Optimal excess-of-loss reinsurance and investment polices under the CEV model

Abstract

This paper focuses on risk control problem of the insurance company in enterprise risk management. The insurer manages its financial risk through purchasing excess-of-loss reinsurance, and investing its wealth in the constant elasticity of variance stock market. We model risk process by Brownian motion with drift, and study the optimization problem of maximizing the exponential utility of terminal wealth under the controls of reinsurance and investment. Using stochastic control theory, we obtain explicit expressions for optimal polices and value function. We also show that the optimal excess-of-loss reinsurance is always better than optimal proportional reinsurance. And some numerical examples are given.

Appendix

In this appendix, we prove Theorem 4.2.

Proof

Step 1: Propose the ansatz. Mimicking the method of Browne (1995, p.942) or Liang et al. (2012, p.588), we want to find a solution to (4.2) with the form

$$\begin{aligned} V(t,x,s)={\uplambda }_{0} -\frac{\gamma }{m}\exp \left[ -mx\mathrm{{e}}^{r(T-t)}+G(t,s)\right] , \end{aligned}$$

(7.1)

where $G(t,s)$ is a suitable function to be determined. And, the boundary condition $V(T,x,s)=u(x)$ implies that

$$\begin{aligned} G(T,s)=0. \end{aligned}$$

(7.2)

Let $G_t , G_s , G_{ss}$, be the partial derivatives of $G(t,s)$. Note that

$$\begin{aligned} \left\{ {{\begin{array}{l} {V_t =[V-\uplambda _0 ][mxr\mathrm{{e}}^{r(T-t)}+G_t ]} \\ {V_x =[V-\uplambda _0 ][-m\mathrm{{e}}^{r(T-t)}]} \\ {V_s =[V-\uplambda _0 ][G_s ]} \\ {V_{xx} =[V-\uplambda _0 ][m^{2}\mathrm{{e}}^{2r(T-t)}]} \\ {V_{ss} =[V-\uplambda _0 ][G_s^2 +G_{ss} ]} \\ {V_{xs} =[V-\uplambda _0 ][-m\mathrm{{e}}^{r(T-t)}G_s ]} \\ \end{array} }} \right. , \end{aligned}$$

(7.3)

Step 2: Derive the optimal control. Substituting (7.3) back into the HJB Eq. (4.2), since $V-\uplambda _0 <0$, we get

$$\begin{aligned}&G_t -m\mu _\infty (\eta -\theta )\mathrm{{e}}^{r(T-t)}+\mu sG_s +\frac{1}{2}\sigma ^{2}s^{2\beta +2}(G_s^2 +G_{ss} )\nonumber \\&\quad +\mathop {\inf }\limits _{\pi \in R} \left\{ \left[ -(\mu -r)\pi -\pi \sigma ^{2}s^{2\beta +1}G_s +\frac{1}{2}\sigma ^{2}\pi ^{2}s^{2\beta }m\mathrm{{e}} ^{r(T-t)}\right] m\mathrm{{e}}^{r(T-t)}\right\} \nonumber \\&\quad +\mathop {\inf }\limits _{a\in [0,N]} \left\{ \left[ -\mu (a)\theta +\frac{1}{2}\sigma ^{2}(a)m\mathrm{{e}}^{r(T-t)}\right] m\mathrm{{e}}^{r(T-t)}\right\} =0. \end{aligned}$$

(7.4)

Let

$$\begin{aligned} f_1 (\pi ,t)=\left[ -(\mu -r)\pi -\pi \sigma ^{2}s^{2\beta +1}G_s +\frac{1}{2}\sigma ^{2}\pi ^{2}s^{2\beta }m\mathrm{{e}}^{r(T-t)}\right] m\mathrm{{e}}^{r(T-t)}, \end{aligned}$$

(7.5)

and

$$\begin{aligned} f_2 (a,t)=\left[ -\mu (a)\theta +\frac{1}{2}\sigma ^{2}(a)m\mathrm{{e}}^{r(T-t)}\right] m\mathrm{{e}}^{r(T-t)}. \end{aligned}$$

(7.6)

Differentiating $f_1 (\pi ,t)$with respect to $\pi$ yields the minimizer

$$\begin{aligned} \pi ^{*}(t)=\frac{(\mu -r)+\sigma ^{2}s^{2\beta +1}G_s }{\sigma ^{2}s^{2\beta }}\frac{\mathrm{{e}}^{-r(T-t)}}{m}, \end{aligned}$$

(7.7)

and the value of $f_1 (\pi ,t)$ at this minimizer is

$$\begin{aligned} f_1 (\pi ^{*},t)=-\frac{1}{2}\frac{\left[ (\mu -r)+\sigma ^{2}s^{2\beta +1}G_s \right] ^{2}}{\sigma ^{2}s^{2\beta }}. \end{aligned}$$

