# Predicting the Last Zero before an exponential time of a Spectrally Negative
Lévy Process

Erik J. Baurdoux111Department of Statistics, London School of Economics and
Political Science. Houghton street, London, WC2A 2AE, United Kingdom. E-mail:
<EMAIL_ADDRESS>& José M. Pedraza222Department of Statistics, London
School of Economics and Political Science. Houghton street, London, WC2A 2AE,
United Kingdom. E-mail<EMAIL_ADDRESS>
###### Abstract

Given a spectrally negative Lévy process, we predict, in a $L_{1}$ sense, the
last passage time of the process below zero before an independent exponential
time. This optimal prediction problem generalises Baurdoux and Pedraza, (2020)
where the infinite horizon problem is solved. Using a similar argument as that
in Urusov, (2005), we show that this optimal prediction problem is equivalent
to solving an optimal prediction problem in a finite horizon setting.
Surprisingly (unlike the infinite horizon problem) an optimal stopping time is
based on a curve that is killed at the moment the mean of the exponential time
is reached. That is, an optimal stopping time is the first time the process
crosses above a non-negative, continuous and non-increasing curve depending on
time. This curve and the value function are characterised as a solution of a
system of non-linear integral equations which can be understood as a
generalisation of the free boundary equations (see e.g. Peskir and Shiryaev,
(2006) Chapter IV.14.1) in the presence of jumps. As an example, we calculate
numerically such curve in the Brownian motion case and a compound Poisson
process with exponential sized jumps perturbed by a Brownian motion.

Keywords: Lévy processes, optimal prediction, optimal stopping.

Mathematics Subject Classification (2000): 60G40, 62M20

##  1 Introduction

The study of last exit times has received much attention in several areas of
applied probability, e.g. risk theory, finance and reliability in the past few
years. Consider the Cramér–Lundberg process, a process consisting of a
deterministic drift and a compound Poisson process with only negative jumps
(see Figure 1), which is typically used to model the capital of an insurance
company. Of particular interest is the moment of ruin, $\tau_{0}$ which is
defined to refer to the first moment when the process becomes negative. Within
the framework of the insurance company having sufficient funds to endure
negative capital for a considerale amount of time, another quantity of
interest is the last time, $g$ that the process is below zero. In a more
general setting, we can consider a spectrally negative Lévy process instead of
the classical risk process. Several studies, for example Baurdoux, (2009) and
Chiu and Yin, (2005) studied the Laplace transform of the last time before an
exponential time that a spectrally negative Lévy process is below some given
level.

$X_{t}$$t$x$\tau_{0}$$g$ Figure 1: Cramér–Lundberg process with $\tau_{0}$,
the moment of ruin and $g$, the last zero.

Last passage time is increasingly becoming a vital factor in financial
modeling as shown in Madan et al., 2008a  and Madan et al., 2008b  where the
authors concludes that the price of a European put and call options, modelled
by non-negative and continuous martingales that vanish at infinity, can be
expressed in terms of the probability distributions of some last passage
times.

Another application of last passage times is in degradation models. Paroissin
and Rabehasaina, (2013) propose a spectrally positive Lévy process to model
the ageing of a device in which they consider a subordinator perturbed by an
independent Brownian motion. A motivation for considering this model is that
the presence of a Brownian motion can model small repairs of the device and
the jumps represent major deterioration. In the literature, the failure time
of a device is defined as the first hitting time of a critical level $b$. An
alternative approach is to consider instead, the last time that the process is
under the level $b$ since the paths of this process are not necessarily
monotone and this allows the process to return below the level $b$ after it
goes above $b$.

The aim of this work is to predict the last time a spectrally negative Lévy
process is below zero before an independent exponential time where the terms
“to predict” are understood to mean to find a stopping time that is closest
(in $L^{1}$ sense) to this random time. This problem is an example of the
optimal prediction problems which have been widely investigated by many.
Graversen et al., (2001) predicted the value of the ultimate maximum of a
Brownian motion in a finite horizon setting whereas Shiryaev, (2009) focused
on the last time of the attainment of the ultimate maximum of a (driftless)
Brownian motion and proceeded to show that it is equivalent to predicting the
last zero of the process in this setting. The work of the latter was
generalised by du Toit et al., (2008) for a linear Brownian motion. Bernyk et
al., (2011) studied the time at which a stable spectrally negative Lévy
process attains its ultimate supremum in a finite horizon of time and this was
later generalised by Baurdoux and van Schaik, (2014) for any Lévy process in
infinite horizon of time. Investigations on the time of the ultimate minimum
and the last zero of a transient diffusion process were carried out by Glover
et al., (2013) and Glover and Hulley, (2014) respectively within a subclass of
functions.

In Baurdoux and Pedraza, (2020) the last zero of a spectrally negative Lévy
process in a infinite horizon setting is predicted. It is shown that the an
optimal stopping time that minimises the last zero of a spectrally negative
Lévy process with drift is the first time the Lévy process crosses above a
fixed level $a^{*}\geq 0$ which is characterised in terms of the cumulative
distribution of the overall infimum of the process. As is in the case in the
Canadisation of American type options (see e.g. Carr, (1998)), given the
memoryless property of the exponential distribution, one would expect that the
generalisation of the aforementioned problem to an exponential time horizon
would result in an infinite horizon optimal prediction problem, and hence have
a non time dependent solution. However, it turns out that this is not the
case. Indeed, we show the existence of a continuous, non-increasing and non-
negative boundary such that an optimal stopping time is given by the first
passage time, before the median of the exponential time, above such curve. The
proof relies on solving an equivalent (finite horizon) optimal stopping
problem that depends on time and the process itself. Moreover, based on the
ideas of du Toit et al., (2008) we characterise the boundary and the value
function as the unique solutions of a system of non-linear integral equations.
Such system can be thought as a generalisation of the free boundary equation
(see e.g. Peskir and Shiryaev, (2006), Section 14) allowing for the presence
of jumps. We consider two examples where numerical calculations are
implemented to find the optimal boundary.

This paper is organised as follows. In Section 2 we introduce some important
notation regarding Lévy processes and we outline some known fluctuation
identities that will be useful later. We then formulate the optimal prediction
problem and we prove that it is equivalent to an optimal stopping problem.
Section 3 is dedicated to the solution of the optimal stopping problem. The
main result of this paper is stated in Theorem 3.9 and its proof is detailed
in Section 3.1. The last section makes use of Theorem 3.9 to find numerical
solution of the optimal stopping problem for the Brownian motion with drift
case and a compound Poisson process perturbed by a Brownian motion.

##  2 Prerequisites and Formulation of the Problem

We start this section by introducing some important notations and we give an
overview of some fluctuation identities of spectrally negative Lévy processes.
Readers can refer to Bertoin, (1998), Sato, (1999) or Kyprianou, (2014) for
more details about Lévy processes.

A Lévy process $X=\\{X_{t},t\geq 0\\}$ is an almost surely càdlàg process that
has independent and stationary increments such that $\mathbb{P}(X_{0}=0)=1$.
Every Lévy process $X$ is also a strong Markov $\mathbb{F}$-adapted process.
For all $x\in\mathbb{R}$, denote $\mathbb{P}_{x}$ as the law of $X$ when
started at the point $x\in\mathbb{R}$, that is,
$\mathbb{E}_{x}(\cdot)=\mathbb{E}(\cdot|X_{0}=x)$. Due to the spatial
homogeneity of Lévy processes, the law of $X$ under $\mathbb{P}_{x}$ is the
same as that of $X+x$ under $\mathbb{P}$.

Let $X$ be a spectrally negative Lévy process, that is, a Lévy process
starting from $0$ with only negative jumps and non-monotone paths, defined on
a filtered probability space $(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$
where $\mathbb{F}=\\{\mathcal{F}_{t},t\geq 0\\}$ is the filtration generated
by $X$ which is naturally enlarged (see Definition 1.3.38 in Bichteler,
(2002)). We suppose that $X$ has Lévy triplet $(\mu,\sigma,\Pi)$ where
$\mu\in\mathbb{R}$, $\sigma\geq 0$ and $\Pi$ is a measure (Lévy measure)
concentrated on $(-\infty,0)$ satisfying $\int_{(-\infty,0)}(1\wedge
x^{2})\Pi(\textup{d}x)<\infty$.

Let $\psi$ be the Laplace exponent of $X$ defined as

$\displaystyle\psi(\beta):=\log(\mathbb{E}(e^{\beta X_{1}})).$

Then $\psi$ exists in $\mathbb{R}_{+}$, it is strictly convex and infinitely
differentiable with $\psi(0)=0$ and $\psi(\infty)=\infty$. From the
Lévy–Khintchine formula, we know that $\psi$ takes the form

$\displaystyle\psi(\beta)=-\mu\beta+\frac{1}{2}\sigma^{2}\beta^{2}+\int_{(-\infty,0)}(e^{\beta
x}-1-\beta x\mathbb{I}_{\\{x>-1\\}})\Pi(\textup{d}x)$

for all $\beta\geq 0$. Moreover, from Lévy–Itô decomposition, we know that
there exists a Brownian Motion $B$ and an independent Poisson random measure
on $\mathbb{R}_{+}\times\mathbb{R}$ with intensity
$\textup{d}t\times\Pi(\textup{d}y)$ such that for each $t\geq 0$,

$\displaystyle X_{t}=-\mu t+\sigma
B_{t}+\int_{[0,t]}\int_{(-\infty,-1)}yN(\textup{d}s,\textup{d}y)+\int_{[0,t]}\int_{(-1,0)}y[N(\textup{d}s,\textup{d}y)-\Pi(\textup{d}y)\textup{d}s].$

Denote $\tau_{a}^{+}$ as the first time the process $X$ is above the level
$a\in\mathbb{R}$, i.e.,

$\displaystyle\tau_{a}^{+}=\inf\\{t>0:X_{t}>a\\}.$

Then it can be shown that its Laplace transform is given by

$\displaystyle\mathbb{E}(e^{-q\tau_{a}^{+}}\mathbb{I}_{\\{\tau_{a}^{+}<\infty\\}})=e^{-\Phi(q)a},$
(1)

where $\Phi$ corresponds to the right inverse of $\psi$, which is defined by

$\displaystyle\Phi(q)=\sup\\{\theta\geq 0:\psi(\theta)=q\\}$

for any $q\geq 0$.

Now we introduce the scale functions. This family of functions is the key to
the derivation of fluctuation identities for spectrally negative Lévy
processes. The notation used is mainly based on Kuznetsov et al., (2013) and
Kyprianou, (2014) (see Chapter 8). For $q\geq 0$, the function $W^{(q)}$ is
such that $W^{(q)}=0$ for $x<0$ and $W^{(q)}$ is characterised on $[0,\infty)$
as a strictly increasing and continuous function whose Laplace transform
satisfies

$\displaystyle\int_{0}^{\infty}e^{-\beta
x}W^{(q)}(x)\textup{d}x=\frac{1}{\psi(\beta)-q},\qquad\text{for
}\beta>\Phi(q).$

We further define the function $Z^{(q)}$ by

$\displaystyle Z^{(q)}(x)=1+q\int_{0}^{x}W^{(q)}(y)\textup{d}y.$

Denote $\tau_{0}^{-}$ as the first passage time of $X$ of the set
$(-\infty,0)$, that is,

$\displaystyle\tau_{0}^{-}=\inf\\{t>0:X_{t}<0\\}.$

It turns out that the Laplace transform of $\tau_{0}^{-}$ can be written in
terms of the scale functions. Specifically,

$\displaystyle\mathbb{E}_{x}(e^{-q\tau_{0}^{-}}\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})=Z^{(q)}(x)-\frac{q}{\Phi(q)}W^{(q)}(x)$
(2)

for all $q\geq 0$ and $x\in\mathbb{R}$. It can be shown that the paths of $X$
are of finite variation if only if

$\displaystyle\sigma=0\qquad\text{and}\qquad\int_{(-1,0)}y\Pi(\textup{d}y)<\infty.$

In such case, we may write

$\displaystyle\psi(\lambda)=\delta\lambda-\int_{(-\infty,0)}(1-e^{\lambda
y})\Pi(dy),$

where

$\displaystyle\delta:=-\mu-\int_{(-1,0)}x\Pi(\textup{d}x).$ (3)

Note that monotone processes are excluded from the definition of spectrally
negative Lévy processes so we assume that $\delta>0$ when $X$ is of finite
variation. The value of $W^{(q)}$ at zero depends on the path variation of
$X$. In the case where $X$ is of infinite variation we have that
$W^{(q)}(0)=0$, otherwise

$\displaystyle W^{(q)}(0)=\frac{1}{\delta}.$ (4)

For any $a\in\mathbb{R}$ and $q\geq 0$, the $q$-potential measure of $X$
killed upon entering the set $[a,\infty)$ is absolutely continuous with
respect to the Lebesgue measure. A density is given for all $x,y\leq a$ by

$\displaystyle\int_{0}^{\infty}e^{-qt}\mathbb{P}_{x}\left(X_{t}\in\textup{d}y,t<\tau_{a}^{+}\right)\textup{d}t=[e^{-\Phi(q)(a-x)}W^{(q)}(a-y)-W^{(q)}(x-y)]\textup{d}y.$
(5)

Let $g_{\theta}$ be the last passage time below zero before an exponential
time, i.e.

$\displaystyle g_{\theta}=\sup\\{0\leq t\leq\widetilde{e}_{\theta}:X_{t}\leq
0\\},$ (6)

where $\widetilde{e}$ is an exponential random variable with parameter
$\theta\geq 0$. Here, we use the convention that an exponential random
variable with parameter $0$ is taken to be infinite with probability $1$. In
the case of $\theta=0$, we simply denote $g=g_{0}$.

Note that $g_{\theta}\leq\widetilde{e}_{\theta}<\infty$ $\mathbb{P}$-a.s. for
all $\theta>0$. However, in the case where $\theta=0$, $g$ could be infinite.
Therefore, we assume that $\theta>0$ throughout this paper. Moreover, we have
that $g_{\theta}$ has finite moments for all $\theta>0$.

######  Remark 2.1.

Since $X$ is a spectrally negative Lévy process, we can exclude the case of a
compound Poisson process and hence the only way of exiting the set
$(-\infty,0]$ is by creeping upwards. This tells us that
$X_{g_{\theta}-}=X_{g_{\theta}}=0$ in the event of
$\\{g_{\theta}<\widetilde{e}_{\theta}\\}$ and that $g_{\theta}=\sup\\{0\leq
t\leq\widetilde{e}_{\theta}:X_{t}<0\\}$ holds $\mathbb{P}$-a.s.

