On the Optimal Dividend Problem in the Dual Model with Surplus-Dependent Premiums

Abstract

This paper concerns the dual risk model, dual to the risk model for insurance applications, where premiums are surplus-dependent. In such a model, premiums are regarded as costs and claims refer to profits. We calculate the mean of the cumulative discounted dividends paid until the time of ruin, if the barrier strategy is applied. We formulate the associated Hamilton–Jacobi–Bellman equation and identify sufficient conditions for a barrier strategy to be optimal. Numerical examples are provided.

Appendices

Appendix A: Proof of the Verification Theorem 4.1

Fix an arbitrary $x\in ]0,\infty [$ and $\pi \in \Pi$. Let $\{t_n\}_{n=1}^{\infty }$ denote all jumping times for $L^{\pi }$. Since m is $C^1$ on $]0,\infty [$ and $X^{\pi }_{t\wedge \sigma ^\pi }\in [0,\infty [$, we are allowed to apply Itô’s formula to the process

$$\begin{aligned} Y_t:=e^{-q(t\wedge \sigma ^\pi )}m(X^\pi _{t\wedge \sigma ^\pi }), \end{aligned}$$

which gives:

$$\begin{aligned} Y_t-Y_0= & {} \int _0^{t\wedge \sigma ^\pi }e^{-qs}({\mathcal {A}}-q{\mathbf {I}})m(X^\pi _{s-})\mathrm{d}s+M_t-\int _0^{t\wedge \sigma ^\pi }e^{-qs}\mathrm{d}L^{\pi ,c}_s\\&+\sum _{0\le t_n\le t\wedge \sigma ^\pi }e^{-qt_n}\left[ m\left( X^\pi _{t_n-}+C_{N_{t_n}}\varDelta N_{t_n}-\varDelta L^\pi _{t_n}\right) -m(X^\pi _{t_n-}+C_{N_{t_n}}\varDelta N_{t_n})\right] , \end{aligned}$$

where $L^{\pi ,c}$ denotes the continuous part of $L^{\pi }$ and $M_t$ is a local martingale with $M_0=0$. Since m satisfies (12), we have

$$\begin{aligned} Y_t-Y_0\le & {} \int _0^{t\wedge \sigma ^\pi }e^{-qs}\left( {\mathcal {A}}-q{\mathbf {I}}\right) m(X^\pi _{s-})\mathrm{d}s+M_t-\int _0^{t\wedge \sigma ^\pi }e^{-qs}\mathrm{d}L^{\pi ,c}_s\\&-\sum _{0\le t_n\le t\wedge \sigma ^\pi }e^{-qt_n}\varDelta L^\pi _{t_n}\\\le & {} -\int _0^{t\wedge \sigma ^\pi }e^{-qs}\mathrm{d}L^{\pi }_s+M_t. \end{aligned}$$

Let $\{s_n\}_{n=1}^{\infty }$ be a localizing sequence of M. Taking expectations and using the fact that $m\ge 0$, we obtain

$$\begin{aligned} m(x)\ge & {} {\mathbb {E}}_x \big [e^{-q(t\wedge \sigma ^\pi \wedge s_n)}m(X^\pi _{t\wedge \sigma ^\pi \wedge s_n})\big ]+{\mathbb {E}}_x\int _0^{t\wedge \sigma ^\pi \wedge s_n}e^{-qs}\mathrm{d}L^\pi _{s}\\\ge & {} {\mathbb {E}}_x\int _0^{t\wedge \sigma ^\pi \wedge s_n}e^{-qs}\mathrm{d}L^\pi _{s}. \end{aligned}$$

Letting first $n\rightarrow \infty$, and then $t\rightarrow \infty$ and applying the Monotone Convergence Theorem yields:

$$\begin{aligned} m(x)\ge v_\pi (x). \end{aligned}$$

Therefore, since both $\pi \in \Pi$ and x are arbitrary, we proved the desired inequality $m(x)\ge v(x)$ for all $x\in [0,\infty [$. $\square$

Appendix B: Additional Facts

Before we prove the main results of this paper, we demonstrate a few auxiliary facts used in these proofs.

