最优寿险消费投资是寿险研究中的一个重要问题. Richard^[1]首次将人寿保险与著名的Merton连续时间模型^{[2, 3]}相结合, 研究了最优消费和投资问题. 从那时起, 许多学者对这一问题进行了深入研究. 分析该优化问题的经典方法是随机动态规划, 例如, Pliska和Ye^[4]使用动态规划研究了具有随机寿命的有薪雇员的最优人寿保险和消费规则. Huang等^[5]研究了具有随机收益情况下的最优人寿保险, 消费和投资选择问题. Mousa等^[6]研究了存在多家人寿保险提供商的市场中, 消费, 投资和人寿保险选择的最佳决策.

研究优化问题的另一种方法是基于等价鞅测度和鞅表示定理的鞅方法. Ye^[7]考虑了具有随机寿命的最优人寿保险, 消费和投资组合选择问题, 并用鞅方法进行了求解. Michelbrink和Le^[8]使用鞅方法研究了跳-扩散模型下使消费和终端财富的预期效用最大化的最优投资-消费策略, 并获得了幂和对数效用函数情形下最优策略的显式解. Kwak等^[9]探讨了一个家庭的最优投资, 消费和人寿保险购买问题, 利用鞅方法找到了价值函数和最优策略的显式解. Liang和Guo^[10]研究了不完备市场中工资型企业的最优保险消费投资问题, 利用鞅方法得到了最优策略. Guambe和Kufakunesu^[11]研究了当投资者受限于资本担保时的最优投资, 消费和人寿保险问题, 并得到了幂效用函数下的最优策略的显式解.

该文采用最小最大鞅测度的纯概率方法研究Lévy模型中的最优寿险消费投资问题. 最小最大局部鞅测度由He和Pearson^[12]引入, Karatzas等^[13], Bellini和Frittelli^{[14, 15]}和Goll和Rüschendorf^[16]对该方法的理论和应用进行多方面的研究. 基于最小最大鞅测度, 该文给出了最优寿险消费投资问题在对数效用函数, 指数效用函数和幂效用函数下的显式解.

$ (\Omega, {\cal F}, ({\cal F}_{t})_{0\leq t\leq T}, {\mathbb{P}}) $为一域流概率空间, 该文中的所有随机过程均定义在该空间中. 假设金融市场中存在一个无风险资产和一个风险资产, 其运动状态由下面的随机微分方程刻画

(2.1) $ \begin{equation} {\rm d}S_{0}(t)=S_{0}(t)r{\rm d}t, \end{equation} $

(2.2) $ \begin{equation} {\rm d}S_{1}(t)=S_{1}(t) \bigg[\mu{\rm d}t+\sigma{\rm d}W(t)+\int_{{\mathbb{R}}}({\rm e}^{y}-1)N({\rm d}t, {\rm d}y)\bigg], \end{equation} $

令$ Y(t)=\ln\frac{S_{1}(t)}{S_{1}(0)} $, 则$ Y(t)=\mu t-\frac{1}{2}\sigma^{2}t+\sigma W(t)+\int_{0}^{t}\int_{{\mathbb{R}}}yN({\rm d}u, {\rm d}y) $, 其中$ \mu $, $ \sigma $和$ r $为正常数. 在概率空间$ (\Omega, {\cal F}, ({\cal F}_{t})_{0\leq t\leq T}, {\mathbb{P}}) $中, $ W(t) $是一标准布朗运动, $ N({\rm d}t, {\rm d}y) $点过程$ \triangle Y(t)=Y(t)-Y(t-) $的Poisson测度, $ Y(t-)=\lim\limits_{u\uparrow t}Y(u) $, $ N({\rm d}t, {\rm d}y) $的补偿为$ \nu({\rm d}y){\rm d}t $, 其中$ \nu({\rm d}y) $是$ Y(t) $的Lévy测度, 其满足条件

(2.3) $ \begin{equation} \int_{-\infty}^{\infty}\frac{y^{2}}{1+y^{2}}\nu({\rm d}y)<\infty. \end{equation} $

该文假定$ \int_{|y|\leq1}|y|\nu({\rm d}y)<\infty $.

在保险市场中, 令$ \lambda(t) $为瞬时死亡力, 由下式定义

(2.4) $ \begin{equation} \lambda(t)=\lim\limits_{\varepsilon\rightarrow0}\frac{P(t<\tau\leq t+\varepsilon|\tau>t)}{\varepsilon}, \end{equation} $

其中$ \tau\geq0 $是随机生存时间. 另外有$ \tau $的条件分布

(2.5) $ \begin{equation} {\mathbb{P}}(\tau<s|\tau>t)=1-\exp\bigg\{-\int_{t}^{s}\lambda(u){\rm d}u\bigg\}, \end{equation} $

(2.6) $ \begin{equation} {\mathbb{P}}(\tau>T|\tau>t)=\exp\bigg\{-\int_{t}^{T}\lambda(u){\rm d}u\bigg\}. \end{equation} $

假设一职员在退休时间$ T $之前有一个非随机的收入率函数$ i(t) $. 令$ c(t):\Omega\times [0, T]\rightarrow {\mathbb{R}}^{+} $为$ ({\cal F})_{t} $ -循序可测的消费过程, 其满足$ \int_{0}^{t}c(s){\rm d}s<\infty $ a.s.; $ \theta(t):\Omega\times [0, T]\rightarrow {\mathbb{R}} $是一$ ({\cal F})_{t} $ -循序可测投资组合过程, 其满足条件$ \int_{0}^{t}\theta^{2}(s){\rm d}s<\infty $ a.s.; $ p(t):\Omega\times [0, T]\rightarrow {\mathbb{R}} $是一$ ({\cal F})_{t} $-循序可测保费过程, 其满足条件$ \int_{0}^{t}p(s){\rm d}s<\infty $ a.s.; 则该职员在$ \pi(t)=(c(t), p(t), \theta(t)) $的控制过程下的财富过程为

(2.7) $ \begin{eqnarray} {\rm d}X(t)&=&[rX(t)+\theta(t)(\mu-r)+i(t)-p(t)-c(t)]{\rm d}t{}\\ &&+\theta(t)\sigma{\rm d}W(t)+\theta(t)\int_{{\mathbb{R}}}({\rm e}^{y}-1)N({\rm d}t, {\rm d}y), \end{eqnarray} $

且$ X(0)=x_{0}>0 $.

在死亡时刻$ \tau=t $, 保险公司支付给投保人保险金$ \frac{p(t)}{\eta(t)} $, 其中$ \eta:[0, T]\rightarrow {\mathbb{R}}^{+} $是一个连续的确定函数, 其满足条件$ \eta(t)\geq\lambda(t) $.

定义准则函数

(2.8) $ \begin{eqnarray} J(X(t), \pi(t))&=&E_{t}\bigg[\int_{t}^{T\wedge\tau}{\rm e}^{-\delta(s-t)}U_{1}(c(s)){\rm d}s {}\\ &&+{\rm e}^{-\delta(\tau-t)}U_{2}(M(\tau))I_{\{\tau\leq T\}}+{\rm e}^{-\delta(T-t)}U_{3}(X(T))I_{\{\tau>T\}}\bigg], \end{eqnarray} $

其中$ \delta>0 $是折现率, $ E_{t}[.]=E[.|{\cal F}_{t}] $表示概率测度$ {\mathbb{P}} $下的条件期望, $ M(t)=X(t)+\frac{p(t)}{\eta(t)} $, $ I_{\{.\}} $为示性函数. 效用函数$ U_{i}: {\mathbb{R}}_{+}\rightarrow {\mathbb{R}}, i=1, 2, 3 $是$ {\rm dom}(U_{i}):=\{x\in {\mathbb{R}}, U_{i}(x)>-\infty\} $上连续可微, 严格增的凹函数.假设$ U'_{i}(\infty)=\lim\limits_{x\rightarrow \infty}U'_{i}(x)=0 $, 且$ U'_{i}(\bar{x})=\lim\limits_{x\downarrow \bar{x}} U'_{i}(x)=\infty $, 其中$ \bar{x}=\inf\{x\in{\mathbb{R}}|U_{i}(x)>-\infty\} $. 令$ I_{i}(.):=(U_{i}')^{-1} $. $ U_{i} $的凸共轭函数$ \tilde{U_{i}}: {\mathbb{R}}_{+}\rightarrow {\mathbb{R}} $定义为$ \tilde{U_{i}}(y):=\sup_{x\in{\mathbb{R}}}\{U_{i}(x)-xy\}=U_{i}(I_{i}(y))-yI_{i}(y) $.

