# Time-inconsistent view on a dividend problem with penalty

Josef Anton Strini Graz University of Technology, Institute of Statistics,
Kopernikusgasse 24/III, 8010 Graz, AustriaContact<EMAIL_ADDRESS>Stefan
Thonhauser11footnotemark: 1 Stefan Thonhauser received support by the Austrian
Science Fund (FWF) Project P 33317.

(2022
April)

###### Abstract

We consider the dividend maximization problem including a ruin penalty in a
diffusion environment. The additional penalty term is motivated by a
constraint on dividend strategies. Intentionally, we use different discount
rates for the dividends and the penalty, which causes time-inconsistency. This
allows to study different types of constraints. For the diffusion
approximation of the classical surplus process we derive an explicit
equilibrium dividend strategy and the associated value function. Inspired by
duality arguments, we can identify a particular equilibrium strategy such that
for a given initial surplus the imposed constraint is fulfilled. Furthermore,
we illustrate our findings with a numerical example.

Keywords: Ruin theory; dividends; time-inconsistent stochastic control;
extended Hamilton-Jacobi-Bellman equation

##  1 Introduction

The dividend maximization problem and many extensions of it are widely
investigated in actuarial science and in particular in its risk theoretic
branch. Under suitable conditions one can detect the optimal control in
prespecified models. For example, in classical diffusion settings optimal
dividend strategies are of constant barrier type, cf. [Shreve et al., 1984] or
[Asmussen and Taksar, 1997]. If using such a reflecting barrier strategy, this
has the consequence of almost sure ruin.
On the other hand, one way to totally avoid ruin is the usage of capital
injections, cf. [Kulenko and Schmidli, 2008], one of the most renowned recent
extensions of the problem. While these results are theoretically remarkable,
they seem not to be satisfying if one aims at a balanced situation between
risk and profit. So there exist two extremal cases: on the one hand maximizing
the output regardless of consequences - always having in mind that this will
cause negative social impacts on the insured, employees, the public reputation
of the insurance company and its management. On the other hand, keeping a
business uncompromisingly alive using capital injections, may be in conflict
with owner interest.
For this reason, we want to identify a dividend strategy which suffices a
particular tradeoff between ruin and profit. Naturally, one way of tackling
this problem is to consider the dividend problem under a ruin constraint. That
is maximizing expected discounted dividends over a set of strategies for which
the ruin probability for fixed initial surplus stays below a given level. This
approach goes back to [Hipp, 2003]. Recent results in this direction are
presented in [Hipp, 2018b, Hipp, 2018a, Hipp, 2020] and mentioned in a talk by
Brandon Israel García Flores111 _Genetic Algorithms in Risk Theory: Optimizing
Dividend Strategies_ at IME conference 2021. On a finite time horizon such a
constrained dividend maximization problem is analyzed in [Grandits, 2015].
For dealing with the ruin probability constraint in the first place and for
achieving a better understanding of the evolution of such constraints in the
second place, we present a time-inconsistent variation or extension of
[Hernández and Junca, 2015] in the following.
There are different origins of time-inconsistency in the dividend maximization
problem. For example, [Zhao et al., 2014] deal with the problem using non-
exponential discounting, whereas [Christensen and Lindensjö, 2019] arrive at
time(space)-inconsistency when adding a moment constraint on the number of
expected dividend payments.

###  1.1 Specification of the model

In the following we will consider the surplus process $X=(X_{t})_{t\geq 0}$
with the dynamics

$dX_{t}=\mu(t,X_{t})dt+\sigma(t,X_{t})dW_{t},$ (1)

where $\mu$ and $\sigma$ are deterministic functions such that (1) is well-
posed and $W=(W_{t})_{t\geq 0}$ is a standard Brownian motion on a probability
space $(\Omega,\mathcal{F},\mathcal{P})$. The flow of information is given by
the $\mathcal{P}$-completed Brownian history, denoted by
$\\{\mathcal{F}_{t}\\}_{t\geq 0}$. Exemplary conditions which ensure that (1)
admits a strong solution are the classical linear growth and Lipschitz
conditions for $\mu$ and $\sigma$, see [Karatzas and Shreve, 1991, p. 289,
Theorem 2.9]. Or, the more general requirements from [Zvonkin, 1974, p. 142,
Theorem 4], which are:

  * •

$\mu$ is bounded and measurable,

  * •

$\sigma$ is bounded from above and away from zero, is continuous in $(t,x)$
and is Hölder continuous with parameter larger or equal than $1/2$ with
respect to $x$ for every $t$.

We subsequently refer to these latter conditions as conditions $(Z)$.
The task in the classical problem, cf. [Shreve et al., 1984], is to choose the
dividend process $L=(L_{t})_{t\geq 0}$ in an optimal way, such that the
discounted expected future dividends are maximized. The resulting value
function is

$\sup_{L}\mathbb{E}_{x}\left[\int\limits_{0}^{\tau^{L}}e^{-\delta
s}dL_{s}\right],$ (2)

where the supremum is taken over all processes $L$ which are non-decreasing,
adapted, càglàd with $L_{0}=0$ and $\Delta L_{t}\leq X^{L}_{t-}$. The
controlled surplus is

$dX^{L}_{t}=\mu(t,X^{L}_{t})dt+\sigma(t,X^{L}_{t})dW_{t}-dL_{t}.$ (3)

For assumptions on $\mu$ and $\sigma$ such that (2) is well-posed, we refer to
[Shreve et al., 1984].

In order to penalize early ruin [Thonhauser and Albrecher, 2007] added the
term $\Lambda e^{-\delta\tau^{L}}$ to the reward function. Based on their
work, [Hernández and Junca, 2015] linked the penalty term to a constraint on
$\mathbb{E}[e^{-\delta\tau^{L}}]$ in order to solve the associated constrained
problem for a given initial surplus. Motivated by these previous results and
by a comment in the recent contribution [Junca et al., 2019, p. 636, Remark
2.1] on an extension with different discount rates, we want to analyze the
following reward function for a $\lambda\leq 0$

$\mathbb{E}_{t,x}\left[\int\limits_{t}^{\tau^{L}}e^{-\delta(s-t)}dL_{s}\right]+\lambda\left(\mathbb{E}_{t,x}\left[e^{-\beta(\tau^{L}-t)}\right]-\alpha\right).$

This integrates the constraint

$\mathbb{E}_{t,x}\left[e^{-\beta(\tau^{L}-t)}\right]\leq\alpha.$ (4)

Here $\beta$ and $\delta$ are positive, not necessarily equal constants,
$0<\alpha\leq 1$ and $\tau^{L}=\tau^{L}(t,x)=\inf\\{s\geq t|X^{L}_{s}\leq
0\,,X^{L}_{t}=x\\}$, for a process $X^{L}$ starting in $x$ at time $t\geq 0$.
We consider the following variant of this problem. We assume that
$dL_{t}=l(t,X^{L}_{t})dt$ and we define the set of admissible controls
$\Theta$ to be the measurable controls of feedback type
$l:[0,\infty)^{2}\to[0,L_{max}]$. The bound $L_{max}>0$ is a prespecified
constant. The main purpose of considering controls of feedback type is due to
the fact that we are facing a time-inconsistent stochastic control problem in
the framework of [Björk et al., 2017, p. 334] and [Christensen and Lindensjö,
2019]. Note that as in the classical time-consistent case, one could try to
approach the singular control problem, where the dividend rate is unbounded
enabling lump sum payments, by considering the limit of $L_{max}$ to infinity
in the problem with restricted dividend payments, see [Schmidli, 2008, p.
102]. But in the present case the underlying situation is different and the
concept of equilibrium controls probably has to be adapted for the singular
case. In order to keep our discussion as condensed as possible, we leave this
task for future research.

######  Remark 1.

Notice that our approach is also able to cover the situation with a constraint
on the ruin probability. For this reason, we consider for the above
expectation (4) the limit $\beta\to 0$. We have that
$\mathbb{E}_{t,x}\left[e^{-\beta(\tau^{L}-t)}\right]=\mathbb{E}_{t,x}\left[e^{-\beta(\tau^{L}-t)}\left(1_{\\{\tau^{L}<\infty\\}}+1_{\\{\tau^{L}=\infty\\}}\right)\right]=\mathbb{E}_{t,x}\left[e^{-\beta(\tau^{L}-t)}1_{\\{\tau^{L}<\infty\\}}\right]$
and $\lim\limits_{\beta\to
0}\mathbb{E}_{t,x}\left[e^{-\beta(\tau^{L}-t)}1_{\\{\tau^{L}<\infty\\}}\right]=P_{t,x}(\tau^{L}<\infty).$

######  Remark 2.

If we set $t=0$ and send $\beta$ to zero in (4), we are able to identify the
minimal argument $x$ for which a ruin probability constraint can be fulfilled
for an $\alpha<1$ and constant $\mu$ and $\sigma$. This minimal $x$ appears,
since the control process can only reduce the current reserve. Hence, if the
constraint cannot be reached from the level $x$ without dividends, it is not
possible to satisfy (4) with a control action. In order to determine this
minimal level, we consider (3) with $dL_{t}=ldt$, where $l$ is a positive
constant. Consequently, we consider the hitting time of zero of a Brownian
motion with drift $\mu-l>0$. The corresponding probability of ruin for this
process is then given by

$\psi(x,l)=e^{\frac{-2(\mu-l)}{\sigma^{2}}x},$ (5)

see [Schmidli, 2017, p. 109, Lemma 5.21]. Certainly, if the drift $\mu-l$ is
negative, we cannot fulfil the constraint (4). In the extremal case $l=0$, we
can satisfy (4) only if

$x\geq-\frac{\sigma^{2}}{2}\frac{\ln(\alpha)}{\mu}>0.$

###  1.2 Reformulation of the reward

Optimization problems with constraints can usually be handled using the
Lagrange multiplier approach. For $\lambda\leq 0$ we define the reward
function to be:

$J(t,x,l)=\mathbb{E}_{t,x}\left[\int\limits_{t}^{\tau^{l}}e^{-\delta(s-t)}l(s,X^{l}_{s})ds+\lambda\left(e^{-\beta(\tau^{l}-t)}-\alpha\right)\right],$
(6a) and for $\beta\to 0$
$J_{0}(t,x,l)=\mathbb{E}_{t,x}\left[\int\limits_{t}^{\tau^{l}}e^{-\delta(s-t)}l(s,X^{l}_{s})ds+\lambda\left(1_{\\{\tau^{l}<\infty\\}}-\alpha\right)\right],$
(6b)

subject to

$dX^{l}_{t}=(\mu(t,X^{l}_{t})-l(t,X^{l}_{t}))dt+\sigma(t,X^{l}_{t})dW_{t}.$
(7)

Moreover, let $\mu(t,x)-l(t,x)$ and $\sigma(t,x)$ be such that a solution to
(7) exists. In our case we assume that $l(\cdot,\cdot)\in\Theta$ and that
$\mu(\cdot,\cdot)$ and $\sigma(\cdot,\cdot)$ fulfil the previously stated
conditions $(Z)$. Further, we obtain that the solution to (7) is a strong
Markov process with infinitesimal generator

$\mathcal{A}^{l}g(t,x)=\frac{\partial}{\partial
t}g(t,x)+(\mu(t,x)-l(t,x))\frac{\partial}{\partial
x}g(t,x)+\frac{\sigma^{2}(t,x)}{2}\frac{\partial^{2}}{\partial x^{2}}g(t,x),$

for a suitable function $g$ and $(t,x)\in\mathbb{R}^{+}\times\mathbb{R}$, see
[Karatzas and Shreve, 1991, p. 322].

