}}-(1+\delta)\overline{}\overrightarrow{\mu} $.从而总损失的部分和为

$ \overrightarrow{S_{n}} = \sum\limits^{n}_{i = 1}(I_{i}\overrightarrow{X^{i}}-(1+\delta)\overline{}\overrightarrow{\mu}), n\geq1, $

直到时刻$ t $总损失的随机和为

$ \overrightarrow{S}_{N(t)} = \sum\limits^{N(t)}_{i = 1}(I_{i}\overrightarrow{X^{i}}-(1+\delta)\overline{}\overrightarrow{\mu}), t\geq 0. $

文献[6]首次提出了基于客户来到过程的风险模型,文献[7]将其推广到负相关随机变量的情形,文献[8]将其模型推广到泊松发射噪声过程,分别得到了损失过程的精细大偏差结果.本文将上述文献中讨论的模型推广到二维风险模型的情形,且每张保单发生实际索赔的概率不同,并且用copula函数表示索赔额之间的相依结构.

若非负随机变量$ X $不存在任何指数阶矩,则称相应的分布$ F $为重尾分布.

定义2.1  $ C $族:若$ F $满足

$ \lim\limits_{\upsilon\downarrow1}\liminf\limits_{x\rightarrow\infty}\frac{\overline{F}(\upsilon x)}{\overline{F}(x)} = 1, {\quad}\mbox{等价地}{\quad} \lim\limits_{\upsilon\uparrow1}\limsup\limits_{x\rightarrow\infty}\frac{\overline{F}(\upsilon x)}{\overline{F}(x)} = 1. $

定义2.2^[9]  对于任意一个非负随机变量$ X $的分布函数$ F $,我们记

$ J_{F}^{+} = -\lim\limits_{y\rightarrow\infty}\frac{\log\overline{F}_{\ast}(y)}{\log y}, J_{F}^{-} = -\lim\limits_{y\rightarrow\infty}\frac{\log\overline{F}^{\ast}(y)}{\log y}, $

其中

$ \overline{F}_{\ast}(y) = \liminf\limits_{x\rightarrow\infty}\frac{\overline{F}(xy)}{\overline{F}(x)}, \overline{F}^{\ast}(y) = \limsup\limits_{x\rightarrow\infty}\frac{\overline{F}(xy)}{\overline{F}(x)}, y>0, $

称$ J_{F}^{+}, J_{F}^{-} $为分布函数$ F $的上下Matuszewska指数.

如果$ F\in C $,对任意$ \rho>J_{F}^{+} $存在正常数$ C $和$ D $,对所有的$ x\geq y\geq D $,有

(2.1) $ \begin{align} \frac{\overline{F}(y)}{\overline{F}(x)}\leq C(\frac{x}{y})^{\rho}. \end{align} $

下面给出本文的一些假设条件.

A1  设存在$ \theta>0 $,使得

(3.1) $ \begin{align} \lim\limits_{n\rightarrow\infty}\frac{1}{n}\sum\limits_{i = 1}^{n}\theta_{i} = \theta. \end{align} $

A2  随机向量$ (X_{1}, X_{2})^{\top} $的边缘分布之间通过一个copula函数$ C(\cdot, \cdot) $连接.存在$ M>0 $满足

(3.2) $ \begin{align} C(\overline{F}_{1}(x_{1}), \overline{F}_{2}(x_{2}))\leq M\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2}), \end{align} $

且存在$ 1\leq r\leq\infty $,对所有的正函数$ f(x) $与$ g(x) $,当满足$ \lim\limits_{x\rightarrow0}f(x) = \lim\limits_{x\rightarrow\infty}g(x) = 0 $时,使得

(3.3) $ \begin{align} \lim\limits_{x\rightarrow 0}\frac{C(\upsilon f(x), \upsilon g(x))}{C(f(x), g(x))} = \upsilon^{r} \end{align} $

成立.

A3  设$ F_{1}\in C $,存在常数$ c\in(0, +\infty) $有

(3.4) $ \begin{align} \lim\limits_{x\rightarrow\infty}\frac{\overline{F}_{2}(x)}{\overline{F}_{1}(x)} = c. \end{align} $

A4  设存在$ \rho>J_{F}^{+} $且任意$ \delta>0 $有

(3.5) $ \begin{align} EN^{2\rho}(t)I_{\{N(t)>(1+\delta)\lambda(t)\}} = O(\lambda^{2}(t)). \end{align} $

注3.1  根据Sklar's Theorem,由假设A2很容易发现$ \overline{F}_{12}(x_{1}, x_{2}) = C(\overline{F}_{1}(x_{1}), \overline{F}_{2}(x_{2})) $.关系式$ (3.2) $不仅仅反映了随机变量之间的负相依性,而且反映了更广泛的相依结构.由假设A3和定义1可知边缘分布$ F_{1}, F_{2}\in C $.

注3.2  下面我们列举两个满足假设A2的copula函数族.

1)如果$ C(\cdot, \cdot) $是来自于Farlie-Gumbel-Morgenstern族的copula函数,即

$ C(x_{1}, x_{2}) = x_{1}x_{2}[1+p(1-x_{1})(1-x_{2})], |p|\leq 1. $

则存在常数$ 1\leq M\leq 2 $成立$ C(x_{1}, x_{2}) = Mx_{1}x_{2} $且

$ \lim\limits_{x\rightarrow 0}\frac{C[\upsilon f(x), \upsilon g(x)]}{C[f(x), g(x)]} = \upsilon^{2}, \upsilon>0. $

2)如果$ C(\cdot, \cdot) $是来自于Ali-Mikhail-Haq族的copula函数,即

$ C(x_{1}, x_{2}) = \frac{x_{1}x_{2}}{1-p(1-x_{1})(1-x_{2})}, p\neq 1, $

则存在常数$ M\geq\max\{1, 1/(1-p)\} $, $ C(x_{1}, x_{2}) = Mx_{1}x_{2} $成立且$ \lim\limits_{x\rightarrow 0}\frac{C[\upsilon f(x), \upsilon g(x)]}{C[f(x), g(x)]} = \upsilon^{2}, \upsilon>0. $

注3.3  由不等式$ (E|X|^{r})^{1/r}\leq (E|X|^{s})^{1/s}, 0<r<s $.得假设A4等价于

(3.6) $ \begin{align} EN^{2\rho}(t)I_{\{N(t)>(1+\delta)\lambda(t)\}} = o(\lambda^{2}(t)), \end{align} $

故由文献[10,引理3.5]得假设A4满足更新计数过程.又根据文献[3,引理2.4],由假设A4,有

(3.7) $ \begin{align} \frac{N(t)}{\lambda(t)}\mathop{\longrightarrow}\limits^{P}1, t\rightarrow\infty. \end{align} $

下面给出本文在证明主要结果过程中用到的一些引理.

引理3.1  假设$ \{\overrightarrow{X^{i}}, i\geq 1\} $是一列非负随机向量,满足假设A2–A3,则下式成立.

