均值-方差框架下保险公司和再保险公司的 Stackelberg 随机微分博弈

均值-方差框架下保险公司和再保险公司的 Stackelberg 随机微分博弈

温玉珍^,^*, 王绍琳^,^*

曲阜师范大学统计与数据科学学院山东曲阜 273165

Stackelberg Stochastic Differential Game of Insurer and Reinsurer Under Mean-Variance Framework

Wen Yuzhen^,^*, Wang Shaolin^,^*

School of Statistics and Data Science, Qufu Normal University, Shandong Qufu 273165

通讯作者: *E-mail: wenyzhen@163.com

收稿日期: 2024-09-7 修回日期: 2025-02-6

基金资助:

山东省自然科学基金(ZR2023MA093)

Received: 2024-09-7 Revised: 2025-02-6

Fund supported:

Natural Science Foundation of Shandong Province(ZR2023MA093)

作者简介 About authors

E-mail:17861318483@163.com

摘要

该文研究保险公司和再保险公司在 Heston 随机波动率风险模型中基于 Stackelberg 随机微分博弈框架下的最优投资再保险问题. 假定保险公司的索赔过程由泊松跳模型来描述, 通过期望值保费原则计算保费, 保险公司购买比例再保险将部分风险转移给再保险公司, 再保险公司接受转移的风险并选择合适的再保费策略, 保险公司和再保险公司还可将盈余投资于无风险资产和 Heston 随机波动率风险模型中. 以最大化终端财富值的均值-方差效用为目标, 利用随机控制理论建立相应的扩展 Hamilton-Jacobi-Bellman 方程, 给出验证定理, 得到均衡策略和值函数的具体形式, 最后通过数值模拟分析了模型参数对均衡策略的影响.

关键词： 均值-方差准则; 投资再保险策略; Heston 随机波动率; Stackelberg 随机微分博弈; 扩展 Hamilton-Jacobi-Bellman 方程

Abstract

This paper studies the optimal reinsurance-investment problem of an insurer and a reinsurer under Heston's stochastic volatility model in the framework of Stackelberg stochastic differential game. We assume that the insurer purchases proportional reinsurance to transfer part of the risk to the reinsurer, and the reinsurer accepts the transferred risk and chooses an appropriate reinsurance premium strategy. At the same time, the insurer and the reinsurer can also invest their surplus in risk-free assets and Heston's stochastic volatility risk model. Under the criterion of maximizing the expected mean-variance utility of terminal wealth, the corresponding extended Hamilton-Jacobi-Bellman equations are developed using stochastic control theory, and verification theorems are given. We obtain specific forms for the equilibrium strategy and the value function. Finally, we analyze the effects of the model parameters on the equilibrium strategies through numerical simulations.

Keywords： mean-variance criterion; reinsurance and investment strategy; Heston's SV; Stackelberg stochastic differential game; extended Hamilton-Jacobi-Bellman equation

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本文引用格式

温玉珍, 王绍琳. 均值-方差框架下保险公司和再保险公司的 Stackelberg 随机微分博弈[J]. 数学物理学报, 2025, 45(4): 1327-1353

Wen Yuzhen, Wang Shaolin. Stackelberg Stochastic Differential Game of Insurer and Reinsurer Under Mean-Variance Framework[J]. Acta Mathematica Scientia, 2025, 45(4): 1327-1353

1 引言

近年来, 投资和再保险是保险公司提高承保能力、控制风险、保证保险业务持续健康发展的重要业务活动. 因此, 有关保险公司最优投资和再保险问题在保险精算领域引起了广泛的关注, 对不同优化准则下的保险公司的最优再保险投资策略进行了大量的研究, 其中应用最广泛的准则包括最小化破产概率^[1-3]和最大化终端财富期望效用^[4-7]以及均值-方差准则^[8-10].

在上述文章中, 大多数只从保险公司的角度考虑最优策略而忽略了再保险公司. 1969 年, Borch^[11]表明双方在再保险保费上存在利益冲突, 再保险保费应该是双方共同达成的协议, 只涉及一方利益的再保险策略可能不会被另一方所接受. 一般来说, 从保险公司和再保险公司的共同角度来研究再保险有两种方法: 一种方法是研究保险公司和再保险公司的加权目标,Zhang 等^[12]在五种不同的标准下, 研究了基于保险公司和再保险公司的共同利益的最优份额再保险问题.Yang 和 Chen^[13]考虑了保险公司和再保险公司之间的共同利益, 保险公司和再保险公司分别从各自角度出发, 在最大化终端时刻期望值同时减小其方差的目标下, 得到了最优再保险合同及投资策略. Yang^[14]在均值-方差准则下研究了保险集团的最优投资再保险策略, 通过对保险公司和再保险公司的财富赋予不同的权重, 研究了模型参数对最优策略的影响.Cai 等^[15]考虑了保险公司和再保险公司的利益, 研究了保险公司与再保险公司的共同生存和盈利概率, 在一般再保险保费原则下, 得到了在一类保险政策中存在最优再保险条约的充分条件.Zhao 等^[16]在 CEV 模型下研究了兼顾保险公司和再保险公司利益的最优投资再保险策略.

另一种方法是制定保险公司和再保险公司之间的 Stackelberg 博弈, 在博弈中再保险公司是领导者, 保险公司是追随者. 再保险公司首先给出任意的再保险保费策略, 保险公司根据再保险公司给定的保费策略得到最优的保险策略, 最后再保险公司根据保险公司的最优策略调整其再保险保费, 从而得到 Stackelberg 均衡策略.事实上, 保险业务中的随机微分博弈已经在几个竞争的保险公司或保险公司和市场之间进行了广泛的研究. 在保险公司与市场零和博弈的框架下,2011 年, Elliott 和 Siu^[17]通过求解倒向随机微分方程解决了最优投资问题,它给出了一种简单自然的方法, 在没有马尔可夫假设的情况下验证了最优投资策略的存在性和唯一性. 在均值-方差效用准则下,Wang 等^[18]研究了两家竞争保险公司的最优策略, 这两家保险公司的索赔过程由复合泊松风险模型和扩散近似模型描述, 两者的目标都是最大化相对于对手的预期终端盈余, 得到了均衡再保险投资策略和相应的值函数. Yan 等^[19]研究了两家保险公司之间的再保险和投资博弈, 两者对开始时获得的索赔信息采取不同态度, 以最大化终端时刻与竞争对手盈余差额为目标, 在指数效用下得到了均衡策略. 然而, 这些文章没有涉及到保险公司和再保险公司之间的博弈. Chen^[20]首先从保险公司和再保险公司相互影响的角度, 运用 Stackelberg 微分博弈分析了最优再保险问题, 在期望效用最大化准则下, 得到了最优保费定价策略和特殊情况下的最优再保险策略. Chen 与 Shen^[21]在均值-方差框架下研究了保险公司和再保险公司之间的 Stackelberg 微分博弈, 在均值保费准则和方差保费准则下分别得到了再保险策略和值函数. Li 与 Young^[22]进一步推广了文献[21], 在随机时刻下研究了保险公司和再保险公司之间的 Stackelberg 微分博弈, 在均值-方差保费准则下得到最优再保险策略为超额损失再保险策略. 同时, 还分析了不同风险分布对均衡再保险和保费策略的影响, 如果索赔分布是轻尾的, 那么均衡再保险就是纯粹的超额损失再保险,对损失的方差没有负荷; 如果索赔分布是重尾的, 那么均衡再保险就有一个非平凡共保, 对损失的方差有相应的正负荷. Bai^[23]考虑到保险公司和再保险公司在金融市场上信息不对称, 还考虑到当前财富受到过去业绩的影响, 在常数绝对风险厌恶效用函数下得到最优再保险和投资策略.

在上述的大多数文献中, 大多数假设风险资产的价格过程遵循几何布朗运动, 这意味着风险资产的波动率是恒定的. 然而, 这与波动率为随机的实证研究结果相冲突, 比如文献[24]和[25]. 因此, 我们有必要考虑风险资产的波动率为随机的模型. Heston 随机波动率模型是最受欢迎的随机波动率模型之一, 它考虑了波动率的均值回归特性、长时期和短时期的波动率特性, 这使得模型在捕捉市场现实波动的特点时表现得更为合理. 同时, Heston 模型可以解释很多众所周知的发现, 比如波动率微笑, 波动率聚类, 收益分布的重尾性质等. Heston 模型还为欧式看涨和看跌期权提供了半解析闭式解,在某些情况下, Heston 模型可以在没有计算机模拟或复杂数值方法的情况下得到期权价格, 这对于市场参与者来说是一个很大的优点. 为了使模型更加贴近现实, Heston^[26]提出了风险资产的波动率是平方根扩散过程驱动的模型, 该模型具有一定的计算优势. 在保险公司和再保险公司对金融市场中的信息了解程度不同时, Guan 等^[27]在 Heston 随机波动率模型和 Alpha-maxmin 均值-方差准则下研究了保险公司和再保险公司的 $\alpha$-Robust 策略.Zhou 等^[28]在 Heston 随机波动率模型下研究了保险公司和再保险公司的 Stackelberg 随机微分博弈, 在均值-方差效用准则下得到了保险公司和再保险公司的均衡策略和值函数. 后来许多学者将其引入到优化问题中, 用 Heston 随机波动率来描述风险资产.

综上所述, 保险公司的最优投资再保险策略在国内外已经有了全面的研究, 但仍有一些可以完善的地方. 大多数的文章中没有考虑再保险公司的利益, 没有将保险公司和再保险公司的共同优化问题反映在模型中. 在本文中我们同时考虑了保险公司和再保险公司的利益, 在 Stackelberg 随机微分博弈下研究了保险公司和再保险公司的最优策略. 不同于以往大多数文章的是在本文中我们引入了突发事件对于保险公司的影响, 在保险公司的索赔过程中引入泊松跳, 以此来描述生活中突发事件对保险公司的影响. 同时, 对风险资产的模型由传统的 Black-Scholes 模型拓宽到 Heston 随机波动率模型, 此模型更能符合现实生活中风险资产的波动情形, 使模型更加贴切现实. 之后文章在均值-方差框架下研究了保险公司和再保险公司之间的 Stackelberg 随机微分博弈, 在此博弈中依次求解保险公司和再保险公司的优化问题, 从而得到保险公司和再保险公司的均衡策略.

2 模型构造

在本章, 我们假设 $(\Omega,\mathcal{F},P)$ 为一个概率空间, $\sigma$-代数 $\mathcal{F}_{t}$ 表示直到 $t$ 时刻可获取的信息, 它满足通常条件. $T>0$ 是固定的常数, $P$ 是实概率测度, 我们假定以下所有的随机过程定义在该空间并且适应于 $\{\mathcal{F}_{t}\}_{t\in [T]}$.

1) 盈余过程

在本节中, 我们考虑保险市场包含一个保险公司和一个再保险公司. 考虑到突发事件发生的随机性, 我们假设保险公司的索赔满足跳扩散过程

(2.1)$\begin{equation} {\rm d} C( t ) = a{\rm d} t-b{\rm d}W_{0}(t) + d \Bigg(\sum _ { i = 1 } ^ { N ( t ) } Z _ { i }\Bigg), \end{equation}$

其中 $a$ 和 $b$ 是两个正常数, $W_{0}(t)$ 是一维标准布朗运动. $\{N(t)\}_{t\geq 0}$ 为齐次的 Poisson 过程, 强度为 $\lambda_{0}>0$, $N(t)$表示在时间间隔 $[t]$ 上发生的索赔次数, $Z_{i}$ 是第 $i$ 次的索赔数额, $Z_{i},i=1,2,\cdots$ 是一列独立同分布的 (i.i.d) 正随机变量, 并且有有限的一阶矩 $\mu_{z}=E[Z_{i}]$ 与二阶矩 $\sigma_{z}^{2}=E[Z_{i}^{2}]$. 进一步, 假设保险公司根据期望值保费原则收取保费, 其安全载荷因子 $\theta>0$, 因此保险公司收取的保费为 $c=({1+\theta})a$, 其盈余过程为

(2.2)$\begin{equation} {\rm d} R ( t ) = \theta a {\rm d} t + b {\rm d} W _ { 0 } ( t ) - d\Bigg( \sum _ { i = 1 } ^ { N ( t ) } Z _ { i }\Bigg), \end{equation}$

这里

(2.3)$\begin{equation} \theta a = ( 1 + \theta )\lambda_{0}\mu_{z}. \end{equation}$

对于每一个 $t\in[T]$, 我们进一步假设保险公司可以购买比例再保险合同, 将索赔 $Z_{i}$ 的 $(1-q(t))\in[0, 1]$ 转移给再保险公司. 这里 $q(t)$ 表示保险公司购买比例再保险后的保留水平, 这里有两种特殊情况: $q(t)=0$ 表示购买全额再保险, $q(t)=1$ 表示没有购买比例再保险. $t$ 时刻的再保费价格为 $p(t)\in[c,\bar{c}]$, 其中 $\bar{c}=(1+\bar{\theta})a$. 这里 $p(t)\geq c$ 表示不允许保险公司不承担风险而获取利润, $\bar{\theta}\geq \theta$ 表示再保险公司相对安全载荷的上界, 不包括过高的再保险保费. 因此, 在比例再保险策略下, 保险公司和再保险公司的盈余过程分别为

(2.4)$\begin{equation} {\rm d} X_{I}(t)=(1+\theta)a{\rm d}t-q(t)\Big(a{\rm d}t-b{\rm d}W_{0}(t)+d \sum _ { i = 1 } ^ { N ( t ) } Z _ { i }\Big)-\Big(1-q(t)\Big)p(t){\rm d}t \end{equation}$

和

(2.5)$\begin{equation} {\rm d} X_{R}(t)=(1-q(t))p(t){\rm d}t-(1-q(t))\Big(a{\rm d}t-b{\rm d}W_{0}(t)+d \sum _ { i = 1 } ^ { N ( t ) } Z _ { i }\Big). \end{equation}$

2) 金融市场

假定金融市场由一个无风险资产和一个风险资产组成. 此外, 我们假设交易可以在金融市场上在没有交易成本或税收的情况下持续进行, 并且所有的金融资产都是无限可分割的. 无风险资产 $S_{0}(t)$ 的价格过程为 ${\rm d}S_{0}(t)=rS_{0}(t){\rm d}t,\, S_{0}(0)=1,$这里 $r$ 为正常数表示无风险资产的利率. 我们假定风险资产 $S(t)$ 的价格过程遵循 Heston 随机波动率模型, $S(t)$ 可以表示为

(2.6)$\begin{equation} \begin{cases} {\rm d}S(t)=S(t)[(r+\alpha L(t)){\rm d}t+\sqrt{L(t)}{\rm d}W_{S}(t)],\quad S(0)=s_{0}>0,\\ {\rm d}L(t)=\beta(\delta-L(t)){\rm d}t+\sigma \sqrt{L(t)}{\rm d}W_{L}(t),\quad L(0)=l_{0}>0, \end{cases} \end{equation}$

其中, $\alpha, \beta, \delta, \sigma$ 都是正常数, $\alpha$ 代表风险资产的预期收益率, $\beta$ 代表 $L(t)$ 的均值回复率, $\delta$ 表示 $L(t)$ 的长期均值, $\sigma$ 表示 $L(t)$ 的波动率. $L(t)$ 为平方根扩散过程, $W_{S}(t)$ 和 $W_{L}(t)$ 为标准的布朗运动并且 $E[W_{S}(t)W_{L}(t)]=\rho t$. 此外, 假设 $W_{0}(t)$ 和 $W_{S}(t)$、$W_{L}(t)$ 是独立的. 同时假设 $2\beta \delta>\sigma^{2}$, 这样可以保证 $L(t)$ 是非负的.

3) 财富过程

保险公司和再保险公司可以投资于风险资产, 用 $\pi_{I}(t)$ 和 $\pi_{R}(t)$ 分别表示保险公司和再保险公司在 $t$ 时刻投资于风险资产的金额. 保险公司在策略 $(q(t),\ \pi_{I}(t))$ 下的财富过程为

(2.7)$\begin{equation} \begin{aligned} {\rm d}X(t)&=(1+\theta)a{\rm d}t-q(t)\Big(a{\rm d}t-b{\rm d}W_{0}(t)+d \sum _ { i = 1 } ^ { N ( t ) } Z _ { i }\Big)-(1-q(t))p(t){\rm d}t\\ & +\pi_{I}(t)\Big[(r+\alpha L(t)){\rm d}t+\sqrt{L(t)}{\rm d}W_{S}(t)\Big]+r\Big(X(t)-\pi_{I}(t)\Big){\rm d}t. \end{aligned} \end{equation}$

同时, 再保险公司在策略 $(p(t), \pi_{R}(t))$ 的盈余过程为

(2.8)$\begin{equation} \begin{aligned} {\rm d}Y(t)&=(1-q(t))p(t) {\rm d}t-(1-q(t))\Big(a{\rm d}t-b{\rm d}W_{0}(t)+d \sum _ { i = 1 } ^ { N ( t ) } Z _ { i }\Big)\\ & +\pi_{R}(t)\Big[(r+\alpha L(t)){\rm d}t+\sqrt{L(t)}{\rm d}W_{S}(t)\Big]+r\Big(Y(t)-\pi_{R}(t)\Big){\rm d}t. \end{aligned} \end{equation}$

4) Stackelberg 微分博弈

目前, 保险公司和再保险公司在保险市场上的地位并不平等, 因为在保险市场上保险公司的数量很多, 保险公司之间竞争很激烈, 而再保险公司的数量却很少, 再保险业务被少量再保险公司所垄断. 因此, 在保险市场中有一个领导者. 在下面, 我们将再保险公司和保险公司分别作为博弈的领导者和追随者. 根据文献[20]和[23], 我们的目标是依次求解优化问题. 更具体地来说, 我们可以将求解 Stackelberg 随机微分优化问题分为以下三个步骤

步骤一 再保险公司作为领导者, 首先给定任意的策略 $(p(\cdot),\ \pi_{R}(\cdot))$;

步骤二 在了解到再保险公司的策略后, 保险公司根据再保险公司的策略解决其优化问题, 找到最优的策略 $q^{\ast}(\cdot)=f^{\ast}(\cdot,\ p(\cdot),\ \pi_{R}(\cdot)), \pi^{\ast}_{I}(\cdot)=g^{\ast}(\cdot,\ p(\cdot),\ \pi_{R}(\cdot))$;

步骤三 知道保险公司的执行策略 $f^{\ast}(\cdot,\ p(\cdot),\ \pi_{R}(\cdot))$ 和 $g^{\ast}(\cdot,\ p(\cdot),\ \pi_{R}(\cdot))$ 后, 再保险公司通过求解其优化问题得到最优的可容许策略$(p^{\ast}(\cdot),\ \pi_{R}^{\ast}(\cdot)).$

注2.1 博弈的双方 $($即保险公司和再保险公司$)$ 可以清楚地观察彼此在保险市场上的策略. 事实上, 他们更加专注于实际策略的形式而不是它是如何产生的. 因此, 我们可以将再保险公司的策略 $(p(\cdot),\ \pi_{R}(\cdot))$ 带入到保险公司的 HJB 方程中. 同样, 我们可以将保险公司的策略 $(q(\cdot),\ \pi_{I}(\cdot))$ 带入到再保险公司的HJB方程中, 然后我们可以求解优化问题.

