## α-Robust Optimal Investment Strategy for Target Benefit Pension Plans Under Default Risk

Shi Yuan,, Zhao Yongxia,

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

 基金资助: 国家自然科学基金.  11501321山东省自然科学基金.  ZR2020MA035

 Fund supported: the NSFC.  11501321the NSF of Shandong Province.  ZR2020MA035

Abstract

This paper considers the optimal investment and benefit payment problem for target benefit pension plan with default risk and model uncertainty. We assume that pension funds are invested in a risk-free asset, a defaultable bond and a stock satisfied a constant elasticity of variance(CEV) model. The payment of pensions depends on the financial status of the plan, with risk sharing between different generations. At the same time, in order to protect the rights of pension holders who dies before retirement, the return of premiums clauses is added to the model. In addition, our model allows the pension manager to have different levels of ambiguity aversion, instead of only considering extremely ambiguity-averse attitude. Using the stochastic optimal control approach, we establish the Hamilton-Jacobi-Bellman equations for both the post-default case and the pre-default case, respectively. We derive the closed-form solutions for α-robust optimal investment strategies and optimal benefit payment adjustment strategies. Finally, numerical analyses illustrate the influence of financial market parameters on optimal control problems.

Keywords： Target benefit plan ; α-robust ; Intergenerational risk sharing ; Default risk ; Return of premiums

Shi Yuan, Zhao Yongxia. α-Robust Optimal Investment Strategy for Target Benefit Pension Plans Under Default Risk. Acta Mathematica Scientia[J], 2022, 42(3): 943-960 doi:

## 1 引言

$$${\rm{d}}P(t, T_1)=P(t-, T_1)[r{\rm{d}}t+(1-Z(t))\delta(1-\Delta){\rm{d}}t-(1-Z(t-))\zeta{\rm{d}}M^p(t)],$$$

2) 成员和计划条款

$t$时刻退休成员的总人数可表示为

$t$时刻所有退休成员的实际总支付率$B(t)$

$$$B(t)=\int_r^\omega n(t-x+a)s(x)B(x, t){\rm{d}}x=I(t)f(t)L(t), \qquad 0\leq t\leq T,$$$

$$$C(t)=\int_a^r n(t-x+a)s(x)c_0e^{\xi t}{\rm{d}}x=C_1(t)\cdot e^{\xi t}, \qquad 0\leq t\leq T,$$$

3) 养老金财富过程

$$$\left\{\begin{array}{ll} {\rm{d}}X(t)=[r_0X(t)+(\mu-r_0)\pi_1(t)+\pi_2(t)(1-Z(t))\delta+C_1(t)e^{\xi t}-I(t)f(t)L(t)\\ \; \; \; \; \; \; \; \; \; \; \; -b(A+D\theta^{a+t})C_1(t)e^{\xi t}-\pi_1(t)\sigma S_t^\beta\phi_W(t)]{\rm{d}}t\\ \; \; \; \; \; \; \; \; \; \; \; +\pi_1(t)\sigma S_t^\beta{\rm{d}}W^\phi(t)-\pi_2(t)(1-Z(t-))\zeta{\rm{d}}Z^\phi(t), \\ X(0)=x_0. \end{array}\right.$$$

$\begin{eqnarray} J_\alpha^{\pi}(t, x, s, z):&=&\alpha\inf\limits_{\phi\in\Theta}\underline{J}^{\pi, \phi}(t, x, s, z)+\hat{\alpha}\sup\limits_{\phi\in\Theta}\overline{J}^{\pi, \phi}(t, x, s, z) {}\\ &=&\alpha\underline{J}^{\pi, \underline{\phi}^{\pi}}(t, x, s, z)+\hat{\alpha}\overline{J}^{\pi, \overline{\phi}^{\pi}}(t, x, s, z), \end{eqnarray}$