令$ {\cal A}(t) $表示使得(2.8)式成立的可允许策略集, 即满足

(2.9) $ \begin{equation} E_{t}\bigg[\int_{t}^{T\wedge\tau}{\rm e}^{-\delta(s-t)}U^{-}_{1}(c(s)){\rm d}s +{\rm e}^{-\delta(\tau-t)}U^{-}_{2}(M(\tau))I_{\{\tau\leq T\}}+{\rm e}^{-\delta(T-t)}U^{-}_{3}(X(T))I_{\{\tau>T\}}\bigg]<\infty, \end{equation} $

其中$ U^{-}_{i}(x)=\max(-U_{i}(x), 0) $.

值函数定义为

(2.10) $ \begin{equation} V(t, X(t))=\sup\limits_{\pi\in {\cal A}(t)}J(X(t), \pi). \end{equation} $

由$ \tau $的条件分布(1.5)和(1.6), 值函数可表示为

(2.11) $ \begin{equation} V(t, X(t))=\sup\limits_{\pi\in {\cal A}(t, X(t))}E_{t}\bigg[\int_{t}^{T}\xi'_{ts}U_{1}(c(s)){\rm d}s +\int_{t}^{T}\lambda(s)\xi'_{ts}U_{2}(M(s)){\rm d}s+\xi'_{tT}U_{3}(X(T))\bigg], \end{equation} $

其中$ \xi'_{ts}={\rm e}^{-\int_{t}^{s}(\lambda(u)+\delta){\rm d}u}, t\leq s\leq T $.

令$ {\cal P} $表示$ \Omega\times[0, T] $上的可料$ \sigma $ -代数, $ {\cal K} $表示$ v=(\phi_{1}(t), \phi_{2}(t, y)) $的集合, 其中$ \phi_{1}(t) $是可料实值过程, $ \phi_{2}(t, y) $是正的$ {\cal P}\times {\cal B}(R) $可测过程, 其满足如下两个条件

(3.1) $ \begin{equation} \int_{0}^{T}\phi_{1}^{2}(t){\rm d}t<\infty, \end{equation} $

(3.2) $ \begin{equation} \int_{0}^{T}\int_{{\mathbb{R}}}|\phi_{2}(t, y)-1|\nu({\rm d}y){\rm d}t<\infty. \end{equation} $

对$ v=(\phi_{1}, \phi_{2})\in{\cal K} $考虑一组随机指数鞅

(3.3) $ \begin{eqnarray} Z^{v}(t)&=&{\cal E}\bigg(\int_{0}^{t}\phi_{1}(u){\rm d}W(u) +\int_{0}^{t}\int_{{\mathbb{R}}}(\phi_{2}(u, y)-1)(N({\rm d}u, {\rm d}y)-\nu({\rm d}y){\rm d}u)\bigg){}\\ &=&\exp\bigg\{\int_{0}^{t}\phi_{1}(u){\rm d}W(u)+\int_{0}^{t}\int_{{\mathbb{R}}}\ln\phi_{2}(u, y)N({\rm d}u, {\rm d}y){}\\ &&+\int_{0}^{t}\bigg[-\frac{1}{2}\phi_{1}(u)^{2}-\int_{{\mathbb{R}}}(\phi_{2}(u, y)-1)\nu({\rm d}y)\bigg]{\rm d}u\bigg\}, \end{eqnarray} $

其是下面方程的解

$ {\rm d}Z^{v}(t)=Z^{v}(t-)\bigg[\phi_{1}(t){\rm d}W(t) +\int_{{\mathbb{R}}}(\phi_{2}(t, y)-1)(N({\rm d}t, {\rm d}y)-\nu({\rm d}y){\rm d}t)\bigg], $

且$ Z^{v}(0)=1 $. 其中$ {\cal E}(.) $表示随机指数. 令$ {\cal K}_{m}=\{v\in{\cal K};Z^{v} $是$ {\mathbb{P}} $下的鞅$ \} $. 则对任意的$ v\in{\cal K}_{m} $, 可以在$ (\Omega, {\cal F}) $上定义密度函数为$ Z^{v}(t) $的概率测度$ {\mathbb{P}}^{v}\sim{\mathbb{P}} $. 令$ {\cal Z}=\{Z^{v};Z^{v} $满足(3.3)式及$ \int_{0}^{T}\phi_{1}^{2}(t){\rm d}t<\infty, \ \int_{0}^{T}\int_{{\mathbb{R}}}|\phi_{2}(t, y)-1|\nu({\rm d}y){\rm d}t<\infty\ {\mathbb{P}} $-a.s.}. 由无套利原理, 对$ \forall v\in {\cal K}_{m} $, $ {\rm e}^{-rt}S_{1}(t) $是$ {\mathbb{P}}^{v} $下的鞅, 因此

(3.4) $ \begin{equation} \mu-r+\sigma\phi_{1}(t)+\int_{{\mathbb{R}}}({\rm e}^{y}-1)\phi_{2}(t, y)\nu({\rm d}y)=0. \end{equation} $

(3.4)式被称为鞅条件, 关于鞅条件的详细讨论可参见文献[17]. 由Girsanov定理, 有

(3.5) $ \begin{equation} W^{v}(t)=W(t)-\int_{0}^{t}\phi_{1}(u){\rm d}u \end{equation} $

是测度$ {\mathbb{P}}^{v} $下的布朗运动,

(3.6) $ \begin{equation} \nu^{v}({\rm d} y){\rm d}t=\phi_{2}(t, y)\nu({\rm d}y){\rm d}t \end{equation} $

是$ N({\rm d}t, {\rm d}y) $的$ {\mathbb{P}}^{v} $补偿. 对$ \forall v\in {\cal K}_{m} $, 定义

(3.7) $ \begin{eqnarray} U(t, Z^{v}):&=&\sup\limits_{\pi\in{\cal A}(t)}\bigg\{J(X(t), \pi):{}\\ &&E_{t}^{v}\bigg[\int_{t}^{T}\xi_{0s}(c(s)+\eta(s)M(s)){\rm d}s+\xi_{0T}X(T)\bigg]\leq(X(t)+b(t))\xi_{0t}\bigg\}, \end{eqnarray} $

其中$ b(t)=\int_{t}^{T}\xi_{ts}i(s){\rm d}s $, $ \xi_{ts}=\exp\{-\int_{t}^{s}(\eta(u)+r){\rm d}u\} $, $ E_{t}^{v}[.]={\rm e}^{v}[.|{\cal F}_{t}] $表示$ {\mathbb{P}}^{v} $下的条件期望.

定义3.1^{[12, 定义1]}  若(3.7)式的解存在且与(2.10) 式的解吻合, 则称$ Z^{v^{*}}\in{\cal Z} $定义了一个局部最小最大鞅测度.

