2. Problem formulation

To give a mathematical formulation of the optimization problem, we start from a one-dimensional Lévy process $X=\{X(t);\ t\geq0\}$ defined on $ (\Omega, \mathcal{F}, \mathbb{F}=\{\mathcal{F}_t,\, t\geq 0\}, \mathbb{P}) $ , a filtered probability space satisfying the usual conditions. Throughout this paper, we assume that X is a spectrally negative Lévy process with the usual exclusion of pure increasing linear drift and the negative of a subordinator. We denote by $\mathbb{P}_x$ the law of X with $X_0=x\geq0$ , and let $\mathbb{E}_x$ denote the corresponding expectation operator. Let $\{\bar{X}(t);\ t\geq0\}$ be the running supremum process of X, where $\bar{X}(t) : =\sup_{0\leq s\leq t}X(s) $ for $t\geq0$ . We assume that in the case of no control, the surplus process for a company evolves as $\{X(t);\ t\geq0\}$ .

A loss-carry-forward tax strategy is described by a one-dimensional stochastic process $\{\gamma(\bar{X}(t)),\,t\geq0\}$ , where $\gamma\colon [0,\infty)\rightarrow[\gamma_{1},\gamma_{2}]$ is a measurable function with $0\leq\gamma_{1}\leq\gamma_{2}\lt 1$ . Since there is a one-to-one correspondence between the function $\gamma$ and the tax strategy $\{\gamma(\bar{X}(t)),\,t\geq0\}$ , we will also denote the corresponding loss-carry-forward tax strategy by $\gamma$ for short. For the case of an insurance company, $\gamma(\bar{X}(t)) $ denotes the fraction of the insurer’s income that is paid out as tax at time t if it is in a profitable situation. Here we say that the company is in a profitable situation at time t if we have $X(t)=\bar{X}(t) $ . Thus, when applying strategy $\gamma$ , the cumulative tax until time t is

$$ \int_{0}^{t}\gamma(\bar{X}(s)){{\rm d}}\,\bar{X}(s), $$

and the controlled surplus process is given by

$$ U^{\gamma}(t)=X(t)-\int_{0}^{t}\gamma(\bar{X}(s)) {{\rm d}}\,\bar{X}(s). $$

The strategy $\gamma$ is said to be admissible if the process $\{\gamma(\bar{X}(t));\ t\geq0\}$ is adapted to the filtration $\{\mathcal{F}_{t};\ t\geq0\}$ and $\gamma(\bar{X}(t))\in[\gamma_{1},\gamma_{2}]$ . By $\Gamma$ we denote the set of all measurable functions $\gamma\colon [0,\infty)\rightarrow[\gamma_{1},\gamma_{2}]$ , i.e. $\Gamma$ consists of all admissible tax strategies in the sense of equating a tax strategy with its corresponding function $\gamma$ . For a given admissible strategy $\gamma\in\Gamma$ , we define the tax return function $f_{\gamma}$ by

\begin{equation} f_{\gamma}(x) =\mathbb{E}_{x}\bigg(\int_{0}^{\tau_{\xi}^{\gamma}}\mathrm{e}^{-qt} \gamma(\bar{X}(t)){{\rm d}}\ \bar{X}(t)\bigg), \qquad x\in[0,\infty), \end{equation}

where $q>0$ is a discount factor, and

(1) \begin{equation} \tau^{\gamma}_{\xi}=\mbox{inf}\{t\geq0;\ U^{\gamma}(t)\lt \xi(\bar{U}^{\gamma}(t))\} \label{eqn1} \end{equation}

with $\bar{U}^{\gamma}(t)\,\, :\!=\sup_{0\leq s\leq t}U^{\gamma}(s) $ , is the general draw-down time associated with the general draw-down function $\xi$ for the process $U^{\gamma}$ . The function $\xi\colon [0,\infty)\rightarrow(\!-\infty,+\infty) $ is called a general draw-down function if it is a continuously differentiable function satisfying $\bar{\xi}(y)=y-\xi(y)>0$ for all $y\in[0,\infty) $ .

It is interesting to see why the time defined by (Section 1) is called the general draw-down time. In the literature, the reflected process of a Markov process Y at its supremum, defined as $\smash{\sup_{s\in[0,t]}}Y(s)-Y(t) $ , is called the draw-down process of Y (see, for example, [Reference Landriault, Li and Zhang21]). Hence, it is natural to name the first time when the magnitude of the draw-down process exceeds a given threshold $a>0$ , i.e.

$$\ell _\alpha ^ + : = \inf \left\{ {t \ge 0;\mathop {\sup }\limits_{s \in [0,t]} \,Y(s) - Y(t) \gt \alpha } \right\} = \inf \left\{ {t \ge 0;Y(t) \lt \mathop {\sup }\limits_{s \in [0,t]} Y(s) - \alpha } \right\},$$

as the draw-down time. By comparing the definitions of $\smash{\ell_{a}^{+}}$ and $\smash{\tau^{\gamma}_{\xi}}$ , we may observe that $\smash{\ell_{a}^{+}}$ is a special case of the general draw-down time for the process Y, in that its general draw-down function is specialized to $\xi_{a}(x)=x-a$ . This explains the origin of the name general draw-down time. In addition, we find that the classical ruin time is recovered if we choose $\xi\equiv0$ in (1), while draw down provides greater reserve-level-related information compared to ruin. In this sense, draw down shall be an efficient tool of characterizing extreme risks from the point of view of risk management.

The optimal tax return function is defined by

(2) \begin{equation} f(x)=\sup_{\gamma\in\Gamma}f_{\gamma}(x)= \sup\limits_{\gamma\in\Gamma}\mathbb{E}_{x}\bigg( \int_{0}^{\tau^{\gamma}_{\xi}}\mathrm{e}^{-qt}\gamma (\bar{X}(t)) {{\rm d}}\ \bar{X}(t)\bigg), \qquad x\in[0,\infty). \label{eqn2} \end{equation}

The objectives of this paper are to find the optimal tax return function and the optimal tax strategy $\gamma^{\ast}$ that satisfies $f(x)=f_{\gamma^{\ast}}(x) $ for all $x\in[0,\infty) $ .

Now we present some useful concepts associated with X. The Laplace exponent of X is defined by

\begin{equation} \psi(\theta)=\mbox{log}\mathbb{E}_{x}(\mathrm{e}^{\theta (X(1)-x)}), \end{equation}

which is known to be finite for at least $\theta\in[0,\infty) $ in which case it is strictly convex and infinitely differentiable. As in [Reference Bertoin14], the q-scale functions $\{\smash{W^{(q)}};\ q\geq0\}$ of X are defined as follows. For each $q\geq0$ , $\smash{W^{(q)}}\colon [0,\infty)\rightarrow[0,\infty) $ is the unique strictly increasing and continuous function with Laplace transform

\begin{equation} \int_{0}^{\infty}\mathrm{e}^{-\theta x}W^{(q)}(x){{\rm d}}x =\frac{1}{\psi(\theta)-q},\qquad\theta>\Phi(q), \end{equation}

where $\Phi(q) $ is the right inverse of $\psi$ , i.e. the largest root of the equation (in $\theta$ ) $\psi(\theta)=q$ . For simplicity, we write W(x) for $\smash{W^{(0)}(x)}$ . In addition, it follows from [Reference Zhou40] that

(3) \begin{equation} \lim_{x\rightarrow\infty}\frac{W^{(q)\prime}(x)}{W^{(q)}(x)}= \Phi(q)\lt 0 \quad\mbox{for }q \gt 0. \label{eqn3} \end{equation}

For any $x\in\mathbb{R}$ and $\vartheta\geq0$ , there exists a well-known exponential change of measure that one may perform for spectrally negative Lévy processes:

(4) \begin{equation} \frac{\mathbb{P}_{x}^{\vartheta}}{\mathbb{P}_{x}}\bigg|_{\mathcal{F}_{t}} =\mathrm{e}^{\vartheta(X(t)-x)-\psi(\vartheta)t}. \label{eqn4} \end{equation}

Furthermore, under the probability measures $\smash{\mathbb{P}_{x}^{\vartheta}}$ , X remains within the class of spectrally negative Lévy processes. Here and after, we shall refer to the functions $\smash{W_{\vartheta}^{(q)}}$ and $W_{\vartheta}$ as the functions that play the role of the q-scale functions and 0-scale function considered under the measure $\smash{\mathbb{P}_{x}^{\vartheta}}$ .

We also briefly recall concepts in excursion theory for the reflected Lévy process $\{\bar{X}(t)-X(t);\ t\geq0\}$ , and we refer the reader to [Reference Bertoin14] for more details. For $x\in[0,\infty) $ , the process $\{L(t)\ :\!= \bar{X}(t)-x,\, t\geq0\}$ serves as a local time at 0 for the Markov process $\{\bar{X}(t)-X(t);\ t\geq0\}$ under $\mathbb{P}_{x}$ . Let the corresponding inverse local time be defined as

$$L^{ - 1} (t): = \inf \{ s \ge 0|L(s) \gt t\} = \sup \{ s \ge 0|L(s) \le t\}.$$

Furthermore, $L^{-1}(t-)=\lim_{s\uparrow t}L^{-1}(s) $ . The Poisson point process of excursions indexed by this local time is denoted by $\{(t, \varepsilon_{t});\ t\geq0\}$ ,

$$\varepsilon _t (s): = X(L^{ - 1} (t)) - X(L^{ - 1} (t - ) + s),s \in (0,L^{ - 1} (t) - L^{ - 1} (t - 1)],$$

whenever $L^{-1}(t)-L^{-1}(t-)>0$ . For the case when $L^{-1}(t)-L^{-1}(t-)=0 $ , define $\varepsilon_{t}=\Upsilon$ with $\Upsilon$ being an additional isolated point. Accordingly, we denote a generic excursion as $\varepsilon(\cdot) $ (or $\varepsilon$ for short) belonging to the space $\mathcal{E}$ of canonical excursions. The intensity measure of the process $\{(t, \varepsilon_{t});\ t\geq0\}$ is given by $\mathrm{d}t\times \mathrm{d}n,$ where n is a measure on the space of excursions. In particular, $\bar{\varepsilon}=\sup_{s\geq0} \varepsilon(s) $ serves as an example of the n-measurable functional of the canonical excursion. Recalling the definition of $L^{-1}(t) $ , we can verify that

\begin{equation} L^{-1}(t-)=\lim_{u\uparrow t}L^{-1}(u)=\inf\{s\geq0\mid L(s)\geq t\} =\sup\{s\geq0\mid L(s){\lt}t\} \end{equation}

and

(5) \begin{equation} s{\lt}L^{-1}(t-)\quad\Longleftrightarrow\quad L(s){\lt}t. \label{eqn5} \end{equation}

Throughout this paper, to avoid the trivial case that X drifts to $-\infty$ (draw down is certain), we assume that $\psi^{\prime}(0+)\geq0$ . It is also assumed that each scale function has continuous derivative of second order.

3. Two-sided exit problem and expected accumulated discounted tax

In this section, solutions of the two-sided exit problem and expected accumulated discounted tax payments are given, in preparation for motivating the HJB equation (see Proposition 3) and the verification proposition (see Proposition 4) in Section 4. The solutions are explicitly expressed by the scale functions associated with the driving spectrally negative Lévy process.

Define the first up-crossing time of level b as

(6) \begin{equation} \tau^{+}_{b}=\inf\{t\geq0;\,U^{\gamma}(t)>b\}, \label{eqn6} \end{equation}

with the convention that $\inf\emptyset=\infty$ . For $\gamma\in\Gamma$ , define

\begin{equation} \bar{\gamma}_{x}(z)\ :\!=x+\int_{x}^{z}(1-\gamma(y)){{\rm d}}y, \qquad z\in[x,\infty). \end{equation}

By Lemma 2.1 of [Reference Kyprianou and Zhou20], we know that the random times $\{t\geq0:\ U^{\gamma}(t) = \bar{U}^{\gamma}(t)\}$ agree precisely with $\{t\geq0\colon X(t) = \bar{X}(t)\}$ . Hence, we have $\smash{X(\tau^{+}_{x+h})=\bar{X}(\tau^{+}_{x+h})}$ , which together with the definitions of $\bar{\gamma}_{x}(z) $ and $U^{\gamma}$ yields, for $h\geq0$ ,

(7) \begin{equation} x+h=U^{\gamma}(\tau^{+}_{x+h})= \bar{X}(\tau^{+}_{x+h})- \int_{0}^{\tau^{+}_{x+h}}\gamma(\bar{X}(s)) {{\rm d}}\ \bar{X}(s) =\bar{\gamma}_{x}(\bar{X} (\tau^{+}_{x+h})), \qquad x\in[0,\infty). \label{eqn7} \end{equation}

Furthermore, using the fact that $\bar{\gamma}_{x}$ is strictly increasing and continuous, we obtain

\begin{equation} \bar{X}(\tau^{+}_{x+h})=\bar{\gamma}_{x}^{-1}(x+h), \qquad x\in[0,\infty), \end{equation}

where $\smash{\bar{\gamma}_{x}^{-1}}$ denotes the well-defined inverse function of $\bar{\gamma}_{x}$ .