Lemma B.1

Let $\beta \ge 0$. Assume that we have a function $h_\beta :{\mathbb {R}}\rightarrow [0,\infty [$, such that $h_\beta (x)=x-\beta +h_\beta (\beta )$ for all $x>\beta$ and $h_\beta (x)=0$ for all $x\le 0$. If the function $h_\beta (x)$ solves the equation

$$\begin{aligned} ({\mathcal {A}}-q{\mathbf {I}})h_\beta (x)=0\quad \text { for } 0<x<\beta \end{aligned}$$

with the boundary conditions $h_\beta '(\beta )=1$, then

$$\begin{aligned} h_\beta (x)=v_\beta (x)\quad \text { for all }x\ge 0. \end{aligned}$$

Proof

Let $X^\beta :=X^{\pi _\beta }$. Take an arbitrary $x\in [0,\beta ]$. Applying Itô’s formula to $e^{-qt\wedge s_n}h_\beta (X^\beta _{t\wedge s_n})$, where $\{s_n\}_{n=1}^{\infty }$ is a localizing sequence of M, we obtain

$$\begin{aligned}&{\mathbb {E}}_x\left[ e^{-q(t\wedge \sigma ^\beta \wedge s_n)}h_\beta (X^\beta _{t\wedge \sigma ^\beta \wedge s_n})\right] \\&\quad =h_\beta (x)+{\mathbb {E}}_x\left[ \int _0^{t\wedge \sigma ^\beta \wedge s_n} e^{-qs}({\mathcal {A}}-q{\mathbf {I}})h_\beta (X_s^\beta )\mathrm{d}s\right] \\&\quad \quad +{\mathbb {E}}_x\left[ \int _0^{t\wedge \sigma ^\beta \wedge s_n}e^{-qs}h_\beta (X^\beta _s)\mathrm{d}L^{\beta ,c}_s\right] \\&\quad \quad +{\mathbb {E}}_x\left[ \sum _{0\le s\le {t\wedge \sigma ^\beta \wedge s_n}}e^{-qs}\left( h_\beta (X_{s-}^\beta -\varDelta L_s^\beta )-h_\beta (X_{s-}^\beta )\right) {\mathbb {I}}_{\varDelta L_s^\beta>0}\right] \\&\quad \quad =h_\beta (x)+{\mathbb {E}}_x\left[ \int _0^{t\wedge \sigma ^\beta \wedge s_n} e^{-qs}({\mathcal {A}}-q{\mathbf {I}})h_\beta (X_s^\beta )\mathrm{d}s\right] \\&\quad \quad \quad -{\mathbb {E}}_x\left[ \sum _{0\le s\le {t\wedge \sigma ^\beta \wedge s_n}}e^{-qs}\varDelta L_s^\beta {\mathbb {I}}_{\varDelta L_s^\beta >0}\right] . \end{aligned}$$

Note that the process $X^{\beta }$ is decreasing between the positive jumps, hence $L^{\beta ,c}\equiv 0$. Moreover, $X_{s-}^\beta >\beta$ on $\{\varDelta L^\beta _s>0\}$ and $h_\beta (X_{s-}^\beta -\varDelta L_s^\beta )=h_\beta (\beta )$. Thus, after rearranging and letting first $n\rightarrow \infty$ and then $t\rightarrow \infty$, using the dominated convergence theorem yields:

$$\begin{aligned} h_\beta (x)={\mathbb {E}}_x\left[ \int _0^{\sigma ^\beta }e^{-qs}\mathrm{d}L^\beta _s\right] =v_\beta (x). \end{aligned}$$

This completes the proof. $\square$

Lemma B.2

Assume that p is $C^1$ on $]0,\infty [$ and that $-p'(x) -q<0$ for all $x\in ]0,\beta ]$. If $v_\beta '(\beta -)\ge 1$, then $v_{\beta }(x)$ is in $C^2$ on $]0,\beta [$ and it is increasing and concave on $]0,\infty [$.