Clearly, up to any time $t\geq 0$ the value of $g_{\theta}$ is unknown (unless
$X$ is trivial), and it is only with the realisation of the whole process that
we know that the last passage time below $0$ has occurred. However, this is
often too late: typically, at any time $t\geq 0$, we would like to know how
close we are to the time $g_{\theta}$ so we can take some actions based on
this information. We search for a stopping time $\tau_{*}$ of $X$ that is as
“close” as possible to $g_{\theta}$. Consider the optimal prediction problem

$\displaystyle V_{*}=\inf_{\tau\in\mathcal{T}}\mathbb{E}(|g_{\theta}-\tau|),$
(7)

where $\mathcal{T}$ is the set of all stopping times.

Note that the random time $g^{(\theta)}_{\theta}$ is only $\mathbb{F}$
measurable so it is not immediately obvious how to solve the optimal
prediction problem by using the theory of optimal stopping. Hence, in order to
find the solution (and hence prove Theorem 2.3) we solved an equivalent
optimal stopping problem. In the next Lemma we establish an equivalence
between the optimal prediction problem (7) and an optimal stopping problem.
This equivalence is mainly based on the work of Urusov, (2005).

######  Lemma 2.2.

Suppose that $\\{X_{t},t\geq 0\\}$ is a spectrally negative Lévy process. Let
$g_{\theta}$ be the last time that $X$ is below the level zero before an
exponential time $\widetilde{e}_{\theta}$ with $\theta>0$, as defined in (6).
Consider the optimal stopping problem given by

$\displaystyle
V=\inf_{\tau\in\mathcal{T}}\mathbb{E}\left(\int_{0}^{\tau}G^{(\theta)}(s,X_{s})\textup{d}s\right),$
(8)

where the function $G^{(\theta)}$ is given by $G^{(\theta)}(s,x)=1+2e^{-\theta
s}\left[\frac{\theta}{\Phi(\theta)}W^{(\theta)}(x)-Z^{(\theta)}(x)\right]$ for
all $x\in\mathbb{R}$. Then the stopping time which minimises (7) is the same
which minimises (8). In particular,

$\displaystyle V_{*}=V+\mathbb{E}(g_{\theta}).$ (9)

###### Proof.

Fix any stopping time $\tau\in\mathcal{T}$. We have that

$\displaystyle|g_{\theta}-\tau|=\int_{0}^{\tau}[2\mathbb{I}_{\\{g_{\theta}\leq
s\\}}-1]\textup{d}s+g_{\theta}.$

From Fubini’s theorem and the tower property of conditional expectations, we
obtain

$\displaystyle\mathbb{E}\left[\int_{0}^{\tau}\mathbb{I}_{\\{g_{\theta}\leq
s\\}}\textup{d}s\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{\infty}\mathbb{I}_{\\{s<\tau\\}}\mathbb{E}[\mathbb{I}_{\\{g_{\theta}\leq
s\\}}|\mathcal{F}_{s}]\textup{d}s\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{\tau}\mathbb{P}(g_{\theta}\leq
s|\mathcal{F}_{s})\textup{d}s\right].$

Note that in the event of $\\{\widetilde{e}_{\theta}\leq s\\}$, we have
$g_{\theta}\leq s$ so that

$\displaystyle\mathbb{P}(g_{\theta}\leq s|\mathcal{F}_{s})=1-e^{-\theta
s}+\mathbb{P}(g_{\theta}\leq s,\widetilde{e}_{\theta}>s|\mathcal{F}_{s}).$

On the other hand for $\\{\widetilde{e}_{\theta}>s\\}$, as a consequence of
Remark 2.1, the event $\\{g_{\theta}\leq s\\}$ is equal to $\\{X_{u}\geq
0\text{ for all }u\in[s,\widetilde{e}_{\theta}]\\}$ (up to a $\mathbb{P}$-null
set). Hence, we get that for all $s\geq 0$ that

$\displaystyle\mathbb{P}(g_{\theta}\leq
s,\widetilde{e}_{\theta}>s|\mathcal{F}_{s})$
$\displaystyle=\mathbb{P}(X_{u}\geq 0\text{ for all
}u\in[s,\widetilde{e}_{\theta}],\widetilde{e}_{\theta}>s|\mathcal{F}_{s})$
$\displaystyle=\mathbb{P}\left(\inf_{0\leq
u\leq\widetilde{e}_{\theta}-s}X_{u+s}\geq
0,\widetilde{e}_{\theta}>s|\mathcal{F}_{s}\right)$ $\displaystyle=e^{-\theta
s}\mathbb{P}_{X_{s}}\left(\underline{X}_{\widetilde{e}_{\theta}}\geq
0\right),$

where the last equality follows from the lack of memory property of the
exponential distribution and the Markov property for Lévy process. Hence, we
have that

$\displaystyle\mathbb{P}(g_{\theta}\leq
s,\widetilde{e}_{\theta}>s|\mathcal{F}_{s})$ $\displaystyle=e^{-\theta
s}F^{(\theta)}(X_{s}),$

where for all $x\in\mathbb{R}$,
$F^{(\theta)}(x)=\mathbb{P}_{x}(\underline{X}_{\widetilde{e}_{\theta}}\geq
0)$. Then, since $\widetilde{e}_{\theta}$ is independent of $X$, we have that
for $x\in\mathbb{R}$,

$\displaystyle
F^{(\theta)}(x)=\mathbb{P}_{x}(\underline{X}_{\widetilde{e}_{\theta}}\geq
0)=\mathbb{P}_{x}(\widetilde{e}_{\theta}<\tau_{0}^{-})=1-\mathbb{E}_{x}(e^{-\theta\tau_{0}^{-}}\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})=\frac{\theta}{\Phi(\theta)}W^{(\theta)}(x)-Z^{(\theta)}(x)+1,$

where the last equality follows from equation (2). Thus,

$\displaystyle\mathbb{P}(g_{\theta}\leq s|\mathcal{F}_{s})$
$\displaystyle=1-e^{-\theta s}+e^{-\theta
s}\left[\frac{\theta}{\Phi(\theta)}W^{(\theta)}(X_{s})-Z^{(\theta)}(X_{s})+1\right]$
$\displaystyle=1+e^{-\theta
s}\left[\frac{\theta}{\Phi(\theta)}W^{(\theta)}(X_{s})-Z^{(\theta)}(X_{s})\right].$

Therefore,

$\displaystyle V_{*}$
$\displaystyle=\inf_{\tau\in\mathcal{T}}\mathbb{E}(|g_{\theta}-\tau|)$
$\displaystyle=\mathbb{E}(g_{\theta})+\inf_{\tau\in\mathcal{T}}\mathbb{E}\left(\int_{0}^{\tau}[2\mathbb{P}(g_{\theta}\leq
s|\mathcal{F}_{s})-1]\textup{d}s\right)$
$\displaystyle=\mathbb{E}(g_{\theta})+\inf_{\tau\in\mathcal{T}}\mathbb{E}\left(\int_{0}^{\tau}\left(1+2e^{-\theta
s}\left[\frac{\theta}{\Phi(\theta)}W^{(\theta)}(X_{s})-Z^{(\theta)}(X_{s})\right]\right)\textup{d}s\right).$

The conclusion holds. ∎

Note that evaluating $\theta=0$, the function $G^{(0)}$ coincides with the
gain function found in Baurdoux and Pedraza, (2020) (see Lemma 3.2 and Remark
3.3). In order to find the solution to the optimal stopping problem (8) (and
hence (7)), we extend its definition to Lévy process (and hence strong Markov
process) $\\{(t,X_{t}),t\geq 0\\}$ in the following way. Define the function
$V:\mathbb{R}_{+}\times\mathbb{R}\mapsto\mathbb{R}$ as

$\displaystyle
V^{(\theta)}(t,x)=\inf_{\tau\in\mathcal{T}}\mathbb{E}_{t,x}\left(\int_{0}^{\tau}G^{(\theta)}(s+t,X_{s+t})\textup{d}s\right)=\inf_{\tau\in\mathcal{T}}\mathbb{E}\left(\int_{0}^{\tau}G^{(\theta)}(s+t,X_{s}+x)\textup{d}s\right).$
(10)

So that

$\displaystyle V_{*}=V^{(\theta)}(0,0)+\mathbb{E}(g_{\theta}).$

The next theorem states the solution of the optimal stopping theorem (10) and
hence the solution of (7).

######  Theorem 2.3.

Let $\\{X_{t},t\geq 0\\}$ be any spectrally negative Lévy process and
$\widetilde{e}_{\theta}$ an exponential random variable with parameter
$\theta>0$ independent of $\mathbb{F}$. There exists a non increasing and
continuous curve $b^{(\theta)}:[0,m_{\theta}]\mapsto\mathbb{R}_{+}$ such that
$b^{(\theta)}\geq h^{(\theta)}$, where
$h^{(\theta)}(t):=\inf\\{x\in\mathbb{R}:G^{(\theta)}(t,x)\geq 0\\}$ and the
infimum in (10) is attained by the stopping time

$\displaystyle\tau_{D}=\inf\\{t\in[0,m_{\theta}]:X_{t}\geq
b^{(\theta)}(t)\\},$

where $m_{\theta}=\log(2)/\theta$. Moreover, the function $b^{(\theta)}$ is
uniquely characterised as in Theorem 3.9.

Note that the proof of Theorem 2.3 is rather long and hence is split in a
series of lemmas. We dedicate Section 3 for that purpose.

##  3 Solution to the optimal stopping problem

In this section we solve the optimal stopping problem (10). The proof relies
on showing that $\tau_{D}$ defined in Theorem 2.3 is indeed an optimal
stopping time by using the general theory of optimal stopping and properties
of the function $V^{(\theta)}$. Hence, some properties of $b^{(\theta)}$ are
derived. The main contribution of this section (Theorem 3.9) characterises
$V^{(\theta)}$ and $b^{(\theta)}$ as the unique solution of a non-linear
system of integral equations within a certain family of functions.

Recall that the $V^{(\theta)}$ is given by

$\displaystyle
V^{(\theta)}(t,x)=\inf_{\tau\in\mathcal{T}}\mathbb{E}_{t,x}\left(\int_{0}^{\tau}G^{(\theta)}(s+t,X_{s+t})\textup{d}s\right).$

From the proof of Lemma 2.2 we note that $G^{(\theta)}$ can be written as,

$\displaystyle G^{(\theta)}(s,x)=1+2e^{-\theta s}[F^{(\theta)}(x)-1],$

where $F^{(\theta)}$ is the distribution function of the positive random
variable $-\underline{X}_{\widetilde{e}_{\theta}}$ given by

$\displaystyle
F^{(\theta)}(x)=\frac{\theta}{\Phi(\theta)}W^{(\theta)}(x)-Z^{(\theta)}(x)+1$
(11)

for all $x\in\mathbb{R}$. Now we give some intuitions about the function
$G^{(\theta)}$. Recall that for all $\theta\geq 0$, $W^{\theta}$ and
$Z^{(\theta)}$ are continuous and strictly increasing functions on
$[0,\infty)$ such that $W^{(\theta)}(x)=0$ and $Z^{(\theta)}(x)=1$ for
$x\in(-\infty,0)$. From the above, equation (11) and from the fact that
$F^{(\theta)}$ is a distribution function, we have that for a fixed $t\geq 0$,
the function $x\mapsto G^{(\theta)}(t,x)$ is strictly increasing and
continuous in $[0,\infty)$ with a possible discontinuity at $0$ depending on
the path variation of $X$. Moreover, we have that
$\lim_{x\rightarrow\infty}G^{(\theta)}(t,x)=1$ for all $t\geq 0$. For $x<0$
and $t\geq 0$, we have that the function $G^{(\theta)}$ takes the form
$G^{(\theta)}(t,x)=1-2e^{-\theta s}$. Similarly, from the fact that
$F^{(\theta)}(x)-1\leq 0$ for all $x\in\mathbb{R}$, we have that for a fixed
$x\in\mathbb{R}$ the function $t\mapsto G^{(\theta)}(t,x)$ is continuous and
strictly increasing on $[0,\infty)$. Furthermore, from the fact that $0\leq
F^{(\theta)}(x)\leq 1$, we have that the function $G$ is bounded by

$\displaystyle 1-2e^{-\theta t}\leq G^{(\theta)}(x,t)\leq 1$ (12)

which implies that $|G^{(\theta)}|\leq 1$. Recall that $m_{\theta}$ is defined
as the median of the random variable $\widetilde{e}_{\theta}$, that is,

$\displaystyle m_{\theta}=\frac{\log(2)}{\theta}.$

Hence from (12) we have that $G^{(\theta)}(t,x)\geq 0$ for all
$x\in\mathbb{R}$ and $t\geq m_{\theta}$. The above observations tell us that,
to solve the optimal stopping problem (10), we are interested in a stopping
time such that before stopping, the process $X$ spends most of its time in the
region where $G^{(\theta)}$ is negative, taking into account that $(t,X)$ can
live in the set
$\\{(s,x)\in\mathbb{R}_{+}\times\mathbb{R}:G^{(\theta)}(s,x)>0\\}$ and then
return back to the set
$\\{(s,x)\in\mathbb{R}_{+}\times\mathbb{R}:G^{(\theta)}(s,x)\leq 0\\}$. The
only restriction that applies is that if a considerable amount of time has
passed, then $\\{x\in\mathbb{R}:G^{(\theta)}(s,x)>0\\}=\mathbb{R}$ for all
$s\geq m_{\theta}$.

Recall that the function $h^{(\theta)}:\mathbb{R}_{+}\mapsto\mathbb{R}$ is
defined as

$\displaystyle h^{(\theta)}(t)=\inf\\{x\in\mathbb{R}:G^{(\theta)}(t,x)\geq
0\\}=\inf\\{x\in\mathbb{R}:F^{(\theta)}(x)\geq 1-\frac{1}{2}e^{\theta
t}\\}\qquad t\geq 0.$ (13)

Hence, we can see that the function $h^{(\theta)}$ is a non-increasing
continuous function on $[0,m_{\theta})$ such that $\lim_{t\uparrow
m_{\theta}}h^{(\theta)}(t)=0$ and $h^{(\theta)}(t)=-\infty$ for
$t\in[m_{\theta},\infty)$. Moreover, from the fact that $G^{(\theta)}(t,x)<0$
for $(t,x)\in[0,m_{\theta})\times(-\infty,0)$, we have that
$h^{(\theta)}(t)\geq 0$ for all $t\in[0,m_{\theta})$.