(3.8) $ \begin{align} \lim\limits_{\upsilon\uparrow 1}\limsup\limits_{x_{1}\rightarrow\infty, x_{2}\rightarrow\infty}\frac{C(\overline{F}_{1}(\upsilon x_{1}), \overline{F}_{2}(\upsilon x_{2}))}{C(\overline{F}_{1}( x_{1}), \overline{F}_{2}(x_{2}))} = 1, \lim\limits_{\upsilon\downarrow 1}\liminf\limits_{x_{1}\rightarrow\infty, x_{2}\rightarrow\infty}\frac{C(\overline{F}_{1}(\upsilon x_{1}), \overline{F}_{2}(\upsilon x_{2}))}{C(\overline{F}_{1}( x_{1}), \overline{F}_{2}(x_{2}))} = 1. \end{align} $

证  由$ F_{i}\in C $,则对任意$ \varepsilon, \upsilon\in(0, 1) $且$ x_{i} $足够大,有$ \overline{F}_{i}(\upsilon x_{i})\leq(1+\varepsilon)\overline{F}_{i}(x_{i}), i = 1, 2. $由copula函数的单调增加性及假设A2有

$ \limsup\limits_{x_{1}\rightarrow\infty, x_{2}\rightarrow\infty}\frac{C(\overline{F}_{1}(\upsilon x_{1}), \overline{F}_{2}(\upsilon x_{2}))}{C(\overline{F}_{1}( x_{1}), \overline{F}_{2}(x_{2}))}\leq\limsup\limits_{x_{1}\rightarrow\infty, x_{2}\rightarrow\infty}\frac{C[(1+\varepsilon)\overline{F}_{1}(x_{1}), (1+\varepsilon)\overline{F}_{2}(x_{2})]}{C(\overline{F}_{1}(x_{1}), \overline{F}_{2}(x_{2}))} = (1+\varepsilon)^{r}, $

因此我们有

$ \lim\limits_{\upsilon\uparrow 1}\limsup\limits_{x_{1}\rightarrow\infty, x_{2}\rightarrow\infty}\frac{C(\overline{F}_{1}(\upsilon x_{1}), \overline{F}_{2}(\upsilon x_{2}))}{C(\overline{F}_{1}( x_{1}), \overline{F}_{2}(x_{2}))}\leq\lim\limits_{\varepsilon\downarrow 0}(1+\varepsilon)^{r} = 1. $

另一方面对于任意的常数$ \upsilon\in(0, 1) $同样地有

$ \lim\limits_{\upsilon\uparrow 1}\limsup\limits_{x_{1}\rightarrow\infty, x_{2}\rightarrow\infty}\frac{C(\overline{F}_{1}(\upsilon x_{1}), \overline{F}_{2}(\upsilon x_{2}))}{C(\overline{F}_{1}( x_{1}), \overline{F}_{2}(x_{2}))}\geq 1. $

我们用相同的方法可以证明对于任意$ \upsilon>1 $有

$ \lim\limits_{\upsilon\uparrow 1}\limsup\limits_{x_{1}\rightarrow\infty, x_{2}\rightarrow\infty}\frac{C(\overline{F}_{1}(\upsilon x_{1}), \overline{F}_{2}(\upsilon x_{2}))}{C(\overline{F}_{1}( x_{1}), \overline{F}_{2}(x_{2}))} = 1. $

引理3.1证毕.

引理3.2^[11]  如果$ F_{1}, F_{2}\in C $,则

$ \lim\limits_{x\rightarrow\infty}\frac{\overline{F}_{i}(x+o(x))}{\overline{F}_{i}(x)} = 1, \lim\limits_{x_1, x_2\rightarrow\infty}\frac{\overline{F}_{12}(x_{1}+o(x_{1}), x_{2}+o(x_{2}))}{\overline{F}_{12}(x_{1}, x_{2})} = 1. $

引理3.3  设$ \{X_{i}, i\geq1\} $是独立同分布非负随机变量序列,其共同分布为$ F $,均值为$ \mu = EX_{i}<\infty $, $ \{I_{i}, i\geq 1\} $是相互独立的随机变量序列,那么对任意的$ \upsilon>0, x>0 $有

$ P\bigg(\sum\limits_{i = 1}^{n}I_{i}X_{i}>x\bigg)\leq n\theta\overline{F}(\frac{x}{\upsilon})+(\frac{e n\mu\theta}{x})^{\upsilon}. $

证  对任意$ \upsilon>0 $和$ x>0 $,记$ I_{i}\widetilde{X}_{i} = \min \{I_{i}X_{i}, \frac{x}{\upsilon}\} $, $ S_{n} = \sum\limits_{i = 1}^{n}I_{i}\widetilde{X}_{i} $则有

$ P\bigg(\sum\limits_{i = 1}^{n}I_{i}X_{i}>x\bigg)\leq P(\max\limits_{1\leq i\leq n}I_{i}X_{i}>\frac{x} {\upsilon})+P(S_{n}>x)\leq n\theta\overline{F}(\frac{x}{\upsilon})+P(S_{n}>x). $

对任意的$ h = h(x)>0 $,有$ P(S_{n}>x)\leq e^{-hx}Ee^{hS_{n}} = e^{-hx}\prod\limits_{i = 1}^{n}Ee^{hI_{i}\widetilde{X}_{i}}. $当$ s>0 $时,函数$ \frac{e^{hs}-1}{s} $单调递增,当$ y>0 $时不等式$ 1+y\leq e^{y} $成立,故

$ \begin{eqnarray*} \prod\limits_{i = 1}^{n}Ee^{hI_{i}\widetilde{X}_{i}}&\leq&\prod\limits_{i = 1}^{n} \bigg(1+\int_{0}^{\frac{x}{ \upsilon}}e^{hs}-1{\rm d}F_{i}{\rm d}s+(e^{h\frac{x}{\upsilon}}-1)\overline{F_{i}}(\frac{x}{\upsilon})\bigg)\\ &\leq&\prod\limits_{i = 1}^{n}\bigg(1+\frac{e^{h\frac{x}{\upsilon}-1}}{\frac{x}{\upsilon}}\mu\theta_{i}\bigg) \leq e^{n\theta\mu\frac{e^{h\frac{x}{\upsilon}-1}}{\frac{x}{\upsilon}}}. \end{eqnarray*} $

令$ h = h(x) = \frac{\upsilon}{x}\log(\frac{x}{n\theta\mu}+1) $有

$ P\bigg(\sum\limits_{i = 1}^{n}I_{i}X_{i}>x\bigg)\leq n\theta\overline{F}(\frac{x}{\upsilon}) +\exp\bigg \{\upsilon-\upsilon\log(\frac{x}{n\theta\mu}+1)\bigg\}\leq n\theta\overline{F}(\frac{x}{\upsilon})+(\frac{e n\mu\theta}{x})^{\upsilon}. $

引理3.3证毕.

定理3.4  假设$ \{\overrightarrow{X}_{i}, i\geq 1\} $是一列非负随机向量,满足假设A1–A3,对任意固定的$ \overrightarrow{\gamma}>\overrightarrow{0} $,当$ \overrightarrow{x}\geq \overrightarrow{\gamma}n $,一致地有

(3.9) $ \begin{align} P(\overrightarrow{S}_{n}-E\overrightarrow{S}_{n}>\overrightarrow{x}) = P(\overrightarrow{Y}_{n}-E\overrightarrow{Y}_{n}>\overrightarrow{x})\sim((\theta n)^{2}-\theta n)\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})+\theta n\overline{F}_{12}(x_{1}, x_{2}), \end{align} $

记$ \overrightarrow{Y}_{n} = \sum\limits_{i = 1}^{n}I_{i}\overrightarrow{X^{i}}, n\geq 1. $

定理3.5  假设$ \{\overrightarrow{X}_{i}, i\geq 1\} $是一个非负随机向量,满足假设A1–A4,对任意固定的$ \overrightarrow{\gamma}>\overrightarrow{0} $,当$ \overrightarrow{x}\geq \overrightarrow{\gamma}\lambda(t) $,一致地有

(3.10) $ \begin{align} P(\overrightarrow{S}_{N(t)}-E\overrightarrow{S}_{N(t)}>\overrightarrow{x})\sim((\theta \lambda(t))^{2}-\theta \lambda(t))\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})+\theta \lambda(t)\overline{F}_{12}(x_{1}, x_{2}), \end{align} $

记$ \overrightarrow{Y}_{N(t)} = \sum\limits_{i = 1}^{N(t)}I_{i}\overrightarrow{X^{i}}, t\geq 0. $