######  Remark 3.

According to [Hernández and Junca, 2015, p. 137, eq. (3)] and
[Hernández et al., 2018] we set $\lambda\leq 0$ in the function (6a), such
that, if the constraint is fulfilled, the term on the right hand side is
positive and corresponds to a penalization. We restrict ourselves to a fixed
value $\lambda\leq 0$. As proposed in [Øksendal, 2000, p. 259-260], we can
first solve the problem for every fixed $\lambda$, and then choose $\lambda$
such that the constraint (4) is fulfilled. This is treated in detail in
Section 5. Nevertheless, the duality approach, especially as used in
[Hernández and Junca, 2015, p. 139, Th. 4.2], is not completely applicable in
our setting. This is because the game-theoretic approach, which we are going
to use, is different to a classical maximization of a reward function.

Since $\delta\neq\beta$, the objective (6a) (or (6b)) is time-inconsistent and
the dynamic programming principle is not valid anymore. Therefore, we make use
of the theory of time-inconsistent stochastic optimal control, see [Ekeland
and Pirvu, 2008], [Ekeland et al., 2012] and [Björk et al., 2017]. For this
purpose, we modify the reward function from (6a) and introduce a new variable
$r\in[0,\infty)$, some artificial reference point in time. For $\beta>0$, we
define the function
$H:\mathbb{R}_{0}^{+}\times\mathbb{R}_{0}^{+}\times[0,L_{max}]\to\mathbb{R}$,
to be

$H\left(r,s,l\right)=e^{-\delta(s-r)}l-\lambda\beta e^{-\beta(s-r)},$

such that we arrive at a representation only using a running reward and a
constant term. We denote this auxiliary reward function by:

$\displaystyle\begin{split}J(r,t,x,l)&=\mathbb{E}_{t,x}\left[\int\limits_{t}^{\tau^{l}}e^{-\delta(s-r)}l(s,X_{s}^{l})-\lambda\beta
e^{-\beta(s-r)}ds+\lambda\left(1-\alpha\right)\right]\\\
&=\mathbb{E}_{t,x}\left[\int\limits_{t}^{\tau^{l}}H\left(r,s,l(s,X_{s}^{l})\right)ds+\lambda\left(1-\alpha\right)\right].\end{split}$
(8a) For $\beta\to 0$ we have:
$\displaystyle\begin{split}J_{0}(r,t,x,l)&=\mathbb{E}_{t,x}\left[\int\limits_{t}^{\tau^{l}}e^{-\delta(s-r)}l(s,X_{s})ds+\lambda\left(1_{\\{\tau^{l}<\infty\\}}-\alpha\right)\right].\end{split}$
(8b)

For $r=t$ we obtain that the auxiliary reward function coincides with the
original reward function $J(t,t,x,l)=J(t,x,l)$ and
$J_{0}(t,t,x,l)=J_{0}(t,x,l)$, respectively.

######  Remark 4.

Note that we have the following integrability property of the reward. For all
fixed $r\in\mathbb{R}_{0}^{+}$ and for all
$(t,x)\in\mathbb{R}_{0}^{+}\times\mathbb{R}$ and each admissible control $l$,
it holds that

$E_{t,x}\left[\int\limits_{t}^{\tau^{l}}\left|H\left(r,s,l(s,X_{s}^{l})\right)\right|ds\right]<\infty.$
(9)

Using the special form of $H$ and the fact that controls are bounded by
$L_{max}>0$, we obtain that the auxiliary reward function and the original
reward function are both bounded in $t$ and $x$. Namely, it holds that

$\mathbb{E}_{t,x}\left[\int\limits_{t}^{\tau^{l}}\left|H\left(r,s,l(s,X_{s}^{l})\right)\right|ds\right]\leq\\\
\frac{L_{max}}{\delta}e^{-\delta(t-r)}+|\lambda|e^{-\beta(t-r)}\leq\frac{L_{max}}{\delta}e^{\delta
r}+|\lambda|e^{\beta r}.$

Moreover, in the special case $r=t$ we obtain that

$\mathbb{E}_{t,x}\left[\int\limits_{t}^{\tau^{l}}\left|H\left(t,s,l(s,X_{s}^{l})\right)\right|ds\right]\leq\frac{L_{max}}{\delta}+|\lambda|.$

##  2 Equilibrium control and associated equilibrium value function

The time-inconsistencies in (8a) and (8b) prevent us from using the dynamic
programming approach for deriving a maximizing dividend strategy. Instead we
rely on the methodology introduced by [Björk et al., 2017]. More precisely, we
are guided by [Björk et al., 2017, p. 345], [Yong, 2012, p. 23] and [Zhao et
al., 2014, p. 3]. Analogously, as discussed in [Björk et al., 2017] the
subsequent definition is motivated by an intrapersonal game of the management
in force with its future composition. In our problem the executive management
has to choose the dividend strategy, but their associated reward function is
of the form (8a) or (8b), so their preferences are time-inconsistent and hence
vary over time. The management of the company is naturally not appointed for
life, hence it will happen that single executives or even the whole leadership
changes over time. Of course, they should always act in favour of the
shareholders of the company. Aiming at a time-consistent management decision
results in various economic benefits. Such a line of action will improve both
the stability inside the company, regarding affiliates and employees, and the
general external assessment, affecting future investors and clients. In the
light of these economic considerations, the present constraint (4), discussed
in Section 5, also can be viewed as a comprehensive alternative to classical
one-year risk considerations, e.g. conditions on the value at risk.

######  Definition 1 ([Björk et al., 2017, Def. 3.4, p. 336]).

Consider an admissible control $\hat{l}\in\Theta$ and choose an other
arbitrary admissible control $l\in\Theta$, a fixed value $h>0$ and an
arbitrary but fixed initial point $(t,x)$. Consider the control $l_{h}$
defined by

$l_{h}(s,y)=\begin{cases}l(s,y),\text{ for }\ t\leq s<t+h,\
y\in\mathbb{R}^{+},\\\ \hat{l}(s,y),\text{ for }\ s\geq t+h,\
y\in\mathbb{R}^{+}.\end{cases}$ (10)

Now, if for all $l\in\Theta$

$\liminf\limits_{h\to 0}\frac{J(t,x,\hat{l})-J(t,x,l_{h})}{h}\geq 0,$

$\hat{l}$ is called equilibrium control and the associated (equilibrium) value
function is defined via

$V(t,x)\coloneqq J(t,x,\hat{l}).$ (11)

######  Definition 2.

The extended Hamilton-Jacobi-Bellman system of equations for a function
$g(r,t,x)=g^{r}(t,x)$ and the given objective (6a) (and for $\beta\to 0$ (6b))
reads as follows for all $t,x\in[0,\infty)$

$\displaystyle\sup\limits_{l\in[0,L_{max}]}\\{\mathcal{A}^{l}g^{t}(t,x)+H\left(t,t,l\right)\\}=0,$
(12a)
$\displaystyle\sup\limits_{l\in[0,L_{max}]}\\{\mathcal{A}^{l}g^{t}(t,x)+l\\}=0,$
(12b)

and for all $r,t,x\in[0,\infty)$

$\displaystyle\mathcal{A}^{\hat{l}}g^{r}(t,x)+H\left(r,t,\hat{l}\right)=0$
(13a) $\displaystyle\mathcal{A}^{\hat{l}}g^{r}(t,x)+e^{-\delta(t-r)}\hat{l}=0$
(13b)
$\displaystyle\lim\limits_{w\to\infty}\mathbb{E}_{t,x}\left[g^{r}(w\wedge\tau^{\hat{l}},X^{\hat{l}}_{w\wedge\tau^{\hat{l}}})\right]=\lambda\left(1-\alpha\right).$
(14a)
$\displaystyle\lim\limits_{w\to\infty}\mathbb{E}_{t,x}\left[g^{r}(w\wedge\tau^{\hat{l}},X^{\hat{l}}_{w\wedge\tau^{\hat{l}}})\right]=\lambda\left(\mathbb{E}_{t,x}\left[1_{\\{\tau^{\hat{l}}<\infty\\}}\right]-\alpha\right).$
(14b)

Moreover, $\hat{l}$ denotes the control for which the supremum is attained.
Where (12a), (13a) and (14a) correspond to the case $\beta>0$ and (12b), (13b)
and (14b) belong to the case $\beta\to 0$.

######  Remark 5.

Notice that the condition (14a) (or (14b)) is the natural one in order to
obtain the representation (16a) (or (16b)). In this instance we are conscious
that the representation (14b) is not exactly in line with [Björk et al.,
2017], since following their representation, we would have to consider the
function

$h_{0}(t,x)=\lambda\left(\mathbb{E}_{t,x}\left[1_{\\{\tau^{\hat{l}}<\infty\\}}\right]-\alpha\right)$

with the associated equation $\mathcal{A}^{\hat{l}}h_{0}(t,x)=0$ and the
corresponding boundary conditions ($\lambda(1-\alpha)$ at $x=0$ and
$-\lambda\alpha$ for $x$ to infinity). But since we want to use a verification
argument to confirm our solution as the value function, we prefer boundary
conditions in terms of deterministic functions. Hence, we impose the following
conditions which imply (14a) (or (14b)):

$\displaystyle\begin{split}g^{r}(t,0)&=\lambda(1-\alpha)\quad(r,t)\in[0,\infty)\times[0,\infty),\\\
\lim\limits_{x\to\infty}g^{r}(t,x)&=e^{-\delta(t-r)}\frac{L_{max}}{\delta}-\lambda
e^{-\beta(t-r)}+\lambda(1-\alpha)\quad(r,t)\in[0,\infty)\times[0,\infty),\end{split}$
(15)

in addition to the requirement that $g^{r}$ is bounded. The above requirements
(15) for $\beta\to 0$ also yield (14b), since

$\displaystyle\lim\limits_{w\to\infty}\mathbb{E}_{t,x}\left[g^{r}(w\wedge\tau^{\hat{l}},X^{\hat{l}}_{w\wedge\tau^{\hat{l}}})\right]=$
$\displaystyle\mathbb{E}_{t,x}\left[1_{\\{\tau^{\hat{l}}=\infty\\}}(-\lambda\alpha)\right]+\mathbb{E}_{t,x}\left[1_{\\{\tau^{\hat{l}}<\infty\\}}\lambda(1-\alpha)\right]=\lambda\left(\mathbb{E}_{t,x}\left[1_{\\{\tau^{\hat{l}}<\infty\\}}\right]-\alpha\right).$

##  3 Verification Theorem

In this section we state and prove the associated _verification theorem_. The
required regularity of the involved function is due to the applicability of a
suitable version of Itô’s formula and the properties of a constructible
explicit solution.