In order to characterise the stopping time that minimises (10), we first
derive some properties of the function $V^{(\theta)}$.

######  Lemma 3.1.

The function $V^{(\theta)}$ is non-drecreasing in each argument. Moreover,
$V^{(\theta)}(t,x)\in(-m_{\theta},0]$ for all $x\in\mathbb{R}$ and $t\geq 0$.
In particular, $V^{(\theta)}(t,x)<0$ for any $t\geq 0$ with
$x<h^{(\theta)}(t)$ and $V^{(\theta)}(t,x)=0$ for all
$(t,x)\in[m_{\theta},\infty)\times\mathbb{R}$.

###### Proof.

First, note that $V^{(\theta)}\leq 0$ follows by taking $\tau\equiv 0$ in the
definition of $V^{(\theta)}$. Moreover, since $G^{(\theta)}\geq 0$ on
$[m_{\theta},\infty)\times\mathbb{R}$ we have that $V^{(\theta)}$ vanishes on
$[m_{\theta},\infty)\times\mathbb{R}$ . The fact that $V^{(\theta)}$ is non-
decreasing in each argument follows from the non-decreasing property of the
functions $t\mapsto G^{(\theta)}(t,x)$ and $x\mapsto G^{(\theta)}(t,x)$ as
well as the monotonicity of the expectation. Moreover, using standard
arguments we can see that
$\\{(t,x)\in\mathbb{R}_{+}\times\mathbb{R}:x<h^{(\theta)}(t)\\}=\\{(t,x)\in\mathbb{R}_{+}\times\mathbb{R}:G^{(\theta)}(t,x)<0\\}\subset\\{(t,x)\in\mathbb{R}_{+}\times\mathbb{R}:V^{(\theta)}(t,x)<0\\}$.

Next we will show that $V^{(\theta)}(t,x)>-m_{\theta}$ for all
$(t,x)\in[0,m_{\theta})\times\mathbb{R}$ and for all $\theta>0$. Note that
$t<m_{\theta}$ if and only if $1-2e^{-\theta t}<0$. Then for all
$(s,x)\in\mathbb{R}_{+}\times\mathbb{R}$ we have that

$\displaystyle G^{(\theta)}(s,x)\geq 1-2e^{-\theta s}\geq(1-2e^{-\theta
s})\mathbb{I}_{\\{s<m_{\theta}\\}}.$

Hence, for all $x\in\mathbb{R}$ and $t<m_{\theta}$

$\displaystyle
V^{(\theta)}(t,x)\geq\inf_{\tau\in\mathcal{T}}\mathbb{E}\left(\int_{0}^{\tau}(1-2e^{-\theta(s+t)})\mathbb{I}_{\\{t+s<m_{\theta}\\}}\textup{d}s\right)=-\sup_{\tau\in\mathcal{T}}\mathbb{E}\left(\int_{0}^{\tau}(2e^{-\theta(s+t)}-1)\mathbb{I}_{\\{t+s<m_{\theta}\\}}\textup{d}s\right).$

The term in the last integral is non-negative, so we obtain for all
$t<m_{\theta}$ and $x\in\mathbb{R}$ that

$\displaystyle
V^{(\theta)}(t,x)\geq-\left(\int_{0}^{\infty}(2e^{-\theta(s+t)}-1)\mathbb{I}_{\\{t+s<m_{\theta}\\}}\textup{d}s\right)=-\left(\int_{0}^{m_{\theta}-t}(2e^{-\theta(s+t)}-1)\textup{d}s\right)>-m_{\theta}.$

∎

By using a dynamic programming argument and the fact that $V^{(\theta)}$
vanishes on the set $[0,m_{\theta})\times\mathbb{R}$ we can see that

$\displaystyle V^{(\theta)}(t,x)$
$\displaystyle=\inf_{\tau\in\mathcal{T}}\mathbb{E}_{t,x}\left(\int_{0}^{\tau\wedge(m_{\theta}-t)}G^{(\theta)}(s+t,X_{s+t})\textup{d}s\right).$

so that (since $|G^{(\theta)}|\leq 1$) we have that for all $t\geq 0$ and
$x\in\mathbb{R}$,

$\displaystyle\mathbb{E}_{t,x}\left(\sup_{s\geq
0}\left|\int_{0}^{s\wedge(m_{\theta}-t)}G^{(\theta)}(r+t,X_{r+t})\textup{d}r\right|\right)<\infty.$

Moreover, as a consequence of the properties of $F^{(\theta)}$ we have that
the function $G^{(\theta)}$ is upper semi-continuous we can see that
$V^{(\theta)}$ is upper semi-continuous (since $V^{(\theta)}$ is the infimum
of upper semi-continuous functions). Therefore, from the general theory of
optimal stopping (see Peskir and Shiryaev, (2006) Corollary 2.9) we have that
an optimal stopping time for (10) is given by

$\displaystyle\tau_{D}=\inf\\{t\geq 0:(t,X_{t})\in D\\},$ (14)

where $D=\\{(t,x)\in\mathbb{R}_{+}\times\mathbb{R}:V^{(\theta)}(t,x)=0\\}$ is
a closed set.

Hence, from Lemma 3.1, we derive that
$D=\\{(t,x)\in\mathbb{R}_{+}\times\mathbb{R}:x\geq b^{(\theta)}(t)\\}$, where
the function $b^{(\theta)}:\mathbb{R}_{+}\mapsto\mathbb{R}$ is given by

$\displaystyle b^{(\theta)}(t)=\inf\\{x\in\mathbb{R}:(t,x)\in D\\},$

for each $t\geq 0$. It follows from Lemma 3.1 that $b^{(\theta)}$ is non-
increasing and $b^{(\theta)}(t)\geq h^{(\theta)}(t)\geq 0$ for all $t\geq 0$.
Moreover, $b^{(\theta)}(t)=-\infty$ for $t\in[m_{\theta},\infty)$, since
$V^{(\theta)}(t,x)=0$ for all $t\geq m_{\theta}$ and $x\in\mathbb{R}$, giving
us $\tau_{D}\leq m_{\theta}$. In the case that $t<m_{\theta}$, we have that
$b^{(\theta)}(t)$ is finitely valued as we will prove in the following Lemma.

######  Lemma 3.2.

Let $\theta>0$. The function $b^{(\theta)}$ is finitely valued for all
$t\in[0,m_{\theta})$.

###### Proof.

For any $\theta>0$ and fix $t\geq 0$, consider the optimal stopping problem,

$\displaystyle\mathcal{V}_{t}^{(\theta)}(x)=\inf_{\tau\in\mathcal{T}_{m_{\theta}-t}}\mathbb{E}_{x}\left(\int_{0}^{\tau}[1+2e^{-\theta
t}(F^{(\theta)}(X_{s})-1)]\textup{d}s\right),\qquad x\in\mathbb{R},$

where $\mathcal{T}_{m_{\theta}-t}$ is the set of all stopping times of
$\mathbb{F}$ bounded by $m_{\theta}-t$. From the fact that for all $s\geq 0$
and $x\in\mathbb{R}$, $G(s+t,x)\geq 1+2e^{-\theta t}(F^{(\theta)}(x)-1)$ and
that $\tau_{D}\in\mathcal{T}_{m_{\theta}-t}$ (under $\mathbb{P}_{t,x}$ for all
$x\in\mathbb{R}$), we have that

$\displaystyle V^{(\theta)}(t,x)\geq\mathcal{V}_{t}^{(\theta)}(x)$ (15)

for all $x\in\mathbb{R}$. Hence it suffices to show that there exists
$\tilde{x}_{t}$ (finite) sufficiently large such that
$\mathcal{V}_{t}^{(\theta)}(x)=0$ for all $x\geq\tilde{x}_{t}$.

It can be shown that an optimal stopping time for $\mathcal{V}_{t}^{(\theta)}$
is $\tau_{\mathcal{D}_{t}}$, the first entry time before $m_{\theta}-t$ to the
set $\mathcal{D}_{t}=\\{x\in\mathbb{R}:\mathcal{V}_{t}^{(\theta)}(x)=0\\}$. We
proceed by contradiction, assume that $\mathcal{D}_{t}=\emptyset$, then
$\tau_{\mathcal{D}_{t}}=m_{\theta}-t$. Hence, by the dominated convergence
theorem and the spatial homogeneity of Lévy processes we have that

$\displaystyle
0\geq\lim_{x\rightarrow\infty}\mathcal{V}_{t}^{(\theta)}(x)=\mathbb{E}\left(\int_{0}^{m_{\theta}-t}\lim_{x\rightarrow\infty}[1+2e^{-\theta
t}(F^{(\theta)}(X_{s}+x)-1)]\textup{d}s\right)=m_{\theta}-t>0$

which is a contradiction. Therefore, we conclude that for each $t\geq 0$,
there exists a finite value $\tilde{x}_{t}$ such that
$b^{(\theta)}(t)\leq\tilde{x}_{t}$. ∎

######  Remark 3.3.

From the proof of Lemma 3.2, we find an upper bound of the boundary
$b^{(\theta)}$. Define, for each $t\in[0,m_{\theta})$,
$u^{(\theta)}(t)=\inf\\{x\in\mathbb{R}:\mathcal{V}^{(\theta)}_{t}(x)=0\\}$.
Then it follows that $u^{(\theta)}$ is a non-increasing finite function such
that

$\displaystyle u^{(\theta)}(t)\geq b^{(\theta)}(t)$

for all $t\in[0,m_{\theta})$.

Next we show that the function $V^{(\theta)}$ is continuous.

######  Lemma 3.4.

The function $V^{(\theta)}$ is continuous. Moreover, for each
$x\in\mathbb{R}$, $t\mapsto V^{(\theta)}(t,x)$ is Lipschitz on
$\mathbb{R}_{+}$ and for every $t\in\mathbb{R}_{+}$, $x\mapsto
V^{(\theta)}(t,x)$ is Lipschitz on $\mathbb{R}$.

###### Proof.

First, we are showing that, for a fixed $t\geq 0$, the function $x\mapsto
V^{(\theta)}(t,x)$ is Lipschitz on $\mathbb{R}$. Recall that if $t\geq
m_{\theta}$, $V^{(\theta)}(t,x)=0$ for all $x\in\mathbb{R}$ so the assertion
is clear. Suppose that $t<m_{\theta}$. Let $x,y\in\mathbb{R}$ and define
$\tau_{x}^{*}=\tau_{D(t,x)}=\inf\\{s\geq 0:X_{s}+x\geq b^{(\theta)}(s+t)\\}$.
Since $\tau_{x}^{*}$ is optimal in $V^{(\theta)}(t,x)$ (under $\mathbb{P}$) we
have that

$\displaystyle V^{(\theta)}(t,y)-V^{(\theta)}(t,x)$
$\displaystyle\leq\mathbb{E}\left(\int_{0}^{\tau_{x}^{*}}G^{(\theta)}(s+t,X_{s}+y)\textup{d}s\right)-\mathbb{E}\left(\int_{0}^{\tau_{x}^{*}}G^{(\theta)}(s+t,X_{s}+x)\textup{d}s\right)$
$\displaystyle=\mathbb{E}\left(\int_{0}^{\tau_{x}^{*}}2e^{-\theta(s+t)}[F^{(\theta)}(X_{s}+y)-F^{(\theta)}(X_{s}+x)]\textup{d}s\right).$

Define the stopping time

$\displaystyle\tau_{b(0)-x}^{+}=\inf\\{t\geq 0:X_{t}\geq
b^{(\theta)}(0)-x\\}.$

Then we have that $\tau_{x}^{*}\leq\tau_{b^{(\theta)}(0)-x}^{+}$ (since
$b^{(\theta)}$ is a non-increasing function). From the fact that
$F^{(\theta)}$ is non-decreasing, we obtain that for $b^{(\theta)}(0)\geq
y\geq x$,

$\displaystyle V^{(\theta)}(t,y)-V^{(\theta)}(t,x)$ $\displaystyle\leq
2\mathbb{E}\left(\int_{0}^{\tau_{b(0)-x}^{+}}e^{-\theta
s}[F^{(\theta)}(X_{s}+y)-F^{(\theta)}(X_{s}+x)]\textup{d}s\right).$

Using Fubini’s theorem and a density of the potential measure of the process
killed upon exiting $(-\infty,b^{(\theta)}(0)]$ (see equation (5)) we get that

$\displaystyle V^{(\theta)}(t,y)-V^{(\theta)}(t,x)$ $\displaystyle\leq
2\int_{-\infty}^{b^{(\theta)}(0)}[F^{(\theta)}(z+y-x)-F^{(\theta)}(z)]\int_{0}^{\infty}e^{-\theta
s}\mathbb{P}_{x}(X_{s}\in\textup{d}z,\tau_{b(0)}^{+}>s)\textup{d}s$
$\displaystyle=2\int_{-\infty}^{b^{(\theta)}(0)}[F^{(\theta)}(z+y-x)-F^{(\theta)}(z)]\left[e^{-\Phi(\theta)(b^{(\theta)}(0)-x)}W^{(\theta)}(b^{(\theta)}(0)-z)-W^{(\theta)}(x-z)\right]\textup{d}z$
$\displaystyle\leq
2e^{-\Phi(\theta)(b^{(\theta)}(0)-x)}W^{(\theta)}(b^{(\theta)}(0)-x+y)\int_{x-y}^{b^{(\theta)}(0)}[F^{(\theta)}(z+y-x)-F^{(\theta)}(z)]\textup{d}z,$

where in the last inequality, we used the fact that $W^{(\theta)}$ is strictly
increasing and non-negative and that $F^{(\theta)}$ vanishes at $(-\infty,0)$.
By an integration by parts argument, we obtain that

$\displaystyle\int_{x-y}^{b^{(\theta)}(0)}[F^{(\theta)}(z+y-x)-F^{(\theta)}(z)]\textup{d}z=(y-x)F^{(\theta)}(b^{(\theta)}(0)+y-x).$

Moreover, it can be checked that (see Kuznetsov et al., (2013) lemma 3.3) the
function $z\mapsto e^{-\Phi(\theta)(z)}W^{(\theta)}(z)$ is a continuous
function in the interval $[0,\infty)$ such that

$\displaystyle\lim_{z\rightarrow\infty}e^{-\Phi(\theta)(z)}W^{(\theta)}(z)=\frac{1}{\psi^{\prime}(\Phi(\theta))}<\infty.$

This implies that there exist a constant $M>0$ such that for every
$z\in\mathbb{R}$, $0\leq e^{-\Phi(\theta)(z)}W^{(\theta)}(z)<M$. Then we
obtain that for all $x\leq y\leq b^{(\theta)}(0)$,

$\displaystyle 0\leq V^{(\theta)}(t,y)-V^{(\theta)}(t,x)\leq
2M(y-x)e^{\Phi(\theta)y}\leq 2M(y-x)e^{\Phi(\theta)b^{(\theta)}(0)}.$

On the other hand, since $b^{(\theta)}(0)\geq b^{(\theta)}(t)$ for all
$t\in[0,m_{\theta})$ we have that for all
$(t,x)\in[0,m_{\theta})\times[b^{(\theta)}(0),\infty)$, $V^{(\theta)}(t,x)=0$.
Hence we obtain that for all $x,y\in\mathbb{R}$ and $t\geq 0$,

$\displaystyle|V^{(\theta)}(t,y)-V^{(\theta)}(t,x)|\leq
2M|y-x|e^{\Phi(\theta)b^{(\theta)}(0)}.$ (16)

Therefore we conclude that for a fixed $t\geq 0$, the function $x\mapsto
V^{(\theta)}(t,x)$ is Lipschitz on $\mathbb{R}$.