定理3.4的证明  记$ (I_{i}X_{1}^{i}, I_{i}X_{2}^{i})^{\top} $的联合分布函数为$ F_{12}^{i}(x_{1}, x_{2}) $,边缘分布函数为$ F_{1}^{i}(x_{1}), F_{2}^{i}(x_{2}) $,则有

$ \overline{F^{i}}_{12}(x_{1}, x_{2}) = 1-F_{12}^{i}(x_{1}, x_{2}) = \theta_{i}\overline{F}_{12}(x_{i}, x_{2}), $

$ \overline{F^{i}}_{1}(x_{1}) = 1-F_{1}^{i}(x_{1}) = \theta_{i}F_{1}(x_{1}), {\quad} \overline{F^{i}}_{2}(x_{2}) = 1-F_{2}^{i}(x_{2}) = \theta_{i}F_{2}(x_{2}). $

记

$ Y_{1n} = \sum\limits_{i = 1}^{n}I_{i}X_{1}^{i}, Y_{2n} = \sum\limits_{i = 1}^{n}I_{i}X_{2}^{i}, {\quad} i\geq 1. $

首先估计(3.9)式的下界,对任意$ \lambda>1 $有

$ \begin{eqnarray*} && P(\overrightarrow{Y}_{n}-E\overrightarrow{Y}_{n}>\overrightarrow{x})\\ &\geq& P(Y_{1n}-EY_{1n}>x_{1}, Y_{2n}-EY_{2n}>x_{2}, \max\limits_{1\leq i\leq n}I_{i}X_{1}^{i}>\lambda x_{1}, \max\limits_{1\leq j\leq n}I_{j}X_{2}^{j}>\lambda x_{2})\\ &\geq&\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{n}P(Y_{1n}-EY_{1n}>x_{1}, Y_{2n}-EY_{2n}>x_{2}, I_{i}X_{1}^{i}>\lambda x_{1}, I_{j}X_{2}^{j}>\lambda x_{2})\\ &&-\sum\limits_{i = 1}^{n}\sum\limits_{j_{1}\neq j_{2}}P(I_{i}X_{1}^{i}>\lambda x_{1}, I_{j_{1}}X_{2}^{j_{1}}>\lambda x_{2}, I_{j_{2}}X_{2}^{j_{2}}>\lambda x_{2})\\ &&-\sum\limits_{i_{1}\neq i_{2}}\sum\limits_{j = 1}^{n}P(I_{i_{1}}X_{1}^{i_{1}}>\lambda x_{1}, I_{i_{2}}X_{1}^{i_{2}}>\lambda x_{1}, I_{j}X_{2}^{j}>\lambda x_{2})\\ & = :&J_{1}-J_{2}-J_{3}. \end{eqnarray*} $

对$ J_{1} $进行估计

$ \begin{eqnarray*} J_{1}& = &\sum\limits_{i = 1}^{n}P(Y_{1n}-EY_{1n}>x_{1}, Y_{2n}-EY_{2n}>x_{2}, I_{i}X_{1}^{i}>\lambda x_{1}, I_{i}X_{2}^{i}>\lambda x_{2})\\ &&+\sum\limits_{i\neq j}P(Y_{1n}-EY_{1n}>x_{1}, Y_{2n}-EY_{2n}>x_{2}, I_{i}X_{1}^{i}>\lambda x_{1}, I_{j}X_{2}^{j}>\lambda x_{2})\\ &\geq&\sum\limits_{i = 1}^{n}P(\sum\limits_{s\neq i}I_{s}X_{1}^{s}-EY_{1n}>(1-\lambda)x_{1}, \sum\limits_{l\neq i}I_{l}X_{2}^{l}-EY_{2n}>(1-\lambda)x_{2}, \\ && I_{i}X_{1}^{i}>\lambda x_{1}, I_{i}X_{2}^{i}>\lambda x_{2})\\ &&+\sum\limits_{i\neq j}P(\sum\limits_{s\neq i, j}I_{s}X_{1}^{s}-EY_{1n}>(1-\lambda)x_{1}, \sum\limits_{l\neq i, j}I_{l}X_{2}^{l}-EY_{2n}>(1-\lambda)x_{2}, \\ && I_{i}X_{1}^{i}>\lambda x_{1}, I_{j}X_{2}^{j}>\lambda x_{2})\\ & = &\sum\limits_{i = 1}^{n}P(Y_{1, n-1}-EY_{1n}>(1-\lambda)x_{1}, Y_{2, n-1}-EY_{2n}>(1-\lambda)x_{2})\\ &&\times P(I_{i}X_{1}^{i}>\lambda x_{1}, I_{i}X_{2}^{i}>\lambda x_{2})\\ &&+\sum\limits_{i\neq j}P(Y_{1, n-2}-EY_{1n}>(1-\lambda)x_{1}, Y_{2, n-2}-EY_{2n}>(1-\lambda)x_{2}) \\ &&\times P(I_{i}X_{1}^{i}>\lambda x_{1}) P(I_{j}X_{2}^{j}>\lambda x_{2}). \end{eqnarray*} $

由强大数定律,我们得到

$ \lim\limits_{n\rightarrow\infty}\inf\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}P(\overrightarrow{Y}_{n-1}-E\overrightarrow{Y}_{n}>(1-\lambda)\overrightarrow{x}) = 1, $

且

$ \lim\limits_{n\rightarrow\infty}\inf\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}P(\overrightarrow{Y}_{n-2}-E\overrightarrow{Y}_{n}>(1-\lambda)\overrightarrow{x}) = 1. $

因此,我们得到

(4.1) $ \begin{eqnarray} &&\liminf\limits_{n\rightarrow\infty}\inf\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{J_{1}}{((\theta n)^{2}-\theta n)\overline{F}_{1}(\lambda x_{1})\overline{F}_{2}(\lambda x_{2})+\theta n\overline{F}_{12}(\lambda x_{1}, \lambda x_{2})}{}\\ & = &\liminf\limits_{n\rightarrow\infty}\inf\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n} \frac{\sum\limits_{i = 1}^{n}P(I_{i}X_{1}^{i}>\lambda x_{1}, I_{i}X_{2}^{i}>\lambda x_{2})+\sum\limits_{i\neq j}P(I_{i}X_{1}^{i}>\lambda x_{1})P(I_{j}X_{2}^{j}>\lambda x_{2})}{((\theta n)^{2}-\theta n)\overline{F}_{1}(\lambda x_{1})\overline{F}_{2}(\lambda x_{2})+\theta n\overline{F}_{12}(\lambda x_{1}, \lambda x_{2})}{}\\ & = &\liminf\limits_{n\rightarrow\infty}\inf\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n} \frac{\overline{F}_{12}(\lambda x_{1}, \lambda x_{2})\sum\limits_{i = 1}^{n}\theta_{i}+\overline{F}_{1}(\lambda x_{1})\overline{F}_{2}(\lambda x_{2})(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{n}\theta_{i}\theta_{j}-\sum\limits_{i = 1}^{n}\theta_{i})}{[(\theta n)^{2}-\theta n]\overline{F}_{1}(\lambda x_{1})\overline{F}_{2}(\lambda x_{2})+\theta n\overline{F}_{12}(\lambda x_{1}, \lambda x_{2})}{}\\ & = &1. \end{eqnarray} $

对$ J_{2} $进行估计

$ \begin{eqnarray*} J_{2}& = &\sum\limits_{j_{1}\neq j_{2}}P(I_{j_{1}}X_{1}^{j_{1}}>\lambda x_{1}, I_{j_{1}}X_{2}^{j_{1}}>\lambda x_{2})P(I_{j_{2}}X_{2}^{j_{2}}>\lambda x_{2})\\ &&+\sum\limits_{i\neq j_{1}}\sum\limits_{j_{1}\neq j_{2}}P(I_{i}X_{1}^{i}>\lambda x_{1}, I_{j_{2}}X_{2}^{j_{2}}>\lambda x_{2})P(I_{j_{1}}X_{2}^{j_{1}}>\lambda x_{2})\\ &\leq & n\overline{F}_{2}(\lambda x_{2})n\theta\overline{F}_{12}(\lambda x_{1}, \lambda x_{2})\\ &&+n\overline{F}_{2}(\lambda x_{2})\{((\theta n)^{2}-\theta n)\overline{F}_{1}(\lambda x_{1})\overline{F}_{2}(\lambda x_{2}) +\theta n\overline{F}_{12}(\lambda x_{1}, \lambda x_{2})\}. \end{eqnarray*} $