Proof

We begin by proving that $v_{\beta }$ is increasing on $]0,\beta [$. Let

$$\begin{aligned} \tau ^+_a:=\inf \{t\ge 0:R_t\ge a\}\quad \text {and} \quad \tau ^-_0:=\inf \{t\ge 0:R_t<0\}. \end{aligned}$$

By the Strong Markov property of the PDMP, $R_t$ for all $0<y<x<\beta$

$$\begin{aligned} v_{\beta }(y)=v_{\beta }(x){\mathbb {E}}_y\left[ e^{-q\tau _x^+};\tau _x^+<\tau _0^-\right] <v_{\beta (x)}, \end{aligned}$$

which completes the proof of this statement. Furthermore, since p is $C^1$ by (15), we have that $v_{\beta }$ is $C^2$ on $]0,\beta [$ and $v_{\beta }$ is left-continuous at $\beta$. Moreover, if we differentiate (15) with respect to x, then we get:

$$\begin{aligned} p(x)v_{\beta }''(x)= & {} (-p'(x)-\lambda -q)v_{\beta }'(x)+\lambda \int _0^{\beta -x}v_{\beta }'(x+z)f(z)\, \mathrm{d}z\nonumber \\&+\lambda \int _{\beta -x}^\infty f(z)\, \mathrm{d}z. \end{aligned}$$

(19)

Consequently, since $v_{\beta }'(\beta -)\ge 1$, we obtain

$$\begin{aligned} p(\beta )v_{\beta }''(\beta -)\le -p'(\beta )-\lambda -q+\lambda \int _{0}^\infty f(z)\, dz=-p'(\beta )-q<0. \end{aligned}$$

We prove the final part by contradiction. Assume that $v_{\beta }$ is not concave. Then, by continuity of $v_{\beta }''$, there exists ${\hat{x}}\in ]0,\beta [$, such that $v_{\beta }''({\hat{x}})=0$ and $v_{\beta }''(x)<0$ for all $x\in ]{\hat{x}},\beta [$. Hence, from the assumption that $v_\beta '(\beta -)\ge 1$, it follows

$$\begin{aligned} 0=p({\hat{x}})v_{\beta }''({\hat{x}})&=\left( -p'({\hat{x}})-\lambda -q\right) v_{\beta }'({\hat{x}})+\lambda \int _0^{\beta -{\hat{x}}}v_{\beta }'({\hat{x}}+z)f(z)\, \mathrm{d}z\nonumber \\ {}&\qquad +\,\lambda \int _{\beta -{\hat{x}}}^\infty f(z)\, \mathrm{d}z\\&< (-p'({\hat{x}})-\lambda -q)v_{\beta }'({\hat{x}})\\&\qquad +\,\lambda v_{\beta }'({\hat{x}})\int _0^{\beta -{\hat{x}}}f(z)\, \mathrm{d}z+\lambda v_{\beta }'({\hat{x}})\int _{\beta -{\hat{x}}}^\infty f(z)\, \mathrm{d}z\\&= (-p'({\hat{x}})-q)v_{\beta }'({\hat{x}})<0. \end{aligned}$$

This leads to the contradiction. Therefore, $v_{\beta }''(x)<0$ for all $x\in ]0,\beta [$. $\square$

Now, we are ready to prove the main results.