定义2.1 策略 $u_{I}(\cdot)=(q(\cdot),\ \pi_{I}(\cdot))$ 和 $u_{R}(\cdot)=(p(\cdot),\ \pi_{R}(\cdot))$ 称为可容许策略, 如果它满足下列条件

(1) $u_{I}(\cdot)$和$u_{R}(\cdot)$是 $\mathcal F_{t} \}_{t\in [T]}$ 循序可测的;

(2) 对于任意的 $t\in[T],\ q(t)\in[0, 1],\ p(t)\in[c,\bar{c}];$

(3) 对于任意的 $t\in[T],$ 满足 $E_{t,x,l}\left[\int_{t}^{T}(q^{4}(s)+\pi_{I}^{4}(s)){\rm d}s\right]<+\infty,$ 并且$E_{t,y,l}\left[\int_{t}^{T}(p^{4}(s)+\pi_{R}^{4}(s)){\rm d}s\right]<+\infty;$

(4) 对于任意的 $(t,x)\in[T]\times R,\ (t,y)\in[T]\times R$, 方程 (2.7) 和 (2.8) 有唯一的强解.

令 $U:=U_{I}\times U_{R}$ 表示所有的可容许策略, $U_{I}$ 和 $U_{R}$ 分别表示保险公司和再保险公司的所有可容许策略.

类似文献[23], 作为 Stackelberg 博弈中的追随者, 保险公司的财富远低于再保险公司. 由于比较心理, 保险公司既关心自己的最终财富, 也关心自己与再保险公司之间的财富差距, 即保险人的目标是寻找一种再保险投资策略, 使下面定义的均值-方差成本函数在其相对绩效下最大化. 这里我们将保险公司的相对绩效定义为 $ \widehat{X}(t):=(1-k)X(t)+k(X(t)-Y(t))=X(t)-kY(t),$这里常数 $k\in[0, 1]$ 表示保险公司对再保险公司绩效的敏感性, $k$ 越大表示保险公司越在意与再保险公司的财富差距, 保险公司越有竞争力. 当 $k=0$ 时, 最优问题就变为没有竞争的传统的优化问题.

保险公司在策略 $(q(t),\ \pi_{I}(t))$ 下的财富过程为

(2.9)$\begin{equation} \begin{aligned} d\widehat{X}(t)&=\Big[r\widehat{X}(t)+\alpha L(t)(\pi_{I}(t)-k\pi_{R}(t))+\theta a-(1+k)(1-q(t))(p(t)-a)\Big]{\rm d}t\\ & +b\Big(q(t)-k+kq(t)\Big){\rm d}W_{0}(t)+\Big[\pi_{I}(t)-k\pi_{R}(t)\Big]{\rm d}W_{S}(t)\\ & -\Big[kq(t)+q(t)-k\Big]d\sum_{i=1}^{N(t)}Z_{i}, \end{aligned} \end{equation}$

初始条件 $ \widehat{X}(0)= \widehat{x}_{0}=x_{0}-ky_{0}.$

在本文, 我们假定保险公司和再保险公司的目标是最大化终端财富效用, 其中效用函数均为均值- 方差准则. 记 $E_{t,\widehat{x},l}[\cdot]=E[\cdot|\widehat{X}(t)=\widehat{x}, L(t)=l]$, $Var_{t,\widehat{x},l}[\cdot]=Var[\cdot|\widehat{X}(t)=\widehat{x}, L(t)=l]$. 那么在均值-方差框架下, 保险公司和再保险公司的 Stackelberg 微分博弈有以下表述.

定义2.2 保险公司的优化问题为

(2.10)$\begin{equation} \begin{cases} \underset{u_{I}(\cdot)\in U_{I}}{\rm{sup}}J_{I}(t,\widehat{x},l ;u_{I}(\cdot),u_{R}(\cdot)):=\underset{u_{I}(\cdot)\in U_{I}}{\rm{sup}}\Big\{E_{t,\widehat{x},l}[\widehat{X}(T)]-\frac{\gamma_{1}}{2}Var_{t,\widehat{x},l}[\widehat{X}(T)]\Big\}\\ \mbox{对于任意的} u_{R}(\cdot)\in U_{R},\quad(\widehat{x},l;\ u_{I}(\cdot),\ u_{R}(\cdot)) \mbox{满足} (2.9)\mbox{式}, \end{cases} \end{equation}$

再保险公司的优化问题为

(2.11)$\begin{equation} \begin{cases} \underset{u_{R}(\cdot)\in U_{R}}{\rm{sup}}J_{R}(t,y,l ;u_{I}^{\ast}(\cdot,u_{R}(\cdot)),u_{R}(\cdot)):=\underset{u_{R}(\cdot)\in U_{R}}{\rm{sup}}\Big\{E_{t,y,l}[Y(T)]-\frac{\gamma_{2}}{2}Var_{t,y,l}[Y(T)]\Big\}\\ \mbox{对于任意的} u_{I}^{\ast}(\cdot,u_{R}(\cdot)),\quad (y,l;\ u_{I}^{\ast}(\cdot,\ u_{R}(\cdot)),\ u_{R}(\cdot)) \mbox{满足} (2.8) \mbox{式}, \end{cases} \end{equation}$

其中正常数 $\gamma_{1}$ 和 $\gamma_{2}$ 分别表示保险公司和再保险公司的风险厌恶系数, $u_{I}^{\ast}(\cdot,u_{R}(\cdot))$为 (2.10) 式中求得的最优策略.

众所周知, 均值-方差准则下的再保险投资问题是时间不一致的. 按照文献[29]的方法, 我们寻求问题 (2.10) 和 (2.11) 的均衡策略和均衡值函数, 定义如下

定义2.3 对任意的再保险公司的策略 $u_{R}(\cdot)\in U_{R}$, 考虑保险公司的一个可容许策略 $u^{\ast}_{I}(\cdot,\ u_{R}(\cdot))\in U_{I}$, 该策略可以非正式地视为保险公司的候选均衡策略. 任意选择$\widehat{u}_{I}(\cdot, u_{R}(\cdot))\in U_{I}$, 任意的正实数 $\epsilon$ 和任意的初始点 $(t,\ \widehat{x},\ l)\in [T]\times R\times R^{+}$, 我们定义策略 $u^{\epsilon}_{I}(\cdot, u_{R}(\cdot))$ 如下

$u_{I}^{\epsilon}(s,\widehat{x},l,u_{R}(\cdot)):= \left\{ \begin{matrix} \widehat{u}_{I}(s,\widehat{x},l,u_{R}(\cdot)),\,\,\,\,\,\quad(s,\widehat{x},l)\in[t,t+ \epsilon)\times R\times R^{+}, \\ u_{I}^{*}(s,\widehat{x},l,u_{R}(\cdot)), \,\,\,\,\,\quad(s,\widehat{x},l)\in[t+ \epsilon, T]\times R\times R^{+}.\\ \end{matrix} \right.$

如果有

$ \lim _{\epsilon \downarrow 0}{\rm{inf}} \frac{J_{I}(t,\widehat{x},l;u_{I}^{*}(\cdot,u_{R}(\cdot)),u_{R}(\cdot))-J_{I}(t,\widehat{x},l;u_{I}^{\epsilon}(\cdot,u_{R}(\cdot)),u_{R}(\cdot))}{\epsilon}\geq 0$

对任意的 $\widehat{u}_{I}(\cdot, u_{R}(\cdot))\in U_{I}$ 都成立, 我们称 $u _ { I } ^ { * } ( \cdot, \ u_ { R }(\cdot))$ 是保险公司的一个均衡策略并且均衡值函数 $V_{I}(t,\widehat{x},l)$ 表示为$V_{I}(t,\widehat{x},l;u_{R}(\cdot)):=J_{I}(t,\widehat{x},l;u_{I}^{*}(\cdot,u_{R}(\cdot)),u_{R}(\cdot)).$

令 $u^{*}_{R}(\cdot)\in U_{R}$ 是再保险公司的策略并且 $u^{*}_{I}(\cdot,\ u_{R}(\cdot))\in U_{I}$ 是上面保险公司的最优策略. 任意选择 $\widehat{u}_{R}(\cdot)\in U_{R}$ 和一个正实数 $\epsilon$ 并且对任意起始点 $(t,y,l)\in [T]\times R\times R^{+}$, 我们定义策略 $u^{\epsilon}_{R}(\cdot)$ 如下

$u_{R}^{\epsilon}(s,y,l):= \left\{ \begin{matrix} \widehat{u}_{R}(s,y,l), \quad(s,y,l)\in[t,t+ \epsilon)\times R\times R^{+}, \\ u_{R}^{*}(s,y,l), \quad(s,y,l)\in[t+ \epsilon,T]\times R\times R^{+}. \\ \end{matrix} \right.$

如果我们有

$ \lim _{\epsilon \downarrow 0}{\rm{inf}} \frac{J_{R}(t,y,l;u_{I}^{*}(\cdot,u_{R}^{*}(\cdot)),u_{R}^{*}(\cdot))-J_{R}(t,y,l;u_{I}^{*}(\cdot,u_{R}^{\epsilon}(\cdot)),u_{R}^{\epsilon}(\cdot))}{\epsilon}\geq 0$

对任意的 $\widehat{u}_{R}(\cdot)\in U_{R}$ 都成立, 我们称 $u^{*}_{R}(\cdot)$ 是再保险公司的一个均衡策略并且均衡值函数 $V_{R}(t,y,l)$ 可以表示为$V_{R}(t,y,l):=J_{R}(t,y,l;u_{I}^{*}(\cdot,u^{*}_{R}(\cdot)),u^{*}_{R}(\cdot)).$

根据定义2.3, 上述均衡策略是时间一致的. 我们的目的是寻求均衡策略 $u^{*}_{I}\times u^{*}_{R}$ 和相应的均衡值函数. 我们有下面的命题.

命题2.1 均值-方差问题 (2.10) 和 (2.11) 的值函数对于他们自己的盈余 (即 $\widehat{x}$ 和 $y$) 是可分离的,并且独立于对手的盈余. 此外, 保险公司和再保险公司的均衡策略都不依赖于状态变量.

根据命题2.1, 我们将保险公司和再保险公司的值函数简化为 $V_{I}(t,\widehat{x},l)$ 和 $V_{R}(x,y,l)$.令 $C^{1,2,2}(\left[ 0,T \right] \times R \times R^{+})$ 表示满足关于 $t$ 一次可微, 关于 $x$ 和 $l$ 二次可微的函数 $\psi(t,x,l)$ 的全体, 其中 $\psi(t,x,l)$ 和 $\psi_{t}(t,x,l),\psi_{x}(t,x,l),\ \psi_{xx}(t,x,l),\ \psi_{l}(t,x,l),\ \psi_{ll}(t,x,l),\ \psi_{xl}(t,x,l)$ 在 $[T]\times R \times R^{+}$ 上是连续的.令 $D_{p}^{1,2,2}(\left[ 0,T \right] \times R \times R^{+})$ 表示 $C^{1,2,2}(\left[ 0,T \right] \times R \times R^{+})$ 中一阶偏导数关于参数 $x$ 和 $l$ 满足多项式增长条件的函数 $\psi(t,x,l)$ 的全体(同文献[30]和[31]). 对于任意的函数 $\psi(t,\widehat{x},l)\in C^{1,2,2}$, 变分算子 $\mathcal{L}^{u_{I},u_{R}}_{1}$ 定义为以下形式

(2.12)$\begin{align*} \mathcal{L}^{u_{I},u_{R}}_{1} \psi(t,\widehat{x},l) =\,&\frac{\partial \psi}{\partial t}+\Big[r\widehat{x}+\alpha l (\pi_{I}(t)-k\pi_{R}(t))+\theta a -(1+k)(1-q(t))(p(t)-a)\Big]\frac{\partial \psi}{\partial \widehat{x}} \\[2mm] &+\frac{1}{2}\Big[(bq(t)-kb+kbq(t))^{2}+(\pi_{I}(t)-k\pi_{R}(t))^{2}l\Big]\frac{\partial ^{2}\psi}{\partial \widehat{x}^{2}} \\[2mm] &+\beta (\delta -l)\frac{\partial \psi}{\partial l}+\frac{1}{2}\sigma^{2}l\frac{\partial ^{2}\psi}{\partial l^{2}}+(\pi_{I}(t)-k\pi_{R}(t))\sigma l \rho \frac{\partial ^{2}\psi}{\partial \widehat{x} \partial l} \\[2mm] &+\lambda_{0}E\Big[\psi(t,\widehat{x}-Z(kq(t)+q(t)-k),l)-\psi(t,\widehat{x},l)\Big]. \end{align*}$

对于任意的函数 $\phi(t,y,l)\in C^{1,2,2}$, 变分算子 $\mathcal{L}^{u_{I},u_{R}}_{2}$ 定义为以下形式

(2.13)$\begin{align*} \mathcal{L}^{u_{I},u_{R}}_{2} \phi(t,y,l) =\,&\frac{\partial \phi}{\partial t}+\Big[ry+\alpha l \pi_{R}(t)+(1-q(t))(p(t)-a)\Big]\frac{\partial \phi}{\partial y} \\ &+\frac{1}{2}\Big[(1-q(t))^{2}b^{2}+l\pi_{R}^{2}(t)\Big]\frac{\partial ^{2}\phi}{\partial {y}^{2}}+\beta (\delta-l)\frac{\partial \phi}{\partial l} \\ &+\frac{1}{2}\sigma^{2}l \frac{\partial^{2} \phi}{\partial {l}^{2}}+\sigma l \rho \pi_{R}(t)\frac{\partial^{2}\phi}{\partial y \partial l}+\lambda_{0}E\Big[\phi(t,y-Z(1-q(t)),l)-\phi(t,y,l)\Big]. \end{align*}$

类似文献[30]和[31], 下面给出假设.

假设2.1 给定多项式增长条件中的常数 $p\geq 1$, 对任意可容许策略$(u_{I}, u_{R})$, 假设对任意的 $(t, \widehat{x})\in [T]\times R$, $(t, y)\in [T]\times R$, 有

$ E_{t,x,l} \left[ \underset{s \in \left[ t, T \right]}{ \rm{sup} } |\widehat{X} (s)|^{ 4p } \right] < + \infty,\,\,\,\,\,\,\,\, E_{t,y,l} \left[ \underset{s \in \left[ t, T \right]}{ \rm{sup} } |Y (s)|^{ 4p } \right] < + \infty.$

下面的定理提供了与均值-方差问题 (2.10) 和 (2.11) 相关的扩展HJB方程的验证定理.

(2.15)定理2.1 [验证定理]对于问题 (2.10), 我们假设存在实值函数 $W_{I}(t,\widehat{x},l), g_{I}(t,\widehat{x},l) \in D_{p}^{1,2,2}([T]\times R \times R^{+})$ 分别满足下面扩展的 HJB 方程 (2.14) 和

(2.14)$\begin{align*} & \underset{u_{I}\in U_{I}}{\rm{max}}\Big\{\mathcal{L}_{1}^{u_{I},u_{R}}W_{I}(t, \widehat{x},l)-\mathcal{L}_{1}^{u_{I},u_{R}}\frac{\gamma_{1}}{2}g_{I}^{2}(t, \widehat{x},l) \\ & +\gamma_{1}g_{I}(t, \widehat{x},l)\times \mathcal{L}_{1}^{u_{I},u_{R}}g_{I}(t, \widehat{x},l) \Big\}=0,\,\,\,0\leq t\leq T, \end{align*}$

(2.15)$\begin{equation} \mathcal{L}_{1}^{u^{*}_{I},u_{R}}g_{I}(t, \widehat{x},l)=0,\ W_{I}(T,\widehat{x},l)=\widehat{x},\ g_{I}(T,\widehat{x},l)=\widehat{x}, \,\,\, 0\leq t\leq T,\end{equation}$

其中

$\begin{equation} \nonumber \begin{aligned} u_{I}^{*}:={\rm{arg}}\underset{u_{I}\in U_{I}}{\rm{max}}\Big\{\mathcal{L}_{1}^{u_{I},u_{R}}W_{I}(t, \widehat{x},l)-\mathcal{L}_{1}^{u_{I},u_{R}}\frac{\gamma_{1}}{2}g_{I}^{2}(t, \widehat{x},l)+\gamma_{1}g_{I}(t, \widehat{x},l)\times \mathcal{L}_{1}^{u_{I},u_{R}}g_{I}(t, \widehat{x},l) \Big\}. \end{aligned} \end{equation}$

那么 $W_{I}(t, \widehat{x},l)=V_{I}(t, \widehat{x},l),\ E_{t, \widehat{x},l}\left[ \widehat{X}^{u_{I}^{*},u_{R}}(T)\right] =g_{I}(t, \widehat{x},l),$ 并且 $u^{*}_{I}$ 是保险公司的均衡策略.

(2.17)(验证定理) 对于问题 $(2.11)$, 我们假设存在实值函数 $ W_{R}(t,y,l),\ g_{R}(t,y,l)\in D_{p}^{1,2,2}([T]\times R \times R^{+})$ 满足下面扩展的 HJB 方程 (2.16) 和

(2.16)$\begin{equation} \begin{aligned} &\underset{u_{R}\in U_{R}}{\rm{max}}\Big\{\mathcal{L}_{2}^{u^{*}_{I},u_{R}}W_{R}(t, y,l)-\mathcal{L}_{2}^{u^{*}_{I},u_{R}}\frac{\gamma_{2}}{2}g_{R}^{2}(t, y,l) \\ &+\gamma_{2}g_{R}(t, y,l)\times \mathcal{L}_{2}^{u^{*}_{I},u_{R}}g_{R}(t, y,l) \Big\}=0,\,\,\, 0\leq t\leq T, \end{aligned} \end{equation}$

(2.17)$\begin{equation} \mathcal{L}_{2}^{u^{*}_{I},u^{*}_{R}}g_{R}(t, y,l)=0,\ W_{R}(T,y,l)=y,\ g_{R}(T,y,l)=y,\,\,\, 0\leq t\leq T, \end{equation}$

其中

$\begin{equation} \nonumber \begin{aligned}u_{R}^{*}:={\rm{arg}} \underset{u_{R}\in U_{R}}{\rm{max}}\Big\{\mathcal{L}_{2}^{u^{*}_{I},u_{R}}W_{R}(t, y,l)-\mathcal{L}_{2}^{u^{*}_{I},u_{R}}\frac{\gamma_{2}}{2}g_{R}^{2}(t, y,l)+\gamma_{2}g_{R}(t, y,l)\times \mathcal{L}_{2}^{u^{*}_{I},u_{R}}g_{R}(t,y,l) \Big\}.\end{aligned} \end{equation}$

那么 $W_{R}(t, y,l)=V_{R}(t, y,l),\ E_{t, y,l}\left[ Y^{u_{I}^{*},u^{*}_{R}}(T)\right] =g_{R}(t,y,l),$ 并且 $u^{*}_{R}$ 是再保险公司的均衡策略.