因此可以得到

(4.2) $ \begin{eqnarray} &&\limsup\limits_{n\rightarrow\infty}\sup\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{J_{2}}{((\theta n)^{2}-\theta n)\overline{F}_{1}(\lambda x_{1})\overline{F}_{2}(\lambda x_{2})+\theta n\overline{F}_{12}(\lambda x_{1}, \lambda x_{2})}{}\\ &\leq&\limsup\limits_{n\rightarrow\infty}\sup\limits_{x_{2}\geq\gamma_{2}n}2n\overline{F}_{2}(\lambda x_{2}){}\\ & = &0. \end{eqnarray} $

类似地可以得到

(4.3) $ \begin{equation} \limsup\limits_{n\rightarrow\infty}\sup\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{J_{3}}{((\theta n)^{2}-\theta n)\overline{F}_{1}(\lambda x_{1})\overline{F}_{2}(\lambda x_{2})+\theta n\overline{F}_{12}(\lambda x_{1}, \lambda x_{2})} = 0. \end{equation} $

由(4.1)–(4.3)式得

$ \begin{eqnarray*} &&\liminf\limits_{n\rightarrow\infty}\inf\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{P(\overrightarrow{Y}_{n }-E\overrightarrow{Y}_{n}>\overrightarrow{x})}{((\theta n)^{2}-\theta n)\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})+\theta n\overline{F}_{12}(x_{1}, x_{2})}\\ &\geq&\bigg(\liminf\limits_{n\rightarrow\infty}\inf\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{P(Y_{1n}-EY_{1n} >x_{1}, Y_{2n}-EY_{2n}>x_{2})}{((\theta n)^{2}-\theta n)\overline{F}_{1}(\lambda x_{1})\overline{F}_{2}(\lambda x_{2})+\theta n\overline{F}_{12}(\lambda x_{1}, \lambda x_{2})}\bigg)\\ &&\times \bigg(\liminf\limits_{n\rightarrow\infty}\inf\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{((\theta n)^{2}-\theta n)\overline{F}_{1}(\lambda x_{1})\overline{F}_{2}(\lambda x_{2})+\theta n\overline{F}_{12}(\lambda x_{1}, \lambda x_{2})}{((\theta n)^{2}-\theta n)\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})+\theta n\overline{F}_{12}(x_{1}, x_{2})}\bigg)\\ & = &\liminf\limits_{n\rightarrow\infty}\inf\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{((\theta n)^{2}-\theta n)\overline{F}_{1}(\lambda x_{1})\overline{F}_{2}(\lambda x_{2})+\theta n\overline{F}_{12}(\lambda x_{1}, \lambda x_{2})}{((\theta n)^{2}-\theta n)\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})+\theta n\overline{F}_{12}(x_{1}, x_{2})}. \end{eqnarray*} $

由假设A3及引理3.1得,对于$ \lambda>1 $有

(4.4) $ \begin{equation} \lim\limits_{\lambda\downarrow 1 }\liminf\limits_{x_{1}\rightarrow\infty, x_{2}\rightarrow\infty}\frac{P(X_{1}^{i}>\lambda x_{1}, X_{2}^{j}>\lambda x_{2})}{P(X_{1}^{i}>x_{1}, X_{2}^{j}>x_{2})} = 1, \end{equation} $

因此我们得到

(4.5) $ \begin{eqnarray} &&\liminf\limits_{n\rightarrow\infty}\inf\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{P(\overrightarrow{Y}_{n}-E\overrightarrow{Y}_{n}>\overrightarrow{x})}{((\theta n)^{2}-\theta n)\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})+\theta n\overline{F}_{12}(x_{1}, x_{2})}{}\\ &\geq&\lim\limits_{\lambda\downarrow 1}\bigg(\liminf\limits_{n\rightarrow\infty}\inf\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{((\theta n)^{2}-\theta n)\overline{F}_{1}(\lambda x_{1})\overline{F}_{2}(\lambda x_{2})+\theta n\overline{F}_{12}(\lambda x_{1}, \lambda x_{2})}{((\theta n)^{2}-\theta n)\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})+\theta n\overline{F}_{12}(x_{1}, x_{2})}\bigg){}\\ & = &1. \end{eqnarray} $

估计(3.9)式上界,对任意的$ 0<\upsilon<1 $,记$ I_{i}\widetilde{X}_{1}^{i} = \min\{I_{i}X_{1}^{i}, \upsilon x_{1}\}, I_{i}\widetilde{X}_{2}^{i} = \min\{I_{i}X_{2}^{i}, \upsilon x_{2}\}, $$ \widetilde{S}_{1, n} = \sum\limits_{i = 1}^{n}I_{i}\widetilde{X}_{1}^{i}, \widetilde{S}_{2, n} = \sum\limits_{i = 1}^{n}I_{i}\widetilde{X}_{2}^{i}, \widetilde{x}_{1} = x_{1}+n\theta\mu_{1}, \widetilde{x}_{2} = x_{2}+n\theta\mu_{2}, $我们得到

(4.6) $ \begin{eqnarray} && P(\overrightarrow{Y}_{n}-E\overrightarrow{Y}_{n}>\overrightarrow{x}){}\\ & = &P(Y_{1n}-EY_{1n}>x_{1}, Y_{2n}-EY_{2n}>x_{2}){}\\ &\leq&\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{n}P(I_{i}X_{1}^{i}>\upsilon x_{1}, I_{j}X_{2}^{j}>\upsilon x_{2}){}\\ &&+P(Y_{1n}-EY_{1n}>x_{1}, Y_{2n}-EY_{2n}>x_{2}, \max\limits_{1\leq i\leq n}I_{i}X_{1}^{i}\leq\upsilon x_{1}, \max\limits_{1\leq j\leq n}I_{j}X_{2}^{j}\leq\upsilon x_{2}){}\\ &&+P(Y_{1n}-EY_{1n}>x_{1}, Y_{2n}-EY_{2n}>x_{2}, \max\limits_{1\leq i\leq n}I_{i}X_{1}^{i}>\upsilon x_{1}, \max\limits_{1\leq j\leq n}I_{j}X_{2}^{j}\leq\upsilon x_{2}){}\\ &&+P(Y_{1n}-EY_{1n}>x_{1}, Y_{2n}-EY_{2n}>x_{2}, \max\limits_{1\leq i\leq n}I_{i}X_{1}^{i}\leq\upsilon x_{1}, \max\limits_{1\leq j\leq n}I_{j}X_{2}^{j}>\upsilon x_{2}){}\\ & = :&K_{1}+K_{2}+K_{3}+K_{4}. \end{eqnarray} $

估计$ K_{1} $

$ \begin{eqnarray*} K_{1}& = &\sum\limits_{i = 1}^{n}P(I_{i}X_{1}^{i}>\upsilon x_{1}, I_{i}X_{2}^{i}>\upsilon x_{2})+\sum\limits_{i\neq j}P(I_{i}X_{1}^{i}>\upsilon x_{1}P(I_{j}X_{2}^{j}>\upsilon x_{2})\\ & = &((\theta n)^{2}-\theta n)\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})+\theta n\overline{F}_{12}(x_{1}, x_{2}) . \end{eqnarray*} $

