Equilibrium Dividend Strategy with Non-exponential Discounting in a Dual Model

Abstract

This paper studies an optimal dividend problem for a company with non-exponential discounting. The surplus process is described by a dual model, and the target is to find a dividend strategy that maximizes the expected discounted value of dividends until ruin. The non-exponential discount function leads to a time-inconsistent problem. We aim at seeking the equilibrium strategy derived by taking our problem as a non-cooperate game, which is a time-consistent strategy. An extended Hamilton–Jacobi–Bellman equation system and a verification theorem are provided to derive the equilibrium strategy and the equilibrium value function. For the case of pseudo-exponential discount function, closed-form expressions for the equilibrium strategy and the equilibrium value function are derived. In addition, some numerical illustrations of our results are showed.

Appendices

Appendix 1: Proof of Theorem 3.1

We first show that V is the value function corresponding to $\hat{\ell }$, i.e., that $V (t, x) = J(t, x, \hat{\ell })$. Note that

$$\begin{aligned} \mathcal {A}^{\hat{\ell }}u^{rs}(t,x) + U(r, t, \hat{l}) = 0. \end{aligned}$$

(34)

By Dynkin’s Theorem for $u^{rs}$, we have

$$\begin{aligned} E_{t,x}\bigg [\int _t^{\tau _{t,x}^{\hat{\ell }}}\mathcal {A}^{\hat{\ell }}u^{rs}(s, R_s^{\hat{\ell }})\hbox {d}s \bigg ] = E_{t,x}[u^{rs}(\tau _{t,x}^{\hat{\ell }}, R_{\tau _{t,x}^{\hat{\ell }}}^{\hat{\ell }})] - u^{rs}(t,x). \end{aligned}$$

Thus, using (34) and boundary condition $u^{rs}(t,0)=0$, we have

$$\begin{aligned} u^{rs}(t,x)= & {} E_{t,x}\bigg [\int _t^{\tau _{t,x}^{\hat{\ell }}}U(r, s, \hat{l}(s, R_s^{\hat{\ell }}))\hbox {d}s \bigg ]= E_{t,x}\bigg [\int _t^{\tau _{t,x}^{\hat{\ell }}}\varphi (s-r)\hat{l}(s, R_s^{\hat{\ell }})\hbox {d}s \bigg ] \end{aligned}$$

and

$$\begin{aligned} u^{s}(t,t,x) = u^{ts}(t,x) = E_{t,x}\bigg [\int _t^{\tau _{t,x}^{\hat{\ell }}}\varphi (s-t)\hat{l}(s, R_s^{\hat{\ell }})\hbox {d}s \bigg ]. \end{aligned}$$

Next, by the extended HJB equation system for V, we have

$$\begin{aligned} \mathcal {A}^{\hat{\ell }}V(t, x) + U(t, t, \hat{l}) - \mathcal {A}^{\hat{\ell }}u^s(t, t, x) + \mathcal {A}^{\hat{\ell }}u^{ts}(t, x)= & {} 0 \end{aligned}$$

and (34). Thus we have equation

$$\begin{aligned} \mathcal {A}^{\hat{\ell }}V(t, x) - \mathcal {A}^{\hat{\ell }}u^s(t, t, x)= & {} 0. \end{aligned}$$

Using Dynkin’s Theorem for V, we thus have

$$\begin{aligned} E_{t,x}\left[ V \left( \tau _{t,x}^{\hat{\ell }}, R_{\tau _{t,x}^{\hat{\ell }}}^{\hat{\ell }}\right) \right] = V(t, x) + E_{t,x}\bigg [\int _t^{\tau _{t,x}^{\hat{\ell }}}\mathcal {A}^{\hat{\ell }}V(s, R_s^{\hat{\ell }})\hbox {d}s \bigg ]. \end{aligned}$$

Hence, we get

$$\begin{aligned} E_{t,x}\left[ V\left( \tau _{t,x}^{\hat{\ell }}, R_{\tau _{t,x}^{\hat{\ell }}}^{\hat{\ell }}\right) \right]= & {} V(t, x) + E_{t,x}\bigg [\int _t^{\tau _{t,x}^{\hat{\ell }}}\mathcal {A}^{\hat{\ell }}u^s(r, r, R_r^{\hat{\ell }})dr \bigg ]. \end{aligned}$$