证因为定理的两部分的证明相似, 下面仅证明第一部分. 假设 $W_{I}(t,\widehat{x},l)$ 和 $g_{I}(t,\widehat{x},l)$ 满足验证定理, HJB 方程中的最优策略在 $u^{*}_{I}$ 处取得, 我们分两步来证明.

步骤一 首先证明以下两个等式

(2.18)$\begin{equation} g_{I}(t,\widehat{x},l)=E_{t,\widehat{x},l}\left[ \widehat{X}^{u^{*}_{I},u_{R}}(T)\right],\\ \ W_{I}(t,\widehat{x},l)=V_{I}(t,\widehat{x},l). \end{equation}$

根据验证定理的条件

$\mathcal{L}_{1}^{u^{*}_{I},u_{R}}g_{I}(t, \widehat{x},l)=0,\ g_{I}(T,\widehat{x},l)=\widehat{x}. $

类似文献[31] 可得, 积分

$\begin{align*} & \int _{t}^{T}\frac{\partial g_{I}(s,\widehat{X}(s),l(s))}{\partial \widehat{x}}(\pi_{I}(s)-k\pi_{R}(s)){\rm d}W_{s}(s), \\ & \int _{t}^{T}\frac{\partial g_{I}(s,\widehat{X}(s),l(s))}{\partial \widehat{x}}b(q(t)-k+kq(t)){\rm d}W_{0}(s), \\ & \int _{t}^{T}\int_{0}^{\infty}\left[g_{I}(s,\widehat{X}(s)-z(kq(t)+q(t)-k),l(s))-g_{I}(s,\widehat{X}(s),l(s))\right]\widetilde{N} ({\rm d}s, {\rm d}z) \end{align*}$

为鞅, 其中 $\widetilde{N}({\rm d}s, {\rm d}z)=N({\rm d}s, {\rm d}z)-\nu({\rm d}z){\rm d}s$ 是泊松测度 $N({\rm d}s, {\rm d}z)$ 的补偿测度. 由 Ito 公式可得

$\begin{equation}\notag \begin{aligned} E_{t,\widehat{x},l}\left[ g_{I}(T,\widehat{X}^{u^{*}_{I},u_{R}}(T),l(T))\right] &=g_{I}(t,\widehat{x},l)+E_{t,\widehat{x},l}\left[\int _{t}^{T}\mathcal{L}_{1}^{u^{*}_{I},u_{R}}g_{I}(s,\widehat{X}(s),l(s)){\rm d}t\right]=g_{I}(t,\widehat{x},l). \end{aligned} \end{equation}$

进一步, 我们有

(2.19)$\begin{equation} g_{I}(t,\widehat{x},l)=E_{t,\widehat{x},l}\left[ g_{I}(T,\widehat{X}^{u_{I}^{*},u_{R}}(T),l(T))\right] =E_{t,\widehat{x},l}\left[ \widehat{X}^{u^{*}_{I},u_{R}}(T)\right]. \end{equation}$

接下来, 证明 $W_{I}(t,\widehat{x},l)=V_{I}(t,\widehat{x},l)$. 最优策略在 (2.14) 式左侧取得, 同时与 (2.15) 式结合起来可得

(2.20)$\begin{equation} \mathcal{L}_{1}^{u_{I}^{*},u_{R}}W_{I}(t, \widehat{x},l)-\mathcal{L}_{1}^{u_{I}^{*},u_{R}}\frac{\gamma_{1}}{2}g_{I}^{2}(t, \widehat{x},l)=0. \end{equation}$

根据条件 $W_{I}(T,\widehat{x},l)=\widehat{x}$, 再次利用 Ito 公式, 及积分

$\begin{equation}\notag \begin{aligned} & \int _{t}^{T}\frac{\partial W_{I}(s,\widehat{X}(s),l(s))}{\partial \widehat{x}}(\pi_{I}(s)-k\pi_{R}(s)){\rm d}W_{s}(s), \\[2mm] & \int _{t}^{T}\frac{\partial W_{I}(s,\widehat{X}(s),l(s))}{\partial \widehat{x}}b(q(t)-k+kq(t)){\rm d}W_{0}(s), \\[2mm] & \int _{t}^{T}\int_{0}^{\infty}\left[W_{I}(s,\widehat{X}(s)-z(kq(t)+q(t)-k),l(s))-W_{I}(s,\widehat{X}(s),l(s))\right]\widetilde{N} ({\rm d}s, {\rm d}z) \end{aligned} \end{equation}$

为鞅, 可得

(2.21)$\begin{equation} \begin{aligned} E_{t, \widehat{x},l}\left[\widehat{X}^{u^{*}_{I},u_{R}}(T)\right] &=E_{t,\widehat{x},l}\left[ W_{I}(T, \widehat{X}^{u_{I}^{*},u_{R}}(T),l(T))\right] \\ &=W_{I}(t, \widehat{x},l)+E_{t,\widehat{x},l}\left[ \int _{t}^{T}\mathcal{L}_{1}^{u^{*}_{I},u_{R}}W_{I}(s, \widehat{X}^{u^{*}_{I}}(s),l(s)){\rm d}s \right]. \end{aligned} \end{equation}$

把 (2.20) 式代入到 (2.21) 式可得

(2.22)$\begin{equation} W_{I}(t, \widehat{x},l)=E_{t, \widehat{x},l}\left[\widehat{X}^{u^{*}_{I},u_{R}}(T)\right]- \frac{\gamma_{1}}{2}E_{t,\widehat{x},l}\left[\int_{t}^{T}\mathcal{L}_{1}^{u^{*}_{I},u_{R}}g^{2}_{I}(s,\widehat{X}^{u^{*},u_{R}}(s),l(s)){\rm d}s\right]. \end{equation}$

此外, 根据终端条件, Ito 公式和 (2.19) 式, 可得

(2.23)$\begin{align*} & E_{t, \widehat{x},l}\left[(\widehat{X}^{u_{I}^{*},u_{R}}(T))^{2}\right] =E_{t,\widehat{x},l}\left[ g_{I}^{2}(T, \widehat{X}^{u_{I}^{*},u_{R}}(T),l(T))\right] \\ &=g_{I}^{2}(t, \widehat{x},l)+E_{t,\widehat{x},l}\left[ \int _{t}^{T}\mathcal{L}_{1}^{u_{I}^{*},u_{R}}g_{I}^{2}(s, \widehat{X}^{u_{I}^{*},u_{R}}(s),l(s)){\rm d}s\right] \\ &=\left(E_{t, \widehat{x},l}\left[ \widehat{X}^{u_{I}^{*},u_{R}}(T)\right]\right)^{2}+E_{t, \widehat{x},l}\left[ \int _{t}^{T}\mathcal{L}_{1}^{u_{I}^{*},u_{R}}g_{I}^{2}(s, \widehat{X}^{u^{*}_{I},u_{R}},l(s)){\rm d}s \right]. \end{align*}$

进一步可得

(2.24)$\begin{equation} Var_{t,\widehat{x},l} \left[ \widehat{X}^{u_{I}^{*},u_{R}}(T)\right]=E_{t,\widehat{x},l}\left[ \int _{t}^{T}\mathcal{L}_{1}^{u^{*}_{I},u_{R}}g^{2}_{I}(s, \widehat{X}^{u_{I}^{*},u_{R}}(s),l(s)){\rm d}s\right]. \end{equation}$

最后, 把 (2.24) 式代入到 (2.22) 式可得

(2.25)$\begin{equation} W_{I}(t,\widehat{x},l) =E_{t,\widehat{x},l}[\widehat{X}^{u^{*}_{I},u_{R}}(T)]-\frac{\gamma_{1}}{2}Var_{t,\widehat{x},l}[\widehat{X}^{u^{*}_{I},u_{R}}(T)]=J_{I}^{u^{*}_{I},u_{R}}(t,\widehat{x},l)=V_{I}(t,\widehat{x},l). \end{equation}$

步骤二 我们证明 $u^{*}_{I}$ 是本文定义2.3中的均衡策略, 对于$J_{I}^{u^{\epsilon}_{I},u_{R}}(t,\widehat{x},l)=E_{t,\widehat{x},l}[\widehat{X}^{u^{\epsilon}_{I},u_{R}}(T)]-\frac{\gamma_{1}}{2}Var_{t,\widehat{x},l}[\widehat{X}^{u^{\epsilon}_{I},u_{R}}(T)]$,其中 $u^{\epsilon}_{I}$ 是定义 2.3 中的策略, 可以得到

(2.26)$\begin{align*} J_{I}^{u^{\epsilon}_{I},u_{R}}(t, \widehat{x},l) &=E_{t, \widehat{x},l}\left[ \widehat{X}^{u^{\epsilon}_{I},u_{R}}(T)\right] - \frac{\gamma_{1}}{2}Var_{t,\widehat{x},l}\left[ \widehat{X}^{u_{I}^{\epsilon},u_{R}}(T)\right] \\ &=E_{t, \widehat{x},l}\left[ \widehat{X}^{u_{I}^{\epsilon},u_{R}}(T)- \frac{\gamma_{1}}{2}(X^{u^{\epsilon}_{I},u_{R}}(T))^{2}\right] + \frac{\gamma_{1}}{2}\left(E_{t,\widehat{x},l}\left[ \widehat{X}^{u^{\epsilon}_{I},u_{R}}_{I}(T)\right]\right)^{2} \\ &=E_{t,\widehat{x},l}\left[E_{t+\epsilon,\widehat{X}^{\widehat{u}_{I},u_{R}}(t+\epsilon),l(t+\epsilon)}\left[ \widehat{X}^{u_{I}^{*},u_{R}}(T)- \frac{\gamma_{1}}{2}\left(X^{u_{I}^{*},u_{R}}(T)\right)^{2}\right]\right] \\ &\,\,\,\,\,\,\,+\frac{\gamma_{1}}{2}\left(E_{t, \widehat{x},l} \left[E_{t+\epsilon,\widehat{X}^{\widehat{u}_{I},u_{R}}(t+\epsilon),l(t+\epsilon)}\left[\widehat{X}^{u^{*}_{I},u_{R}}(T)\right]\right]\right)^{2} \\ &=E_{t, \widehat{x},l}\left[J^{{u}^{*}_{I},u_{R}}_{I}\left( t+\epsilon,\widehat{X}^{\widehat{u}_{I},u_{R}}(t+\epsilon),l(t+\epsilon) \right)\right] \\ &\,\,\,\,\,\,\,-\frac{\gamma_{1}}{2}E_{t, \widehat{x},l}\left[ \left( E_{t+\epsilon,\widehat{X}^{\widehat{u}_{I},u_{R}}(t+\epsilon),l(t+\epsilon)}\left[\widehat{X}^{u^{*}_{I},u_{R}}(T) \right] \right)^{2}\right] \\ &\,\,\,\,\,\,\,+\frac{\gamma_{1}}{2}\left(E_{t, \widehat{x},l}\left[ E_{t+\epsilon,\widehat{X}^{\widehat{u}_{I},u_{R}}(t+\epsilon),l(t+\epsilon)}\left[\widehat{X}^{u^{*}_{I},u_{R}}(T) \right] \right]\right)^{2} \\ &=E_{t, \widehat{x},l}\left[ J_{I}^{u_{I}^{*},u_{R}}\left(t+ \epsilon, \widehat{X}^{\widehat{u}_{I},u_{R}}(t+ \epsilon),l(t+\epsilon)\right)\right] \\ &\,\,\,\,\,\,\,- \frac{\gamma_{1}}{2}E_{t, \widehat{x},l}\left[ g_{I}^{2}\left(t+ \epsilon, \widehat{X}^{\widehat{u}_{I},u_{R}}(t+ \epsilon),l(t+\epsilon)\right)\right] \\ &\,\,\,\,\,\,\,+\frac{\gamma_{1}}{2}\left(E_{t, \widehat{x},l}\left[ g_{I}\left(t+ \epsilon, \widehat{X}^{\widehat{u}_{I},u_{R}}(t+ \epsilon),l(t+\epsilon)\right)\right]\right)^{2}. \end{align*}$

然后我们把 $g^{2}_{I}(t,\widehat{x},l)$ 和 $W_{I}(t,\widehat{x},l)$ 代入到 (2.26) 式并且利用第一步的结论, 可得

(2.27)$\begin{equation} \begin{aligned} J_{I}^{u_{I}^{\epsilon},u_{R}}(t, \widehat{x},l) &=E_{t, \widehat{x},l}\left[ W_{I}^{u_{I}^{*},u_{R}}\left(t+ \epsilon, \widehat{X}^{\widehat{u}_{I},u_{R}}(t+ \epsilon),l(t+\epsilon)\right)\right]\\ &\,\,\,\,\,\,\,- \frac{\gamma_{1}}{2}\left\{ E_{t, \widehat{x},l }\left[ g_{I}^{2}(t+ \epsilon, \widehat{X}^{\widehat{u}_{I},u_{R}}(t+ \epsilon),l(t+\epsilon) )\right] -g^{2}_{I}(t, \widehat{x},l )\right\}\\ &\,\,\,\,\,\,\,+ \frac{\gamma_{1}}{2}\left\{\left(E_{t,\widehat{x},l}\left[ g_{I}\left(t+ \epsilon, \widehat{X}^{\widehat{u}_{1},u_{R}}(t+ \epsilon),l(t+\epsilon)\right)\right]\right)^{2}-g^{2}_{I}(t, \widehat{x},l)\right\}. \end{aligned} \end{equation}$

对于任意的 $u_{I}\in U_{I}$, 足够小的 $\epsilon>0$ 和 $W_{I}(t,\widehat{x},l)\in D_{p}^{1,2,2}([T]\times R \times R^{+})$, 我们可以定义一个算子

(2.28)$\begin{equation} \mathcal{L}_{1 \epsilon}^{u_{I},u_{R}}W_{I}(t, \widehat{x},l):=E_{t,\widehat{x},l}\left[ W_{I}\left(t+ \epsilon, \widehat{X}^{u_{I},u_{R}}(t+ \epsilon),l(t+\epsilon)\right)\right] -W_{I}(t, \widehat{x},l). \end{equation}$

对于任意的 $\epsilon>0$ 有

(2.29)$\begin{equation} \mathcal{L}_{1}^{\widehat{u}_{I},u_{R}}W_{I}(t, \widehat{x},l)= \lim _{\epsilon \downarrow 0}\frac{\mathcal{L} _{1\epsilon}^{\widehat{u}_{I},u_{R}}W_{I}(t, \widehat{x},l)}{\epsilon}. \end{equation}$

根据 (2.28), (2.27) 式变为

(2.30)$\begin{equation} \begin{aligned} J_{I}^{u_{I}^{\epsilon},u_{R}}(t, \widehat{x},l) &=W_{I}(t,\widehat{x},l)+\mathcal{L}^{\widehat{u}_{I},u_{R}}_{1\epsilon}W_{I}(t,\widehat{x},l)-\frac{\gamma_{1}}{2}\mathcal{L}^{\widehat{u}_{I},u_{R}}_{1\epsilon}g^{2}_{I}(t,\widehat{x},l)\\ &\,\,\,\,\,\,\,+ \frac{\gamma_{1}}{2}\left\{\left(E_{t,\widehat{x},l}\left[ g_{I}\left(t+ \epsilon, \widehat{X}^{\widehat{u}_{1},u_{R}}(t+ \epsilon),l(t+\epsilon)\right)\right]\right)^{2}-g^{2}_{I}(t, \widehat{x},l)\right\}. \end{aligned} \end{equation}$

同样地利用 Ito 公式, 可得

(2.31)$\begin{equation} \begin{aligned} & E_{t, \widehat{x},l}\left[ g_{I}\left(t+ \epsilon, \widehat{X}^{\widehat{u}_{I},u_{R}}(t+ \epsilon),l(t+\epsilon)\right)\right]\\ &=g_{I}(t, \widehat{x},l)+E_{t,\widehat{x},l} \left[ \int _{t}^{t+ \epsilon}\mathcal{L}_{1}^{\widehat{u}_{I},u_{R}}g_{I}\left(s, \widehat{X}^{\widehat{u}_{I},u_{R}}(s),l(t+\epsilon)\right){\rm d}s \right], \end{aligned} \end{equation}$

这表明

(2.32)$\begin{equation} \begin{aligned} & \left\{ E_{t, \widehat{x},l}\left[ g_{I}\left(t+ \epsilon, \widehat{X}^{\widehat{u}_{I},u_{R}}(t+ \epsilon),l(t+\epsilon)\right)\right] \right\} ^{2}-g^{2}(t, \widehat{x},l )\\ &=2g_{I}(t, \widehat{x},l)E_{t, \widehat{x},l}\left[ \int _{t}^{t+ \epsilon}\mathcal{L}_{1}^{\widehat{u}_{I},u_{R}}g_{I}\left(s, \widehat{X}^{\widehat{u}_{I},u_{R}}(s),l(s)\right){\rm d}s\right]\\ &\,\,\,\,\,\,\,+ \left\{ E_{t, \widehat{x},l}\left[ \int _{t}^{t+ \epsilon}\mathcal{L}_{1}^{\widehat{u}_{I},u_{R}}g_{I}\left(s, \widehat{X}^{\widehat{u}_{I},u_{R}}(s),l(s)\right){\rm d}s \right] \right\} ^{2}. \end{aligned} \end{equation}$

将 (2.32) 式代入到 (2.30) 式中, 可得

(2.33)$\begin{equation} \begin{aligned} J_{I}^{u_{I}^{\epsilon},u_{R}}(t, \widehat{x},l) &=W_{I}(t,\widehat{x},l)+\mathcal{L}^{\widehat{u}_{I},u_{R}}_{1\epsilon}W_{I}(t,\widehat{x},l)-\frac{\gamma_{1}}{2}\mathcal{L}^{\widehat{u}_{I},u_{R}}_{1\epsilon}g^{2}_{I}(t,\widehat{x},l)\\ &\,\,\,\,\,\,\,+\gamma_{1}g_{I}(t, \widehat{x})E_{t, \widehat{x}}\left[ \int _{t}^{t+ \epsilon}\mathcal{L}_{1}^{\widehat{u}_{I},u_{R}}g_{I}\left(s, \widehat{X}^{\widehat{u}_{I},u_{R}}(s)\right){\rm d}s\right]\\ &\,\,\,\,\,\,\,+\frac{\gamma_{1}}{2}\left\{ E_{t, \widehat{x},l}\left[ \int _{t}^{t+ \epsilon}\mathcal{L}_{1}^{\widehat{u}_{I},u_{R}}g_{I}\left(s, \widehat{X}^{\widehat{u}_{I},u_{R}}(s),l(s)\right){\rm d}s \right] \right\} ^{2}. \end{aligned} \end{equation}$