由假设A3及引理3.1,对任意$ \upsilon\in(0, 1) $,有

(4.7) $ \begin{equation} \lim\limits_{\lambda\uparrow 1 }\limsup\limits_{x_{1}\rightarrow\infty, x_{2}\rightarrow\infty}\frac{P(X_{1}^{i}>\upsilon x_{1}, X_{2}^{j}>\upsilon x_{2})}{P(X_{1}^{i}>x_{1}, X_{2}^{j}>x_{2})} = 1. \end{equation} $

因此

(4.8) $ \begin{equation} \lim\limits_{\lambda\uparrow 1 }\limsup\limits_{n\rightarrow\infty}\sup\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{K_{1}}{((\theta n)^{2}-\theta n)\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})+\theta n\overline{F}_{12}(x_{1}, x_{2})} = 1. \end{equation} $

估计$ K_{2} $:我们让$ a_{1} = \max\{-\log n\overrightarrow{F}_{1}(\upsilon x_{1}), 1\}, a_{2} = \max\{-\log n\overrightarrow{F}_{2}(\upsilon x_{2}), 1\} $,当$ x_{1}\geq\gamma_{1}n, $$ x_{2}\geq\gamma_{2}n $时,有$ a_{1} $, $ a_{2} $分别一致地趋于无穷大,对任意的$ h_{1}>0, h_{2}>0 $有

$ \begin{eqnarray*} &&\frac{K_{2}}{((\theta n)^{2}-\theta n)\overline{F}_{1}(\upsilon x_{1})\overline{F}_{2}(\upsilon x_{2})+\theta n\overline{F}_{12}(\upsilon x_{1}, \upsilon x_{2})}\\ &\leq&\frac{n}{n\theta^{2}-\theta}\frac{P(\widetilde{S}_{1n}>\widetilde{x}_{1}, \widetilde{S}_{2n}>\widetilde{x}_{2})}{n\overline{F}_{1}(\upsilon x_{1})n\overline{F}_{2}(\upsilon x_{2})}\\ &\leq&\frac{n}{n\theta^{2}-\theta}\frac{e^{-h_{1}\widetilde{x}_{1}-h_{2}\widetilde{x}_{2}}\prod\limits_{i = 1}^{n}Ee^{h_{1}I_{i}\widetilde{X}_{1}^{i}+h_{2}I_{i}\widetilde{X}_{2}^{i}}}{n\overline{F}_{1}(\upsilon x_{1})n\overline{F}_{2}(\upsilon x_{2})}\\ &\leq&\frac{n}{n\theta^{2}-\theta}e^{-h_{1}\widetilde{x}_{1}-h_{2}\widetilde{x}_{2}+a_{1}+a_{2}}\prod\limits_{i = 1}^{n} \bigg(1+\int_{0}^{\upsilon x_{2}}\int_{0}^{\upsilon x_{1}}(e^{h_{1}s_{1}+h_{2}s_{2}}-1)F_{12}^{i}({\rm d}s_{1}, {\rm d}s_{2})\\ &&+\int_{0}^{\upsilon x_{2}}\int_{\upsilon x_{1}}^{\infty}(e^{h_{1}\upsilon x_{1}+h_{2}s_{2}}-1)F_{12}^{i}({\rm d}s_{1}, {\rm d}s_{2})+\int_{\upsilon x_{2}}^{\infty}\int_{0}^{\upsilon x_{1}}(e^{h_{1}s_{1}+h_{2}\upsilon x_{2}}-1)F_{12}^{i}({\rm d}s_{1}, {\rm d}s_{2})\\ &&+\int_{\upsilon x_{2}}^{\infty}\int_{\upsilon x_{1}}^{\infty}(e^{h_{1}\upsilon x_{1}+h_{2}\upsilon x_{2}}-1)F_{12}^{i}({\rm d}s_{1}, {\rm d}s_{2})\bigg)\\ &\leq&\frac{n}{n\theta^{2}-\theta}e^{-h_{1}\widetilde{x}_{1}-h_{2}\widetilde{x}_{2}+a_{1}+a_{2}} \exp\bigg\{\sum\limits_{i = 1}^{n}\theta_{i}\bigg(\int_{0}^{\upsilon x_{2}}\int_{0}^{\upsilon x_{1}}(e^{h_{1}s_{1}+h_{2}s_{2}}-1)F_{12}({\rm d}s_{1}, {\rm d}s_{2})\\ &&+\int_{0}^{\upsilon x_{2}}\int_{\upsilon x_{1}}^{\infty}(e^{h_{1}\upsilon x_{1}+h_{2}s_{2}}-1)F_{12}({\rm d}s_{1}, {\rm d}s_{2})+\int_{\upsilon x_{2}}^{\infty}\int_{0}^{\upsilon x_{1}}(e^{h_{1}s_{1}+h_{2}\upsilon x_{2}}-1)F_{12}({\rm d}s_{1}, {\rm d}s_{2})\\ &&+\int_{\upsilon x_{2}}^{\infty}\int_{\upsilon x_{1}}^{\infty}(e^{h_{1}\upsilon x_{1}+h_{2}\upsilon x_{2}}-1)F_{12}({\rm d}s_{1}, {\rm d}s_{2})\bigg)\bigg\}\\ & = &\frac{n}{n\theta^{2}-\theta} \exp \bigg\{\sum\limits_{i = 1}^{n}\theta_{i}(L_{1}+L_{2}+L_{3}+L_{4})-h_{1}\widetilde{x}_{1}-h_{2}\widetilde{x}_{2}+a_{1}+a_{2}\bigg\}, \end{eqnarray*} $

其中地四个不等式用到基本不等式$ 1+x\leq e^{x} $.对积分$ L_{1} $进行划分得

$ \begin{eqnarray*} L_{1}& = &\bigg(\int_{0}^{\upsilon x_{2}/a_{2}^{2}}\int_{0}^{\upsilon x_{1}/a_{1}^{2}}+\int_{0}^{\upsilon x_{2}/a_{2}^{2}}\int_{\upsilon x_{1}/a_{1}^{2}}^{\upsilon x_{1}}+\int_{\upsilon x_{2}/a_{2}^{2}}^{\upsilon x_{2}}\int_{0}^{\upsilon x_{1}/a_{1}^{2}}+\int_{\upsilon x_{2}/a_{2}^{2}}^{\upsilon x_{2}}\int_{\upsilon x_{1}/a_{1}^{2}}^{\upsilon x_{1}}\bigg)\\ &&\times (e^{h_{1}s_{1}+h_{2}s_{2}}-1)F_{12}({\rm d}s_{1}, {\rm d}s_{2})\\ &\leq&e^{h_{1}\upsilon x_{1}/a_{1}^{2}+h_{2}\upsilon x_{2}/a_{2}^{2}}(h_{1}\mu_{1}+h_{2}\mu_{2})+e^{h_{1}\upsilon x_{1}+h_{2}\upsilon x_{2}/a_{2}^{2}}\overline{F}_{1}(\upsilon x_{1}/a_{1}^{2})\\ &\quad&+e^{h_{1}{\upsilon x_{1}/a_{1}^{2}}+h_{2}\upsilon x_{2}}\overline{F}_{2}(\upsilon x_{2}/a_{2}^{2})+e^{h_{1}\upsilon x_{1}+h_{2}\upsilon x_{2}}\overline{F}_{12}(\upsilon x_{1}/a_{1}^{2}, \upsilon x_{2}/a_{2}^{2}). \end{eqnarray*} $