######  Theorem 1 (Verification Theorem).

Given a real-valued function
$g\in\mathcal{C}^{0,1,1}\left([0,\infty)^{3}\right)$ $\cap$
$\mathcal{C}^{0,1,2}\left([0,\infty)\times[0,\infty)\times([0,\infty)\backslash\\{b\\})\right)$
for a value $b\geq 0$, which is bounded for all $t\in[0,\infty)$ and for all
$x\in[0,\infty)$. In addition, suppose that $\frac{\partial}{\partial
x}g(r,t,x)$, $\frac{\partial^{2}}{\partial x^{2}}g(r,t,x)$ and
$\frac{\partial}{\partial t}g(r,t,x)$ are bounded for all $t\in[0,\infty)$ and
for all $x\in[0,\infty)$. Furthermore, let $g$ solve the extended HJB system
in Definition 2, where $\hat{l}$ is an admissible control for which the
supremum is attained.
Then,

  * -

$\hat{l}$ is an equilibrium control according to Definition 1,

  * -

$V(t,x)=g(t,t,x)$ is the associated value function and

  * -

$g^{r}(t,x)=g(r,t,x)$ has the representation, for all $r\in[0,\infty)$:

$g^{r}(t,x)=\mathbb{E}_{t,x}\left[\int\limits_{t}^{\tau^{\hat{l}}}e^{-\delta(s-r)}\hat{l}(s,X_{s}^{\hat{l}})-\lambda\beta
e^{-\beta(s-r)}ds+\lambda\left(1-\alpha\right)\right]\text{ for }\beta>0,$
(16a) and
$g^{r}(t,x)=\mathbb{E}_{t,x}\left[\int\limits_{t}^{\tau^{\hat{l}}}e^{-\delta(s-r)}\hat{l}(s,X_{s}^{\hat{l}})ds+\lambda\left(1_{\\{\tau^{\hat{l}}<\infty\\}}-\alpha\right)\right]\text{
for }\beta\to 0.$ (16b)

###### Proof.

At first, we are going to show that $g(t,t,x)=J(t,x,\hat{l})$. We are
certainly guided by the work of [Björk et al., 2017, p. 345], [Yong, 2012, p.
23] and [Zhao et al., 2014, p. 3].
We start with showing the stochastic representation of $g^{r}$, for that
reason we use a modified version of Itô’s formula, cf. [Leobacher et al.,
2015, p. 7, Thm. 2.9], and take expectations. Let $r,t,x\in[0,\infty)$ and
$w>t$ be fixed, further we set $v=w\wedge\tau^{\hat{l}}$ and use the bounded
derivatives of $g$ in addition to (13a) to obtain

$\displaystyle\mathbb{E}_{t,x}\left[g^{r}(v,X^{\hat{l}}_{v})-g^{r}(t,x)\right]=$
$\displaystyle\mathbb{E}_{t,x}\left[\int\limits_{t}^{v}\mathcal{A}^{\hat{l}}g^{r}(s,X^{\hat{l}}_{s})ds\right]=\mathbb{E}_{t,x}\left[\int\limits_{t}^{v}-H\left(r,s,\hat{l}(s,X^{\hat{l}}_{s})\right)ds\right].$

Now the usage of boundary conditions from (14a), the boundedness as discussed
in Remark 4 and the dominated convergence theorem (or even monotone
convergence if $\lambda\leq 0$ because $H\geq 0$) leads to the desired form

$\displaystyle g^{r}(t,x)=$
$\displaystyle\mathbb{E}_{t,x}\left[\int\limits_{t}^{\tau^{\hat{l}}}H\left(r,s,\hat{l}(s,X^{\hat{l}}_{s})\right)ds+\lambda\left(1-\alpha\right)\right]$
$\displaystyle=$
$\displaystyle\mathbb{E}_{t,x}\left[\int\limits_{t}^{\tau^{\hat{l}}}e^{-\delta(s-r)}\hat{l}(s,X_{s}^{\hat{l}})-\lambda\beta
e^{-\beta(s-r)}ds+\lambda\left(1-\alpha\right)\right].$

This gives indeed

$g^{t}(t,x)=g(t,t,x)=J(t,x,\hat{l}).$

In the second part we prove that $\hat{l}$ is indeed an equilibrium control.
Therefore, we consider, as proposed in Definition 1, an arbitrary admissible
control $l\in\Theta$, $h>0$ and arbitrary but fixed $t,x\in[0,\infty)$ in
order to define $l_{h}(s,y)$ for $s\in[t,\infty)$ and $y\in[0,\infty)$ as in
(10). Set $v=(t+h)\wedge\tau^{l_{h}}(t,x)$, then we have

$\displaystyle J(t,x,\hat{l})-J(t,x,l_{h})$
$\displaystyle=g(t,t,x)-\mathbb{E}_{t,x}\left[\int\limits_{t}^{v}H\left(t,s,l_{h}(s,X^{l_{h}}_{s})\right)ds+\lambda\left(1-\alpha\right)\right]$
$\displaystyle-\mathbb{E}_{t,x}\left[\mathbb{E}_{v,X^{l_{h}}_{v}}\left[\int\limits_{v}^{\tau^{l_{h}}(t,x)}H\left(t,s,l_{h}(s,X^{l_{h}}_{s})\right)ds\right]\right].$

The first subsequent equality holds due to the definition of $l_{h}$ and the
strong Markov property. For the second one we need that (16a) holds true for
different values in the first and second argument of $g$ ,

$\displaystyle\mathbb{E}_{t,x}\left[\mathbb{E}_{v,X^{l_{h}}_{v}}\left[\int\limits_{v}^{\tau^{l_{h}}(t,x)}H\left(t,s,l_{h}(s,X^{l_{h}}_{s})\right)ds+\lambda(1-\alpha)\right]\right]$
$\displaystyle=\mathbb{E}_{t,x}\left[\mathbb{E}_{v,X^{l_{h}}_{v}}\left[\int\limits_{v}^{\tau^{\hat{l}}(v,X^{l_{h}}_{v})}H\left(t,s,\hat{l}(s,X^{\hat{l}}_{s})\right)ds+\lambda(1-\alpha)\right]\right]$
$\displaystyle=\mathbb{E}_{t,x}\left[g(t,v,X^{l_{h}}_{v})\right].$

Moreover, taking this into account and using the version of Itô’s formula from
[Leobacher et al., 2015, p. 7, Thm. 2.9], we end up with

$\displaystyle\begin{split}&J(t,x,\hat{l})-J(t,x,l_{h})\\\
&=\mathbb{E}_{t,x}\left[g(t,t,x)-g(t,v,X^{l_{h}}_{v})-\int\limits_{t}^{v}H\left(t,s,l_{h}(s,X^{l_{h}}_{s})\right)ds\right]\\\
&=-\mathbb{E}_{t,x}\left[\int\limits_{t}^{v}\mathcal{A}^{l_{h}}g^{t}(s,X^{l_{h}}_{s})+H\left(t,s,l_{h}(s,X^{l_{h}}_{s})\right)ds\right].\end{split}$
(17)

The expectation of the stochastic integral part of (17) is zero. This holds
true since $\frac{\partial}{\partial x}g$ and $\sigma$ are continuous in $t$
and $x$ and the controlled process is adapted and continuous, we get that the
integrand of the stochastic integral is predictable. Furthermore,
$\frac{\partial}{\partial x}g$ and $\sigma$ are bounded in $t$ and $x$ and
hence the stochastic integral is indeed a martingale for general admissible
controls $l$.
Further, we divide (17) by $h$ and consider the limit inferior for $h$ to
zero. Using bounded convergence to interchange limit and expectation and
Lebesgue’s differentiation theorem [Wheeden and Zygmund, 1977, p. 100,
108-109], we obtain the second equality in (18)

$\displaystyle\begin{split}&\liminf\limits_{h\to
0}\frac{J(t,x,\hat{l})-J(t,x,l_{h})}{h}\\\ &=\liminf\limits_{h\to
0}\mathbb{E}_{t,x}\left[-\frac{1}{h}\int\limits_{t}^{v}\mathcal{A}^{l_{h}}g^{t}(s,X^{l_{h}}_{s})+H\left(t,s,l_{h}(s,X^{l_{h}}_{s})\right)ds\right]\\\
&=-\left(\mathcal{A}^{l}g^{t}(t,x)+H\left(t,t,l(t,x)\right)\right)\geq
0.\end{split}$ (18)

Moreover, we use (12a) for the last inequality in (18) to finally obtain that
$\hat{l}$ is an equilibrium control. This gives that
$V(t,x)=J(t,x,\hat{l})=g(t,t,x)$. ∎

######  Remark 6.

In the theory of stochastic optimal control, the classical solution approach
suggests to prove a verification theorem and present a suitable solution
satisfying the needed regularity requirements. For the sake of completeness,
we want to mention the result by [Lindensjö, 2019], where it is shown that,
under regularity assumptions, it is even necessary for the equilibrium to
solve the extended HJB system. We adopt the notation of the following theorem
and just mention that our defined extended HJB equation is a special case of
the extended HJB system II in [Lindensjö, 2019]. We omit further details and
refer to the original source.

######  Theorem 2 ( [Lindensjö, 2019, p. 431, Theorem 3.12] ).

A regular equilibrium $(\hat{u},V_{\hat{u}},f_{\hat{u}},g_{\hat{u}})$ solves
the extended HJB system II.

##  4 Classical solution approach

In this section we are able to reveal an explicit solution to the above stated
time-inconsistent problem under the assumption of constant coefficients, i.e.,
$\mu(t,x)\equiv\mu$ and $\sigma(t,x)\equiv\sigma$.

######  Remark 7.

Please notice that in the subsequent analysis we can send $\beta\to 0$ and do
not violate its general validity if $\mu>L_{max}$. Since in this instance the
function $V_{2}(x;b)$, which will be constructed subsequently, solves the
associated ODE (39) also in the limiting case.
On the contrary, if $\mu\leq L_{max}$ this function $V_{2}(x;b)$ is zero in
the limiting case and cannot fulfil the condition
$\lim\limits_{x\to\infty}V_{2}(x;b)=-\lambda$ in (39). Therefore, the present
problem for $\mu\leq L_{max}$ reduces, as $\beta\to 0$, to the unconstrained
classical dividend problem with the value function shifted by the constant
$\lambda(1-\alpha)$ and optimal threshold height $b^{*}=\bar{b}$.

