G-布朗运动驱动的随机金融风险系统模型的渐近行为

Asymptotic Behavior of Stochastic Financial Risk System Models Driven by G-Brownian Motion

摘要: 本文主要研究了G-布朗运动驱动的金融风险系统模型的渐近行为。利用李雅普诺夫函数和Gronwall不等式，证明了G-布朗运动驱动的随机金融风险系统模型解的存在唯一性。运用G-伊藤公式和G期望不等式等相关知识，研究了G-布朗运动驱动的随机金融风险系统模型解在不同平衡点的有界性与全局指数吸引集。

Abstract: This paper mainly studies the asymptotic behavior of the financial risk system model driven by G-Brownian motion. Using Lyapunov function and Gronwall inequality, the existence and uniqueness of the solution of the stochastic financial risk system model driven by G-Brownian motion are proved. Using G-Itô formula and G-expectation inequality, the boundedness and global exponential attraction set of the solution of the stochastic financial risk system model driven by G-Brownian motion at different equilibrium points are discussed.

1. 引言

人们试图使用混沌理论的知识和方法来探究金融模型和经济相关问题，是因为诸如股价、金融风险、量化数据等从数据表面看是极不规律的，而混沌正是系统极不规律的一种表现形式 [1] [2] [3]。2016年，徐玉华等人提出金融系统风险的演化分为三个阶段：第一阶段，金融系统受到来自外部或内部的冲击；第二阶段，金融系统间的传染效应，使得系统风险再一次改变，从而破坏系统稳定性；第三阶段，面对系统风险，金融机构、监管部门和货币政策作出调整和管控。因此，他们提出了金融风险系统模型 [4]，

$\{\begin{array}{l}\stackrel{\dot{}}{x}=yz-\left(a+1\right)x,\hfill \\ \stackrel{\dot{}}{y}=xz-\left(b+1\right)y,\hfill \\ \stackrel{\dot{}}{z}=\left(c-1\right)z-xy,\hfill \end{array}$ (1.1)

其中，x表示第一阶段来自金融系统外部或内部对系统冲击的总风险值，y表示第二阶段经过金融系统传染效应后的总风险值，z表示第三阶段金融系统受到调控后的总风险值。a，b表示风险程度，c表示控制强度， $a,b\in R^{+}$， $c>1$。三个阶段的风险相互作用，相互影响。金融风险系统模型(1.1)存在一个无风险均衡点 $E_0=\left(0,0,0\right)$ 和四个有风险均衡点。

非线性期望理论发展非常迅速，与概率理论所推导出的线性期望理论不同的是，线性期望理论主要擅长处理那些其相应的概率模型能够通过数理统计方法和数据分析予以确定的情形，而真实世界与人有关的各项数据并不能确定其分布情况 [5]。将非线性期望理论运用在金融系统具有现实意义和应用基础 [6]，如次线性期望的平移不变性，意味着为了降低风险，可以在风险位置增加现金流；次线性期望的次可加性，意味着投资组合可以在很大程度上降低风险，这是符合人们认知的。且由G-布朗运动驱动的随机动力系统的动力学性质研究甚少，包括随机吸引子、随机有界性、随机分岔和随机混沌等。

因此，本文不再局限于经典的标准布朗运动，反之引入由次线性期望的体系中建立的非线性布朗运动，即G-布朗运动，从而进一步将金融风险系统模型(1.1)转化成如下随机金融风险系统模型：

$\{\begin{array}{l}\text{d}x\left(t\right)=\left(y\left(t\right)z\left(t\right)-\left(a+1\right)x\left(t\right)\right)\text{d}t+c_{1}x\left(t\right)\text{d}{\langle B_1,B_1\rangle }_t+{\sigma }_{1}x\left(t\right)\text{d}{B}_1\left(t\right),\hfill \\ \text{d}y\left(t\right)=\left(x\left(t\right)z\left(t\right)-\left(b+1\right)y\left(t\right)\right)\text{d}t+c_{2}y\left(t\right)\text{d}{\langle B_2,B_2\rangle }_t+{\sigma }_{2}y\left(t\right)\text{d}{B}_2\left(t\right),\hfill \\ \text{d}z\left(t\right)=\left(\left(c-1\right)z\left(t\right)-x\left(t\right)y\left(t\right)\right)\text{d}t+c_{3}z\left(t\right)\text{d}{\langle B_3,B_3\rangle }_t+{\sigma }_{3}z\left(t\right)\text{d}{B}_3\left(t\right),\hfill \end{array}$ (1.2)

其中， $B_i$ 是两两相互独立的G-布朗运动，它们都满足 $\frac{B_i\left(t\right)}{\sqrt{t}}\stackrel{d}{=}N\left(0,\left[{\underset{\_}{\sigma }}^2,{\bar{\sigma }}^2\right]\right),i=1,2,3$， $\langle B_i,B_i\rangle$ 是G-布朗运动二次变差过程。 ${\sigma }_1,{\sigma }_2,{\sigma }_3$ 是G-布朗运动二次变差过程对应的扰动强度， $c_1,c_2,c_3$ 是G-布朗运动对应的扰动强度。

本文的组织结构如下：在第2节中，给出一些准备知识，供后续章节使用。在第3节中，利用李雅普诺夫函数，证明了G-布朗运动驱动的随机金融风险系统模型(1.2)解的存在唯一性。在第4节中，运用G-伊藤公式和G期望不等式等相关知识，研究了G-布朗运动驱动的随机金融风险系统模型(1.2)解在不同平衡点的有界性与全局指数吸引集。

2. 准备知识

本节，我们介绍一些关于次线性期望和G-布朗运动的基本概念和性质 [7] - [18]。

定义2.1 [7] 一个非线性期望 $\hat{\mathbb{E}}$ 是定义在 $\mathcal{H}$ 上的实值函数， $\mathcal{H}$ 是定义在 $\Omega$ 上的实质函数所组成的一个线性空间， $\hat{\mathbb{E}}:\mathcal{H}\to ℝ$，并满足以下四个条件：

1) 单调性：对于所有的 $X,Y\in \mathcal{H}$，且满足 $X\left(\omega \right)\ge Y\left(\omega \right)$，那么 $\hat{\mathbb{E}}\left(X\right)\ge \hat{\mathbb{E}}\left(Y\right)$ ；

2) 保常数性：对于常数 $c\in ℝ$，则有 $\hat{\mathbb{E}}\left(X+c\right)=\hat{\mathbb{E}}\left(X\right)+c$ ；

3) 次可加性：对任意的 $X,Y\in \mathcal{H}$，有 $\hat{\mathbb{E}}\left(X+Y\right)\le \hat{\mathbb{E}}\left(X\right)+\hat{\mathbb{E}}\left(Y\right)$ ；

4) 正齐次性：对于任意的 $X\in \mathcal{H}$ 和数 $\lambda >0$，有 $\hat{\mathbb{E}}\left(\lambda X\right)=\lambda \hat{\mathbb{E}}\left(X\right)$，

称二元组 $\left(\Omega ,\mathcal{H},\hat{\mathbb{E}}\right)$ 为次线性期望空间。

定义2.2 [7] (G-布朗运动的定义)定义在一个次线性期望 $\left(\Omega ,\mathcal{H},\hat{\mathbb{E}}\right)$ 空间中的随机过程 $B_t\left(\omega \right)\left(t\ge 0\right)$ 为关于 $\hat{\mathbb{E}}$ 的布朗运动，如果对于每一个 $n\in \mathbb{N}$ 和 $0\le t_1,\cdots ,t_n\le \infty$，都有

1) $B_0\left(\omega \right)=0$ ；

2) $B_t$ 的增量平稳且独立，即对于每一个 $t,s\ge 0$， $B_{t+s}-B_t$ 服从G分布；

3) 对于所有的 $n\in \mathbb{N}$ 和 $t_1,\cdots ,t_n\in \left[0,t\right]$， $B_{t+s}-B_t$ 独立于 $B_{t_1},B_{t_2},\cdots ,B_{t_n}$。

定理2.3 [7] 在 $\left(\Omega ,L_G^P\left(\Omega \right),\hat{\mathbb{E}}\right)$ 中的一个完备的伊藤过程为