(35)

By using Dynkin’s Theorem for $u^s$, we have

$$\begin{aligned} E_{t,x}\bigg [\int _t^{\tau _{t,x}^{\hat{\ell }}}\mathcal {A}^{\hat{\ell }}u^s\left( r, r, R_r^{\hat{\ell }}\right) dr \bigg ] = E_{t,x}\left[ u^s\left( \tau _{t,x}^{\hat{\ell }}, \tau _{t,x}^{\hat{\ell }}, R_{\tau _{t,x}^{\hat{\ell }}}^{\hat{\ell }}\right) \right] - u^s(t,t,x). \end{aligned}$$

(36)

Using (35), (36) and boundary conditions for V and $u^s$, we can derive

$$\begin{aligned} V(t, x) = E_{t,x}\bigg [\int _t^{\tau _{t,x}^{\hat{\ell }}}U\left( t, s, \hat{l}\left( s,R_s^{\hat{\ell }}\right) \right) \hbox {d}s \bigg ] = J\left( t, x, \hat{\ell }\right) . \end{aligned}$$

We now show that $\hat{\ell }$ is indeed an equilibrium strategy. For any $h > 0$ and an arbitrary $l\in [0, M]$, define the strategy $\ell _h$ as in Definition 2.2. Without loss of generality, suppose that $t+h<\tau _{t,x}^{\ell _h}$. Then we have

$$\begin{aligned} J(t,x,\ell _h)= & {} E_{t,x} \bigg [\int _t^{\tau _{t,x}^{\ell _h}}U(t, s, l_h(s,R_s^{\ell _h}))\hbox {d}s \bigg ] \\= & {} E_{t,x} \bigg [\int _t^{t+h}U(t, s, l_h(s,R_s^{\ell _h}))\hbox {d}s \bigg ] + E_{t,x} \bigg [\int _{t+h}^{\tau _{t,x}^{\ell _h}}U(t, s, l_h(s,R_s^{\ell _h}))\hbox {d}s \bigg ] \\= & {} E_{t,x} \bigg [\int _t^{t+h}U(t, s, l_h(s,R_s^{\ell _h}))\hbox {d}s \bigg ] + E_{t,x} \bigg [V(t+h, R_{t+h}^{\ell _h}) \bigg ] \\&\quad +\,E_{t,x} \bigg [\int _{t+h}^{\tau _{t,x}^{\ell _h}}U(t, s, l_h(s,R_s^{\ell _h}))\hbox {d}s - V(t+h, R_{t+h}^{\ell _h})\bigg ]. \end{aligned}$$

Using Dynkin’s Theorem for V, we have

$$\begin{aligned} E_{t,x}\left[ V\left( t+h, R_{t+h}^{\ell _h}\right) \right] = V(t, x) + E_{t,x}\bigg [\int _t^{t+h}\mathcal {A}^{l_h}V\left( s, R_s^{\ell _h}\right) \hbox {d}s \bigg ]. \end{aligned}$$

Then

$$\begin{aligned} J(t,x,\ell _h)= & {} E_{t,x} \bigg [\int _t^{t+h}U(t, s, l_h(s,R_s^{\ell _h}))\hbox {d}s \bigg ] + E_{t,x}\bigg [\int _t^{t+h}\mathcal {A}^{l_h}V(s, R_s^{\ell _h})\hbox {d}s \bigg ] \\+ & {} V(t, x) + E_{t,x} \bigg [\int _{t+h}^{\tau _{t,x}^{\ell _h}} [\varphi (s-t)-\varphi (s-t-h)]l_h(s,R_s^{\ell _h})\hbox {d}s \bigg ]. \end{aligned}$$

Noting the fact that $V (t, x) = J(t, x, \hat{\ell })$, we obtain

$$\begin{aligned} J(t, x, \hat{\ell }) - J(t,x,\ell _h)= & {} -E_{t,x} \bigg [\int _t^{t+h}U(t, s, l(s,R_s^{\ell _h}))\hbox {d}s \bigg ] \\&- E_{t,x}\bigg [\int _t^{t+h}\mathcal {A}^{l_h}V(s, R_s^{\ell _h})\hbox {d}s \bigg ] \\&- E_{t,x} \bigg [\int _{t+h}^{\tau _{t,x}^{\ell _h}} [\varphi (s-t)-\varphi (s-t-h)]l_h(s,R_s^{\ell _h})\hbox {d}s \bigg ]. \end{aligned}$$