The following result gives the solution to the two-sided exit problem in terms of scale functions via an excursion argument. The excursion theory has also been used in some recent works; see [Reference Avram, Vu and Zhou12], [Reference Kyprianou and Zhou20], and [Reference Li, Vu and Zhou24].

Proposition 1. For $a\in(0,\infty) $ , tax strategy $\gamma,$ and draw-down function $\xi,$ we have

(8) \begin{equation} \mathbb{E}_{x}(\mathrm{e}^{-q \tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+} \lt \tau^{\gamma}_{\xi}\}}) =\exp\bigg(\!-\int_{x}^{a}\frac{W^{(q)\prime}(\bar{\xi}(y))} {W^{(q)}(\bar{\xi}(y))}\frac{1}{1-\gamma (\bar{\gamma}_{x}^{-1}(y))} {{\rm d}}y\bigg), \qquad x\in[0,a], \label{eqn8} \end{equation}

with $\bar{\xi}(y)\ :\!=y-\xi(y) $ .

Proof. On $\{\tau_{a}^{+}\lt \tau^{\gamma}_{\xi}\}$ , we have

\begin{equation} U^{\gamma}(t)\geq \xi(\bar{U}^{\gamma}(t))= \xi\Big(\sup_{0\leq s\leq t}U^{\gamma}(s)\Big)\quad\text{for all } t\in[0,\tau_{a}^{+}]. \end{equation}

Hence, we can rewrite the set $\smash{\{\tau_{a}^{+}\lt \tau^{\gamma}_{\xi}\}}$ as

(9) \begin{equation} \{\tau_{a}^{+}\lt \tau^{\gamma}_{\xi}\}=\{\bar{\varepsilon}_{t}\leq \bar{\xi}(\bar{\gamma}_{x}(x+t)) \text{ for all } t\in[0,\bar{\gamma}_{x}^{-1}(a)-x]\}. \label{eqn9} \end{equation}

Here we have used the fact that, for any $ s\in[0,L^{-1}(t)-L^{-1}(t-)]$ ,

\begin{align*} U^{\gamma}(L^{-1}(t-)+s)&= X(L^{-1}(t-)+s)-\int_{0}^{L^{-1}(t-)+s}\gamma(\bar{X}(u) ){{\rm d}}\ \bar{X}(u) \\ & = X(L^{-1}(t-)+s)-\int_{0}^{L^{-1}(t-)}\gamma(\bar{X}(u) ){{\rm d}}\ \bar{X}(u) , \end{align*}

which implies that $U^{\gamma}(L^{-1}(t-)+s)\geq \xi(\sup_{u\leq L^{-1}(t-)+s}U^{\gamma}(u))=\xi(U^{\gamma}(L^{-1}(t))) $ and is equivalent to

\begin{align*}\\[-24pt] \varepsilon_{t}(s)&= X(L^{-1}(t))-X(L^{-1}(t-)+s) \\ &=U^{\gamma}(L^{-1}(t))-U^{\gamma}(L^{-1}(t-)+s) \\ &\leq U^{\gamma}(L^{-1}(t))-\xi(U^{\gamma}(L^{-1}(t))) \\ &=\bar{\xi}(\bar{\gamma}_{x}(x+t)), \qquad s\in[0,L^{-1}(t)-L^{-1}(t-)]. \end{align*}

By (9) we have

(10) \begin{align} \mathbb{P}_{x}(\{\tau_{a}^{+}\lt \tau^{\gamma}_{\xi}\}) &= \mathbb{P}_{x}(\{\bar{\varepsilon}_{t}\leq \bar{\xi}(\bar{\gamma}_{x}(x+t)), \text{ for all } t\in[0,\bar{\gamma}_{x}^{-1}(a)-x]\})\nonumber \\ &= \exp\bigg(\!-\int_{0}^{\bar{\gamma}_{x}^{-1}(a)-x} n(\bar{\varepsilon}>\bar{\xi}(\bar{\gamma}_{x}(x+t) )){{\rm d}}t\bigg) \nonumber \\ &= \exp\bigg(\!-\int_{x}^{a}\frac{W^{\prime}(\bar{\xi}(y))} {W(\bar{\xi}(y))} \frac{1}{1-\gamma(\bar{\gamma}_{x}^{-1}(y))} {{\rm d}}y\bigg), \label{eqn10}\end{align}

where in the last equality we have changed variables via $y=\bar{\gamma}_{x}(x+t) $ and used $n(\bar{\varepsilon}>x)={W^{\prime}(x)}/{W(x)},$ providing that x is not a point of discontinuity in the derivative of W (see [Reference Kyprianou19]).

By (4) we have

(11) \begin{align} \mathbb{P}_{x}^{\Phi(q)}(\tau_{a}^{+}\lt \tau^{\gamma}_{\xi}) &= \mathbb{E}_{x}(\mathrm{e}^{\Phi(q)(X(\tau_{a}^{+})-x)-q\tau_{a}^{+}} \mathbf{1}_{\{\tau_{a}^{+}\lt \tau^{\gamma}_{\xi}\}}) \nonumber \\ &= \mathrm{e}^{\Phi(q)(\bar{\gamma}_{x}^{-1}(a)-x)} \mathbb{E}_{x}(\mathrm{e}^{-q\tau_{a}^{+}}\mathbf{1}_{\{\tau_{a}^{+} \lt \tau^{\gamma}_{\xi}\}}), \label{eqn11}\end{align}

while, on the other hand, we also have

(12) \begin{align} \mathbb{P}_{x}^{\Phi(q)}(\tau_{a}^{+}\lt \tau^{\gamma}_{\xi}) & = \exp\bigg(\!-\int_{x}^{a}\frac{W_{\Phi(q)}^{\prime}(\bar{\xi} (y))} {W_{\Phi(q)}(\bar{\xi}(y))} \frac{1}{1-\gamma(\bar{\gamma}_{x}^{-1}(y))} {{\rm d}}y\bigg) \nonumber \\ &= \exp\bigg(\!-\int_{x}^{a}\frac{W^{(q)\prime}(\bar{\xi}(y))} {W^{(q)}(\bar{\xi}(y))} \frac{1}{1-\gamma(\bar{\gamma}_{x}^{-1}(y))}{{\rm d}}y +\Phi(q)(\bar{\gamma}_{x}^{-1}(a)-x)\bigg), \label{eqn12}\end{align}

where a similar argument leading to (10) and $W_{\Phi(q)}(x)=\smash{\mathrm{e}^{-\Phi(q)x}W^{(q)}(x)}$ are used. Combining (11) and (12), we arrive at the desired result (8).

The following result solves the expected accumulated discounted tax payments (tax return function) for a given tax strategy $\gamma$ .

Proposition 2. For $a\in(0,\infty) $ , tax strategy $\gamma,$ and draw-down function $\xi,$ we have

(13) \begin{align} &\mathbb{E}_{x}\bigg(\int_{0}^{\tau_{a}^{+} \wedge \tau^{\gamma}_{\xi}}\mathrm{e}^{-q t}\gamma(\bar{X}(t)){{\rm d}} (\bar{X}(t)-x)\bigg) \nonumber \\ &\qquad= \int_{x}^{\bar{\gamma}_{x}^{-1}(a)} \exp\bigg(\!-\int_{x}^{y} \frac{W^{(q)\prime}(\bar{\xi}(\bar{\gamma}_{x}(s) ))}{W^{(q)}(\bar{\xi}(\bar{\gamma}_{x}(s) ))}{{\rm d}}s\bigg) \gamma(y){{\rm d}}y, \qquad x\in[0,a]. \label{eqn13}\end{align}

Proof. See Appendix B.

Remark 1

Letting $a\rightarrow \infty$ in (13) we obtain the solution of the expected total discounted tax payments paid until the general draw-down time

\begin{align} &\mathbb{E}_{x}\bigg(\int_{0}^{\tau^{\gamma}_{\xi}}\mathrm{e}^{-q t}\gamma(\bar{X}(t)){{\rm d}}(\bar{X}(t) -x)\bigg)\nonumber \\ &\qquad= \int_{x}^{\infty} \exp\bigg(\!-\int_{x}^{y} \frac{W^{(q)\prime}(\bar{\xi}(\bar{\gamma}_{x}(s)))} {W^{(q)}(\bar{\xi} (\bar{\gamma}_{x}(s) ))}{{\rm d}}s\bigg) \gamma(y){{\rm d}}y,\qquad x\geq0. \end{align}

Note that, since ${W^{(q)\prime}(x)}/{W^{(q)}(x)}\geq \Phi(q)>0$ (see (29)), we can obtain an upper bound for $f_{\gamma}(x) $ as which yields

\begin{equation} f(x)=\sup_{\gamma\in\Gamma}f_{\gamma}(x)\leq\frac{\gamma_{2}}{\Phi(q)}, \qquad x\in[0,\infty). \end{equation}

4. HJB equation and verification arguments

In this section, the HJB equation and the verification proposition are presented. It turns out that the optimal tax return function (i.e. the expected accumulated discounted tax payments corresponding to the optimal tax strategy) satisfies the HJB equation. Since a solution to the HJB equation is not necessarily the optimal tax return function, then a verification proposition is in need to guarantee that the solution to the HJB equation is indeed the optimal tax return function.

The following result gives the HJB equation satisfied by the optimal tax return function defined by (2).

Proposition 3. Assume that f(x) defined by (2) is once continuously differentiable over $[0,\infty) $ . Then f(x) satisfies the following HJB equation

(14) \begin{equation} \sup_{\gamma\in[\gamma_{1},\gamma_{2}]}\bigg(\frac{\gamma}{1-\gamma}- \frac{1}{1-\gamma} \frac{W^{(q)\prime}(\bar{\xi} (x))}{W^{(q)} (\bar{\xi}(x))}\ f(x) +f^{\prime}(x)\bigg)=0, \qquad x\in[0,\infty). \label{eqn14} \end{equation}

Proof. See Appendix C.

In the following result, we argue that a solution to the HJB equation (14) serves as the optimal tax return function f(x) defined in (2).

Proposition 4. (Verification proposition.) Assume that f(x) is a once continuously differentiable solution to (14), and let $\gamma^{*}$ be the function that maximizes the left-hand side of (14), i.e.

(15) \begin{equation} \gamma^{*}(z)=\mathop{\arg\max}_{\gamma\in[\gamma_{1},\gamma_{2}]}\bigg[ \frac{\gamma}{1-\gamma}-\frac{1}{1-\gamma} \frac{W^{(q)\prime} (\bar{\xi}(\bar{\gamma}_{x}(z)))} {W^{(q)}(\bar{\xi}(\bar{\gamma}_{x}(z)))}\ f(\bar{\gamma}_{x}(z))+f'(\bar{\gamma}_{x}(z))\bigg] \label{eqn15} \end{equation}

for $z\in [x,\infty) $ . Then $\gamma^{*}$ serves as an optimal tax strategy.

Proof. Since $\gamma(z)\in[\gamma_{1},\gamma_{2}]$ for all $z\in[x,\infty) $ and $\gamma\in\Gamma$ , it can be checked from (14) that, for any $\gamma\in\Gamma$ ,

\begin{equation} \gamma(z)\leq -\bigg((1-\gamma(z))\ f'(\bar{\gamma}_{x}(z)) -\frac{W^{(q)\prime}(\bar{\xi}(\bar{\gamma}_{x}(z) ))} {W^{(q)}(\bar{\xi} (\bar{\gamma}_{x}(z)))}\ f(\bar{\gamma}_{x}(z))\bigg),\qquad z\in [x,\infty), \end{equation}

and the equality holds when $\gamma\equiv\gamma^{*}$ . Replacing z by $\bar{X}(t) $ in the above inequality we should have, for all $s\geq0,$

(16) \begin{equation} \gamma(\bar{X}(s))\leq \frac{W^{(q)\prime}(\bar{\xi}(\bar{\gamma}_{x} (\bar{X}(s))))} {W^{(q)}(\bar{\xi}(\bar{\gamma}_{x} (\bar{X}(s))))}\ f (\bar{\gamma}_{x}(\bar{X}(s))) -(1-\gamma(\bar{X}(s)))\ f'(\bar{\gamma}_{x}(\bar{X}(s))) , \label{eqn16} \end{equation}

and the equality holds when $\gamma\equiv\gamma^{*}$ .

