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2023 Volume 48 Issue 6
Article Contents

LIAO Juan, PENG Zuoxiang. The Asymptotic Expansion of Extreme Tail Dependence Copula Based on Archimedean Copula[J]. Journal of Southwest China Normal University(Natural Science Edition), 2023, 48(6): 49-53. doi: 10.13718/j.cnki.xsxb.2023.06.007
Citation: LIAO Juan, PENG Zuoxiang. The Asymptotic Expansion of Extreme Tail Dependence Copula Based on Archimedean Copula[J]. Journal of Southwest China Normal University(Natural Science Edition), 2023, 48(6): 49-53. doi: 10.13718/j.cnki.xsxb.2023.06.007

The Asymptotic Expansion of Extreme Tail Dependence Copula Based on Archimedean Copula

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  • Corresponding author: PENG Zuoxiang
  • Received Date: 06/09/2022
    Available Online: 20/06/2023
  • MSC: O211.4

  • Based on the extreme tail dependence copula and its convergence theorems under the condition of first-order regular variation proposed by Juri et al, this paper proposes the extreme tail dependence copula's asymptotic expansion under the condition of second-order regular variation.
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The Asymptotic Expansion of Extreme Tail Dependence Copula Based on Archimedean Copula

    Corresponding author: PENG Zuoxiang

Abstract: Based on the extreme tail dependence copula and its convergence theorems under the condition of first-order regular variation proposed by Juri et al, this paper proposes the extreme tail dependence copula's asymptotic expansion under the condition of second-order regular variation.

  • 连接函数Copula[1]可以将多个边缘分布函数结合成一个联合分布函数,其中边缘分布是随机变量的分布,因此可以借助连接函数来刻画变量之间的相依关系. 关于Copula函数的性质及其应用的更多研究,见文献[2-4].

    Copula函数族中有种类众多的Copula函数,包括阿基米德Copula、椭圆Copula、极值Copula等. 阿基米德Copula是一种性质优良的Copula函数,具有构造简单、计算容易且便于应用的优点.

    ψ:[0, 1]→[0, ∞]为连续的、严格单调递减的凸函数,满足ψ(1)=0. 且

    则称Copula

    为阿基米德Copula且函数ψC的生成元. 当ψ(0)=∞时,则称生成元ψ和其对应的阿基米德Copula是严格的[5].

    当阿基米德Copula的生成元为$ \psi(t)=\frac{1}{\alpha}\left(t^{-\alpha}-1\right)$,α>0时,由ψ(t)生成的Copula为Clayton Copula[6],其表达形式为

    利用Copula可以解决许多重要问题,其中就有极值问题. 极值理论需要估计比以往所观测到的现象更极端的事件的发生概率,因此引发了对条件随机向量相依结构的研究. 文献[7]提出了Copula C在水平u的极尾相依Copula的概念.

    对于一个Copula C并且u∈(0,1)使得C(uu)>0,令

    关于C在水平u的极尾相依Copula为

    Fu是连续的分布函数,$F_u(x)=\mathrm{P}[X \leqslant x \mid X \leqslant u, Y \leqslant u] $,并且由(5)式可得${C_u}(x, y) = P[X \le \left. {F_u^{ - 1}(x), Y \le F_u^{ - 1}(y)\mid X \le u, Y \le u} \right] $,即对于较小的u值,Cu用Copula C描述了两个随机变量尾部的条件依赖结构,关于极尾相依Copula的研究见文献[8]和文献[9].

    文献[8]讨论了当阿基米德Copula的生成元$\psi \in R V_{-\alpha}\left(0^{+}\right) $α>0,即生成元ψ在一阶正规变换的条件下,$ u \rightarrow 0^{+}$时,极尾相依Copula Cu收敛到Clayton Copula.

    文献[10-13]讨论了正规变换函数和二阶正规变换函数,用以研究某个估计量的收敛速度,本文讨论生成元ψ在原点处满足二阶正规变换的条件下,即$ \psi \in 2 R V_{-\alpha, \beta}\left(0^{+}\right)$,其中

    辅助函数A(t)是定号的(见文献[14]),得到极尾相依Copula Cu的渐近展开.