上式用到基本不等式$ e^{x}-1\leq xe^{x} $.下面我们考虑$ L_{2} $

$ \begin{eqnarray*} L_{2}& = &\bigg(\int_{0}^{\upsilon x_{2}/a_{2}^{2}}\int_{\upsilon x_{1}}^{\infty}+\int_{\upsilon x_{2}/a_{2}^{2}}^{\upsilon x_{2}}\int_{\upsilon x_{1}}^{\infty}\bigg)(e^{h_{1}\upsilon x_{1}+h_{2}s_{2}}-1)F_{12}({\rm d}s_{1}, {\rm d}s_{2})\\ &\leq&e^{h_{1}\upsilon x_{1}+h_{2}\upsilon x_{2}/a_{2}^{2}}\overline{F}_{1}(\upsilon x_{1})+e^{h_{1}\upsilon x_{1}+h_{2}\upsilon x_{2}}\overline{F}_{12}(\upsilon x_{1}, \upsilon x_{2}/a_{2}^{2}). \end{eqnarray*} $

类似于$ L_{2} $的方法处理$ L_{3} $有

$ L_{3}\leq e^{h_{1}\upsilon x_{1}/a_{1}^{2}+h_{2}\upsilon x_{2}}\overline{F}_{2}(\upsilon x_{2})+e^{h_{1}\upsilon x_{1}+h_{2}\upsilon x_{2}}\overline{F}_{12}(\upsilon x_{1}/a_{1}^{2}, \upsilon x_{2}). $

对于$ L_{4} $有

$ L_{4}\leq e^{h_{1}\upsilon x_{1}+h_{2}\upsilon x_{2}}\overline{F}_{12}(\upsilon x_{1}, \upsilon x_{2}). $

让$ h_{1} = \frac{a_{1}-2\rho\log a_{1}}{\upsilon x_{1}}, h_{2} = \frac{a_{2}-2\rho\log a_{2}}{\upsilon x_{2}} $,通过文献[3,引理2.2]及(3.2)式.由$ F\in C $,则存在$ M, C_{1}, C_{2} $使得

$ \begin{eqnarray*} &&\frac{K_{2}}{((\theta n)^{2}-\theta n)\overline{F}_{1}(\upsilon x_{1})\overline{F}_{2}(\upsilon x_{2})+\theta n\overline{F}_{12}(\upsilon x_{1}, \upsilon x_{2})}\\ &\leq&\frac{n}{n\theta^{2}-\theta}{\rm exp}(e^{a_{1}^{-1}}e^{a_{2}^{-1}}n(\theta+o(1))(h_{1}\mu_{1}+h_{2}\mu_{2})+\frac{1}{n}\sum\limits_{i = 1}^{n}\theta_{i}C_{1}e^{a_{2}^{-1}}+\frac{1}{n}\sum\limits_{i = 1}^{n}\theta_{i}C_{2}e^{a_{1}^{-1}}\\ &&+\frac{1}{n}\sum\limits_{i = 1}^{n}\theta_{i}\frac{MC_{1}C_{2}}{n}+\frac{1}{n}\sum\limits_{i = 1}^{n}\theta_{i}o(1)e^{a_{2}^{-1}}+\frac{1}{n}\sum\limits_{i = 1}^{n}\theta_{i}o(1)\frac{MC_{2}}{n}+\frac{1}{n}\sum\limits_{i = 1}^{n}\theta_{i}o(1)e^{a_{1}^{-1}}\\ &&+\frac{1}{n}\sum\limits_{i = 1}^{n}\theta_{i}o(1)\frac{MC_{1}}{n}+\frac{1}{n}\sum\limits_{i = 1}^{n}\theta_{i}o(1)\frac{M}{n}-h_{1}\widetilde{x}_{1}-h_{2}\widetilde{x}_{2}+a_{1}+a_{2})\\ &\leq& O(1){\rm exp}\{e^{a_{1}^{-1}}e^{a_{2}^{-1}}n(\theta+o(1))(h_{1}\mu_{1}+h_{2}\mu_{2}) -h_{1}(x_{1}+n\theta\mu_{1})-h_{2}(x_{2}+n\theta\mu_{2}) +a_{1}+a_{2}\}\\ &\leq& O(1){\rm exp}\{(e^{a_{1}^{-1}}e^{a_{2}^{-1}}-1)(\theta h_{1}\mu_{1}n+\theta h_{2}\mu_{2}n)+(o(n)h_{1}\mu_{1}+o(n)h_{2}\mu_{2})e^{a_{1}^{-1}}e^{a_{2}^{-1}}\\ &&-h_{1}x_{1}-h_{2}x_{2}+a_{1}+a_{2}\}\\ &\leq &O(1)\exp\bigg\{o(n)\bigg(\frac{a_{1}-2\rho\log a_{1}}{\upsilon x_{1}}+\frac{a_{2}-2\rho\log a_{2}}{\upsilon x_{2}}\bigg)\\ && -\frac{a_{1}-2\rho\log a_{1}}{\upsilon}-\frac{a_{2}-2\rho\log a_{2}}{\upsilon}+a_{1}+a_{2}\bigg\}\\ & = &O(1)\{o(\theta_{1})+o(\theta_{2})+(1-\frac{1}{\upsilon})\theta_{1}+(1-\frac{1}{\upsilon})\theta_{2}\}, \end{eqnarray*} $

因此

(4.9) $ \begin{equation} \limsup\limits_{n\rightarrow\infty}\sup\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{K_{2}}{((\theta n)^{2}-\theta n)\overline{F}_{1}(\upsilon x_{1})\overline{F}_{2}(\upsilon x_{2})+\theta n\overline{F}_{12}(\upsilon x_{1}, \upsilon x_{2})} = 0. \end{equation} $

估计$ K_{3} $,由于函数$ g(X_{2}^{i}) = \widetilde{X}_{2}^{i} $是单调增函数,因此(3.2)式对$ X_{1}^{i} $与$ \widetilde{X}_{2}^{i} $也满足.所以

$ \begin{eqnarray*} K_{3}& = &\sum\limits_{i = 1}^{n}P(I_{i}X_{1}^{i}>\upsilon x_{1}, \widetilde{S}_{2n}>\widetilde{x}_{2})\\ &\leq&\sum\limits_{i = 1}^{n}\int_{0}^{\upsilon x_{2}}\cdots\int_{0}^{\upsilon x_{2}}P(I_{i}X_{1}^{i}>\upsilon x_{1}, I_{i}\widetilde{X}_{2}^{i}>\widetilde{x}_{2}-x_{21}\cdots x_{2, i-1}, x_{2, i+1}\cdots x_{2n})\\ &\quad&\times G_{i}({\rm d}x_{21}\cdots {\rm d}x_{2, i-1}, {\rm d}x_{2, i+1}\cdots {\rm d}x_{2n})\\ & = &\sum\limits_{i = 1}^{n}\theta_{i}\int_{0}^{\upsilon x_{2}}\cdots\int_{0}^{\upsilon x_{2}}P(X_{1}^{i}>\upsilon x_{1}, \widetilde{X}_{2}^{i}>\widetilde{x}_{2}-x_{21}\cdots x_{2, i-1}, x_{2, i+1}\cdots x_{2n})\\ &\quad&\times G_{i}({\rm d}x_{21}\cdots {\rm d}x_{2, i-1}, {\rm d}x_{2, i+1}\cdots {\rm d}x_{2n})\\ &\leq&M\sum\limits_{i = 1}^{n}\int_{0}^{\upsilon x_{2}}\cdots\int_{0}^{\upsilon x_{2}}P(X_{1}^{i}>\upsilon x_{1})P(I_{i}\widetilde{X}_{2}^{i}>\widetilde{x}_{2}-x_{21}\cdots x_{2, i-1}, x_{2, i+1}\cdots x_{2n})\\ &\quad&\times G_{i}({\rm d}x_{21}\cdots {\rm d}x_{2, i-1}, {\rm d}x_{2, i+1}\cdots {\rm d}x_{2n})\\ & = &Mn\overline{F}_{1}(\upsilon x_{1})P(\widetilde{S}_{2n}>\widetilde{x}_{2}), \end{eqnarray*} $