We follow the classical line of attack as in [Zhao et al., 2014]. Since we
have to solve the equation (12a),

$\displaystyle\sup\limits_{l\in[0,L_{max}]}\left\\{\frac{\partial}{\partial
t}g^{t}(t,x)+(\mu-l)\frac{\partial}{\partial
x}g^{t}(t,x)+\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial
x^{2}}g^{t}(t,x)+l-\lambda\beta\right\\}=$
$\displaystyle\frac{\partial}{\partial
t}g^{t}(t,x)+\mu\frac{\partial}{\partial
x}g^{t}(t,x)+\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial
x^{2}}g^{t}(t,x)-\lambda\beta+\sup\limits_{l\in[0,L_{max}]}\left\\{l\left(1-\frac{\partial}{\partial
x}g^{t}(t,x)\right)\right\\}=0,$

we obtain as in the classical problem for the maximizing argument

$\displaystyle l(t,x)=\left\\{\begin{array}[]{lr}0,&\text{if
}\frac{\partial}{\partial x}g^{t}(t,x)>1,\\\ L_{max},&\text{if
}\frac{\partial}{\partial x}g^{t}(t,x)\leq 1.\end{array}\right.$

Moreover, if we assume that our function is increasing and concave in $x$,
there exists a threshold level $b$ such that $\frac{\partial}{\partial
x}g(t,t,x)>1$ for $0\leq x<b$ and $\frac{\partial}{\partial x}g(t,t,x)\leq 1$
for $x\geq b$. Under this premise and taking account of Remark 5, fulfilling
the following equation is sufficient for the solution of the extended HJB
equation in Definition 2:

$\displaystyle\left\\{\begin{array}[]{ll}0=\frac{\partial}{\partial
t}g(r,t,x)+\mu\frac{\partial}{\partial
x}g(r,t,x)+\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial
x^{2}}g(r,t,x)-\lambda\beta e^{-\beta(t-r)},\\\ \quad\text{ for
}(r,t,x)\in[0,\infty)\times[0,\infty)\times[0,b),\\\
0=\frac{\partial}{\partial t}g(r,t,x)+(\mu-L_{max})\frac{\partial}{\partial
x}g(r,t,x)+\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial
x^{2}}g(r,t,x)+e^{-\delta(t-r)}L_{max}\\\ -\lambda\beta e^{-\beta(t-r)},\\\
\quad\text{ for }(r,t,x)\in[0,\infty)\times[0,\infty)\times[b,\infty),\\\
g(r,t,0)=\lambda(1-\alpha),\\\ \quad\text{ for
}(r,t)\in[0,\infty)\times[0,\infty),\\\
\lim\limits_{x\to\infty}g(r,t,x)=e^{-\delta(t-r)}\frac{L_{max}}{\delta}-\lambda
e^{-\beta(t-r)}+\lambda(1-\alpha),\\\ \quad\text{ for
}(r,t)\in[0,\infty)\times[0,\infty).\end{array}\right.$ (28)

In order to take advantage of Theorem 1 we make the following well guessed
ansatz:

$\nu(r,t,x;b)=e^{-\delta(t-r)}V_{1}(x;b)+e^{-\beta(t-r)}V_{2}(x;b)+\lambda(1-\alpha),$
(29)

where $V_{1}(x;b)$ is the solution of

$\displaystyle\left\\{\begin{array}[]{lr}-\delta
v_{1}(x)+\mu\frac{\partial}{\partial
x}v_{1}(x)+\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial
x^{2}}v_{1}(x)=0,&0\leq x<b\\\ -\delta v_{1}(x)+(\mu-
L_{max})\frac{\partial}{\partial
x}v_{1}(x)+\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial
x^{2}}v_{1}(x)+L_{max}=0,&x\geq b,\\\ v_{1}(0)=0,\\\
\lim\limits_{x\to\infty}v_{1}(x)=\frac{L_{max}}{\delta},\end{array}\right.$
(34)

and $V_{2}(x;b)$ solves

$\displaystyle\left\\{\begin{array}[]{lr}-\beta
v_{2}(x)+\mu\frac{\partial}{\partial
x}v_{2}(x)+\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial
x^{2}}v_{2}(x)-\lambda\beta=0,&0\leq x<b\\\ -\beta v_{2}(x)+(\mu-
L_{max})\frac{\partial}{\partial
x}v_{2}(x)+\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial
x^{2}}v_{2}(x)-\lambda\beta=0,&x\geq b,\\\ v_{2}(0)=0,\\\
\lim\limits_{x\to\infty}v_{2}(x)=-\lambda.\end{array}\right.$ (39)

The above equations lead to the following solutions

$\displaystyle
V_{1}(x;b)=\left\\{\begin{array}[]{lr}A_{1}(b)e^{a_{1}x}+A_{2}(b)e^{a_{2}x},&\text{if
}0\leq x<b\\\
\frac{L_{max}}{\delta}+B_{1}(b)e^{b_{1}x}+B_{2}(b)e^{b_{2}x},&\text{if }x\geq
b,\end{array}\right.$

where

$\displaystyle
a_{1,2}=-\frac{\mu}{\sigma^{2}}\pm\sqrt{\frac{\mu^{2}}{\sigma^{4}}+\frac{2\delta}{\sigma^{2}}},\quad\quad
b_{1,2}=-\frac{(\mu-L_{max})}{\sigma^{2}}\pm\sqrt{\frac{(\mu-
L_{max})^{2}}{\sigma^{4}}+\frac{2\delta}{\sigma^{2}}}.$

Due to the conditions $v_{1}(0)=0$ and
$\lim\limits_{x\to\infty}v_{1}(x)=L_{max}/\delta$, we obtain
$A_{2}(b)=-A_{1}(b)$ and $B_{1}(b)=0$. On the other hand we have for $V_{2}$:

$\displaystyle
V_{2}(x;b)=\left\\{\begin{array}[]{lr}-\lambda+C_{1}(b)e^{c_{1}x}+C_{2}(b)e^{c_{2}x},&\text{if
}0\leq x<b\\\ -\lambda+D_{1}(b)e^{d_{1}x}+D_{2}(b)e^{d_{2}x},&\text{if }x\geq
b,\end{array}\right.$

where

$\displaystyle
c_{1,2}=-\frac{\mu}{\sigma^{2}}\pm\sqrt{\frac{\mu^{2}}{\sigma^{4}}+\frac{2\beta}{\sigma^{2}}},\quad\quad
d_{1,2}=-\frac{(\mu-L_{max})}{\sigma^{2}}\pm\sqrt{\frac{(\mu-
L_{max})^{2}}{\sigma^{4}}+\frac{2\beta}{\sigma^{2}}}.$

From the conditions $v_{2}(0)=0$ and
$\lim\limits_{x\to\infty}v_{2}(x)=-\lambda$, we obtain $C_{2}(b)=\lambda-
C_{1}(b)$ and $D_{1}(b)=0$. Furthermore, we ask for the following smooth fit
conditions for $i=1,2$,

$\displaystyle V_{i}(b-;b)=V_{i}(b+;b),$ (40) $\displaystyle
V_{i}^{\prime}(b-;b)=V_{i}^{\prime}(b+;b)\text{ and }$ (41)
$\displaystyle\frac{\partial}{\partial x}\nu(t,t,b-;b)=1.$ (42)

Note that the condition (42) is in contrast to the smooth fit condition of the
classical dividend problem. In our case, we have to make sure that
$V^{\prime}_{1}(b-;b)+V^{\prime}_{2}(b-;b)=1$.

For $V_{1}$ the first two conditions (40) and (41) immediately yield

$\displaystyle A_{1}(b)$
$\displaystyle=-\frac{L_{max}}{\delta}\frac{b_{2}}{(a_{1}-b_{2})e^{a_{1}b}+(b_{2}-a_{2})e^{a_{2}b}},$
$\displaystyle B_{2}(b)$
$\displaystyle=\frac{A_{1}(b)}{b_{2}e^{b_{2}b}}(a_{1}e^{a_{1}b}-a_{2}e^{a_{2}b}).$

Since $b_{2}<0$, $a_{1}-b_{2}>0$ and $b_{2}-a_{2}>0$, we get that $A_{1}(b)>0$
and $B_{2}(b)<0$. Analogously, (40) and (41) give for $V_{2}$ that

$\displaystyle C_{1}(b)$
$\displaystyle=\frac{\lambda(c_{2}-d_{2})e^{c_{2}b}}{(d_{2}-c_{1})e^{c_{1}b}+(c_{2}-d_{2})e^{c_{2}b}},$
$\displaystyle D_{2}(b)$ $\displaystyle=\frac{C_{1}(b)e^{c_{1}b}+(\lambda-
C_{1}(b))e^{c_{2}b}}{e^{d_{2}b}}.$

Because $c_{2}-d_{2}<0$ and $\lambda<0$, we obtain $\lambda-C_{1}(b)<0$,
$C_{1}(b)<0$ and $D_{2}(b)<0$. The last condition (42) reveals the exact
candidate equilibrium threshold level $b$. Indeed, we are looking for a value
$b$ such that

$\displaystyle
A_{1}(b)(a_{1}e^{a_{1}b}-a_{2}e^{a_{2}b})+C_{1}(b)c_{1}e^{c_{1}b}+(\lambda-
C_{1}(b))c_{2}e^{c_{2}b}=$
$\displaystyle\frac{L_{max}}{\delta}\frac{(-b_{2})(a_{1}e^{a_{1}b}-a_{2}e^{a_{2}b})}{(a_{1}-b_{2})e^{a_{1}b}+(b_{2}-a_{2})e^{a_{2}b}}+\lambda\frac{(c_{2}-c_{1})d_{2}e^{(c_{1}+c_{2})b}}{(d_{2}-c_{1})e^{c_{1}b}+(c_{2}-d_{2})e^{c_{2}b}}=1.$

Therefore, we analyze the following function

$\displaystyle
G(b)=\frac{L_{max}}{\delta}\frac{(-b_{2})(a_{1}e^{a_{1}b}-a_{2}e^{a_{2}b})}{(a_{1}-b_{2})e^{a_{1}b}+(b_{2}-a_{2})e^{a_{2}b}}+\lambda\frac{(c_{2}-c_{1})d_{2}e^{(c_{1}+c_{2})b}}{(d_{2}-c_{1})e^{c_{1}b}+(c_{2}-d_{2})e^{c_{2}b}}-1.$
(43)

We have to find a $b>0$ such that $G(b)=0$. If this root exists, we denote it
by $b^{*}$.