Using a similar argument and the fact that the function $t\mapsto e^{-\theta
t}$ is Lipschitz continuous on $[0,\infty)$ we can show that for any
$s,t<m_{\theta}$,

$\displaystyle|V^{(\theta)}(s,x)-V^{(\theta)}(t,x)|\leq 2\theta
m_{\theta}|s-t|$

and therefore $t\mapsto V^{(\theta)}(t,x)$ is Lipschitz continuous for all
$x\in\mathbb{R}$. ∎

In order to derive more properties of the boundary $b^{(\theta)}$, we first
state some auxiliary results. Recall that if $f\in
C_{b}^{1,2}(\mathbb{R}_{+}\times\mathbb{R})$, the set of real bounded
$C^{1,2}$ functions on $\mathbb{R}_{+}\times\mathbb{R}$ with bounded
derivatives, the infinitesimal generator of $(t,X)$ is given by

$\displaystyle\mathcal{A}_{(t,X)}(f)(t,x)$
$\displaystyle=\frac{\partial}{\partial t}f(t,x)-\mu\frac{\partial}{\partial
x}f(t,x)+\frac{1}{2}\sigma^{2}\frac{\partial^{2}}{\partial x^{2}}f(t,x)$
$\displaystyle\qquad+\int_{(-\infty,0)}[f(t,x+y)-f(t,x)-y\mathbb{I}_{\\{y>-1\\}}\frac{\partial}{\partial
x}f(t,x)]\Pi(\textup{d}y).$ (17)

Let $C=\mathbb{R}_{+}\times\mathbb{R}\setminus
D=\\{(t,x)\in\mathbb{R}_{+}\times\mathbb{R}:x<b^{(\theta)}(t)\\}$ be the
continuation region. Then we have that the value function $V^{(\theta)}$
satisfies a variational inequality in the sense of distributions. The proof is
analogous to the one presented in Lamberton and Mikou, (2008) (see Proposition
2.5) so the details are omitted.

######  Lemma 3.5.

Fix $\theta>0$. The distribution
$\mathcal{A}_{(t,X)}V^{(\theta)}+G^{(\theta)}$ is non-negative on
$\mathbb{R}_{+}\times\mathbb{R}$. Moreover, we have that
$\mathcal{A}_{(t,X)}V^{(\theta)}+G^{(\theta)}=0$ on $C$.

We define a special function which is useful to prove the left-continuity of
the boundary $b^{(\theta)}$. For $\theta>0$, we define an auxiliary function
in the set $D$. Let

$\displaystyle\varphi^{(\theta)}(t,x)=\int_{(-\infty,0)}V^{(\theta)}(t,x+y)\Pi(\textup{d}y)+G^{(\theta)}(t,x),\qquad(t,x)\in
D.$ (18)

From the fact that $V^{(\theta)}$ vanishes on $D$ and that $\Pi$ is finite on
sets of the form $(-\infty,-\varepsilon)$ for $\varepsilon>0$ we can see that
$|\varphi^{(\theta)}(t,x)|<\infty$ for all $(t,x)\in D$. Moreover, by the
Lemma above and the properties of $V^{(\theta)}$ and $G^{(\theta)}$, it can be
shown that $\varphi$ is strictly positive, continuous and strictly increasing
(in each argument) in the interior of $D$.

Now we are ready to give further properties of the curve $b^{(\theta)}$ in the
set $[0,m_{\theta})$.

######  Lemma 3.6.

The function $b^{(\theta)}$ is continuous on $[0,m_{\theta})$. Moreover we
have that $\lim_{t\uparrow m_{\theta}}b^{(\theta)}(t)=0$ .

###### Proof.

The method proof of the continuity of $b^{(\theta)}$ in $[0,m_{\theta})$ is
heavyly based on the work of Lamberton and Mikou, (2008) (see Theorem 4.2,
where the continuity of the boundary is shown in the American option context)
so is omitted.

We then show that the limit holds. Define
$b^{(\theta)}(m_{\theta}-):=\lim_{t\uparrow m_{\theta}}b^{(\theta)}(t)$. We
obtain $b^{(\theta)}(m_{\theta}-)\geq 0$ since $b^{(\theta)}(t)\geq
h^{(\theta)}(t)\geq 0$ for all $t\in[0,m_{\theta})$. The proof is by
contradiction so we assume that $b^{(\theta)}(m_{\theta}-)>0$. Note that for
all $x\in\mathbb{R}$, we have that $V^{(\theta)}(m_{\theta},0)=0$ and
$G^{(\theta)}(m_{\theta},x)=F^{(\theta)}(x)$. Moreover, we have that

$\displaystyle\mathcal{A}_{X}(V^{(\theta)})+G^{(\theta)}=-\partial_{t}V^{(\theta)}\leq
0$

in the sense of distributions on
$(0,m_{\theta})\times(0,b^{(\theta)}(m_{\theta}-))$. Hence, by continuity, we
can derive, for $t\in[0,m_{\theta})$, that
$\mathcal{A}_{X}(V^{(\theta)})(t,\cdot)+G^{(\theta)}(t,\cdot)\leq 0$ on the
interval $(0,b^{(\theta)}(m_{\theta}-))$. Hence, by taking $t\uparrow
m_{\theta}$ we obtain that

$\displaystyle 0\geq\lim_{t\uparrow
m_{\theta}}\mathcal{A}_{X}(V^{(\theta)})(t,\cdot)+G^{(\theta)}(t,\cdot)=F^{(\theta)}>0$

in the sense of distributions, where we used the continuity if $V^{(\theta)}$
and $G^{(\theta)}$, the fact that $V^{(\theta)}(m_{\theta},x)=0$ for all
$x\in\mathbb{R}$ and that $F^{(\theta)}(x)>0$ for all $x>0$. Note that we have
got a contradiction and we conclude that $b^{(\theta)}(m_{\theta})=0$. ∎

Define the value

$\displaystyle t_{b}:=\inf\\{t\geq 0:b^{(\theta)}(t)\leq 0\\}.$ (19)

Note that in the case where $X$ is a process of infinite variation, we have
that the distribution function of the random variable
$-\underline{X}_{\widetilde{e}_{\theta}}$, $F^{(\theta)}$, is continuous on
$\mathbb{R}$, strictly increasing and strictly positive in the open set
$(0,\infty)$ with $F^{(\theta)}(0)=0$. This fact implies that the inverse
function of $F^{(\theta)}$ exists on $(0,\infty)$ and then function
$h^{(\theta)}$ can be written for $t\in[0,m_{\theta})$ as

$\displaystyle h^{(\theta)}(t)=(F^{(\theta)})^{-1}\left(1-\frac{1}{2}e^{\theta
t}\right).$

Hence we conclude that $h^{(\theta)}(t)>0$ for all $t\in[0,m_{\theta})$.
Therefore, when $X$ is a process of infinite variation, we have
$b^{(\theta)}(t)>0$ for all $t\in[0,m_{\theta})$ and hence $t_{b}=m_{\theta}$.
For the case of finite variation, we have that $t_{b}\in[0,m_{\theta})$ which
implies that $b^{(\theta)}(t)=0$ for all $t\in[t_{b},m_{\theta})$ and
$b^{(\theta)}(t)>0$ for all $t\in[0,t_{b})$. In the next lemma, we
characterise its value.

######  Lemma 3.7.

Let $\theta>0$ and $X$ be a process of finite variation. We have that for all
$t\geq 0$ and $x\in\mathbb{R}$,

$\displaystyle\int_{(-\infty,0)}[V^{(\theta)}(t,x+y)-V^{(\theta)}(t,x)]\Pi(\textup{d}y)>-\infty.$

Moreover, for any Lévy process, $t_{b}$ is given by

$\displaystyle
t_{b}=\inf\left\\{t\in[0,m_{\theta}]:\int_{(-\infty,0)}V^{(\theta)}_{B}(t,y)\Pi(\textup{d}y)+G^{(\theta)}(t,0)\geq
0\right\\},$ (20)

where $V^{(\theta)}_{B}$ is given by

$\displaystyle
V^{(\theta)}_{B}(t,y)=\mathbb{E}_{y}(\tau_{0}^{+}\wedge(m_{\theta}-t))-\frac{2}{\theta}e^{-\theta
t}[1-\mathbb{E}_{y}(e^{-\theta(\tau_{0}^{+}\wedge(m_{\theta}-t))})]$

for all $t\in[0,m_{\theta})$ and $y\in\mathbb{R}$.

###### Proof.

Assume that $X$ is a process of finite variation. We first show that

$\displaystyle\int_{(-\infty,0)}[V^{(\theta)}(t,x+y)-V^{(\theta)}(t,x)]\Pi(\textup{d}y)>-\infty$

for all $t\geq 0$ and $x\in\mathbb{R}$. The case $t\geq m_{\theta}$ is
straightforward since $V^{(\theta)}(t,x)=0$ for all $x\in\mathbb{R}$. The case
$t<m_{\theta}$ follows from the Lipschitz continuity of the mapping $x\mapsto
V^{(\theta)}(t,x)$, that $\Pi$ is finite on intervals away from zero and since
$\int_{(-1,0)}y\Pi(\textup{d}y)>-\infty$ when $X$ is of finite variation.
Moreover, from Lemma 3.5, we obtain that

$\displaystyle\int_{(-\infty,0)}[V^{(\theta)}(t,x+y)-V^{(\theta)}(t,x)]\Pi(\textup{d}y)+G^{(\theta)}(t,x)=-\frac{\partial}{\partial
t}V^{(\theta)}(t,x)-\delta\frac{\partial}{\partial x}V^{(\theta)}(t,x)\leq 0$

on $C$ in the sense of distributions, where the last inequality follows since
$V^{(\theta)}$ is non-decreasing in each argument and $\delta>0$ is defined in
(3). Then by continuity of the functions $V^{(\theta)}$ and $G^{(\theta)}$
(recall that $G^{(\theta)}$ is at least continuous on
$(0,\infty)\times(0,\infty)$ and right-continuous at points of the form
$(t,0)$ for $t\geq 0$) we can derive

$\displaystyle\int_{(-\infty,0)}[V^{(\theta)}(t,y)-V^{(\theta)}(t,0)]\Pi(\textup{d}y)+G^{(\theta)}(t,0)\leq
0$ (21)

for all $t\in[0,t_{b})$.

Next, we show that the set $\\{t\in[0,m_{\theta}):b^{(\theta)}(t)=0\\}$ is non
empty. We proceed by contradiction, assume that $b^{(\theta)}(t)>0$ for all
$t\in[0,m_{\theta})$ so that $t_{b}=m_{\theta}$. Taking $t\uparrow m_{\theta}$
in (21) and applying dominated convergence theorem, we obtain that

$\displaystyle 0\geq\lim_{t\uparrow
m_{\theta}}\left\\{\int_{(-\infty,0)}[V^{(\theta)}(t,y)-V^{(\theta)}(t,0)]\Pi(\textup{d}y)+G^{(\theta)}(t,0)\right\\}=G^{(\theta)}(m_{\theta},0)=F^{(\theta)}(0)>0,$

where the strict inequality follows from
$F^{(\theta)}(0)=\frac{\theta}{\Phi(\theta)}W^{(\theta)}(0)=\frac{\theta}{\delta\Phi(\theta)}>0$
since $X$ is of finite variation. Therefore, we observe a contradiction which
shows that $\\{t\in[0,m_{\theta}):b^{(\theta)}(t)=0\\}\neq\emptyset$.
Moreover, by the definition, we have that
$t_{b}=\inf\\{t\in[0,m_{\theta}):b^{(\theta)}(t)=0\\}$.

Next we find an expression for $V^{(\theta)}(t,x)$ when $t\in(0,m_{\theta})$
and $x<0$. Since $b^{(\theta)}(t)\geq 0$ for all $t\in[0,m_{\theta})$, we have
that

$\displaystyle V^{(\theta)}(t,x)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{+}\wedge(m_{\theta}-t)}(1-2e^{-\theta(t+s)})\textup{d}s\right)+\mathbb{E}_{x}\left(\mathbb{I}_{\\{\tau_{0}^{+}<m_{\theta}-t\\}}V^{(\theta)}(t+\tau_{0}^{+},0)\right)$
$\displaystyle=V^{(\theta)}_{B}(t,x)+\mathbb{E}_{x}\left(\mathbb{I}_{\\{\tau_{0}^{+}<m_{\theta}-t\\}}V^{(\theta)}(t+\tau_{0}^{+},0)\right),$
(22)

where the first equality follows since $X_{s}\leq 0$ for all
$s\leq\tau_{0}^{+}$ and $G(t,x)=1-2e^{-\theta t}$ for all $x<0$. Hence, in
particular, we have that $V^{(\theta)}(t,x)=V^{(\theta)}_{B}(t,x)$ for all
$t\in[t_{b},m_{\theta})$ and $x\in\mathbb{R}$.