由 (2.14) 式中的扩展 HJB 方程, 可得

$\begin{equation} \mathcal{L}_{1}^{\widehat{u}_{I},u_{R}}W_{I}(t, \widehat{x},l)-\mathcal{L}_{1}^{\widehat{u}_{I},u_{R}}\frac{\gamma_{1}}{2}g_{I}^{2}(t, \widehat{x},l)+\gamma_{1}g_{I}(t, \widehat{x},l)\mathcal{L}_{1}^{\widehat{u}_{I},u_{R}}g_{I}(t, \widehat{x},l)\leq 0.\nonumber \end{equation}$

结合 (2.29) 式, 可得

$\begin{equation} \begin{aligned} &\mathcal{L}_{1\epsilon}^{\widehat{u}_{I},u_{R}}W_{I}(t, \widehat{x},l)-\mathcal{L}_{1\epsilon}^{\widehat{u}_{I},u_{R}}\frac{\gamma_{1}}{2}g_{I}^{2}(t, \widehat{x},l)+\gamma_{1}g_{I}(t,\widehat{x},l)\\ &\times E_{t,\widehat{x},l}\left[\int_{t}^{t+\epsilon}\mathcal{L}^{\widehat{u}_{I},u_{R}}_{1}g_{I}\left(s,\widehat{X}^{\widehat{u}_{I},u_{R}}(s),l(s) \right)\right]\leq o(\epsilon). \end{aligned}\nonumber \end{equation}$

把上式代入到 (2.33) 式可得

$\begin{equation} \begin{aligned} J_{I}^{u_{I}^{\epsilon},u_{R}}(t, \widehat{x},l) &\leq W_{I}(t, \widehat{x},l)+ \frac{\gamma_{1}}{2}\left\{ E_{t, \widehat{x},l}\left[ \int _{t}^{t+ \epsilon}\mathcal{L}_{1}^{u_{I}^{\epsilon},u_{R}}g_{I}\left(s, \widehat{X}^{\widehat{u}_{I},u_{R}}(s),l(s)\right){\rm d}s \right] \right\}^{2}+o(\epsilon)\\ &=J_{I}^{u^{*}_{I},u_{R}}(t, \widehat{x}, l )+ \frac{\gamma_{1}}{2}\left\{ E_{t, \widehat{x},l }\left[ \int _{t}^{t+ \epsilon}\mathcal{L}_{1}^{\widehat{u}_{I},u_{R}}g_{I}\left(s, \widehat{X}^{\widehat{u}_{I},u_{R}}(s),l(s)\right){\rm d}s \right] \right\}^{2}+o(\epsilon).\nonumber \end{aligned} \end{equation}$

由此便可得到结论

$\begin{equation} \lim _{\epsilon \downarrow 0}{\rm{inf}} \frac{J_{I}^{u^{*}_{I},u_{R}}(t, \widehat{x},l)-J_{I}^{u_{I}^{\epsilon},u_{R}}(t, \widehat{x},l)}{\epsilon}\geq 0. \nonumber\end{equation}$

因此, $u^{*}_{I}$ 是一个均衡的再保险-投资策略. 再保险公司的策略的证明和保险公司的一样, 这里就省略了证明.

3 主要结果

我们依次求解问题的纳什均衡策略和相应的值函数, 为了简化计算和表达, 我们记

(3.1)$\begin{equation} \begin{aligned} &\Theta= \frac{\alpha ^{4}\sigma ^{2}(1- \rho ^{2})}{2\gamma_{1}(\beta + \alpha \rho \sigma)^{2}},\quad \Phi= \frac{\alpha ^{3}\rho \sigma}{\gamma_{1}(\beta + \alpha \rho \sigma)},\quad \Psi= \frac{\alpha ^{2}}{2\gamma_{1}}-(\Theta + \Phi),\\ &\xi= \frac{2 \Theta + \Phi}{\alpha \rho \sigma}- \frac{\Theta}{\beta +2 \alpha \rho \sigma}- \frac{\Psi}{\beta},\quad H=\frac{\gamma_{1}(1+k)+\gamma_{2}}{2\gamma_{1}(1+k)+\gamma_{2}},\\ &H^{\theta}=\frac{\theta \lambda_{0}\mu_{z}} {\gamma_{1}(b^{2}+\lambda_{0} \sigma^{2}_{z}) } {\rm e}^{r(t-T)},\quad H^{\bar{\theta}}=\frac{\bar{\theta}\lambda_{0}\mu_{z}}{\gamma_{1}(b^{2}+\lambda_{0}\sigma^{2}_{z})} {\rm e}^{r(t-T)}. \end{aligned} \end{equation}$

步骤一 在 Stackelberg 博弈中, 再保险公司 (即领导者) 首先宣布其任何可接受的策略 $(p(\cdot),\ \pi_{R}(\cdot)) \in U_{R};$

步骤二 在观测到再保险公司的策略后, 保险公司 (即追随者) 在均值方差准则下解决优化问题 (2.10). 首先利用变分算子 (2.12), 我们可以把等式 (2.14) 简化为

(3.2)$\begin{equation} \begin{aligned} &\underset{u_{I}\in U_{I}}{\rm{max}} \left \{\frac{\partial W_{I}}{\partial t}+\Big[r\widehat{x}+\alpha l (\pi_{I}(t)-k\pi_{R}(t))+\theta a -(1+k)(1-q(t))\right.\\ &\left.(p(t)-a)\Big]\frac{\partial W_{I}}{\partial \widehat{x}}+\frac{1}{2}\Big[(bq(t)-kb+kbq(t))^{2}+(\pi_{I}(t)-k\pi_{R}(t))^{2}l\Big]\frac{\partial ^{2} W_{I}} {\partial \widehat{x}^{2}}\right.\\ &\left.+\beta(\delta-l)\frac{\partial W_{I}}{\partial l}+\frac{1}{2}\sigma^{2}l\frac{\partial^{2} W_{I} }{\partial {l}^{2}}+(\pi_{I}(t)-k\pi_{R}(t))\sigma l \rho \frac{\partial^{2}W_{I}}{\partial \widehat{x} \partial l}\right.\\ &\left.+\lambda_{0}E\Big[ W_{I} (t,\widehat{x}-z(kq(t)+q(t)-k),l)-W_{I}(t,\widehat{x},l)\Big]\right.\\ &\left.-\frac{\gamma_{1}}{2}\Big[(bq(t)-kb+kbq(t))^{2}+(\pi_{I}(t)-k\pi_{R}(t))^{2}l\Big]\left(\frac{\partial g_{I}}{\partial \widehat{x}}\right)^{2}\right.\\ &\left.-\frac{\gamma_{1}}{2}\sigma^{2}l\left(\frac{\partial g_{I}}{\partial l}\right)^{2}-\gamma_{1} (\pi_{I}(t)-k\pi_{R}(t) )\sigma l \rho \left(\frac{\partial g_{I}}{\partial \widehat{x}}\right)\left(\frac{\partial g_{I}}{\partial l}\right)\right.\\ &\left.-\frac{\gamma_{1}}{2}\lambda_{0}E\Big[g_{I}(t,\widehat{x}-z(kq(t)+q(t)-k),l)-g_{I}(t,\widehat{x},l)\Big]^{2} \right\}=0.\\ \end{aligned} \end{equation}$

为了求解 (2.15) 和 (3.2) 式, 我们推测保险公司的解有如下形式

$W_{I}(t,\widehat{x},l)=A_{1}(t)\widehat{x}+B_{1}(t)l+C_{1}(t),\quad A_{1}(T)=1,\quad B_{1}(T)=C_{1}(T)=0,$

$g_{I}(t,\widehat{x},l)=A_{2}(t)\widehat{x}+B_{2}(t)l+C_{2}(t),\quad A_{2}(T)=1,\quad B_{2}(T)=C_{2}(T)=0.$

根据简单的计算, 我们可以很容易得到以下偏导数

$\begin{align*} & \frac{\partial W_{I}}{\partial t}=A^{\prime}_{1}(t)\widehat{x} +{B^{\prime}_{1}(t)}l+C^{\prime}_{1}(t),\quad \frac{\partial W_{I}}{\partial \widehat{x}}=A_{1}(t), \quad \frac{\partial W_{I}}{\partial l}=B_{1}(t),\\[2mm] & \frac{\partial g_{I}}{\partial t}=A^{\prime}_{2}(t)\widehat{x} +{B^{\prime}_{2}(t)}l+C^{\prime}_{2}(t),\quad \frac{\partial g_{I}}{\partial \widehat{x}}=A_{2}(t), \quad \frac{\partial g_{I}}{\partial l}=B_{2}(t). \end{align*}$

把上面的结果代入到(3.2)式, 可得

(3.3)$\begin{align*} &\underset{u_{I}\in U_{I}}{\rm{max}} \Big\{A^{\prime}_{1}(t)\widehat{x}+B^{\prime}_{1}(t)l+C^{\prime}_{1}(t)+[r\widehat{x}+\alpha l (\pi_{I}(t)-k\pi_{R}(t))+\theta a-(1+k)(1-q(t)) \\ &(p(t)-a)]A_{1}(t)+\beta(\delta-l)B_{1}(t)-\lambda_{0}(kq(t)+q(t)-k)\mu_{z}A_{1}(t) \\ &-\frac{\gamma_{1}}{2} [(bq(t)-kb+kbq(t))^{2}+(\pi_{I}(t)-k \pi_{R}(t))^{2}l ]A^{2}_{2}(t) \\ &-\frac{\gamma_{1}}{2}\sigma^{2}lB^{2}_{2}(t)-\gamma_{1}(\pi_{I}(t)-k\pi_{R}(t))\sigma l \rho A_{2}(t)B_{2}(t) \\ &-\frac{\gamma_{1}}{2}\lambda_{0}\sigma^{2}_{z}\Big(kq(t)+q(t)-k\Big)^{2}A^{2}_{2}(t) \Big\}=0. \end{align*}$

由上式最大化的一阶条件, 可得

$\begin{align*} & (1+k)(p(t)-a)A_{1}(t)-(1+k)\lambda_{0}\mu_{z}A_{1}(t)-\gamma_{1}(1+k)b^{2}(kq(t)+q(t)-k)A_{2}^{2}(t)\\ &-\gamma_{1}(1+k)\lambda_{0}\sigma_{z}^{2}(kq(t)+q(t)-k)A_{2}^{2}(t)=0, \\ & \alpha l A_{1}(t)-\gamma_{1}(\pi_{I}(t)-k\pi_{R}(t))lA_{2}^{2}(t)-\gamma_{1}\sigma l\rho A_{2}(t)B_{2}(t)=0. \end{align*}$

通过上面两个等式, 可知投资策略和再保险策略是独立的. 因此, $f^{*}(\cdot,\ p(\cdot), \ \pi _{R}(\cdot))$ 和 $g^{*}(\cdot,\ p(\cdot),$$ \pi _{R}(\cdot))$

可以分别写为 $f^{*}(\cdot,\ p(\cdot))$

和 $g^{*}(\cdot,\ \pi _{R}(\cdot)),$ 由于 $q(t)\in[0, 1]$, 我们有

(3.4)$q^{*}(t,p(t))=f^{*}(t,p(t))= \left[ \frac{(p(t)- \lambda_{0} \mu_{z}-a)A_{1}(t)}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma^{2}_{z}){A}^{2}_{2}(t)}+ \frac{k}{1+k}\right] \vee 0\wedge 1,$

(3.5)$\pi _{I}^{*}(t)=g^{*}(t, \pi _{R}(t))= \frac{\alpha A_{1}(t)-\gamma_{1}\sigma \rho A_{2}(t)B_{2}(t)}{\gamma_{1} A^{2}_{2}(t)}+k \pi _{R}(t).$

这里很容易证明

$\begin{equation}\nonumber\frac{(p(t)- \lambda_{0} \mu_{z}-a)A_{1}(t)}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma^{2}_{z}){A}^{2}_{2}(t)}+ \frac{k}{1+k}>0,\end{equation}$

我们需要讨论以下两种情况

情况 (Ia) 如果

$\begin{equation}\nonumber\frac{(p(t)- \lambda_{0} \mu_{z}-a)A_{1}(t)}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma^{2}_{z}){A}^{2}_{2}(t)}+ \frac{k}{1+k}\geq1,\end{equation}$

那么 $q^{*}(t,p(t))=1$, 我们把 $q^{*}(t,p(t))=1$ 和 (3.5) 式中的 $\pi^{*}_{I}(t)$ 代入到等式 (2.15) 和 (3.3) 中, 可得

$\begin{align*}\nonumber &A^{\prime}_{1}(t)\widehat{x}+B^{\prime}_{1}(t)l+C_{1}^{\prime}(t)+\left[r\widehat{x}+\alpha l\left(\frac{\alpha A_{1}(t)-\gamma_{1}\sigma \rho A_{2}(t)B_{2}(t)}{\gamma_{1}A_{2}^{2}(t)}\right) +\theta a\right]A_{1}(t)\\ &+\beta (\delta-l)B_{1}(t)-\lambda_{0}\mu_{z}A_{1}(t)-\frac{\gamma_{1}}{2}\left[b^{2}+\left(\frac{\alpha A_{1}(t)-\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t)}{\gamma_{1}A_{2}^{2}(t)}\right)^{2}l\right]A_{2}^{2}(t)\\[2mm] &-\frac{\gamma_{1}}{2}\sigma^{2}l B_{2}^{2}(t)-\gamma_{1}\sigma l\rho A_{2}(t)B_{2}(t)\left(\frac{\alpha A_{1}(t)-\gamma_{1}\sigma \rho A_{2}(t)B_{2}(t)}{\gamma_{1}A_{2}^{2}(t)}\right) -\frac{\gamma_{1}}{2}\lambda_{0}A_{2}^{2}(t)\sigma_{z}^{2}=0, \\[1mm] &A^{\prime}_{2}(t)\widehat{x}+B^{\prime}_{2}(t)l+C_{2}^{\prime}(t)-\lambda_{0}\mu_{z}A_{2}(t)\\[1mm] &+\left[r\widehat{x}+\alpha l\left(\frac{\alpha A_{1}(t)-\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t)}{\gamma_{1}A_{2}^{2}(t)}\right)+ \theta a \right]A_{2}(t) \beta(\delta-l)B_{2}(t)=0. \end{align*}$

通过分离变量 $\widehat{x}$, $l$, 我们可以得到下面等式

$\begin{equation} \nonumber \begin{cases} A_{1}^{\prime}(t)+rA_{1}(t)=0,\quad A_{2}^{\prime}(t)+rA_{2}(t)=0,\\ B^{\prime}_{1}(t)+\alpha \left(\frac{\alpha A_{1}(t)-\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t)}{\gamma_{1}A_{2}^{2}(t)}\right)A_{1}(t)-\beta B_{1}(t) -\frac{\gamma_{1}}{2}\left(\frac{\alpha A_{1}(t)-\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t)}{\gamma_{1}A_{2}^{2}(t)}\right)^{2}A_{2}^{2}(t)\\ -\frac{\gamma_{1}}{2}\sigma^{2} B_{2}^{2}(t) -\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t)\left(\frac{\alpha A_{1}(t)-\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t)}{\gamma_{1}A_{2}^{2}(t)}\right)=0,\\ B_{2}^{\prime}(t)+\alpha \left(\frac{\alpha A_{1}(t)-\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t)}{\gamma_{1}A_{2}^{2}(t)}\right)A_{2}(t)-\beta B_{2}(t)=0,\\ C_{1}^{\prime}(t)+\theta a A_{1}(t)+\beta \delta B_{1}(t)-\lambda_{0}A_{1}(t)\mu_{z}-\frac{\gamma_{1}}{2}b^{2}A_{2}^{2}(t)-\frac{\gamma_{1}}{2}\lambda_{0}A_{2}^{2}(t)\sigma_{z}^{2}=0,\\ C_{2}^{\prime}(t)+\theta a A_{2}(t)+\beta \delta B_{2}(t)-\lambda_{0}\mu_{z}A_{2}(t)=0. \end{cases} \end{equation}$

结合边值条件 $A_{1}(T)=A_{2}(T)=1,\ B_{1}(T)=B_{2}(T)=0,\ C_{1}(T)=C_{2}(T)=0$, 可得

(3.6)$ A_{1}(t)={\rm e}^{r(T-t)},\quad A_{2}(t)={\rm e}^{r(T-t)},$

(3.7)$B_{1}(t)=\frac{\Theta {\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}}{\beta+2\alpha\sigma\rho}-\frac{(2\Theta+\Phi){\rm e}^{(\beta+\alpha \sigma \rho)(t-T)}}{\alpha \sigma \rho}+\xi {\rm e}^{\beta(t-T)}+\frac{\Psi}{\beta},$

(3.8)$B_{2}(t)=\frac{\alpha^{2}}{\gamma_{1}(\beta+\alpha \sigma \rho)}\left[1-{\rm e}^{(\beta+\alpha \sigma \rho)(t-T)}\right],$

(3.9)$\begin{align*}& C_{1}^{Ia}(t)=\delta \Psi(T-t)+\frac{\theta \lambda_{0}\mu_{z}}{r}[{\rm e}^{r(T-t)-1}]-\frac{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})}{4r}[{\rm e}^{2r(T-t)}-1] \\ &\qquad\qquad+\delta \xi[1-{\rm e}^{\beta(t-T)}]-\frac{\beta\delta(2\Theta+\Phi)}{\alpha\sigma\rho(\beta+\alpha\sigma\rho)}[1-{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}] \\ &\qquad\qquad+\frac{\beta\delta\theta}{2(\beta+\alpha\sigma\rho)(\beta+2\alpha\sigma\rho)}[1-{\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}], \end{align*}$

(3.10)$\begin{align*}& C_{2}^{Ia}(t)=\frac{\theta\lambda_{0}\mu_{z}}{r}[{\rm e}^{r(T-t)}-1]+\frac{\beta\delta\alpha^{2}}{\gamma_{1}(\beta+\alpha\sigma\rho)} +\frac{\beta\delta\alpha^{2}}{\gamma_{1}(\beta+\alpha\sigma\rho)}(T-t) \\[2mm] &\qquad\qquad-\frac{\beta\delta\alpha^{2}}{\gamma_{1}(\beta+\alpha\sigma\rho)^{2}}[1-{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}]. \end{align*}$

把等式 (3.6) 代入到 (3.5) 式, 我们可以得到保险公司的最优策略

(3.11)$\begin{equation} q^{*}(t,p(t))=1,\quad \pi^{*}_{I}(t)=\frac{\alpha-\gamma_{1}\sigma\rho B_{2}(t)}{\gamma_{1}{\rm e}^{r(T-t)}}+k\pi_{R}(t). \end{equation}$