其中$ G_{i} $是$ I_{1}\widetilde{X}_{2}^{1}, \cdots I_{i-1}\widetilde{X}_{2}^{i-1}, I_{i+1}\widetilde{X}_{2}^{i+1}\cdots I_{n}\widetilde{X}_{2}^{n} $的联合分布,因此

$ \frac{K_{3}}{((\theta n)^{2}-\theta n)\overline{F}_{1}(\upsilon x_{1})\overline{F}_{2}(\upsilon x_{2})+\theta n\overline{F}_{12}(\upsilon x_{1}, \upsilon x_{2})}\leq\frac{nM}{n\theta^{2}-\theta}\frac{P(\widetilde{S}_{2n}>\widetilde{x}_{2})}{n\overline{F}_{1}(\upsilon x_{1})}. $

由前面用到的$ a_{2} = \max\{-\log n\overline{F}_{2}(\upsilon x_{2}), 1\}, h_{2} = \frac{a_{2}-2\rho\log a_{2}}{\upsilon x_{2}} $,同样地,对$ K_{3} $有

$ \begin{eqnarray*} &&\frac{K_{3}}{((\theta n)^{2}-\theta n)\overline{F}_{1}(\upsilon x_{1})\overline{F}_{2}(\upsilon x_{2})+\theta n\overline{F}_{12}(\upsilon x_{1}, \upsilon x_{2})}\\ &\leq&\frac{nM}{n\theta^{2}-\theta} \exp\bigg\{\frac{1}{n}\sum\limits_{i = 1}^{n}\theta_{i}\{h_{2}\mu_{2}n(e^{h_{2}\upsilon x_{2}}/a_{2}^{2}-1)+B_{2}e^{h_{2}\upsilon x_{2}}a_{2}^{2\rho}n\overline{F}_{2}(\upsilon x_{2})\} -h_{2}x_{2}+a_{2}\bigg\}. \end{eqnarray*} $

即

(4.10) $ \begin{equation} \limsup\limits_{n\rightarrow\infty}\sup\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{K_{3}}{((\theta n)^{2}-\theta n)\overline{F}_{1}(\upsilon x_{1})\overline{F}_{2}(\upsilon x_{2})+\theta n\overline{F}_{12}(\upsilon x_{1}, \upsilon x_{2})} = 0. \end{equation} $

由于$ K_{3} $与$ K_{4} $具有相同结构,则

(4.11) $ \begin{equation} \limsup\limits_{n\rightarrow\infty}\sup\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{K_{4}}{((\theta n)^{2}-\theta n)\overline{F}_{1}(\upsilon x_{1})\overline{F}_{2}(\upsilon x_{2})+\theta n\overline{F}_{12}(\upsilon x_{1}, \upsilon x_{2})} = 0. \end{equation} $

由(4.6)–(4.11)式得

(4.12) $ \begin{equation} \limsup\limits_{n\rightarrow\infty}\sup\limits_{\overrightarrow{x}\geq\overrightarrow{\gamma}n}\frac{P(\overrightarrow{Y}_{n}-E\overrightarrow{Y}_{n}>\overrightarrow{x})}{((\theta n)^{2}-\theta n)\overline{F}_{1}(\upsilon x_{1})\overline{F}_{2}(\upsilon x_{2})+\theta n\overline{F}_{12}(\upsilon x_{1}, \upsilon x_{2})}\leq1. \end{equation} $

由(4.5)与(4.12)式,定理3.4得证.

定理3.5的证明  注意到

$ \begin{eqnarray*} P(\overrightarrow{S}_{N(t)}-E\overrightarrow{S}_{N(t)}>\overrightarrow{x})& = &\sum\limits_{n = 1}^{\infty}P(\overrightarrow{S}_{n}-E\overrightarrow{S}_{N(t)}>\overrightarrow{x})P(N(t) = n)\\ & = &\bigg(\sum\limits_{n<(1-\delta)\lambda(t)}+\sum\limits_{(1-\delta)\lambda(t)\leq n\leq(1+\delta)\lambda(t)}+\sum\limits_{n>(1+\delta)\lambda(t)}\bigg)\\ &\quad&\times P(\overrightarrow{S}_{n}-E\overrightarrow{S}_{N(t)}>\overrightarrow{x})P(N(t) = n)\\ & = &L_{1}+L_{2}+L_{3}, \end{eqnarray*} $

这儿当$ t\rightarrow\infty $时$ E\overrightarrow{S}_{N(t)}\sim(\theta-1-\delta)\overrightarrow{\mu}\lambda(t), 0<\delta<\min\{\gamma_{1}/\mu_{1}, \gamma_{2}/\mu_{2}\} $的任意常数,下面估计$ L_{1} $,由(3.7)式与定理3.4得

(4.13) $ \begin{eqnarray} L_{1}&\leq&\sum\limits_{n<(1-\delta)\lambda(t)}P(\overrightarrow{S}_{[(1-\delta)\lambda(t)]} -E\overrightarrow{S}_{N(t)}>\overrightarrow{x})P(N(t) = n) {}\\ &&\sim([(1-\delta)\lambda(t)]^{2}\theta^{2}-[(1-\delta)\lambda(t)]\theta)\overline{F}_{1}(x'_{1}) \overline{F}_{2}(x'_{2}){}\\ && +[(1-\delta)\lambda(t)]\theta\overline{F}_{12}(x'_{1}, x'_{2}) P(N(t)\leq(1-\delta)\lambda(t)){}\\ & = &o\{((\theta\lambda(t))^{2}-\theta\lambda(t))\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2}) +\theta\lambda(t)\overline{F}_{1, 2}(x_{1}, x_{2})\}, \end{eqnarray} $

这里$ x'_{i} = x_{i}+(\theta-1-\delta)\overrightarrow{\mu}\lambda(t)-[(1-\delta)\lambda(t)](\theta-1-\delta)\overrightarrow{\mu}>x_{i}, i = 1, 2 $下面我们估计$ L_{2} $,注意当$ n\rightarrow\infty $时$ E\overrightarrow{Y}_{n} = E(\sum\limits_{i = 1}^{n}I_{i}\overrightarrow{X}_{1}^{i})\sim\theta\overrightarrow{\mu}n $,当$ n, t\rightarrow\infty $时$ (1+\delta)\overrightarrow{\mu}n+E\overrightarrow{S}_{N(t)}-E\overrightarrow{Y}_{n}\sim(1+\delta-\theta)\overrightarrow{\mu}(n-\lambda(t)) $,由(3.7)式,引理3.2及定理3.4,对任意固定的$ \overrightarrow{\gamma}>\overrightarrow{0} $,当$ \overrightarrow{x}\geq\overrightarrow{\gamma}\lambda(t) $时,对$ (1-\delta)\lambda(t)\leq n\leq(1+\delta)\lambda(t) $一致地有