下面讨论优化问题(2.10)的对偶问题

(3.8) $ \begin{equation} \inf\limits_{Z^{v}\in{\cal Z}}U(t, Z^{v}). \end{equation} $

利用Lagrange乘子$ \lambda>0 $, 问题(3.8)可退化为无限制的最大化问题,

(3.9) $ \begin{eqnarray} L(\lambda, v)&=&E_{t}\bigg[\int_{t}^{T}\xi_{ts}'U_{1}(c(s)){\rm d}s +\int_{t}^{T}\xi_{ts}'\lambda(s)U_{2}(M(s)){\rm d}s+\xi'_{tT}U_{3}(X(T))\bigg]{}\\ &&+\lambda\xi_{0t}'^{-1}\bigg\{(X(t)+b(t))\xi_{0t}Z^{v}(t){}\\ &&-E_{t}\bigg[Z^{v}(T)\xi_{0T}X(T) +\int_{t}^{T}\xi_{0s}Z^{v}(s)(c(s)+\eta(s)M(s)){\rm d}s\bigg]\bigg\}{}\\ &=&\lambda\xi_{0t}'^{-1}\xi_{0t}(X(t)+b(t))Z^{v}(t){}\\ && +E_{t}\bigg[\int_{t}^{T}(\xi_{ts}'U_{1}(c(s))-\lambda\xi_{0s}\xi'^{-1}_{0t}Z^{v}(s)c(s)){\rm d}s\bigg]{}\\ &&+E_{t}\bigg[\int_{t}^{T}(\xi_{ts}'\lambda(s)U_{2}(M(s))-\lambda\xi_{0s}\xi'^{-1}_{0t}\eta(s)Z^{v}(s)M(s)){\rm d}s\bigg]{}\\ &&+E_{t}\bigg[\xi_{tT}'U_{3}(X(T))-\lambda\xi_{0T}\xi'^{-1}_{0t}Z^{v}(T)X(T)\bigg]{}\\ &\leq&\lambda\xi_{0t}'^{-1}\xi_{0t}(X(t)+b(t))Z^{v}(t){}\\ &&+E_{t}\bigg[\int_{t}^{T}(\xi_{ts}'\tilde{U}_{1}(\frac{\lambda\xi_{0s}Z^{v}(s)}{\xi'_{0s}}) +\lambda(s)\xi_{ts}'\tilde{U}_{2}(\frac{\lambda\xi_{0s}\eta(s)Z^{v}(s)}{\lambda(s)\xi'_{0s}})){\rm d}s {}\\ &&+\xi_{tT}'\tilde{U}_{3}(\frac{\lambda\xi_{0T}Z^{v}(T)}{\xi'_{0T}})\bigg], \end{eqnarray} $

等号成立当且仅当$ c(s)=I_{1}(\frac{\lambda\xi_{0s}Z^{v}(s)}{\xi'_{0s}}), M(s)=I_{2}(\frac{\lambda\xi_{0s}\eta(s)Z^{v}(s)}{\lambda(s)\xi'_{0s}}) $且$ X(T)=I_{3}(\frac{\lambda\xi_{0T}Z^{v}(T)}{\xi'_{0T}}) $. 则问题(2.10)的对偶问题定义为

(3.10) $ \begin{eqnarray} &&\inf\limits_{\lambda>0}\bigg\{\inf\limits_{v\in{\cal K}_{m}}E_{t}\bigg[\int_{t}^{T}(\xi_{ts}'\tilde{U}_{1}(\frac{\lambda\xi_{0s}Z^{v}(s)}{\xi'_{0s}}) +\lambda(s)\xi_{ts}'\tilde{U}_{2}(\frac{\lambda\xi_{0s}\eta(s)Z^{v}(s)}{\lambda(s)\xi'_{0s}})){\rm d}s{}\\ &&+\xi_{tT}'\tilde{U}_{3}(\frac{\lambda\xi_{0T}Z^{v}(T)}{\xi'_{0T}})\bigg]+\lambda\xi_{0t}'^{-1}\xi_{0t}(X(t)+b(t))Z^{v}(t)\bigg\}. \end{eqnarray} $

对任意$ Z^{v}\in{\cal Z} $, 定义

$ H_{v}(\lambda)=E_{t}\bigg[\int_{t}^{T}(\xi_{ts}'\tilde{U}_{1}(\frac{\lambda\xi_{0s}Z^{v}(s)}{\xi'_{0s}}) +\lambda(s)\xi_{ts}'\tilde{U}_{2}(\frac{\lambda\xi_{0s}\eta(s)Z^{v}(s)} {\lambda(s)\xi'_{0s}})){\rm d}s+\xi_{tT}'\tilde{U}_{3}(\frac{\lambda\xi_{0T}Z^{v}(T)}{\xi'_{0T}})\bigg]. $

并假定$ \forall\tilde{\lambda}>0 $, 若$ H_{v}(\tilde{\lambda})<\infty $, 则存在一个常数$ \delta_{\lambda}\in(0, \tilde{\lambda}) $使得$ H_{v}(\tilde{\lambda}-\delta_{\lambda})<\infty $.

定理3.1  若(3.10)式有解, 则(3.8)和(3.7)式有解. 用$ (\lambda_{0}, v^{*}) $表示(3.10) 式的解, 则$ Z^{v^{*}} $是(3.8)式的解, $ Z^{v^{*}} $定义了一个最小最大局部鞅值测度. $ (c^{*}, p^{*}, \theta^{*}) $是(3.7)式的解, 其中$ c^{*}(t)=I_{1}(\frac{\lambda_{0}\xi_{0t}Z^{v^{*}}(t)}{\xi'_{0t}}), X(t)+\frac{p^{*}(t)}{\eta(t)}=I_{2}(\frac{\lambda_{0}\xi_{0t}\eta(t)Z^{v^{*}}(t)}{\lambda(t)\xi'_{0t}}) $, $ \lambda_{0} $是关于(3.7)式的Lagrange乘子.

证  考虑函数

(3.11) $ \begin{eqnarray} h(\lambda)&=&E_{t}\bigg[\int_{t}^{T}(\xi_{ts}'\tilde{U}_{1}(\frac{\lambda\xi_{0s}Z^{v^{*}}(s)}{\xi'_{0s}}) +\lambda(s)\xi_{ts}'\tilde{U}_{2}(\frac{\lambda\xi_{0s}\eta(s)Z^{v^{*}}(s)}{\lambda(s)\xi'_{0s}})){\rm d}s{}\\ &&+\xi_{tT}'\tilde{U}_{3}(\frac{\lambda\xi_{0T}Z^{v^{*}}(T)}{\xi'_{0T}})\bigg]+\lambda\xi_{0t}'^{-1}\xi_{0t}(X(t)+b(t))Z^{v^{*}}(t). \end{eqnarray} $

当$ \lambda\in(\lambda_{0}-\delta_{\lambda_{0}}, +\infty) $时, $ h $关于$ \lambda $可微, 其中$ \delta_{\lambda_{0}}>0 $. 由于$ h(\lambda) $$ \lambda=\lambda_{0} $达到最小值, $ h'_{\lambda}(\lambda_{0})=0 $; i.e.

(3.12) $ \begin{eqnarray} &&E_{t}\bigg[\int_{t}^{T}Z{v^{*}}(\xi_{ts}'\tilde{U}_{1}(\frac{\lambda_{0}\xi_{0s}Z^{v^{*}}(s)}{\xi'_{0s}}) +\lambda(s)\xi_{ts}'\tilde{U}_{2}(\frac{\lambda_{0}\xi_{0s}\eta(s)Z^{v^{*}}(s)}{\lambda(s)\xi'_{0s}})){\rm d}s{}\\ &&+\xi_{tT}'\tilde{U}_{3}(\frac{\lambda_{0}\xi_{0T}Z^{v^{*}}(T)}{\xi'_{0T}})\bigg] +\lambda_{0}\xi_{0t}'^{-1}\xi_{0t}(X(t)+b(t))Z^{v^{*}}(t)=0. \end{eqnarray} $

由于$ c^{*}(t)=I_{1}(\frac{\lambda_{0}\xi_{0t}Z^{v^{*}}(t)}{\xi'_{0t}}) $, 且$ X(t)+\frac{p^{*}(t)}{\eta(t)}=I_{2}(\frac{\lambda_{0}\xi_{0t}\eta(t)Z^{v^{*}}(t)}{\lambda(t)\xi'_{0t}}) $, 故$ (c^{*}, p^{*}, \theta^{*}) $是(3.7)式的解.