情况 (Ib) 如果

$\begin{equation} \nonumber\frac{(p(t)- \lambda_{0} \mu_{z}-a)A_{1}(t)}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma^{2}_{z}){A}^{2}_{2}(t)}+ \frac{k}{1+k}<1,\end{equation}$

那么取

$\begin{equation} \nonumber q^{*}=\frac{(p(t)- \lambda_{0} \mu_{z}-a)A_{1}(t)}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma^{2}_{z}){A}^{2}_{2}(t)}+ \frac{k}{1+k},\end{equation}$

我们把 $q^{*}(t,p(t))$ 和 $\pi^{*}_{I}(t)$ 代入到等式 (2.15) 和 (3.3) 中, 可得

$\begin{align*} &A^{\prime}_{1}(t)\widehat{x}+B^{\prime}_{1}(t)l+C_{1}^{\prime}(t)+[r\widehat{x}+\theta a -(1+k)(p(t)-a)]A_{1}(t)+\beta(\delta-l)B_{1}(t)\\ &+k\lambda_{0}\mu_{z}A_{1}(t)-\frac{\gamma_{1}}{2}\sigma^{2}lB_{2}^{2}(t)-\frac{\gamma_{1}}{2}k^{2}A_{2}^{2}(t)(b^{2}+\lambda_{0}\sigma_{z}^{2})\\ &+\frac{[(1+k)(p(t)-a-\lambda_{0}\mu_{z})A_{1}(t)+\gamma_{1}k(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}^{2}(t)]^{2}}{2\gamma_{1}(1+k)^{2}(b^{2}+\lambda_{0}\sigma^{2}_{z})A_{2}^{2}(t)}\\ &+\alpha l A_{1}(t)\frac{\alpha A_{1}(t)-\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t)}{\gamma_{1}A_{2}^{2}(t)}-\gamma_{1}\sigma\rho l B_{2}(t)\frac{\alpha A_{1}(t)-\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t)}{\gamma_{1} A_{2}(t)}\\ &-\frac{(\alpha A_{1}(t)-\gamma_{1}\sigma \rho A_{2}(t)B_{2}(t))^{2}l}{2\gamma_{1}A_{2}^{2}(t)}=0, \\ &A^{\prime}_{2}(t)\widehat{x}+B^{\prime}_{2}(t)l+C_{2}^{\prime}(t)+(r\widehat{x}+\theta a)A_{2}(t)+\beta(\delta-l)B_{2}(t)\\ &+\alpha l A_{2}(t)\frac{\alpha A_{1}(t)-\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t)}{\gamma_{1}A_{2}^{2}(t)}-(p(t)-a)A_{2}(t)\\ &+\frac{(p(t)-a-\lambda_{0}\mu_{z})^{2}A_{1}(t)}{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)}=0. \end{align*}$

通过分离变量 $\widehat{x}$, $l$, 我们可以得到下面等式

$\begin{equation} \nonumber \begin{cases} A_{1}^{\prime}(t)+rA_{1}(t)=0,\quad A_{2}^{\prime}(t)+rA_{2}(t)=0,\\[2mm] B^{\prime}_{1}(t)-\beta B_{1}(t)-\frac{\gamma_{1}}{2}\sigma^{2}B_{2}^{2}(t)+\alpha A_{1}(t)\frac{\alpha A_{1}(t)-\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t)}{\gamma_{1} A_{2}(t)}\\[2mm] -\sigma\rho B_{2}(t)\frac{\alpha A_{1}(t)-\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t)}{ A_{2}(t)}-\frac{(\alpha A_{1}(t)-\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t))^{2}}{2\gamma_{1} A_{2}^{2}(t)}=0,\\[2mm] B^{\prime}_{2}(t)-\beta B_{2}(t)+\alpha\frac{\alpha A_{1}(t)-\gamma_{1}\sigma\rho A_{2}(t)B_{2}(t)}{\gamma_{1}A_{2}(t)}=0,\\[2mm] C_{1}^{\prime}(t)+\theta aA_{1}(t)-(1+k)(p(t)-a)A_{1}(t)+\beta\delta B_{1}(t)+k\lambda_{0}\mu_{z}A_{1}(t)\\[2mm] -\frac{\gamma_{1}}{2}k^{2}(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}^{2}(t)+\frac{[(1+k)(p(t)-a-\lambda_{0}\mu_{z})A_{1}(t)+\gamma_{1}k(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}^{2}(t)]^{2}} {2\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}^{2}(t)}=0,\\[2mm] C_{2}^{\prime}(t)+\theta a A_{2}(t)+\beta \delta B_{2}(t)-(p(t)-a)A_{2}(t)+\frac{(p(t)-a-\lambda_{0}\mu_{z})^{2}A_{1}(t)}{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)}=0. \end{cases} \end{equation}$

结合边值条件 $A_{1}(T)=A_{2}(T)=1,\ B_{1}(T)=B_{2}(T)=0,\ C_{1}(T)=C_{2}(T)=0$, 可得

(3.12)$A_{1}(t)={\rm e}^{r(T-t)},\quad A_{2}(t)={\rm e}^{r(T-t)},$

(3.13)$B_{1}(t)=\frac{\Theta {\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}}{\beta+2\alpha\sigma\rho}-\frac{(2\Theta+\Phi){\rm e}^{(\beta+\alpha \sigma \rho)(t-T)}}{\alpha \sigma \rho}+\xi {\rm e}^{\beta(t-T)}+\frac{\Psi}{\beta},$

(3.14)$B_{2}(t)=\frac{\alpha^{2}}{\gamma_{1}(\beta+\alpha \sigma \rho)}\left[1-{\rm e}^{(\beta+\alpha \sigma \rho)(t-T)}\right],$

(3.15)$\begin{align*} C_{1}^{Ib}(t)&=\int_{t}^{T}\frac{(p(t)-a-\lambda_{0}\mu_{z})^{2}}{2\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})}{\rm d}s-\int_{t}^{T}(p(t)-a)A_{1}(s){\rm d}s \\ & +\frac{\beta \delta\theta[1-{\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}]}{2(\beta+\alpha\sigma\rho)(\beta+2\alpha\sigma\rho)}-\frac{\beta\delta(2\Theta+\Phi)[1-{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}]} {\alpha\sigma\rho (\beta+\alpha\sigma\rho)} \\ & +\delta\Psi(T-t)+\delta\xi[1-{\rm e}^{\beta(t-T)}]+\frac{\theta a }{r}[{\rm e}^{r(T-t)}-1],\end{align*}$

(3.16)$\begin{align*} C_{2}^{Ib}(t)&=\frac{\theta a }{r}[{\rm e}^{r(T-t)}-1]+\frac{\alpha^{2}\beta\delta(T-t)}{\gamma_{2}(\beta+\alpha\sigma\rho)}+\frac{\alpha^{2}\beta\delta}{\gamma_{2}(\beta+\alpha\sigma\rho)^{2}}[{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}-1] \\ & -\int_{t}^{T}(p(s)-a)A_{2}(s){\rm d}s+\int_{t}^{T}\frac{(p(s)-a-\lambda_{0}\mu_{z})^{2}}{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})}{\rm d}s. \end{align*}$

同样的, 保险公司的最优策略为

(3.17)$\begin{align*} &q^{*}(t,p(t))=\frac{(p(t)- \lambda_{0} \mu_{z}-a)}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma^{2}_{z})A_{2}(t)}+ \frac{k}{1+k}, \\ &\pi^{*}_{I}(t)=\frac{\alpha-\gamma_{1}\sigma\rho B_{2}(t)}{\gamma_{1}A_{2}(t)}+k\pi_{R}(t). \end{align*}$

步骤三 再保险公司现在知道保险公司将根据等式 (3.4) 和 (3.5) 采取策略, 再保险公司将决定最优策略 $(p^{*}(t),\ \pi^{*}_{R}(t))\in U_{R}$ 响应保险公司的决定. 通过简化等式 (2.16), 可得

(3.18)$\begin{aligned} &\underset{u_{R}\in U_{R}}{\rm{max}} \Big\{\frac{\partial W_{R}}{\partial t}+[ry+\alpha l\pi_{R}(t)+(1-q^{*}(t))(p(t)-a)]\frac{\partial W_{R}}{\partial y}+\frac{1}{2}[(1-q^{*}(t))^{2}b^{2}\\ &+\pi^{2}_{R}(t)l]\frac{\partial ^{2} W_{R}} {\partial y^{2}}+\beta(\delta-l)\frac{\partial W_{R}}{\partial l}+\frac{1}{2}\sigma^{2}l\frac{\partial^{2} W_{R}}{\partial {l}^{2}} +\sigma\rho l \pi_{R}(t)\frac{\partial^{2} W_{R}}{\partial y \partial l}\\ &+\lambda_{0}E\left[ W_{R} (t,y-z(1-q^{*}(t)),l)-W_{R}(t,y,l)\right]-\frac{\gamma_{2}}{2}[(1-q^{*}(t))^{2}b^{2}\\ &+\pi^{2}_{R}(t)l] \left(\frac{\partial g_{R}}{\partial y}\right)^{2}-\frac{\gamma_{2}}{2}\sigma^{2}l\left(\frac{\partial g_{R}}{\partial l} \right)^{2}-\gamma_{2}\sigma\rho l\pi_{R}(t)\frac{\partial g_{R}}{\partial y}\frac{\partial g_{R} }{\partial l}\\ &-\frac{\gamma_{2}}{2}\lambda_{0}E[g_{R}(t,y-z(1-q^{*}(t)),l)-g_{R}(t,y,l)]^{2} \Big\}=0. \end{aligned}$

同样地, 我们猜测再保险公司的解有如下形式

$W_{R}(t,\widehat{x},l)=D_{1}(t)\widehat{x}+E_{1}(t)l+F_{1}(t),\quad D_{1}(T)=1,\quad E_{1}(T)=F_{1}(T)=0,$

$g_{R}(t,\widehat{x},l)=D_{2}(t)\widehat{x}+E_{2}(t)l+F_{2}(t),\quad D_{2}(T)=1,\quad E_{2}(T)=F_{2}(T)=0.$

根据简单的计算, 我们可以很容易得到以下偏导数

$ \frac{\partial W_{R}}{\partial t}=D^{\prime}_{1}(t)\widehat{x} +{E^{\prime}_{1}(t)}l+F^{\prime}_{1}(t),\quad \frac{\partial W_{R}}{\partial \widehat{x}}=D_{1}(t), \quad \frac{\partial W_{R}}{\partial l}=E_{1}(t),$

$ \frac{\partial g_{R}}{\partial t}=D^{\prime}_{2}(t)\widehat{x} +{E^{\prime}_{2}(t)}l+F^{\prime}_{2}(t),\quad \frac{\partial g_{R}}{\partial \widehat{x}}=D_{2}(t), \quad \frac{\partial g_{R}}{\partial l}=E_{2}(t).$

我们把以上偏导数代入到等式 (3.18), 可得

(3.19)$\begin{aligned} &\underset{u_{R}\in U_{R}}{\rm{max}} \Big\{D^{\prime}_{1}(t)y+E_{1}^{\prime}(t)l+F_{1}^{\prime}(t)+[ry+\alpha l \pi_{R}(t)+(1-q^{*}(t))(p(t)-a)]D_{1}(t)\\ &+\beta(\delta-l)E_{1}(t)-\lambda_{0}\mu_{z}(1-q^{*}(t))D_{1}(t)-\frac{\gamma_{2}}{2}[(1-q^{*}(t))^{2}b^{2}+\pi_{R}^{2}(t)l]D_{2}^{2}(t)\\ &-\frac{\gamma_{2}}{2}\sigma^{2} l E_{2}^{2}(t)-\gamma_{2}\sigma l \rho \pi_{R}(t)D_{2}(t)E_{2}(t)-\frac{\gamma_{2}}{2}\lambda_{0}\sigma_{z}^{2}(1-q^{*}(t))^{2}D_{2}^{2}(t)\Big\}=0. \end{aligned}$

在情况 (Ia) 下, 我们把 $q^{*}(t,p(t))=1$ 代入到等式 (3.19), 可得

(3.20)$\begin{equation} \begin{aligned} &\underset{u_{R}\in U_{R}}{\rm{max}} \Big\{D^{\prime}_{1}(t)y+E_{1}^{\prime}(t)l+F_{1}^{\prime}(t)+[ry+\alpha l\pi_{R}(t)]D_{1}(t) +\beta(\delta-l)E_{1}(t)\\ &-\frac{\gamma_{2}}{2}\pi_{R}^{2}(t)lD_{2}^{2}(t)-\frac{\gamma_{2}}{2}\sigma^{2} l E_{2}^{2}(t)-\gamma_{2}\sigma l \rho \pi_{R}(t)D_{2}(t)E_{2}(t) \Big\}=0. \end{aligned} \end{equation}$

上式对 $\pi_{R}(t)$ 求微分, 可得再保险公司的最优策略为

(3.21)$\begin{equation} \pi^{*}_{R}(t)=\frac{\alpha D_{1}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t)}{\gamma_{2}D_{2}^{2}(t)},\quad p^{*}(t)=p, \end{equation}$

这里 $p$ 是 $[c,\ \bar{c}]$ 中的任意值. 把上面的投资策略和保费策略代入到等式 (2.17) 和 (3.20) 中, 得

$\begin{align*} &D^{\prime}_{1}(t)y+E^{\prime}_{1}(t)l+F_{1}^{\prime}(t)+ryD_{1}(t)+\beta(\delta-l)E_{1}(t)-\frac{\gamma_{2}}{2}\sigma^{2}lE_{2}^{2}(t)\\ &+\frac{(\alpha l D_{1}(t)-\gamma_{2}\sigma l \rho D_{2}(t)E_{2}(t))^{2}}{2\gamma_{2}l D_{2}^{2}(t)}=0, \\ &D^{\prime}_{2}(t)y+E^{\prime}_{2}(t)l+F^{\prime}_{2}(t)+ryD_{2}(t)+\alpha l D_{2}(t)\frac{\alpha D_{1}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t)}{\gamma_{2}D_{2}^{2}(t)}+\beta(\delta-l)E_{2}(t)=0. \end{align*}$

通过分离变量 $y$, $l$, 我们可以得到以下等式

$\begin{equation}\nonumber \begin{cases} &D_{1}^{\prime}(t)+rD_{1}(t)=0,\quad D_{2}^{\prime}(t)+rD_{2}(t)=0,\\[2mm] &E^{\prime}_{1}-\beta E_{1}(t)-\frac{\gamma_{2}}{2}\sigma^{2}E_{2}^{2}(t)+\frac{(\alpha D_{1}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t))^{2}}{2\gamma_{2}D_{2}^{2}(t)}=0,\\[2mm] &E^{\prime}_{2}(t)+\alpha D_{2}(t)\frac{\alpha D_{1}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t)}{\gamma_{2}D_{2}^{2}(t)}-\beta E_{2}(t)=0,\\[2mm] &F_{1}^{\prime}(t)+\beta \delta E_{1}(t)=0,\\[1mm] &F_{2}^{\prime}(t)+\beta\delta E_{2}(t)=0. \end{cases} \end{equation}$

结合边值条件 $D_{1}(T)=D_{2}(T)=1,\ E_{1}(T)=E_{2}(T)=0,\ F_{1}(T)=F_{2}(T)=0$, 可得

(3.22)$D_{1}(t)={\rm e}^{r(T-t)},\quad D_{2}(t)={\rm e}^{r(T-t)},$

(3.23)$E_{1}(t)=\frac{\Theta {\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}}{\beta+2\alpha\sigma\rho}-\frac{(2\Theta+\Phi){\rm e}^{(\beta+\alpha \sigma \rho)(t-T)}}{\alpha \sigma \rho}+\xi {\rm e}^{\beta(t-T)}+\frac{\Psi}{\beta},$

(3.24)$E_{2}(t)=\frac{\alpha^{2}}{\gamma_{2}(\beta+\alpha \sigma \rho)}\left[1-{\rm e}^{(\beta+\alpha \sigma \rho)(t-T)}\right],$

(3.25)$\begin{align*} &F_{1}^{Ia}(t)=\frac{\beta\delta\Theta[1-{\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}]}{2(\beta+2\alpha\sigma\rho)(\beta+\alpha\sigma\rho)}- \frac{\beta\delta(2\Theta+\Phi)[1-{\rm e}^{(\beta+\alpha \sigma\rho)(t-T)}]}{\alpha\sigma\rho(\beta+\alpha\sigma\rho)} \\ &\qquad\qquad+\delta\Psi(T-t)+\delta\xi[1-{\rm e}^{\beta(t-T)}], \end{align*}$

(3.26)$F_{2}^{Ia}(t)=\frac{\alpha^{2}\beta\delta(T-t)}{\gamma_{2}(\beta+\alpha\sigma\rho)}+\frac{\alpha^{2}\beta\delta}{\gamma_{2}(\beta+\alpha\sigma\rho)^{2}}[{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}-1].$

这里先验条件 $\frac{(p(t)- \lambda_{0} \mu_{z}-a)A_{1}(t)}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma^{2}_{z}){A}^{2}_{2}(t)}+ \frac{k}{1+k}\geq1$ 变为 $H^{\theta} \geq 1$.把等式 (3.22) 代入到等式 (3.21) 并结合等式 (3.11), 可以得到最优策略

$\begin{align*} & p^{*}(t)=p,\quad q^{*}(t)=1,\\[1mm] & \pi_{I}^{*}(t)=\frac{\alpha-\gamma_{1}\sigma\rho B_{2}(t)}{\gamma_{1}{\rm e}^{r(T-t)}}+k\frac{\alpha-\gamma_{2}\sigma\rho E_{2}(t)}{\gamma_{2}{\rm e}^{r(T-t)}},\quad \pi_{R}^{*}(t)=\frac{\alpha-\gamma_{2}\sigma\rho E_{2}(t)}{\gamma_{2}{\rm e}^{r(T-t)}}, \end{align*}$

这里 $p$ 是区间 $[c,\ \bar{c}]$ 中的任意值.

