Consider the classical compound Poisson risk model

$ \begin{eqnarray*} X(t)=x+ct-S(t)=x+ct-\sum\limits_{i=1}^{N(t)} Z_{i}, \ t\geq0, \end{eqnarray*} $

(1.1)

where $X(0)=x\geq 0$ is the initial surplus, $c>0$ is the premium rate, and $\{S(t);\ t\geq0\}$ represents the aggregate claims process. More specifically, $\{N(t);\ t\geq0\}$ is a Poisson process with intensity $\lambda>0$, which denotes the number of claims up to time $t$, i.e., the interclaim times $\{T_i;\ i\geq 1\}$ form a sequence of independent and identically distributed(i.i.d) positive random variables(r.v.s) and have a common exponential distribution with expectation $\frac{1}{\lambda}$. The claim sizes $\{Z_i;\ i\geq 1\}$ form a sequence of i.i.d mixed exponentially distributed r.v.s with a common density function $f_{Z}(z)=(1-\theta)\beta e^{-\beta z}+k\theta\beta e^{-k\beta z}$( $0<\theta<1$), we consider for simplicity $k=2$.

The dividend problem for an insurance risk model was first proposed by Finetti [1] who proposed to look for the expected discounted sum of dividend payments until the time of ruin. Since then the risk model in the presence of dividend payments has become a more and more popular topic in risk theory. Two recent survey papers are [2] and [3].

For classical compound Poisson risk model, the problem of looking for a strategy which maxizmizes the cumulative expected discounted dividend payments was first studied by Gerber [4]. Under the exponentially distributed assumption, the explicit expression for the value function is given and the optimal dividend strategy is proved to be a barrier strategy. Azcue and Muler [5] studied the optimal dividend problem in the compound Poisson model again by using viscosity solution method. Thonhauser and Albrecher [6] considered the model with time value of ruin and proved that the optimal dividend strategy is also a barrier strategy. Zhang and Liu [7] studied the optimal dividend payment and capital injection problem for the classical compound Poisson risk model with both proportional and fixed costs. Yao et al. [8] considered the combined optimal dividend, capital injection and reinsurance strategies for the classical compound Poisson risk model. Other interesting works can be found in Yang and Hua [9] and Wu and Wang [10]. In this paper, assuming that the surplus process is described by the classical compound Poisson risk model and the claim sizes are mixed exponentially distributed, we prove that the optimal dividend strategy is a barrier strategy. In addition, the explicit expression for the expected discounted dividend payments until ruin is obtained.

This paper is organized as follows. In Section 2, the model we discussed is introduced. In Section 3, the HJB equation for the value function is given and solved explicitly.

Let $L(t)$ be the accumulated dividends paid up to time $t$. So the controlled process $\{U^{L}(t);\ t\geq0\}$ is defined by

$ \begin{eqnarray*} U^{L}(t)=x+ct-\sum\limits_{i=1}^{N(t)} Z_{i}-L(t), \ t\geq0. \end{eqnarray*} $

(2.1)

Let $\tau=\text{inf}\{t\geq0, \ U^{L}(t)<0\}$ be the ruin time. A dividend process $L=\{L(t);\ t\geq0\}$ is called admissible if it is an adapted càglàd(previsible, $L(t-)=L(t)$) and non-decreasing process, the paying dividends cannot cause ruin, i.e., $L(t)-L(t-)\leq U^{L}(t-)$ and $L(0-)=0$. In addition, no dividend is paid after ruin, i.e., $\text{d}L(s)=0$ for $s\geq \tau$.

Assume that the dividend intensity is unbounded and dividends are discounted at a constant force of interest $\delta$. In this paper, we aim to identify the dividend payment strategy $L=\{L(t);\ t\geq0\}$ which maximizes the expected discounted dividend payments until ruin

$ \begin{align}\label{eq1} V(x, L)=E\Big{[}\int_{0}^{\tau}e^{-\delta t}\text{d}L(t)|U^{L}(0)=x\Big{]}, \end{align} $

(2.2)

i.e., we are looking for the value function

$ \begin{align} V(x)=\text{sup}_{L}V(x, L), \end{align} $

(2.3)

where the supremum is taken over all admissible strategies.