We show that (20) holds. From the discussion after Lemma 3.5, we know that

$\displaystyle\varphi^{(\theta)}(t,x)=\int_{(-\infty,0)}V^{(\theta)}(t,x+y)\Pi(\textup{d}y)+G^{(\theta)}(t,x)>0$

for all $x>0$ and $t\geq t_{b}$. Then by taking $x\downarrow 0$, making use of
the right continuity of $x\mapsto G(t,x)$, continuity of $V^{(\theta)}$ (see
Lemma 3.4) and applying dominated convergence theorem, we derive that

$\displaystyle\int_{(-\infty,0)}V^{(\theta)}(t_{b},y)\Pi(\textup{d}y)+G^{(\theta)}(t_{b},0)\geq
0.$

In particular, if $t_{b}=0$, (20) holds since $t\mapsto V^{(\theta)}(t,y)$
(for all $y\in\mathbb{R}$) and $G^{(\theta)}(t,0)$ are non-decreasing
functions. If $t_{b}>0$, taking $t\uparrow t_{b}$ in (21) gives us

$\displaystyle\int_{(-\infty,0)}V^{(\theta)}(t_{b},y)\Pi(\textup{d}y)+G^{(\theta)}(t,0)\leq
0.$

Hence, we have that
$\int_{(-\infty,0)}V^{(\theta)}_{B}(t_{b},y)\Pi(\textup{d}y)+G^{(\theta)}(t_{b},0)=0$
with (20) becoming clear due to the fact that $t\mapsto V^{(\theta)}_{B}(t,x)$
is non-decreasing. If $X$ is a process of infinite variation, we have that
$h^{(\theta)}(t)>0$ for all $t\in[0,m_{\theta})$ and therefore
$G^{(\theta)}(t,x)<0$ for all $t\in[0,m_{\theta})$ and $x\leq 0$ which implies
that

$\displaystyle
t_{b}=m_{\theta}=\inf\left\\{t\in[0,m_{\theta}]:\int_{(-\infty,0)}V^{(\theta)}_{B}(t,y)\Pi(\textup{d}y)+G^{(\theta)}(t,0)\geq
0\right\\}.$

∎

Now we prove that the derivatives of $V^{(\theta)}$ exist at the boundary
$b^{(\theta)}$ for those points in which $b^{(\theta)}$ is strictly positive.

######  Lemma 3.8.

For all $t\in[0,t_{b})$, the partial derivatives of $V^{(\theta)}(t,x)$ at the
point $(t,b^{(\theta)}(t))$ exist and are equal to zero, i.e.,

$\displaystyle\frac{\partial}{\partial
t}V^{(\theta)}(t,b^{(\theta)}(t))=0\qquad\text{and}\qquad\frac{\partial}{\partial
x}V^{(\theta)}(t,b^{(\theta)}(t))=0.$

###### Proof.

First, we prove that the assertion in the first argument. Using a similar idea
as in Lemma 3.4, we have that for any $t<t_{b}$, $x\in\mathbb{R}$ and $h>0$,

$\displaystyle
0\leq\frac{V^{(\theta)}(t,b^{(\theta)}(t))-V^{(\theta)}(t-h,b^{(\theta)}(t))}{h}$
$\displaystyle\leq
2\mathbb{E}_{b^{(\theta)}(t)}\left(\int_{0}^{\tau^{*}_{h}}\frac{[e^{-\theta(r+t-h)}-e^{-\theta(r+t)}]}{h}\textup{d}r\right)$
$\displaystyle\leq
2\mathbb{E}_{b^{(\theta)}(t)}\left(\int_{0}^{\tau^{+}_{b^{(\theta)}(t-h)}}\frac{[e^{-\theta(r+t-h)}-e^{-\theta(r+t)}]}{h}\textup{d}r\right)$
$\displaystyle=\frac{[e^{-\theta(t-h)}-e^{-\theta
t}]}{h}\frac{1}{\theta}\mathbb{E}_{b^{(\theta)}(t)}\left(1-e^{-\theta\tau^{+}_{b^{(\theta)}(t-h)}}\right)$

where $\tau^{*}_{h}=\inf\\{r\in[0,m_{\theta}-t+h]:X_{r}\geq
b^{(\theta)}(r+t-h)\\}$ is the optimal stopping time for $V^{(\theta)}(t-h,x)$
and the second inequality follows since $b$ is non increasing. Hence, by (1)
and continuity of $b^{(\theta)}$ we obtain that

$\displaystyle\lim_{h\downarrow
0}\frac{V^{(\theta)}(t,b^{(\theta)}(t))-V^{(\theta)}(t-h,b^{(\theta)}(t))}{h}=0.$

Now we show that the partial derivative of the second argument exists at
$b^{(\theta)}(t)$ and is equal to zero. Fix any time $t\in[0,t_{b})$,
$\varepsilon>0$ and $x\leq b^{(\theta)}(t)$ (without loss of generality, we
assume that $\varepsilon<x$). By a similar argument that as provided in Lemma
3.4, we obtain that

$\displaystyle V^{(\theta)}$
$\displaystyle(t,x)-V^{(\theta)}(t,x-\varepsilon)$ $\displaystyle\leq
2\int_{-\infty}^{b^{(\theta)}(t)}[F^{(\theta)}(z+\varepsilon)-F^{(\theta)}(z)]\left[e^{-\Phi(\theta)(b^{(\theta)}(t)-x+\varepsilon)}W^{(\theta)}(b^{(\theta)}(t)-z)-W^{(\theta)}(x-\varepsilon-z)\right]\textup{d}z$
$\displaystyle=2e^{-\Phi(\theta)(b^{(\theta)}(t)-x+\varepsilon)}\int_{x-\varepsilon}^{b^{(\theta)}(t)}[F^{(\theta)}(z+\varepsilon)-F^{(\theta)}(z)]W^{(\theta)}(b^{(\theta)}(t)-z)\textup{d}z$
$\displaystyle\qquad+2\int_{0}^{x-\varepsilon}[F^{(\theta)}(z+\varepsilon)-F^{(\theta)}(z)]\left[e^{-\Phi(\theta)(b^{(\theta)}(t)-x+\varepsilon)}W^{(\theta)}(b^{(\theta)}(t)-z)-W^{(\theta)}(x-\varepsilon-z)\right]\textup{d}z$
$\displaystyle\qquad+2\int_{-\varepsilon}^{0}F^{(\theta)}(z+\varepsilon)\left[e^{-\Phi(\theta)(b^{(\theta)}(t)-x+\varepsilon)}W^{(\theta)}(b^{(\theta)}(t)-z)-W^{(\theta)}(x-\varepsilon-z)\right]\textup{d}z.$

Dividing by $\varepsilon$, we have that for $t\in[0,m_{\theta})$ and
$\varepsilon<x$ that

$\displaystyle
0\leq\frac{V^{(\theta)}(t,x)-V^{(\theta)}(t,x-\varepsilon)}{\varepsilon}\leq
R_{1}^{(\varepsilon)}(t,x)+R_{2}^{(\varepsilon)}(t,x)+R_{3}^{(\varepsilon)}(t,x),$

where

$\displaystyle R_{1}^{(\varepsilon)}(t,x)$
$\displaystyle=2e^{-\Phi(\theta)(b^{(\theta)}(t)-x+\varepsilon)}\frac{1}{\varepsilon}\int_{x-\varepsilon}^{b^{(\theta)}(t)}[F^{(\theta)}(z+\varepsilon)-F^{(\theta)}(z)]W^{(\theta)}(b^{(\theta)}(t)-z)\textup{d}z,$
$\displaystyle R_{2}^{(\varepsilon)}(t,x)$
$\displaystyle=2\frac{1}{\varepsilon}\int_{0}^{x-\varepsilon}[F^{(\theta)}(z+\varepsilon)-F^{(\theta)}(z)]\left[e^{-\Phi(\theta)(b^{(\theta)}(t)-x+\varepsilon)}W^{(\theta)}(b^{(\theta)}(t)-z)-W^{(\theta)}(x-\varepsilon-z)\right]\textup{d}z,$
$\displaystyle R_{3}^{(\varepsilon)}(t,x)$
$\displaystyle=2\frac{1}{\varepsilon}\int_{-\varepsilon}^{0}F^{(\theta)}(z+\varepsilon)\left[e^{-\Phi(\theta)(b^{(\theta)}(t)-x+\varepsilon)}W^{(\theta)}(b^{(\theta)}(t)-z)-W^{(\theta)}(x-\varepsilon-z)\right]\textup{d}z.$

By using that $W$ and $F$ are non-decreasing, that $W^{(\theta)}$ (and hence
$F^{(\theta)}$) has left and right derivatives and the dominated convergence
theorem we can show that for $t\in[0,t_{b})$, $\lim_{\varepsilon\downarrow
0}R_{i}^{(\varepsilon)}(t,b^{(\theta)}(t))=0$ for each $i=1,2,3$. Hence, we
have that

$\displaystyle\lim_{\varepsilon\downarrow
0}\frac{V^{(\theta)}(t,b^{(\theta)}(t))-V^{(\theta)}(t,b^{(\theta)}(t)-\varepsilon)}{\varepsilon}=0$

proving that $x\mapsto V^{(\theta)}(x,t)$ is differentiable at
$b^{(\theta)}(t)$ with $\partial/\partial xV^{(\theta)}(t,b^{(\theta)}(t))=0$
for $t\in[0,t_{b})$.

∎

The next theorem looks at how the value function $V^{(\theta)}$ and the curve
$b^{(\theta)}$ can be characterised as a solution of non-linear integral
equations within a certain family of functions. These equations are in fact
generalisations of the free boundary equation (see e.g. Peskir and Shiryaev,
(2006) Section 14.1 in a diffusion setting) in the presence of jumps. It is
important to mention that the proof of Theorem 3.9 is mainly inspired by the
ideas of du Toit et al., (2008) with some extensions to allow for the presence
of jumps.

######  Theorem 3.9.

Let $X$ be a spectrally negative Lévy process and let $t_{b}$ be as
characterised in (20). For all $t\in[0,t_{b})$ and $x\in\mathbb{R}$, we have
that

$\displaystyle V^{(\theta)}(t,x)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}G^{(\theta)}(r+t,X_{r})\mathbb{I}_{\\{X_{r}<b^{(\theta)}(r+t)\\}}\textup{d}r\right)$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}\int_{(-\infty,b^{(\theta)}(r+t)-X_{r})}V^{(\theta)}(r+t,X_{r}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{r}>b^{(\theta)}(r+t)\\}}\textup{d}r\right)$
(23)

and $b^{(\theta)}(t)$ solves the equation

$\displaystyle\mathbb{E}_{b^{(\theta)}(t)}$
$\displaystyle\left(\int_{0}^{m_{\theta}-t}G^{(\theta)}(r+t,X_{r})\mathbb{I}_{\\{X_{r}<b^{(\theta)}(r+t)\\}}\textup{d}r\right)$
$\displaystyle\qquad-\mathbb{E}_{b^{(\theta)}(t)}\left(\int_{0}^{m_{\theta}-t}\int_{(-\infty,b^{(\theta)}(r+t)-X_{r})}V^{(\theta)}(r+t,X_{r}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{r}>b^{(\theta)}(r+t)\\}}\textup{d}r\right)=0.$
(24)

If $t\in[t_{b},m_{\theta})$, we have that $b^{(\theta)}(t)=0$ and

$\displaystyle
V^{(\theta)}(t,x)=\mathbb{E}_{x}(\tau_{0}^{+}\wedge(m_{\theta}-t))-\frac{2}{\theta}e^{-\theta
t}[1-\mathbb{E}_{x}(e^{-\theta(\tau_{0}^{+}\wedge(m_{\theta}-t))})]$ (25)

for all $x\in\mathbb{R}$. Moreover, the pair $(V^{(\theta)},b^{(\theta)})$ is
uniquely characterised as the solutions to equations (3.9)-(25) in the class
of continuous functions in $\mathbb{R}_{+}\times\mathbb{R}$ and
$\mathbb{R}_{+}$, respectively, such that $b^{(\theta)}\geq h^{(\theta)}$,
$V^{(\theta)}\leq 0$ and
$\int_{(-\infty,0)}V^{(\theta)}(t,x+y)\Pi(\textup{d}y)+G^{(\theta)}(t,x)\geq
0$ for all $t\in[0,t_{b})$ and $x\geq b^{(\theta)}(t)$.

###  3.1 Proof of Theorem 3.9

Since the proof of Theorem 3.9 is rather long, we split it into a series of
Lemmas. This subsection is entirely dedicated for this purpose. With the help
of Itô formula and following an analogous argument as in Lamberton and Mikou,
(2013) (in the infinite variation case), we prove that $V^{(\theta)}$ and
$b^{(\theta)}$ are solutions to the integral equations listed above. The
finite variation case is proved using an argument that considers the
consecutive times in which $X$ hits the curve $b^{(\theta)}$.

######  Lemma 3.10.

The pair $(V^{(\theta)},b^{(\theta)})$ are solutions to the equations
(3.9)-(25).

###### Proof.

Recall from Lemma 3.7 that, when $t_{b}<m_{\theta}$, the value function
$V^{(\theta)}(t,x)$ satisfies equation (25) for $t\in[t_{b},m_{\theta})$ and
$x\in\mathbb{R}$. We also have that equation (3.9) follows from (3.9) by
letting $x=b^{(\theta)}(t)$ and using that
$V^{(\theta)}(t,b^{(\theta)}(t))=0$.