$\begin{eqnarray} \underline{J}^{\pi, \phi}(t, x, s, z)&=&E_{t, x, s, z}^\phi\bigg[-\frac{\lambda}{m}e^{-m(X(T)-x_0e^{r_0T})}\cdot e^{-r_0T} {} \\ &&-\int_t^T\frac{1}{m}e^{-m(B(v)-B^*e^{\beta_0v})}\cdot e^{-r_0v}{\rm{d}}v+\int_t^Tg_\beta(\phi(v)){\rm{d}}v\bigg], \end{eqnarray}$

$\begin{eqnarray} \overline{J}^{\pi, \phi}(t, x, s, z)&=&E_{t, x, s, z}^\phi\bigg[-\frac{\lambda}{m}e^{-m(X(T)-x_0e^{r_0T})}\cdot e^{-r_0T} {} \\ &&-\int_t^T\frac{1}{m}e^{-m(B(v)-B^*e^{\beta_0v})}\cdot e^{-r_0v}{\rm{d}}v-\int_t^Tg_\beta(\phi(v)){\rm{d}}v\bigg], \end{eqnarray}$

$({\rm{i}}) $$\pi$$ {\cal F}_t$ -可测的;

$({\rm{ii}}) $$\forall\;\mu\in[t, T], f(\mu)\geq 0 并且 ({\rm{iii}})$$ \forall(t, x, s, z)\in[0, T]\times {{\Bbb R}} \times {{\Bbb R}} ^+\times\{0, 1\}$, 有$(3.1)$式的唯一解.

$\Pi$表示所有可容许策略的集合.

## 4 模型求解

$\begin{eqnarray} &&\sup\limits_{\pi\in\Pi}\Big\{W_t+[r_0x+(\mu-r_0)\pi_1+\pi_2(1-z)\delta+C_1e^{\xi t}-IfL{}\\ &&-b(A+D\theta^{a+t})C_1e^{\xi t}]W_x+\frac{1}{2}\pi_1^2\sigma^2s^{2\beta}W_{xx} +\mu sW_s{}\\ &&+\frac{1}{2}\sigma^2s^{2\beta+2}W_{ss}+\pi_1\sigma^2s^{2\beta+1}W_{xs}-\frac{1}{m}e^{-m(IfL-B^*e^{\beta_0t})}e^{-r_0t}{}\\ &&+\alpha\inf\limits_{\phi\in\Theta}[H^{\pi, \phi}(W)+g_\beta(\phi)]+\hat{\alpha}\sup\limits_{\phi\in\Theta}[H^{\pi, \phi}(W)-g_\beta(\phi)]\Big\}=0, \end{eqnarray}$

$$$W(T, x, s, z)=-\frac{\lambda}{m}e^{-m(x-x_0e^{r_0T})}\cdot e^{-r_0T},$$$

$\begin{eqnarray} (\pi^*, \underline{\phi}^*, \overline{\phi}^*):&=&\arg\sup\limits_{\pi\in\Pi} \Big\{W_t+[r_0x+(\mu-r_0)\pi_1+\pi_2(1-z)\delta{}\\ &&+C_1e^{\xi t}-IfL-b(A+D\theta^{a+t})C_1e^{\xi t}]W_x +\frac{1}{2}\pi_1^2\sigma^2s^{2\beta}W_{xx}{}\\ &&+\mu sW_s+\frac{1}{2}\sigma^2s^{2\beta+2}W_{ss} +\pi_1\sigma^2s^{2\beta+1}W_{xs} -\frac{1}{m}e^{-m(IfL-B^*e^{\beta_0t})}e^{-r_0t}{}\\ &&+\alpha\inf\limits_{\phi\in\Theta}[H^{\pi, \phi}(W)+g_\beta(\phi)]+\hat{\alpha}\sup\limits_{\phi\in\Theta}[H^{\pi, \phi}(W)-g_\beta(\phi)]\Big\}, \end{eqnarray}$

$(3.5)$式给出的目标函数$V(t, x, s, z)=W(t, x, s, z)$, $\pi^*$为最优策略.