Define $\{\eta(s);\ s\geq0\}$ as

(17) \begin{align} \eta(s)&= \lim_{h\downarrow0}\frac{1}{h}\,\mathbb{E}_{x}\Big(\Big( \mathrm{e}^{-q\tau_{\bar{\gamma}_{x}(\bar{X}(s)) +(1-\gamma(\bar{X}(s)))h}^{+}}\mathbf{1}_{\{ \tau_{\bar{\gamma}_{x}(\bar{X}(s))+(1- \gamma(\bar{X}(s)))h}^{+}\lt \tau^{\gamma}_{\xi}\}}\ f(\bar{\gamma}_{x}(\bar{X}(s))\nonumber \\ & +(1\!-\!\gamma(\bar{X}(s)))h) -\!\mathrm{e}^{-q\tau_{_{\bar{\gamma}_{x} (\bar{X}(s))}}^{+}} \mathbf{1}_{\{\tau_{_{\bar{\gamma}_{x}(\bar{X}(s) )}}^{+} \lt \tau^{\gamma}_{\xi}\}}\ f(\bar{\gamma}_{x}(\bar{X}(s))) \!\Big)\Bigm|\mathcal{F}_{\tau^{+}_{\bar{\gamma}_{x} (\bar{X}(s))}}\Big). \label{eqn17}\end{align}

Then, by mimicking the arguments in the proof of the martingale property of the process (4.3) in [Reference Gerber and Shiu17] or the process (3.5) in [Reference Wang and Hu32], we can verify that the compensated process

\begin{equation} Z(t)=\mathrm{e}^{-q\tau_{_{\bar{\gamma}_{x}(\bar{X}(t) )}}^{+}} \mathbf{1}_{\{\tau_{_{\bar{\gamma}_{x}(\bar{X}(t) )}}^{+} \lt \tau^{\gamma}_{\xi}\}} f(\bar{\gamma}_{x}(\bar{X}(t))) -\int_{0}^{\tau_{_{\bar{\gamma}_{x}(\bar{X}(t) )}}^{+}}\eta(s){{\rm d}}\ \bar{X}(s),\qquad t\geq0, \end{equation}

is a martingale with respect to the filtration $\smash{\{\mathcal{F}_{\tau_{\bar{\gamma}_{x}(\bar{X}(t))}^{+}}};\ t\geq0\}$ . In addition, the right-hand side of (17) can be rewritten as

(18) \begin{align} \eta(s) &=\lim_{h\downarrow0}\frac{1}{h}\Big(\mathbb{E}_{x}\Big( \mathrm{e}^{-q\tau_{_{\bar{\gamma}_{x}(\bar{X}(s)) +(1-\gamma(\bar{X}(s)))h}}^{+}}\mathbf{1}_{\{ \tau_{_{\bar{\gamma}_{x}(\bar{X}(s)) +(1-\gamma(\bar{X}(s)))h}}^{+}\lt \tau^{\gamma}_{\xi}\}} \nonumber \\ & \times f(\bar{\gamma}_{x}(\bar{X}(s)) +(1-\gamma(\bar{X}(s)))h) \bigm|\mathcal{F}_{\tau_{\bar{\gamma}_{x}(\bar{X}(s) )}^{+}}\Big) \nonumber \\ & -\mathrm{e}^{-q\tau_{\bar{\gamma}_{x}(\bar{X}(s) )}^{+}}\mathbf{1}_{\{\tau_{\bar{\gamma}_{x}(\bar{X}(s))}^{+} \lt \tau^{\gamma}_{\xi}\}} f(\bar{\gamma}_{x}(\bar{X}(s)))\Big) \nonumber \\ &=\lim_{h\downarrow0}\frac{1}{h}\Big(\mathbb{E}_{x} \Big(\mathrm{e}^{-q\tau_{_{\bar{\gamma}_{x} (\bar{X}(s))+(1-\gamma(\bar{X}(s) ))h}}^{+}}\mathbf{1}_{\{ \tau_{_{\bar{\gamma}_{x}(\bar{X}(s))+(1 -\gamma(\bar{X}(s)))h}}^{+}\lt \tau^{\gamma}_{\xi}\}} \Bigm|\mathcal{F}_{\tau_{\bar{\gamma}_{x}(\bar{X}(s) )}^{+}}\Big) \nonumber \\ & \times (f(\bar{\gamma}_{x}(\bar{X}(s)))+ f'(\bar{\gamma}_{x} (\bar{X}(s)))(1- \gamma(\bar{X}(s)))h+o(h)) \nonumber \\ & -\mathrm{e}^{-q\tau_{\bar{\gamma}_{x}(\bar{X}(s))}^{+}} \mathbf{1}_{\{\tau_{\bar{\gamma}_{x}(\bar{X}(s))}^{+} \lt \tau^{\gamma}_{\xi}\}} f(\bar{\gamma}_{x}(\bar{X}(s))) \Big) \nonumber \\ &=\lim_{h\downarrow0} \frac{1}{h}\bigg(\bigg(1-\frac{(W^{(q)})'(\bar{\xi} (\bar{\gamma}_{x}(\bar{X}(s))))} {W^{(q)}(\bar{\xi}(\bar{\gamma}_{x} (\bar{X}(s))))}h+o(h)\bigg)\nonumber \\ & \times(f(\bar{\gamma}_{x}(\bar{X}(s)))+ f'(\bar{\gamma}_{x}(\bar{X}(s)))(1-\gamma(\bar{X}(s) ))h+o(h)) \nonumber \\ & -f(\bar{\gamma}_{x}(\bar{X}(s)))\bigg) \mathrm{e}^{-q\tau_{\bar{\gamma}_{x} (\bar{X}(s))}^{+}}\mathbf{1}_{\{\tau_{\bar{\gamma}_{x} (\bar{X}(s))}^{+} \lt \tau^{\gamma}_{\xi}\}} \nonumber \\ &= \bigg( -\frac{W^{(q)'}(\bar{\xi}(\bar{\gamma}_{x} (\bar{X}(s))))} {W^{(q)}(\bar{\xi}(\bar{\gamma}_{x} (\bar{X}(s))))} f(\bar{\gamma}_{x}(\bar{X}(s))) \nonumber \\ & +(1-\gamma(\bar{X}(s))) f'(\bar{\gamma}_{x}(\bar{X}(s)))\bigg) \mathrm{e}^{-q\tau_{\bar{\gamma}_{x}(\bar{X}(s) )}^{+}}\mathbf{1}_{\{\tau_{\bar{\gamma}_{x}(\bar{X}(s) )}^{+} \lt \tau^{\gamma}_{\xi}\}}, \label{eqn18}\end{align}

where in the third equality of (18) we have used Lemma 1 (see Appendix A) and the fact that

$$z - \xi _1 (z) = \bar y _x (z) - \xi (\bar y _x (z)) = \bar \xi (\bar y _x (z)) $$

Combining (16) and (18) yields

(19) \begin{equation} \gamma(\bar{X}(s))\mathrm{e}^{-q\tau_{\bar {\gamma}_{x}(\bar{X}(s))}^{+}} \mathbf{1}_{\{\tau_{\bar{\gamma}_{x}(\bar{X}(s))}^{+} \lt \tau^{\gamma}_{\xi}\}}\leq-\eta(s), \label{eqn19} \end{equation}

where the equality holds when $\gamma=\gamma^{*}$ . It is also seen from (19) that $\{-\eta(s);\ s\geq0\}$ is a nonnegative-valued process. Noting that $\{Z(t);\ t\geq0\}$ is a martingale with respect to $\smash{\{\mathcal{F}_{\tau_{\bar{\gamma}_{x}(\bar{X}(t))}^{+}}} ;\ t\geq0\}$ , we have

(20) \begin{equation} \mathbb{E}_{x}\bigg[\mathrm{e}^{-q\tau_{_{\bar{\gamma}_{x} (\bar{X}(t))}}^{+}} \mathbf{1}_{\{\tau_{_{\bar{\gamma}_{x} (\bar{X}(t) )}}^{+} \lt \tau^{\gamma}_{\xi}\}} f(\bar{\gamma}_{x}(\bar{X}(t))) -\int_{0}^{\tau_{\bar{\gamma}_{x}(\bar{X}(t) )}^{+}}\eta(s){{\rm d}}\ \bar{X}(s)\bigg]=f(x). \label{eqn20} \end{equation}

Now note that $\limsup_{t\rightarrow\infty}X(t)=\infty$ almost surely (a.s.) implies that $\lim_{t\rightarrow\infty}\bar{X}(t)=\infty$ a.s., which further implies that $\lim_{t\rightarrow\infty}\tau_{\bar{\gamma}_{x} (\bar{X}(t))}^{+}=\infty$ a.s. Combining the facts that f(x) is bounded over $[0,\infty) $ (see Remark 1) and $\smash{\lim_{t\rightarrow\infty}\tau_{\bar{\gamma}_{x} (\bar{X}(t))}^{+}}=\infty$ a.s., we obtain

\begin{equation} \lim_{t\rightarrow\infty}\mathbb{E}_{x}\Big(\mathrm{e}^{-q \tau_{_{\bar{\gamma}_{x}(\bar{X}(t))}}^{+}} \mathbf{1}_{\{\tau_{_{\bar{\gamma}_{x}(\bar{X}(t) )}}^{+} \lt \tau^{\gamma}_{\xi}\}} f(\bar{\gamma}_{x} (\bar{X}(t)))\Big)=0, \end{equation}

due to the bounded convergence theorem. Letting $t\rightarrow\infty$ in (20) and using the monotone convergence theorem, we arrive at

(21) \begin{equation} \mathbb{E}_{x}\bigg[ -\int_{0}^{\infty}\eta(s){{\rm d}}\ \bar{X}(s)\bigg] = f(x). \label{eqn21} \end{equation}

Since (19) becomes equality when $\gamma=\gamma^{*}$ , by (21) we obtain

\begin{equation} \mathbb{E}_{x}\bigg[ \int_{0}^{\infty}\mathrm{e}^{-q\tau_{\bar{\gamma^{*}} (\bar{X}(s))}^{+}} \mathbf{1}_{\{\tau_{\bar{\gamma^{*}} (\bar{X}(s))}^{+} \lt \tau^{\gamma^{*}}_{\xi}\}} \gamma^{*} (\bar{X}(s)){{\rm d}}\ \bar{X}(s)\bigg]= f(x). \end{equation}

Because tax would be paid when and only when $\bar{X}(s) $ genuinely increases (intuitively, when ‘ $\mathrm{d}\bar{X}(s)>0$ ’ or $\smash{s\in\{\tau_{\bar{\gamma^{*}}(\bar{X}(s))}^{+};\ s\geq0\}}$ ), we have

\begin{align} &\mathbb{E}_{x}\bigg[ \int_{0}^{\infty}\mathrm{e}^{-qs} \mathbf{1}_{\{s \lt \tau^{\gamma^{*}}_{\xi}\}} \gamma^{*} (\bar{X}(s)){{\rm d}}\ \bar{X}(s)\bigg] \\ &\qquad= \mathbb{E}_{x}\bigg[ \int_{0}^{\infty}\mathrm{e}^{-q\tau_{\bar{\gamma^{*}} (\bar{X}(s))}^{+}} \mathbf{1}_{\{\tau_{\bar{\gamma^{*}}(\bar{X}(s))}^{+} \lt \tau^{\gamma^{*}}_{\xi}\}} \gamma^{*} (\bar{X}(s)){{\rm d}}\ \bar{X}(s)\bigg] \\ &\qquad=f(x), \end{align}

that is, $ f_{\gamma^{*}}(x)= f(x) $ .

While, for arbitrary $\gamma\in\Gamma$ , by (19) and similar arguments, we can obtain

\begin{equation} \mathbb{E}_{x}\bigg[ \int_{0}^{\infty}\mathrm{e}^{-qs} \mathbf{1}_{\{s \lt \tau^{\gamma}_{\xi}\}} \gamma (\bar{X}(s)){{\rm d}}\ \bar{X}(s)\bigg]\leq f(x),\nonumber \end{equation}

that is, $ f_{\gamma}(x)\leq f(x) $ for all $\gamma\in\Gamma$ .