(7.8)

Similarly, from the first order condition

$$\begin{aligned} \frac{\partial f_2 (a,t)}{\partial a}=\left[ -\theta \bar{{F}}(a)+a\bar{{F}}(a)m\mathrm{{e}}^{r(T-t)}\right] m\mathrm{{e}}^{r(T-t)}=0, \end{aligned}$$

we know that, without restriction with respect to $a$,

$$\begin{aligned} \tilde{a}(t)=\frac{\theta \mathrm{{e}}^{-r(T-t)}}{m}, \end{aligned}$$

(7.9)

which leads to

$$\begin{aligned} f_2 (\tilde{a},t)=-\theta m\mathrm{{e}}^{r(T-t)}\int \limits _0^{{\theta \mathrm{{e}}^{-r(T-t)}}/m} {\bar{{F}}(y)dy} +m^{2}\mathrm{{e}}^{2r(T-t)}\int \limits _0^{{\theta \mathrm{{e}}^{-r(T-t)}}/m} {y\bar{{F}}(y)dy} . \end{aligned}$$

(7.10)

Step 3: Separate the variables. We need to discuss the two cases according to the value of $\tilde{a}(t)$.

Case 1 $({ {Nm>\theta }})$ If $t<\bar{{T}}{:=}T+\frac{\ln (Nm)-\ln \theta }{r}$, then $\tilde{a}(t)\in [0,\;N)$. So,

$$\begin{aligned} (\pi ^{*}(t),a^{*}(t))=\left( {\frac{(\mu -r)+\sigma ^{2}s^{2\beta +1}G_s }{\sigma ^{2}s^{2\beta }}\frac{\mathrm{{e}}^{-r(T-t)}}{m},\frac{\theta \mathrm{{e}}^{-r(T-t)}}{m}} \right) \end{aligned}$$

(7.11)

coincides with the optimal policy. Since $T<\bar{{T}}, \left( {\pi ^{*}(t),a^{*}(t)} \right)$ is optimal policy on $\left[ {0,\;T} \right]$.

Up to now, we still need to solve $G(t,s)$ to find the optimal investment $\pi ^{*}(t)$ and the value function $V(t,x,s)$ in this case. Substituting $\left( {\pi ^{*}(t),a^{*}(t)} \right)$ (i.e. expression (7.11)) back to (7.4), we can get

$$\begin{aligned} G_t -m\mu _\infty (\eta -\theta )\mathrm{{e}}^{r(T-t)}+rsG_s +\frac{1}{2}\sigma ^{2}s^{2\beta +2}G_{ss} -\frac{(\mu -r)^{2}}{2\sigma ^{2}s^{2\beta }}+f_2 (a^{*},t)=0. \end{aligned}$$

(7.12)

We appeal to power transformation technique and variable change method to solve the problem.

Let

$$\begin{aligned} G(t,s)=h(t,y),\;\;y=s^{-2\beta }, \end{aligned}$$

(7.13)

with boundary condition

$$\begin{aligned} h(T,y)=0, \end{aligned}$$

(7.14)

and

$$\begin{aligned} \left\{ {{\begin{array}{l} {G_t =h_t}\\ {\;G_s =-2\beta s^{-2\beta -1}h_y}\\ {G_{ss} =2\beta (2\beta +1)s^{-2\beta -2}h_y +4\beta ^{2}s^{-4\beta -2}h_{yy}} \\ \end{array} }} \right. , \end{aligned}$$

(7.15)

where $h_t , h_y , h_{yy}$ are partial derivatives of $h(t,y)$.

Putting the partial derivatives of $G(t,s)$ into the Eq. (7.12), we obtain equation

$$\begin{aligned} h_t +\left[ \sigma ^{2}(2\beta +1)-2ry\right] \beta h_y +2\sigma ^{2}\beta ^{2}yh_{yy} -\frac{(\mu -r)^{2}}{2\sigma ^{2}}y+M_1 (t)=0, \end{aligned}$$

(7.16)

where $M_1 (t)=-m\mu _\infty (\eta -\theta )\mathrm{{e}}^{r(T-t)}+f_2 (a^{*},t)$. We try to find a solution of the above equation with the following form

$$\begin{aligned} h(t,y)=K_1 (t)+L_1 (t)y, \end{aligned}$$

(7.17)

with boundary condition

$$\begin{aligned} \left\{ {{\begin{array}{l} {K_1 (T)=0} \\ {L_1 (T)=0\;} \\ \end{array} }} \right. , \end{aligned}$$

(7.18)

and

$$\begin{aligned} h_t =K_1 ^{\prime }+L_1 ^{\prime }y,\;\;h_y =L_1 ,\;\;h_{yy} =0, \end{aligned}$$