$$$V(t, x, s, z)=W(t, x, s, z)=(1-z)W(t, x, s, 0)+zW(t, x, s, 1),$$$

$\begin{eqnarray} & &-mP_1^\prime(t)x+Q_1^\prime(t)+U_1^\prime(t)s-mP_1(t)[r_0x+C_1e^{\xi t}-b(A+D\theta^{a+t})C_1e^{\xi t}]{}\\ &&+r_0sU_1(t)-mP_1(t)\bigg[\frac{\ln \lambda P_1(t)-mP_1(t)x+Q_1(t)+U_1(t)s+r_0t}{m}-B^*e^{\beta_0t}-\frac{1}{m}\bigg]{}\\ &&-\frac{1}{2}\bigg[\frac{(\mu-r_0)^2}{\sigma^2s^{2\beta}(1-(1-2\alpha)\frac{\rho_1}{m})}-(1-2\alpha)\frac{\rho_1}{m}\sigma^2s^{2\beta+2}U_1^2(t)\bigg]=0. \end{eqnarray}$

$x, s, 1$的系数为0, 则有

$\begin{eqnarray} && -mP_1^\prime(t)-mP_1(t)r_0+mP_1^2(t)=0, {}\\ && U_1^\prime(t)+(r_0-P_1(t))U_1(t)=0, {}\\ && Q_1^\prime(t)-P_1(t)Q_1(t)-P_1(t)[mC_1e^{\xi t}-mb(A+D\theta^{a+t})C_1e^{\xi t}+\ln\lambda P_1(t)+r_0t\\ && -mB^*e^{\beta_0t}-1]-\frac{(\mu-r_0)^2}{2\sigma^2s^{2\beta}(1-(1-2\alpha)\frac{\rho_1}{m})}+\frac{(1-2\alpha)\rho_1\sigma^2s^{2\beta+2}U_1^2(t)}{2m}=0. {} \end{eqnarray}$

$\begin{eqnarray} P_1(t)&=&\frac{r_0}{1+(r_0-1)\exp\{-r_0(T-t)\}}, \qquad U_1(t)=0, {}\\ Q_1(t)&=&(mx_0e^{r_0T}-r_0T)h_1(t)-h_2(t)\int_t^T \bigg[P_1(v)[m(C_1(v)e^{\xi v}{}\\ &&-b(A+D\theta^{a+v}) C_1(v)e^{\xi v}-B^*e^{\beta_0v})+\ln\lambda P_1(v)+r_0v-1]\\ &&+\frac{m(\mu-r_0)^2}{2\sigma^2s^{2\beta}(m-(1-2\alpha)\rho_1)}\bigg]h_3(v){\rm{d}}v, {} \end{eqnarray}$

$$$\left\{\begin{array}{ll} h_1(t)=\exp\{r_0(t-T)+\ln r_0-\ln(1+(r_0-1)e^{r_0(t-T)})\}, \\ h_2(t)=\exp\{r_0t-\ln(1+(r_0-1)e^{r_0(t-T)})+\ln(1+(r_0-1)e^{-r_0T})\}, \\ h_3(t)=\exp\{-r_0t+\ln(1+(r_0-1)e^{r_0(t-T)})-\ln(1+(r_0-1)e^{-r_0T})\}. \end{array}\right.$$$