$X_t^{\mu }=X_0^{\mu }+{\displaystyle {\int }_0^t{\alpha }_s^{\mu }}\text{d}s+{\displaystyle {\int }_0^t{\eta }_s^{\mu }}\text{d}{\langle B\rangle }_s+{\displaystyle {\int }_0^t{\beta }_s^{\mu }}\text{d}{B}_s$

设 ${\alpha }^{\mu },{\eta }^{\mu }\in M_G^1\left(0,T\right)$ 和 ${\beta }^{\mu }\in M_G^2\left(0,T\right)$， $\mu =1,\cdots ,n$，则对于每一个 $t\in \left[0,T\right]$ 和函数 $\Phi \in C^{1,2}\left(\left[0,T\right]\times {ℝ}^n\right)$，有

$\begin{array}{c}\Phi \left(t,X_t\right)-\Phi \left(s,X_s\right)={\displaystyle \underset{\mu =1}{\overset{n}{\sum }}{\displaystyle {\int }_s^t{\partial }_{x^{\mu }}\Phi \left(u,X_u\right){\beta }_u^{\mu }\text{d}{B}_u}+{\displaystyle {\int }_s^t\left[{\partial }_u\Phi \left(u,X_u\right)+{\partial }_{x^{\mu }}\Phi \left(u,X_u\right){\alpha }_u^{\mu }\right]\text{d}u}}\\ +{\displaystyle {\int }_s^t\left\{{\displaystyle \underset{\mu =1}{\overset{n}{\sum }}{\partial }_{x^{\mu }}\Phi \left(u,X_u\right){\eta }_u^{\mu }}+\frac{1}{2}{\displaystyle \underset{\mu =1}{\overset{n}{\sum }}{\partial }_{x^{\mu }{x}^{\mu }}^2\Phi \left(u,X_u\right){\beta }_u^{\mu }{\beta }_u^{\mu }}\right\}\text{d}{\langle B\rangle }_u}.\end{array}$

3. 全局解的存在唯一性

在本小节文，主要证了G-布朗运动驱动的明随机金融风险系统模型(1.2)解的存在唯一性。

定理3.1对任意给定的初值 $\left(x\left(0\right),y\left(0\right),z\left(0\right)\right)\in {ℝ}^3$，则G-布朗运动驱动的随机金融风险系统模型(1.2)存在唯一解 $\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)$。

证明：由于G-布朗运动驱动的随机金融风险系统模型(1.2)的系数是局部Lipschitz连续的，因此对给定的初值 $\left(x\left(0\right),y\left(0\right),z\left(0\right)\right)\in {ℝ}^3$，存在一个局部解 $\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)$， $t\in \left[0,{\tau }_e\right]$，其中 ${\tau }_e$ 为爆破时刻。为了证明解是全局的，我们只需要证明 ${\tau }_e=\infty$，q.s.。

令 $k_0\ge 1$ 充分大，使得 $x_0,y_0,z_0\in \left[-k_0,k_0\right]$，对于每个整数 $k\ge k_0$，定义停时：

${\tau }_k=\inf \left\{t\in \left[0,{\tau }_e\right);\min \left\{x\left(t\right),y\left(t\right),z\left(t\right)\right\}\le -k\&\max \left\{x\left(t\right),y\left(t\right),z\left(t\right)\right\}\ge k\right\}.$

显然，随着 $k\to \infty$， ${\tau }_k$ 不断递增。令 ${\tau }_{\infty }=\underset{k\to \infty }{\lim }{\tau }_k$，有 ${\tau }_{\infty }\le {\tau }_e$，q.s.如果能证明 ${\tau }_e=\infty$，q.s.，则有 ${\tau }_e=\infty$，q.s.，并且对所有的 $t\ge 0$， $\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)\in {ℝ}^3$，q.s.。

为了证明该结论，我们先定义一个 $C^3$ 上的函数 $V:R^3\to R$，如下：

$V\left(x,y,z\right)=m{x}^2+n{y}^2+\left(m+n\right)z^2,$

$m,n,l$ 为常数。令 $k>k_0,T>0$ 为任意数，当 $0\le t\le {\tau }_k\wedge T$，对函数V运用G-伊藤公式，得到

$\begin{array}{l}V\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)\\ =V\left(x\left(0\right),y\left(0\right),z\left(0\right)\right)+{\displaystyle {\int }_0^{T\wedge {\tau }_k}2}mx\left(t\right)\cdot \left(y\left(t\right)z\left(t\right)-\left(a+1\right)x\left(t\right)\right)\\ +2ny\left(t\right)\cdot \left(x\left(t\right)z\left(t\right)-\left(b+1\right)y\left(t\right)\right)+2\left(m+n\right)z\left(t\right)\cdot \left(\left(c-1\right)z\left(t\right)-x\left(t\right)y\left(t\right)\right)\text{d}t\\ +{\displaystyle {\int }_0^{T\wedge {\tau }_k}2m{\sigma }_1\cdot x^2\left(t\right)\text{d}{B}_1\left(t\right)}+2n{\sigma }_2\cdot y^2\left(t\right)\text{d}{B}_2\left(t\right)+2\left(m+n\right){\sigma }_3\cdot z^2\left(t\right)\text{d}{B}_3\left(t\right)\\ +{\displaystyle {\int }_0^{T\wedge {\tau }_k}2m{c}_1{x}^2\left(t\right)}+m{\sigma }_1^2{x}^2\left(t\right)\text{d}{\langle B_1,B_1\rangle }_t+2n{c}_2{y}^2\left(t\right)+n{\sigma }_2^2{y}^2\left(t\right)\text{d}{\langle B_2,B_2\rangle }_t\\ +2\left(m+n\right)c_3{z}^2\left(t\right)+\left(m+n\right){\sigma }_3^2{z}^2\left(t\right)\text{d}{\langle B_3,B_3\rangle }_t.\end{array}$ (3.1)

因此，上述等式(3.1)可以写成：

$\begin{array}{l}V\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)-V\left(x\left(0\right),y\left(0\right),z\left(0\right)\right)\\ ={\displaystyle {\int }_0^{T\wedge {\tau }_k}2}mx\left(t\right)\cdot \left(y\left(t\right)z\left(t\right)-\left(a+1\right)x\left(t\right)\right)+2ny\left(t\right)\cdot \left(x\left(t\right)z\left(t\right)-\left(b+1\right)y\left(t\right)\right)\\ +2\left(m+n\right)z\left(t\right)\cdot \left(\left(c-1\right)z\left(t\right)-x\left(t\right)y\left(t\right)\right)\text{d}t+{\displaystyle {\int }_0^{T\wedge {\tau }_k}2}m{\sigma }_1\cdot x^2\left(t\right)\text{d}{B}_1\left(t\right)\\ +2n{\sigma }_2\cdot y^2\left(t\right)\text{d}{B}_2\left(t\right)+2\left(m+n\right){\sigma }_3\cdot z^2\left(t\right)\text{d}{B}_3\left(t\right)+{\displaystyle {\int }_0^{T\wedge {\tau }_k}{\bar{\sigma }}_1^2}\left(2m{c}_1{x}^2\left(t\right)+m{\sigma }_1^2{x}^2\left(t\right)\right)\text{d}t\\ +{\bar{\sigma }}_2^2\left(2n{c}_2{y}^2\left(t\right)+n{\sigma }_2^2{y}^2\left(t\right)\right)\text{d}t+{\bar{\sigma }}_3^2\left(2\left(m+n\right)c_3{z}^2\left(t\right)+\left(m+n\right){\sigma }_3^2{z}^2\left(t\right)\right)\text{d}t\end{array}$