Appendix C: Proof of the Theorem 4.4

From (13) and (15), we can conclude that $v_\beta$ on $]0,\beta [$ satisfies the equation:

$$\begin{aligned}&-p(x)v_\beta '(x)-(\lambda +q)v_\beta (x)+\lambda \int _0^{\beta -x}v_\beta (x+z)f(z)\,\mathrm{d}z\nonumber \\&\quad +\lambda \int _{\beta -x}^\infty \left( x+z-\beta +v_\beta (\beta )\right) f(z)\,\mathrm{d}z=0 \end{aligned}$$

(20)

with the initial condition $v_\beta (0)=0$. We recall that we are looking for $\beta ^*$ satisfying $v_{\beta ^*}'(\beta ^*)=1$. Let

$$\begin{aligned} u_\beta (x):=v_{\beta }'(x). \end{aligned}$$

Transforming equation (20), we obtain the following Fredholm Equation for $u_\beta$:

$$\begin{aligned} u_\beta (x)=G_\beta (x)+\int _0^\beta K(x,y)u_\beta (y)\,\mathrm{d}y, \end{aligned}$$

(21)

where

$$\begin{aligned} G_\beta (x):= & {} \frac{\lambda }{p(x)}\left( \int _{\beta -x}^\infty \left( x+z-\beta \right) f(z)\mathrm{d}z\right) ,\\ K(x,y):= & {} \left\{ \begin{array}{lll} \frac{-q}{p(x)}&{}\text { for }&{} 0\le y\le x,\\ \frac{\lambda }{p(x)}\int _{y-x}^\infty f(z)\,\mathrm{d}z&{}\text { for }&{} y>x\ge 0. \end{array} \right. \end{aligned}$$

Taking $x=\beta$ in (21) leads to the equation:

$$\begin{aligned} u_\beta (\beta )=\frac{\lambda }{p(\beta )}{\mathbb {E}}(C_1)-\frac{q}{p(\beta )}\int _0^\beta u_\beta (y)\,\mathrm{d}y. \end{aligned}$$

(22)

Let:

$$\begin{aligned} \gamma (\beta ):=u_\beta (\beta ). \end{aligned}$$

We want to prove the existence of $\beta ^*$ with the constraint:

$$\begin{aligned} \gamma (\beta ^*)=1. \end{aligned}$$

(23)

By our assumption:

$$\begin{aligned} \gamma (0)=\frac{\lambda }{p(0)}{\mathbb {E}}(C_1)>1. \end{aligned}$$

(24)

We will show that

$$\begin{aligned} \gamma ({\hat{x}})\le 1. \end{aligned}$$

(25)

By contradiction, assume that $\gamma ({\hat{x}})>1$. Then, by Lemma B.2, we have that $u_{{\hat{x}}}$ is increasing and consequently $u_{{\hat{x}}}(y)>1$ for all $y\in ]0,{\hat{x}}[$. Thus,

$$\begin{aligned} \gamma ({\hat{x}})= & {} \frac{\lambda }{p({\hat{x}})}{\mathbb {E}}(C_1)-\frac{q}{p({\hat{x}})}\int _0^{{\hat{x}}} u_{{\hat{x}}}(y)\,\mathrm{d}y\\\le & {} \frac{\lambda }{p({\hat{x}})}{\mathbb {E}}(C_1)-\frac{q}{p({\hat{x}})}{\hat{x}}<1. \end{aligned}$$

In this way, we derived a contradiction. To get (23) by (24) and (25), it suffices to prove that $\gamma$ is a continuous function. The latter follows from the following observation.

We denote $\varDelta _\beta b_\beta (x)=b_{\beta }(x)-b_{\beta -}(x)$ for a general function $b_\beta$. From (21), it follows that

$$\begin{aligned} \varDelta _\beta u_\beta (\beta )=\int _0^\beta K(\beta ,y)\varDelta _\beta u_\beta (y)\,\mathrm{d}y, \end{aligned}$$

and hence, function $\beta \rightarrow \varDelta _\beta u_\beta (\beta )$ is continuous. Note that from the last conclusion, it follows that

$$\begin{aligned} \varDelta _\beta \varDelta _\beta u_\beta (\beta )=\varDelta _\beta u_\beta (\beta )=0, \end{aligned}$$

and function $\beta \rightarrow u_\beta (\beta )=\gamma (\beta )$ is also continuous. This completes the proof. $\square$