5. Characterizing the optimal return function and strategy

As usual in stochastic control problems, we guess the qualitative nature of the optimal return function and make some assumptions. Based on those assumptions we find the solution to the HJB equation and later we need to verify that the obtained solution satisfies the assumptions. To be more specific, we proceed as follows.

• Guess in advance that ‘ ${W^{(q)\prime}(\bar{\xi}(x))}f(x)/{W^{(q)} (\bar{\xi}(x))}\leq1$ for all x’, which is equivalent to (32), to arrive at a candidate solution (28) to (14). Then we should prove that (28) satisfies ‘ ${W^{(q)\prime}(\bar{\xi}(x))}f(x)/{W^{(q)}(\bar{\xi} (x))}\leq1$ for all x’, to guarantee that this candidate solution (28) is indeed a solution to (14). See Proposition 5 and its proof.

• Guess in advance that ‘ ${W^{(q)\prime}(\bar{\xi}(x))}f(x)/{W^{(q)} (\bar{\xi}(x))}\geq1$ for all x’, which is equivalent to (35), to arrive at a candidate solution (34) to (14). Then we should prove that (34) satisfies ‘ ${W^{(q)\prime}(\bar{\xi}(x))}f(x)/{W^{(q)} (\bar{\xi}(x))}\geq1$ for all x’, to guarantee that this candidate solution (34) is indeed a solution to (14). See Proposition 6 and its proof.

• Guess in advance that ‘ ${W^{(q)\prime}(\bar{\xi}(x))}f(x)/{W^{(q)} (\bar{\xi}(x))}\geq1$ for all $x\in(0,x_{1}) $ and ${W^{(q)\prime}(\bar{\xi}(x))}f(x)/ {W^{(q)}(\bar{\xi}(x))}\leq1$ for all $x\in[x_{1},\infty) $ ’, to arrive at a candidate solution (38) to (14). Then we should prove that (38) satisfies ‘ ${W^{(q)\prime}(\bar{\xi}(x))}f(x)/{W^{(q)} (\bar{\xi}(x))}\geq1$ for all $x\in(0,x_{1}) $ and ${W^{(q)\prime}(\bar{\xi}(x))}f(x)/{W^{(q)} (\bar{\xi}(x))}\leq1$ for all $x\in[x_{1},\infty) $ ’, to guarantee that this candidate solution (38) is indeed a solution to (14). See case (v) of Proposition 7 and its proof. Or, vice versa for case (vi) of Proposition 7.

It is already argued in Proposition 4 that the solution to (14) serves as the optimal tax return function of the taxation optimization problem (2), and the optimal tax strategy is then obtained using such a solution through (15). In order to solve (14), one may eliminate the operator ‘ $\sup_{\gamma\in[\gamma_{1},\gamma_{2}]}$ ’ on the right-hand side of (14) beforehand. To this end, we first guess that the solution f(x) of (14) satisfies ${W^{(q)\prime}(\bar{\xi}(x))}f(x)/{W^{(q)} (\bar{\xi}(x))}\leq1$ for all x, in which case

$${y \over {1 - y}} - {1 \over {1 - y}}{{W^{(q)prime} (\bar \xi (x))} \over {W^{(q)} (\bar \xi (x))}}f(x) $$

is nondecreasing with respect to $\gamma$ for all given x. Then, using this fact, we may rewrite (14) as

(23) \begin{equation} 0=\gamma_{2}-\frac{W^{(q)\prime}(\bar{\xi}(x))}{W^{(q)} (\bar{\xi}(x))}f(x)+(1-\gamma_{2})f^{\prime}(x) \quad\mbox{for all }x\geq0. \label{eqn22} \end{equation}

Instead of solving (23), we first turn to the homogeneous differential equation of (23), which can be rewritten as

\begin{equation} \frac{\mathrm{d}}{\mathrm{d}x}\ln f(x)=\frac{f^{\prime}(x)}{f(x)} =\frac{1}{1-\gamma_{2}}\frac{W^{(q)\prime}(\bar{\xi}(x))} {W^{(q)}(\bar{\xi}(x))},\qquad x\geq0. \end{equation}

The solution of the above homogeneous differential equation is given by

\begin{equation} f(x)=C\exp\bigg(\frac{1}{1-\gamma_{2}}\int_{0}^{x}\frac{W^{(q)\prime} (\bar{\xi}(y))}{W^{(q)}(\bar{\xi}(y))} {{\rm d}}y\bigg),\qquad x\geq0, \end{equation}

with C being some constant. By the standard method of the variation of constants, we guess that the solution to (23) is

(24) \begin{equation} f(x)=C(x)\exp\bigg(\frac{1}{1-\gamma_{2}}\int_{0}^{x}\frac{W^{(q)\prime} (\bar{\xi}(y))}{W^{(q)}(\bar{\xi}(y))} {{\rm d}}y\bigg),\qquad x\geq0, \label{eqn23} \end{equation}

with C(x) being some unknown function to be determined later. Substituting (24) into (23) we obtain

\begin{align} 0&= \gamma_{2}-\frac{W^{(q)\prime}(\bar{\xi}(x))} {W^{(q)}(\bar{\xi}(x))}\,C(x)\exp\bigg(\frac{1} {1-\gamma_{2}} \int_{0}^{x}\frac{W^{(q)\prime}(\bar{\xi}(y))}{W^{(q)} (\bar{\xi}(y))}{{\rm d}}y\bigg) \\ & +(1-\gamma_{2}) \exp\bigg(\frac{1}{1-\gamma_{2}}\int_{0}^{x}\frac{W^{(q)\prime} (\bar{\xi}(y))} {W^{(q)}(\bar{\xi}(y))}{{\rm d}}y\bigg) \bigg(C^{\prime}(x)+\frac{C(x)}{1-\gamma_{2}}\frac{W^{(q)\prime} (\bar{\xi}(x))}{W^{(q)}(\bar{\xi}(x))} \bigg), \end{align}

which can be simplified as

\begin{equation} C^{\prime}(x)= -\frac{\gamma_{2}}{1-\gamma_{2}} \exp\bigg(\!-\frac{1}{1-\gamma_{2}}\int_{0}^{x}\frac{W^{(q)\prime} (\bar{\xi}(y))} {W^{(q)}(\bar{\xi}(y))} {{\rm d}}y\bigg). \end{equation}

The solution of the above differential equation is given by

(25) \begin{equation} C(x)= C_{0}-\frac{\gamma_{2}}{1-\gamma_{2}}\int_{0}^{x} \exp\bigg(\!-\frac{1}{1-\gamma_{2}}\int_{0}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))} {W^{(q)}(\bar{\xi}(u))}{{\rm d}}u\bigg){{\rm d}}y, \label{eqn24} \end{equation}

where $C_{0}$ is some constant. Combining (24) and (25) we know that the solution of (23) is

(26) \begin{align} f(x)&=\bigg(C_{0}-\frac{\gamma_{2}}{1-\gamma_{2}}\int_{0}^{x} \exp\bigg(\!-\frac{1}{1-\gamma_{2}}\int_{0}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\bigg){{\rm d}}y\bigg) \nonumber \\ & \times\exp\bigg(\frac{1}{1-\gamma_{2}}\int_{0}^{x} \frac{W^{(q)\prime} (\bar{\xi}(y))}{W^{(q)} (\bar{\xi}(y))}{{\rm d}}y\bigg),\qquad x\geq0. \label{eqn25}\end{align}

It remains to determine the unknown constant $C_{0}$ . For this purpose, we first deduce that

(27) \begin{equation} \lim_{x\rightarrow\infty}\exp\bigg(\frac{1}{1-\gamma_{2}}\int_{0}^{x} \frac{W^{(q)\prime}(\bar{\xi}(y))}{W^{(q)} (\bar{\xi}(y))}{{\rm d}}y\bigg)=\infty, \label{eqn26} \end{equation}

which together with the fact that $f(\infty)\lt \infty$ implies that (26) can be rewritten as

(28) \begin{equation} f(x)=\frac{\gamma_{2}}{1-\gamma_{2}}\int_{x}^{\infty} \exp\bigg(\!-\frac{1}{1-\gamma_{2}}\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))} {W^{(q)}(\bar{\xi}(u))}{{\rm d}}u\bigg){{\rm d}}y, \qquad x\geq0. \label{eqn27} \end{equation}

We prove (27) in the sequel. Note that the scale function $\smash{W^{(q)}(x)}$ is continuously differentiable, positive valued, and strictly increasing over $ (0,\infty) $ . By the equality $\smash{W^{(q)}(x)}=\smash{\mathrm{e}^{\Phi(q)x}W_{\Phi(q)}(x)}$ we can conclude that the function $W^{\prime}(x) $ must also be a continuous function over $ (0,\infty) $ . Furthermore, we have

(29) \begin{align} \frac{W^{(q)\prime}(x)}{W^{(q)}(x)}&= \frac{\Phi(q)\mathrm{e}^{\Phi(q)x}W_{\Phi(q)}(x)+\mathrm{e}^{\Phi(q)x} W_{\Phi(q)}^{\prime}(x)}{\mathrm{e}^{\Phi(q)x}W_{\Phi(q)}(x)}\nonumber \\ &=\Phi(q)+ \frac{W_{\Phi(q)}^{\prime}(x)}{W_{\Phi(q)}(x)}\nonumber \\ &=\Phi(q)+n_{\Phi(q)} (\bar{\varepsilon}>x)\nonumber \\ &\geq\Phi(q)>0 \quad\text{for all }x\in(0,\infty) \text{ and all } q>0. \label{eqn28}\end{align}

Hence, recalling that $\bar{\xi}(y)>0$ for all $y\in[0,\infty) $ , we can find that

\begin{equation} \lim_{x\rightarrow\infty} \exp\bigg(\frac{1}{1-\gamma_{2}}\int_{0}^{x}\frac{W^{(q)\prime} (\bar{\xi}(y))}{W^{(q)}(\bar{\xi}(y))} {{\rm d}}y\bigg) \geq \lim_{x\rightarrow\infty}\exp\bigg(\frac{1}{1-\gamma_{2}}\Phi(q)x\bigg) =\infty, \end{equation}

as is required in (27).

For the candidate optimal solution given by (28), we need to justify that (for all $x>0$ ) to assure its optimality. Using integration by parts, we can rewrite (28) as,

(30) \begin{align} f(x)&= \int_{x}^{\infty}\mathrm{d}\left(\!-\frac{\exp\Big(\!-\frac{\gamma_{2}} {1-\gamma_{2}}\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi} (u))}{W^{(q)}(\bar{\xi}(u))}{{\rm d}}u\Big)} {\exp\Big(\int_{x}^{y}\frac{W^{(q)\prime}(\bar{\xi}(u))} {W^{(q)}(\bar{\xi}(u))}{{\rm d}}u\Big) \frac{W^{(q)\prime}(\bar{\xi}(y))}{W^{(q)} (\bar{\xi}(y))}}\right)\nonumber \\ & -\int_{x}^{\infty}\frac{\Big[\exp\Big(\int_{x}^{y}\frac{W^{(q)\prime}(\bar{\xi} (u))}{W^{(q)}(\bar{\xi}(u))}{{\rm d}}u\Big) \frac{W^{(q)\prime}(\bar{\xi}(y))}{W^{(q)}(\bar{\xi} (y))}\Big]' \exp\Big(\!-\frac{\gamma_{2}}{1-\gamma_{2}} \int_{x}^{y}\frac{W^{(q)\prime}(\bar{\xi}(u))}{W^{(q)} (\bar{\xi}(u))}{{\rm d}}u\Big)} {\Big[\exp\Big(\int_{x}^{y}\frac{W^{(q)\prime}(\bar{\xi} (u))}{W^{(q)}(\bar{\xi}(u))}{{\rm d}}u\Big) \frac{W^{(q)\prime}(\bar{\xi}(y))}{W^{(q)} (\bar{\xi}(y))}\Big]^{2}}\,\,{{\rm d}}y \nonumber \\ &= -\frac{\exp\Big(\!-\frac{\gamma_{2}}{1-\gamma_{2}}\int_{x}^{y} \frac{W^{(q)\prime}(\bar{\xi}(u))}{W^{(q)}(\bar{\xi} (u))}{{\rm d}}u\Big)} {\exp\Big(\int_{x}^{y} \frac{W^{(q)\prime}(\bar{\xi}(u))}{W^{(q)} (\bar{\xi}(u))}{{\rm d}}u\Big)\frac{W^{(q)\prime} (\bar{\xi}(y))}{W^{(q)}(\bar{\xi}(y))}} \Bigg\vert_{x}^{_{\infty}} \nonumber \\ & -\int_{x}^{\infty}\frac{\Big[\exp\Big(\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\Big)\frac{W^{(q)\prime}(\bar{\xi}(y))} {W^{(q)}(\bar{\xi}(y))}\Big]' \exp\Big(\!-\frac{\gamma_{2}}{1-\gamma_{2}}\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\Big)} {\Big[\exp\Big(\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi} (u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\Big)\frac{W^{(q)\prime}(\bar{\xi}(y))} {W^{(q)}(\bar{\xi}(y))}\Big]^{2}}{{\rm d}}y \nonumber \\ &= \frac{W^{(q)}(\bar{\xi}(x))}{W^{(q)\prime} (\bar{\xi}(x))} -G_2(x), \qquad x\geq0, \label{eqn29}\end{align}

where

\begin{equation} G_{2}(x)=\int_{x}^{\infty} \exp\bigg(\!-\frac{1}{1-\gamma_{2}}\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\bigg)g(y){{\rm d}}y,\qquad x\geq 0. \end{equation}