######  Remark 8.

Concerning the dependence of the equilibrium threshold level $b^{*}$ on
$\beta$ we obtain that: if $\beta$ tends to zero, then $c_{1}=0$,
$c_{2}=-(2\mu)/{\sigma^{2}}$, and if $\mu-L_{max}>0$, we obtain $d_{1}=0$ and
$d_{2}=-2(\mu-L_{max})/{\sigma^{2}}<0$.
If $\mu-L_{max}\leq 0$ we obtain $d_{1}=-2(\mu-L_{max})/{\sigma^{2}}\geq 0$
and $d_{2}=0$. Which means that the function $G(b)$ reduces to

$\displaystyle
G_{0}(b)=\left\\{\begin{array}[]{lr}\frac{L_{max}}{\delta}\frac{(-b_{2})(a_{1}e^{a_{1}b}-a_{2}e^{a_{2}b})}{(a_{1}-b_{2})e^{a_{1}b}+(b_{2}-a_{2})e^{a_{2}b}}+\lambda\frac{2\frac{\mu}{\sigma^{2}}(\mu-
L_{max})}{-(\mu-L_{max})e^{\frac{2\mu}{\sigma^{2}}b}-L_{max}}-1,&\text{if
}\mu>L_{max}\\\
\frac{L_{max}}{\delta}\frac{(-b_{2})(a_{1}e^{a_{1}b}-a_{2}e^{a_{2}b})}{(a_{1}-b_{2})e^{a_{1}b}+(b_{2}-a_{2})e^{a_{2}b}}-1,&\text{if
}\mu\leq L_{max}.\end{array}\right.$ (46)

Therefore, the threshold level tends to the solution of the equation
$G_{0}(b)=0$, as $\beta$ tends to zero. Moreover, if $\mu\leq L_{max}$, we
arrive at the optimal threshold of the classical problem

$\bar{b}=\frac{1}{a_{1}-a_{2}}\ln\left(\frac{a_{2}(b_{2}-a_{2})}{a_{1}(b_{2}-a_{1})}\right),$
(47)

since ruin is for sure. This corresponds to the same control strategy as if
$\lambda$ equals zero. Furthermore, the statement of Lemma 1 also holds true
for the special case $\beta\to 0$, which just means that $G$ is replaced by
$G_{0}$.

######  Lemma 1.

If

$(-b_{2})\frac{L_{max}}{\delta}+\lambda d_{2}-1>0$

then G(b)=0 admits a unique positive solution denoted by $b^{*}$.

###### Proof.

Analogously as in [Zhao et al., 2014, p. 6, Lemma 4.1], we know that
$G(0)=(-b_{2})L_{max}/{\delta}+\lambda d_{2}-1>0$ by assumption and $G$ is
strictly monotonically decreasing since $G^{\prime}(b)<0$ for all $b\geq 0$.
This can be seen using $\lambda<0$. Finally, we have that

$\lim\limits_{b\to\infty}G(b)=\frac{a_{1}b_{2}}{b_{2}-a_{1}}\frac{L_{max}}{\delta}-1<0,$

since

$\frac{a_{1}b_{2}}{b_{2}-a_{1}}\frac{L_{max}}{\delta}-1=\frac{a_{1}(-b_{2})}{a_{1}-b_{2}}\left(\frac{L_{max}}{\delta}+\frac{1}{b_{2}}-\frac{1}{a_{1}}\right)<0.$

The last inequality is true because the first factor is positive and
$\left(L_{max}/{\delta}+1/{b_{2}}-1/{a_{1}}\right)<0$, see also [Asmussen and
Taksar, 1997, p. 4, Lemma 2.1]. Altogether, we obtain the desired result. ∎

Finally, we get the following candidate solution:

$\displaystyle\nu(r,t,x)=\left\\{\begin{array}[]{lr}e^{-\delta(t-r)}A_{1}(b^{*})(e^{a_{1}x}-e^{a_{2}x})\\\
+e^{-\beta(t-r)}(-\lambda+C_{1}(b^{*})e^{c_{1}x}+(\lambda-
C_{1}(b^{*}))e^{c_{2}x})\\\ +\lambda(1-\alpha),&\text{if }0\leq x<b^{*}\\\
e^{-\delta(t-r)}(\frac{L_{max}}{\delta}+B_{2}(b^{*})e^{b_{2}x})\\\
+e^{-\beta(t-r)}(-\lambda+D_{2}(b^{*})e^{d_{2}x})+\lambda(1-\alpha),&\text{if
}x\geq b^{*},\end{array}\right.$ (53)

with
$\nu(r,t,x)=\nu(r,t,x;b^{*})\in\mathcal{C}^{0,1,1}\left([0,\infty)\times[0,\infty)\times[0,\infty)\right)$
on the whole domain since the coefficients are chosen such that (40) and (41)
are fulfilled.

######  Theorem 3.

For the time-inconsistent stochastic optimal control problem with objective
(6a) and state process (7) the function $\nu$ defined in (53) solves the
equilibrium HJB equation in Definition 2 in case that

$(-b_{2})\frac{L_{max}}{\delta}+\lambda d_{2}-1>0.$ (54)

Furthermore, the function $V(t,x)=\nu(t,t,x)$ is an equilibrium value function
where $b^{*}$ is the unique positive solution of the equation $G(b)=0$ and

$\displaystyle\hat{l}(t,x)=\left\\{\begin{array}[]{lr}0,&\text{if }0\leq
x<b^{*},\\\ L_{max},&\text{if }x\geq b^{*},\end{array}\right.$ (57)

is the equilibrium control according to Definition 1.

###### Proof.

We have to make sure that $\nu$ in (53) and $\hat{l}$ in (57) satisfy the
assumptions for the usage of Theorem 1. Using the explicit representation (57)
we immediately observe that $\hat{l}\in\Theta$. Hence, we start with showing
that $\nu$ in (53) together with (40), (41) and (42) solves the extended HJB
equation. For that reason it is straightforward to insert $\nu$ into the
equation (28) and observe that it is a solution (using (40) and (41)).
Further, using the assumption (54), Lemma 1 yields that there exists a unique
positve $b^{*}$ such that (42) is fulfilled. Moreover, this gives that
$\frac{\partial^{2}}{\partial
x^{2}}\nu(t,t,b^{*}-)=\frac{\partial^{2}}{\partial x^{2}}\nu(t,t,b^{*}+)$.
Using (28) at $(t,t,b^{*}-)$ and $(t,t,b^{*}+)$ yields that
$-\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial
x^{2}}\nu(t,t,b^{*}-)=\mu\frac{\partial}{\partial
x}\nu(t,t,b^{*}-)-\lambda\beta$ and
$-\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial x^{2}}\nu(t,t,b^{*}+)=(\mu-
L_{max})\frac{\partial}{\partial x}\nu(t,t,b^{*}-)+L_{max}-\lambda\beta$.
Together with (42), we obtain that $\nu(t,t,x)$ is twice continuously
differentiable. Summing up, we have that
$\nu(r,t,x)\in\mathcal{C}^{0,1,1}\left([0,\infty)^{3}\right)\ \cap\
\mathcal{C}^{0,1,2}\left([0,\infty)\times[0,\infty)\times([0,\infty)\backslash\\{b^{*}\\})\right)$
in addition to the fact that $\nu(t,t,x)$ is twice continuously differentiable
in $x$.

Furthermore, we show that $\nu(t,t,x)$ is concave for $x\in[0,\infty)$, i.e.
$\frac{\partial^{2}}{\partial x^{2}}\nu(t,t,x)<0$. First of all, we
immediately observe that $\frac{\partial^{2}}{\partial x^{2}}\nu(t,t,x)<0$ for
$x\geq b^{*}$, since $B_{2}(b^{*})<0$ and $D_{2}(b^{*})<0$. On the other hand,
we have that

$\frac{\partial^{3}}{\partial
x^{3}}V_{2}(b^{*};b^{*})=(-\lambda)e^{(c_{1}+c_{2})b^{*}}(c_{1}-c_{2})\frac{c_{1}c_{2}(c_{1}+c_{2})-(c_{1}^{2}+c_{1}c_{2}+c_{2}^{2})d_{2}}{{}^{(}c_{1}-d_{2})e^{c_{1}b^{*}}+(d_{2}-c_{2})e^{c_{2}b^{*}}}>0,$

and since $\frac{\partial^{4}}{\partial x^{4}}V_{2}(x;b^{*})<0$ for $0\leq
x<b^{*}$, we obtain that $\frac{\partial^{3}}{\partial x^{3}}V_{2}(x;b^{*})>0$
for $0\leq x<b^{*}$. This, together with $\frac{\partial^{3}}{\partial
x^{3}}V_{1}(x;b^{*})>0$, yields that $\frac{\partial^{3}}{\partial
x^{3}}\nu(t,t,x)>0$ for $0\leq x<b^{*}$. This in turn implies for $0\leq
x<b^{*}$ that $\frac{\partial^{2}}{\partial
x^{2}}\nu(t,t,x)<\frac{\partial^{2}}{\partial x^{2}}\nu(t,t,b^{*})$. Since
$\nu(t,t,x)$ is twice continuously differentiable in $x$ we obtain
$\frac{\partial^{2}}{\partial
x^{2}}\nu(t,t,b^{*}-)=\frac{\partial^{2}}{\partial x^{2}}\nu(t,t,b^{*}+)<0$.
Finally, this yields that $\nu(t,t,x)$ is concave.

We already know that $\frac{\partial}{\partial x}\nu(t,t,b^{*})=1$ for
$b^{*}>0$. Combining this with concavity, we obtain that
$\frac{\partial}{\partial x}\nu(t,t,b^{*})>1$ for $0\leq x<b^{*}$ and
$\frac{\partial}{\partial x}\nu(t,t,b^{*})\leq 1$ for $x\geq b^{*}$. On top of
that, the function $\nu(t,t,x)$ is strictly monotonically increasing since
$\frac{\partial}{\partial x}\nu(t,t,x)>0$. For the case $0\leq x<b^{*}$ this
holds because the first derivative is even larger than one. For $x\geq b^{*}$
we have the desired monotonicity from $\frac{\partial}{\partial
x}\nu(t,t,x)=B_{2}(b^{*})b_{2}e^{b_{2}x}+D_{2}(b^{*})d_{2}e^{d_{2}x}>0$.