In this section, some basic properties of the value function are given and the corresponding HJB equation is derived and solved.

The next proposition is stated in Lemma 2.37 of Schmidli [11].

Proposition 3.1  The function $V(x)$ is increasing and locally Lipschitz continuous over $[0, \infty]$, and therefore absolutely continuous. For any $x\geq 0$, we have $x+\frac{c}{\lambda+\delta}\leq V(x)\leq x+\frac{c}{\delta}$ and for any $y>x$, we have $V(y)-V(x)\geq y-x$.

The next proposition gives the HJB equation which is proved in Theorem 2.39 of Schmidli [11].

Proposition 3.2  The function $V(x)$ satisfies the HJB equation

$ \begin{equation} \text{max}\Big{\{}1-V'(x), \mathcal{A}V(x)\Big{\}}=0, \end{equation} $

(3.1)

where $\mathcal{A}V(x)=cV'(x)-(\lambda+\delta)V(x)+\lambda\displaystyle\int_{0}^{x}V(x-z)f_{Z}(z)\text{d}z$.

Assume that (3.1) has a concave differentiable solution. The crucial point where the first derivative of the value function becomes smaller than one is denoted by $x_{0}$. For $x>x_{0}$, we have $1-V'(x)=0$, which immediately gives $V_{2}(x)= x+B_{1}$ for some constant $B_{1}$. For $x\leq x_{0}$, we have to solve

$ \begin{equation} cV'(x)-(\lambda+\delta)V(x)+\lambda\int_{0}^{x}V(x-z)f_{Z}(z)\text{d}z=0. \end{equation} $

(3.2)

Plugging $f_{Z}(z)=(1-\theta)\beta e^{-\beta z}+2\theta\beta e^{-2\beta z}$ into (3.2), changing the integration variable, we get

$ \begin{equation} cV'(x)-(\lambda+\delta)V(x)+ (1-\theta)\lambda\beta e^{-\beta x}\int_{0}^{x}V(z)e^{\beta z}\text{d}z+\\ 2\theta\lambda\beta e^{-2\beta x}\int_{0}^{x}V(z)e^{2\beta z}\text{d}z=0. \end{equation} $

(3.3)

Applying the operator $\big{(}\frac{\text{d}}{\text{d}x}+\beta\big{)}$ to (3.3), we have

$ \begin{equation} cV''(x)+(\beta c-\lambda-\delta)V'(x)+\beta(\theta\lambda-\delta)V(x)-2\theta\lambda\beta^{2} e^{-2\beta x}\int_{0}^{x}V(z)e^{2\beta z}\text{d}z =0. \end{equation} $

(3.4)

Applying the operator $\big{(}\frac{\text{d}}{\text{d}x}+2\beta\big{)}$ to (3.4), we have

$ \begin{equation} cV'''(x)+(3\beta c-\lambda-\delta)V''(x)+\beta(2\beta c+\theta\lambda-2\lambda-3\delta)V'(x)-2\beta^{2}\delta V(x)=0. \end{equation} $

(3.5)

It is well known that the solution of (3.5) is of the form

$ \begin{equation} V(x)=A_{1}e^{r_{1}x}+A_{2}e^{r_{2}x}+A_{3}e^{r_{3}x} \end{equation} $

(3.6)

for some constants $A_1, \ A_2, \ A_3, $ where $r_{1}, \ r_{2}, \ r_3\ (r_{1}>0>r_{2}>r_3)$ are three real roots of the characteristic equation in $\xi$:

$ \begin{equation} c \xi^{3}+(3\beta c-\lambda-\delta)\xi^{2}+\beta(2\beta c+\theta\lambda-2\lambda-3\delta)\xi-2\beta^{2}\delta =0. \end{equation} $

(3.7)

Remark 3.3  It is easy to see that

$ \begin{align*} &r_1+r_2+r_3=\frac{\lambda+\delta}{c}-3\beta, \\ &r_1r_2r_3=\frac{2\beta^{2}\delta}{c}, \\ &r_1r_2+r_2r_3+r_3r_1=\frac{\beta(2\beta c+\theta\lambda-2\lambda-3\delta)}{c}. \end{align*} $