$(4.14)$式代入$(4.13)$式, 得到违约后由$(4.5) $$(4.7) 式给出的最优策略 \pi_1^*(t) , f^*(t) . (ⅱ) 违约前情况( z=0 ) 接下来推导违约前的 \alpha -鲁棒最优策略. (4.1) 式为 \begin{eqnarray} &&\sup\limits_{\pi\in\Pi}\bigg\{W_t+[r_0x+(\mu-r_0)\pi_1+\pi_2\delta+C_1e^{\xi t}-IfL-b(A+D\theta^{a+t})C_1e^{\xi t}]W_x {}\\ &&+\frac{1}{2}\pi_1^2\sigma^2s^{2\beta}W_{xx}+\mu sW_s+\frac{1}{2}\sigma^2s^{2\beta+2}W_{ss} +\pi_1\sigma^2s^{2\beta+1}W_{xs}-\frac{1}{m}e^{-m(IfL-B^*e^{\beta_0t})}e^{-r_0t} {}\\ &&+\alpha\inf\limits_{\phi\in\Theta}\bigg[-\sigma s^{\beta+1}\phi_WW_s-\pi_1\sigma s^\beta\phi_WW_x +[W(t, x-\pi_2\zeta, s, 1)-W(t, x, s, 0)]\phi_Nh^p{}\\ &&+\frac{\phi_W^2}{2\beta_W} +\frac{h^p(\phi_N\ln\phi_N-\phi_N+1)}{\beta_N}\bigg] +\hat{\alpha}\sup\limits_{\phi\in\Theta}\bigg[-\sigma s^{\beta+1}\phi_WW_s -\pi_1\sigma s^\beta\phi_WW_x{}\\ &&+[W(t, x-\pi_2\zeta, s, 1)-W(t, x, s, 0)]\phi_Nh^p -\frac{\phi_W^2}{2\beta_W}-\frac{h^p(\phi_N\ln\phi_N-\phi_N+1)}{\beta_N} \bigg]\bigg\}=0, {\qquad} \end{eqnarray} 这里 W_t, W_x, W_{xx}, W_s, W_{ss}, W_{xs} 表示 W(t, x, s, 0) 关于 t, x, s 的一阶和二阶偏导. 假设 由边界条件 (4.2) 知, P_2(T)=1,$$ Q_2(T)=mx_0e^{r_0T}-r_0T, U_2(T)=0$.

$\begin{eqnarray} &&W_t=[-mP_2^\prime(t)x+Q_2^\prime(t)+U_2^\prime(t)s]W(t, x, s, 0), \quad W_x=-mP_2(t)W(t, x, s, 0), {}\\ &&W_s=U_2(t)W(t, x, s, 0), \qquad W_{xx}=m^2P_2^2(t)W(t, x, s, 0), {}\\ &&W_{xs}=-mP_2(t)U_2(t)W(t, x, s, 0), \qquad W_{ss}=U_2^2(t)W(t, x, s, 0), \\ && W(t, x-\pi_2\zeta, s, 1)-W(t, x, s, 0)=W(t, x, s, 0)[\exp\{-m(P_1(t)-P_2(t))x{}\\ && +Q_1(t)-Q_2(t)+(U_1(t)-U_2(t))s+mP_1(t)\pi_2\zeta\}-1].{} \end{eqnarray}$

$\pi_1^*, \pi_2^*, f^*$的一阶可微条件, 得到

$\begin{eqnarray} \begin{array}{ll} { } \pi_1^*=\frac{\alpha\sigma s^\beta\underline{\phi}_WW_x+\hat{\alpha}\sigma s^\beta\overline{\phi}_WW_x-\sigma^2s^{2\beta+1}W_{xs}-(\mu-r_0)W_x}{W_{xx}\sigma^2s^{2\beta}}, \\ { } \pi_2^*=\frac{1}{mP_1(t)\zeta}\bigg[\ln\frac{-\delta W_x}{(\alpha\underline{\phi}_N+\hat{\alpha}\overline{\phi}_N)h^pWmP_1(t)\zeta}+m(P_1(t)-P_2(t))x\\ {\qquad}{\quad} -(Q_1(t)-Q_2(t))-(U_1(t)-U_2(t))s\bigg], \\ { } f^*=\frac{1}{IL}\cdot\Big(-\frac{\ln W_x+r_0t}{m}+B^*e^{\beta_0t}\Big). \end{array} \end{eqnarray}$

$\pi_1^*, \pi_2^*, f^*$代入$(4.19)$式, 根据$\phi_W, \phi_N$的一阶最优条件, 找到$(4.19)$式取下确界和上确界的$\phi_W, \phi_N$:

$\begin{eqnarray} &&\underline{\phi}_W^*=\frac{\beta_WW_x(-\sigma^2s^{2\beta+1}W_{xs}-(\mu-r_0)W_x)+\beta_W\sigma^2s^{2\beta+1}W_sW_{xx}}{(W_{xx}+(1-2\alpha)W_x^2\beta_W)\sigma s^\beta}, {}\\ & &\overline{\phi}_W^*=-\frac{\beta_WW_x(-\sigma^2s^{2\beta+1}W_{xs}-(\mu-r_0)W_x)+\beta_W\sigma^2s^{2\beta+1}W_sW_{xx}}{(W_{xx}+(1-2\alpha)W_x^2\beta_W)\sigma s^\beta}, {}\\ &&\alpha\underline{\phi}_N^*h^p-\frac{\alpha h^p\underline{\phi}_N^*}{\beta_NW}\ln\underline{\phi}_N^*+\frac{W_x\delta \alpha\underline{\phi}_N^*}{mP_1(t)\zeta W(\alpha\underline{\phi}_N^*+\hat{\alpha}\overline{\phi}_N^*)}=0, \\ &&\hat{\alpha}\overline{\phi}_N^*h^p+\frac{\hat{\alpha} h^p\overline{\phi}_N^*}{\beta_NW}\ln\overline{\phi}_N^*+\frac{W_x\delta \hat{\alpha}\overline{\phi}_N^*}{mP_1(t)\zeta W(\alpha\underline{\phi}_N^*+\hat{\alpha}\overline{\phi}_N^*)}=0.{} \end{eqnarray}$

$\begin{eqnarray} & &x\bigg[-mP_2^\prime(t)-mP_2(t)r_0+mP_2^2(t)-\frac{mP_2(t)\delta}{P_1(t)\zeta}(P_1(t)-P_2(t))\bigg]{}\\ &&s\bigg[U_2^\prime(t)+\mu U_2(t)-(\mu-r_0)U_2(t)-P_2(t)U_2(t)+\frac{P_2(t)\delta(U_1(t)-U_2(t))}{P_1(t)\zeta}\bigg]{}\\ &&+Q_2^\prime(t)-(P_2(t)+\frac{\delta P_2(t)}{\zeta P_1(t)})Q_2(t)-P_2(t)[mC_1e^{\xi t}-mb(A+D\theta^{a+t})C_1e^{\xi t}{}\\ & &+\ln\lambda P_2(t)+r_0t-mB^*e^{\beta_0t}-1]-\frac{m(\mu-r_0)^2}{2\sigma^2s^{2\beta}(m-(1-2\alpha)\rho_1)}{}\\ &&-\frac{P_2(t)\delta}{P_1(t)\zeta} \bigg[\ln\frac{\delta P_2(t)}{(\alpha\underline{\phi}_N^*+\hat{\alpha}\overline{\phi}_N^*)h^pP_1(t)\zeta}-Q_1(t)-1\bigg] -h^p(\alpha\underline{\phi}_N^*+\hat{\alpha}\overline{\phi}_N^*){}\\ & &+\frac{mh^p}{\rho_2}[-\alpha(\underline{\phi}_N^*\ln\underline{\phi}_N^*-\underline{\phi}_N^*+1)+\hat{\alpha}(\overline{\phi}_N^*\ln\overline{\phi}_N^*-\overline{\phi}_N^*+1)]=0. \end{eqnarray}$

$x, s$的系数及剩余项为0, 有

$\begin{eqnarray} & &-mP_2^\prime(t)-mP_2(t)(r_0+\frac{\delta}{\zeta})+m(1+\frac{\delta}{P_1(t)\zeta})P_2^2(t)=0, \end{eqnarray}$

$\begin{eqnarray} &&U_2^\prime(t)+r_0U_2(t)-P_2(t)U_2(t)+\frac{P_2(t)\delta(U_1(t)-U_2(t))}{P_1(t)\zeta}=0, \end{eqnarray}$