We proceed to show that $(V^{(\theta)},b^{(\theta)})$ solves equation (3.9).
First, we assume that $X$ is a process of infinite variation. We follow an
analogous argument as Lamberton and Mikou, (2013) (see Theorem 3.2). Consider
a regularized sequence $\\{\rho_{n}\\}_{n\geq 1}$ of non-negative
$C^{\infty}(\mathbb{R}_{+}\times\mathbb{R})$ functions with support in
$[-1/n,0]\times[-1/n,0]$ such that
$\int_{-\infty}^{0}\int_{-\infty}^{0}\rho_{n}(s,y)\textup{d}s\textup{d}y=1$.
For every $n\geq 1$, define the function $V^{(\theta)}_{n}$ by

$\displaystyle
V^{(\theta)}_{n}(t,x)=(V^{(\theta)}\ast\rho_{n})(t,x)=\int_{-\infty}^{0}\int_{-\infty}^{0}V^{(\theta)}(t+s,x+y)\rho_{n}(s,y)\textup{d}s\textup{d}y.$

for any $(t,x)\in[1/n,\infty)\times\mathbb{R}$. Then for each $n\geq 1$, the
function $V^{(\theta)}_{n}$ is a $C^{1,2}(\mathbb{R}_{+}\times\mathbb{R})$
bounded function (since $V^{(\theta)}$ is bounded). Moreover, it can be shown
that $V^{(\theta)}_{n}\uparrow V$ on $\mathbb{R}_{+}\times\mathbb{R}$ when
$n\rightarrow\infty$ and that (see Lamberton and Mikou, (2008), proof of
Proposition 2.5),

$\displaystyle\frac{\partial}{\partial
t}V^{(\theta)}_{n}(t,x)+\mathcal{A}_{X}(V^{(\theta)}_{n})(t,x)=-(G^{(\theta)}*\rho_{n})(u,x)\qquad\text{for
all }(t,x)\in[1/n,\infty)\times\mathbb{R}\cap C,$ (26)

where $\mathcal{A}_{X}$ is the infinitesimal generator of $X$ given in (3) and
$C=\mathbb{R}_{+}\times\mathbb{R}\setminus D$. Let $t\in(0,t_{b}]$, $m>0$ such
that $t>1/m$ and $x\in\mathbb{R}$. Applying Itô formula to
$V^{(\theta)}_{n}(t+s,X_{s}+x)$, for $s\in[0,m_{\theta}-t]$, we obtain that
for any $n\geq m$,

$\displaystyle V^{(\theta)}_{n}(s+t,X_{s}+x)$
$\displaystyle=V^{(\theta)}_{n}(t,x)+M_{s}^{t,n}+\int_{0}^{s}\left[\frac{\partial}{\partial
t}V^{(\theta)}_{n}(r+t,X_{r}+x)+\mathcal{A}_{X}(V^{(\theta)}_{n})(r+t,X_{r}+x)\right]\textup{d}r,$

where $\\{M_{s}^{t,n},t\geq 0\\}$ is a zero mean martingale. Hence, taking
expectation and using (26), we derive that

$\displaystyle\mathbb{E}(V^{(\theta)}_{n}(s+t,X_{s}+x))$
$\displaystyle=V^{(\theta)}_{n}(t,x)+\mathbb{E}\left(\int_{0}^{s}\left[\frac{\partial}{\partial
t}V^{(\theta)}_{n}(r+t,X_{r}+x)+\mathcal{A}_{X}(V^{(\theta)}_{n})(r+t,X_{r}+x)\right]\textup{d}r\right)$
$\displaystyle=V^{(\theta)}_{n}(t,x)-\mathbb{E}\left(\int_{0}^{s}(G^{(\theta)}*\rho_{n})(r+t,X_{r}+x)\mathbb{I}_{\\{X_{r}<b^{(\theta)}(r+t)\\}}\textup{d}r\right)$
$\displaystyle\qquad+\mathbb{E}\left(\int_{0}^{s}\int_{(-\infty,0)}V^{(\theta)}_{n}(r+t,X_{r}+x+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{r}>b^{(\theta)}(r+t)\\}}\textup{d}r\right),$

where we used the fact that $b^{(\theta)}(s)$ is finite for all $s\geq 0$ and
that $\mathbb{P}(X_{s}+x=b(t+s))=0$ for all $s>0$ and $x\in\mathbb{R}$ when
$X$ is of infinite variation (see Sato, (1999)). Taking $s=m_{\theta}-t$,
using the fact that $V^{(\theta)}(m_{\theta},x)=0$ for all $x\in\mathbb{R}$
and letting $n\rightarrow\infty$ (by the dominated convergence theorem), we
obtain that (3.9) holds for any $(t,x)\in(0,t_{b})\times\mathbb{R}$. The case
when $t=0$ follows by continuity.

For the finite variation case, we define the auxiliary function

$\displaystyle R^{(\theta)}(t,x)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}G^{(\theta)}(r+t,X_{r})\mathbb{I}_{\\{X_{r}<b^{(\theta)}(r+t)\\}}\textup{d}r\right)$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}\int_{(-\infty,0)}V^{(\theta)}(r+t,X_{r}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{r}>b^{(\theta)}(r+t)\\}}\textup{d}r\right)$

for all $(t,x)\in\mathbb{R}_{+}\times\mathbb{R}$. We then prove that
$R^{(\theta)}=V^{(\theta)}$. First, note that from the discussion after Lemma
3.5 we have that $\int_{(-\infty,0)}V^{(\theta)}(t,x+y)+G^{(\theta)}(t,x)\geq
0$ for all $(t,x)\in D$. Then we have that for all
$(t,x)\in[0,m_{\theta}]\times\mathbb{R}$,

$\displaystyle|R^{(\theta)}(t,x)|\leq\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}|G^{(\theta)}(r+t,X_{r})|\textup{d}r\right)\leq
m_{\theta}-t,$

where we used that $|G^{(\theta)}|\leq 1$ in the last inequality. For each
$(t,x)\in\mathbb{R}_{+}\times\mathbb{R}$, we define the times at which the
process $X$ hits the curve $b^{(\theta)}$. Let
$\tau_{b}^{(1)}=\inf\\{s\in[0,m_{\theta}-t]:X_{s}\geq b^{(\theta)}(s+t)\\}$
and for $k\geq 1$,

$\displaystyle\sigma_{b}^{(k)}$
$\displaystyle=\inf\\{s\in[\tau_{b}^{k},m_{\theta}-t]:X_{s}<b^{(\theta)}(s+t)\\}$
$\displaystyle\tau_{b}^{(k+1)}$
$\displaystyle=\inf\\{s\in[\sigma_{b}^{k},m_{\theta}-t]:X_{s}\geq
b^{(\theta)}(s+t)\\},$

where in this context, we understand that $\inf\emptyset=m_{\theta}-t$. Taking
$t\in[0,m_{\theta}]$ and $x>b^{(\theta)}(t)$ and gives us

$\displaystyle R^{(\theta)}(t,x)$
$\displaystyle=-\mathbb{E}_{x}\left(\int_{0}^{\sigma_{b}^{(1)}}\int_{(-\infty,0)}V^{(\theta)}(r+t,X_{r}+y)\Pi(\textup{d}y)\textup{d}r\right)+\mathbb{E}_{x}\left(\int_{\sigma_{b}^{(1)}}^{\tau_{b}^{(2)}}G^{(\theta)}(r+t,X_{r})\textup{d}r\right)$
$\displaystyle\qquad+\mathbb{E}_{x}\left(\mathbb{I}_{\\{\tau_{b}^{(2)}<m_{\theta}-t\\}}\int_{\tau_{b}^{(2)}}^{m_{\theta}-t}G^{(\theta)}(r+t,X_{r})\mathbb{I}_{\\{X_{r}<b^{(\theta)}(r+t)\\}}\textup{d}r\right)$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\mathbb{I}_{\\{\tau_{b}^{(2)}<m_{\theta}-t\\}}\int_{\tau_{b}^{(2)}}^{m_{\theta}-t}\int_{(-\infty,0)}V^{(\theta)}(r+t,X_{r}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{r}>b^{(\theta)}(r+t)\\}}\textup{d}r\right)$
$\displaystyle=-\mathbb{E}_{x}\left(\int_{0}^{\sigma_{b}^{(1)}}\int_{(-\infty,0)}V^{(\theta)}(r+t,X_{r}+y)\Pi(\textup{d}y)\textup{d}r\right)+\mathbb{E}_{x}(V^{(\theta)}(t+\sigma_{b}^{(1)},X_{\sigma_{b}^{(1)}})\mathbb{I}_{\\{\sigma_{b}^{(1)}<m_{\theta}-t\\}})$
$\displaystyle\qquad+\mathbb{E}_{x}(R^{(\theta)}(t+\tau_{b}^{(2)},X_{\tau_{b}^{(2)}})\mathbb{I}_{\\{\tau_{b}^{(2)}<m_{\theta}-t\\}}),$

where the last equality follows from the strong Markov property applied at
time $\sigma_{b}^{(1)}$ and $\tau_{b}^{(2)}$, respectively, and the fact that
$\tau_{D}$ is optimal for $V^{(\theta)}$. Using the compensation formula for
Poisson random measures (see Kyprianou, (2014) Theorem 4.4), it can be shown
that

$\displaystyle\mathbb{E}_{x}\left(\int_{0}^{\sigma_{b}^{(1)}}\int_{(-\infty,0)}V^{(\theta)}(r+t,X_{r}+y)\Pi(\textup{d}y)\textup{d}r\right)=\mathbb{E}_{x}(V^{(\theta)}(t+\sigma_{b}^{(1)},X_{\sigma_{b}^{(1)}})\mathbb{I}_{\\{\sigma_{b}^{(1)}<m_{\theta}-t\\}}).$

Hence, for all $(t,x)\in D$, we have that

$\displaystyle
R^{(\theta)}(t,x)=\mathbb{E}_{x}(R^{(\theta)}(t+\tau_{b}^{(2)},X_{\tau_{b}^{(2)}})\mathbb{I}_{\\{\tau_{b}^{(2)}<m_{\theta}-t\\}}).$

Using an induction argument, it can be shown that for all $(t,x)\in D$ and
$n\geq 2$,

$\displaystyle
R^{(\theta)}(t,x)=\mathbb{E}_{x}(R^{(\theta)}(t+\tau_{b}^{(n)},X_{\tau_{b}^{(n)}})\mathbb{I}_{\\{\tau_{b}^{(n)}<m_{\theta}-t\\}})=\mathbb{E}_{x}(R^{(\theta)}(t+\tau_{b}^{(n)},X_{\tau_{b}^{(n)}})),$
(27)

where the last equality follows since $R^{(\theta)}(m_{\theta},x)=0$ for all
$x\in\mathbb{R}$. Since $X$ is of finite variation, it can be shown that for
all $x\in\mathbb{R}$, $\lim_{n\rightarrow\infty}\tau_{b}^{(n)}=m_{\theta}-t$
$\mathbb{P}_{x}$-a.s. Therefore, from (27) and taking $n\rightarrow\infty$, we
conclude that for all $(t,x)\in D$,

$\displaystyle|R^{(\theta)}(t,x)|\leq\lim_{n\rightarrow\infty}\mathbb{E}_{x}\left(|R^{(\theta)}(t+\tau_{b}^{(n)},X_{\tau_{b}^{(n)}})|\right)\leq\lim_{n\rightarrow\infty}\mathbb{E}_{x}(m_{\theta}-t-\tau_{b}^{(n)})=0,$

where the last inequality follows from the dominated convergence theorem. On
the other hand, if we take $t\in[0,m_{\theta}]$ and $x<b^{(\theta)}(t)$ we
have, by the strong Markov property applied to the filtration at time
$\tau_{b}^{(1)}$, that

$\displaystyle
R^{(\theta)}(t,x)=\mathbb{E}_{x}\left(\int_{0}^{\tau_{b}^{(1)}}G^{(\theta)}(r+t,X_{r})\textup{d}r\right)+\mathbb{E}_{x}(R^{(\theta)}(t+\tau_{b}^{(1)},X_{\tau_{b}^{(1)}}))=V^{(\theta)}(t,x),$

where we used the fact that $\tau_{b}^{(1)}$ is an optimal stopping time for
$V^{(\theta)}$ and that $R^{(\theta)}$ vanishes on $D$. So then (3.9) also
holds in the finite variation case.

∎

Next we proceed to show the uniqueness result. Suppose that there exist a non-
positive continuous function
$U^{(\theta)}:[0,m_{\theta}]\times\mathbb{R}\mapsto(-\infty,0]$ and a
continuous function $c^{(\theta)}$ on $[0,m_{\theta})$ such that
$c^{(\theta)}\geq h^{(\theta)}$ and $c^{(\theta)}(t)=0$ for all
$t\in[t_{b},m_{\theta})$. We assume that the pair
$(U^{(\theta)},c^{(\theta)})$ solves the equations

$\displaystyle U^{(\theta)}(t,x)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}G^{(\theta)}(r+t,X_{r})\mathbb{I}_{\\{X_{r}<c^{(\theta)}(r+t)\\}}\textup{d}r\right)$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}\int_{(-\infty,c^{(\theta)}(r+t)-X_{r})}U^{(\theta)}(r+t,X_{r}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{r}>c^{(\theta)}(r+t)\\}}\textup{d}r\right)$
(28)

and

$\displaystyle\mathbb{E}_{c^{(\theta)}(t)}$
$\displaystyle\left(\int_{0}^{m_{\theta}-t}G^{(\theta)}(r+t,X_{r})\mathbb{I}_{\\{X_{r}<c^{(\theta)}(r+t)\\}}\textup{d}r\right)$
$\displaystyle\qquad-\mathbb{E}_{c^{(\theta)}(t)}\left(\int_{0}^{m_{\theta}-t}\int_{(-\infty,c^{(\theta)}(r+t)-X_{r})}U^{(\theta)}(r+t,X_{r}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{r}>c^{(\theta)}(r+t)\\}}\textup{d}r\right)=0$
(29)

when $t\in[0,t_{b})$ and $x\in\mathbb{R}$. For $t\in[t_{b},m_{\theta})$ and
$x\in\mathbb{R}$, we assume that

$\displaystyle
U^{(\theta)}(t,x)=\mathbb{E}_{x}(\tau_{0}^{+}\wedge(m_{\theta}-t))-\frac{2}{\theta}e^{-\theta
t}[1-\mathbb{E}_{x}(e^{-\theta(\tau_{0}^{+}\wedge(m_{\theta}-t))})].$ (30)

In addition, we assume that

$\displaystyle\int_{(-\infty,c^{(\theta)}(t)-x)}U^{(\theta)}(t,x+y)\Pi(\textup{d}y)+G^{(\theta)}(t,x)\geq
0\qquad\text{for all }t\in[0,t_{b})\text{ and }x>c^{(\theta)}(t).$ (31)

Note that $(U^{(\theta)},c^{(\theta)})$ solving the above equations means that
$U^{(\theta)}(t,c^{(\theta)}(t))=0$ for all $t\in[0,m_{\theta})$ and
$U^{(\theta)}(m_{\theta},x)=0$ for all $x\in\mathbb{R}$. Denote $D_{c}$ as the
“stopping region” under the curve $c^{(\theta)}$, i.e.,
$D_{c}=\\{(t,x)\in[0,m_{\theta}]\times\mathbb{R}:x\geq c^{(\theta)}(t)\\}$ and
recall that $D=\\{(t,x)\in[0,m_{\theta}]\times\mathbb{R}:x\geq
b^{(\theta)}(t)\\}$ is the “stopping region” under the curve $b^{(\theta)}$.
We show that $U^{(\theta)}$ vanishes on $D_{c}$ in the next Lemma.