Let

$ h_1(\xi)=c \xi^{3}+(3\beta c-\lambda-\delta)\xi^{2}+\beta(2\beta c+\theta\lambda-2\lambda-3\delta)\xi-2\beta^{2}\delta, $

then $r_{1}, \ r_{2}, \ r_3$ are three real zeros of $h_1(\xi)$. Because $h_1(-2\beta)<0$, $h_1(-\beta)>0$ and $h_1(0)<0$, we have $-\beta<r_2<0$ and $-2\beta<r_3<-\beta$.

Plugging (3.6) into (3.3) and (3.4), then letting $x$ tend to 0 from the left, we have

$ \begin{equation} [cr_{1}-\lambda-\delta]A_{1}+[cr_{2}-\lambda-\delta]A_{2}+[cr_{3}-\lambda-\delta]A_{3}=0 \end{equation} $

(3.8)

and

$ \begin{align} &\big{[}cr_{1}^{2}-(\lambda+\delta)r_{1}+(1+\theta)\lambda\beta\big{]}A_{1}+\big{[}cr_{2}^{2}-(\lambda+\delta)r_{2}+(1+\theta)\lambda\beta\big{]}A_{2}\notag\\ &+\big{[}cr_{3}^{2}-(\lambda+\delta)r_{3}+(1+\theta)\lambda\beta\big{]}A_{3}=0. \end{align} $

(3.9)

Equations (3.8) and (3.9) imply that

$ \begin{equation} A_2=A_1R_{2}, \ A_3=A_1R_3, \end{equation} $

(3.10)

where

$ \begin{align*} &R_2=\frac{R_{31}}{R_{23}}, ~~~~ R_3=\frac{R_{12}}{R_{23}}, \\ &R_{31}=[cr_{3}-\lambda-\delta][cr_{1}^{2}-(\lambda+\delta)r_{1}+(1+\theta)\lambda\beta]-\\ &[cr_{1}-\lambda-\delta][cr_{3}^{2}-(\lambda+\delta)r_{3}+(1+\theta)\lambda\beta], \\ &R_{12}=[cr_{1}-\lambda-\delta][cr_{2}^{2}-(\lambda+\delta)r_{2}+(1+\theta)\lambda\beta]-\\ &[cr_{2}-\lambda-\delta][cr_{1}^{2}-(\lambda+\delta)r_{1}+(1+\theta)\lambda\beta], \\ &R_{23}=[cr_{2}-\lambda-\delta][cr_{3}^{2}-(\lambda+\delta)r_{3}+(1+\theta)\lambda\beta]-\\ &[cr_{3}-\lambda-\delta][cr_{2}^{2}-(\lambda+\delta)r_{2}+(1+\theta)\lambda\beta]. \end{align*} $

We need to find a differentiable solution, so the differentiability of $V(x)$ over $x=x_{0}$ gives that $B_1=-x_{0}+V_{1}(x_{0})$ and $V'_{1}(x_{0})=1$, hence we have

$ A_{1}=\frac{1}{r_1e^{r_1x_{0}}+R_2r_2e^{r_2x_{0}}+R_3r_3e^{r_3}x_0}. $

Therefore we get the form of $V(x)$ that

$ \begin{equation} V(x)= \begin{cases} \displaystyle\frac{g(x)}{g'(x_0)},& x\leq x_0, \\ x-x_{0}+V(x_{0}),& x>x_0, \\ \end{cases} \end{equation} $

(3.11)

where $g(t)=e^{r_1t}+R_2e^{r_2t}+R_3e^{r_3t}$.

In order to determine $V(x)$, we are still short of an additional condition to determine $x_0$. Noting that $\{A_1, A_2, A_3\}$ are the functions of the barrier $x_0$, it is easy to see that the optimal barrier $x_0$ can be determined by minimizing $g'(t)=r_1e^{r_1t}+R_2r_2e^{r_2t}+R_3r_3e^{r_3t}$, i.e., if $x_0>0$, then $x_0$ supplies the equation

$ \begin{equation} g''(t)=r_1^{2}e^{r_1t}+R_2r_2^{2}e^{r_2t}+R_3r_3^{2}e^{r_3t}=0. \end{equation} $

(3.12)

In the following we show that the equation $g''(t)=0$ has a unique root if $x_0>0$.