下面证明对$ \forall\lambda>0 $, $ Z^{v}\in{\cal Z} $, 有

(3.13) $ \begin{eqnarray} &&\inf\limits_{\lambda>0}\bigg\{\inf\limits_{v\in{\cal K}_{m}}E_{t}\bigg[\int_{t}^{T}(\xi_{ts}'\tilde{U}_{1}(\frac{\lambda\xi_{0s}Z^{v}(s)}{\xi'_{0s}}) +\lambda(s)\xi_{ts}'\tilde{U}_{2}(\frac{\lambda\xi_{0s}\eta(s)Z^{v}(s)}{\lambda(s)\xi'_{0s}})){\rm d}s{}\\ &&+\xi_{tT}'\tilde{U}_{3}(\frac{\lambda\xi_{0T}Z^{v}(T)}{\xi'_{0T}})\bigg]+\lambda\xi_{0t}'^{-1}\xi_{0t}(X(t)+b(t))Z^{v}(t)\bigg\} {}\\&\leq& E_{t}\bigg[\int_{t}^{T}(\xi_{ts}'\tilde{U}_{1}(\frac{\lambda\xi_{0s}Z^{v}(s)}{\xi'_{0s}}) +\lambda(s)\xi_{ts}'\tilde{U}_{2}(\frac{\lambda\xi_{0s}\eta(s)Z^{v}(s)}{\lambda(s)\xi'_{0s}})){\rm d}s{}\\ &&+\xi_{tT}'\tilde{U}_{3}(\frac{\lambda\xi_{0T}Z^{v}(T)}{\xi'_{0T}})\bigg]+\lambda\xi_{0t}'^{-1}\xi_{0t}(X(t)+b(t))Z^{v}(t). \end{eqnarray} $

当$ \lambda=\lambda_{0} $和$ Z^{v}=Z^{v^{*}} $时, 上式两边均等于$ U(t, Z^{v^{*}}) $, 故$ Z^{v^{*}} $定是(3.8) 式的解. 证毕.

假设职员的效用函数为$ U_{i}(x)=k_{i}\ln x $, 其中$ k_{i}>0, i=1, 2, 3 $为常数. 则$ I_{i}(y)=\frac{k_{i}}{y}, i=1, 2, 3. $对(3.9)式, 由一阶条件可得

(4.1) $ \begin{equation} c(s)=\frac{k_{1}\xi'_{0s}}{\lambda_{0}\xi_{0s}Z^{v}(s)}, \ \ M(s)=X(t)+\frac{p(t)}{\eta(t)}=\frac{k_{2}\lambda(s)\xi'_{0s}}{\lambda_{0}\xi_{0s}\eta(s)Z^{v}(s)}, \ \ X(T)=\frac{k_{3}\xi'_{0T}}{\lambda_{0}\xi_{0T}Z^{v}(T)}, \end{equation} $

其中$ \lambda_{0} $满足

(4.2) $ \begin{equation} E_{t}^{v}\bigg[\int_{0}^{T}\xi_{0s}(c(s)+\eta(s)M(s)){\rm d}s+\xi_{0T}X(T)\bigg]=(X(t)+b(t))\xi_{0t}. \end{equation} $

将(4.1)式代入(4.2)式可得

(4.3) $ \begin{equation} \xi_{tT}E_{t}[Z^{v}(T)X(T)]+E_{t}\bigg[\int_{t}^{T}\xi_{ts}Z^{v}(s)(c(s)+\eta(s)M(s)){\rm d}s\bigg]=(X(t)+b(t))Z^{v}(t), \end{equation} $

解该方程得到

(4.4) $ \begin{equation} \frac{X(t)+b(t)}{G_{t, T}}=\lambda_{0}^{-1}\xi_{0t}^{-1}Z^{v}(t)^{-1}, \end{equation} $

其中$ G_{t, T}=\int_{t}^{T}(k_{1}\xi'_{0s}+k_{2}\lambda(s)\xi'_{0s}){\rm d}s+k_{3}\xi'_{0T} $. 对$ X(t)+b(t) $采用Itô's公式, 可得

(4.5) $ \begin{eqnarray} {\rm d}(X(t)+b(t))&=&G_{t, T}\lambda_{0}^{-1}\xi_{0t}^{-1}{\rm d}Z^{v}(t)^{-1}+\lambda_{0}Z^{v}(t)^{-1}{\rm d}(G_{t, T}\xi_{0t}^{-1}){}\\ &=&-(X(t)+b(t))\phi_{1}(t){\rm d}W(t){}\\ &&+(X(t)+b(t))\int_{{\mathbb{R}}}(\frac{1}{\phi_{2}(t, y)}-1)N({\rm d}t, {\rm d}y){}\\ &&-(X(t)+b(t))\bigg[\int_{{\mathbb{R}}}(1-\phi_{2}(t, y))\nu({\rm d}y)-\phi_{1}^{2}(t)-\eta(t)-r\bigg]{\rm d}t{}\\ &&-\lambda_{0}^{-1}(k_{1}+k_{2}\lambda(t))Z^{v}(t)^{-1}\xi'_{0t}\xi_{0t}^{-1}{\rm d}t. \end{eqnarray} $

由$ b(t) $的定义, 有

(4.6) $ \begin{equation} {\rm d}b(t)=-i(t){\rm d}t+(\eta(t)+r)b(t){\rm d}t. \end{equation} $

将(4.6)式代入(4.5)式可得

(4.7) $ \begin{eqnarray} {\rm d}X(t)&=&-(X(t)+b(t))\phi_{1}(t){\rm d}W(t)\\ &&+(X(t)+b(t))\int_{{\mathbb{R}}}(\frac{1}{\phi_{2}(t, y)}-1)N({\rm d}t, {\rm d}y)\\ & &+\{(X(t)+b(t))\bigg[\int_{{\mathbb{R}}}(\phi_{2}(t, y)-1)\nu({\rm d}y)+\phi_{1}^{2}(t)\bigg]+(\eta(t)+r)X(t)+i(t)\\ &&-\lambda_{0}^{-1}\xi_{0t}^{-1}\xi'_{0t}(k_{1}+k_{2}\lambda(t))Z^{v}(t)^{-1}\}{\rm d}t. \end{eqnarray} $

将(4.1)式代入(2.7)式可得

(4.8) $ \begin{eqnarray} {\rm d}X(t)&=&\theta(t)\sigma{\rm d}W(t)+\theta(t)\int_{{\mathbb{R}}}({\rm e}^{y}-1)N({\rm d}t, {\rm d}y){}\\ &&+\bigg[(\eta(t)+r)X(t)+i(t)-\lambda_{0}^{-1}\xi_{0t}^{-1}\xi'_{0t}(k_{1}+k_{2}\lambda(t))Z^{v}(t)^{-1}+\theta(t)(\mu-r)\bigg]{\rm d}t. \end{eqnarray} $

比较(4.7)和(4.8)式中$ {\rm d}W $, $ N({\rm d}t, {\rm d}y) $的$ {\rm d}t $系数可得

(4.9) $ \begin{equation} -(X(t)+b(t))\phi_{1}(t)=\theta(t)\sigma, \end{equation} $

(4.10) $ \begin{equation} (X(t)+b(t))(\frac{1}{\phi_{2}(t, y)}-1)=\theta(t)({\rm e}^{y}-1), \end{equation} $

(4.11) $ \begin{equation} (X(t)+b(t))\bigg[\int_{{\mathbb{R}}}(\phi_{2}(t, y)-1)\nu({\rm d}y)+\phi_{1}^{2}(t)\bigg]=\theta(t)(\mu-r). \end{equation} $

可验证当$ \theta(t)\neq0 $和(4.9)–(4.11)式成立时鞅条件(3.4)满足.

将(4.1)式代入(3.8)式可得

(4.12) $ \begin{eqnarray} \inf\limits_{Z^{v}\in {\cal Z}}U(t, Z^{v})&=&\inf\limits_{Z^{v}\in {\cal Z}}E_{t}\bigg[\int_{t}^{T}\xi'_{ts}(k_{1}\ln\frac{k_{1}\xi'_{0s}}{\lambda_{0}\xi_{0s}Z^{v}(s)} +\lambda(s)k_{2}\ln\frac{k_{2}\lambda(s)\xi'_{0s}}{\lambda_{0}\xi_{0s}\eta(s)Z^{v}(s)}){\rm d}s\\ &&+\xi'_{tT}k_{3}\ln\frac{k_{3}\xi'_{0T}}{\lambda_{0}\xi_{0T}Z^{v}(T)}\bigg]\\ &=&\inf\limits_{Z^{v}\in {\cal Z}}\bigg\{H_{t, T}-E_{t}\bigg[\int_{t}^{T}\xi'_{ts}(k_{1}\ln Z^{v}(s) +\lambda(s)k_{2}\ln Z^{v}(s)){\rm d}s\\ & &+k_{3}\xi'_{tT}\ln Z^{v}(T)\bigg]\bigg\}, \end{eqnarray} $