Here the function g is defined as

(31) \begin{equation} g(x)=\xi^{\prime}(x)+(1-\xi^{\prime}(x))\frac{W^{(q)\prime \prime}(\bar{\xi}(x))W^{(q)}(\bar{\xi}(x))} {(W^{(q)\prime}(\bar{\xi}(x)))^{2}},\qquad x\geq 0. \label{eqn30} \end{equation}

With the above alternative representation of f(x), we conclude that (for all $x>0$ ) + is equivalent to the inequality

(32) \begin{equation} G_2(x)\geq0 \quad\text{for all }x\lt0. \label{eqn31} \end{equation}

In order to characterize the optimal tax strategy and the optimal tax return function explicitly, we need the following assumption, under which the optimal tax return function and the optimal tax strategy can be obtained. As can be seen, the extension from the ruin-based tax optimization problem to our general draw-down-based tax optimization problem incurs in greater difficulties and complexity, especially in finding the optimal tax return function and the optimal tax strategy, while characterizing them is very critical in addressing optimal control problems. When our model and tax structure reduce to those in the existing literature, our results coincide with the corresponding results.

Assumption 1. Let g be defined by (31). Assume that there exists an $x_{0}\in[0,\infty) $ such that the function g changes its sign at the point $x_{0}$ . Here, we say that the function g(x) changes its sign at the point $x_{0}$ if and only if $g(x_{1})g(x_{2})\leq0$ for all $x_{1}\leq x_{0},\,x_{2}\geq x_{0}$ .

We proceed to characterize the optimal solution for (14). This time we guess that the solution f(x) of (14) satisfies ${W^{(q)\prime}(\bar{\xi}(x))}f(x)/{W^{(q)}(\bar{\xi}(x))} \geq1$ for all x. Then, for all $x\geq0$ , the function f satisfies

(33) \begin{equation} 0 =\gamma_{1}-\frac{W^{(q)\prime}(\bar{\xi}(x))} {W^{(q)}(\bar{\xi}(x))}f(x)+(1-\gamma_{1})f^{\prime}(x). \label{eqn32} \end{equation}

Solving (33) by similar arguments as in solving (23), we obtain

(34) \begin{align} f(x)&= \frac{\gamma_{1}}{1-\gamma_{1}}\int_{x}^{\infty} \exp\bigg(\!-\frac{1}{1-\gamma_{1}}\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\bigg){{\rm d}}y \nonumber \\ &= \frac{W^{(q)}(\bar{\xi}(x))}{W^{(q)\prime} (\bar{\xi}(x))}-G_1(x), \qquad x\geq0, \label{eqn33}\end{align}

where

\begin{equation} G_{1}(x)=\int_{x}^{\infty}\exp\bigg(\!-\frac{1}{1-\gamma_{1}}\int_{x}^{y} \frac{W^{(q)\prime}(\bar{\xi}(u))}{W^{(q)}(\bar{\xi} (u))}{{\rm d}}u\bigg)g(y){{\rm d}}y. \end{equation}

By (34), it is obvious that is equivalent to

(35) \begin{equation} G_1(x)\leq0 \quad\text{for all } x\lt0. \label{eqn34} \end{equation}

Aside from cases (i)–(iv), we also consider the following two cases.

• (v) $x_{0}\in(0,\infty) $ , $g(x)\leq0$ for $x\in[0,x_{0}]$ , $g(x)\geq0$ for $x\in(x_{0},\infty) $ , and there is an $\bar{x}>0$ such that $G_{2}(\bar{x})\lt0$ .

• (vi) $x_{0}\in(0,\infty) $ , $g(x)\geq0$ for $x\in[0, x_{0}]$ , $g(x)\leq0$ for $x\in(x_{0},\infty) $ , and there is an $\bar{\bar{x}}>0$ such that $G_{1}(\bar{\bar{x}})>0$ .

In case (v), we must have $\bar{x}{\lt}x_{0}$ . By the definition of $x_{0}$ we know that $G_{2}(x_{0})\geq0$ . Then from the intermediate value theorem for the continuous function $G_{2}(x) $ we claim that there must exist some $x\in(\bar{x},x_{0}]$ such that $G_{2}(x)=0$ . Let

(36) \begin{equation} x_{1}=\inf\{x\in(\bar{x},x_{0}]\mid G_{2}(x)=0\}. \label{eqn35} \end{equation}

Then $G_{2}(x_{1})=0$ , $G_{2}(x)\lt0$ for all $0{\lt}x{\lt}x_{1}$ , and $G_{2}(x)\geq0$ for all $x\in[x_{1},\infty) $ . Hence, this time we guess that the solution f(x) of (14) satisfies

$${{{W^{(q)t}}(\bar \xi (x))} \over {{W^{(q)t}}(\bar \xi (x))}}f(x) \ge 1\,{\rm{for}}\,{\rm{all}}\,x\, \in (0,{x_1}),\,{\rm{and}}\,{{{W^{(q)t}}(\bar \xi (x))} \over {{W^{(q)t}}(\bar \xi (x))}}f(x) \le 1\,{\rm{for}}\,{\rm{all}}\,x \in [{x_1},\infty ). $$

That is, f(x) satisfies the system of differential equations

(37) \begin{equation} \begin{gathered} 0=\gamma_{1}-\frac{W^{(q)\prime}(\bar{\xi}(x))} {W^{(q)}(\bar{\xi}(x))}f(x)+(1-\gamma_{1})f^{\prime}(x) \quad\text{for all } x\in(0,x_{1}), \\ 0=\gamma_{2}-\frac{W^{(q)\prime} (\bar{\xi}(x))}{W^{(q)}(\bar{\xi}(x))} f(x)+(1-\gamma_{2})f^{\prime}(x) \quad\text{for all } x\in[x_{1},\infty). \end{gathered} \label{eqn36} \end{equation}

Solving the above system of equations yields

\begin{align} f(x)&= \bigg(C-\frac{\gamma_{1}}{1-\gamma_{1}}\int_{0}^{x} \exp\bigg(\!-\frac{1}{1-\gamma_{1}}\int_{0}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} \mathrm{d}u\bigg){{\rm d}}y\bigg) \nonumber \\ & \times \exp\bigg(\frac{1}{1-\gamma_{1}}\int_{0}^{x}\frac{W^{(q)\prime} (\bar{\xi}(y))}{W^{(q)}(\bar{\xi}(y))} {{\rm d}}y\bigg) \quad\text{for all }x\in(0,x_{1}), \nonumber \\ f(x)&=\frac{\gamma_{2}}{1-\gamma_{2}} \int_{x}^{\infty}\exp\bigg(\!-\frac{1}{1-\gamma_{2}}\int_{x}^{y} \frac{W^{(q)\prime}(\bar{\xi}(u))}{W^{(q)} (\bar{\xi}(u))}{{\rm d}}u\bigg){{\rm d}}y \quad\text{for all } x\in[x_{1},\infty).\nonumber \end{align}

Using the continuity condition $f(x_{1}-)=\ f(x_{1}+)=\ f(x_{1}),$ we can determine the constant C as

\begin{align} C&=\frac{\gamma_{1}}{1-\gamma_{1}} \int_{0}^{x_{1}} \exp\bigg(\!-\frac{1}{1-\gamma_{1}} \int_{0}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))} {W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\bigg){{\rm d}}y \nonumber \\ & +\frac{\gamma_{2}}{1-\gamma_{2}} \exp\bigg(\frac{-1}{1- \gamma_{1}}\int_{0}^{x_{1}}\!\frac{W^{(q)\prime}(\bar{\xi} (y))}{W^{(q)}(\bar{\xi}(y))}{{\rm d}}y\bigg) \int_{x_{1}}^{\infty}\!\! \exp\bigg(\frac{-1}{1-\gamma_{2}}\int_{x_{1}}^{y}\! \frac{W^{(q)\prime}(\bar{\xi}(u))}{W^{(q)}(\bar{\xi} (u))}{{\rm d}}u\bigg){{\rm d}}y.\nonumber \end{align}

Therefore, we can write down the solution to the equations in (37) as

(38) \begin{align} f(x)&= \frac{\gamma_{1}}{1-\gamma_{1}}\int_{x}^{x_{1}} \exp\bigg(\frac{-1}{1-\gamma_{1}}\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}\left(u\right))}{W^{(q)}(\bar{\xi}\left(u\right))} {{\rm d}}u\bigg) {{\rm d}}y \nonumber \\ & +\frac{\gamma_{2}}{1-\gamma_{2}} \exp\bigg(\frac{-1}{1-\gamma_{1}} \int_{x}^{x_{1}}\frac{W^{(q)\prime}(\bar{\xi}(y))} {W^{(q)}(\bar{\xi}(y))}{{\rm d}}y\bigg) \nonumber \\ & \times \int_{x_{1}}^{\infty} \exp\bigg(\frac{-1} {1-\gamma_{2}}\int_{x_{1}}^{y}\frac{W^{(q)\prime}(\bar{\xi} (u))}{W^{(q)}(\bar{\xi}(u))}{{\rm d}}u\bigg) {{\rm d}}y \quad\text{for all }x\in(0,x_{1}), \nonumber \\ f(x)&=\frac{\gamma_{2}}{1-\gamma_{2}} \int_{x}^{\infty}\exp\bigg(\!-\frac{1}{1-\gamma_{2}}\int_{x}^{y} \frac{W^{(q)\prime}(\bar{\xi}(u))}{W^{(q)} (\bar{\xi}(u))}{{\rm d}}u\bigg){{\rm d}}y \quad\text{for all }x\in[x_{1},\infty). \label{eqn37}\end{align}

We need to only further prove that the function given by (38) satisfies

$$\frac{{W^{(q)'} (\bar \xi (x))}} {{W^{(q)} (\bar \xi (x))}}f(x) \ge 1\quad {\text{for}}\,{\text{all}}\,x \in (0,x_1 ) $$

to guarantee itself a solution to the HJB equation (14). By some algebraic manipulations we obtain, for $x\in(0,x_{1}) $ ,