Lemma 3.4   $R_{31}>0$, $R_{12}>0$ and $R_{23}<0$, and therefore $R_2<0$ and $R_3<0$.

Proof  Because

$ \begin{align} R_{31}=&c^{2}r_1r_3(r_1-r_3)+c(1+\theta)\lambda\beta(r_3-r_1)\notag\\ &+(\lambda+\delta)\big{[}c(r_3-r_1)(r_3+r_1)+(\lambda+\delta)(r_1-r_3)\big{]}\notag\\ =&(r_1-r_3)\big{[}c^{2}r_1r_3-c(\lambda+\delta)(r_1+r_3)-c(1+\theta)\lambda\beta+(\lambda+\delta)^{2}\big{]}\notag\\ =&\frac{c(r_1-r_3)}{r_2}\Big{[}2\beta^{2}\delta+(\lambda+\delta)r_2^{2}+\beta[3(\lambda+\delta)-(1+\theta)\lambda]r_2\Big{]}\\ =&\frac{c(r_1-r_3)}{r_2}\big{[}cr_2^{3}+3\beta cr_{2}^{2}+\beta[2\beta c+\theta\lambda-2\lambda-3\delta]r_2\notag+\beta[3(\lambda+\delta)-\\ &(1+\theta)\lambda]r_2\Big{]}\notag\\ =&c^{2}(r_1-r_3)(r_2+\beta)(r_2+2\beta), \end{align} $

(3.13)

$ \begin{align} R_{12}=&\frac{c(r_2-r_1)}{r_3}\Big{[}2\beta^{2}\delta+(\lambda+\delta)r_3^{2}+\beta[3(\lambda+\delta)-(1+\theta)\lambda]r_3\Big{]}\\ =&c^{2}(r_2-r_1)(r_3+\beta)(r_3+2\beta), \end{align} $

(3.14)

$ \begin{align} R_{23}=&\frac{c(r_3-r_2)}{r_1}\Big{[}2\beta^{2}\delta+(\lambda+\delta)r_1^{2}+\beta[3(\lambda+\delta)-(1+\theta)\lambda]r_1\Big{]}\\ =&c^{2}(r_3-r_2)(r_1+\beta)(r_1+2\beta), \end{align} $

(3.15)

we have $R_{31}>0$, $R_{12}>0$ and $R_{23}<0$, and therefore $R_2<0$ and $R_3<0$.

Lemma 3.5   $g'''(t)>0$ for any $t\geq 0$.

Proof  As

$ g'''(t)=r_1^{3}e^{r_1t}+R_2r_2^{3}e^{r_2t}+R_3r_3^{3}e^{r_3t}, $

we know that $g'''(t)>0$ for any $t\geq 0$ by Lemma 3.4.

Lemma 3.6  If

$ (\lambda+\delta)^{2}\geq (1+\theta)\lambda\beta c, $

then the equation $g''(t)=0$ has no positive root.

If

$ (\lambda+\delta)^{2}<(1+\theta)\lambda\beta c, $

then the equation $g''(t)=0$ has a unique positive root.