其中$ H_{t, T}=\int_{t}^{T}\xi'_{ts}(k_{1}\ln\frac{k_{1}\xi'_{0s}}{\lambda_{0}\xi_{0s}} +\lambda(s)k_{2}\ln\frac{k_{2}\lambda(s)\xi'_{0s}}{\lambda_{0}\xi_{0s}\eta(s)}){\rm d}s +\xi'_{tT}k_{3}\ln\frac{k_{3}\xi'_{0T}}{\lambda_{0}\xi_{0T}} $. 由于$ k_{i}>0, i=1, 2, 3 $, 且$ \xi'_{ts}>0 $, (3.12)式等价于

(4.13) $ \begin{equation} \sup\limits_{Z^{v}\in{\cal Z}}E[\ln Z^{v}(s)] =\ln Z^{v}(t)+\sup\limits_{Z^{v}\in{\cal Z}}\int_{t}^{s}(-\frac{1}{2}\phi_{1}(u)^{2}+\int_{{\mathbb{R}}}(\ln\phi_{2}(u, y)-\phi_{2}(u, y)+1)\nu({\rm d}y)){\rm d}u. \end{equation} $

因此, 只需计算

(4.14) $ \begin{equation} \sup\limits_{v\in {\cal K}_{m}}\bigg\{-\frac{1}{2}\phi_{1}(u)^{2}+\int_{{\mathbb{R}}}(\ln\phi_{2}(u, y)-\phi_{2}(u, y)+1)\nu({\rm d}y)\bigg\}, \forall u>0, \end{equation} $

其中$ \phi_{1}, \phi_{2} $应当满足(3.9)–(3.10)式. 由(4.9)式可得

(4.15) $ \begin{equation} \phi_{1}(t)=-\frac{\theta(t)\sigma}{X(t)+b(t)}. \end{equation} $

由(4.10)式可得

(4.16) $ \begin{equation} \phi_{2}(t)=\frac{X(t)+b(t)}{X(t)+b(t)+({\rm e}^{y}-1)\theta(t)}. \end{equation} $

将(4.15)和(4.16)式代入(4.11)式可得

(4.17) $ \begin{equation} (X(t)+b(t))\bigg\{\frac{\sigma^{2}\theta(t)^{2}}{(X(t)+b(t))^{2}} +\int_{{\mathbb{R}}}(\frac{X(t)+b(t)}{X(t)+b(t)+({\rm e}^{y}-1)\theta(t)}-1)\nu({\rm d}y)\bigg\}=\theta(t)(\mu-r). \end{equation} $

令$ {\cal D}(\theta) $表示(4.17)式所有解的集合. 定义

(4.18)

问题(4.14)可表示为

(4.19)

定理4.1  对$ U_{i}=k_{i}\ln x, i=1, 2, 3 $, 值函数$ V(t, X(t)) $由下式给出

(4.20) $ \begin{equation} V(t, X(t))=H_{t, T}-\int_{t}^{T}(\xi'_{st}k_{1}+\lambda(s)\xi'_{ts}k_{2})\int_{t}^{s}g(u){\rm d}u{\rm d}s-\xi'_{tT}k_{3}\int_{t}^{T}g(s){\rm d}s, \end{equation} $

其中$ g(.) $表示(4.19)式的上确界,

$ H_{t, T}=\int_{t}^{T}\xi'_{ts}(k_{1}\ln\frac{k_{1}\xi'_{0s}}{\lambda_{0}\xi_{0s}} +\lambda(s)k_{2}\ln\frac{k_{2}\lambda(s)\xi'_{0s}}{\lambda_{0}\eta(s)\xi_{0s}}){\rm d}s +\xi'_{tT}k_{3}\ln\frac{k_{3}\xi'_{0T}}{\lambda_{0}\xi_{0T}}. $

最优保险、消费和投资策略为

(4.21) $ \begin{equation} \begin{array}{ll} { } \theta^{*}(t)=-\frac{\phi_{1}^{*}(t)(X(t)+b(t))}{\sigma}, \ \ c^{*}(t)=\frac{k_{1}\xi'_{0t}(X(t)+b(t))}{G_{t, T}}, \\ { } p^{*}(t)=\frac{k_{2}\lambda(t)\xi'_{0t}(X(t)+b(t))}{G_{t, T}}-X(t)\eta(t), \end{array} \end{equation} $

其中$ v=(\phi_{1}^{*}, \phi_{2}^{*}) $是(3.19)式的解, $ G_{t, T}=\int_{t}^{T}(k_{1}\xi'_{0s}+k_{2}\lambda(s)\xi'_{0s}){\rm d}s+k_{3}\xi'_{0T} $.

本节讨论效用函数为指数效用函数时(2.10)的解, 即$ U_{i}(x)=-\frac{1}{\alpha_{i}}\exp\{-\alpha_{i} x\}, \alpha_{i}>0, i=1, 2, 3. $此时$ I_{i}(y)=-\frac{1}{\alpha_{i}}\ln y, i=1, 2, 3. $

对(3.9)式, 由一阶条件

(5.1) $ \begin{eqnarray} c(s)&=&-\frac{1}{\alpha_{1}}\bigg[\ln\frac{\lambda_{0}\xi_{0s}}{\xi'_{0s}}+\ln Z^{v}(s)\bigg], {}\\ M(s)&=&X(s)+\frac{p(s)}{\eta(s)}=-\frac{1}{\alpha_{2}} \bigg[\ln\frac{\lambda_{0}\xi_{0s}\eta(s)}{\lambda(s)\xi'_{0s}}+\ln Z^{v}(s)\bigg], \\ X(T)&=&-\frac{1}{\alpha_{3}}\bigg[\ln\frac{\lambda_{0}\xi_{0T}}{\xi'_{0T}}+\ln Z^{v}(T)\bigg], {} \end{eqnarray} $

其中$ \lambda_{0} $满足

(5.2) $ \begin{equation} \xi_{tT}E_{t}[Z^{v}(T)X(T)]+E_{t}\bigg[\int_{t}^{T}\xi_{ts}Z^{v}(s)(c(s)+\eta(s)M(s)){\rm d}s\bigg]=(X(t)+b(t))Z^{v}(t). \end{equation} $

将(5.1)式代入(5.3)式可得

(5.3) $ \begin{eqnarray} &&\ln \lambda_{0}\bigg\{-\frac{\xi_{0T}}{\alpha_{3}}E_{t}[Z^{v}(T)] -\int_{t}^{T}(\frac{\xi_{0s}}{\alpha_{1}}+\frac{\eta(s)\xi_{0s}}{\alpha_{2}})E_{t}[Z^{v}(s)]{\rm d}s\bigg\} -\frac{\xi_{0T}}{\alpha_{3}}\ln\frac{\xi_{0T}}{\xi'_{0T}}E_{t}[Z^{v}(T)]{}\\ &&-\int_{t}^{T}(\frac{\xi_{0s}}{\alpha_{1}}\ln\frac{\xi_{0s}}{\xi'_{0s}}+ \frac{\eta(s)\xi_{0s}}{\alpha_{2}}\ln\frac{\xi_{0s}\eta(s)}{\lambda(s)\xi'_{0s}})E_{t}[Z^{v}(s)]{\rm d}s -\frac{\xi_{0T}}{\alpha_{3}}E_{t}[Z^{v}(T)\ln Z^{v}(T)]{}\\ &&-\int_{t}^{T}(\frac{\xi_{0s}}{\alpha_{1}}+\frac{\eta(s)\xi_{0s}}{\alpha_{2}})E_{t}[Z^{v}(s)\ln Z^{v}(s)]{\rm d}s=\xi_{0t}(X(t)+b(t))Z^{v}(t). \end{eqnarray} $

由Bayes'公式, (3.5)式和(3.6)式可得

(5.4) $ \begin{eqnarray} \frac{1}{Z^{v}(t)}E_{t}[Z^{v}(s)\ln Z^{v}(s)]&=&E_{t}^{v}[\ln Z^{v}(s)]{}\\ &=&\ln Z^{v}(t)+\int_{t}^{s}\int_{{\mathbb{R}}}g_{1}(u, y)\nu({\rm d}y){\rm d}u+\int_{t}^{s}\frac{1}{2}\phi_{1}(u)^{2}{\rm d}u, \end{eqnarray} $

其中$ g_{1}(u, y)=\phi_{2}(u, y)\ln\phi_{2}(u, y)-\phi_{2}(u, y)+1 $.