Appendix D: Proof of the Theorem 4.5

By Lemma B.2, we have $v_{\beta ^*}'(x)\ge v_{\beta ^*}'(\beta ^*)=1$ for all $x\in ]0,\beta ^*[$. Moreover, $({\mathcal {A}}-q{\mathbf {I}})v_{\beta ^*}(x)=0$ for $x\in ]0,\beta ^*[$. It remains to prove that $({\mathcal {A}}-q{\mathbf {I}})v_{\beta ^*}(x)\le 0$ for $x>\beta ^*$.

Note that $v_{\beta ^*}$ is $C^1$, and therefore, by assumption that $p\in C^1$, the function $({\mathcal {A}}-q{\mathbf {I}})v_{\beta ^*}(x)$ is continuous at $x=\beta ^*$. Thus, we have $({\mathcal {A}}-q{\mathbf {I}})v_{\beta ^*}(\beta ^*)=0$. Furthermore, since

$$\begin{aligned} v_{\beta ^*}(x)=x-\beta ^*+v_{\beta ^*}(\beta ^*)\quad \text { for }x>\beta ^*, \end{aligned}$$

and $({\mathcal {A}}-q{\mathbf {I}})v_{\beta ^*}(x)=-p(x)+\lambda \int _0^\infty zdF(z)-q(x-\beta ^*+v_{\beta ^*}(\beta ^*))$, we have:

$$\begin{aligned} ({\mathcal {A}}-q{\mathbf {I}})v_{\beta ^*}(x)= & {} \left( {\mathcal {A}}-q{\mathbf {I}}\right) \left( v_{\beta ^*}(x)-v_{\beta ^*}(\beta ^*)\right) \nonumber \\ {}= & {} -p(x)-q(x-\beta ^*)+p(\beta ^*)<0. \end{aligned}$$

(26)

Hence, by Theorem 4.1, $v_{\beta ^*}(x)\ge v(x)$ for all $x\ge 0$. At the same time, by the definition of the value function, we have that $v_{\beta ^*}(x)\le v(x)$ for all $x\ge 0$. Consequently, $v_{\beta ^*}(x)=v(x)$ and by Lemma B.1 v must uniquely solve the equation (15) with the boundary condition $v^\prime (\beta ^*)=1$. This completes the proof. $\square$

Appendix E: Proof of the Lemma 4.1

Assume first that Condition (i) holds. In this case, we will follow the same arguments as in the proof of Theorem 4.5 to get the opposite to (26) inequality $({\mathcal {A}}-q{\mathbf {I}})v_{\beta ^*}({\check{x}})>0$. Indeed, by the continuity of the function $x\mapsto ({\mathcal {A}}-q)v_{\beta ^*}(x)$, there exists an open and bounded interval $\mathrm {J}\subset ]\beta ^*,\infty [$ such that ${\check{x}}\in \mathrm {J}$ and $({\mathcal {A}}-q)v_{\beta ^*}(x)> 0$ for all $x\in \mathrm {J}$. Define a strategy ${\tilde{\pi }}$ as follows: whenever the process $X^{{\tilde{\pi }}}$ takes values in the interval $\mathrm {J}$, do not pay any dividends, otherwise operate according to $\pi _\beta ^*$. If we extend $v_{\beta ^*}$ to the negative half-axis by $v_{\beta ^*}(x)=0$ for $x<0$, then we have

$$\begin{aligned} v_{{\tilde{\pi }}}(x) = {\left\{ \begin{array}{ll} {\mathbb {E}}_x[e^{-qT_\mathrm {J}}v_{\beta ^*}(R_{T_\mathrm {J}})] &{}\quad \text { for }x\in \mathrm {J}, \\ v_{\beta ^*}(x) &{} \quad \text { for } x\not \in \mathrm {J}, \end{array}\right. } \end{aligned}$$