So

$$\begin{aligned} \liminf _{h\rightarrow 0} \frac{J(t, x, \hat{\ell }) - J(t, x, \ell _h)}{h}= & {} \liminf _{h\rightarrow 0} \Bigg \{-\mathcal {A}^{\hat{\ell }}V(t, x) - U(t, t, \hat{l}(t, x)) \\&- E_{t,x} \bigg [\int _{t}^{\tau _{t,x}^{\hat{\ell }}} \varphi ^{\prime }(s-t)\hat{l}(s,R_s^{\hat{\ell }})\hbox {d}s \bigg ] \Bigg \}\ge 0. \end{aligned}$$

The last inequality follows from

$$\begin{aligned} \sup _{l\in [0,M]} \Big \{ \mathcal {A}^{\ell }V(t, x) + U(t, t, l) - \mathcal {A}^{\ell }u^s(t, t, x) + \mathcal {A}^{\ell }u^{ts}(t, x) \Big \}= & {} 0. \end{aligned}$$

$\square$

Appendix 2: Proof of Proposition 3.1

Note that

$$\begin{aligned} u^{rs}(t,x) = E_{t,x} \bigg [\int _{t}^{\tau _{t,x}^{\hat{\ell }}} \varphi (s-r)\hat{l}(s,R_s^{\hat{\ell }})\hbox {d}s \bigg ] \end{aligned}$$

and

$$\begin{aligned} u^{s}(t,t,x) = E_{t,x} \bigg [\int _{t}^{\tau _{t,x}^{\hat{\ell }}} \varphi (s-t)\hat{l}(s,R_s^{\hat{\ell }})\hbox {d}s \bigg ]. \end{aligned}$$

Then

$$\begin{aligned} \mathcal {A}^{\ell }u^{rs}(t,x)= & {} -\varphi (t-r)\hat{l}(t,R_t^{\hat{\ell }})-(\mu +l(t,x))u^{rs}_x(t,x)\\&+\, \lambda \delta \bigg [\int _0^{\infty }(u^{rs}(t,x+y)-u^{rs}(t,x))\hbox {e}^{-\delta y}\hbox {d}y \bigg ]. \end{aligned}$$

Thus,

$$\begin{aligned} \mathcal {A}^{\ell }u^{ts}(t,x)= & {} -\varphi (t-t)\hat{l}(t,R_t^{\hat{\ell }})-(\mu +l(t,x))u^{ts}_x(t,x) \nonumber \\&+\, \lambda \delta \bigg [\int _0^{\infty }(u^{ts}(t,x+y)-u^{ts}(t,x))\hbox {e}^{-\delta y}\hbox {d}y \bigg ] \end{aligned}$$

(37)

and

$$\begin{aligned}&\mathcal {A}^{\ell }u^{s}(t,t,x) = -\varphi (t-t)\hat{l}(t,R_t^{\hat{\ell }})-E_{t,x} \bigg [\int _{t}^{\tau _{t,x}^{\hat{\ell }}} \varphi ^{\prime }(s-t)\hat{l}(s,R_s^{\hat{\ell }})\hbox {d}s \bigg ] \nonumber \\&\quad -(\mu +l(t,x))u^{s}_x(t,t,x) + \lambda \delta \bigg [\int _0^{\infty }(u^{s}(t,t,x+y)-u^{s}(t,t,x))\hbox {e}^{-\delta y}\hbox {d}y \bigg ].\nonumber \\ \end{aligned}$$

(38)

Inserting (37)–(38) into equations of Definition 3.1, we complete the proof. $\square$

Appendix 3: Proof of Lemma 4.1

Note that

$$\begin{aligned} A_1 = \frac{M\xi _5}{\rho _1}\left[ \left( \xi _5-\xi _1\right) \hbox {e}^{\xi _1b} + \left( \xi _2-\xi _5\right) \hbox {e}^{\xi _2b}\right] ^{-1}. \end{aligned}$$

(i):

If $\xi _5-\xi _1>0$, then $A_1<0$.