$\begin{eqnarray} &&Q_2^\prime(t)-(P_2(t)+\frac{\delta P_2(t)}{\zeta P_1(t)})Q_2(t)-P_2(t)[mC_1e^{\xi t}-mb(A+D\theta^{a+t})C_1e^{\xi t}{} \\ &&+\ln\lambda P_2(t)+r_0t-mB^*e^{\beta_0t}-1]-\frac{m(\mu-r_0)^2}{2\sigma^2s^{2\beta}(m-(1-2\alpha)\rho_1)}{} \\ & &-\frac{P_2(t)\delta}{P_1(t)\zeta} \bigg[\ln\frac{\delta P_2(t)}{(\alpha\underline{\phi}_N^*+\hat{\alpha}\overline{\phi}_N^*)h^pP_1(t)\zeta}-Q_1(t)-1\bigg] -h^p(\alpha\underline{\phi}_N^*+\hat{\alpha}\overline{\phi}_N^*){}\\ &&+\frac{mh^p}{\rho_2}[-\alpha(\underline{\phi}_N^*\ln\underline{\phi}_N^*-\underline{\phi}_N^*+1)+\hat{\alpha}(\overline{\phi}_N^*\ln\overline{\phi}_N^*-\overline{\phi}_N^*+1)]=0. \end{eqnarray}$

$\begin{eqnarray} P_2(t)&=&\frac{r_0}{1+(r_0-1)\exp\{-r_0(T-t)\}}, \qquad\qquad U_2(t)=0, {} \\ Q_2(t)&=&(mx_0e^{r_0T}-r_0T)h_4(t)-h_5(t)\int_t^T \bigg\{P_2(v)[m(C_1(v)e^{\xi v}-b(A+D\theta^{a+v})C_1(v)e^{\xi v}{} \\ & &-B^*e^{\beta_0v})+\ln\lambda P_2(v)+r_0v-1]+\frac{m(\mu-r_0)^2}{2\sigma^2s^{2\beta}(m-(1-2\alpha)\rho_1)}{} \\ &&+\frac{\delta}{\zeta} \bigg[\ln\frac{\delta}{(\alpha\underline{\phi}_N^*+\hat{\alpha}\overline{\phi}_N^*)h^p\zeta}-Q_1(v)-1\bigg] +h^p(\alpha\underline{\phi}_N^*+\hat{\alpha}\overline{\phi}_N^*)\\ &&-\frac{mh^p}{\rho_2}[-\alpha(\underline{\phi}_N^*\ln\underline{\phi}_N^*-\underline{\phi}_N^*+1) +\hat{\alpha}(\overline{\phi}_N^*\ln\overline{\phi}_N^*-\overline{\phi}_N^*+1)]\bigg\}h_6(v){\rm{d}}v, {} \end{eqnarray}$

$$$\left\{\begin{array}{ll} { } h_4(t)=\exp\bigg\{(r_0+\frac{\delta}{\zeta})(t-T)+\ln r_0-\ln(1+(r_0-1)e^{r_0(t-T)})\bigg\}, \\ { } h_5(t)=\exp\bigg\{(r_0+\frac{\delta}{\zeta})t-\ln(1+(r_0-1)e^{r_0(t-T)})+\ln(1+(r_0-1)e^{-r_0T})\bigg\}, \\ { } h_6(t)=\exp\bigg\{-(r_0+\frac{\delta}{\zeta})t+\ln(1+(r_0-1)e^{r_0(t-T)})-\ln(1+(r_0-1)e^{-r_0T})\bigg\}. \end{array}\right.$$$

$(4.20) $$(4.22) 式代入 (4.21) 式, 得到违约前由 (4.5)$$ (4.6)$式给出的最优策略$\pi_1^*(t)$, $\pi_2^*(t)$, 以及$(4.7)$式给出的$f^*(t)$.

$({\rm{ii}}) $$\alpha=0 , 并且 \frac{m}{\rho_2}e^{\frac{\rho_2-m}{m}}\geq\frac{1}{\Delta} 成立, 方程组 (4.9) 存在一组正解; ({\rm{iii}})$$ \alpha\in(0, 1)$, 并且$h^p\geq\frac{2mh^p}{\rho_2}, \frac{\delta m}{\zeta\rho_2h^p}>1$成立, 方程组$(4.9)$存在一组正解.