$\begin{array}{l}={\displaystyle {\int }_0^{T\wedge {\tau }_k}\left(2m{c}_1{\bar{\sigma }}_1^2+m{\sigma }_1^2{\bar{\sigma }}_1^2-2m\left(a+1\right)\right)x^2\left(t\right)1_{0,{\tau }_k}\text{d}t}\\ +\left(2n{c}_2{\bar{\sigma }}_2^2+n{\sigma }_2^2{\bar{\sigma }}_2^2-2n\left(b+1\right)\right)y^2\left(t\right)1_{0,{\tau }_k}\text{d}t\\ +\left(2\left(m+n\right)c_3{\bar{\sigma }}_3^2+\left(m+n\right){\sigma }_3^2{\bar{\sigma }}_3^2+2\left(m+n\right)\left(c-1\right)\right)z^2\left(t\right)1_{0,{\tau }_k}\text{d}t\\ +{\displaystyle {\int }_0^{T\wedge {\tau }_k}2}m{\sigma }_1\cdot x^2\left(t\right)1_{0,{\tau }_k}\text{d}{B}_1\left(t\right)+2n{\sigma }_2\cdot y^2\left(t\right)1_{0,{\tau }_k}\text{d}{B}_2\left(t\right)\\ +2\left(m+n\right){\sigma }_3\cdot z^2\left(t\right)1_{0,{\tau }_k}\text{d}{B}_3\left(t\right).\end{array}$ (3.2)

对式子(3.2)，两边取G-期望后，有

$\hat{\mathbb{E}}{\displaystyle {\int }_0^{T\wedge {\tau }_k}2}m{\sigma }_1\cdot x^2\left(t\right)1_{0,{\tau }_k}\text{d}{B}_1\left(t\right)=0,$

$\hat{\mathbb{E}}{\displaystyle {\int }_0^{T\wedge {\tau }_k}2}n{\sigma }_2\cdot y^2\left(t\right)1_{0,{\tau }_k}\text{d}{B}_2\left(t\right)=0,$

$\hat{\mathbb{E}}{\displaystyle {\int }_0^{T\wedge {\tau }_k}2}l{\sigma }_3\cdot z^2\left(t\right)1_{0,{\tau }_k}\text{d}{B}_3\left(t\right)=0.$

因此，可以得到

$\begin{array}{c}\hat{\mathbb{E}}V\left(X_{T\wedge {\tau }_k},Y_{T\wedge {\tau }_k},Z_{T\wedge {\tau }_k}\right)\le V\left(x_0,y_0,z_0\right)+\hat{\mathbb{E}}{\displaystyle {\int }_0^{T\wedge {\tau }_k}C}V\left(t,x,y,z\right)\text{d}t\\ \le V\left(x_0,y_0,z_0\right)+\hat{\mathbb{E}}{\displaystyle {\int }_0^{T\wedge {\tau }_k}C}V\left(X_{T\wedge {\tau }_k},Y_{T\wedge {\tau }_k},Z_{T\wedge {\tau }_k}\right)\text{d}t.\end{array}$

其中， $C=\max \{2m{c}_1{\bar{\sigma }}_1^2+m{\sigma }_1^2{\bar{\sigma }}_1^2-2m\left(a+1\right)$， $2n{c}_2{\bar{\sigma }}_2^2+n{\sigma }_2^2{\bar{\sigma }}_2^2-2n\left(b+1\right)$， $2\left(m+n\right)c_3{\bar{\sigma }}_3^2+\left(m+n\right){\sigma }_3^2{\bar{\sigma }}_3^2+2\left(m+n\right)\left(c-1\right)\}$，因此，由Gronwall不等式可以得到：

$\hat{\mathbb{E}}V\left(X_{T\wedge {\tau }_k},Y_{T\wedge {\tau }_k},Z_{T\wedge {\tau }_k}\right)\le V\left(x_0,y_0,z_0\right){\text{e}}^{CT}.$ (3.3)

对于每一个 $\omega \in {\tau }_i\le T$，至少存在 $X_{0,{\tau }_k}$ 、 $Y_{0,{\tau }_k}$ 、 $Z_{0,{\tau }_k}$ 中的一项等于 $-k$ 或k。记 $\hat{c}=\min \left\{m,n,m+n\right\}$，于是

$V\left(X_{T\wedge {\tau }_k},Y_{T\wedge {\tau }_k},Z_{T\wedge {\tau }_k}\right)\le \hat{c}{k}^2.$ (3.4)

因此，可以得到：

$\begin{array}{c}V\left(x_0,y_0,z_0\right){\text{e}}^{CT}\ge \hat{\mathbb{E}}\left[1_{{\tau }_k\le T}V\left(X_{T\wedge {\tau }_k},Y_{T\wedge {\tau }_k},Z_{T\wedge {\tau }_k}\right)\right]\\ \ge \bar{C}\left(\left\{\omega :{\tau }_k\le T\right\}\right)V\left(X_{T\wedge {\tau }_k},Y_{T\wedge {\tau }_k},Z_{T\wedge {\tau }_k}\right)\\ \ge \bar{C}\left(\left\{\omega :{\tau }_k\le T\right\}\right)\hat{c}{k}^2,\end{array}$ (3.5)

其中 $1_{{\tau }_k\le T}$ 是 ${\tau }_k\le T$ 的示性函数。令 $k\to \infty$ 有

$\underset{k\to \infty }{\lim }\bar{C}\left\{\omega :{\tau }_k\le T\right\}=0.$

从而，有

$\bar{C}\left\{{\tau }_{\infty }\le T\right\}=0.$

因为 $T>0$ 是任意的，我们可以得出：

$\bar{C}\left\{{\tau }_{\infty }=\infty \right\}=1.\text{q}\text{.s}.$

则在 ${ℝ}^3$ 上，随机金融风险系统模型(1.2)存在全局唯一解。证明完毕。

4. 随机吸引子与有界性

本小节主要研究G-布朗运动驱动的随机金融风险系统模型(1.2)的渐近行为，包括随机吸引子和有界性。

定理4.1如果 $2\left(a+1\right)>{\bar{\sigma }}_1^2\left(2c_1+{\sigma }_1^2\right)$， $2\left(b+1\right)>{\bar{\sigma }}_2^2\left(2c_2+{\sigma }_2^2\right)$， $K_1>0$ 成立。则对给定初解 $E_0=\left(0,0,0\right)\in {ℝ}^3$，G-布朗运动驱动的随机金融风险系统模型(1.2)的解在无风险均衡点有以下性质：

$\begin{array}{l}\underset{t\to \infty }{\lim \sup }\frac{1}{t}\hat{\mathbb{E}}{\displaystyle {\int }_0^t{x}^2}\left(s\right)+y^2\left(s\right)+{\left(z\left(s\right)+\frac{1}{K_1}\right)}^2\text{d}t\\ \le \frac{1}{K_2}\left(a+b+2+\left(\frac{{\sigma }_1^2}{2}-c_1\right){\bar{\sigma }}_1^2+\left(\frac{{\sigma }_2^2}{2}-c_2\right){\bar{\sigma }}_2^2+\frac{1}{K_1}\right).\end{array}$

其中，

$K_1=4\left(1-c\right)-{\bar{\sigma }}_3^2\left(4c_3+2{\sigma }_3^2\right),$

$K_2=\min \left\{2\left(a+1\right)-{\bar{\sigma }}_1^2\left(2c_1+{\sigma }_1^2\right),2\left(b+1\right)-{\bar{\sigma }}_2^2\left(2c_2+{\sigma }_2^2\right),K_1\right\}.$

证明：构造以下Lyapunov函数：

$V\left(x,y,z\right)=x^2-\ln x+y^2-\ln y+2z^2$

根据G-伊藤公式，可以得到：

$\begin{array}{c}\text{d}V=\left(\left(2x-\frac{1}{x}\right)\frac{\text{d}x}{\text{d}t}+\left(2y-\frac{1}{y}\right)\frac{\text{d}y}{\text{d}t}+4z\frac{\text{d}z}{\text{d}t}\right)\text{d}t\\ +\left(\left(2x-\frac{1}{x}\right)\frac{\text{d}x}{\text{d}{B}_t}+\left(2y-\frac{1}{y}\right)\frac{\text{d}y}{\text{d}{B}_t}+4z\frac{\text{d}z}{\text{d}{B}_t}\right)\text{d}{B}_t\\ +\left(\left(2x-\frac{1}{x}\right)\frac{\text{d}x}{\text{d}{\langle B_1,B_1\rangle }_t}+\left(2y-\frac{1}{y}\right)\frac{\text{d}y}{\text{d}{\langle B_2,B_2\rangle }_t}+4z\frac{\text{d}z}{\text{d}{\langle B_3,B_3\rangle }_t}\right)\text{d}{\langle B,B\rangle }_t\\ +\left(\left(2+\frac{1}{x^2}\right)\frac{{\text{d}}^{2}x}{2\text{d}{B}_t^2}+\left(2+\frac{1}{y^2}\right)\frac{{\text{d}}^{2}y}{2\text{d}{B}_t^2}+4\frac{{\text{d}}^{2}z}{2\text{d}{B}_t^2}\right)\text{d}{\langle B,B\rangle }_t.\end{array}$ (4.1)