where $T_\mathrm {J}:=\inf \left\{ t\ge 0:R_t\notin \mathrm {J}\right\}$. Note that by Davis [22] and Rolski et al. [23], the process $R_t$ is a piecewise deterministic Markov process, and the function $v_{\beta ^*}(x)$ is in a domain of extended generator ${\mathcal {A}}-q{\mathbf {I}}$ of exponentially killed dual process R. Thus, the stopped process

$$\begin{aligned} K_{t}:=e^{-q t\wedge T_\mathrm {J}}v_{\beta ^*}(R_{t\wedge T_\mathrm {J}})-v_{\beta ^*}(x)-{\mathbb {E}}_x\bigg [\int _0^{t\wedge T_\mathrm {J}}({\mathcal {A}}-q{\mathbf {I}})v_{\beta ^*}(R_s)\,\mathrm{d}s\bigg ] \end{aligned}$$

is a Dynkin mean-zero martingale for $R_0=x\in \mathrm {J}$. In fact, it is uniformly integrable since using (26) we have

$$\begin{aligned} K_t\le T_{\mathrm {J}} \sup _{x\in \mathrm {J}}\left| -p(x)-q(x-\beta ^*)+p(\beta ^*)\right| + (-\beta ^*+v_{\beta ^*}(\beta ^*))+ R_{T_{\mathrm {J}}} \end{aligned}$$

and ${\mathbb {E}}T_{\mathrm {J}}<\infty$ because of the positive drift assumption and ${\mathbb {E}}R_{T_{\mathrm {J}}}\le \sup _{x\in \mathrm {J}}x +{\mathbb {E}}C_{N(T_{\mathrm {J}})}<\infty$. Now, we apply the Optional Stopping Theorem to $K_{t}$ to get, for all $x\in \mathrm {J}$,

$$\begin{aligned} v_{{\tilde{\pi }}}(x)={\mathbb {E}}_x[e^{-q T_\mathrm {J}}v_{\beta ^*}(R_{ T_\mathrm {J}})]=v_{\beta ^*}(x)+{\mathbb {E}}_x\bigg [\int _0^{ T_\mathrm {J}}({\mathcal {A}}-q{\mathbf {I}})v_{\beta ^*}(R_s)\,\mathrm{d}s\bigg ]> v_{\beta ^*}(x). \end{aligned}$$

Therefore, $\pi _\beta ^*$ is not optimal among all admissible strategies.

Assuming Condition (ii), we can follow the proof of Lemma B.2. Taking $\beta =\beta ^*$ and $x=\beta ^*$ in (19), we have

$$\begin{aligned}&p(\beta ^*)v_{\beta }''(\beta ^*)=(-p'(\beta ^*)-\lambda -q)v_{\beta }'(\beta ^*)\\&\quad \quad +\lambda \int _0^\infty f(z)\,\mathrm{d}z=(-p'(\beta ^*)-\lambda -q)+\lambda {\mathbb {E}}C_1>0. \end{aligned}$$

By the continuity of $v_{\beta }''$ and the fact that $v_{\beta }'(\beta ^*)=1$, we can conclude that there exists an interval $]{\dot{x}},\beta ^*[$ for which $v_{\beta }'(x)<1$. We now consider a strategy ${\tilde{\pi }}$, which pays immediately anything above the level ${\dot{x}}$ if the reserve process takes values in $]{\dot{x}},\beta ^*[$, and then follows the strategy $\pi _\beta ^*$. Then, for $x\in ]{\dot{x}},\beta ^*[$ , we have

$$\begin{aligned} v_\pi (x)=x-{\dot{x}}+v_\beta ^*({\dot{x}}). \end{aligned}$$

Since $v_{\beta }'(x)<1$ for all $x\in ]{\dot{x}},\beta ^*[$, hence $v_\pi (x)>v_\beta ^*(x)$. Thus, $\pi _\beta ^*$ is not optimal. $\square$