在情况 (Ib) 中, 我们把$ q^{*}(t,p(t))=\frac{p(t)-a-\lambda_{0}\mu_{z}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma^{2}_{z})A_{2}(t)}+\frac{k}{1+k}$代入到等式 (3.19) 中, 可得

(3.27)$\begin{equation} \begin{aligned} &\underset{u_{R}\in U_{R}}{\rm{max}}\bigg\{ D^{\prime}_{1}(t)y+E^{\prime}(t)l+F^{\prime}_{1}(t)+ryD_{1}(t)+\alpha l \pi_{R}(t)D_{1}(t)+\beta(\delta-l)E_{1}(t)\\[2mm] &-\frac{\gamma_{2}}{2}\sigma^{2}lE_{2}^{2}(t)-\frac{\gamma_{2}}{2}\pi_{R}^{2}(t)lD_{2}^{2}(t)-\gamma_{2}\pi_{R}(t)\sigma l \rho D_{2}(t)E_{2}(t)\\[2mm] &+(p(t)-a-\lambda_{0}\mu_{z})D_{1}(t)\left(\frac{1}{1+k}-\frac{p(t)-a-\lambda_{0}\mu_{z}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)}\right)\\[2mm] &-\frac{\gamma_{2}}{2}D_{2}^{2}(t)(b^{2}+\lambda_{0}\sigma_{z}^{2})\left(\frac{1}{1+k}-\frac{p(t)- a-\lambda_{0}\mu_{z}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)} \right)^{2} \bigg \}=0. \end{aligned} \end{equation}$

通过对 $p(t)$ 和 $\pi_{R}(t)$ 求微分, 可得

$\begin{align*} & \pi_{R}^{*}(t)=\frac{\alpha D_{1}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t)}{\gamma_{2}D_{2}^{2}(t)},\\[2mm] & p^{*}(t)=\left[(a+\lambda_{0}\mu_{z})+\frac{\gamma_{1}(1+k)D_{1}(t)A_{2}(t)+\gamma_{2}D^{2}_{2}(t)}{2\gamma_{1}(1+k)D_{1}(t)A_{2}(t) +\gamma_{2}D^{2}_{2}(t)}(b^{2}+\lambda_{0}\sigma^{2}_{z})\gamma_{1}A_{2}(t)\right]\vee c \wedge \bar{c}. \end{align*}$

子情况 (Ib1) 如果

$\begin{equation}\nonumber\left[(a+\lambda_{0}\mu_{z})+\frac{\gamma_{1}(1+k)D_{1}(t)A_{2}(t)+\gamma_{2}D^{2}_{2}(t)}{2\gamma_{1}(1+k)D_{1}(t)A_{2}(t) +\gamma_{2}D^{2}_{2}(t)}(b^{2}+\lambda_{0}\sigma^{2}_{z})\gamma_{1}A_{2}(t)\right]\geq \bar{c},\end{equation}$

我们取 $p^{*}(t)=\bar{c}$, 那么$ q^{*}(t)=\frac{\bar{\theta}\lambda_{0}\mu_{z}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma^{2}_{z})A_{2}(t)}+\frac{k}{1+k}=\frac{H^{\bar{\theta}}+k}{1+k},$把它们代入到等式 (2.17) 和 (3.27) 可得

$\begin{equation}\nonumber \begin{aligned} & D^{\prime}_{1}(t)y+E^{\prime}_{1}(t)l+F^{\prime}_{1}(t)+ryD_{1}(t)+\beta(\delta-l)E_{1}(t)-\frac{\gamma_{2}}{2}\sigma^{2}lE_{2}^{2}(t)\\[2mm] &+\frac{(\alpha D_{1}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t))^{2}}{2\gamma_{2}D_{2}^{2}(t)}l+\frac{\bar{\theta}\lambda_{0}\mu_{z}D_{1}(t)}{1+k} -\frac{\bar{\theta}^{2}\lambda_{0}^{2}\mu_{z}^{2}D_{1}(t)}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)}\\[2mm] &- \frac{\gamma_{2}}{2}(b^{2}+\lambda_{0}\sigma_{z}^{2})D_{2}^{2}(t)\left[\frac{1}{1+k}-\frac{\bar{\theta}\lambda_{0}\mu_{z}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)}\right]^{2}=0, \\[2mm] &D^{\prime}_{2}(t)y+E^{\prime}_{2}(t)l+F^{\prime}_{2}(t)+ryD_{2}(t)+\beta(\delta-l)E_{2}(t)+\alpha l\frac{\alpha D_{1}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t)}{\gamma_{2}D_{2}(t)}\\[2mm] &+\left(\frac{1}{1+k}-\frac{\bar{\theta}\lambda_{0}\mu_{z}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)} \right)\bar{\theta}aD_{2}(t)\\[2mm] &-\lambda_{0}\mu_{z}D_{2}(t)\left[\frac{1}{1+k}-\frac{\bar{\theta}\lambda_{0}\mu_{z}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)} \right]=0. \end{aligned} \end{equation}$

通过分离变量 $y$, $l$, 我们可以得到以下等式

结合边值条件 $D_{1}(T)=D_{2}(T)=1,\ E_{1}(T)=E_{2}(T)=0,\ F_{1}(T)=F_{2}(T)=0$, 可得

(3.28)$\begin{align*} &D_{1}(t)={\rm e}^{r(T-t)},\quad D_{2}(t)={\rm e}^{r(T-t)}, \nonumber\\[1mm] &E_{1}(t)=\frac{\Theta {\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}}{\beta+2\alpha\sigma\rho}-\frac{(2\Theta+\Phi){\rm e}^{(\beta+\alpha \sigma \rho)(t-T)}}{\alpha \sigma \rho}+\xi {\rm e}^{\beta(t-T)}+\frac{\Psi}{\beta}, \nonumber\\[2mm] &E_{2}(t)=\frac{\alpha^{2}}{\gamma_{2}(\beta+\alpha \sigma \rho)}\left[1-{\rm e}^{(\beta+\alpha \sigma \rho)(t-T)}\right], \nonumber\\[2mm] &F_{1}^{Ib1}(t)=\frac{\bar{\theta}\lambda_{0}\mu_{z}}{r(1+k)}[{\rm e}^{r(T-t)}-1]-\frac{\bar{\theta}^{2}\lambda_{0}^{2}\mu_{z}^{2}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})}(T-t)\nonumber\\[2mm] &\qquad\qquad-\frac{\gamma_{2}}{4r(1+k)^{2}}(b^{2}+\lambda_{0}\sigma_{z}^{2})[{\rm e}^{2r(T-t)}-1]+\frac{\gamma_{2}}{\gamma_{1}}\frac{\bar{\theta}\lambda_{0}\mu_{z}}{r(1+k)^{2}}[{\rm e}^{r(T-t)}-1]\nonumber\\[2mm] &\qquad\qquad-\frac{\gamma_{2}\bar{\theta}^{2}\lambda_{0}^{2}\mu_{z}^{2}}{2\gamma_{1}^{2}(1+k)^{2}(b^{2}+\lambda_{0}\sigma_{z}^{2})}[T-t] +\frac{\beta\delta\Theta[1-{\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}]}{2(\beta+\alpha\sigma\rho)(\beta+2\alpha\sigma\rho)}\nonumber\\[2mm] &\qquad\qquad-\frac{\beta\delta(2\Theta+\Phi)[1-{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}]} {\alpha\sigma\rho(\beta+\alpha\sigma\rho)} +\delta\Psi(T-t)+\delta\xi[1-{\rm e}^{\beta(t-T)}],\end{align*}$

(3.29)$\begin{align*} &F_{2}^{Ib1}(t)=\frac{\bar{\theta}\lambda_{0}\mu_{z}}{r(1+k)}[{\rm e}^{r(T-t)}-1]-\frac{\bar{\theta}^{2}\lambda_{0}^{2}\mu_{z}^{2}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})}(T-t)\nonumber\\[2mm] &\qquad\qquad+\frac{\beta\delta\alpha^{2}}{\gamma_{2}(\beta+\alpha\sigma\rho)}(T-t)-\frac{\beta\delta\alpha^{2}}{\gamma_{2}(\beta+\alpha\sigma\rho)^{2}}[1-{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}].\end{align*}$

因此, 条件

$\begin{equation} \nonumber \left[(a+\lambda_{0}\mu_{z})+\frac{\gamma_{1}(1+k)D_{1}(t)A_{2}(t)+\gamma_{2}D^{2}_{2}(t)}{2\gamma_{1}(1+k)D_{1}(t)A_{2}(t)+\gamma_{2}D^{2}_{2}(t)}(b^{2}+\lambda_{0}\sigma^{2}_{z})\gamma_{1}A_{2}(t)\right]\geq \bar{c},\end{equation}$

变为 $H^{\bar{\theta}}\leq H$,然后可得 $H^{\theta}<H^{\bar{\theta}}\leq H<1$,这也表明先验条件

$\begin{equation} \nonumber q^{*}=\frac{p(t)-a-\lambda_{0}\mu_{z}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma^{2}_{z})A_{2}(t)}+\frac{k}{1+k}<1\end{equation}$

自然就是正确的.等式 (3.15) 和 (3.16) 变为

(3.30)$\begin{align*} C^{Ib1}_{1}(t)&=\frac{\beta\delta\Theta[1-{\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}]}{2(\beta+\alpha\sigma\rho)(\beta+2\alpha\sigma\rho)}-\frac{\beta\delta(2\Theta+\Phi)[1-{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}]} {\alpha\sigma\rho(\beta+\alpha\sigma\rho)} \\ & +\frac{(\theta-\bar{\theta})a}{r}[{\rm e}^{r(T-t)}-1]+\frac{\bar{\theta}^{2}\lambda_{0}^{2}\mu_{z}^{2}}{2\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})}(T-t)\\ & +\delta\Psi(T-t)+\delta\xi[1-{\rm e}^{\beta(t-T)}], \end{align*}$

(3.31)$\begin{equation} \begin{aligned} C^{Ib1}_{2}(t)&=\frac{(\theta-\bar{\theta})a}{r}[{\rm e}^{r(T-t)}-1]+\frac{\alpha^{2}\beta\delta(T-t)}{\gamma_{2}(\beta+\alpha\sigma\rho)}\\ & +\frac{\alpha^{2}\beta\delta}{\gamma_{2}(\beta+\alpha\sigma\rho)^{2}}[{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}-1] +\frac{\bar{\theta}^{2}\lambda_{0}^{2}\mu_{z}^{2}}{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})}(T-t). \end{aligned} \end{equation}$

因此最优策略变为

$\begin{align*} p^{*}(t)\!=\!\bar{c}, q^{*}(t)\!=\!\frac{H^{\bar{\theta}}+k}{1+k}, \pi_{I}^{*}(t)\!=\!\frac{\alpha-\gamma_{1}\sigma\rho B_{2}(t)}{\gamma_{1}{\rm e}^{r(T-t)}}\!+\!k\frac{\alpha-\gamma_{2}\sigma\rho E_{2}(t)}{\gamma_{2}{\rm e}^{r(T-t)}}, \pi_{R}^{*}(t)\!=\!\frac{\alpha-\gamma_{2}\sigma\rho E_{2}(t)}{\gamma_{2}{\rm e}^{r(T-t)}}. \end{align*}$

子情况 (Ib2) 如果

$\begin{equation}\nonumber\left[(a+\lambda_{0}\mu_{z})+\frac{\gamma_{1}(1+k)D_{1}(t)A_{2}(t)+\gamma_{2}D^{2}_{2}(t)}{2\gamma_{1}(1+k)D_{1}(t)A_{2}(t)+\gamma_{2}D^{2}_{2}(t)}(b^{2} +\lambda_{0}\sigma^{2}_{z})\gamma_{1}A_{2}(t)\right]\leq c,\end{equation}$

我们取 $p^{*}(t)=c$, 那么$ q^{*}(t)=\frac{\theta\lambda_{0}\mu_{z}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma^{2}_{z})A_{2}(t)}+\frac{k}{1+k}=\frac{H^{\theta}+k}{1+k},$把它们代入到等式 (2.17) 和 (3.27) 可得

$\begin{equation} \nonumber \begin{aligned} &D^{\prime}_{1}(t)y+E^{\prime}_{1}(t)l+F^{\prime}_{1}(t)+ryD_{1}(t)+\beta(\delta-l)E_{1}(t)-\frac{\gamma_{2}}{2}\sigma^{2}lE_{2}^{2}(t)\\ &+\frac{(\alpha D_{1}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t))^{2}}{2\gamma_{2}D_{2}^{2}(t)}l+\frac{\theta\lambda_{0}\mu_{z}D_{1}(t)}{1+k} -\frac{\theta^{2}\lambda_{0}^{2}\mu_{z}^{2}D_{1}(t)}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)}\\ &-\frac{\gamma_{2}}{2}(b^{2}+\lambda_{0}\sigma_{z}^{2})D_{2}^{2}(t)\left[\frac{1}{1+k}-\frac{\theta\lambda_{0}\mu_{z}}{\gamma_{1}(+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)}\right]^{2}=0, \\ &D^{\prime}_{2}(t)y+E^{\prime}_{2}(t)l+F^{\prime}_{2}(t)+ryD_{2}(t)+\beta(\delta-l)E_{2}(t)+\alpha l\frac{\alpha D_{1}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t)}{\gamma_{2}D_{2}(t)}\\ &+\left(\frac{1}{1+k}-\frac{\theta\lambda_{0}\mu_{z}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)} \right)\theta aD_{2}(t)\\ &-\lambda_{0}\mu_{z}D_{2}(t)\left[\frac{1}{1+k}-\frac{\theta\lambda_{0}\mu_{z}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)} \right]=0. \end{aligned} \end{equation}$

通过分离变量 $y$, $l$, 我们可以得到以下等式

$\begin{equation} \nonumber \begin{cases} D_{1}^{\prime}(t)+rD_{1}(t)=0,\quad D_{2}^{\prime}(t)+rD_{2}(t)=0,\\ E^{\prime}_{1}(t)-\beta E_{1}(t)-\frac{\gamma_{2}}{2}\sigma^{2}E_{2}^{2}(t)+\frac{(\alpha D_{1}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t))^{2}}{2\gamma_{2}D_{2}^{2}(t)}=0,\\ E^{\prime}_{2}(t)-\beta E_{2}(t)+\alpha \frac{\alpha D_{1}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t)}{\gamma_{2}D_{2}(t)}=0,\\ F^{\prime}_{1}(t)+\beta\delta E_{1}(t)+\frac{\theta\lambda_{0}\mu_{z}D_{1}(t)}{1+k}-\frac{\theta^{2}\lambda_{0}^{2}\mu_{z}^{2}D_{1}(t)}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)}\\ -\frac{\gamma_{2}}{2}(b^{2}+\lambda_{0}\sigma_{z}^{2})D_{2}^{2}(t)\left(\frac{1}{1+k}-\frac{\theta\lambda_{0}\mu_{z}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)} \right)^{2}=0,\\ F_{2}^{\prime}(t)+\beta\delta E_{2}(t)+\left(\frac{1}{1+k}-\frac{\theta\lambda_{0}\mu_{z}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)}\right)\theta \lambda_{0}\mu_{z}D_{2}(t)=0. \end{cases} \end{equation}$

结合边值条件 $D_{1}(T)=D_{2}(T)=1,\ E_{1}(T)=E_{2}(T)=0,\ F_{1}(T)=F_{2}(T)=0$, 可得

(3.32)$\begin{align*} &D_{1}(t)={\rm e}^{r(T-t)},\quad D_{2}(t)={\rm e}^{r(T-t)}, \nonumber\\ &E_{1}(t)=\frac{\Theta {\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}}{\beta+2\alpha\sigma\rho}-\frac{(2\Theta+\Phi){\rm e}^{(\beta+\alpha \sigma \rho)(t-T)}}{\alpha \sigma \rho}+\xi {\rm e}^{\beta(t-T)}+\frac{\Psi}{\beta}, \nonumber\\ &E_{2}(t)=\frac{\alpha^{2}}{\gamma_{2}(\beta+\alpha \sigma \rho)}\left[1-{\rm e}^{(\beta+\alpha \sigma \rho)(t-T)}\right], \nonumber\\ &F_{1}^{Ib2}(t)=\frac{\theta\lambda_{0}\mu_{z}}{r(1+k)}[{\rm e}^{r(T-t)}-1]-\frac{\theta^{2}\lambda_{0}^{2}\mu_{z}^{2}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})}(T-t)\nonumber\\ &\qquad\qquad-\frac{\gamma_{2}}{4r(1+k)^{2}}(b^{2}+\lambda_{0}\sigma_{z}^{2})[{\rm e}^{2r(T-t)}-1]+\frac{\gamma_{2}}{\gamma_{1}}\frac{\theta\lambda_{0}\mu_{z}}{r(1+k)^{2}}[{\rm e}^{r(T-t)}-1]\nonumber\\ &\qquad\qquad-\frac{\gamma_{2}\theta^{2}\lambda_{0}^{2}\mu_{z}^{2}}{2\gamma_{1}^{2}(1+k)^{2}(b^{2}+\lambda_{0}\sigma_{z}^{2})}[T-t] +\frac{\beta\delta\Theta[1-{\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}]}{2(\beta+\alpha\sigma\rho)(\beta+2\alpha\sigma\rho)}\nonumber\\ &\qquad\qquad-\frac{\beta\delta(2\Theta+\Phi)[1-{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}]} {\alpha\sigma\rho(\beta+\alpha\sigma\rho)} +\delta\Psi(T-t)+\delta\xi[1-{\rm e}^{\beta(t-T)}],\end{align*}$

(3.33)$\begin{align*} &F_{2}^{Ib2}(t)=\frac{\theta\lambda_{0}\mu_{z}}{r(1+k)}[{\rm e}^{r(T-t)}-1]-\frac{\theta^{2}\lambda_{0}^{2}\mu_{z}^{2}}{\gamma_{1}(1+k)(b^{2}+\lambda_{0}\sigma_{z}^{2})}(T-t)\nonumber\\ &\qquad\qquad+\frac{\beta\delta\alpha^{2}}{\gamma_{2}(\beta+\alpha\sigma\rho)}(T-t)-\frac{\beta\delta\alpha^{2}}{\gamma_{2}(\beta+\alpha\sigma\rho)^{2}}[1-{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}]. \end{align*}$

因此, 条件

$\begin{equation}\nonumber\left[(a+\lambda_{0}\mu_{z})+\frac{\gamma_{1}(1+k)D_{1}(t)A_{2}(t)+\gamma_{2}D^{2}_{2}(t)}{2\gamma_{1}(1+k)D_{1}(t)A_{2}(t)+\gamma_{2}D^{2}_{2}(t)}(b^{2}+\lambda_{0}\sigma^{2}_{z})\gamma_{1}A_{2}(t)\right]\leq c,\end{equation}$

意味着 $H\leq H^{\theta}$, 结合先验条件 $H^{\theta}<1$, 条件就变为 $H\leq H^{\theta}<1$.因此等式 (3.15) 和 (3.16) 变为

(3.34)$\begin{align*} C^{Ib2}_{1}(t)&=\frac{\beta\delta\Theta[1-{\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}]}{2(\beta+\alpha\sigma\rho)(\beta+2\alpha\sigma\rho)}-\frac{\beta\delta(2\Theta+\Phi)[1-{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}]} {\alpha\sigma\rho(\beta+\alpha\sigma\rho)} \\ & +\frac{\theta^{2}\lambda_{0}^{2}\mu_{z}^{2}}{2\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})}(T-t) +\delta\Psi(T-t)+\delta\xi[1-{\rm e}^{\beta(t-T)}],\end{align*}$