(ⅰ) $\alpha=1$时, 只需证明存在一个$\underline{\phi}_N^*>0$即可. 令$F_1(\underline{\phi}_N)= h^p\underline{\phi}_N+\frac{mh^p\underline{\phi}_N\ln\underline{\phi}_N}{\rho_2}-\frac{\delta}{\zeta}$, 则$F_1^\prime(\underline{\phi}_N)= h^p+\frac{mh^p}{\rho_2}(\ln\underline{\phi}_N+1),$从而得到$F_1(\underline{\phi}_N) $$(0, e^{-\frac{m+\rho_2}{m}}) 递减, 在 (e^{-\frac{m+\rho_2}{m}}, +\infty) 递增, 且有 \lim\limits_{\underline{\phi}_N\rightarrow0^+}F_1(\underline{\phi}_N) =-\frac{\delta}{\zeta}<0, \, \lim\limits_{\underline{\phi}_N\rightarrow +\infty}F_1(\underline{\phi}_N)=+\infty. 因此存在一个 \underline{\phi}_N^*>0 . (ⅱ) \alpha=0 时, 只需证明存在一个 \overline{\phi}_N^*>0 即可. 令 F_2(\overline{\phi}_N)= h^p\overline{\phi}_N-\frac{mh^p\overline{\phi}_N\ln\overline{\phi}_N}{\rho_2}-\frac{\delta}{\zeta} , 则 F_2^\prime(\overline{\phi}_N)= h^p-\frac{mh^p}{\rho_2}(\ln\overline{\phi}_N+1) , 从而得到 F_2(\overline{\phi}_N)$$ (0, e^{\frac{\rho_2-m}{m}})$递增, 在$(e^{\frac{\rho_2-m}{m}}, +\infty)$递减, 且有$F_2(e^{\frac{\rho_2-m}{m}})=\frac{mh^p}{\rho_2}e^{\frac{\rho_2-m}{m}}-\frac{\delta}{\zeta}\geq0$, 因此存在一个$\overline{\phi}_N^*>0$.

(ⅲ) $\alpha\in(0, 1)$时, 方程组$(4.9)$可写为

$\begin{eqnarray} \overline{\phi}_N=\frac{\delta}{\hat{\alpha}\zeta(h^p+\frac{mh^p\ln\underline{\phi}_N}{\rho_2})} -\frac{\alpha\underline{\phi}_N}{\hat{\alpha}}, \end{eqnarray}$

$\begin{eqnarray} \underline{\phi}_N=\frac{\delta}{\alpha\zeta (h^p-\frac{mh^p\ln\overline{\phi}_N}{\rho_2})}-\frac{\hat{\alpha}\overline{\phi}_N}{\alpha}. \end{eqnarray}$

$F_3(\overline{\phi}_N)=\delta\overline{\phi}_N-\alpha h^p\zeta+\frac{mh^p\alpha\zeta\ln\overline{\phi}_N} {\rho_2}-h^p\zeta\hat{\alpha}(\overline{\phi}_N)^2+\frac{mh^p\hat{\alpha}\zeta(\overline{\phi}_N)^2\ln\overline{\phi}_N}{\rho_2},$则有$F_3(1)=\delta-h^p\zeta>0, $$\lim\limits_{\overline{\phi}_N\rightarrow0^+}F_3(\overline{\phi}_N)=-\infty<0 , 则由零点存在定理, 存在 d_1\in(0, 1) , 使得 F_3(d_1)=0 , 故 F_3(\overline{\phi}_N)=0 存在一个正根 \overline{\phi}_N^*. (4.30) 式两边求一阶导, 有 \underline{\phi}_N^\prime=\frac{\delta mh^p}{\alpha\zeta\rho_2 \overline{\phi}_N(h^p-\frac{mh^p\ln\overline{\phi}_N}{\rho_2})^2}-\frac{\hat{\alpha}}{\alpha} ; 对 (4.30) 式两边求二阶导, 有 \underline{\phi}_N^{\prime\prime}=\frac{-\delta mh^p (h^p-\frac{mh^p\ln\overline{\phi}_N}{\rho_2})(h^p-\frac{mh^p\ln\overline{\phi}_N}{\rho_2} -\frac{2mh^p}{\rho_2})}{\alpha\zeta\rho_2\overline{\phi}_N^2(h^p-\frac{mh^p\ln\overline{\phi}_N}{\rho_2})^4}.$$ \overline{\phi}_N^*\in(0, 1)$时, $\underline{\phi}_N^{\prime\prime}<0$, 则$\underline{\phi}_N^\prime $$(0, 1) 区间上递减. 当 \overline{\phi}_N^*\rightarrow1 时, \underline{\phi}_N^\prime=\frac{\delta m}{\alpha\zeta\rho_2 h^p}-\frac{\hat{\alpha}}{\alpha}>0 , 则 \underline{\phi}_N^\prime$$ (0, 1)$区间上恒大于0, 故$\underline{\phi}_N $$(0, 1) 区间上递增. 当 \overline{\phi}_N^*\rightarrow0 时, \underline{\phi}_N\rightarrow0^+>0 , 则 \underline{\phi}_N$$ (0, 1)$区间上恒大于0. 结论得证.