(7.19)

where $K_1^{\prime }, L_1^{\prime }$ are the derivatives of $K_1 (t),\;L_1 (t)$ respectively. Putting (7.19) into (7.16), we derive

$$\begin{aligned} {K}'_1 +\sigma ^{2}\beta (2\beta +1)L_1 +\left[ {L}'_1 -2rL_1 -\frac{(\mu -r)^{2}}{2\sigma ^{2}}\right] y+M_1 (t)=0. \end{aligned}$$

(7.20)

By matching coefficients, we have

$$\begin{aligned} \left\{ {{\begin{array}{l} {{K}'_1 +\sigma ^{2}\beta (2\beta +1)L_1 +M_1 (t)=0} \\ {{L}'_1 -2rL_1 -\frac{(\mu -r)^{2}}{2\sigma ^{2}}=0} \\ \end{array} }} \right. , \end{aligned}$$

(7.21)

Taking into boundary condition, we have the solution of Eq. (7.21):

$$\begin{aligned} L_1 (t)&= -\frac{(\mu -r)^{2}}{4r\sigma ^{2}}(1-\mathrm{{e}}^{-2r(T-t)}), \end{aligned}$$

(7.22)

$$\begin{aligned} K_1 (t)&= \int \limits _t^T {\sigma ^{2}\beta (2\beta +1)L_1 (z)+M_1 (z)dz}\nonumber \\&= -\frac{\beta (2\beta +1)(\mu -r)^{2}}{4r}\left[ (T-t)-\frac{1-\mathrm{{e}}^{-2r(T-t)}}{2r}\right] \nonumber \\&\quad -\,\frac{m\mu _\infty (\eta -\theta )}{r}(\mathrm{{e}}^{r(T-t)}-1)\nonumber \\&\quad +\,\int \limits _t^T \left\{ -\theta m\mathrm{{e}}^{r(T-z)}\int \limits _0^{{\theta \mathrm{{e}}^{-r(T-z)}}/m} {\bar{{F}}(y)dy} +m^{2} \mathrm{{e}}^{2r(T-z)}\int \limits _0^{{\theta \mathrm{{e}}^{-r(T-z)}}/m} {y\bar{{F}}(y)dy} \right\} dz.\nonumber \\ \end{aligned}$$

(7.23)

Putting these parameters into $G(t,s)$, we obtain

$$\begin{aligned} G(t,s)=K_1 (t)+L_1 (t)s^{-2\beta }. \end{aligned}$$

(7.24)

Therefore the optimal investment policy is

$$\begin{aligned} \pi ^{*}(t)=\frac{2r(\mu -r)+\beta (\mu -r)^{2}(1-\mathrm{{e}}^{-2r(T-t)})}{2r\sigma ^{2}s^{2\beta }}\frac{\mathrm{{e}}^{-r(T-t)}}{m}, \end{aligned}$$

(7.25)

and the corresponding value function has the form

$$\begin{aligned} V(t,x,s)={\uplambda }_{0} -\frac{\gamma }{m}\exp (-mx\mathrm{{e}}^{r(T-t)}+K_1 (t)+L_1 (t)s^{-2\beta }), \end{aligned}$$

(7.26)

where $L_1 (t)$ and $K_1 (t)$ is determined by (7.22) and (7.23), respectively.

Case 2 $({ {Nm\le \theta }})$ If $t<\bar{{T}}{:=}T+\frac{\ln (Nm)-\ln \theta }{r}$ (noting that $\bar{{T}}\le T)$, then $\tilde{a}(t)\in [0,\;N)$ from expression (7.9). Similar to case 1, incorporating the constants of the calculations, we get the optimal value function

$$\begin{aligned} V(t,x,s)={\uplambda }_{0} -\frac{\gamma }{m}\exp (-mx\mathrm{{e}}^{r(T-t)}+K_1 (t)+L_1 (t)s^{-2\beta }+k), \end{aligned}$$

(7.27)

where the constant $k$ will be determined from the following (7.37), and the optimal policies are

$$\begin{aligned} (\pi ^{*}(t),a^{*}(t))=\left( {\frac{(\mu -r)+\sigma ^{2}s^{2\beta +1}G_s }{\sigma ^{2}s^{2\beta }}\frac{\mathrm{{e}}^{-r(T-t)}}{m},\frac{\theta \mathrm{{e}}^{-r(T-t)}}{m}} \right) , \end{aligned}$$

(7.28)

where $G(t,s)=K_1 (t)+L_1 (t)s^{-2\beta }+k.$ If $\bar{{T}}\le t\le T$, then If $\tilde{a}(t)\ge N$. We get that the optimal retention level is $a^{*}(t)=N$. In this case, $\mu (N)=\mu _\infty$, and $\sigma ^{2}(N)=\sigma _\infty ^2$.