(3.35)$\begin{align*} C^{Ib2}_{2}(t)&=\frac{\alpha^{2}\beta\delta(T-t)}{\gamma_{2}(\beta+\alpha\sigma\rho)} +\frac{\alpha^{2}\beta\delta}{\gamma_{2}(\beta+\alpha\sigma\rho)^{2}}[{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}-1] \\ & +\frac{\theta^{2}\lambda_{0}^{2}\mu_{z}^{2}}{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})}(T-t). \end{align*}$

因此最优策略变为

$\begin{align*} p^{*}(t)\!=\!c, q^{*}(t)\!=\!\frac{H^{\theta}+k}{1+k}, \pi_{I}^{*}(t)\!=\!\frac{\alpha-\gamma_{1}\sigma\rho B_{2}(t)}{\gamma_{1}{\rm e}^{r(T-t)}}+k\frac{\alpha-\gamma_{2}\sigma\rho E_{2}(t)}{\gamma_{2}{\rm e}^{r(T-t)}}, \pi_{R}^{*}(t)\!=\!\frac{\alpha-\gamma_{2}\sigma\rho E_{2}(t)}{\gamma_{2}{\rm e}^{r(T-t)}}. \end{align*}$

子情况 (Ib3) 如果

$\begin{equation} \nonumber c<\left[(a+\lambda_{0}\mu_{z})+\frac{\gamma_{1}(1+k)D_{1}(t)A_{2}(t)+\gamma_{2}D^{2}_{2}(t)}{2\gamma_{1}(1+k)D_{1}(t)A_{2}(t)+\gamma_{2}D^{2}_{2}(t)}(b^{2}+\lambda_{0}\sigma^{2}_{z})\gamma_{1}A_{2}(t)\right]\\< \bar{c},\end{equation}$

我们取

$\begin{align*} p^{*}(t)&=\left[(a+\lambda_{0}\mu_{z})+\frac{\gamma_{1}(1+k)D_{1}(t)A_{2}(t)+\gamma_{2}D^{2}_{2}(t)}{2\gamma_{1}(1+k)D_{1}(t)A_{2}(t)+\gamma_{2}D^{2}_{2}(t)}(b^{2} +\lambda_{0}\sigma^{2}_{z})\gamma_{1}A_{2}(t)\right], \\[2mm] q^{*}(t)&=\frac{\gamma_{1}(1+2k)D_{1}(t)A_{2}(t)+\gamma_{2}D^{2}_{2}(t)}{2\gamma_{1}(1+k)D_{1}(t)A_{2}(t)+\gamma_{2}D_{2}^{2}(t)}, \end{align*}$

把它们代入到等式 (2.17) 和 (3.27) 可得

$\begin{align*} &D^{\prime}_{1}(t)y+E_{1}^{\prime}(t)l+F^{\prime}(t)+rD_{1}(t)y+\beta(\delta-l)E_{1}(t)-\frac{\gamma_{2}}{2}\sigma^{2}lE_{2}^{2}(t)\\ &+\frac{(\alpha D_{1}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t))^{2}}{2\gamma_{2}D_{2}^{2}(t)}l+\frac{\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)} {2\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)}\\ &\times\frac{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)D_{1}(t)}{1+k} -\frac{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)D_{1}(t)}{1+k}\\ &\times\left(\frac{\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)}{2\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)}\right)^{2} -\frac{\gamma_{2}(b^{2}+\lambda_{0}\sigma_{z}^{2})D_{2}^{2}(t)}{2}\\ &\times\frac{\gamma_{1}^{2}A_{2}^{2}(t)D_{1}^{2}(t)}{[2\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)]^{2}}=0, \\ &D_{2}^{\prime}(t)y+E_{2}^{\prime}(t)l+F^{\prime}_{2}(t)+rD_{2}(t)y+\beta(\delta-l)E_{2}(t) +\alpha l D_{2}(t)\frac{\alpha-\gamma_{2}\sigma\rho E_{2}(t)}{\gamma_{2}A_{2}(t)}\\ &+\frac{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)D_{2}(t)}{1+k} \times\frac{\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)}{2\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D^{2}_{2}(t)}\\ &-\frac{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)D_{2}(t)}{1+k} \times\left[\frac{\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)}{2\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}}\right]^{2}=0. \end{align*}$

通过分离变量 $y$, $l$ 我们可以得到以下等式

$\begin{equation} \nonumber \begin{cases} D_{1}^{\prime}(t)+rD_{1}(t)=0,\quad D_{2}^{\prime}(t)+rD_{2}(t)=0,\\[1mm] E^{\prime}_{1}(t)-\beta E_{1}(t)-\frac{\gamma_{2}}{2}\sigma^{2}E_{2}^{2}(t)+\frac{(\alpha D_{1}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t))^{2}}{2\gamma_{2}D_{2}^{2}(t)}=0,\\[1mm] E^{\prime}_{2}(t)-\beta E_{2}(t)+\alpha \frac{\alpha D_{2}(t)-\gamma_{2}\sigma\rho D_{2}(t)E_{2}(t)}{\gamma_{2}D_{2}(t)}=0,\\[1mm] F^{\prime}_{1}(t)+\beta\delta E_{1}(t)+\frac{\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)}{2\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)} \frac{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)D_{1}(t)}{1+k}\\[1mm] -\frac{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)D_{1}(t)}{1+k} \times\left[\frac{\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)}{2\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D^{2}_{2}(t)} \right]^{2}\\[1mm] -\frac{\gamma_{2}(b^{2}+\lambda_{0}\sigma_{z}^{2})D_{2}^{2}(t)}{2}\left[\frac{\gamma_{1}A_{2}(t)D_{1}(t)}{2\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)}\right]^{2}=0,\\[1mm] F_{2}^{\prime}(t)+\beta\delta E_{2}(t)+\frac{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t))D_{2}(t)}{1+k}\frac{\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)} {2\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)}\\[1mm] -\frac{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})A_{2}(t)D_{2}(t)}{1+k}\left[\frac{\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)}{2\gamma_{1}(1+k)A_{2}(t)D_{1}(t)+\gamma_{2}D_{2}^{2}(t)} \right]^{2}=0. \end{cases} \end{equation}$

结合边值条件 $D_{1}(T)=D_{2}(T)=1,\ E_{1}(T)=E_{2}(T)=0\ F_{1}(T)=F_{2}(T)=0$, 可得

(3.36)$\begin{align*} &D_{1}(t)={\rm e}^{r(T-t)},\quad D_{2}(t)={\rm e}^{r(T-t)}, \\ &E_{1}(t)=\frac{\Theta {\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}}{\beta+2\alpha\sigma\rho}-\frac{(2\Theta+\Phi){\rm e}^{(\beta+\alpha \sigma \rho)(t-T)}}{\alpha \sigma \rho}+\xi {\rm e}^{\beta(t-T)}+\frac{\Psi}{\beta}, \\ &E_{2}(t)=\frac{\alpha^{2}}{\gamma_{2}(\beta+\alpha \sigma \rho)}\left[1-{\rm e}^{(\beta+\alpha \sigma \rho)(t-T)}\right], \\ &F_{1}^{Ib3}(t)=\frac{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})}{2r(1+k)}(H-H^{2}){\rm e}^{2r(T-t)}-\frac{\gamma_{1}^{2}\gamma_{2}(b^{2}+\lambda_{0}\sigma_{z}^{2})}{4r[2\gamma_{1}(1+k)+\gamma_{2}]^{2}} {\rm e}^{2r(T-t)} \\ &\qquad\qquad+\frac{\beta\delta\Theta[1-{\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}]}{2(\beta+\alpha\sigma\rho)(\beta+2\alpha\sigma\rho)} -\frac{\beta\delta(2\Theta+\Phi)[1-{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}]} {\alpha\sigma\rho(\beta+\alpha\sigma\rho)} \\ &\qquad\qquad+\delta\Psi(T-t)+\delta\xi[1-{\rm e}^{\beta(t-T)}],\end{align*}$

(3.37)$\begin{align*}&F_{2}^{Ib3}(t)=\frac{\gamma_{1}(b^{2}+\lambda_{0}\sigma_{z}^{2})}{2r(1+k)}(H-H^{2}){\rm e}^{2r(T-t)}+\frac{\beta\delta\alpha^{2}}{\gamma_{2}(\beta+\alpha\sigma\rho)}(T-t) \\ &\qquad\qquad-\frac{\beta\delta\alpha^{2}}{\gamma_{2}(\beta+\alpha\sigma\rho)^{2}}[1-{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}]. \end{align*}$

因此, 条件

变为 $H^{\theta}<H<H^{\bar{\theta}}$, 并且 $ p^{*}(t)=a+\lambda_{0}\mu_{z}+(b^{2}+\lambda_{0}\sigma^{2}_{z})\gamma_{1}A_{2}(t)H, $$q^{*}(t)=\frac{(1+2k)\gamma_{1}+\gamma_{2}}{2(1+k)\gamma_{1}+\gamma_{2}}$,因此等式 (3.15) 和 (3.16) 变为

(3.38)$\begin{align*} C_{1}^{Ib3}(t)=\,&\frac{\gamma_{1}H^{2}(b^{2}+\lambda_{0}\sigma_{z}^{2})}{4r}[{\rm e}^{2r(T-t)}-1]-\frac{\lambda_{0}\mu_{z}}{r}[{\rm e}^{r(T-t)}-1] -\frac{\gamma_{1}H(b^{2}+\lambda_{o}\sigma_{z}^{2})} {2r}[{\rm e}^{2r(T-t)}-1] \\ &+\frac{\beta \delta\theta[1-{\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}]}{2(\beta+\alpha\sigma\rho)(\beta+2\alpha\sigma\rho)}-\frac{\beta\delta(2\Theta+\Phi)[1-{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}]} {\alpha\sigma\rho (\beta+\alpha\sigma\rho)} \\ &+\delta\Psi(T-t)+\delta\xi[1-{\rm e}^{\beta(t-T)}]+\frac{\theta a }{r}[{\rm e}^{r(T-t)}-1],\end{align*}$

(3.39)$\begin{align*}C_{2}^{Ib3}(t)=\,&\frac{\theta a }{r}[{\rm e}^{r(T-t)}-1]+\frac{\alpha^{2}\beta\delta(T-t)}{\gamma_{2}(\beta+\alpha\sigma\rho)}+\frac{\alpha^{2}\beta\delta}{\gamma_{2}(\beta+\alpha\sigma\rho)^{2}}[{\rm e}^{(\beta+\alpha\sigma\rho)(t-T)}-1] \\ &+\frac{\gamma_{1}H^{2}(b^{2}+\lambda_{0}\sigma_{z}^{2})}{2r}[{\rm e}^{2r(T-t)}-1]-\frac{\lambda_{0}\mu_{z}}{r}[{\rm e}^{r(T-t)}-1] \\ &-\frac{\gamma_{1}H(b^{2}+\lambda_{o}\sigma_{z}^{2})} {2r}[{\rm e}^{2r(T-t)}-1]. \end{align*}$

因此最优策略为

$\begin{align*} &p^{*}(t)=a+\lambda_{0}\mu_{z}+(b^{2}+\lambda_{0}\sigma^{2}_{z})\gamma_{1}A_{2}(t)H,\quad q^{*}(t)=\frac{(1+2k)\gamma_{1}+\gamma_{2}}{2(1+k)\gamma_{1}+\gamma_{2}},\\ &\pi_{I}^{*}(t)=\frac{\alpha-\gamma_{1}\sigma\rho B_{2}(t)}{\gamma_{1}{\rm e}^{r(T-t)}}+k\frac{\alpha-\gamma_{2}\sigma\rho E_{2}(t)}{\gamma_{2}{\rm e}^{r(T-t)}}, \pi_{R}^{*}(t)=\frac{\alpha-\gamma_{2}\sigma\rho E_{2}(t)}{\gamma_{2}{\rm e}^{r(T-t)}}. \end{align*}$

总结以上各种情况, 我们可以得到下面的定理.

定理3.1 对于问题 (2.10) 和 (2.11), 保险公司和再保险公司的 Stackelberg 微分博弈的纳什均衡策略由($q^{*}(t),\ \pi_{I}^{*}(t),\ p^{*}(t),\ \pi_{R}^{*}(t)$)给出, 其中 $q^{*}(t)$ 和 $p^{*}(t)$ 由下面的表格1分情况给出, $\pi_{I}^{*}(t)$ 和 $\pi_{R}^{*}(t)$ 为

(3.40)$\pi_{I}^{*}(t)=\frac{\alpha-\gamma_{1}\sigma\rho B_{2}(t)}{\gamma_{1}{\rm e}^{r(T-t)}}+k\frac{\alpha-\gamma_{2}\sigma\rho E_{2}(t)}{\gamma_{2}{\rm e}^{r(T-t)}},$

(3.41)$\pi_{R}^{*}(t)=\frac{\alpha-\gamma_{2}\sigma\rho E_{2}(t)}{\gamma_{2}{\rm e}^{r(T-t)}},$

其中 $H$, $H^{\bar{\theta}}$ 和 $H^{\theta}$ 由等式 (3.1) 给出, $B_{2}(t)$ 和 $E_{2}(t)$ 由等式 (3.8) 和 (3.24) 给出.

表1 在保险市场中不同情况下的最优保费策略和再保险策略

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保险公司和再保险公司的值函数分别由以下等式给出

(3.42)$V_{I}(t,\widehat{x})=A_{1}(t)\widehat{x}+B_{1}(t)l+C_{1}(t),$

(3.43)$V_{R}(t,y)=D_{1}(t)y+E_{1}(t)l+F_{1}(t),$

其中

$\begin{align*} & A_{1}(t)=D_{1}(t)={\rm e}^{r(T-t)},\\ & B_{1}(t)=E_{1}(t)=\frac{\Theta {\rm e}^{2(\beta+\alpha\sigma\rho)(t-T)}}{\beta+2\alpha\sigma\rho}-\frac{(2\Theta+\Phi){\rm e}^{(\beta+\alpha \sigma \rho)(t-T)}}{\alpha \sigma \rho}+\xi {\rm e}^{\beta(t-T)}+\frac{\Psi}{\beta},\\ & C_{1}(t)=\begin{cases} C_{1}^{Ia}(t), \quad H^{\theta}\geq 1; \\ C_{1}^{Ib1}(t), \quad H^{\bar{\theta}}\leq H; \\ C_{1}^{Ib2 }(t), \quad H \leq H^{\theta}<1;\\ C_{1}^{Ib3}(t), \quad H^{\theta}<H<H^{\bar{\theta}}. \end{cases}\\ & F_{1}(t)=\begin{cases} F_{1}^{Ia}(t), \quad H^{\theta}\geq 1; \\ F_{1}^{Ib1}(t), \quad H^{\bar{\theta}}\leq H; \\ F_{1}^{Ib2 }(t), \quad H \leq H^{\theta}<1;\\ F_{1}^{Ib3}(t), \quad H^{\theta}<H<H^{\bar{\theta}}. \end{cases} \end{align*}$

$C_{1}^{Ia}(t),\ C_{1}^{Ib1}(t),\ C_{1}^{Ib2}(t),\ C_{1}^{Ib3}(t),\ F_{1}^{Ia}(t),\ F_{1}^{Ib1}(t),\ F_{1}^{Ib2}(t),\ F_{1}^{Ib3}(t)$ 分别由等式 (3.9), (3.30),

(3.34), (3.38), (3.25),(3.28),(3.32),(3.36)给出.

命题3.1 定理 3.1中的 $q^{*}(t)$ 除了情况 (1), 在其余情况下都是 $k$ 的增函数.

证在 (2), (3), (4) 分别对 $k$ 求导

$q^{*\prime}(t)= \begin{cases} \frac{1-H^{\bar{\theta}}}{(1+k)^{2}},\quad &(Ib1); \\ \frac{1-H^{\theta}}{(1+k)^{2}},\quad &(Ib2);\\ \frac{2\gamma_{1}^{2}}{[2(1+k)\gamma_{1}+\gamma_{2}]^{2}}, \quad &(Ib3), \\ \end{cases} $

很容易看出 $q^{* \prime}(t)$ 为正数, 因此除了情况 (1), 在其余情况下 $q^{*}(t)$ 是 $k$ 的增函数.

注3.1 从命题 3.1 中我们可以看出均衡保留水平 $q^{*}(t)$ 除了在情况 (1) 下 $q^{*}(t)=1$, 在其余情况下是 $k$ 的增函数. 正如我们所知, 更大的 $k$ 意味着保险公司更加有竞争力, 更加关心自己与再保险公司财富的差距, 竞争使得保险公司更愿意冒险. 一种可能的解释是, 竞争敏感性参数 $k$ 越大的保险公司相对于再保险公司更关注其预期终端盈余, 因此保险公司倾向于通过保留更多的风险来支付更少的再保险保费, 从而在终端时刻积累更多的财富.

注3.2 从定理 3.1, 我们可以得到以下两个结论: 均衡再保险-投资策略与当前财富无关; 保险公司的投资策略独立于再保险策略. 这与大多数文献的结论相一致,如文献[23,32,33].

注3.3 与文献[20]和[23]相似, 在定理3.1中的情况 (Ib3), 最优再保险保费遵循方差保费原则. 也就是说, 对于每一单位的风险, 与让出比例 $1-q^{*}(t)$ 相关的总瞬时再保险保费可写成

$p^{*}(t)(1-q^{*}(t))=(a+\lambda_{0} \mu_{z})(1-q^{*}(t))+[(1+k)\gamma_{1}+\gamma_{2}]A_{2}(t)(b^{2}+\lambda_{0}\sigma^{2}_{z})(1-q^{*}(t))^{2},$

第一部分是均值分量, 第二部分是方差分量.

4 数值分析

在本节中, 我们将给出一些数值例子, 对第 3 节推导的均衡再保险投资策略进行敏感性分析. 下面作为基准使用的基本模型参数如表2 所示, 除非另有说明, 否则这些参数不变.

表2 模型参数

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图1 显示了敏感性系数 $k$, 保险公司风险厌恶系数 $\gamma_{1}$ 对保险公司均衡投资策略的影响, 其中 $\gamma_{1}$ 取 0.3, 0.6, 0.9, 我们可以观察到, 敏感性系数 $k$ 越大的保险公司会增加对风险资产的投资. 也就是说, 随着敏感性系数 $k$ 的增大, 竞争的存在使保险公司更加追求风险, 保险公司将更多的资金投入到风险资产中. 这是因为, 如果保险公司增加了对风险资产的敞口, 那么他在最后时刻比再保险公司创造更多财富的机会就会增加. 另外, 当 $k = 0$ 时, 风险厌恶系数 $\gamma_{2}$ 不影响保险公司的投资策略, 因为保险公司不考虑与再保险公司之间的财富差距, 保险公司只考虑自己的投资策略. 此外, 对于固定敏感性系数为正的情况, 保险公司的风险厌恶系数越小, 保险公司对风险资产的投资越多. 图2 描述了敏感性参数 $k$, 再保险公司风险厌恶系数 $\gamma_{2}$ 对再保险公司均衡投资策略的影响, 其中 $\gamma_{2}$ 取 0.1, 0.4, 0.7. 我们很容易发现, 再保险公司的投资策略与敏感度系数 $k$ 无关, 这是因为再保险公司只关心自己的财富. 随着再保险公司风险厌恶系数 $\gamma_{2}$ 的增大, 再保险公司会减少对风险资产的投资.