(39) \begin{align} f(x)&=\int_{x}^{x_{1}}\mathrm{d} \left(\!-\frac{\exp\Big( -\frac{\gamma_{1}}{1-\gamma_{1}} \int_{x}^{y}\frac{W^{(q)\prime}(\bar{\xi}(u))} {W^{(q)}(\bar{\xi}(u))}{{\rm d}}u\Big)} {\exp\Big(\int_{x}^{y}\frac{W^{(q)\prime}(\bar{\xi}(u))} {W^{(q)}(\bar{\xi}(u))}{{\rm d}}u\Big) \frac{W^{(q)\prime}(\bar{\xi}(y))}{W^{(q)} (\bar{\xi}(y))}}\right)\nonumber \\ & - \int_{x}^{x_{1}} \frac{\Big[\exp\Big(\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\Big)\frac{W^{(q)\prime}(\bar{\xi}(y))} {W^{(q)}(\bar{\xi}(y))}\Big]' \exp\Big(\!-\frac{\gamma_{1}} {1-\gamma_{1}}\int_{x}^{y} \frac{W^{(q)\prime} (\bar{\xi}(u))} {W^{(q)}(\bar{\xi}(u))}{{\rm d}}u\Big)} {\Big[\exp\Big(\int_{x}^{y}\frac{W^{(q)\prime}(\bar{\xi} (u))}{W^{(q)}(\bar{\xi}(u))}{{\rm d}}u\Big) \frac{W^{(q)\prime}(\bar{\xi}(y))}{W^{(q)} (\bar{\xi}(y))}\Big]^{2}}{{{\rm d}}y} \nonumber \\ & +\exp\bigg(\frac{-1}{1-\gamma_{1}}\int_{x}^{x_{1}} \frac{W^{(q)\prime}(\bar{\xi}(y))}{W^{(q)} (\bar{\xi}(y))}{{\rm d}}y\bigg) \nonumber \\ & \times \left( \int_{x_{1}}^{\infty}\mathrm{d}\left(\!-\frac{\exp\Big( -\frac{\gamma_{2}}{1-\gamma_{2}}\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\Big)} {\exp\Big(\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\Big)\frac{W^{(q)\prime}(\bar{\xi}(y))} {W^{(q)}(\bar{\xi}(y))}}\right)\right.\nonumber \\ &\left.-\int_{x_{1}}^{\infty}\frac{\Big[\exp\Big(\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi} (u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\Big)\frac{W^{(q)\prime}(\bar{\xi}(y))} {W^{(q)}(\bar{\xi}(y))}\Big]' \exp\Big(\!-\frac{\gamma_{2}}{1-\gamma_{2}}\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\Big)} {\Big[\exp\Big(\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\Big)\frac{W^{(q)\prime}(\bar{\xi}(y))} {W^{(q)}(\bar{\xi}(y))}\Big]^{2}}{{\rm d}}y \right)\nonumber \\ &= \frac{W^{(q)}(\bar{\xi}(x))}{W^{(q)\prime} (\bar{\xi}(x))} -\int_{x}^{x_{1}} \exp\bigg(\!-\frac{1}{1-\gamma_{1}} \int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)} (\bar{\xi}(u))} {{\rm d}}u\bigg)g(y){{\rm d}}y \nonumber \\ & -\exp\bigg(\frac{-1}{1-\gamma_{1}}\int_{x}^{x_{1}} \frac{W^{(q)\prime}(\bar{\xi}(y))}{W^{(q)} (\bar{\xi}(y))}{{\rm d}}y\bigg) G_{2}(x_{1}) \nonumber \\ &\geq\frac{W^{(q)}(\bar{\xi}(x))}{W^{(q)\prime} (\bar{\xi}(x))}, \label{eqn38}\end{align}

since $g(y)\leq0$ for $y\in[x,x_{1}]\subseteq[0,x_{0}]$ and $G_{2}(x_{1})=0$ . Inequality (39) reveals that ${W^{(q)\prime}(\bar{\xi}(x))}f(x)/{W^{(q)}(\bar{\xi} (x))}\geq1$ for all $x\in(0,x_{1}) $ .

Proposition 7. In case (v), the once continuously differentiable solution to (14) (optimal tax function) is defined by (38) with $x_{1}$ determined by (36) and $x_{0}$ given in Assumption 1. In addition, the optimal tax strategy is given by $\gamma(x)= \smash{\gamma_{1}\mathbf{1}_{[0,x_{1}]}(x)+ \gamma_{2}\mathbf{1}_{(x_{1},\infty)}(x)}$ .

In case (vi), the once continuously differentiable solution to (14) (optimal tax function) is defined by,

\begin{align} f(x)&= \frac{\gamma_{2}}{1-\gamma_{2}}\int_{x}^{x_{2}} \exp\bigg(\frac{-1}{1-\gamma_{2}}\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\bigg) {{\rm d}}y \nonumber \\ & +\frac{\gamma_{1}}{1-\gamma_{1}} \exp\bigg(\frac{-1} {1-\gamma_{2}}\int_{x}^{x_{2}}\frac{W^{(q)\prime}(\bar{\xi} (y))}{W^{(q)}(\bar{\xi}(y))}{{\rm d}}y\bigg) \nonumber \\ & \times\int_{x_{2}}^{\infty} \exp(\frac{-1}{1-\gamma_{2}} \int_{x_{2}}^{y}\frac{W^{(q)\prime}(\bar{\xi}(u))} {W^{(q)}(\bar{\xi}(u))}{{\rm d}}u\bigg){{\rm d}}y \quad\text{for all }x\in(0,x_{2}), \nonumber \\ f(x)&=\frac{\gamma_{1}}{1-\gamma_{1}}\int_{x}^{\infty}\exp \bigg(\!-\frac{1}{1-\gamma_{1}}\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\bigg){{\rm d}}y \quad\text{for all } x\in[x_{2},\infty), \end{align}

with $x_{2}$ determined by,

\begin{equation}{x2} x_{2}=\inf\{x\in(\bar{\bar{x}},x_{0}]\mid G_{1}(x)=0\}, \end{equation}

and $x_{0}$ given by Assumption 1. In addition, the optimal tax strategy is given by $\gamma(x)=\smash{\gamma_{2}\mathbf{1}_{[0,x_{2}]}(x)}+ \smash{\gamma_{1}\mathbf{1}_{(x_{2},\infty)}(x)}$ .

Proof. The arguments of the proof of case (v) are given immediately above this proposition. The proof of case (vi) is much similar.

Remark 2. Let $\xi(x)\equiv0$ . Then the HJB equation (14) coincides well with Equation (2.6) of [Reference Wang and Hu32]. Furthermore, if it is assumed that each scale function is three times differentiable and its first derivative is a strictly convex function (as was assumed in [Reference Wang and Hu32] and [Reference Albrecher, Renaud and Zhou4]), then

$$ g(x) = \frac{{W^{(q)''} (x)W^{(q)} (x)}} {{(W^{(q)'} (x))^2 }},\quad x \ge (0,x_1 ), $$

will change its sign at most once. In addition, if g(x) changes its sign once at $x_{0}\in[0,\infty) $ , we must have $g(x)\leq0$ for $x\in[0,x_{0}]$ , and $g(x)\geq0$ for $x\geq x_{0}$ .

That is, only cases (i), (ii), and (v) are possible scenarios. Combined with the last equality in (39), it is obvious that (38) in case (v) coincides with Equation (5.7) of [Reference Wang and Hu32]. Meanwhile, (28) in cases (i) and (ii) coincides with Equation (4.2) of [Reference Wang and Hu32]. Indeed, when $\xi\equiv0,$ we have

\begin{equation} G_{2}(x) = (W^{(q)}(x))^{{1}/{(1-\gamma_{2})}} \int_{x}^{\infty}\frac{W^{(q)\prime\prime}(y)(W^{(q)}(y) )^{1-{1}/{(1-\gamma_{2})}}} {(W^{(q)\prime}(y))^{2}}{{\rm d}}y, \qquad x\geq 0.\nonumber \end{equation}

Hence, $x_{1}$ as defined by (36) in case (v) coincides with $u_{0}$ defined in Equation (5.15) of [Reference Wang and Hu32], which implies that Equation (5.22) of [Reference Wang and Hu32] holds with $\beta=\gamma_{2}$ and $c=q$ , that is,

(40) \begin{equation} \frac{W^{(q)}(x_{1})}{W^{(q)\prime}(x_{1})}=\frac{\gamma_{2}}{1-\gamma_{2}} \int_{x_{1}}^{\infty}\bigg(\frac{W^{(q)}(x_{1})}{W^{(q)}(y)} \bigg)^{{1}/{(1-\gamma_{2})}}\mathrm{d}y. \label{eqn39} \end{equation}

By (40), (38) can be rewritten as

\begin{align} f(x)&= \frac{\gamma_{1}}{1-\gamma_{1}}\int_{x}^{x_{1}} \exp\bigg(\frac{-1}{1-\gamma_{1}}\int_{x}^{y}\frac{W^{(q)\prime}(u)} {W^{(q)}(u)}{{\rm d}}u\bigg){{\rm d}}y \\ & +\exp\bigg(\frac{-1}{1-\gamma_{1}}\int_{x}^{x_{1}} \frac{W^{(q)\prime}(y)} {W^{(q)}(y)}{{\rm d}}y\bigg) \frac{\gamma_{2}}{1-\gamma_{2}}\int_{x_{1}}^{\infty}\exp \bigg(\frac{-1}{1-\gamma_{2}}\int_{x_{1}}^{y}\frac{W^{(q)\prime}(u)} {W^{(q)}(u)}{{\rm d}}u\bigg){{\rm d}}y \\ &= (W^{(q)}(x))^{{1}/{(1-\gamma_{1})}} \bigg(\frac{\gamma_{1}}{1-\gamma_{1}}\int_{x}^{x_{1}} (W^{(q)}(y))^{-{1}/{(1-\gamma_{1})}}\mathrm{d}y +\frac{(W^{(q)}(x_{1}))^{1-{1}/{(1-\gamma_{1})}}} {W^{(q)\prime}(x_{1})}\bigg) \end{align}

for $x\in(0,x_{1}) $ , and

\begin{align} f(x) &=\frac{\gamma_{2}}{1-\gamma_{2}} (W^{(q)}(x))^{{1}/{(1-\gamma_{2})}} \int_{x}^{\infty} (W^{(q)}(y))^{-{1}/{(1-\gamma_{2})}}{{\rm d}}y \\ &= (W^{(q)}(x))^{{1}/{(1-\gamma_{2})}} \bigg( \frac{(W^{(q)}(x_{1}))^{1-{1}/{(1-\gamma_{2})}}} {W^{(q)\prime}(x_{1})} - \frac{\gamma_{2}}{1-\gamma_{2}}\int_{x_{1}}^{x} (W^{(q)}(y))^{-{1}/{(1-\gamma_{2})}}{{\rm d}}y \bigg) \end{align}

for $ x\in[x_{1},\infty) $ , coinciding well with Equation (5.7) of [Reference Wang and Hu32].

6. Examples

In this section, we consider several examples of spectrally negative Lévy processes X and show that Assumption 1 holds for certain choices of the draw-down function $\xi$ . For simplicity of notation, write $\bar{\xi}(x)\ :\!=x-\xi(x) $ , $W_\xi(z)\ :\!={W^{(q)}(\bar{\xi}(z))}/ {W^{(q)\prime}(\bar{\xi}(z))},$ and, hence, $W_0(z)\ :\!={W^{(q)}(z)}/ {W^{(q)\prime}(z)}$ .

Example 1. In this example, let $X(t)=\mu t+B(t) $ for a constant $\mu$ and a Brownian motion B. The q-scale function of X is given by

\begin{equation} W^{(q)}(x)=\frac{1}{\sqrt{\mu^2+2q}\,}(\mathrm{e}^{\theta_1x}- \mathrm{e}^{\theta_2x}),\qquad x\geq0, \end{equation}

with $\theta_1=-\mu+\smash{\sqrt{\mu^2+2q}}$ and $\theta_2=-(\mu+\smash{\sqrt{\mu^2+2q}}) $ . It can be checked that

\begin{gather} W^{(q)\prime}(x)=\frac{1}{\sqrt{\mu^2+2q}\,}(\theta_1 \mathrm{e}^{\theta_1x}-\theta_2\mathrm{e}^{\theta_2x})\gt0, \qquad x\geq0, \\ W^{(q)\prime\prime}(x)=2(qW^{(q)}(x) -\mu W^{(q)\prime}(x)),\qquad x\geq0. \end{gather}

The function g can be rewritten as

\begin{align} g(x) &=\xi^{\prime}(x)+(1-\xi^{\prime}(x))(2qW_{\xi}(x)^2-2\mu W_{\xi}(x)), \end{align}

from which we can deduce that

\begin{align} g^{\prime}(x)=\xi^{\prime\prime}(x)f_1(W_{\xi}(x))+(\xi^{\prime}(x)-1) W_{\xi}^{\prime}(x)f_2(W_{\xi}(x)), \end{align}

with $f_1(x)=1-2qx^2+2\mu x$ and $f_2(x)=-4qx+2\mu$ . We can find that $f_1(x)\gt(\lt)\,0$ on $[0,{(\mu+\sqrt{\mu^2+2q})}/{2q}) $ $ (({(\mu+\sqrt{\mu^2+2q})}/{2q},\infty)) $ , and $f_1(x)=0$ for $x={(\mu+\sqrt{\mu^2+2q})}/{2q}$ ; while $f_2(x)\gt(\lt)\,0$ on $[0,{\mu}/{2q}) $ $ (({\mu}/{2q},\infty)) $ , and $f_2(x)=0$ for $x={\mu}/{2q}$ . There are two cases to be considered.

Case 1: $\mu>0$ , and hence $0\lt {\mu}/{2q}\lt {(\mu+\sqrt{\mu^2+2q})}/{2q}$ . First, choose $\xi$ in such a way that

• $\xi(0)=0, \lim _{x\rightarrow\infty}\bar{\xi}(x) =\infty, \xi^{\prime}(0)\leq0$ ;

• $\xi^{\prime}(x)\lt 1$ for all $x\in[0,\infty) $ ;

• $\xi^{\prime\prime}(x)\leq0$ for all $x\in[0,W_{\xi}^{-1}({\mu}/{2q})]$ and $ \xi^{\prime\prime}(x)\geq0$ for all $x\in(W_{\xi}^{-1}({\mu}/{2q}),\infty) $ .