将(5.4)式代入(5.3)式可得

(5.5) $ \begin{eqnarray} &&-g_{2}(t)\ln(\lambda_{0}Z(t))-\frac{\xi_{0T}}{\alpha_{3}}\ln\frac{\xi_{0T}}{\xi'_{0T}} -\int_{t}^{T}(\frac{\xi_{0s}}{\alpha_{1}}\ln\frac{\xi_{0s}}{\xi'_{0s}}+\frac{\eta(s)\xi_{0s}}{\alpha_{2}}\ln\frac{\xi_{0s}\eta(s)}{\lambda(s)\xi'_{0s}}){\rm d}s{}\\ &&-\frac{\xi_{0T}}{\alpha_{3}}\bigg[\int_{t}^{T}\frac{1}{2}\phi_{1}(u)^{2}{\rm d}u+\int_{t}^{T}\int_{{\mathbb{R}}}g_{1}(u, y)\nu({\rm d}y){\rm d}u\bigg]{}\\ &&-\int_{t}^{T}(\frac{\xi_{0s}}{\alpha_{1}}+\frac{\eta(s)\xi_{0s}}{\alpha_{2}})\bigg[\int_{t}^{s}\frac{1}{2}\phi_{1}(u)^{2}{\rm d}u+\int_{t}^{s}\int_{{\mathbb{R}}}g_{1}(u, y)\nu({\rm d}y){\rm d}u\bigg]{\rm d}s{}\\ &=&\xi_{0t}(X(t)+b(t)), \end{eqnarray} $

其中$ g_{2}(t)=\frac{\xi_{0T}}{\alpha_{3}} +\int_{t}^{T}(\frac{\xi_{0s}}{\alpha_{1}}+\frac{\eta(s)\xi_{0s}}{\alpha_{2}}){\rm d}s $. 对$ (X(t)+b(t))\xi_{0t} $使用Itô公式可得

(5.6) $ \begin{eqnarray} {\rm d}(X(t)+b(t))\xi_{0t}&=&\bigg[\frac{\xi_{0t}}{\alpha_{1}}\ln\frac{\lambda_{0}\xi_{0t}}{\xi'_{0t}} +\frac{\eta(t)\xi_{0t}}{\alpha_{2}}\ln\frac{\lambda_{0}\xi_{0t}\eta(t)}{\lambda(t)\xi'_{0t}}+g_{2}(t)(\phi_{1}(t)^{2}{}\\ &&+\int_{{\mathbb{R}}}\phi_{2}(t, y)\ln\phi_{2}(t, y)\nu({\rm d}y)) +\ln Z^{v}(t)(\frac{\xi_{0t}}{\alpha_{1}}+\frac{\eta(t)\xi_{0t}}{\alpha_{2}})\bigg]{\rm d}t{}\\ & &-g_{2}(t)\phi_{1}(t){\rm d}W(t)-\int_{{\mathbb{R}}}g_{2}(t, y)\ln\phi_{2}(t, y)N({\rm d}t, {\rm d}y). \end{eqnarray} $

另一方面

(5.7) $ \begin{equation} {\rm d}(X(t)+b(t))\xi_{0t}=-(X(t)+b(t))\xi_{0t}(\eta(t)+r){\rm d}t+\xi_{0t}{\rm d}(X(t)+b(t)). \end{equation} $

将(5.1)式代入上面的方程可得

(5.8) $ \begin{eqnarray} {\rm d}(X(t)+b(t))\xi_{0t}&=&\bigg\{\frac{\xi_{0t}}{\alpha_{1}}\ln\frac{\lambda_{0}\xi_{0t}}{\xi'_{0t}} +\frac{\eta(t)\xi_{0t}}{\alpha_{2}}\ln\frac{\lambda_{0}\xi_{0t}\eta(t)}{\lambda(t)\xi'_{0t}} +(\frac{\xi_{0t}}{\alpha_{1}} +\frac{\eta(t)\xi_{0t}}{\alpha_{2}})\ln Z^{v}(t){}\\ &&+\theta(t)(\mu-r)\bigg\}{\rm d}t +\theta(t)\sigma{\rm d}W(t)+\theta(t)\int_{{\mathbb{R}}}({\rm e}^{y}-1)N({\rm d}t, {\rm d}y). \end{eqnarray} $

比较(5.6)式和(5.8)式中$ {\rm d}W $, $ N({\rm d}t, {\rm d}y) $和$ {\rm d}t $的系数可得

(5.9) $ \begin{equation} -g_{2}(t)\phi_{1}(t)=\theta(t)\sigma, \end{equation} $

(5.10) $ \begin{equation} -g_{2}(t)\ln\phi_{2}(t, y)=\theta(t)({\rm e}^{y}-1), \end{equation} $

(5.11) $ \begin{equation} g_{2}(t)[\phi_{1}(t)^{2}+\int_{{\mathbb{R}}}\phi_{2}(t, y)\ln\phi_{2}(t, y)\nu({\rm d}y)]=\theta(t)(\mu-r). \end{equation} $

可验证$ \theta(t)\neq0 $且(5.9)–(5.11)式成立时鞅条件(3.4)满足. 将(5.1)式代入(3.8)式可得

(5.12) $ \begin{eqnarray} \inf\limits_{Z^{v}\in {\cal Z}}U(t, Z^{v}) &=&\inf\limits_{Z^{v}\in {\cal Z}}E_{t}\bigg[-\int_{t}^{T}(\frac{\lambda_{0}\xi_{0s}}{\alpha_{1}\xi'_{0t}} +\frac{\lambda_{0}\xi_{0s}\eta(s)}{\alpha_{2}\xi'_{0t}})Z^{v}(s){\rm d}s-\frac{\lambda_{0}\xi_{0T}}{\alpha_{3}\xi'_{0t}}Z^{v}(T)\bigg]{}\\ &=&-\bigg[\int_{t}^{T}(\frac{\xi_{0s}}{\alpha_{1}\xi'_{0t}} +\frac{\xi_{0s}\eta(s)}{\alpha_{2}\xi'_{0t}}){\rm d}s+\frac{\xi_{0T}}{\alpha_{3}\xi'_{0t}}\bigg]\lambda_{0}Z^{v}(t), \end{eqnarray} $

其是$ {\cal F}_{t} $可测的. 因此最优解$ v^{*} $只需满足(5.9)–(5.10)式. 由(5.9)和(5.10)式可得

(5.13) $ \begin{equation} \phi_{1}(t)=-\frac{\theta(t)\sigma}{g_{2}(t)}, \ \ \phi_{2}(t, y)=\exp\bigg\{\frac{\theta(t)}{g_{2}(t)}(1-{\rm e}^{y})\bigg\}. \end{equation} $

将(5.13)式代入(5.11)式可得

(5.14) $ \begin{equation} \frac{\theta(t)\sigma^{2}}{g_{2}(t)}+\int_{{\mathbb{R}}}\exp\bigg\{\frac{\theta(t)}{g_{2}(t)}(1-{\rm e}^{y})\bigg\}(1-{\rm e}^{y})\nu({\rm d}y)=\mu-r. \end{equation} $

定理5.1  对$ U_{i}=-\frac{1}{\alpha_{i}}\exp\{-\alpha_{i} x\}, \alpha_{i}>0, i=1, 2, 3 $, 值函数$ V(t, X(t)) $为

(5.15) $ \begin{equation} V(t, X(t))=-\bigg[\int_{t}^{T}(\frac{\xi_{0s}}{\alpha_{1}\xi'_{0t}}+\frac{\xi_{0s}\eta(s)}{\alpha_{2}\xi'_{0t}}){\rm d}s +\frac{\xi_{0T}}{\alpha_{3}\xi'_{0t}}\bigg]\exp\bigg\{\frac{\tilde{H}_{t, T}+\xi_{0t}(X(t)+b(t))}{-g_{2}(t)}\bigg\}. \end{equation} $