$$$\tilde{V}(t, x, s, z)=\tilde{W}(t, x, s, z)=(1-z)\tilde{W}(t, x, s, 0)+z\tilde{W}(t, x, s, 1),$$$

$$$\overline{\pi}_1^*(t)=\frac{\mu-r_0}{mP(t)\sigma^2s^{2\beta}}, \quad\;t\in[0, T],$$$

$$$\overline{\pi}_2^*(t)= \left\{\begin{array}{ll} { }\frac{1}{m\zeta P(t)}\cdot \bigg[\ln\frac{\delta}{h^p\zeta}-(\overline{Q}_1(t)-\overline{Q}_2(t))\bigg], \ &t\in[0, \tau\wedge T), \\ 0, &t\in[\tau\wedge T, T], \end{array}\right.$$$

$$$\overline{f}^*(t)= \left\{\begin{array}{ll} { }\frac{1}{I(t)L(t)}\cdot\Big(-\frac{\ln\lambda P(t)-m P(t)x+\overline{Q}_2(t)+r_0t}{m}+B^*e^{\beta_0t}\Big), &t\in[0, \tau\wedge T), \\ { }\frac{1}{I(t)L(t)}\cdot\Big(-\frac{\ln\lambda P(t)-m P(t)x+\overline{Q}_1(t)+r_0t}{m}+B^*e^{\beta_0t}\Big), \ &t\in[\tau\wedge T, T], \end{array}\right.$$$

$\begin{eqnarray} \overline{Q}_1(t)&=&(mx_0e^{r_0T}-r_0T)h_1(t)-h_2(t)\int_t^T\bigg\{P_1(v)[m(C_1(v)e^{\xi v}-b(A+D\theta^{a+v})C_1(v)e^{\xi v}{} \\ & &-B^*e^{\beta_0v})+\ln\lambda P_1(v)+r_0v-1]+\frac{(\mu-r_0)^2}{2\sigma^2s^{2\beta}}\bigg\}h_3(v){\rm{d}}v, \end{eqnarray}$

$\begin{eqnarray} \overline{Q}_2(t)&=&(mx_0e^{r_0T}-r_0T)h_4(t)-h_5(t)\int_t^T\bigg\{P_2(v)[m(C_1(v)e^{\xi v}-b(A+D\theta^{a+v})C_1(v)e^{\xi v}{} \\ &&-B^*e^{\beta_0v})+\ln\lambda P_2(v)+r_0v-1]+\frac{(\mu-r_0)^2}{2\sigma^2s^{2\beta}}{} \\ &&+\frac{\delta}{\zeta}\bigg[\ln\frac{\delta}{h^p\zeta}-\overline{Q}_1(v)-1\bigg]+h^p\bigg\}h_6(v){\rm{d}}v. \end{eqnarray}$

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