We can observe the following facts:

• $\bar{\xi}(x) $ is strictly increasing over $[0,\infty) $ (because $\bar{\xi}^{\prime}(x)=1-\xi^{\prime}(x)>0$ for all $x\in[0,\infty) $ );

• $W_\xi(z) $ is strictly increasing (since ${W^{(q)}(z)}/ {W^{(q)\prime}(z)}$ and $\bar{\xi}(z) $ are both strictly increasing) with supremum $W_\xi(\infty)={(\mu+\sqrt{\mu^2+2q})}/{2q}$ ;

• $\lim_{x\rightarrow\infty}g(x)=\lim_{x\rightarrow\infty} (\xi^{\prime}(x)+1-\xi^{\prime}(x))=1>0, \lim_{x\rightarrow0}g(x)=\xi^{\prime}(0)\leq0$ ;

• $g^{\prime}(x)\!=\xi^{\prime\prime}(x)f_1(W_{\xi}(x))+(\xi^{\prime}(x)-1) W_{\xi}^{\prime}(x)f_2(W_{\xi}(x))\lt 0$ for all $x\in[0,W_{\xi}^{-1}({\mu}/{2q})) $ ;

• $\xi^{\prime}(W_{\xi}^{-1}({\mu}/{2q}))\lt 0, g(W_{\xi}^{-1} ({\mu}/{2q})) =\xi^{\prime}(W_{\xi}^{-1}({\mu}/{2q})) +(1-\xi^{\prime}(W_{\xi}^{-1}({\mu}/{2q})))\times(\!-~{\mu^2}/{2q})\lt 0$ ;

• $g^{\prime}(x)=\xi^{\prime\prime}(x)f_1(W_{\xi}(x))+(\xi^{\prime}(x)-1) W_{\xi}^{\prime}(x)f_2(W_{\xi}(x))>0,\, x\in(W_{\xi}^{-1}({\mu}/{2q}),\infty) $ .

Based on these observations, we can deduce that there should exist an $x_0\in(W_{\xi}^{-1}({\mu}/{2q}),\infty) $ such that $g(x)\leq0$ for $x\leq x_0$ , and $g(x)>0$ for $x>x_0$ . Thus, Assumption 1 holds.

The above class of the general draw-down function however may seem to be restrictive because conditions are imposed on the second derivative of $\xi$ . Now we construct a more general class of the general draw-down functions. It is seen that

(41) \begin{align} g(x)\geq0 &\quad\Longleftrightarrow\quad \frac{1}{1-\xi^{\prime}((\bar{\xi})^{-1}(z))}-1\geq-\frac{W^{(q) \prime\prime}(z)W^{(q)}(z)}{(W^{(q)\prime}(z))^{2}}, \qquad z=\bar{\xi}(x) \nonumber \\ &\quad\Longleftrightarrow\quad ((\bar{\xi})^{-1})^{\prime}(z)-1\geq 2\mu W_{0}(z)-2qW_{0}(z)^2, \qquad z=\bar{\xi}(x), \label{eqn40}\end{align}

provided that $\xi^{\prime}(x)\lt 1$ for all $x\geq0$ (hence, ${(\bar{\xi})^{-1}}$ is well defined). Recalling that $W^{(q)\prime\prime}(z) $ takes positive values for large z (In fact, $W_{\xi}^{-1}({\mu}/{q}) $ is the unique zero of $W^{(q)\prime\prime}(z) $ with $W^{(q)\prime\prime}(z)\lt (\gt)\,0$ over $[0, W_{\xi}^{-1}({\mu}/{q})) \,((W_{\xi}^{-1}({\mu}/{q}),\infty)) $ , we can also choose $\xi$ satisfying

• $0\leq\xi^{\prime}(x)\lt 1$ for all $x\in[0,\infty) $ ;

• $\bar{\xi}(0)=\underline{d}\in(0,\infty), \bar{\xi}(\infty)=\bar{d}\in(0,\infty], a\in[\underline{d},\bar{d}]$ ;

• $ ((\bar{\xi})^{-1})^{\prime}(z)-1\lt 2\mu W_{0}(z)-2qW_{0}(z)^2$ for all $z\in [\underline{d},a) $ ;

• $ ((\bar{\xi})^{-1})^{\prime}(z)-1\geq2\mu W_{0}(z)-2qW_{0}(z)^2$ for all $z\in[a,\bar{d} ) $ ,

so that Assumption 1 holds with $x_{0}=(\bar{\xi})^{-1}(a) $ . In particular, if $a=\bar{d}$ then $g(x)\lt 0$ for all $x\in[0,\infty) $ ; if $a=\underline{d}$ then $g(x)\geq0$ for all $x\in[0,\infty) $ . To see that such a construction is feasible, it would be quite useful to note that $\xi^{\prime}(x)\in[0,1) $ (or, equivalently, $\bar{\xi}^{\prime}(x)\in(0,1]$ ) over $x\in[0,\infty) $ can result in $ ((\bar{\xi})^{-1})^{\prime}(z)\in[1,\infty) $ over $z\in[0,\infty) $ . Note that the function on the right-hand side of the final inequality of (41) does not depend on $\xi$ .

In particular, we can fix $d_{1}\in[0,\infty) $ , $a\in[d_{1},\infty),$ and $M_{1}\in[1,\infty) $ , and then choose $\xi$ satisfying

• $ (\bar{\xi})^{-1}(d_{1})=0$ ;

• $ ((\bar{\xi})^{-1})^{\prime}(z)\in[1,M_{1}), \,z\in[d_{1},\infty) $ ;

• $ ((\bar{\xi})^{-1})^{\prime}(z)-1\lt 2\mu W_{0}(z)-2qW_{0}(z)^2$ for all $z\in[d_{1},a) $ ;

• $ ((\bar{\xi})^{-1})^{\prime}(z)-1\geq2\mu W_{0}(z)-2qW_{0}(z)^2$ for all $z\in[a,\infty),$

to fulfill Assumption 1 with $x_{0}=(\bar{\xi})^{-1}(a) $ . Here, we should note that the function on the right-hand side of the final inequality of (41) is bounded, and $\bar{\xi}(\infty)=\infty$ by construction.

Case 2: $\mu\leq0$ . First, consider the case $\xi$ satisfying

• $\xi^{\prime}(x)\lt 1$ for all $x\in[0,\infty) $ ;

• $\xi(0)=0,\lim_{x\rightarrow\infty}\bar{\xi}(x)=\infty,\xi^{\prime} (0)=0$ ;

• $\xi^{\prime\prime}(x)\geq0$ for all $x\in[0,\infty),$

so that $\bar\xi(x) $ is strictly increasing on $[0,\infty) $ ; $g(0)=0$ ; and $g^{\prime}(x)=\xi^{\prime\prime}(x)f_1(W_{\xi}(x))+(\xi^{\prime} (x)-1)W_{\xi}^{\prime}(x)f_2(W_{\xi}(x))>0$ (and, hence, $g(x)>0$ ) for all $x\in(0,\infty) $ . Hence, Assumption 1 should hold with $x_0=0$ .

Due to the fact that

\begin{align} g(x)\geq0 \quad\Longleftrightarrow\quad \xi^{\prime}(x)+(1- \xi^{\prime}(x)) (2qW_{\xi}(x)^2-2\mu W_{\xi}(x))\geq0, \end{align}

we can also just choose $\xi$ such that $\xi^{\prime}(x)\in[0,1) $ for $x\in[0,\infty) $ , so that Assumption 1 holds with $x_0=0$ .

Example 2. Let $X(t)=x+pt-\smash{\sum_{i=1}^{N(t)}e_{i}}$ , where $p>0$ , $\{e_{i};\ i\geq1\}$ are independent and identically distributed exponential random variables with mean ${1}/{\mu},$ and $\{N(t), t\geq0\}$ is an independent Poisson process with intensity $\lambda$ . The q-scale function of X is given by

\begin{equation} W^{(q)}(x)=p^{-1}(A_{+}{\rm e}^{\theta_{+}x} -A_{-}{\rm e}^{\theta_{-}x}),\qquad x\geq0, \end{equation}

where

$$A_ \pm = \frac{{\mu + \theta _ \pm }} {{\theta _ + - \theta _ - }} > 0,\quad \theta _ \pm = \frac{{q + \lambda - \mu p \pm \sqrt {(q + \lambda - \mu p)^2 + 4pq\mu } }} {{2p}}.$$

Some algebraic manipulations yield

$$ W^{(q)''} (x) = p^{ - 1} (A_ + (\theta _ + )^2 e^{\theta _ + x} - A_ - (\theta _ - )^2 e^{\theta _ - x} ) = - \theta _ - \theta _ + W^{(q)} (x) + (\theta _ + + \theta _ - )W^{(q)'} (x). $$

Hence, g(x) can be rewritten as

\begin{align} g(x) =\xi^{\prime}(x)+(1-\xi^{\prime}(x))(\!-\theta_{+}\theta_{-} W_{\xi}(x)^2+(\theta_{+}+\theta_{-}) W_{\xi}(x)). \end{align}

Let $f_{3}(x)=-\theta_{+}\theta_{-}x^2+(\theta_{+}+\theta_{-})x$ , which is nonpositive and decreasing over $[0,0\vee {(\theta_{+}+\theta_{-})}/{2\theta_{+}\theta_{-}}) $ , and increasing over $[0\vee {(\theta_{+}+\theta_{-})}/{2\theta_{+}\theta_{-}},\infty) $ . Because $W_\xi(z) $ is strictly increasing with upper bound ${1}/{\theta_{+}}$ (cannot be attained) and lower bound 0, $1-f_{3}(W_{\xi}(x))>0$ , leading to $1-{1}/{(1-f_{3}(W_{\xi}(x)))}\lt 1,\,x\in[0,\infty) $ . Hence, we can verify that

(42) \begin{equation} g(x)\geq0 \quad\Longleftrightarrow\quad \xi^{\prime}(x)\geq \frac{-f_{3}(W_{\xi}(x))}{1-f_{3}(W_{\xi}(x))}=1- \frac{1}{1-f_{3}(W_{\xi}(x))}. \label{eqn41} \end{equation}

Then we can choose $\xi$ satisfying

$$ {\text{o}}\,\xi '(x) \ge 1 - \frac{1} {{1 - f_3 (0 \vee (\theta _ + + \theta _ - )/2\theta _ + \theta _ - )}}\quad {\text{for}}\,{\text{all}}\,x \in [0,\infty ), $$

in order to guarantee that $g(x)\geq0$ for all $x\in[0,\infty) $ , in which case Assumption 1 holds with $x_0=0$ .

By imposing conditions on the first and second derivatives of $\xi,$ and using a very similar argument as in the first construction method for the case $\mu>0$ in Example 1, we can construct an appropriate class of the general draw-down functions fulfilling Assumption 1.

Note that (42) is equivalent to

(43) \begin{align} g(x)\geq0 &\quad\Longleftrightarrow\quad \xi'((\bar{\xi})^{-1}(z))\geq\frac{-f_{3}(W_{0}(z))} {1-f_{3}(W_{0}(z))},\qquad z=\bar{\xi}(x) \nonumber \\ &\quad\Longleftrightarrow\quad 1-(((\bar{\xi})^{-1})^{\prime}(z))^{-1} \geq\frac{-f_{3}(W_{0}(z))}{1-f_{3}(W_{0}(z))}, \qquad z=\bar{\xi}(x), \label{eqn42}\end{align}

provided that $ (\bar{\xi})^{-1}$ is well defined. Thus, we can also choose $\xi$ satisfying

• $\xi^{\prime}(x)\lt 1$ for all $x\in[0,\infty) $ ;

• $\bar{\xi}(0)=\underline{d}\in(0,\infty),\bar{\xi}(\infty) =\bar{d}\in(0,\infty], a\in[\underline{d},\bar{d}]$ ;

• $1-(((\bar{\xi})^{-1})^{\prime}(z))^{-1}\lt \displaystyle\frac{-f_{3}(W_{0}(z))}{1-f_{3}(W_{0}(z))}$ for all $z\in[\underline{d},a) $ ;

• $1-(((\bar{\xi})^{-1})^{\prime}(z))^{-1} \geq\displaystyle\frac{-f_{3}(W_{0}(z))}{1-f_{3}(W_{0}(z))}$ for all $z\in[a,\bar{d}),$

so that Assumption 1 holds with $x_{0}=(\bar{\xi})^{-1}(a) $ . To understand the feasibility of such a construction, it would be quite useful to note that $\xi^{\prime}(x)\in(\!-\infty,1) $ (or, equivalently, $\bar{\xi}^{\prime}(x)\in(0,\infty) $ ) over $x\in[0,\infty) $ can lead to $ ((\bar{\xi})^{-1})^{\prime}(z)\in(0,\infty) $ over $z\in[0,\infty) $ . Note that the function on the right-hand side of the final inequality of (43) does not depend on $\xi$ .