(ii):

If $\xi _5-\xi _1<0$, then $(\xi _5-\xi _1)\hbox {e}^{\xi _1b}>\xi _5-\xi _1$ and $(\xi _2-\xi _5)\hbox {e}^{\xi _2b}>\xi _2-\xi _5$.

So

$$\begin{aligned} \left( \xi _5-\xi _1\right) \hbox {e}^{\xi _1b}+\left( \xi _2-\xi _5\right) \hbox {e}^{\xi _2b}> & {} \left( \xi _5-\xi _1\right) +(\xi _2-\xi _5)=\xi _2-\xi _1>0, \end{aligned}$$

and then $A_1<0$.

Similarly, we can prove $A_2<0,\;\; B_1<0$, and $B_2<0.$ $\square$

Appendix 4: Proof of Lemma 4.2

$(\mathbf {i})$ We only prove the first inequality $(P+1)\xi _2 > \xi _5$. The second one is similar. Inequality $(P+1)\xi _2 > \xi _5$ is equivalent to $\frac{M}{\rho _1} < \frac{1}{\xi _2} - \frac{1}{\xi _5}.$

Firstly, note that

$$\begin{aligned} \xi _2= & {} \frac{(\mu \delta -\lambda -\rho _1)+\sqrt{(\mu \delta -\lambda -\rho _1)^2+4\mu \rho _1\delta }}{2\mu },\\ -\xi _5= & {} \frac{-[(\mu +M)\delta -\lambda -\rho _1]+\sqrt{[(\mu +M)\delta -\lambda -\rho _1]^2+4(\mu +M)\rho _1\delta }}{2(\mu +M)}. \end{aligned}$$

In addition, according to Cauchy inequality, we have

$$\begin{aligned} \sqrt{a^2+c}-a < \frac{c}{2a}, \quad a>0 \;\;\hbox {and}\;\; c>0. \end{aligned}$$

Next, we split the problem into the following three situations.

(1) If $(\mu +M)\delta -\lambda -\rho _1>0$ and $\mu \delta -\lambda -\rho _1<0,$ then we have

$$\begin{aligned} -\xi _5 < \frac{\rho _1\delta }{(\mu +M)\delta -\lambda -\rho _1}, \quad \xi _2 < \frac{\rho _1\delta }{-\mu \delta +\lambda +\rho _1}, \end{aligned}$$

i.e.,

$$\begin{aligned}&-\frac{1}{\xi _5} > \frac{(\mu +M)\delta -\lambda -\rho _1}{\rho _1\delta }, \quad \frac{1}{\xi _2} > \frac{-\mu \delta +\lambda +\rho _1}{\rho _1\delta },\\&\quad \frac{1}{\xi _2}-\frac{1}{\xi _5} > \frac{(\mu +M)\delta -\lambda -\rho _1}{\rho _1\delta } + \frac{-\mu \delta +\lambda +\rho _1}{\rho _1\delta } = \frac{M}{\rho _1}. \end{aligned}$$

(2) If $(\mu +M)\delta -\lambda -\rho _1\le 0$, we have $\mu \delta -\lambda -\rho _1<0.$ Hence, we have

$$\begin{aligned} \frac{\mu }{\rho _1} + \frac{M}{\rho _1} - \frac{\lambda }{\rho _1\delta } - \frac{1}{\delta } \le 0, \end{aligned}$$

i.e.,

$$\begin{aligned} \frac{M}{\rho _1} \le \frac{\lambda }{\rho _1\delta } + \frac{1}{\delta } - \frac{\mu }{\rho _1}=\frac{1}{\xi _2}. \end{aligned}$$

(3) If $\mu \delta -\lambda -\rho _1\ge 0,$ we have $(\mu +M)\delta -\lambda -\rho _1>0.$ Thus

$$\begin{aligned} - \frac{1}{\xi _5}> & {} \frac{(\mu +M)\delta -\lambda -\rho _1}{\rho _1\delta } =\frac{M}{\rho _1} + \frac{\mu \delta -\lambda -\rho _1}{\rho _1\delta } ~~\ge ~~ \frac{M}{\rho _1}. \end{aligned}$$

Combining (1), (2) and (3), we complete the proof of $(\mathbf {i})$.