最优保险、消费和投资策略为

(5.16) $ \begin{equation} \begin{array}{ll} { } \theta^{*}(t), \ \ c^{*}(t)=-\frac{1}{\alpha_{1}}\bigg[\ln\frac{\xi_{0t}}{\xi'_{0t}}+\frac{\tilde{H}_{t, T}+\xi_{0t}(X(t)+b(t))}{-g_{2}(t)}\bigg], \\ { } X(t)+\frac{p^{*}(t)}{\eta(t)}=-\frac{1}{\alpha_{2}} \bigg[\ln\frac{\eta(t)\xi_{0t}}{\lambda(t)\xi'_{0t}}+\frac{\tilde{H}_{t, T}+\xi_{0t}(X(t)+b(t))}{-g_{2}(t)}\bigg], \end{array} \end{equation} $

其中

(5.17) $ \begin{eqnarray} \tilde{H}_{t, T}&=&\frac{\xi_{0T}}{\alpha_{3}}\ln\frac{\xi_{0T}}{\xi'_{0T}} +\int_{t}^{T}\bigg(\frac{\xi_{0s}}{\alpha_{1}}\ln\frac{\xi_{0s}}{\xi'_{0s}} +\frac{\xi_{0s}\eta(s)}{\alpha_{2}}\ln\frac{\eta(s)\xi_{0s}}{\lambda(s)\xi'_{0s}}\bigg){\rm d}s{}\\& &+\frac{\xi_{0T}}{\alpha_{3}}\int_{t}^{T}\bigg(\int_{{\mathbb{R}}}g_{1}^{*}(u, y)\nu({\rm d}y)+\frac{1}{2}\phi_{1}^{*}(u)^{2}\bigg){\rm d}u{}\\ &&+\int_{t}^{T}\bigg(\frac{\xi_{0s}}{\alpha_{1}}+\frac{\xi_{0s}\eta(s)}{\alpha_{2}}\bigg)\int_{t}^{s} \bigg(\int_{{\mathbb{R}}}g_{1}^{*}(u, y)\nu({\rm d}y) +\frac{1}{2}\phi_{1}^{*}(u)^{2}\bigg){\rm d}u{\rm d}s, \end{eqnarray} $

(5.18) $ \begin{equation} g_{1}^{*}(u, y)=\phi_{2}^{*}(u, y)\ln\phi_{2}^{*}(u, y)-\phi^{*}_{2}(u, y)+1, \end{equation} $

(5.19) $ \begin{equation} g_{2}(t)=\frac{\xi_{0T}}{\alpha_{3}} +\int_{t}^{T}\bigg(\frac{\xi_{0s}}{\alpha_{1}}+\frac{\eta(s)\xi_{0s}}{\alpha_{2}}\bigg){\rm d}s, \end{equation} $

(5.20) $ \begin{equation} \phi_{1}^{*}(t)=-\frac{\theta^{*}(t)\sigma}{g_{2}(t)}, \ \ \phi_{2}^{*}(t, y)=\exp \bigg\{\frac{\theta^{*}(t)}{g_{2}(t)}(1-{\rm e}^{y})\bigg\}, \end{equation} $

其中$ \theta^{*}(t) $是(5.14)式的实数解.

假设职员的效用函数为幂效用函数, 即$ U_{i}(x)=\beta_{i}\frac{x^{\gamma}}{\gamma}, \beta_{i}>0, \gamma\neq0, \gamma<1, i=1, 2, 3, $其中$ \beta_{i} $为常数, 则$ I_{i}(y)=(\frac{y}{\beta_{i}})^{\frac{1}{\gamma-1}}, i=1, 2, 3. $由一阶条件可得

(6.1) $ \begin{equation} c(t)=\bigg(\frac{\lambda_{0}\xi_{0t}Z^{v}(t)}{\beta_{1}\xi'_{0t}}\bigg)^{\frac{1}{\gamma-1}}, \ M(t)=\bigg(\frac{\lambda_{0}\xi_{0t}\eta(t)Z^{v}(t)}{\beta_{2}\lambda(t)\xi'_{0t}}\bigg)^{\frac{1}{\gamma-1}}, \ X(T)=\bigg(\frac{\lambda_{0}\xi_{0T}Z^{v}(T)}{\beta_{3}\xi'_{0T}}\bigg)^{\frac{1}{\gamma-1}}, \end{equation} $

其中$ \lambda_{0} $满足(5.2)式, 将(6.1)式代入(5.2)式, 可得

(6.2) $ \begin{equation} (\lambda_{0}Z^{v}(t))^{\frac{1}{\gamma-1}}F_{2}(t, T)=X(t)+b(t), \end{equation} $

其中

$ F_{2}(t, T)=\xi_{tT}(\frac{\beta_{3}\xi'_{0T}}{\xi_{0T}})^{\frac{1}{1-\gamma}}F_{1}(t, T) +\int_{t}^{T}(\xi_{ts}(\frac{\beta_{1}\xi'_{0s}}{\xi_{0s}})^{\frac{1}{1-\gamma}} +\xi_{ts}\eta(s)(\frac{\beta_{2}\xi'_{0s}\lambda(s)}{\eta(s)\xi_{0s}})^{\frac{1}{1-\gamma}})F_{1}(t, s){\rm d}s, $

$ F_{1}(t, s)=\exp\bigg\{\int_{t}^{s}(\frac{\gamma\phi_{1}(u)^{2}}{2(\gamma-1)^{2}}+\int_{{\mathbb{R}}}(\phi_{2}(u, y)^{\frac{\gamma}{\gamma-1}} -\frac{\gamma}{\gamma-1}\phi_{2}(u, y)+\frac{1}{\gamma-1})\nu({\rm d}y)){\rm d}u\bigg\}. $

对$ X(t)+b(t) $使用Itô公式可得

(6.3) $ \begin{eqnarray} {\rm d}(X(t)+b(t))&=&(X(t)+b(t))\bigg\{\frac{1}{\gamma-1}\phi_{1}(t){\rm d}W(t)+\int_{{\mathbb{R}}}(\phi_{2}(t, y)^{\frac{1}{\gamma-1}}-1)N({\rm d}t, {\rm d}y){}\\ &&+(\eta(t)+r+\frac{1}{\gamma-1}\phi_{1}(t)^{2}+\int_{{\mathbb{R}}}(\phi_{2}(t, y)-\phi_{2}(t, y)^{\frac{\gamma}{\gamma-1}})\nu({\rm d}y)){\rm d}t\bigg\}{}\\ &&-(\lambda_{0}Z^{v}(t))^{\frac{1}{\gamma-1}}f_{1}(t, t){\rm d}t, \end{eqnarray} $

其中$ f_{1}(t, s)=(\xi_{ts}(\frac{\beta_{1}\xi'_{0s}}{\xi_{0s}})^{\frac{1}{1-\gamma}} +\xi_{ts}\eta(s)(\frac{\beta_{2}\xi'_{0s}\lambda(s)}{\eta(s)\xi_{0s}})^{\frac{1}{1-\gamma}})F_{1}(t, s) $.

另一方面

(6.4) $ \begin{eqnarray} {\rm d}(X(t)+b(t))&=&\{(\eta(t)+r)(X(t)+b(t))+\theta(t)(\mu-r)-(\lambda_{0}Z^{v}(t))^{\frac{1}{\gamma-1}}f_{1}(t, t)\}{\rm d}t{}\\ & &+\theta(t)\sigma{\rm d}W(t)+\int_{{\mathbb{R}}}\theta(t)({\rm e}^{y}-1)N({\rm d}t, {\rm d}y). \end{eqnarray} $

比较(6.3)和(6.4)式中$ {\rm d}W $, $ N({\rm d}t, {\rm d}y) $和$ {\rm d}t $的系数可得

(6.5) $ \begin{equation} \theta(t)\sigma=\frac{(X(t)+b(t))\phi_{1}(t)}{\gamma-1}, \end{equation} $

(6.6) $ \begin{equation} \theta(t)({\rm e}^{y}-1)=(X(t)+b(t))(\phi_{2}(t, y)^{\frac{1}{\gamma-1}}-1), \end{equation} $

(6.7) $ \begin{equation} \theta(t)(\mu-r)=(X(t)+b(t))\bigg\{\frac{1}{\gamma-1}\phi_{1}(t)^{2} +\int_{{\mathbb{R}}}(\phi_{2}(t, y)-\phi_{2}(t, y)^{\frac{\gamma}{\gamma-1}})\nu({\rm d}y)\bigg\}. \end{equation} $