######  Lemma 3.11.

We have that $U^{(\theta)}(t,x)=0$ for all $(t,x)\in D_{c}$.

###### Proof.

Since the statement is clear for $(t,x)\in[t_{b},m_{\theta})\times[0,\infty)$,
we take $t\in[0,t_{b})$ and $x\geq c^{(\theta)}(t)$. Define $\sigma_{c}$ to be
the first time that the process is outside $D_{c}$ before time $m_{\theta}-t$,
i.e.,

$\displaystyle\sigma_{c}=\inf\\{0\leq s\leq
m_{\theta}-t:X_{s}<c^{(\theta)}(t+s)\\},$

where in this context, we understand that $\inf\emptyset=m_{\theta}-t$. From
the fact that $X_{r}\geq c^{(\theta)}(t+r)$ for all $r<\sigma_{c}$ and the
strong Markov property at time $\sigma_{c}$, we obtain that

$\displaystyle U^{(\theta)}(t,x)$
$\displaystyle=\mathbb{E}_{x}(U^{(\theta)}(t+\sigma_{c},X_{\sigma_{c}}))-\mathbb{E}_{x}\left(\int_{0}^{\sigma_{c}}\int_{(-\infty,c^{(\theta)}(r+t)-X_{r})}U^{(\theta)}(r+t,X_{r}+y)\Pi(\textup{d}y)\textup{d}r\right)$
$\displaystyle=\mathbb{E}_{x}(U^{(\theta)}(t+\sigma_{c},X_{\sigma_{c}})\mathbb{I}_{\\{\sigma_{c}<m_{\theta}-t,X_{\sigma_{c}}<c^{(\theta)}(t+\sigma_{c})\\}})$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\int_{0}^{\sigma_{c}}\int_{(-\infty,c^{(\theta)}(r+t)-X_{r})}U^{(\theta)}(r+t,X_{r}+y)\Pi(\textup{d}y)\textup{d}r\right),$

where the last equality follows since $U^{(\theta)}(m_{\theta},x)=0$ for all
$x\in\mathbb{R}$ and $U^{(\theta)}(t,c^{(\theta)}(t))=0$ for all
$t\in[0,t_{b})$. Then, applying the compensation formula for Poisson random
measures (see Kyprianou, (2014) Theorem 4.4) we get

$\displaystyle\mathbb{E}_{x}(U^{(\theta)}(t+\sigma_{c},X_{\sigma_{c}})\mathbb{I}_{\\{\sigma_{c}<m_{\theta}-t,X_{\sigma_{c}}<c^{(\theta)}(t+\sigma_{c})\\}})$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\sigma_{c}}\int_{(-\infty,c^{(\theta)}(t+r)-X_{t+r})}U^{(\theta)}(t+r,X_{r}+y)\Pi(\textup{d}y)\textup{d}r\right).$

Hence $U^{(\theta)}(t,x)=0$ for all $(t,x)\in D_{c}$ as we claimed. ∎

The next lemma shows that $U^{(\theta)}$ can be expressed as an integral
involving only the gain function $G^{(\theta)}$ stopped at the first time the
process enters the set $D_{c}$. As a consequence, $U^{(\theta)}$ dominates the
function $V^{(\theta)}$.

######  Lemma 3.12.

We have that $U^{(\theta)}(t,x)\geq V^{(\theta)}(t,x)$ for all
$(x,t)\in\mathbb{R}\times[0,m_{\theta}]$,

###### Proof.

Note that we can assume that $t\in[0,t_{b})$ because for $(t,x)\in D_{c}$, we
have that $U^{(\theta)}(t,x)=0\geq V^{(\theta)}(t,x)$ and for
$t\in[t_{b},m_{\theta})$, $U^{(\theta)}(t,x)=V^{(\theta)}(t,x)$ for all
$x\in\mathbb{R}$. Consider the stopping time

$\displaystyle\tau_{c}=\inf\\{s\in[0,m_{\theta}-t]:X_{s}\geq
c^{(\theta)}(t+s)\\}.$

Let $x\leq c^{(\theta)}(t)$, using the fact that $X_{r}<c^{(\theta)}(t+r)$ for
all $r\leq\tau_{c}$ and the strong Markov property at time $\tau_{c}$, we
obtain that

$\displaystyle
U^{(\theta)}(t,x)=\mathbb{E}_{x}\left(\int_{0}^{\tau_{c}}G^{(\theta)}(r+t,X_{r})\textup{d}r\right)+\mathbb{E}_{x}(U^{(\theta)}(t+\tau_{c},X_{\tau_{c}}))=\mathbb{E}_{x}\left(\int_{0}^{\tau_{c}}G^{(\theta)}(r+t,X_{r})\textup{d}r\right),$
(32)

where the second equality follows since $X$ creeps upwards and therefore
$X_{\tau_{c}}=c^{(\theta)}(t+\tau_{c})$ for $\\{\tau_{c}<m_{\theta}-t\\}$ and
$U^{(\theta)}(m_{\theta},x)=0$ for all $x\in\mathbb{R}$. Then from the
definition of $V^{(\theta)}$ (see (10)), we have that

$\displaystyle
U^{(\theta)}(t,x)\geq\inf_{\tau\in\mathcal{T}}\mathbb{E}_{t,x}\left(\int_{0}^{\tau}G^{(\theta)}(X_{t+r},t+r)dr\right)=V^{(\theta)}(t,x).$

Therefore $U^{(\theta)}\geq V^{(\theta)}$ on $[0,m_{\theta}]\times\mathbb{R}$.
∎

We proceed by showing that the function $c^{(\theta)}$ is dominated by
$b^{(\theta)}$. In the upcoming lemmas, we show that equality indeed holds.

######  Lemma 3.13.

We have that $b^{(\theta)}(t)\geq c^{(\theta)}(t)$ for all
$t\in[0,m_{\theta})$.

###### Proof.

The statement is clear for $t\in[t_{b},m_{\theta})$. We prove the statement by
contradiction. Suppose that there exists a value $t_{0}\in[0,t_{b})$ such that
$b^{(\theta)}(t_{0})<c^{(\theta)}(t_{0})$ and take
$x\in(b^{(\theta)}(t_{0}),c^{(\theta)}(t_{0}))$. Consider the stopping time

$\displaystyle\sigma_{b}=\inf\\{s\in[0,m_{\theta}-t_{0}]:X_{s}<b^{(\theta)}(t_{0}+s)\\}.$

Applying the strong Markov property to the filtration at time $\sigma_{b}$, we
obtain that

$\displaystyle U^{(\theta)}(t_{0},x)$
$\displaystyle=\mathbb{E}_{x}(U^{(\theta)}(t_{0}+\sigma_{b},X_{\sigma_{b}}))+\mathbb{E}_{x}\left(\int_{0}^{\sigma_{b}}G^{(\theta)}(t_{0}+r,X_{r})\mathbb{I}_{\\{X_{r}<c^{(\theta)}(t_{0}+r)\\}}\textup{d}r\right)$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\int_{0}^{\sigma_{b}}\int_{(-\infty,0)}U^{(\theta)}(t+r,X_{r}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{r}>c^{(\theta)}(t_{0}+r)\\}}\textup{d}r\right),$

where we used the fact that $U^{(\theta)}(t,x)=0$ for all $(t,x)\in D_{c}$.
From Lemma 3.12 and the fact that $U^{(\theta)}\leq 0$ (by assumption), we
have that for all $t\in[0,m_{\theta})$ and $x>b^{(\theta)}(t)$,
$U^{(\theta)}(t,x)=0$. Hence, by the compensation formula for Poisson random
measures, we obtain that

$\displaystyle 0$ $\displaystyle\geq U^{(\theta)}(t_{0},x)$
$\displaystyle=\mathbb{E}_{x}(U^{(\theta)}(t_{0}+\sigma_{b},X_{\sigma_{b}})\mathbb{I}_{\\{\sigma_{b}<m_{\theta}-t,X_{\sigma_{b}}<b^{(\theta)}(t_{0}+\sigma_{b})\\}})+\mathbb{E}_{x}\left(\int_{0}^{\sigma_{b}}G^{(\theta)}(t_{0}+r,X_{r})\mathbb{I}_{\\{X_{r}<c^{(\theta)}(t_{0}+r)\\}}\textup{d}r\right)$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\int_{0}^{\sigma_{b}}\int_{(-\infty,0)}U^{(\theta)}(t+r,X_{r}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{r}>c^{(\theta)}(t_{0}+r)\\}}\textup{d}r\right)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\sigma_{b}}\int_{(-\infty,0)}U^{(\theta)}(t_{0}+r,X_{r}+y)\Pi(\textup{d}y)\textup{d}r\right)+\mathbb{E}_{x}\left(\int_{0}^{\sigma_{b}}G^{(\theta)}(t_{0}+r,X_{r})\mathbb{I}_{\\{X_{r}<c^{(\theta)}(t_{0}+r)\\}}\textup{d}r\right)$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\int_{0}^{\sigma_{b}}\int_{(-\infty,0)}U^{(\theta)}(t+r,X_{r}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{r}>c^{(\theta)}(t_{0}+r)\\}}\textup{d}r\right)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\sigma_{b}}\left[\int_{(-\infty,0)}U^{(\theta)}(t_{0}+r,X_{r}+y)\Pi(\textup{d}y)+G^{(\theta)}(t_{0}+r,X_{r})\right]\mathbb{I}_{\\{X_{r}<c^{(\theta)}(t_{0}+r)\\}}\textup{d}r\right).$

Recall from the discussion after Lemma 3.5 that the function
$\varphi_{t}^{(\theta)}$ is strictly positive on $D$. Hence, we obtain that
for all $(t,x)\in D$,

$\displaystyle\int_{(-\infty,0)}U^{(\theta)}(t,x+y)\Pi(\textup{d}y)+G^{(\theta)}(t,x)\geq\int_{(-\infty,0)}V^{(\theta)}(t,x+y)\Pi(\textup{d}y)+G^{(\theta)}(t,x)=\varphi_{t}^{(\theta)}(t,x)>0.$

The assumption that $b^{(\theta)}(t_{0})<c^{(\theta)}(t_{0})$ together with
the continuity of the functions $b^{(\theta)}$ and $c^{(\theta)}$ mean that
there exists $s_{0}\in(t_{0},m_{\theta})$ such that
$b^{(\theta)}(r)<c^{(\theta)}(r)$ for all $r\in[t_{0},s_{0}]$. Consequently,
the $\mathbb{P}_{x}$ probability of $X$ spending a strictly positive amount of
time (with respect to Lebesgue measure) in this region is strictly positive.
We can then conclude that

$\displaystyle
0\geq\mathbb{E}_{x}\left(\int_{0}^{\sigma_{b}}\left[\int_{(-\infty,0)}U^{(\theta)}(t_{0}+r,X_{r}+y)\Pi(\textup{d}y)+G^{(\theta)}(t_{0}+r,X_{r})\right]\mathbb{I}_{\\{X_{r}<c^{(\theta)}(t_{0}+r)\\}}\textup{d}r\right)>0.$

This is a contradiction and therefore we conclude that $b^{(\theta)}(t)\geq
c^{(\theta)}(t)$ for all $t\in[0,m_{\theta})$. ∎

Note that the definition of $U^{(\theta)}$ on
$[t_{b},m_{\theta})\times\mathbb{R}$ (see equation (30)) together with
condition (31) imply that

$\displaystyle\int_{(-\infty,0)}U^{(\theta)}(t,x+y)\Pi(\textup{d}y)+G^{(\theta)}(t,x)\geq
0$

for all $t\in[0,m_{\theta})$ and $x>c^{(\theta)}(t)$. The next Lemma shows
that $U^{(\theta)}$ and $V^{(\theta)}$ coincide.

######  Lemma 3.14.

We have that $b^{(\theta)}(t)=c^{(\theta)}(t)$ for all $t\geq 0$ and hence
$V^{(\theta)}=U^{(\theta)}$.

###### Proof.

We prove that $b^{(\theta)}=c^{(\theta)}$ by contradiction. Assume that there
exists $s_{0}$ such that $b^{(\theta)}(s_{0})>c^{(\theta)}(s_{0})$. Since
$c^{(\theta)}(t)=b^{(\theta)}(t)=0$ for all $t\in[t_{b},m_{\theta})$, we
deduce that $s_{0}\in[0,t_{b})$. Let $\tau_{b}$ be the stopping time

$\displaystyle\tau_{b}=\inf\\{t\geq 0:X_{s}\geq b^{(\theta)}(s_{0}+t)\\}.$

With the Markov property applied to the filtration at time $\tau_{b}$, we
obtain that for any $x\in(c^{(\theta)}(s_{0}),b^{(\theta)}(s_{0}))$

$\displaystyle\mathbb{E}_{x}(U^{(\theta)}(s_{0}+\tau_{b},X_{\tau_{b}}))$
$\displaystyle=U^{(\theta)}(s_{0},x)-\mathbb{E}_{x}\left(\int_{0}^{\tau_{b}}G^{(\theta)}(r+s_{0},X_{r})\mathbb{I}_{\\{X_{r}<c^{(\theta)}(r+s_{0})\\}}\textup{d}r\right)$
$\displaystyle\qquad+\mathbb{E}_{x}\left(\int_{0}^{\tau_{b}}\int_{(-\infty,0)}U^{(\theta)}(r+s_{0},X_{r}+y)\mathbb{I}_{\\{X_{r}>c^{(\theta)}(r+s_{0})\\}}\Pi(\textup{d}y)\textup{d}r\right)$
$\displaystyle\geq
V^{(\theta)}(s_{0},x)-\mathbb{E}_{x}\left(\int_{0}^{\tau_{b}}G^{(\theta)}(r+s_{0},X_{r})\mathbb{I}_{\\{X_{r}<c^{(\theta)}(r+s_{0})\\}}\textup{d}r\right)$
$\displaystyle\qquad+\mathbb{E}_{x}\left(\int_{0}^{\tau_{b}}\int_{(-\infty,0)}U^{(\theta)}(r+s_{0},X_{r}+y)\mathbb{I}_{\\{X_{r}>c^{(\theta)}(r+s_{0})\\}}\Pi(\textup{d}y)\textup{d}r\right)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\tau_{b}}G^{(\theta)}(r+s_{0},X_{r})\mathbb{I}_{\\{X_{r}\geq
c^{(\theta)}(r+s_{0})\\}}\textup{d}r\right)$
$\displaystyle\qquad+\mathbb{E}_{x}\left(\int_{0}^{\tau_{b}}\int_{(-\infty,0)}U^{(\theta)}(r+s_{0},X_{r}+y)\mathbb{I}_{\\{X_{r}>c^{(\theta)}(r+s_{0})\\}}\Pi(\textup{d}y)\textup{d}r\right),$

where the second inequality follows from the fact that $U^{(\theta)}\geq
V^{(\theta)}$ (see Lemma 3.12) and the last equality follows as $\tau_{b}$ is
the optimal stopping time for $V^{(\theta)}(s_{0},x)$. Note that since $X$
creeps upwards, we have that
$U^{(\theta)}(s_{0}+\tau_{b},X_{\tau_{b}})=U^{(\theta)}(s_{0}+\tau_{b},b^{(\theta)}(s_{0}+\tau_{b}))=0$.
Hence,