(4.14) $ \begin{eqnarray} L_{2}& = &\sum\limits_{(1-\delta)\lambda(t)\leq n\leq(1+\delta)\lambda(t)}P(\overrightarrow{S}_{n}-E\overrightarrow{S}_{N(t)}>\overrightarrow{x})P(N(t) = n){}\\ & = &\sum\limits_{(1-\delta)\lambda(t)\leq n\leq(1+\delta)\lambda(t)}P(\overrightarrow{Y}_{n}>\overrightarrow{x}+(1+\delta)\overrightarrow{\mu}n+E\overrightarrow{S}_{N(t)})P(N(t) = n){}\\ & = &\sum\limits_{(1-\delta)\lambda(t)\leq n\leq(1+\delta)\lambda(t)}P(\overrightarrow{Y}_{n}-E\overrightarrow{Y}_{n}>\overrightarrow{x}+(1+\delta)\overrightarrow{\mu}n+E\overrightarrow{S}_{N(t)}-E\overrightarrow{Y}_{n}){}\\ && P(N(t) = n){}\\ & &\sim\sum\limits_{(1-\delta)\lambda(t)\leq n\leq(1+\delta)\lambda(t)}P(\overrightarrow{Y}_{n}-E\overrightarrow{Y}_{n}>\overrightarrow{x}+(1+\delta-\theta)\overrightarrow{\mu}(n-\lambda(t))P(N(t) = n){}\\ &&\sim\sum\limits_{(1-\delta)\lambda(t)\leq n\leq(1+\delta)\lambda(t)}P(N(t) = n)(n\theta\overrightarrow{F}_{1, 2}(x''_{1}, x''_{2})+((n\theta)^{2}-n\theta)\overrightarrow{F}_{1}(x''_{1})\overrightarrow{F}_{2}(x''_{2})){}\\ &&\sim P((1-\delta)\lambda(t)\leq N(t)\leq(1+\delta)\lambda(t))\theta\lambda(t)\overrightarrow{F}_{1, 2}(x_{1}, x_{2}) {}\\ &&+((\theta\lambda(t))^{2}-\theta\lambda(t)) \overrightarrow{F}_{1}(x_{1})\overrightarrow{F}_{2}(x_{2}){}\\ &&\sim\theta\lambda(t)\overrightarrow{F}_{1, 2}(x_{1}, x_{2})+((\theta\lambda(t))^{2}-\theta\lambda(t))\overrightarrow{F}_{1}(x_{1})\overrightarrow{F}_{2}(x_{2}), \end{eqnarray} $

这里$ x''_{i} = x_{i}+(1+\delta-\theta)\mu_{i}(n-\lambda(t)) = x_{i}+o(x_{i}) $,下面估计$ L_{3} $,结合引理3.3,存在$ \rho>J^{+}_{F} $,我们得到

$ \begin{eqnarray*} L_{3}&\leq&\sum\limits_{n>(1+\delta)\lambda(t)}P(\overrightarrow{Y}_{n}>\overrightarrow{x})P(N(t) = n)\\ &\leq&\sum\limits_{n>(1+\delta)\lambda(t)}\sup\limits_{\overrightarrow{\gamma}\lambda(t)\leq\overrightarrow{x}\leq2n\overrightarrow{\mu}\theta}(\frac{4n^{2}\theta^{2}\mu_{1}\mu_{2}}{x_{1}x_{1}})^{\rho}P(N(t) = n)\\ &&+\sum\limits_{n>(1+\delta)\lambda(t)}\sup\limits_{\overrightarrow{x}>2n\overrightarrow{\mu}\theta}P(\overrightarrow{Y}_{n}>\overrightarrow{x})P(N(t) = n)\\ &&+\sum\limits_{n>(1+\delta)\lambda(t)}\sup\limits_{\gamma_{1}\lambda(t)\leq x_{1}\leq2n\mu_{1}\theta, x_{2}>2n\mu_{2}\theta}(\frac{2n\mu_{1}\theta}{x_{1}})^{\rho}P(Y_{2n}>x_{2})P(N(t) = n)\\ &&+\sum\limits_{n>(1+\delta)\lambda(t)}\sup\limits_{\gamma_{2}\lambda(t)\leq x_{2}\leq2n\mu_{2}\theta, x_{1}>2n\mu_{1}\theta}(\frac{2n\mu_{2}\theta}{x_{2}})^{\rho}P(Y_{1n}>x_{1})P(N(t) = n)\\ & = &M_{1}+M_{2}+M_{3}+M_{4}, \end{eqnarray*} $

对于$ M_{1} $,由(3.6)式及文献[12,引理1]得存在$ C_{1}>0 $使得下式成立

(4.15) $ \begin{eqnarray} M_{1}&\leq&\sum\limits_{n>(1+\delta)\lambda(t)}C_{1}\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})(4n^{2}\theta^{2}\mu^{1}\mu_{2})^{\rho}P(N(t) = n){}\\ & = &(4\theta^{2}\mu_{1}\mu_{2})^{\rho}C_{1}\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})EN^{2\rho}(t)I_{\{N(t)>(1+\delta)\lambda(t)\}}{}\\ & = &o(\theta^{2}\lambda^{2}(t)-\theta\lambda(t))\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2}), \end{eqnarray} $

对于$ M_{2} $,由(2.1)式, (3.2)式及定理3.4得:存在正常数$ M $和$ C_{2} $有

$ \begin{eqnarray*} M_{2}&\leq&\sum\limits_{n>(1+\delta)\lambda(t)}\sup\limits_{\overrightarrow{x}>2n\overrightarrow{\mu}\theta}P(\overrightarrow{Y}_{n}-E\overrightarrow{Y}_{n}>\frac{\overrightarrow{x}}{2})P(N(t) = n)\\ &&\sim\sum\limits_{n>(1+\delta)\lambda(t)}((n\theta)^{2}-(n\theta))\overline{F}_{1}(\frac{x_{1}}{2})\overline{F}_{2}(\frac{x_{2}}{2})+n\theta\overline{F}_{12}(\frac{x_{1}}{2}, \frac{x_{2}}{2})P(N(t) = n)\\ &\leq&\sum\limits_{n>(1+\delta)\lambda(t)}C_{2}2^{2\rho}((n\theta)^{2}-(n\theta))\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})+MC_{2}2^{2\rho}n\theta\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})\\ & = &C_{2}2^{2\rho}\theta^{2}\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})EN^{2}(t)I_{\{N(t)>(1+\delta)\lambda(t)\}} \\ && -C_{2}2^{2\rho}\theta\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2}) EN(t)I_{\{N(t)>(1+\delta)\lambda(t)\}}\\ &&+MC_{2}2^{2\rho}\theta\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})EN(t)I_{\{N(t)>(1+\delta)\lambda(t)\}}, \end{eqnarray*} $

由(3.6)式我们得到

(4.16) $ \begin{equation} M_{2} = o(\theta^{2}\lambda^{2}(t)-\theta\lambda(t))\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2}). \end{equation} $

对$ M_{3} $,通过(2.1)式及文献[12,引理1],存在常数$ C_{3}, C_{4} $得

(4.17) $ \begin{eqnarray} M_{3}&\leq&\sum\limits_{n>(1+\delta)\lambda(t)}C_{3}\overline{F}_{1}(x_{1})(2\mu_{1}n\theta)^{p}\sup\limits_{x_{2}>2n\mu_{2}\theta}P(Y_{2n}-EY_{2n}>\frac{x_{2}}{2})P(N(t) = n){}\\ &&\sim\sum\limits_{n>(1+\delta)\lambda(t)}C_{3}\overline{F}_{1}(x_{1})(2\mu_{1}n\theta)^{p}n\theta\overline{F}_{2}(\frac{x_{2}}{2})P(N(t) = n){}\\ & = &C_{4}C_{4}2^{2\rho}\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2})EN^{\rho+1}(t)I_{\{N(t)>(1+\delta)\lambda(t)\}}{}\\ & = &o(\theta^{2}\lambda^{2}(t)-\theta\lambda(t))\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2}). \end{eqnarray} $

同样地我们能得到

(4.18) $ \begin{equation} M_{2} = o(\theta^{2}\lambda^{2}(t)-\theta\lambda(t))\overline{F}_{1}(x_{1})\overline{F}_{2}(x_{2}). \end{equation} $

因此由(4.15)–(4.18)式我们得到

(4.19) $ \begin{equation} L_{3} = o\{\theta\lambda(t)\overrightarrow{F}_{1, 2}(x_{1}, x_{2})+((\theta\lambda(t))^{2}-\theta\lambda(t))\overrightarrow{F}_{1}(x_{1})\overrightarrow{F}_{2}(x_{2})\}. \end{equation} $

结合(4.13), (4.14)和(4.19)式,定理3.5得证.

本文在重尾分布族$ C $下,得到了二维风险模型的部分和与随机和的精细大偏差.