In particular, we can fix $d_{2}\in[0,\infty) $ , $a\in[d_{2},\infty),$ and $M_{2}\in(0,\infty) $ , and then choose $\xi$ satisfying

• $ (\bar{\xi})^{-1}(d_{2})=0$ ;

• $ ((\bar{\xi})^{-1})^{\prime}(z)\in(0,M_{2}), \,z\in[d_{2},\infty) $ ;

• $1-(((\bar{\xi})^{-1})^{\prime}(z))^{-1}\lt \displaystyle\frac{-f_{3}(W_{0}(z))} {1-f_{3}(W_{0}(z))}$ for all $z\in[d_{2},a) $ ;

• $1-(((\bar{\xi})^{-1})^{\prime}(z))^{-1}\geq \displaystyle\frac{-f_{3}(W_{0}(z))}{1-f_{3}(W_{0}(z))}$ for all $z\in[a,\infty) $ ,

to fulfill Assumption 1 with $x_{0}=(\bar{\xi})^{-1}(a) $ . Here, we should note that the function on the right-hand side of the final inequality of (43) is bounded from above, and $\bar{\xi}(\infty)=\infty$ by construction.

Example 3. In this example we consider a general spectrally negative Lévy process X. It still holds that

\begin{equation} g(x)\geq0 \quad\Longleftrightarrow\quad ((\bar{\xi})^{-1})^{\prime}(z)-1\geq-\frac{W^{(q) \prime\prime}(z)W^{(q)}(z)}{(W^{(q)\prime}(z))^{2}},\qquad z= \bar{\xi}(x), \end{equation}

provided that $\xi^{\prime}(x)\lt 1$ for all $x\geq0$ . We can choose $\xi$ satisfying

• $\xi^{\prime}(x)\in(\!-\infty,1) $ for all $x\in[0,\infty) $ ;

• $\bar{\xi}(0)=\underline{d}\in(0,\infty),\bar{\xi}(\infty)= \bar{d}\in(0,\infty], \,a\in[\underline{d},\bar{d}]$ ;

• $ ((\bar{\xi})^{-1})^{\prime}(z)-1\lt(\gt)-\displaystyle\frac{W^{(q)\prime\prime} (z)W^{(q)}(z)}{(W^{(q)\prime}(z))^{2}}$ for all $z\in[\underline{d},a) $ ;

• $ ((\bar{\xi})^{-1})^{\prime}(z)-1\geq(\leq)-\displaystyle\frac{W^{(q) \prime\prime}(z)W^{(q)}(z)}{(W^{(q)\prime}(z))^{2}}$ for all $z\in[a,\bar{d}),$

to guarantee that Assumption 1 holds with $x_{0}=(\bar{\xi})^{-1}(a) $ . It would also be useful to note that $\xi^{\prime}(x)\in(\!-\infty,1) $ (or, equivalently, $\bar{\xi}^{\prime}(x)\in(0,\infty) $ ) over $x\in[0,\infty) $ can result in $ ((\bar{\xi})^{-1})^{\prime}(z)\in(0,\infty) $ over $z\in[0,\infty) $ . In essence, an appropriate class of general draw-down functions satisfying Assumption 1 are constructed via imposing conditions on the inverse function $ (\bar{\xi})^{-1} $ instead of on $\xi$ or $\bar{\xi}$ .

For the spectrally negative Lévy process whose Lévy measure has a completely monotone density, we can also follow the method in [Reference Wang and Zhou34] to construct the class of the general draw-down functions fulfilling Assumption 1 with $x_{0}=(\bar{\xi})^{-1}(a) $ and $a\ :\!=\inf\{z\geq 0\colon \smash{W^{(q)\prime\prime}(z)> 0 \}}$ .

7. Numerical analysis

In this section we provide some numerical examples to illustrate the theoretical results obtained in the previous sections. We consider a linear draw-down function

$$\xi (x) = kx - d,\quad k \lt 1,d \ge 0. $$

Under this assumption, we have, for $x{\lt}y$ ,

$$ \int_x^y {\frac{{W^{(q)'} (\bar \xi (u))}} {{W^{(q)} (\bar \xi (u))}}du = \frac{1} {{1 - K}}\ln \left( {\frac{{W^{(q)} ((1 - k)y + d)}} {{W^{(q)} ((1 - k)x + d)}}} \right)} , $$

which has frequently appeared in the formulae in Section 4.

For the risk process, it is assumed to be a linear Brownian motion, i.e.

(44) \begin{equation} X(t)=\mu t+\sigma B(t),\qquad t\geq 0, \label{eqn43} \end{equation}

where $\mu\in \mathbb{R}$ , $\sigma>0$ , and $\{B(t)\}$ is a standard Brownian motion. Note that our main results are all expressed in terms of the q-scale function $W^{(q)}$ . By [Reference Kyprianou19] we know that the q-scale function for the linear Brownian motion (44) is given by

\begin{equation} W^{(q)}(x)=\frac{2}{\sigma^2\Xi} \mathrm{e}^{-{\mu}x/{\sigma^2} } \sinh(\Xi x),\qquad x\geq 0, \end{equation}

where $ \Xi={\sqrt{ \mu^2+2q\sigma^2 } }/{ \sigma^2}.$ Throughout this section, we set $\mu=0.03, \sigma=0.4,$ and $q=0.01$ .

It follows from Propositions 5 and 6 that Assumption 1 plays an important role in characterizing the optimal return function. It is seen that

$$ \mathop {\lim }\limits_{x \to \infty } \frac{{W^{(1)''} (\bar \xi (x))W^{(q)} (\bar \xi (x))}} {{(W^{(q)'} (\bar \xi (x)))^2 }} = 1, $$

which yields $\lim_{x\rightarrow \infty}g(x)=1$ . In Figure 1, we plot the behavior of g(x) for $d=1$ and $k=-10, -5, -1, 0, 0.1$ . It follows that the function g indeed converges to 1 as $x\rightarrow \infty$ , and we also find that the function g indeed changes its sign at some point $x_0$ .

Figure 1:

Behavior of the function g for $d=1$ and $k=-10, -5, -1, 0, 0.1$ .

Now for the draw-down parameters (k, d), we shall consider two cases: $ (0.1, 1) $ and $ (\!-1, 1) $ . Furthermore, we set $ (\gamma_1, \gamma_2)=(0.2, 0.6) $ . First, we consider the conditions in Proposition 5. Besides Assumption 1, we also need to check inequality (32). It follows from Figure 1 that $g(y)>0$ for $y>x_0$ . Hence, to check inequality (32), we need to only check the condition

\begin{equation} G_2(x)\geq 0 \quad\text{for all } 0{\lt}x\leq x_0. \end{equation}

Figure 2:

The function $G_2$ for $d=1$ and $k=-1, 0.1$ .

For $ (k, d)=(0.1, 1) $ , we can obtain $x_0=1.360$ ; for $ (k,d)=(\!-1,1) $ , we can obtain $x_0=1.443$ . In Figure 2, we plot $G_2$ as a function of x for $0\leq x \leq 1.5$ . When $k=0.1$ , we find that $G_2(x) $ is strictly positive for $0\leq x\leq x_0$ . It follows from Proposition 5 that the optimal strategy function is $\gamma(x)\equiv\gamma_{2}$ for all $x\geq0$ , and the optimal return function is given by (28). In Figure 3, we plot the optimal return value as a function of the initial surplus level x. We observe that f(x) is an increasing function and converges to a constant value $2.8209$ . The limit can be checked mathematically as follows. Using (30), we have

$$f(x) = \frac{{W^{(q)} (\bar \xi (x))}} {{W^{(q)'} (\bar \xi (x))}} - \int_x^\infty {\exp \left( { - \frac{1} {{1 - \gamma _2 }}\int_x^y {\frac{{W^{(q)'} (\bar \xi (u))}} {{W^{(q)} (\bar \xi (u))}}} } \right)g(y)dy.} $$

Figure 3:

The optimal return function when $d=1$ , $k=0.1$ , $\gamma_1=0.2,$ and $\gamma_2=0.6$ .

which, together with $g(y)\rightarrow 1$ as $y\rightarrow \infty$ , yields, for large x,

\begin{equation} f(x) \approx \frac{W^{(q)}(\bar{\xi}(x))}{W^{(q)\prime} (\bar{\xi}(x))}-\frac{1-\gamma_2}{\gamma_2}f(x), \end{equation}

where we have used (3), (27), (28), and l’Hôpital’s rule to deduce that

\begin{align} & \lim_{x\rightarrow\infty}\frac{\int_{x}^{\infty}\exp \Big(\!-\frac{1}{1-\gamma_{2}}\int_{x}^{y}\frac{W^{(q)\prime} (\bar{\xi}(u))}{W^{(q)}(\bar{\xi}(u))} {{\rm d}}u\Big) g(y){{\rm d}}y}{f(x)} \\ &\qquad = \lim_{x\rightarrow\infty} \frac{-g(x)+\frac{1}{1-\gamma_{2}}\frac{W^{(q)\prime} (\bar{\xi}(x))}{W^{(q)}(\bar{\xi}(x))} \int_{x}^{\infty}\exp\Big(\!-\frac{1}{1-\gamma_{2}}\int_{x}^{y} \frac{W^{(q)\prime}(\bar{\xi}(u))}{W^{(q)} (\bar{\xi}(u))}{{\rm d}}u\Big) g(y){{\rm d}}y} {-\frac{\gamma_{2}}{1-\gamma_{2}}+\frac{\gamma_{2}}{1-\gamma_{2}} \frac{1}{1-\gamma_{2}}\frac{W^{(q)\prime}(\bar{\xi}(x))} {W^{(q)}(\bar{\xi}(x))} \int_{x}^{\infty}\exp\Big(\!-\frac{1}{1-\gamma_{2}}\int_{x}^{y} \frac{W^{(q)\prime}(\bar{\xi}(u))}{W^{(q)} (\bar{\xi}(u))}{{\rm d}}u\Big) {{\rm d}}y} \\ &\qquad =\frac{1-\gamma_{2}}{\gamma_{2}}. \end{align}

Hence, for large x, we have

$$f(x) \approx \gamma _2 \frac{{W^{(q)} (\bar \xi (x))}} {{W^{(q)'} (\bar \xi (x))}} = 0.6\frac{{W^{(q)} (0.9x + 1)}} {{W^{(q)'} (0.9x + 1)}} \to \frac{{0.6}} {{\Phi (0.01)}} = 2.8209\quad {\text{as}}\,x \to \infty , $$

where the last step follows from (3).

As for $k=-1$ , after observing Figure 2 we find that $G_2(x) $ is not always positive for $0\leq x\leq x_0$ . Hence, the optimal return function is not characterized by Proposition 5. However, it is seen that the conditions in case (v) are satisfied; then the optimal return function is given by formulae in (38). In Figure 4 we plot the optimal return function for $0\leq x\leq 20$ . In this case, we find that f(x) converges to the same limit as the case when $k=0.1$ , which is due to the fact that the optimal return function has the same form when x is large, and the corresponding limit $2.8209$ is independent of the parameter k.

Figure 4:

The optimal return function when $d=1$ , $k=-1$ , $\gamma_1=0.2,$ and $\gamma_2=0.6$ .

The patterns of the optimal return functions shown in Figure 3 and Figure 4 reveal that the smaller the parameter k, the larger the optimal return function. This is because, for each admissible tax strategy $\gamma\in\Gamma$ , the draw-down time $\tau^{\gamma}_{\xi}$ decreases a.s. as k increases, bearing in mind (1) and $\xi(x)=kx-d$ with $d=1$ set in Figure 3 and Figure 4. These reasonings together with the definition of the optimal return function given by (2) explain why smaller k gives a larger optimal return function value, i.e. $f(x)|_{k=k_{1}}\leq f(x)|_{k=k_{2}}$ for $x\in[0,\infty) $ and $k_{2}\leq k_{1}\lt 1$ .