因此，获得

$\begin{array}{l}V\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)\\ =V\left(x\left(0\right),y\left(0\right),z\left(0\right)\right)+{\displaystyle {\int }_0^t\left(2x\left(s\right)-\frac{1}{x\left(s\right)}\right)}\left(y\left(s\right)z\left(s\right)-\left(a+1\right)x\left(s\right)\right)\\ +\left(2y\left(s\right)-\frac{1}{y\left(s\right)}\right)\left(x\left(s\right)z\left(s\right)-\left(b+1\right)y\left(s\right)\right)+4z\left(s\right)\left(\left(c-1\right)z\left(s\right)-x\left(s\right)y\left(s\right)\right)\text{d}s\\ +{\displaystyle {\int }_0^t\left(2x\left(s\right)-\frac{1}{x\left(s\right)}\right)c_{1}x\left(s\right)}+\frac{1}{2}\left(2+\frac{1}{x{\left(s\right)}^2}\right){\sigma }_1^2{x}^2\left(s\right)\text{d}{\langle B_1,B_1\rangle }_s\\ +\left(2y\left(s\right)-\frac{1}{y\left(s\right)}\right)c_{2}y\left(s\right)+\frac{1}{2}\left(2+\frac{1}{y{\left(s\right)}^2}\right){\sigma }_2^2{y}^2\left(s\right)\text{d}{\langle B_2,B_2\rangle }_s\\ +4z\left(s\right)c_{3}z\left(s\right)+4\frac{1}{2}{\sigma }_3^2{z}^2\left(s\right)\text{d}{\langle B_3,B_3\rangle }_s.\end{array}$ (4.2)

对于等式(4.2)，两边取G-期望后，得到：

$\begin{array}{c}0\le \hat{\mathbb{E}}V\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)\\ \le \hat{\mathbb{E}}V\left(x\left(0\right),y\left(0\right),z\left(0\right)\right)+\hat{\mathbb{E}}{\displaystyle {\int }_0^t-}2\left(a+1\right)x^2\left(s\right)-\frac{y\left(t\right)z\left(t\right)}{x\left(t\right)}+a+1-2\left(b+1\right)y^2\left(s\right)\\ -\frac{z\left(t\right)z\left(t\right)}{y\left(t\right)}+b+1-4\left(1-c\right)z^2\left(s\right)\text{d}t+\hat{\mathbb{E}}{\displaystyle {\int }_0^{t}2c_1{x}^2\left(s\right)}-c_1+{\sigma }_1^2{x}^2\left(s\right)+\frac{{\sigma }_1^2}{2}\text{d}{\langle B_1,B_1\rangle }_s\\ +\hat{\mathbb{E}}{\displaystyle {\int }_0^{t}2c_2{y}^2\left(s\right)}-c_2+{\sigma }_2^2{y}^2\left(s\right)+\frac{{\sigma }_2^2}{2}\text{d}{\langle B_2,B_2\rangle }_s+\hat{\mathbb{E}}{\displaystyle {\int }_0^{t}4c_3{z}^2\left(s\right)}+2{\sigma }_3^2{z}^2\left(s\right)\text{d}{\langle B_3,B_3\rangle }_s,\end{array}$ (4.3)

对不等式(4.3)，通过基本不等式，可以得到：

$\begin{array}{c}0\le \hat{\mathbb{E}}V\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)-\hat{\mathbb{E}}V\left(x\left(0\right),y\left(0\right),z\left(0\right)\right)\\ \le \hat{\mathbb{E}}{\displaystyle {\int }_0^t-}2\left(a+1\right)x^2\left(s\right)+{\bar{\sigma }}_1^2{x}^2\left(s\right)\left(2c_1+{\sigma }_1^2\right)-2\left(b+1\right)y^2\left(s\right)+{\bar{\sigma }}_2^2{y}^2\left(s\right)\left(2c_2+{\sigma }_2^2\right)\\ -4\left(1-c\right)z^2\left(s\right)+{\bar{\sigma }}_3^2{z}^2\left(s\right)\left(4c_3+2{\sigma }_3^2\right)-2z\left(s\right)\text{d}s\\ +\left(a+b+2\right)t+\left(\frac{{\sigma }_1^2}{2}-c_1\right){\bar{\sigma }}_1^{2}t+\left(\frac{{\sigma }_2^2}{2}-c_2\right){\bar{\sigma }}_2^{2}t.\end{array}$ (4.4)

从而，有

$\begin{array}{l}\hat{\mathbb{E}}{\displaystyle {\int }_0^t\left(2\left(a+1\right)-{\bar{\sigma }}_1^2\left(2c_1+{\sigma }_1^2\right)\right)x^2\left(s\right)}+\left(2\left(b+1\right)-{\bar{\sigma }}_2^2\left(2c_2+{\sigma }_2^2\right)\right)y^2\left(s\right)+K_1{\left(z\left(s\right)+\frac{1}{K_1}\right)}^2\text{d}t\\ \le \hat{\mathbb{E}}V\left(x\left(0\right),y\left(0\right),z\left(0\right)\right)+\left(a+b+2\right)t+\left(\frac{{\sigma }_1^2}{2}-c_1\right){\bar{\sigma }}_1^{2}t+\left(\frac{{\sigma }_2^2}{2}-c_2\right){\bar{\sigma }}_2^{2}t+\frac{t}{K_1},\end{array}$

其中， $K_1=4\left(1-c\right)-{\bar{\sigma }}_3^2\left(4c_3+2{\sigma }_3^2\right)$。因此：

$\begin{array}{l}\underset{t\to \infty }{\lim \sup }\frac{1}{t}\hat{\mathbb{E}}{\displaystyle {\int }_0^t\left(2\left(a+1\right)-{\bar{\sigma }}_1^2\left(2c_1+{\sigma }_1^2\right)\right)x^2\left(s\right)}+\left(2\left(b+1\right)-{\bar{\sigma }}_2^2\left(2c_2+{\sigma }_2^2\right)\right)y^2\left(s\right)+K_1{\left(z\left(s\right)+\frac{1}{K_1}\right)}^2\text{d}t\\ \le a+b+2+\left(\frac{{\sigma }_1^2}{2}-c_1\right){\bar{\sigma }}_1^2+\left(\frac{{\sigma }_2^2}{2}-c_2\right){\bar{\sigma }}_2^2+\frac{1}{K_1}.\end{array}$

令

$K_2=\min \left\{2\left(a+1\right)-{\bar{\sigma }}_1^2\left(2c_1+{\sigma }_1^2\right),2\left(b+1\right)-{\bar{\sigma }}_2^2\left(2c_2+{\sigma }_2^2\right),K_1\right\},$

那么有，

$\begin{array}{l}\underset{t\to \infty }{\lim \sup }\frac{1}{t}\hat{\mathbb{E}}{\displaystyle {\int }_0^t{x}^2}\left(s\right)+y^2\left(s\right)+{\left(z\left(s\right)+\frac{1}{K_1}\right)}^2\text{d}t\\ \le \frac{1}{K_2}\left(a+b+2+\left(\frac{{\sigma }_1^2}{2}-c_1\right){\bar{\sigma }}_1^2+\left(\frac{{\sigma }_2^2}{2}-c_2\right){\bar{\sigma }}_2^2+\frac{1}{K_1}\right).\end{array}$

定理4.2假设 $e_1,e_2,e_3>0$ 成立。则对给定初解 $E_1=\left(x^{*},y^{*},z^{*}\right)\in {ℝ}^3$，G-布朗运动驱动的随机金融风险系统模型(1.2)的解在有风险均衡点有以下性质：