1.   渐近展开式
  • 定理1  假设Copula C是严格的阿基米德Copula,其生成元ψ可微并满足(6)式,即$ \psi \in 2R{V_{ - \alpha , \beta }}\left( {{0^ + }} \right)$,其中0 < α < ∞,β>0,辅助函数为A(t),则对任意0≤x,y≤1,$u \rightarrow 0^{+} $时,有

    其中$ B(u) \sim \frac{1}{\alpha \beta}\left(A \circ \psi^{-1}\left(2\left(x^{-\alpha}+y^{-\alpha}-1\right) \psi(u)\right)-A \circ \psi^{-1}(2 \psi(u))\right)$.

      对于生成元为ψ的阿基米德Copula C,由文献[7]命题3.2知

    且根据(2),(5)和(8)式,有

    k=ψ-1(2ψ(u)),那么

    $\psi \in 2 R V_{-\alpha, \beta}\left(0^{+}\right), 0<\alpha<\infty, \beta>0 $,辅助函数为A(t)可得

    由文献[14]命题2.7(i)得,逆函数$ \psi^{-1} \in 2 R V_{-\frac{1}{\alpha}, -\frac{\beta}{\alpha}}$,且辅助函数为$ B(t) = - {\alpha ^{ - 2}}A \circ {\psi ^{ - 1}}(t)$,即ψ-1(t)有下列展开式

    成立,其中c≠0(见文献[15]引理2.2). 当$u \rightarrow 0^{+} $时,$ k \rightarrow 0^{+}, \psi(k) \rightarrow \infty$,由(11)式知,

    由(10)式得,当$k \rightarrow 0^{+} \text {时, } L_{x, y, \alpha}=\frac{\psi(k x)}{\psi(k)}+\frac{\psi(k y)}{\psi(k)}-1=\left(x^{-\alpha}+y^{-\alpha}-1\right)(1+o(A(k))) $,则由(11)式得

    将(12)和(13)式代入(9)式得,当$ u \rightarrow 0^{+}, k=\psi^{-1}(2 \psi(u)) \rightarrow 0^{+}$时,

    其中,令

    则有

    $ u \rightarrow 0^{+}$

    定理证毕.

2.   实例
  • 例1  给定阿基米德Copula C的生成元$ \psi(t)=\left(t^{-\frac{1}{\theta}}-1\right)^\theta, \alpha=1, \beta=\frac{1}{\theta}$,可以得到$ \psi \in 2 R V_{-1, \frac{1}{\theta}}\left(0^{+}\right)$,辅助函数$A(t)=\frac{t^{\frac{1}{\theta}}}{t^{\frac{1}{\theta}}-1}, \psi^{-1}(t)=\left(t^{\frac{1}{\theta}}+1\right)^{-\theta}A(t)=\frac{t^{\frac{1}{\theta}}}{t^{\frac{1}{\theta}}-1}, \psi^{-1}(t)=\left(t^{\frac{1}{\theta}}+1\right)^{-\theta} $.

    ψψ-1的具体表达式代入(14)式,令

    则有

    $ u \rightarrow 0^{+}$时,由泰勒展开式得

    并且

    $ B(u)=\theta\left(\frac{\frac{u^{\frac{1}{\theta}}}{2^{\frac{1}{\theta}}\left(1-u^{\frac{1}{\theta}}\right)}-\frac{u^{\frac{1}{\theta}}}{D_u(x, y)}}{1+\frac{u^{\frac{1}{\theta}}}{D_u(x, y)}}\right) \text {, 当 } u \rightarrow 0^{+} \text {时 }, D_u(x, y) \rightarrow 2^{\frac{1}{\theta}}\left(x^{-1}+y^{-1}-1\right)^{\frac{1}{\theta}}$,则有

    与定理的结论一致.

    例2  给定阿基米德Copula C的生成元$\psi(t)=\left(\frac{1}{t}-1\right)^\theta, \alpha=\theta, \beta=1 $,可以得到$\psi \in 2 R V_{-\theta, 1}\left(0^{+}\right) $,辅助函数$A(t)=\frac{\theta t}{t-1}, \psi^{-1}(t)=\left(t^{\frac{1}{\theta}}+1\right)^{-1} $.

    ψψ-1的具体表达式代入(15)式得

    $u \rightarrow 0^{+} $时,由泰勒展开式得,

    并且,

    $ u \rightarrow 0^{+} \text {时, } D(u) \rightarrow 2^{\frac{1}{\theta}}\left(x^{-\theta}+y^{-\theta}-1\right)^{\frac{1}{\theta}}$,则有

    与定理的结论一致.

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