图1

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图1 $k$ 和 $\gamma_{1}$ 对 $\pi_{I}^{*}(0)$ 的影响

图2

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图2 $k$ 和 $\gamma_{2}$ 对 $\pi_{R}^{*}(0)$ 的影响

图3 和图4 描述了 $t$ 和 $\sigma$ 分别对 $\pi_{I}^{*}(t)$ 和 $\pi_{R}^{*}(t)$ 的影响. 从图3 和图4 可以看出 $\pi_{I}^{*}(t)$ 和 $\pi_{R}^{*}(t)$ 随着 $t$ 的增加而增加, 这是因为随着时间的推移, 保险公司和再保险公司将会获得更多的盈余, 然后他们可以将更多的财富投资于风险资产. $\pi_{I}^{*}(t)$ 和 $\pi_{R}^{*}(t)$ 随着 $\sigma$ 的增加而减少, 这是因为随着风险资产的波动率更大, 保险公司和再保险公司对风险资产的投资将会变得更加保守.

图3

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图3 $t$ 和 $\sigma$ 对 $\pi_{I}^{*}(t)$ 的影响

图4

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图4 $t$ 和 $\sigma$ 对 $\pi_{R}^{*}(t)$ 的影响

图5 和图6 描述了再保险公司的最优保费定价策略 $p^{*}(t)$ 和保险公司的最优保留水平 $q^{*}(t)$ 随时间的变化. 图5 和图6 对应的数值结果如表3 所示, 我们可以看出, 由于保险公司的财富水平不足以支持其早期的所有索赔, 因此保险公司有强烈的愿望与再保险公司签订再保险合同, 以降低其索赔风险. 因此, 由于再保险公司在供需关系中的优势地位, 再保险公司提供最昂贵的再保险保费价格作为其最优保费定价策略. 随着时间的推移, 保险公司也试图通过增加保留水平 $q^{*}(t)$ 来降低再保险成本, 这意味着在这种情况下, 最优保留水平 $q^{*}(t)$ 是时间 $t$ 的递增函数, 同时保险公司积累了财富. 当定理 3.1 中的条件 $H^{\theta}<H<H^{\bar{\theta}}$ 满足时, 保险公司和再保险公司将达到一个新的均衡. 保险公司有足够的财富承担固定比例的索赔 (即 $\frac{(1+2k)\gamma_{1}+\gamma_{2}}{2(1+k)\gamma_{1}+\gamma_{2}}$), 其再保险策略保持不变. 而再保险公司则降低其安全负荷, 以刺激保险公司增加再保险比例, 这是图5 中的 Case(4) 和表3 中 $t>5$ 的情况.

图5

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图5 $t$ 对 $p^{*}(t)$ 的影响

图6

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图6 $t$ 对 $q^{*}(t)$ 的影响

表3 数值结果

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图7 描述了在 Case(4) 中 $\lambda_{0}$ 对 $p^{*}(t)$ 的影响, 随着 $\lambda_{0}$ 的增大, 突发事件的平均索赔次数增加, 再保险公司不得不提高再保险的价格. 同时, 我们可以看到 $p^{*}(t)$ 随着 $t$ 的增加而减小, 这是因为随着 $t$ 的增加, 保险公司的财富积累到了一定的水平, 购买再保险的意愿就会减小, 此时再保险公司不得不降低再保险的价格以刺激保险公司购买再保险.从图8 中我们可以看出, 在 Case(2) 中 $q^{*}(t)$ 随着 $t$ 的增加而增加, 因为随着 $t$ 的增加, 保险公司不断积累财富, 有能力处理一定的风险, 从而将更多的盈余投入到风险资产中. $q^{*}(t)$ 随着 $\lambda_{0}$ 的增加而增加, 一种可能的解释是在这种情况下, 保险公司认为再保险保费已经达到上限, 购买再保险反而会得不偿失, 因此自身保留更多的风险.

图7

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图7 $\lambda_{0}$ 和 $t$ 对 $p^{*}(t)$ 的影响

图8

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图8 $\lambda_{0}$ 和 $t$ 对 $q^{*}(t)$ 的影响

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Optimal investment strategy to minimize the ruin probability of an insurance company under borrowing constraints

2009

... 近年来, 投资和再保险是保险公司提高承保能力、控制风险、保证保险业务持续健康发展的重要业务活动. 因此, 有关保险公司最优投资和再保险问题在保险精算领域引起了广泛的关注, 对不同优化准则下的保险公司的最优再保险投资策略进行了大量的研究, 其中应用最广泛的准则包括最小化破产概率^[1-3]和最大化终端财富期望效用^[4-7]以及均值-方差准则^[8-10]. ...

Minimizing the probability of absolute ruin under the mean-variance premium principles

2021

Minimizing the probability of ruin under interest force

2008

The optimal investment problem for an insurer and a reinsurer under the constant elasticity of variance model

2016

Optimal investment strategies for an insurer and a reinsurer with a jump diffusion risk process under the CEV model

2018

Optimal proportional reinsurance and investment for a constant elasticity of variance model under variance principle

2015

基于模型不确定性的保险人最优投资再保险问题研究

2022

基于模型不确定性的保险人最优投资再保险问题研究

2022

Dynamic mean-variance and optimal reinsurance problems under the no-bankruptcy constraint for an insurer

2014

Optimal investment for the insurer in the Levy market under the mean-variance criterion

2010

Portfolio selection and risk control for an insurer in the Lévy market under mean-variance criterion

2017

The optimal reinsurance treaty

1969

... 在上述文章中, 大多数只从保险公司的角度考虑最优策略而忽略了再保险公司. 1969 年, Borch^[11]表明双方在再保险保费上存在利益冲突, 再保险保费应该是双方共同达成的协议, 只涉及一方利益的再保险策略可能不会被另一方所接受. 一般来说, 从保险公司和再保险公司的共同角度来研究再保险有两种方法: 一种方法是研究保险公司和再保险公司的加权目标,Zhang 等^[12]在五种不同的标准下, 研究了基于保险公司和再保险公司的共同利益的最优份额再保险问题.Yang 和 Chen^[13]考虑了保险公司和再保险公司之间的共同利益, 保险公司和再保险公司分别从各自角度出发, 在最大化终端时刻期望值同时减小其方差的目标下, 得到了最优再保险合同及投资策略. Yang^[14]在均值-方差准则下研究了保险集团的最优投资再保险策略, 通过对保险公司和再保险公司的财富赋予不同的权重, 研究了模型参数对最优策略的影响.Cai 等^[15]考虑了保险公司和再保险公司的利益, 研究了保险公司与再保险公司的共同生存和盈利概率, 在一般再保险保费原则下, 得到了在一类保险政策中存在最优再保险条约的充分条件.Zhao 等^[16]在 CEV 模型下研究了兼顾保险公司和再保险公司利益的最优投资再保险策略. ...

Optimal quota-share reinsurance based on the mutual benefit of insurer and reinsurer

2018

Optimal reinsurance pricing, risk sharing and investment strategies in a joint reinsurer-insurer framework

2023

2022

Optimal reciprocal reinsurance treaties under the joint survival probability and the joint profitable probability

2013

Time-consistent investment-reinsurance strategies towards joint interests of the insurer and the reinsurer under CEV models

2017

A BSDE approach to a risk-based optimal investment of an insurer

2011

... 另一种方法是制定保险公司和再保险公司之间的 Stackelberg 博弈, 在博弈中再保险公司是领导者, 保险公司是追随者. 再保险公司首先给出任意的再保险保费策略, 保险公司根据再保险公司给定的保费策略得到最优的保险策略, 最后再保险公司根据保险公司的最优策略调整其再保险保费, 从而得到 Stackelberg 均衡策略.事实上, 保险业务中的随机微分博弈已经在几个竞争的保险公司或保险公司和市场之间进行了广泛的研究. 在保险公司与市场零和博弈的框架下,2011 年, Elliott 和 Siu^[17]通过求解倒向随机微分方程解决了最优投资问题,它给出了一种简单自然的方法, 在没有马尔可夫假设的情况下验证了最优投资策略的存在性和唯一性. 在均值-方差效用准则下,Wang 等^[18]研究了两家竞争保险公司的最优策略, 这两家保险公司的索赔过程由复合泊松风险模型和扩散近似模型描述, 两者的目标都是最大化相对于对手的预期终端盈余, 得到了均衡再保险投资策略和相应的值函数. Yan 等^[19]研究了两家保险公司之间的再保险和投资博弈, 两者对开始时获得的索赔信息采取不同态度, 以最大化终端时刻与竞争对手盈余差额为目标, 在指数效用下得到了均衡策略. 然而, 这些文章没有涉及到保险公司和再保险公司之间的博弈. Chen^[20]首先从保险公司和再保险公司相互影响的角度, 运用 Stackelberg 微分博弈分析了最优再保险问题, 在期望效用最大化准则下, 得到了最优保费定价策略和特殊情况下的最优再保险策略. Chen 与 Shen^[21]在均值-方差框架下研究了保险公司和再保险公司之间的 Stackelberg 微分博弈, 在均值保费准则和方差保费准则下分别得到了再保险策略和值函数. Li 与 Young^[22]进一步推广了文献[21], 在随机时刻下研究了保险公司和再保险公司之间的 Stackelberg 微分博弈, 在均值-方差保费准则下得到最优再保险策略为超额损失再保险策略. 同时, 还分析了不同风险分布对均衡再保险和保费策略的影响, 如果索赔分布是轻尾的, 那么均衡再保险就是纯粹的超额损失再保险,对损失的方差没有负荷; 如果索赔分布是重尾的, 那么均衡再保险就有一个非平凡共保, 对损失的方差有相应的正负荷. Bai^[23]考虑到保险公司和再保险公司在金融市场上信息不对称, 还考虑到当前财富受到过去业绩的影响, 在常数绝对风险厌恶效用函数下得到最优再保险和投资策略. ...

Reinsurance-investment game between two mean-variance insurers under model uncertainty

2021

A reinsurance and investment game between two insurance companies with the different opinions about some extra information

2017

On a new paradigm of optimal reinsurance: A stochastic Stackelberg differential game between an insurer and a reinsurer

2018

... 目前, 保险公司和再保险公司在保险市场上的地位并不平等, 因为在保险市场上保险公司的数量很多, 保险公司之间竞争很激烈, 而再保险公司的数量却很少, 再保险业务被少量再保险公司所垄断. 因此, 在保险市场中有一个领导者. 在下面, 我们将再保险公司和保险公司分别作为博弈的领导者和追随者. 根据文献[20]和[23], 我们的目标是依次求解优化问题. 更具体地来说, 我们可以将求解 Stackelberg 随机微分优化问题分为以下三个步骤 ...

... 注3.3 与文献[20]和[23]相似, 在定理3.1中的情况 (Ib3), 最优再保险保费遵循方差保费原则. 也就是说, 对于每一单位的风险, 与让出比例

$1-q^{*}(t)$

相关的总瞬时再保险保费可写成 ...

Stochastic Stackelberg differential reinsurance games under time-inconsistent mean-variance framework

2019

... 进一步推广了文献[21], 在随机时刻下研究了保险公司和再保险公司之间的 Stackelberg 微分博弈, 在均值-方差保费准则下得到最优再保险策略为超额损失再保险策略. 同时, 还分析了不同风险分布对均衡再保险和保费策略的影响, 如果索赔分布是轻尾的, 那么均衡再保险就是纯粹的超额损失再保险,对损失的方差没有负荷; 如果索赔分布是重尾的, 那么均衡再保险就有一个非平凡共保, 对损失的方差有相应的正负荷. Bai^[23]考虑到保险公司和再保险公司在金融市场上信息不对称, 还考虑到当前财富受到过去业绩的影响, 在常数绝对风险厌恶效用函数下得到最优再保险和投资策略. ...

Stackelberg differential game for reinsurance: Mean-variance framework and random horizon

2022

A Stackelberg reinsurance-investment game with asymmetric information and delay

2021

... 类似文献[23], 作为 Stackelberg 博弈中的追随者, 保险公司的财富远低于再保险公司. 由于比较心理, 保险公司既关心自己的最终财富, 也关心自己与再保险公司之间的财富差距, 即保险人的目标是寻找一种再保险投资策略, 使下面定义的均值-方差成本函数在其相对绩效下最大化. 这里我们将保险公司的相对绩效定义为

$ \widehat{X}(t):=(1-k)X(t)+k(X(t)-Y(t))=X(t)-kY(t),$

这里常数

$k\in[0, 1]$

表示保险公司对再保险公司绩效的敏感性,

$k$

越大表示保险公司越在意与再保险公司的财富差距, 保险公司越有竞争力. 当

$k=0$

时, 最优问题就变为没有竞争的传统的优化问题. ...

... 注3.2 从定理 3.1, 我们可以得到以下两个结论: 均衡再保险-投资策略与当前财富无关; 保险公司的投资策略独立于再保险策略. 这与大多数文献的结论相一致,如文献[23,32,33]. ...

... 注3.3 与文献[20]和[23]相似, 在定理3.1中的情况 (Ib3), 最优再保险保费遵循方差保费原则. 也就是说, 对于每一单位的风险, 与让出比例

$1-q^{*}(t)$

相关的总瞬时再保险保费可写成 ...

Expected stock returns and volatility

1987

... 在上述的大多数文献中, 大多数假设风险资产的价格过程遵循几何布朗运动, 这意味着风险资产的波动率是恒定的. 然而, 这与波动率为随机的实证研究结果相冲突, 比如文献[24]和[25]. 因此, 我们有必要考虑风险资产的波动率为随机的模型. Heston 随机波动率模型是最受欢迎的随机波动率模型之一, 它考虑了波动率的均值回归特性、长时期和短时期的波动率特性, 这使得模型在捕捉市场现实波动的特点时表现得更为合理. 同时, Heston 模型可以解释很多众所周知的发现, 比如波动率微笑, 波动率聚类, 收益分布的重尾性质等. Heston 模型还为欧式看涨和看跌期权提供了半解析闭式解,在某些情况下, Heston 模型可以在没有计算机模拟或复杂数值方法的情况下得到期权价格, 这对于市场参与者来说是一个很大的优点. 为了使模型更加贴近现实, Heston^[26]提出了风险资产的波动率是平方根扩散过程驱动的模型, 该模型具有一定的计算优势. 在保险公司和再保险公司对金融市场中的信息了解程度不同时, Guan 等^[27]在 Heston 随机波动率模型和 Alpha-maxmin 均值-方差准则下研究了保险公司和再保险公司的

$\alpha$

-Robust 策略.Zhou 等^[28]在 Heston 随机波动率模型下研究了保险公司和再保险公司的 Stackelberg 随机微分博弈, 在均值-方差效用准则下得到了保险公司和再保险公司的均衡策略和值函数. 后来许多学者将其引入到优化问题中, 用 Heston 随机波动率来描述风险资产. ...

Alternative models for conditional stock volatility

1990

$\alpha$

A closed-form solution for options with stochastic volatility with applications to bond and currency options

1993

$\alpha$

A Stackelberg reinsurance-investment game under $\alpha$-maxmin mean-variance criterion and stochastic volatility

2024

$\alpha$

A Stackelberg reinsurance-investment game under Heston's stochastic volatility model

2023

$\alpha$

A general theory of markovian time inconsistent stochastic control problems

2010

... 众所周知, 均值-方差准则下的再保险投资问题是时间不一致的. 按照文献[29]的方法, 我们寻求问题 (2.10) 和 (2.11) 的均衡策略和均衡值函数, 定义如下 ...

Time-consistent equilibrium reinsurance-investment strategy for $n$ competitive insurers under a new interaction mechanism and a general investment framework

2020

... 根据命题2.1, 我们将保险公司和再保险公司的值函数简化为

$V_{I}(t,\widehat{x},l)$

和

$V_{R}(x,y,l)$

.令

$C^{1,2,2}(\left[ 0,T \right] \times R \times R^{+})$

表示满足关于

$t$

一次可微, 关于

$x$

和

$l$

二次可微的函数

$\psi(t,x,l)$

的全体, 其中

$\psi(t,x,l)$

和

$\psi_{t}(t,x,l),\psi_{x}(t,x,l),\ \psi_{xx}(t,x,l),\ \psi_{l}(t,x,l),\ \psi_{ll}(t,x,l),\ \psi_{xl}(t,x,l)$

在

$[T]\times R \times R^{+}$

上是连续的.令

$D_{p}^{1,2,2}(\left[ 0,T \right] \times R \times R^{+})$

表示

$C^{1,2,2}(\left[ 0,T \right] \times R \times R^{+})$

中一阶偏导数关于参数

$x$

和

$l$

满足多项式增长条件的函数

$\psi(t,x,l)$

的全体(同文献[30]和[31]). 对于任意的函数

$\psi(t,\widehat{x},l)\in C^{1,2,2}$

, 变分算子

$\mathcal{L}^{u_{I},u_{R}}_{1}$

定义为以下形式 ...

... 类似文献[30]和[31], 下面给出假设. ...

Time-consistent reinsurance-investment games for multiple mean-variance insurers with mispricing and default risks

2024

... 根据命题2.1, 我们将保险公司和再保险公司的值函数简化为

$V_{I}(t,\widehat{x},l)$

和

$V_{R}(x,y,l)$

.令

$C^{1,2,2}(\left[ 0,T \right] \times R \times R^{+})$

表示满足关于

$t$

一次可微, 关于

$x$

和

$l$

二次可微的函数

$\psi(t,x,l)$

的全体, 其中

$\psi(t,x,l)$

和

$\psi_{t}(t,x,l),\psi_{x}(t,x,l),\ \psi_{xx}(t,x,l),\ \psi_{l}(t,x,l),\ \psi_{ll}(t,x,l),\ \psi_{xl}(t,x,l)$

在

$[T]\times R \times R^{+}$

上是连续的.令

$D_{p}^{1,2,2}(\left[ 0,T \right] \times R \times R^{+})$

表示

$C^{1,2,2}(\left[ 0,T \right] \times R \times R^{+})$

中一阶偏导数关于参数

$x$

和

$l$

满足多项式增长条件的函数

$\psi(t,x,l)$

的全体(同文献[30]和[31]). 对于任意的函数

$\psi(t,\widehat{x},l)\in C^{1,2,2}$

, 变分算子

$\mathcal{L}^{u_{I},u_{R}}_{1}$

定义为以下形式 ...

... 类似文献[30]和[31], 下面给出假设. ...

... 类似文献[31] 可得, 积分 ...

A class of non-zero-sum stochastic differential investment and reinsurance games

2014

Non-zero-sum stochastic differential reinsurance and investment games with default risk

2018

〈

〉