Altogether, the behaviour of the first and second derivatives of $\nu(t,t,x)$
yields that, if this function fulfils the equation (28), it solves equations
(12a) and (13a). On top of that $\nu(r,t,x)$ is bounded in $t$ and $x$, hence
we have that (15) implies (14a) as mentioned in Remark 5. Since

$\lim\limits_{\begin{subarray}{c}x\to\infty\\\
t\to\infty\end{subarray}}\nu(r,t,x)=\lambda(1-\alpha)$

for every fixed $r\in[0,\infty)$ we have:

$\displaystyle\lim\limits_{w\to\infty}\mathbb{E}_{t,x}\left[\nu(r,w\wedge\tau^{\hat{l}},X^{\hat{l}}_{w\wedge\tau^{\hat{l}}})\right]$
$\displaystyle=\lim\limits_{w\to\infty}\mathbb{E}_{t,x}\left[\nu(r,w\wedge\tau^{\hat{l}},X^{\hat{l}}_{w\wedge\tau^{\hat{l}}})1_{\\{\tau^{\hat{l}}<\infty\\}}+\nu(r,w\wedge\tau^{\hat{l}},X^{\hat{l}}_{w\wedge\tau^{\hat{l}}})1_{\\{\tau^{\hat{l}}=\infty\\}}\right]$
$\displaystyle=\mathbb{E}_{t,x}\left[\lambda\left(1-\alpha\right)1_{\\{\tau^{\hat{l}}<\infty\\}}+\lambda\left(1-\alpha\right)1_{\\{\tau^{\hat{l}}=\infty\\}}\right]$
$\displaystyle=\lambda\left(1-\alpha\right).$

The last step, for verifying that $\nu$ and $\hat{l}$ fulfil the requirements
of the verification theorem, is to show that $\nu$ and its relevant
derivatives are bounded. Since,

$\frac{\partial}{\partial x}\nu(r,t,x)\leq
e^{(\delta\vee\beta)r}e^{-(\delta\wedge\beta)t}\frac{\partial}{\partial
x}\nu(t,t,x)\leq e^{(\delta\vee\beta)r}\frac{\partial}{\partial x}\nu(t,t,0),$

and $0<\nu(t,t,0)<\infty$ is a constant, we obtain that the first derivative
in $x$ is bounded. Furthermore, we observe, using $b_{2}<0$ and $d_{2}<0$,
that $\nu$ is bounded, and by building the respective derivatives that also
$\frac{\partial^{2}}{\partial x^{2}}\nu(r,t,x)$ and $\frac{\partial}{\partial
t}\nu(r,t,x)$ are bounded.
Altogether, we can apply Theorem 1 and obtain that $\nu$ is the equilibrium
value function and $\hat{l}$ is the associated equilibrium control. ∎

Analogous to Theorem 3 we can deal with the remaining parameter constellation.

######  Proposition 1.

If $(-b_{2})L_{max}/{\delta}+\lambda d_{2}-1\leq 0,$ then $b^{*}=0$ and the
equilibrium control becomes $\hat{l}(t,x)=L_{max}$. The function $\nu$ from
(53) now reads as

$\nu(r,t,x)=e^{-\delta(t-r)}\frac{L_{max}}{\delta}(1-e^{b_{2}x})+e^{-\beta(t-r)}\lambda(e^{d_{2}x}-1)+\lambda(1-\alpha).$

This function solves the extended HJB equation in Definition 2 and
$V(t,x)=\nu(t,t,x)$ is an equilibrium value function associated to
$\hat{l}(t,x)=L_{max}$.

###### Proof.

We proceed exactly in the same way as in the proof for Theorem 3. It can be
verified that $\nu$ solves the equation (28) with $b=b^{*}=0$. Furthermore,
since $\frac{\partial}{\partial x}\nu(r,t,x)>0$ and
$\frac{\partial^{2}}{\partial x^{2}}\nu(r,t,x)<0$ for all
$(r,t,x)\in[0,\infty)\times[0,\infty)\times[0,\infty)$ in addition to
$\frac{\partial}{\partial x}\nu(r,t,0)\leq 1$ for all
$(r,t)\in\\{(q,s)\in\mathbb{R}^{2}|0\leq q\leq s<\infty\\}$ we obtain that

$\frac{\partial}{\partial x}\nu(r,t,x)\leq 1$

for all $(r,t)\in\\{(q,s)\in\mathbb{R}^{2}|0\leq q\leq s<\infty\\}$ and
$x\in[0,\infty)$. Hence, $\nu$ and $\hat{l}(t,x)$ fulfil (12a) and (13a).
Moreover, $\nu$ is bounded and satisfies the correct boundary condition, which
yields that it fulfils (14a). Finally,
$\nu\in\mathcal{C}^{0,1,2}\left([0,\infty)^{3}\right)$ and it satisfies the
extended HJB equation. Since $\nu$, $\frac{\partial}{\partial t}\nu$,
$\frac{\partial}{\partial x}\nu$ and $\frac{\partial^{2}}{\partial x^{2}}\nu$
are all bounded in $t$ and $x$, the requirements of Theorem 1 are fulfilled.
Consequently, $\nu$ is the equilibrium value function and $\hat{l}(t,x)$ the
associated equilibrium control. ∎

###  4.1 Parameter areas

The arising inequality in Theorem 3 splits the parameter space into different
areas, such that this classification gives rise to an interpretation. First of
all, we know that the classical dividend threshold $\bar{b}$ is positive if
$(-b_{2})L_{max}/{\delta}-1>0$ which is equivalent to

$\frac{\frac{\sigma^{2}}{2}}{\mu}<\frac{L_{max}}{\delta},$ (58)

see Schmidli (2008). Otherwise, if (58) is not fulfilled, we have that
$\bar{b}=0$. Furthermore, note that by the properties of $G(b)$ in (43) we
obtain that the equilibrium dividend threshold $b^{*}$ fulfills
$b^{*}\geq\bar{b}$. As a matter of fact, we obtain the following cases, which
are also illustrated in Table 1.

  1. 1.

If (58) is fulfilled and $\bar{b}>0$, then we obtain also that $b^{*}>0$. The
condition (58) can be viewed as an advantageous parameter configuration which
motivates the controller to take less risk and to payout dividends only at a
surplus level larger than zero in order to prevent ruin for a longer time. The
prospect is advantageous since the maximal reachable value of dividends (the
right-hand side of (58)) is relatively large compared to the variance of the
uncontrolled surplus divided by its mean (the left-hand side of (58)). Note
that $\mathbb{E}(X_{t})=\mu t$ and $Var(X_{t})=\sigma^{2}t$. Moreover, we
observe that if $\lambda$ decreases, then the threshold level increases since
one wants to prevent ruin and the associated larger penalty even more.

  2. 2.

Whereas if (58) is not fulfilled and hence $\bar{b}=0$, the future prospect is
such that the controller in the classical problem ($\lambda=0$) would pay out
dividends at any surplus level. Especially if $\mu-L_{max}$ is negative, this
strategy can be interpreted as the ruin strategy since ruin is certain in this
case. But given that a penalty exists ($\lambda<0$), in case of ruin, we can
distinguish the following two cases:

     * i)

If $\lambda<\Lambda$ for

$\Lambda\coloneqq\left(1+b_{2}\frac{L_{max}}{\delta}\right)\frac{1}{d_{2}},$

the impact of the penalty is too severe and we obtain that $b^{*}>0$. This
yields that dividends are paid out only for positive surplus levels as in the
first case. We observe that the penalty in case of ruin has a stronger
deterrent effect and therefore one wants to prevent ruin.

     * ii)

On the other hand, if $\lambda\geq\Lambda$ meaning that the weight on the
penalty is moderate, we obtain that also $b^{*}=0$. In this case the penalty
is not severe enough and the equilibrium strategy suggests to pay out
dividends at any surplus level even if it is more likely that ruin occurs
compared to the use of a positive threshold level.

$\displaystyle(-b_{2})\frac{L_{max}}{\delta}-1>0\quad\Rightarrow\quad\bar{b}>0$ |  | $\Rightarrow\quad b^{*}>0$
---|---|---
$\displaystyle(-b_{2})\frac{L_{max}}{\delta}-1\leq 0\quad\Rightarrow\quad\bar{b}=0$ |  | if $\lambda<\Lambda\quad\Rightarrow\quad b^{*}>0$
| if $\lambda\geq\Lambda\quad\Rightarrow\quad b^{*}=0$
Table 1: The nature of the strategy depending on the parameter configuration.

##  5 Meeting the constraint

In the following we discuss the motivating problem, i.e. dividend maximization
including the constraint (4). We try to find a solution which fulfils the
constraint for fixed $(t,x)$, this means in a _precommitment sense_. Up to now
we have found for every fixed $\lambda\leq 0$ an equilibrium threshold
strategy $b^{*}_{\lambda}$ and an associated value function. If

$\lambda<\left(1+b_{2}\frac{L_{max}}{\delta}\right)\frac{1}{d_{2}}=\Lambda$
(59)

there exists a unique equilibrium threshold level $b^{*}_{\lambda}>0$ for the
associated time-inconsistent problem. If $\lambda\geq\Lambda$ the equilibrium
threshold is $b^{*}_{\lambda}=0$. Concerning (59) note that $d_{2}<0$.
In the case $\beta>0$, we observe from equation (43), $G(b)=0$, that if
$\lambda\to-\infty$, $b$ tends to $+\infty$. The factor next to $\lambda$ is
negative and consequently monotonically increasing in $b$, so this term
vanishes for $b\to\infty$. The same holds true if $\beta=0$ and $\mu>L_{max}$.
On the other hand, if $\beta=0$ and $\mu\leq L_{max}$, equation (46), now
$G_{0}(b)=0$, yields that $b^{*}=\bar{b}$, where $\bar{b}$ is the classical
threshold level. Hence, we consider in the following $\beta>0$. The case
$\mu>L_{max}$ and $\beta\to 0$ can be analogously treated.
But, for $\mu\leq L_{max}$ and $\beta\to 0$, we have for all $\lambda\leq 0$
that $b^{*}=\bar{b}$. Therefore, the ruin probability is one in this case.
This yields that the constraint cannot be fulfilled and the equilibrium
threshold level coincides with the classical threshold level.
Focusing on the constraint (4), we can rewrite $V_{2}(x;b^{*}_{\lambda})$,
which is constructed to solve (39),

$\tilde{V}_{2}(x;b^{*}_{\lambda}):=\frac{1}{\lambda}\left(V_{2}(x;b^{*}_{\lambda})+\lambda\right)=\mathbb{E}_{t,x}\left[e^{-\beta(\tau^{\hat{l}^{\lambda}}-t)}\right].$
(60)

For an arbitrary threshold level $b\geq 0$ we define:

$w(x,b):=\mathbb{E}_{t,x}\left[e^{-\beta(\tau^{b}-t)}\right]\left(=\mathbb{E}_{0,x}\left[e^{-\beta\tau^{b}}\right]\right).$
(61)

Here $\tau^{b}=\inf\\{s\geq t|X^{b}_{s}\leq 0\,,X^{b}_{t}=x\\}$ and $X^{b}$ is
the process (7) with $dL_{s}=L_{max}1_{\\{X^{b}_{s}\geq b\\}}ds$. A solution
to equation (39), incorporating (40) and (41), for $b\geq 0$ has the form
$(\lambda w(x,b)-\lambda)$. Hence, we have,

$\displaystyle
w(x,b)=\left\\{\begin{array}[]{lr}\frac{(d_{2}-c_{2})e^{-c_{1}(b-x)}+(c_{1}-d_{2})e^{-c_{2}(b-x)}}{(d_{2}-c_{2})e^{-c_{1}b}+(c_{1}-d_{2})e^{-c_{2}b}},&\text{if
}0\leq x<b,\\\
\frac{(c_{1}-c_{2})e^{d_{2}(x-b)}}{(d_{2}-c_{2})e^{-c_{1}b}+(c_{1}-d_{2})e^{-c_{2}b}},&\text{if
}x\geq b.\end{array}\right.$

This can be seen using a Feynman-Kac type argument. According to (40) and
(41), $w(x,b)$ is continuously differentiable in $x$. For $b_{\lambda}^{*}$
the equilibrium threshold with $G(b_{\lambda}^{*})=0$, we have
$\tilde{V}_{2}(x;b^{*}_{\lambda})=w(x,b_{\lambda}^{*})$. On top of this, if
$\mu>L_{max}$, we obtain

$\lim\limits_{\beta\to 0}w(x,0)=\psi(x,L_{max}),$

from (5).