$\begin{array}{l}\underset{t\to \infty }{\lim \sup }\frac{1}{t}\hat{\mathbb{E}}{\displaystyle {\int }_0^t{\left(x\left(s\right)-\left(1+g_1\right)x^{*}\right)}^2}+{\left(y\left(s\right)-\left(1+g_2\right)y^{*}\right)}^2+{\left(z\left(s\right)-\left(1+g_3\right)z^{*}\right)}^2\text{d}t\\ \le \frac{1}{K_3}\left(\frac{1}{2}{\sigma }_1^2{\bar{\sigma }}_1^2{x}^{*}{}^2+\frac{1}{2}{\sigma }_2^2{\bar{\sigma }}_2^2{y}^{*}{}^2+{\sigma }_3^2{\bar{\sigma }}_3^2{z}^{*}+\frac{f_1^2}{4e_1}+\frac{f_2^2}{4e_2}+\frac{f_3^2}{4e_3}\right),\end{array}$

其中，

$e_1=a-z^{*}{}^2-c_1{\bar{\sigma }}_1^2-\frac{1}{2}{\sigma }_1^2{\bar{\sigma }}_1^2,e_2=b-1-c_2{\bar{\sigma }}_2^2-\frac{1}{2}{\sigma }_2^2{\bar{\sigma }}_2^2,$

$e_3=2-2c-\frac{x^{*}{}^2}{4}-\frac{y^{*}{}^2}{4}-2c_3{\bar{\sigma }}_3^2-{\sigma }_3^2{\bar{\sigma }}_3^2,f_1=y^{*}{z}^{*}-\left(a+1\right)x^{*}+c_1{\bar{\sigma }}_1^2{x}^{*}+{\sigma }_1^2{\bar{\sigma }}_1^2{x}^{*},$

$f_2=x^{*}{z}^{*}-\left(b+1\right)y^{*}+c_2{\bar{\sigma }}_2^2{y}^{*}+{\sigma }_2^2{\bar{\sigma }}_2^2{y}^{*},f_3=2\left(c-1\right)z^{*}-2x^{*}{y}^{*}+2c_3{\bar{\sigma }}_3^2{z}^{*}+2{\sigma }_3^2{\bar{\sigma }}_3^2{z}^{*},$

$\left(x^{*},y^{*},z^{*}\right)=\left(\sqrt{\left(b+1\right)\left(c-1\right)},\sqrt{\left(a+1\right)\left(c-1\right)},\sqrt{\left(a+1\right)\left(b+1\right)}\right),$

$\frac{f_1}{2e_1}=g_1{x}^{*},\frac{f_2}{2e_2}=g_2{y}^{*},\frac{f_3}{2e_3}=g_3{z}^{*},K_3=\min \left\{e_1,e_2,e_3\right\}.$

证明：首先我们作一个变换 $u=x-x^{*}$， $v=y-y^{*}$， $w=z-z^{*}$，则模型随机金融风险系统模型(1.2)可以写成

$\{\begin{array}{l}\text{d}u\left(t\right)=\left(\left(v\left(t\right)+y^{*}\right)\left(w\left(t\right)+z^{*}\right)-\left(a+1\right)\left(u\left(t\right)+x^{*}\right)\right)\text{d}t+c_1\left(u\left(t\right)+x^{*}\right)\text{d}{\langle B_1,B_1\rangle }_t+{\sigma }_1\left(u\left(t\right)+x^{*}\right)\text{d}{B}_1\left(t\right),\\ \text{d}v\left(t\right)=\left(\left(u\left(t\right)+x^{*}\right)\left(w\left(t\right)+z^{*}\right)-\left(b+1\right)\left(v\left(t\right)+y^{*}\right)\right)\text{d}t+c_2\left(v\left(t\right)+y^{*}\right)\text{d}{\langle B_2,B_2\rangle }_t+{\sigma }_2\left(v\left(t\right)+y^{*}\right)\text{d}{B}_2\left(t\right),\\ \text{d}w\left(t\right)=\left(\left(c-1\right)\left(w\left(t\right)+z^{*}\right)-\left(u\left(t\right)+x^{*}\right)\left(v\left(t\right)+y^{*}\right)\right)\text{d}t+c_3\left(w\left(t\right)+z^{*}\right)\text{d}{\langle B_3,B_3\rangle }_t+{\sigma }_3\left(w\left(t\right)+z^{*}\right)\text{d}{B}_3\left(t\right).\end{array}$

构造Lyapunov函数：

$V\left(u,v,w\right)=\frac{1}{2}{u}^2+\frac{1}{2}{v}^2+w^2,$

且由G-伊藤公式可得：

$\begin{array}{c}V\left(u,v,w\right)=V\left(u\left(0\right),v\left(0\right),w\left(0\right)\right)+{\displaystyle {\int }_0^{t}L}V\text{d}s+{\displaystyle {\int }_0^t{\sigma }_1}u\left(s\right)\left(u\left(s\right)+x^{*}\right)\\ +{\sigma }_{2}v\left(s\right)\left(v\left(s\right)+y^{*}\right)+2{\sigma }_{3}w\left(s\right)\left(w\left(s\right)+z^{*}\right)\text{d}B\left(s\right)\\ +{\displaystyle {\int }_0^t{c}_1}u\left(s\right)\left(u\left(s\right)+x^{*}\right)+\frac{1}{2}{\sigma }_1^2{\left(u\left(s\right)+x^{*}\right)}^2\text{d}{\langle B_1,B_1\rangle }_s\\ +{\displaystyle {\int }_0^t{c}_2}v\left(s\right)\left(v\left(s\right)+y^{*}\right)+\frac{1}{2}{\sigma }_2^2{\left(v\left(s\right)+y^{*}\right)}^2\text{d}{\langle B_2,B_2\rangle }_s\\ +{\displaystyle {\int }_0^{t}2}{c}_{3}w\left(s\right)\left(w\left(s\right)+z^{*}\right)+{\sigma }_3^2{\left(w\left(s\right)+z^{*}\right)}^2\text{d}{\langle B_3,B_3\rangle }_s,\end{array}$ (4.5)

其中，

$\begin{array}{c}LV=u\left(t\right)\left(\left(v\left(t\right)+y^{*}\right)\left(w\left(t\right)+z^{*}\right)-\left(a+1\right)\left(u\left(t\right)+x^{*}\right)\right)\\ +v\left(t\right)\left(\left(u\left(t\right)+x^{*}\right)\left(w\left(t\right)+z^{*}\right)-\left(b+1\right)\left(v\left(t\right)+y^{*}\right)\right)\\ +w\left(t\right)\left(\left(c-1\right)\left(w\left(t\right)+z^{*}\right)-\left(u\left(t\right)+x^{*}\right)\left(v\left(t\right)+y^{*}\right)\right).\end{array}$

通过基本不等式，可以得到：

$\begin{array}{c}LV=-\left(a+1\right)u^2\left(s\right)-\left(b+1\right)v^2\left(s\right)-2\left(1-c\right)w^2\left(s\right)-y^{*}u\left(t\right)w\left(t\right)-x^{*}v\left(t\right)w\left(t\right)+2z^{*}u\left(t\right)v\left(t\right)\\ +u\left(t\right)\left(y^{*}{z}^{*}-\left(a+1\right)x^{*}\right)+v\left(t\right)\left(x^{*}{z}^{*}-\left(b+1\right)y^{*}\right)+2w\left(t\right)\left(\left(c-1\right)z^{*}-x^{*}{y}^{*}\right)\\ \le -\left(a-z^{*}{}^2\right)u^2\left(s\right)-\left(b-1\right)v^2\left(s\right)-\left(2-2c-\frac{x^{*}{}^2}{4}-\frac{y^{*}{}^2}{4}\right)w^2\left(s\right)\\ +u\left(t\right)\left(y^{*}{z}^{*}-\left(a+1\right)x^{*}\right)+v\left(t\right)\left(x^{*}{z}^{*}-\left(b+1\right)y^{*}\right)+2w\left(t\right)\left(\left(c-1\right)z^{*}-x^{*}{y}^{*}\right).\end{array}$