Proof  By (3.13), (3.14) and (3.15), we have

$ \begin{align*} g''(0)=&r_1^{2}+R_2r_2^{2}+R_3r_3^{2}\notag\\ =&\frac{1}{R_{23}}(R_{23}r_1^{2}+R_{31}r_2^{2}+R_{12}r_3^{2})\notag\\ =&\frac{c}{R_{23}}\Big{[}r_1(r_3-r_2)\Big{[}2\beta^{2}\delta+(\lambda+\delta)r_1^{2}+\beta[3(\lambda+\delta)-(1+\theta)\lambda]r_1\Big{]}\Big{]}\notag\\ &+\Big{[}r_2(r_1-r_3)\Big{[}2\beta^{2}\delta+(\lambda+\delta)r_2^{2}+\beta[3(\lambda+\delta)-(1+\theta)\lambda]r_2\Big{]}\Big{]}\notag\\ &+\Big{[}r_3(r_2-r_1)\Big{[}2\beta^{2}\delta+(\lambda+\delta)r_3^{2}+\beta[3(\lambda+\delta)-(1+\theta)\lambda]r_3\Big{]}\Big{]}\notag\\ =&\frac{c}{R_{23}}\Big{[}(\lambda+\delta)\big{[}r_1r_2(r_2-r_1)(r_2+r_1)+\\ &r_1r_3(r_1-r_3)(r_1+r_3)+r_2r_3(r_3-r_2)(r_3+r_2)\big{]}\\ &+\beta[3(\lambda+\delta)-(1+\theta)\lambda](r_1-r_2)(r_2-r_3)(r_3-r_1)\Big{]}\\ =&\frac{c}{R_{23}}\Big{[}(\lambda+\delta)\big{[}r_1r_2(r_2-r_1)(-\frac{3\beta c-\lambda-\delta}{c}-r_3)\\ &+r_1r_3(r_1-r_3)(-\frac{3\beta c-\lambda-\delta}{c}-r_2)+r_2r_3(r_3-r_2)(-\frac{3\beta c-\lambda-\delta}{c}-r_1)\big{]}\\ &+\beta[3(\lambda+\delta)-(1+\theta)\lambda](r_1-r_2)(r_2-r_3)(r_3-r_1)\Big{]}\\ =&\frac{1}{R_{23}}[(\lambda+\delta)^{2}-(1+\theta)\lambda\beta c](r_1-r_2)(r_2-r_3)(r_3-r_1). \end{align*} $

If $(\lambda+\delta)^{2}\geq (1+\theta)\lambda\beta c, $ we have $g''(0)\geq 0, $ we know that the equation $g''(t)=0$ has no positive root by Lemma 3.5. If $(\lambda+\delta)^{2}<(1+\theta)\lambda\beta c, $ we have $g''(0)<0, $ which together with the fact $\lim\limits_{t\rightarrow \infty}g''(t)=\infty$, implies that the equation $g''(t)=0$ has a unique root.

Lemma 3.7  If $(\lambda+\delta)^{2}\geq (1+\theta)\lambda\beta c, $ then for any $x\geq0$, we have

$ \begin{align} I(x)=&\lambda(1-\theta)\Big{(}\frac{1}{\beta}-\frac{c}{\lambda+\delta}\Big{)}e^{-\beta x}+\lambda\theta\Big{(}\frac{1}{2\beta}-\frac{c}{\lambda+\delta}\Big{)}e^{-2\beta x}-\delta x+\\ &\frac{\lambda c}{\lambda+\delta}+\frac{\lambda\theta}{2\beta}-\frac{\lambda}{\beta}\leq 0. \end{align} $

(3.16)

Proof  It is easy to see that $I(0)=0$ and

$ I'(x)=-\lambda (1-\theta)(1-\frac{\beta c}{\lambda+\delta})e^{-\beta x}-\lambda\theta(1-\frac{2\beta c}{\lambda+\delta})e^{-2\beta x}-\delta. $

If $\beta c\geq \lambda+\delta$, we have

$ \begin{align*} &I(x)\leq\lambda(1-\theta)\Big{(}\frac{1}{\beta}-\frac{c}{\lambda+\delta}\Big{)}(1-\beta x)+\lambda\theta\Big{(}\frac{1}{2\beta}-\frac{c}{\lambda+\delta}\Big{)}(1-2\beta x)-\delta x+\frac{\lambda c}{\lambda+\delta}+\\ &\frac{\lambda\theta}{2\beta}-\frac{\lambda}{\beta}\\ &=\Big{[}(1+\theta)\frac{\lambda\beta c}{\lambda+\delta}-(\lambda+\delta)\Big{]}x\leq 0. \end{align*} $

If $\beta c\leq \frac{\lambda+\delta}{2}, $ we have $I'(x)\leq 0, $ hence

$ I(x)\leq I(0)=0. $

If $\frac{\lambda+\delta}{2}<\beta c<\lambda+\delta, $ setting $t=e^{-\beta x}, $ we have