$(\mathbf {ii})$ If $\omega P+(1-\omega )Q+1<0$, we have $F(0)>0.$ Note

$$\begin{aligned} F(b) = \frac{\omega P[\xi _1\hbox {e}^{\xi _1b}-\xi _2\hbox {e}^{\xi _2b}]}{[(\xi _5-\xi _1)\hbox {e}^{\xi _1b}+(\xi _2-\xi _5)\hbox {e}^{\xi _2b}]} + \frac{(1-\omega )Q\,[\xi _3\hbox {e}^{\xi _3b}-\xi _4\hbox {e}^{\xi _4b}]}{[(\xi _6-\xi _3)\hbox {e}^{\xi _3b}+(\xi _4-\xi _6)\hbox {e}^{\xi _4b}]} - 1. \end{aligned}$$

According to $(\mathbf {i})$, we have

$$\begin{aligned} \lim _{b\rightarrow +\infty }F(b)= & {} \omega P\, \frac{-\xi _2}{\xi _2-\xi _5} + (1-\omega )Q\, \frac{-\xi _4}{\xi _4-\xi _6}-1 \\= & {} -\omega \left[ P\, \frac{\xi _2}{\xi _2-\xi _5}+1\right] - (1-\omega ) \left[ Q\, \frac{\xi _4}{\xi _4-\xi _6}+1\right] \\= & {} -\omega \frac{(P+1)\xi _2-\xi _5}{\xi _2-\xi _5} - (1-\omega ) \frac{(Q+1)\xi _4-\xi _6}{\xi _4-\xi _6}<0. \end{aligned}$$

Furthermore,

$$\begin{aligned} F^\prime (b)= & {} \omega P\, \frac{-\xi _5(\xi _1-\xi _2)^2\hbox {e}^{(\xi _1+\xi _2)b}}{[(\xi _5-\xi _1)\hbox {e}^{\xi _1b}+(\xi _2-\xi _5)\hbox {e}^{\xi _2b}]^2} \\&+\,(1-\omega )Q\, \frac{-\xi _6(\xi _3-\xi _4)^2\hbox {e}^{(\xi _3+\xi _4)b}}{[(\xi _6-\xi _3)\hbox {e}^{\xi _3b}+(\xi _4-\xi _6)\hbox {e}^{\xi _4b}]^2} \\= & {} -\omega \frac{M}{\rho _1}\, \frac{\xi _5^2(\xi _1-\xi _2)^2\hbox {e}^{(\xi _1+\xi _2)b}}{[(\xi _5-\xi _1)\hbox {e}^{\xi _1b}+(\xi _2-\xi _5)\hbox {e}^{\xi _2b}]^2} \\&-\,(1-\omega )\frac{M}{\rho _2}\, \frac{\xi _6^2(\xi _3-\xi _4)^2\hbox {e}^{(\xi _3+\xi _4)b}}{[(\xi _6-\xi _3)\hbox {e}^{\xi _3b}+(\xi _4-\xi _6)\hbox {e}^{\xi _4b}]^2}<0. \end{aligned}$$

Therefore, $F(b)=0$ has a unique positive root. The proof of $(\mathbf {ii})$ is completed. $\square$

Appendix 5: Proof of Theorem 4.1

$(\mathbf {i})$ We can verify that V(x) defined by (32) is a continuously differentiable concave function and satisfies $V(0)=0$. Since

$$\begin{aligned} V^{\prime }(0)=-\left[ \omega \frac{M}{\rho _1}\xi _5 + (1-\omega )\frac{M}{\rho _2}\xi _6\right] \le 1, \end{aligned}$$

$V^{\prime }(x)\le 1$ for all $x\ge 0$ and

$$\begin{aligned} (M-l)(V^{\prime }(x)-1)\le 0,\quad l\in [0,M]. \end{aligned}$$

(39)

In addition, we know that the function V given by (32) satisfies

$$\begin{aligned} -(\mu \!+\!M)V^{\prime }(x)\!+\!\lambda \delta \left( \int _0^{\infty }\!V(x\!+\!y)\hbox {e}^{-\delta y}\hbox {d}y \right) \!-\!\lambda V(x)\!+\!\varPhi (t, x)\!+\!M=0. \end{aligned}$$

(40)

Adding the inequality (39) to the equality (40), we can obtain (12).