###  5.1 Determining the matching $\lambda$ for $\beta>0$

In the following, we can present two propositions which are based on
Proposition 3.2 and Proposition 4.1 from [Hernández and Junca, 2015].
Certainly, our approach also covers the particular situation of
$\beta=\delta$. Therefore, we obtain similar results as in [Hernández and
Junca, 2015], but for the diffusion model with bounded dividend rate.

######  Proposition 2.

If $\Lambda\leq 0$, we have that for all $b>0$ there exists a unique
$\lambda<\Lambda$ such that $b$ is the equilibrium threshold level for (57)
with parameter $\lambda$.
On the other hand, if $\Lambda>0$, we have that for all $b\geq\bar{b}$, (where
$\bar{b}$ is specified in (47)), there exists a unique $\lambda\leq 0$ such
that $b$ is the equilibrium threshold level for (57) with parameter $\lambda$.

###### Proof.

Since the relation between $\lambda$ and the equilibrium threshold results
from equation (43), we define for a given $b\geq 0$ the corresponding value of
$\lambda$ to be

$\lambda(b):=\frac{1-N_{1}(b)}{N_{2}(b)}.$

Where we have that

$N_{1}(b)=\frac{L_{max}}{\delta}\frac{(-b_{2})(a_{1}e^{a_{1}b}-a_{2}e^{a_{2}b})}{(a_{1}-b_{2})e^{a_{1}b}+(b_{2}-a_{2})e^{a_{2}b}}$

and

$N_{2}(b)=\frac{(c_{2}-c_{1})d_{2}e^{(c_{1}+c_{2})b}}{(d_{2}-c_{1})e^{c_{1}b}+(c_{2}-d_{2})e^{c_{2}b}}.$

For these two functions we have that $N_{1}(b)>0$, $N^{\prime}_{1}(b)<0$,
$N_{2}(b)<0$ and $N^{\prime}_{2}(b)>0$.
Moreover, if $\Lambda\leq 0$, then $1+b_{2}L_{max}/{\delta}=1-N_{1}(0)\geq 0$,
which in turn gives that $1-N_{1}(b)\geq 0$. Finally, this yields that
$\lambda^{\prime}(b)<0$ and we obtain the first statement, since
$\lambda(0)=\Lambda\leq 0$ and $\lambda(b)\to-\infty$ for $b\to\infty$.
For the second statement let $\Lambda>0$, we obtain that
$1-N_{1}\left(\bar{b}\right)=0$ and since $N_{1}(b)$ is monotonically
decreasing, we get that $1-N_{1}(b)\geq 0$ for all $b\geq\bar{b}$. This along
the lines of the classical dividend problem. As before, we see that
$\lambda^{\prime}(b)<0$ for all $b\geq\bar{b}$. This, in combination with
$\lambda\left(\bar{b}\right)=0$ and $\lambda(b)\to-\infty$ for $b\to\infty$,
proves the second assertion. ∎

######  Remark 9.

For fulfilling the required constraint, we have to choose the pair
$(b^{*},\lambda^{*})$ dependent on the current state $x$ (setting $t=0$). This
means that we have a solution of the constrained problem in a precommitment
sense. This notion is well explained in [Delong, 2018]. Since we cannot fulfil
the constraint $w(x,b)\leq\alpha$ for a finite $b>0$ if
$x\leq\bar{x}=\ln(\alpha)/{c_{2}}$, we only consider the case $x>\bar{x}$. In
the special case $x=\ln(\alpha)/{c_{2}}$ we are only able to meet the
constraint if we never pay out dividends, which corresponds to a threshold at
infinity.

######  Proposition 3.

For every given $0<\alpha\leq 1$ and every $x>\bar{x}$ there exists a pair
$(b^{*},\lambda^{*})$ with $b^{*}\geq 0$ and $\lambda^{*}\leq 0$, such that
$b^{*}$ is the equilibrium threshold level for (57) with parameter
$\lambda^{*}$ and further for this triple $(x,b^{*},\lambda^{*})$ it holds
that

$\displaystyle w(x,b^{*})\leq\alpha,$ (62)

and

$\displaystyle\lambda^{*}(w(x,b^{*})-\alpha)=0.$ (63)

###### Proof.

In the following we consider the different cases:
1.) $\Lambda\leq 0$ with 1a) $w(x,0)\leq\alpha$ and 1b) $w(x,0)>\alpha$; on
the other hand 2.) $\Lambda>0$ with 2a) $w(x,\bar{b})\leq\alpha$ and 2b)
$w(x,\bar{b})>\alpha$. Moreover, note that since $x>\bar{x}$ we always have
that $\alpha>\lim\limits_{b\to\infty}w(x,b)=e^{c_{2}x}$. Firstly, let
$\Lambda\leq 0$ and $w(x,0)\leq\alpha$, then we can take the equilibrium
threshold to be $b^{*}=0$ since the constraint is already fulfilled. Moreover,
$\lambda^{*}$ has to be zero, such that the second part of the objective
function vanishes. Since $b^{*}=0$, all choices of $\lambda\in[\Lambda,0]$
yield an equilibrium threshold at level zero. Hence, we can take
$\lambda^{*}=0$.
If $w(x,0)>\alpha$, and consequently

$w(x,0)>\alpha>\lim\limits_{b\to\infty}w(x,b)=e^{c_{2}x},$

we have that there exists a unique $b^{*}>0$ such that $w(x,b^{*})=\alpha$
(remember $\frac{\partial}{\partial b}w(x,b)<0$). Using Proposition 2, there
exists a unique $\lambda^{*}<\Lambda$ such that $b^{*}$ is an equilibrium
threshold.
Secondly, if $\Lambda>0$ and $w(x,\bar{b})\leq\alpha$, we have to take
$b^{*}=\bar{b}$ such that we can meet the constraint. As before we can set
$\lambda^{*}=0$ since $\bar{b}$ is the equilibrium threshold level for
$\lambda^{*}=0$. On the other hand, if $w(x,\bar{b})>\alpha$ and hence

$w(x,\bar{b})>\alpha>\lim\limits_{b\to\infty}w(x,b)=e^{c_{2}x},$

then there exists a unique $b^{*}>\bar{b}$ such that $w(x,b^{*})=\alpha$. Due
to Proposition 2, there exists a unique $\lambda^{*}<0$, such that $b^{*}$ is
an associated equilibrium threshold. ∎

Since we have revealed an equilibrium control strategy and the associated
equilibrium value function for the objective (6a), we cannot use the
properties of a maximal reward function as in (2). Hence, standard duality
statements do not apply.
We summarize our achievements as follows. For a given ruin constraint level
$0<\alpha\leq 1$ we can identify the correct threshold level
$b^{*}=b^{*}(\alpha)$, such that the constraint (4) is fulfilled for a given
initial state $x_{0}>\bar{x}$. Subsequently, we can specify a value
$\lambda^{*}$, such that $b^{*}$ is the threshold level of an equilibrium
control strategy (57) for the equilibrium value function
$V(t,x_{0})=\nu(t,t,x_{0})$ as in (53).
Notice that the equilibrium threshold level and the associated $\lambda^{*}$
is depending on the current state $x_{0}$, in such a way that (62) and (63)
for this $\alpha$ is only fulfilled at $x_{0}$.
In contrast to [Hernández and Junca, 2015], we only obtain one part of the
analogous duality statements. Let $\hat{l}^{\lambda}$ be the equilibrium
control for a problem with general parameter $\lambda$, and
$\hat{l}^{\lambda^{*}}$ the corresponding equilibrium control such that the
constraint for $\alpha$ is fulfilled in $x_{0}>\bar{x}:$

$\inf_{\lambda\leq 0}J(t,x_{0},\hat{l}^{\lambda})\leq
J(t,x_{0},\hat{l}^{\lambda^{*}})=\mathbb{E}_{t,x_{0}}\left[\int\limits_{t}^{\tau^{\hat{l}^{\lambda^{*}}}}e^{-\delta(s-t)}\hat{l}^{\lambda^{*}}(s,X_{s}^{\hat{l}^{\lambda^{*}}})ds+\lambda^{*}\left(e^{-\beta(\tau^{\hat{l}^{\lambda^{*}}}-t)}-\alpha\right)\right]\\\
=\mathbb{E}_{t,x_{0}}\left[\int\limits_{t}^{\tau^{\hat{l}^{\lambda^{*}}}}e^{-\delta(s-t)}\hat{l}^{\lambda^{*}}(s,X_{s}^{\hat{l}^{\lambda^{*}}})ds\right]\leq\sup_{L}\mathbb{E}_{t,x_{0}}\left[\int\limits_{t}^{\tau^{L}}e^{-\delta(s-t)}dL_{s}\right].$

The last equality holds true, because the penalty term vanishes in $x_{0}$ for
$\hat{l}^{\lambda^{*}}$ as in (63).

######  Remark 10.

The functions $\tilde{V}_{2}$ and $w$ defined in (60) and (61) characterize
the evolution of the constraint, or more precisely, of the level $\alpha$.
Suppose that the level $\alpha$, the initial surplus $x_{0}>\bar{x}$ and the
corresponding equilibrium strategy $\hat{l}^{\lambda^{*}}$ with threshold
height $b^{*}_{\lambda^{*}}(=b^{*}(x_{0}))$ are initially fixed. Then, at some
point in time $t>0$ with $\tau^{\hat{l}^{\lambda^{*}}}(0,x_{0})>t$, the
constraint level

$\alpha_{t}:=e^{-\beta
t}w\left(X_{t}^{\hat{l}^{\lambda^{*}}},b^{*}(x_{0})\right),$

is reached for $\left(t,X_{t}^{\hat{l}^{\lambda^{*}}}\right)$ and original
$(x_{0},\,\alpha)-$admissible strategy $\hat{l}^{\lambda^{*}}$. From the
Markov property we get that
$\left(\alpha_{t\wedge\tau^{\hat{l}^{\lambda^{*}}}(0,x_{0})}\right)_{t\geq 0}$
is an $\\{\mathcal{F}_{t}\\}_{t\geq 0}$-martingale. Consequently, the strategy
$\hat{l}^{\lambda^{*}}$ is the particular equilibrium control for which the
constraint level $\alpha_{t}$ is sharp if the process is restarted at
$X_{t}^{\hat{l}^{\lambda^{*}}}$. This means that the quality of the strategy
is preserved if the constraint is adapted. The martingale property of
$\left(\alpha_{t\wedge\tau^{\hat{l}^{\lambda^{*}}}(0,x_{0})}\right)_{t\geq 0}$
shows that these adaptions are consistent over time.