可验证鞅条件(3.4)满足. 将(6.1)式代入(3.8)式可得

(6.8) $ \begin{eqnarray} &&\inf\limits_{Z^{v}\in {\cal Z}}U(t, Z^{v}){}\\ &=&\inf\limits_{Z^{v}\in {\cal Z}}E_{t}\bigg[\int_{t}^{T}\frac{\xi'_{ts}}{\gamma} (\beta_{1}^{\frac{1}{1-\gamma}}(\frac{\lambda_{0}\xi_{0s}}{\xi'_{0s}})^{\frac{\gamma}{\gamma-1}} +\lambda(s)\beta_{2}^{\frac{1}{1-\gamma}}(\frac{\lambda_{0}\eta(s)\xi_{0s}}{\lambda(s)\xi'_{0s}})^{\frac{\gamma}{\gamma-1}})Z^{v}(s)^{\frac{\gamma}{\gamma-1}}{\rm d}s{}\\ & &+\frac{\xi'_{tT}}{\gamma} \beta_{3}^{\frac{1}{1-\gamma}}(\frac{\lambda_{0}\xi_{0T}}{\xi'_{0T}})^{\frac{\gamma}{\gamma-1}}Z^{v}(T)^{\frac{\gamma}{\gamma-1}}\bigg], \end{eqnarray} $

等价于

(6.9) $ \begin{eqnarray} \inf\limits_{Z^{v}\in{\cal Z}}E[ Z^{v}(s)^{\frac{\gamma}{\gamma-1}}] &=&Z^{v}(t)^{\frac{\gamma}{\gamma-1}}\inf\limits_{Z^{v}\in{\cal Z}} \exp\bigg\{\int_{t}^{s}(\frac{\gamma}{2(\gamma-1)^{2}}\phi_{1}(u)^{2}{}\\ &&+\int_{{\mathbb{R}}}(\phi_{2}(u, y)^{\frac{\gamma}{\gamma-1}} -\frac{\gamma\phi_{2}(u, y)}{\gamma-1}+\frac{1}{\gamma-1})\nu({\rm d}y)){\rm d}u\bigg\}, \end{eqnarray} $

对$ \forall t<s<T $, 等价于计算

(6.10) $ \begin{equation} \inf\limits_{v\in\tilde{{\cal K}'}_{m}}\bigg\{\frac{\gamma}{2(\gamma-1)^{2}}\phi_{1}(u)^{2}+\int_{{\mathbb{R}}}(\phi_{2}(u, y)^{\frac{\gamma}{\gamma-1}} -\frac{\gamma\phi_{2}(u, y)}{\gamma-1}+\frac{1}{\gamma-1})\nu({\rm d}y)\bigg\}, \end{equation} $

其中

$ \begin{eqnarray} \tilde{{\cal K}'}_{m}&=&\bigg\{v=(\phi_{1}, \phi_{2})\in{\cal K}_{m}, \phi_{1}=\frac{\theta(t)\sigma(\gamma-1)}{X(t)+b(t)}, \\ &&{\quad} \phi_{2}=(\frac{\theta(t)({\rm e}^{y}-1)}{X(t)+b(t)}+1)^{\gamma-1}, \theta(t)\in{\cal D'}(\theta)/\{0\}\bigg\}, \end{eqnarray} $

令$ {\cal D'}(\theta) $表示下面方程的解的集合

(6.11) $ \begin{eqnarray} \theta(t)(\mu-r)&=&(X(t)+b(t))\int_{{\mathbb{R}}}\bigg[(\frac{\theta(t)({\rm e}^{y}-1)}{X(t)+b(t)}+1)^{\gamma-1} -(\frac{\theta(t)({\rm e}^{y}-1)}{X(t)+b(t)}+1)^{\gamma}\bigg]\nu({\rm d}y){}\\ &&+\frac{\sigma^{2}\theta(t)^{2}(1-\gamma)}{X(t)+b(t)}. \end{eqnarray} $

定理6.1  对$ U_{i}=\beta_{i}\frac{x^{\gamma}}{\gamma}, \gamma\neq0, \gamma<1, \beta_{i}>0, i=1, 2, 3 $, 值函数$ V(t, X(t)) $为

(6.12) $ \begin{equation} V(t, X(t))=\frac{\xi_{0t}}{\gamma\xi'_{0t}}(\frac{X(t)+b(t)}{F^{*}_{2}(t, T)})^{\gamma}F^{*}_{2}(t, T), \end{equation} $

最优保险、消费和投资的最优策略为

(6.13) $ \begin{equation} \begin{array}{lll} { } \theta^{*}(t)=\frac{\phi_{1}^{*}(t)(X(t)+b(t))}{\sigma(\gamma-1)}, \ \ c^{*}(t)=(\frac{\beta_{1}\xi'_{0t}}{\xi_{0t}})^{\frac{1}{1-\gamma}}\frac{X(t)+b(t)}{F^{*}_{2}(t, T)}\\ { } X(t)+\frac{p^{*}(t)}{\eta(t)}=(\frac{\beta_{2}\xi'_{0t}\eta(t)}{\lambda(t)\xi_{0t}})^{\frac{1}{1-\gamma}}\frac{X(t)+b(t)}{F^{*}_{2}(t, T)}, \end{array} \end{equation} $

其中$ v=(\phi_{1}^{*}, \phi_{2}^{*}) $是(6.10)式的解, 且

$ F_{2}^{*}(t, T)=\xi_{tT}(\frac{\beta_{3}\xi'_{0T}}{\xi_{0T}})^{\frac{1}{1-\gamma}}F^{*}_{1}(t, T) +\int_{t}^{T}\bigg[\xi_{ts}(\frac{\beta_{1}\xi'_{0s}}{\xi_{0s}})^{\frac{1}{1-\gamma}} +\xi_{ts}\eta(s)(\frac{\beta_{2}\xi'_{0s}\lambda(s)}{\eta(s)\xi_{0s}})^{\frac{1}{1-\gamma}}\bigg]F^{*}_{1}(t, s){\rm d}s, $

$ F_{1}^{*}(t, s)=\exp\bigg\{\int_{t}^{s}\bigg[\frac{\gamma\phi^{*}_{1}(u)^{2}}{2(\gamma-1)^{2}}+\int_{{\mathbb{R}}}(\phi_{2}^{*}(u, y)^{\frac{\gamma}{\gamma-1}} -\frac{\gamma}{\gamma-1}\phi^{*}_{2}(u, y)+\frac{1}{\gamma-1})\nu({\rm d}y)\bigg]{\rm d}u\bigg\}. $

本节给出一个数值例子来研究股票价格的跳如何影响最优策略. 假设Lévy测度$ \nu({\rm d}y) $是如下的离散形式

$ \nu({\rm d}y)=\lambda_{1}\delta_{a_{1}}({\rm d}y)+\lambda_{2}\delta_{-a_{2}}({\rm d}y), a_{1}>0, , a_{2}>0. $

假设职员年龄为25, 她/他将在$ T=65 $岁退休. 金融市场参数为$ r=0.01, \lambda(t)=\eta(t)=\frac{1}{10.54}\exp\{\frac{t-87.24}{10.54}\}, i(s)=5000{\rm e}^{0.02(s-t)} $(当$ s\geq t $时), $ \lambda_{1}=\lambda_{2}=2, X_{t}=100000, \delta=0.05 $和$ \mu=0.1 $.

图 1和图 2展示了对数效用函数情况下股票价格的跳对最优策略$ \theta^{*} $的影响. 图 3–图 8展示了幂效用函数情况下股票价格的跳对最优策略$ \theta^{*}, c^{*} $和$ p^{*} $的影响. 图 9–图 14展示了指数效用函数情况下股票价格的跳对最优策略$ \theta^{*}, c^{*} $和$ p^{*} $的影响. 从图中可以发现对三种效用函数$ \theta^{*} $关于股票价格的正跳是增函数, 关于股票价格的负跳是减函数. 对指数效用函数和幂效用函数, $ c^{*} $和$ p^{*} $关于股票价格的负跳是减函数, 关于股票价格的正跳是凹函数.