对等式(4.5)，两边取G-期望，得到：

$\begin{array}{c}0\le \hat{\mathbb{E}}V\left(u,v,w\right)\\ \le \hat{\mathbb{E}}V\left(u\left(0\right),v\left(0\right),w\left(0\right)\right)+\hat{\mathbb{E}}{\displaystyle {\int }_0^t-}\left(a-z^{*}{}^2-c_1{\bar{\sigma }}_1^2-\frac{1}{2}{\sigma }_1^2{\bar{\sigma }}_1^2\right)u^2\left(s\right)\\ -\left(b-1-c_2{\bar{\sigma }}_2^2-\frac{1}{2}{\sigma }_2^2{\bar{\sigma }}_2^2\right)v^2\left(s\right)-\left(2-2c-\frac{x^{*}{}^2}{4}-\frac{y^{*}{}^2}{4}-2c_3{\bar{\sigma }}_3^2-{\sigma }_3^2{\bar{\sigma }}_3^2\right)w^2\left(s\right)\\ +\left(y^{*}{z}^{*}-\left(a+1\right)x^{*}+c_1{\bar{\sigma }}_1^2{x}^{*}+{\sigma }_1^2{\bar{\sigma }}_1^2{x}^{*}\right)u\left(s\right)+\left(x^{*}{z}^{*}-\left(b+1\right)y^{*}+c_2{\bar{\sigma }}_2^2{y}^{*}+{\sigma }_2^2{\bar{\sigma }}_2^2{y}^{*}\right)v\left(s\right)\\ +\left(2\left(c-1\right)z^{*}-2x^{*}{y}^{*}+2c_3{\bar{\sigma }}_3^2{z}^{*}+2{\sigma }_3^2{\bar{\sigma }}_3^2{z}^{*}\right)w\left(s\right)\text{d}t+\frac{1}{2}{\sigma }_1^2{\bar{\sigma }}_1^2{x}^{*}{}^{2}t+\frac{1}{2}{\sigma }_2^2{\bar{\sigma }}_2^2{y}^{*}{}^{2}t+{\sigma }_3^2{\bar{\sigma }}_3^2{z}^{*}{}^{2}t.\end{array}$ (4.6)

令

$e_1=a-z^{*}{}^2-c_1{\bar{\sigma }}_1^2-\frac{1}{2}{\sigma }_1^2{\bar{\sigma }}_1^2,e_2=b-1-c_2{\bar{\sigma }}_2^2-\frac{1}{2}{\sigma }_2^2{\bar{\sigma }}_2^2,$

$e_3=2-2c-\frac{x^{*}{}^2}{4}-\frac{y^{*}{}^2}{4}-2c_3{\bar{\sigma }}_3^2-{\sigma }_3^2{\bar{\sigma }}_3^2,f_1=y^{*}{z}^{*}-\left(a+1\right)x^{*}+c_1{\bar{\sigma }}_1^2{x}^{*}+{\sigma }_1^2{\bar{\sigma }}_1^2{x}^{*},$

$f_2=x^{*}{z}^{*}-\left(b+1\right)y^{*}+c_2{\bar{\sigma }}_2^2{y}^{*}+{\sigma }_2^2{\bar{\sigma }}_2^2{y}^{*},f_3=2\left(c-1\right)z^{*}-2x^{*}{y}^{*}+2c_3{\bar{\sigma }}_3^2{z}^{*}+2{\sigma }_3^2{\bar{\sigma }}_3^2{z}^{*}.$

因此，G-期望不等式(4.6)可以写成以下形式：

$\begin{array}{l}\hat{\mathbb{E}}{\displaystyle {\int }_0^t-e_1{u}^2\left(s\right)}-e_2{v}^2\left(s\right)-e_3{w}^2\left(s\right)+f_{1}u\left(s\right)+f_{2}v\left(s\right)+f_{3}w\left(s\right)\text{d}t+\frac{1}{2}{\sigma }_1^2{\bar{\sigma }}_1^2{x}^{*}{}^{2}t+\frac{1}{2}{\sigma }_2^2{\bar{\sigma }}_2^2{y}^{*}{}^{2}t+{\sigma }_3^2{\bar{\sigma }}_3^2{z}^{*}{}^{2}t\\ =\hat{\mathbb{E}}{\displaystyle {\int }_0^t-e_1}{\left(u\left(s\right)-\frac{f_1}{2e_1}\right)}^2-e_2{\left(v\left(s\right)-\frac{f_2}{2e_2}\right)}^2-e_3{\left(w\left(s\right)-\frac{f_3}{2e_3}\right)}^2\text{d}t\\ +\frac{1}{2}{\sigma }_1^2{\bar{\sigma }}_1^2{x}^{*}{}^{2}t+\frac{1}{2}{\sigma }_2^2{\bar{\sigma }}_2^2{y}^{*}{}^{2}t+{\sigma }_3^2{\bar{\sigma }}_3^2{z}^{*}{}^{2}t+\frac{f_1^2}{4e_1}t+\frac{f_2^2}{4e_2}t+\frac{f_3^2}{4e_3}t.\end{array}$

因此，有：

$\begin{array}{l}\underset{t\to \infty }{\lim \sup }\frac{1}{t}\hat{\mathbb{E}}{\displaystyle {\int }_0^t{e}_1}{\left(u\left(s\right)-\frac{f_1}{2e_1}\right)}^2+e_2{\left(v\left(s\right)-\frac{f_2}{2e_2}\right)}^2+e_3{\left(w\left(s\right)-\frac{f_3}{2e_3}\right)}^2\text{d}t\\ \le \frac{1}{2}{\sigma }_1^2{\bar{\sigma }}_1^2{x}^{*}{}^2+\frac{1}{2}{\sigma }_2^2{\bar{\sigma }}_2^2{y}^{*}{}^2+{\sigma }_3^2{\bar{\sigma }}_3^2{z}^{*}+\frac{f_1^2}{4e_1}+\frac{f_2^2}{4e_2}+\frac{f_3^2}{4e_3},\end{array}$

如果令 $K_3=\min \left\{e_1,e_2,e_3\right\}$，则

$\begin{array}{l}\underset{t\to \infty }{\lim \sup }\frac{1}{t}\hat{\mathbb{E}}{\displaystyle {\int }_0^t{\left(u\left(s\right)-\frac{f_1}{2e_1}\right)}^2}+{\left(v\left(s\right)-\frac{f_2}{2e_2}\right)}^2+{\left(w\left(s\right)-\frac{f_3}{2e_3}\right)}^2\text{d}t\\ \le \frac{1}{K_3}\left(\frac{1}{2}{\sigma }_1^2{\bar{\sigma }}_1^2{x}^{*}{}^2+\frac{1}{2}{\sigma }_2^2{\bar{\sigma }}_2^2{y}^{*}{}^2+{\sigma }_3^2{\bar{\sigma }}_3^2{z}^{*}+\frac{f_1^2}{4e_1}+\frac{f_2^2}{4e_2}+\frac{f_3^2}{4e_3}\right).\end{array}$

将原变换代入式子，并有 $\frac{f_1}{2e_1}=g_1{x}^{*}$， $\frac{f_2}{2e_2}=g_2{y}^{*}$， $\frac{f_3}{2e_3}=g_3{z}^{*}$，则有：

$\begin{array}{l}\underset{t\to \infty }{\lim \sup }\frac{1}{t}\hat{\mathbb{E}}{\displaystyle {\int }_0^t{\left(x\left(s\right)-\left(1+g_1\right)x^{*}\right)}^2}+{\left(y\left(s\right)-\left(1+g_2\right)y^{*}\right)}^2+{\left(z\left(s\right)-\left(1+g_3\right)z^{*}\right)}^2\text{d}t\\ \le \frac{1}{K_3}\left(\frac{1}{2}{\sigma }_1^2{\bar{\sigma }}_1^2{x}^{*}{}^2+\frac{1}{2}{\sigma }_2^2{\bar{\sigma }}_2^2{y}^{*}{}^2+{\sigma }_3^2{\bar{\sigma }}_3^2{z}^{*}+\frac{f_1^2}{4e_1}+\frac{f_2^2}{4e_2}+\frac{f_3^2}{4e_3}\right).\end{array}$

定理4.3如果下列条件

1) $2\left(a+1\right)-2c_1{\bar{\sigma }}_1^2-{\bar{\sigma }}_1^2{\sigma }_1^2-k\frac{m+n}{m}>0$，

2) $2\left(b+1\right)-2c_2{\bar{\sigma }}_2^2-{\bar{\sigma }}_2^2{\sigma }_2^2-k\frac{m+n}{n}>0$，