###  5.2 Quality of the solution

As mentioned above we do not obtain classical duality results, nevertheless
using our explicit form we are able to optimize within the class of threshold
strategies. Namely, we can rewrite the constrained optimization problem

$\sup_{b\geq 0}V_{1}(x;b),\quad\text{s.t. }w(x,b)\leq\alpha\quad\text{ for
fixed }x\in(\bar{x},\infty).$ (64)

Using that $\bar{b}$ is the maximizing argument of the classical problem,
together with $\frac{\partial}{\partial b}w(x,b)<0$ for all $b\geq 0$, we
obtain for an arbitrary but fixed $x>\bar{x}$ that (64) simplifies to

$\sup_{b\geq\tilde{b}}V_{1}(x;b).$

Whereas $\tilde{b}$ is defined as follows

$\tilde{b}=\begin{cases}\bar{b},\text{ if }w(x,\bar{b})\leq\alpha,\\\
b^{\prime},\text{ where }w(x,b^{\prime})=\alpha.\end{cases}$ (65)

Finally, the optimal threshold of the constrained problem will be this
$\tilde{b}$. Since $\tilde{b}\geq\bar{b}$ and $\frac{\partial}{\partial
b}V_{1}(x;b)<0$ for all $b>\bar{b}$. Note, that the optimal threshold
$\tilde{b}$ depends on the initial value $x$. If $x\leq\bar{x}$ the constraint
cannot be fulfilled.
Recalling the proof of Proposition 3, we observe that our chosen equilibrium
threshold of the penalized problem coincides with the optimal threshold
$\tilde{b}$. This just means that if $w(x,\bar{b})\leq\alpha$, then we can
take $b^{*}=\bar{b}(=\tilde{b})$ which is the optimal threshold of the
constrained problem and $\lambda^{*}=0$. This yields
$\nu(t,t,x;b^{*})=V_{1}(x;b^{*})$.
If $w(x,\bar{b})>\alpha$, then we take $b^{*}=\tilde{b}(>\bar{b})$, which is
again the optimal threshold in the constrained problem. In this case, we have
to take $\lambda^{*}<0$ for the penalized reward function such that $b^{*}$
corresponds to an equilibrium control, but for the initially fixed
$x>\bar{x}$. Again we obtain $\nu(t,t,x;b^{*})=V_{1}(x;b^{*})$ by Proposition
3.
Note, that the constraint is only fulfilled for this initial $x$ and the
penalized function $\nu$ will change for a different $x$ since $\lambda^{*}$
is not zero and depends on $x$. In the end, $\tilde{b}$ from (64) is optimal
in the family of threshold strategies, but our analysis shows that for a
certain $\lambda^{*}$ it is also an equilibrium control from the general class
of strategies $\Theta$ for the time-inconsistent penalized problem.
On top of this, we have computed the optimal threshold for the penalized
problem with $\lambda\leq 0$ fixed

$\sup_{b\geq 0}\nu(t,t,x;b).$ (66)

The optimal threshold will depend on the initial $x$ and the solution is
therefore in a precommitment sense.
Altogether, we illustrate these findings in Figure 1, where the dashed red
line corresponds to the $x$-dependent optimal threshold of (66) for
$\lambda=-10$. Furthermore, in Figure 1 the green area is limited by the green
line $\bar{b}$ and by the black line which corresponds to those $b^{\prime}$
such that $w(x,b^{\prime})=\alpha$. Consequently, this boundary of the green
area exactly equals our optimal threshold level $\tilde{b}$ as in (65) for
different $x$. We can observe that if $x$ is close to $\bar{x}$ we cannot use
$\bar{b}$ anymore. Hence, we have to take a threshold which corresponds to the
equilibrium threshold for a value $\lambda^{*}<0$. This is for example
illustrated by the blue line for the value $\lambda^{*}=-10$.

Figure 1: The different threshold levels for different initial values.
Parameters: $\mu=2$, $\sigma=1$, $\delta=0.1$, $\beta=0.2$, $L_{max}=4$,
$\alpha=0.01$.

###  5.3 Concerning the ruin probability

As mentioned in Remark 8, if we send $\beta$ to zero, there are some
differences depending on the relation between $\mu$ and $L_{max}$. The above
procedure can be repeated, if we look at the limit $\beta$ to zero for
$\mu>L_{max}$.
But, if $\mu\leq L_{max}$, the equation $G_{0}(b)=0$, see (46), does not
depend on $\lambda$ anymore. The second term in $G_{0}(b)$ vanishes if $\beta$
tends to zero. This yields that for any $\lambda\leq 0$, we obtain the same
equilibrium threshold level, namely, the classical threshold $b=\bar{b}$. This
is because for any finite threshold strategy, the ruin probability is one.
Therefore, $V_{2}(x)=0$ and $\tilde{V}_{2}(x)=1$. Moreover, the classical
threshold is the only reasonable solution, since the reward function differs
only by a constant from the classical reward function in the time-consistent
situation.
We are aware of the fact that the reward function could be increased by using
a precommitment solution which takes account of the potentially high penalty
and has therefore a ruin probability smaller than one. The drawback of such
solutions is that one has to restrict (or even specify) the strategies in
order to verify the desired optimality in that class of controls.

##  6 Graphical illustrations

In this short section we present a numerical example, which displays the
dependence of the equilibrium threshold on the differing discount rate $\beta$
and the parameter $\lambda$. In the corresponding plots one nicely observes
that the proposed _type of duality_ considerations are highly legitimate.

###  6.1 The case $\mu>L_{max}$

In the following, we consider the exemplary parameters displayed in Table 2.

$\mu$ | $\sigma$ | $\delta$ | $L_{max}$ | $\alpha$
---|---|---|---|---
2 | 1 | 0.1 | 1.9 | 0.5
Table 2: Parameter set

If we do not vary $\beta$ and $\lambda$, we set $\beta=0.2$ and $\lambda=-50$.
The change of the threshold, if we alter the underlying discount rate $\beta$,
is illustrated in Figure 2. We can observe that if we send $\beta$ to infinity
the equilibrium threshold tends to the classical optimal threshold level
$\bar{b}$. Furthermore, we are able to reproduce the results from [Thonhauser
and Albrecher, 2007], namely, in our setting we have to take $\beta=\delta$
besides $\lambda(-\delta)=\tilde{\Lambda}$, where $\tilde{\Lambda}$ denotes
the constant in the penalty term of [Thonhauser and Albrecher, 2007]. The
corresponding optimal control threshold is denoted by $b(\delta)$.
On the other hand, the dependence on $\lambda$ for a fixed value of $\beta$ is
depicted in Figure 3. The behaviour of the equilibrium threshold for different
values of $\lambda$ is similar for positive $\beta$ and for the limit $\beta$
to zero. Clearly, this is due to the fact that the equation $G_{0}(b)=0$ still
depends on $\lambda$ if $\mu>L_{max}$, as already mentioned in Remark 8.

Figure 2: We illustrate the dependence of the equilibrium threshold on $\beta$
in the case $\mu>L_{max}$. Figure 3: The threshold dependent on $\lambda$ in
the case $\mu>L_{max}$.

###  6.2 The case $\mu\leq L_{max}$

We consider again the parameters from Table 2 with the exception of
$L_{max}=4$. If we send $\beta$ to zero, we obtain that $b^{*}=\bar{b}$ since
the equation $G_{0}(b)=0$ is independent of $\lambda$ as can be seen in Figure
4. Moreover, this yields that the solution $b^{*}$ equals the classical
threshold $\bar{b}$ for all $\lambda\leq 0$. As in the above case, the
equilibrium threshold tends to the classical optimal threshold level $\bar{b}$
for $\beta$ to infinity. Varying the value of $\lambda$ and keeping $\beta>0$
constant, we obtain a similar behaviour as in the previous case, which can be
observed in Figure 5. Furthermore, in Figure 5 (as in Figure 3 above) we
notice that the equilibrium threshold increases very slowly if we decrease
$\lambda$. In fact, the equilibrium threshold remains close to the classical
threshold $\bar{b}$ even for very small values of $\lambda$, in comparison to
the other parameters. This indicates that, even for relatively small
$\lambda$, the penalty in case of ruin is not deterrent enough to cause a high
threshold level.

Figure 4: Here we illustrate the dependence of the equilibrium threshold on
$\beta$ in the case $\mu\leq L_{max}$. Figure 5: The threshold dependent on
$\lambda$ in the case $\mu\leq L_{max}$ and for $\beta=0.2$.

##  7 Conclusion

In this contribution we solve a time-inconsistent variation of the dividend
maximization problem for a diffusion surplus process. This problem arises from
imposing a constraint on the Laplace transform of the ruin time in the
classical dividend problem. In order to include even a constraint on the ruin
probability, we consider different discount rates in the dividend and the
penalty term, which in fact leads to the time-inconsistency of the problem.
The presented verification theorem, which links an extended system of HJB-
equations to an equilibrium dividend strategy and its value function, is
applicable for state dependent diffusion coefficients. In the particular case
of constant coefficients - the classical diffusion approximation - we are able
to construct an explicit solution. This allows us to link the proposed reward
function to the constrained dividend problem. In fact, we are able to fulfil
the original constraint in a precommitment sense, i.e. for a given initial
surplus level. Moreover, the reward function of the problem including a
penalty coincides with the reward function of the classical dividend problem
at this given initial surplus, since the penalty term vanishes.
Finally, the obtained equilibrium strategy is of threshold type, which is
based on the restriction to Markovian controls and partly by the focus on an
infinite time horizon. This has the consequence that a constraint on the ruin
probability itself can only be feasible if the drift term of the surplus
process is larger than the maximal payout rate. Otherwise, the equilibrium is
achieved by ignoring the constraint. In conclusion, the time-inconsistent
approach allows us to assess the performance of a strategy in terms of an
equilibrium over time.

##  8 Funding Information

This research was funded in whole, or in part, by the Austrian Science Fund
(FWF) P 33317. For the purpose of open access, the author has applied a CC BY
public copyright licence to any Author Accepted Manuscript version arising
from this submission.

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