Putting the optimal policies

$$\begin{aligned} (\pi ^{*}(t),a^{*}(t))=\left( {\frac{(\mu -r)+\sigma ^{2}s^{2\beta +1}G_s }{\sigma ^{2}s^{2\beta }}\frac{\mathrm{{e}}^{-r(T-t)}}{m},N} \right) \end{aligned}$$

(7.29)

into the Eq. (7.4), we obtain

$$\begin{aligned} G_t -m\mu _\infty \eta \mathrm{{e}}^{r(T-t)}+rsG_s +\frac{1}{2}\sigma ^{2}s^{2\beta +2}G_{ss} -\frac{1}{2}\frac{(\mu -r)^{2}}{\sigma ^{2}s^{2\beta }}+\frac{1}{2}\sigma _\infty ^2 m^{2}\mathrm{{e}}^{2r(T-t)}=0.\nonumber \\ \end{aligned}$$

(7.30)

Again, we use the power transformation technique and variable change method to solve the Eq. (7.30) with the boundary condition (4.1).

Similarly, let $G(t,s)=h(t,y),\;\;y=s^{-2\beta },$ we have

$$\begin{aligned}&h_t +[\sigma ^{2}(2\beta +1)-2ry]\beta h_y +2\sigma ^{2}\beta ^{2}yh_{yy} -\frac{(\mu -r)^{2}}{2\sigma ^{2}}y\nonumber \\&\quad -\,m\mu _\infty \eta \mathrm{{e}}^{r(T-t)}+\frac{1}{2}\sigma _\infty ^2 m^{2}\mathrm{{e}}^{2r(T-t)}=0. \end{aligned}$$

(7.31)

And we try to the following form and match coefficients,

$$\begin{aligned} h(t,y)=K_2 (t)+L_2 (t)y, \end{aligned}$$

(7.32)

Therefore, we get

$$\begin{aligned} L_2 (t)=-\frac{(\mu -r)^{2}}{4r\sigma ^{2}}(1-\mathrm{{e}}^{-2r(T-t)}), \end{aligned}$$

which is the same as the expression $L_1 (t)$, denoted by $L(t)$ for simplicity.

$$\begin{aligned} K_2 (t)&= -\frac{\beta (2\beta +1)(\mu -r)^{2}}{4r}[(T-t)-\frac{1-\mathrm{{e}}^{-2r(T-t)}}{2r}]\nonumber \\&\quad -\,\frac{m\mu _\infty \eta }{r}(\mathrm{{e}}^{r(T-t)}-1) -\frac{m\sigma _\infty ^2 }{4r}(1-\mathrm{{e}}^{2r(T-t)}), \end{aligned}$$

(7.33)

and

$$\begin{aligned} G(t,s)=K_2 (t)+L_2 (t)s^{-2\beta }. \end{aligned}$$

(7.34)

Thus, in case 2, the optimal excess-of-loss reinsurance and investment policies are

$$\begin{aligned} (\pi ^{*}(t),a^{*}(t))=\left\{ {{\begin{array}{l@{\quad }l} {\left( {\pi ^{*}(t),\frac{\theta \mathrm{{e}}^{-r(T-t)}}{m}} \right) ,}&{} {0\le t<\bar{{T}},} \\ {\left( {\pi ^{*}(t),N} \right) ,}&{} {\bar{{T}}\le t\le T,} \\ \end{array} }} \right. \end{aligned}$$

(7.35)

where $\pi ^{*}(t)=\frac{2r(\mu -r)+\beta (\mu -r)^{2}(1-\mathrm{{e}}^{-2r(T-t)})}{2r\sigma ^{2}s^{2\beta }}\frac{\mathrm{{e}}^{-r(T-t)}}{m}$. And the corresponding value function has the form

$$\begin{aligned} V(t,x,s)=\left\{ {{\begin{array}{l@{\quad }l} {\uplambda _0 -\frac{\gamma }{m}\exp (-mx\mathrm{{e}}^{r(T-t)}+K_1 (t)+L(t)s^{-2\beta }+k),}&{} {0\le t<\bar{{T}},} \\ {\uplambda _0 -\frac{\gamma }{m}\exp (-mx\mathrm{{e}}^{r(T-t)}+K_2 (t)+L(t)s^{-2\beta }),}&{} {\bar{{T}}\le t<T,} \\ \end{array} }} \right. \end{aligned}$$

(7.36)

where choose $k$ in the way that $V(t,x,s)$ given by (7.37) is continuous at $\bar{{T}}$, that is,

$$\begin{aligned} k=K_2 (\bar{{T}})-K_1 (\bar{{T}}). \end{aligned}$$

(7.37)

Thus, we complete the proof. $\square$