3) ${\bar{\sigma }}_3^2\left(m+n\right)\left(2c_3+{\sigma }_3^2\right)>0$

成立，令 $L_1=\left(2-\frac{c_3{\bar{\sigma }}_3^2+3}{{\bar{\sigma }}_3^2\left(2c_3+{\sigma }_3^2\right)}\right)\left(m+n\right)k^2$， $l_0$ 是正定矩阵M的最小(正)特征值，有 $l=\frac{L_1}{l_0}$，其中，

$M=\left(\begin{array}{ccc}2\left(a+1\right)-2c_1{\bar{\sigma }}_1^2-{\bar{\sigma }}_1^2{\sigma }_1^2-k\frac{m+n}{m}& 0& 0\\ 0& 2\left(b+1\right)-2c_2{\bar{\sigma }}_2^2-{\bar{\sigma }}_2^2{\sigma }_2^2-k\frac{m+n}{n}& 0\\ 0& 0& 2\end{array}\right).$

那么以下估计成立， $\hat{\mathbb{E}}\left(V_{m,n}\left(x,y,z\right)-l\right)\le \left(V_{m,n}\left(x_0,y_0,z_0\right)-l\right){\text{e}}^{-l_0\left(t-t_0\right)}$。特别地， ${\Omega }_1=\left\{\left(x,y,z\right)|\hat{\mathbb{E}}{V}_{m,n}\left(x,y,z\right)\le l\right\}$ 是G-布朗运动驱动的随机金融风险系统模型(1.2)的全局指数吸引集，其中 $V_{m,n}\left(x,y,z\right)=m{x}^2+n{y}^2+\left(m+n\right){\left(z-k\right)}^2$。

证明：构造以下Lyapunov函数

$V_{m,n}\left(x,y,z\right)=m{x}^2+n{y}^2+\left(m+n\right){\left(z-k\right)}^2.$

根据G-伊藤公式，计算 $V_{m,n}\left(x,y,z\right)$ 沿随机金融风险系统模型(1.2)轨迹的导数，可以得到：

$\begin{array}{c}\text{d}V=\left(2mx\frac{\text{d}x}{\text{d}t}+2ny\frac{\text{d}y}{\text{d}t}+2\left(m+n\right)\left(z-k\right)\frac{\text{d}z}{\text{d}t}\right)\text{d}t\\ +\left(2mx\frac{\text{d}x}{\text{d}{B}_t}+2ny\frac{\text{d}y}{\text{d}{B}_t}+2\left(m+n\right)\left(z-k\right)\frac{\text{d}z}{\text{d}{B}_t}\right)\text{d}{B}_t\\ +\left(2mx\frac{\text{d}x}{\text{d}{\langle B_1,B_1\rangle }_t}+2ny\frac{\text{d}y}{\text{d}{\langle B_2,B_2\rangle }_t}+2\left(m+n\right)\left(z-k\right)\frac{\text{d}z}{\text{d}{\langle B_3,B_3\rangle }_t}\right)\text{d}{\langle B,B\rangle }_t\\ +\left(m\frac{{\text{d}}^{2}x}{\text{d}{B}_t^2}+n\frac{{\text{d}}^{2}y}{\text{d}{B}_t^2}+\left(m+n\right)\frac{{\text{d}}^{2}z}{\text{d}{B}_t^2}\right)\text{d}{\langle B,B\rangle }_t.\end{array}$ (4.7)

因此，有

$\begin{array}{l}V\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)\\ =V\left(x\left(0\right),y\left(0\right),z\left(0\right)\right)+{\displaystyle {\int }_{t_0}^{t}2mx}\left(s\right)\left(y\left(s\right)z\left(s\right)-\left(a+1\right)x\left(s\right)\right)\\ +2ny\left(s\right)\left(x\left(s\right)z\left(s\right)-\left(b+1\right)y\left(s\right)\right)+2\left(m+n\right)\left(z\left(s\right)-k\right)\left(\left(c-1\right)z\left(s\right)-x\left(s\right)y\left(s\right)\right)\text{d}s\\ +{\displaystyle {\int }_{t_0}^{t}2m{\sigma }_1{x}^2\left(s\right)}+2n{\sigma }_2{y}^2\left(s\right)+2\left(m+n\right){\sigma }_{3}z\left(s\right)\left(z\left(s\right)-k\right)\text{d}B\left(s\right)\\ +{\displaystyle {\int }_{t_0}^{t}2m{c}_1{x}^2\left(s\right)}+m{\sigma }_1^2{x}^2\left(s\right)\text{d}{\langle B_1,B_1\rangle }_s+2n{c}_2{y}^2\left(s\right)+n{\sigma }_2^2{y}^2\left(s\right)\text{d}{\langle B_2,B_2\rangle }_s\\ +2\left(m+n\right)c_{3}z\left(s\right)\left(z\left(s\right)-k\right)+\left(m+n\right){\sigma }_3^2{z}^2\left(s\right)\text{d}{\langle B_3,B_3\rangle }_s.\end{array}$ (4.8)

对(4.8)式子，两边取G-期望，获得

$\begin{array}{l}\hat{\mathbb{E}}V\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)\\ \le V\left(x\left(0\right),y\left(0\right),z\left(0\right)\right)+\hat{\mathbb{E}}{\displaystyle {\int }_{t_0}^t-}\left(2\left(a+1\right)-2c_1{\bar{\sigma }}_1^2-{\bar{\sigma }}_1^2{\sigma }_1^2\right)m{x}^2\left(s\right)\\ -\left(2\left(b+1\right)-2c_2{\bar{\sigma }}_2^2-{\bar{\sigma }}_2^2{\sigma }_2^2\right)n{y}^2\left(s\right)\\ +2\left(m+n\right)\left(c-1\right)z^2\left(s\right)-2k\left(m+n\right)z\left(s\right)+2k\left(m+n\right)x\left(s\right)y\left(s\right)\\ +{\bar{\sigma }}_3^2\left(2\left(m+n\right)c_3+\left(m+n\right){\sigma }_3^2\right)z^2\left(s\right)-2c_{3}k{\bar{\sigma }}_3^2\left(m+n\right)z\left(s\right)\text{d}s,\end{array}$ (4.9)

根据基本不等式， $2k\left(m+n\right)x\left(s\right)y\left(s\right)\le k\left(m+n\right)x^2\left(s\right)+k\left(m+n\right)y^2\left(s\right)$，因此有

$\begin{array}{l}\hat{\mathbb{E}}V\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)\\ \le V\left(x\left(0\right),y\left(0\right),z\left(0\right)\right)+\hat{\mathbb{E}}{\displaystyle {\int }_{t_0}^t-}\left(2\left(a+1\right)-2c_1{\bar{\sigma }}_1^2-{\bar{\sigma }}_1^2{\sigma }_1^2+k\frac{m+n}{m}\right)m{x}^2\left(s\right)\\ -\left(2\left(b+1\right)-2c_2{\bar{\sigma }}_2^2-{\bar{\sigma }}_2^2{\sigma }_2^2+k\frac{m+n}{n}\right)n{y}^2\left(s\right)-2\left(m+n\right){\left(z\left(s\right)-k\right)}^2+F\left(x,y\right)\text{d}s,\end{array}$

其中，

$\begin{array}{c}F\left(x,y\right)={\bar{\sigma }}_3^2\left(m+n\right)\left(2c_3+{\sigma }_3^2\right){\left(z-\frac{c_3{\bar{\sigma }}_3^2+3}{{\bar{\sigma }}_3^2\left(2c_3+{\sigma }_3^2\right)}\right)}^2\\ +\left(2-\frac{c_3{\bar{\sigma }}_3^2+3}{{\bar{\sigma }}_3^2\left(2c_3+{\sigma }_3^2\right)}\right)\left(m+n\right)k^2\\ \le \sup F\left(x,y\right)\\ =\left(2-\frac{c_3{\bar{\sigma }}_3^2+3}{{\bar{\sigma }}_3^2\left(2c_3+{\sigma }_3^2\right)}\right)\left(m+n\right)k^2\\ =L_1.\end{array}$