$(\mathbf {ii})$ $V(0)=0$ is obvious. The first-order derivative of V(x) given by (33) is:

$$\begin{aligned} V^{\prime }(x)= {\left\{ \begin{array}{ll} \omega A_1(\xi _1\hbox {e}^{\xi _1 x}-\xi _2\hbox {e}^{\xi _2 x}) + (1-\omega )A_2(\xi _3\hbox {e}^{\xi _3 x}-\xi _4\hbox {e}^{\xi _4 x}), \quad 0<x<b, \\ \omega B_1\xi _5\hbox {e}^{\xi _5 x} + (1-\omega )B_2\xi _6\hbox {e}^{\xi _6 x},\qquad \qquad \qquad \qquad \qquad \quad x\ge b. \end{array}\right. } \end{aligned}$$

From Lemma 4.1, we have that $V^{\prime }(x)>0$ for all $x>0$, which implies that V(x) is a strictly increasing function.

In addition, we can derive the second-order derivative of V(x) as follows:

$$\begin{aligned} V^{\prime \prime }(x)= {\left\{ \begin{array}{ll} \omega A_1(\xi _1^2\hbox {e}^{\xi _1 x}-\xi _2^2\hbox {e}^{\xi _2 x}) + (1-\omega )A_2(\xi _3^2\hbox {e}^{\xi _3 x}-\xi _4^2\hbox {e}^{\xi _4 x}), \quad 0<x<b, \\ \omega B_1\xi _5^2\hbox {e}^{\xi _5 x} + (1-\omega )B_2\xi _6^2\hbox {e}^{\xi _6 x},\qquad \qquad \qquad \qquad \qquad \quad x\ge b. \end{array}\right. } \end{aligned}$$

From Lemma 4.1, we have that $V^{\prime \prime }(x)<0$ for all $x\ge b$. Now, we consider the case when $0<x<b$. Note that

$$\begin{aligned} V^{\prime \prime \prime }(x)=\omega A_1(\xi _1^3\hbox {e}^{\xi _1 x}-\xi _2^3\hbox {e}^{\xi _2 x}) + (1-\omega )A_2(\xi _3^3\hbox {e}^{\xi _3 x}-\xi _4^3\hbox {e}^{\xi _4 x}), \quad 0<x<b. \end{aligned}$$

Thus $V^{\prime \prime \prime }(x)>0$ and $V^{\prime \prime }(x)$ is increasing on interval (0, b). Then we have that $V^{\prime \prime }(x)< V^{\prime \prime }(b-)$ for $0<x<b$.

Besides, from (16), (18) and (19), we have

$$\begin{aligned} \mu V^{\prime \prime }(b-)= & {} \omega (\mu \delta -\lambda -\rho _1)V_1^{\prime }(b)+\omega \rho _1 \delta V_1(b) \\&\quad +\, (1-\omega )(\mu \delta -\lambda -\rho _2)V_2^{\prime }(b)+(1-\omega )\rho _2 \delta V_2(b) \end{aligned}$$

and

$$\begin{aligned} (\mu +M)V^{\prime \prime }(b+)= & {} \omega (\mu \delta +M\delta -\lambda -\rho _1)V_1^{\prime }(b)+\omega \rho _1 \delta V_1(b)-M\delta \\&\quad +\, (1-\omega )(\mu \delta +M\delta -\lambda -\rho _2)V_2^{\prime }(b)+(1-\omega )\rho _2 \delta V_2(b). \end{aligned}$$

Noting that $V^{\prime }(b)=1$, we have $\mu V^{\prime \prime }(b-)=(\mu +M)V^{\prime \prime }(b+)\le 0.$ Therefore, $V^{\prime \prime }(x)< 0$ for $0<x<b$.

In conclusion, V(x) defined by (33) is an increasing and continuously differentiable concave function on $]0, +\infty [$. This implies the uniqueness of b.

In addition, we can adopt the same way as (i) to verify that V(x) given by (33) satisfies Eq. (12), and here we omit it. $\square$