$ I'(x):=J(t)=\lambda\theta\Big{(}\frac{2\beta c}{\lambda+\delta}-1\Big{)}t^{2}-\lambda (1-\theta)\Big{(}1-\frac{\beta c}{\lambda+\delta}\Big{)}t-\delta. $

For any $0\leq t\leq 1, $ we have $J(t)\leq \text{max}\{J(0), J(1)\}.$ Since $J(0)=-\delta\leq 0$ and

$ \begin{align*} J(1)=\lambda\theta\Big{(}\frac{2\beta c}{\lambda+\delta}-1\Big{)}-\lambda (1-\theta)\Big{(}1-\frac{\beta c}{\lambda+\delta}\Big{)}-\delta=\Big{[}(1+\theta)\frac{\lambda\beta c}{\lambda+\delta}-(\lambda+\delta)\Big{]}\leq0, \end{align*} $

we get $I'(x)\leq 0$ for any $x\geq 0, $ hence we have $I(x)\leq I(0)=0.$

Theorem 3.8  If

$ (\lambda+\delta)^{2}\geq (1+\theta)\lambda\beta c, $

then $V(x)=x+\frac{c}{\lambda+\delta}$ is a solution to (3.1). If

$ (\lambda+\delta)^{2}<(1+\theta)\lambda\beta c, $

then $V(x)$ defined by (3.11) is a twice continuously differentiable concave solution to (3.1), where $x_0$ is the unique root of the equation (3.12).

Proof  Let's first consider the case $(\lambda+\delta)^{2}\geq (1+\theta)\lambda\beta c$. It is obvious that $V(x)=x+\displaystyle\frac{c}{\lambda+\delta}$ solves $1-V'(x)=0$. Thus we have to show that

$ \begin{equation} cV'(x)-(\lambda+\delta)V(x)+ (1-\theta)\lambda\beta e^{-\beta x}\int_{0}^{x}V(z)e^{\beta z}\text{d}z+2\theta\lambda\beta e^{-2\beta x}\int_{0}^{x}V(z)e^{2\beta z}\text{d}z\leq 0. \end{equation} $

(3.17)

Plugging $V(x)=x+\frac{c}{\lambda+\delta}$ into the left of (3.17), by Lemma 3.7, we have

$ \begin{align*} I(x)=&cV'(x)-(\lambda+\delta)V(x)+(1-\theta)\lambda\beta e^{-\beta x}\int_{0}^{x}V(z)e^{\beta z}\text{d}z\\ &+2\theta\lambda\beta e^{-2\beta x}\int_{0}^{x}V(z)e^{2\beta z}\text{d}z\leq 0. \end{align*} $

If $(\lambda+\delta)^{2}<(1+\theta)\lambda\beta c$. The facts $I(0)=0$ and $I'(0)=\Big{[}(1+\theta)\frac{\lambda\beta c}{\lambda+\delta}-(\lambda+\delta)\Big{]}>0$ imply that there exists some $x_1>0$ such that $I(x_1)>0$, so the second part in the maximum of (3.1) is positive, hence $V(x)=x+\frac{c}{\lambda+\delta}$ does not solve (3.1).

As $V''(x)=\frac{g''(x)}{g'(x_0)}<0$ for $x<x_0$, $V''(x_{0}-)=V''(x_{0}+)=\frac{g''(x_{0}-)}{g'(x_0)}=0$, $V'(x_{0}-)=1$ and $V''(x)=0$ for $x>x_0$, we know that (3.11) is a twice continuously differentiable concave solution to (3.1).

In the end, by Theorem 3.8, we give a verification theorem which tells that the function $V(x)$ defined by (3.11) is the value function. We omit its proof because it is quite similar to Proposition 5 of Thonhauser and Albrecher [6].

Theorem 3.9  For every admissible dividend strategy $L$, $V(x)\geq V(x, L)$, where the function $V(x)$ is defined by (3.11). Let $L_{0}$ be the barrier strategy given by the barrier $x_0$, then $V(x)=V(x, L_{0})$.