因此，获得

$\hat{\mathbb{E}}\left(V_{m,n}\left(x,y,z\right)-l\right)\le V_{m,n}\left(x_0,y_0,z_0\right)+{\displaystyle {\int }_{t_0}^t\left(-l_0\hat{\mathbb{E}}{V}_{m,n}\left(x,y,z\right)+L_1\right)\text{d}s}.$

利用Gronwall不等式，可以得到：

$\begin{array}{c}\hat{\mathbb{E}}\left(V_{m,n}\left(x,y,z\right)-l\right)\le V_{m,n}\left(x_0,y_0,z_0\right){\text{e}}^{-l_0\left(t-t_0\right)}+L_1{\displaystyle {\int }_{t_0}^t{\text{e}}^{-l_0\left(t-t_0\right)}\text{d}s}\\ =V_{m,n}\left(x_0,y_0,z_0\right){\text{e}}^{-l_0\left(t-t_0\right)}+\frac{L_1}{l_0}\left(1-{\text{e}}^{-l_0\left(t-t_0\right)}\right)\end{array}$

令 $l=\frac{L_1}{l_0}$，当 $\hat{\mathbb{E}}\left(V_{m,n}\left(x,y,z\right)-l\right)>0$ 和 $\hat{\mathbb{E}}\left(V_{m,n}\left(x_0,y_0,z_0\right)-l\right)>0$ 时，以下估计成立：

$\hat{\mathbb{E}}\left(V_{m,n}\left(x,y,z\right)-l\right)\le \left(V_{m,n}\left(x_0,y_0,z_0\right)-l\right){\text{e}}^{-l_0\left(t-t_0\right)},$

从而 $\underset{t\to \infty }{\lim }\hat{\mathbb{E}}{V}_{m,n}\left(x,y,z\right)\le l$，所以可得， $\Omega =\left\{\left(x,y,z\right)|\hat{\mathbb{E}}{V}_{m,n}\left(x,y,z\right)\le l\right\}=\left\{\left(x,y,z\right)|\hat{\mathbb{E}}\left[m{x}^2+n{y}^2+\left(m+n\right){\left(z+k\right)}^2\right]\le l\right\}$ 是随机金融风险系统模型(1.2)的全局指数吸引集。

定理4.4对于任意 $\epsilon \in \left(0,1\right)$，存在 $\rho =\rho \left(\epsilon \right)>0$，假设 $2\left(a+1\right)-{\bar{\sigma }}_1^2\left(2c_1+{\sigma }_1^2\right)>0$， $2\left(b+1\right)-{\bar{\sigma }}_2^2\left(2c_2+{\sigma }_2^2\right)>0$， $4\left(1-c\right)-{\bar{\sigma }}_3^2\left(4c_3+2{\sigma }_3^2\right)>0$，且 $\lambda =\max \left\{2\left(a+1\right)-{\bar{\sigma }}_1^2\left(2c_1+{\sigma }_1^2\right),2\left(b+1\right)-{\bar{\sigma }}_2^2\left(2c_2+{\sigma }_2^2\right),4\left(1-c\right)-{\bar{\sigma }}_3^2\left(4c_3+2{\sigma }_3^2\right)\right\}$，使得G-布朗运动驱动的随机金融风险系统模型(1.2)的解 $X\left(t\right)$ 满足

$\underset{t\to \infty }{\lim \sup }P\left\{\left|X\left(t\right)\right|>\rho \right\}<\epsilon ,$

则G-布朗运动驱动的随机金融风险系统模型(1.2)是随机最终有界的，其中 $\left|X\left(t\right)\right|=\sqrt{x^2+y^2+z^2}$。

证明：构造以下Lyapunov函数，

$V\left(x,y,z\right)=x^2+\frac{1}{4}+y^2+\frac{1}{4}+2z^2+\frac{1}{8}.$

根据G-伊藤公式，可以得到：

$\begin{array}{l}\text{d}V\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)\\ =2x\left(t\right)\left(y\left(t\right)z\left(t\right)-\left(a+1\right)x\left(t\right)\right)+2y\left(t\right)\left(x\left(t\right)z\left(t\right)-\left(b+1\right)y\left(t\right)\right)\\ +4z\left(t\right)\left(\left(c-1\right)z\left(t\right)-x\left(t\right)y\left(t\right)\right)\text{d}t+2{\sigma }_1{x}^2\left(t\right)+2{\sigma }_2{y}^2\left(t\right)+4{\sigma }_3{z}^2\left(t\right)\text{d}B\left(t\right)\\ +2c_1{x}^2\left(t\right)+{\sigma }_1^2{x}^2\left(t\right)\text{d}{\langle B_1,B_1\rangle }_t+2c_2{y}^2\left(t\right)+{\sigma }_2^2{y}^2\left(t\right)\text{d}{\langle B_2,B_2\rangle }_t\\ +4c_3{z}^2\left(t\right)+2{\sigma }_3^2{z}^2\left(t\right)\text{d}{\langle B_3,B_3\rangle }_t.\end{array}$ (4.10)

对(4.10)式子两边积分，并取G-期望，得到

$\begin{array}{l}\hat{\mathbb{E}}V\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)\\ \le V\left(x\left(0\right),y\left(0\right),z\left(0\right)\right)+\hat{\mathbb{E}}{\displaystyle {\int }_{t_0}^t-}\left(2\left(a+1\right)-{\bar{\sigma }}_1^2\left(2c_1+{\sigma }_1^2\right)\right)x^2\left(s\right)\\ -\left(2\left(b+1\right)-{\bar{\sigma }}_2^2\left(2c_2+{\sigma }_2^2\right)\right)y^2\left(s\right)-\left(4\left(1-c\right)-{\bar{\sigma }}_3^2\left(4c_3+2{\sigma }_3^2\right)\right)z^2\left(s\right)\text{d}s\\ <-\lambda V\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)+C_0\end{array}$ (4.11)

其中 $C_0>0$，故有

$\begin{array}{c}\hat{\mathbb{E}}\left[{\text{e}}^{\lambda t}V\left(t\right)\right]-\hat{\mathbb{E}}\left[{\text{e}}^{0}V\left(t\right)\right]=\hat{\mathbb{E}}\left[{\displaystyle {\int }_0^t\text{d}}\left({\text{e}}^{\lambda s}V\left(s\right)\right)\right]\\ =\hat{\mathbb{E}}\left[{\displaystyle {\int }_0^t\lambda }{\text{e}}^{\lambda s}V\left(s\right)\text{d}s+{\text{e}}^{\lambda s}\text{d}V\left(s\right)\right]\\ =\hat{\mathbb{E}}{\displaystyle {\int }_0^t\lambda }{\text{e}}^{\lambda s}V\left(s\right)\text{d}s+{\text{e}}^{\lambda s}\hat{\mathbb{E}}V\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)\\ \le C_0\hat{\mathbb{E}}\left[{\displaystyle {\int }_0^t{\text{e}}^{\lambda s}}\text{d}s\right]\\ \le \frac{C_0}{\lambda }\left({\text{e}}^{\lambda t}-1\right),\end{array}$ (4.12)

从而，获得

$\hat{\mathbb{E}}\left[V\left(t\right)\right]\le {\text{e}}^{-\lambda t}\hat{\mathbb{E}}\left[V\left(t\right)\right]+\frac{C_0}{\lambda }\left(1-{\text{e}}^{\lambda t}\right)\le \hat{\mathbb{E}}\left[V\left(t\right)\right]+\frac{C_0}{\lambda }=:L.$

假设 $\rho$ 足够大，使得 $\frac{C_0}{\lambda \rho }<1$。由chebyshev’s不等式，得 $P\left\{V\left(t\right)>\rho \right\}\le \frac{L}{\rho }$。由此可得

$\underset{t\to \infty }{\lim \sup }P\left\{V\left(t\right)>\rho \right\}\le \frac{C_0}{\lambda \rho }=:\epsilon .$

因为 $V\left(t\right)\ge x+y+z\ge \sqrt{x^2+y^2+z^2}$，所以有 $\underset{t\to \infty }{\lim \sup }P\left\{\left|X\left(t\right)\right|>\rho \right\}<\epsilon$，则随机金融风险系统模型(1.2)是随机最终有界的。

NOTES

^*通讯作者。