$\displaystyle\mathbb{E}_{x}$
$\displaystyle\left(\int_{0}^{\tau_{b}}G^{(\theta)}(r+s_{0},X_{r})\mathbb{I}_{\\{X_{r}\geq
c^{(\theta)}(r+s_{0})\\}}\textup{d}r\right)$
$\displaystyle\qquad+\mathbb{E}_{x}\left(\int_{0}^{\tau_{b}}\int_{(-\infty,0)}U^{(\theta)}(r+s_{0},X_{r}+y)\mathbb{I}_{\\{X_{r}>c^{(\theta)}(r+s_{0})\\}}\Pi(\textup{d}y)\textup{d}r\right)\leq
0.$

However, the continuity of the functions $b^{(\theta)}$ and $c^{(\theta)}$
gives the existence of $s_{1}\in(s_{0},m_{\theta})$ such that
$c^{(\theta)}(r)<b^{(\theta)}(r)$ for all $r\in[s_{0},s_{1}]$. Hence, together
with the fact that
$\int_{(-\infty,0)}U^{(\theta)}(x+y,t)\Pi(\textup{d}y)+G^{(\theta)}(x,t)>0$
for all $(t,x)\in D_{c}$ we can conclude that

$\displaystyle\mathbb{E}_{x}$
$\displaystyle\left(\int_{0}^{\tau_{b}}G^{(\theta)}(r+s_{0},X_{r})\mathbb{I}_{\\{X_{r}\geq
c^{(\theta)}(r+s_{0})\\}}\textup{d}r\right)$
$\displaystyle\qquad+\mathbb{E}_{x}\left(\int_{0}^{\tau_{b}}\int_{(-\infty,0)}U^{(\theta)}(r+s_{0},X_{r}+y)\mathbb{I}_{\\{X_{r}>c^{(\theta)}(r+s_{0})\\}}\Pi(\textup{d}y)\textup{d}r\right)>0,$

which shows a contradiction.

∎

##  4 Examples

###  4.1 Brownian motion with drift

Suppose that $X=\\{X_{t},t\geq 0\\}$ is a Brownian motion with drift. That is
for any $t\geq 0$, $X_{t}=\mu t+\sigma B_{t}$, where $\sigma>0$ and
$\mu\in\mathbb{R}$. In this case, we have that

$\displaystyle\psi(\beta)=\mu\beta+\frac{1}{2}\sigma^{2}\beta^{2}$

for all $\beta\geq 0$. Then

$\displaystyle\Phi(q)=\frac{1}{\sigma^{2}}\left[\sqrt{\mu^{2}+2\sigma^{2}q}-\mu\right].$

It is well known that $-\underline{X}_{\widetilde{e}_{\theta}}$ has
exponential distribution (see e.g. Borodin and Salminen, (2002) pp251 or
Kyprianou, (2014) pp 233) with distribution function given by

$\displaystyle
F^{(\theta)}(x)=1-\exp\left(-\frac{x}{\sigma^{2}}\left[\sqrt{\mu^{2}+2\sigma^{2}\theta}+\mu\right]\right)\qquad\text{for
}x>0.$

Denote $\Phi(x;a,b^{2})$ as the distribution function of a Normal random
variable with mean $a\in\mathbb{R}$ and variance $b^{2}$, i.e., for any
$x\in\mathbb{R}$,

$\displaystyle\Phi(x;a,b^{2})=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi
b^{2}}}e^{-\frac{1}{2b^{2}}(y-a)^{2}}\textup{d}y.$

For any $b,s,t\geq 0$ and $x\in\mathbb{R}$, define the function

$\displaystyle K(t,x,s,b)$
$\displaystyle=\mathbb{E}\left(G^{(\theta)}(s+t,X_{s}+x)\mathbb{I}_{\\{X_{s}+x\leq
b\\}}\right).$

Then it can be easily shown that

$\displaystyle K(t,x,s,b)$ $\displaystyle=\Phi(b-x;\mu
s,\sigma^{2}s)-2e^{-\theta(s+t)}\Phi(-x;\mu s,\sigma^{2}s)-2e^{-\theta
t}\exp\left(-\frac{x}{\sigma^{2}}\left[\sqrt{\mu^{2}+2\sigma^{2}\theta}+\mu\right]\right)$
$\displaystyle\qquad\times\left[\Phi(b-x,-s\sqrt{\mu^{2}+2\sigma^{2}\theta},s\sigma^{2})-\Phi(-x,-s\sqrt{\mu^{2}+2\sigma^{2}\theta},s\sigma^{2})\right].$

Thus, we have that $b^{(\theta)}$ satisfies the non-linear integral equation

$\displaystyle\int_{0}^{m_{\theta}-t}K(t,b^{(t)}(t),s,b^{(\theta)}(t+s))\textup{d}s=0$

for all $t\in[0,m_{\theta})$ and the value function $V^{(\theta)}$ is given by

$\displaystyle
V^{(\theta)}(t,x)=\int_{0}^{m_{\theta}-t}K(t,x,s,b^{(\theta)}(t+s))\textup{d}s$

for all $(t,x)\in\mathbb{R}_{+}\times\mathbb{R}$. Note that we can approximate
the integrals above by Riemann sums so a numerical approximation can be
implement. Indeed, take $n\in\mathbb{Z}_{+}$ sufficiently large and define
$h=m_{\theta}/n$. For each $k\in\\{0,1,2,\ldots,n\\}$, we define $t_{k}=kh$.
Then the sequence of times $\\{t_{k},k=0,1,\ldots,n\\}$ is a partition of the
interval $[0,m_{\theta}]$. Then, for any $x\in\mathbb{R}$ and
$t\in[t_{k},t_{k+1})$ for $k\in\\{0,1,\ldots,n-1\\}$ we approximate
$V^{(\theta)}(t,x)$ by

$\displaystyle
V^{(\theta)}_{h}(t_{k},x)=\sum_{i=k}^{n-1}K(t_{k},x,t_{i-k+1},b_{i})h,$

where the sequence $\\{b_{k},k=0,1,\ldots,n-1\\}$ is a solution to

$\displaystyle\sum_{i=k}^{n-1}K(t_{k},x,t_{i-k+1},b_{i})=0$

for each $k\in\\{0,1,\ldots,n-1\\}$. Note that the sequence
$\\{b_{k},k=0,1,\ldots,n\\}$ is a numerical approximation to the sequence
$\\{b^{(\theta)}(t_{k}),k=0,1,\ldots,n-1\\}$ (for $n$ sufficiently large) and
can be calculated by using backwards induction. In the Figure 3, we show a
numerical calculation of the equations above. The parameters used are $\mu=2$
and $\sigma=1$, whereas we chose $m_{\theta}=10$.

Figure 2: Brownian motion with drift $\mu=2$ and $\sigma=1$. Left hand side:
Optimal boundary; Right hand side: Value function fixing $t=1$.

###  4.2 Brownian motion with exponential jumps

Let $X=\\{X_{t},t\geq 0\\}$ be a compound Poisson process perturbed by a
Brownian motion, that is

$\displaystyle X_{t}=\sigma B_{t}+\mu t-\sum_{i=1}^{N_{t}}Y_{i},$ (33)

where $B=\\{B_{t},t\geq 0\\}$ is a standard Brownian motion, $N=\\{N_{t},t\geq
0\\}$ is Poisson process with rate $\lambda$ independent of $B$,
$\mu\in\mathbb{R}$, $\sigma>0$ and the sequence $\\{Y_{1},Y_{2},\ldots\\}$ is
a sequence of independent random variables exponentially distributed with mean
$1/\rho>0$. Then in this case, the Laplace exponent is derived as

$\displaystyle\varphi(\beta)=\frac{\sigma^{2}}{2}\beta^{2}+\mu\beta-\frac{\lambda\beta}{\rho+\beta}.$

Its Lévy measure, given by $\Pi(\textup{d}y)=\lambda\rho e^{\rho
y}\mathbb{I}_{\\{y<0\\}}\textup{d}y$ is a finite measure and $X$ is a process
of infinite variation. According to Kuznetsov et al., (2013), the scale
function in this case is given for $q\geq 0$ and $x\geq 0$ by,

$\displaystyle
W^{(q)}(x)=\frac{e^{\Phi(q)x}}{\psi^{\prime}(\Phi(q))}+\frac{e^{\zeta_{1}(q)x}}{\psi^{\prime}(\zeta_{1}(q))}+\frac{e^{\zeta_{2}(q)x}}{\psi^{\prime}(\zeta_{2}(q))},$

where $\zeta_{2}(q),\zeta_{1}(q)$ and $\Phi(q)$ are the three real solutions
to the equation $\psi(\beta)=q$, which satisfy
$\zeta_{2}(q)<-\rho<\zeta_{1}(q)<0<\Phi(q)$. The second scale function,
$Z^{(q)}$, takes the form

$\displaystyle Z^{(q)}(x)$
$\displaystyle=1+q\left[\frac{e^{\Phi(q)x}-1}{\Phi(q)\psi^{\prime}(\Phi(q))}+\frac{e^{\zeta_{1}(q)x}-1}{\zeta_{1}(q)\psi^{\prime}(\zeta_{1}(q))}+\frac{e^{\zeta_{2}(q)x}-1}{\zeta_{2}(q)\psi^{\prime}(\zeta_{2}(q))}\right].$

Note that since we have exponential jumps (and hence
$\Pi(\textup{d}y)=\lambda\rho e^{\rho y}\mathbb{I}_{\\{y<0\\}}$), we have that
for all $t\in[0,m_{\theta})$ and $x>0$,

$\displaystyle\int_{(-\infty,-x)}V^{(\theta)}(t,b^{(\theta)}(t)+x+y)\Pi(\textup{d}y)$
$\displaystyle=e^{-\rho
x}\int_{(-\infty,0)}V^{(\theta)}(t,b^{(\theta)}(t)+y)\Pi(\textup{d}y).$

Then, for any $(t,x)\in[0,m_{\theta}]\times\mathbb{R}$, equation (3.9) reads
as

$\displaystyle V^{(\theta)}(t,x)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}G^{(\theta)}(r+t,X_{r})\mathbb{I}_{\\{X_{r}<b^{(\theta)}(r+t)\\}}\textup{d}r\right)$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}e^{-\rho(X_{r}-b^{(\theta)}(r+t))}\mathcal{V}(r+t)\mathbb{I}_{\\{X_{r}>b^{(\theta)}(r+t)\\}}\textup{d}r\right)$

where for any $r,s\in[0,m_{\theta})$, $b\geq 0$ and $x\in\mathbb{R}$,

$\displaystyle\mathcal{V}(t)$
$\displaystyle=\int_{(-\infty,0)}V^{(\theta)}(t,b^{(\theta)}(t)+y)\Pi(\textup{d}y).$

Note that the equation above suggest that in order to find a numerical value
of $b^{(\theta)}$ using Theorem 3.9 we only need to know the values of the
function $\mathcal{V}$ and not the values of
$\int_{(-\infty,0)}V^{(\theta)}(t,x+y)\Pi(\textup{d}y)$ for all
$t\in[0,m_{\theta}]$ and $x>b^{(\theta)}(t)$. The next Corollary confirms that
notion.

######  Corollary 4.1.

Let $\theta>0$. Assume that $X=\\{X_{t},t\geq 0\\}$ is of the form (33) with
$\mu\in\mathbb{R}$, $\sigma,\lambda,\rho>0$. Suppose that $c^{(\theta)}$ and
$\mathcal{U}$ are continuous functions on $[0,m_{\theta})$ such that
$c^{(\theta)}\geq h^{(\theta)}$ and
$0\geq\mathcal{U}(t)\geq-G^{(\theta)}(t,c^{(\theta)}(t))$ for all
$t\in[0,m_{\theta})$. For any $(t,x)\in[0,m_{\theta}]\times\mathbb{R}$ we
define the function

$\displaystyle U^{(\theta)}(t,x)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}G^{(\theta)}(r+t,X_{r})\mathbb{I}_{\\{X_{r}<c^{(\theta)}(r+t)\\}}\textup{d}r\right)$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}e^{-\rho(X_{r}-c^{(\theta)}(r+t))}\mathcal{U}(r+t)\mathbb{I}_{\\{X_{r}>c^{(\theta)}(r+t)\\}}\textup{d}r\right).$

Further assume that there exists a value $h>0$ such that $U^{(\theta)}(t,x)=0$
for any $t\in[0,m_{\theta})$ and $x\in[b^{(\theta)}(t),b^{(\theta)}(t)+h]$. If
$U^{(\theta)}$ is a non-positive function, we have that
$c^{(\theta)}=b^{(\theta)}$ and $U^{(\theta)}=V^{(\theta)}$.

###### Proof.

First note that, since $X$ is of infinite variation,
$\mathbb{P}_{x}(X_{r}=c^{(\theta)}(r+t))=0$ for all $r,t\in[0,m_{\theta})$
such that $r+t<m_{\theta}$ and $x\in\mathbb{R}$. Hence, by continuity of
$G^{(\theta)}$ and $\mathcal{U}$, and by dominated convergence theorem, we
have that $U^{(\theta)}$ is continuous. By means of Theorem 3.9 is enough to
show that $U^{(\theta)}$ satisfies the integral equation,

$\displaystyle U^{(\theta)}(t,x)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}G^{(\theta)}(r+t,X_{r})\mathbb{I}_{\\{X_{r}<c^{(\theta)}(r+t)\\}}\textup{d}r\right)$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}\int_{(-\infty,c^{(\theta)}(r+t)-X_{r})}U^{(\theta)}(r+t,X_{r}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{r}>c^{(\theta)}(r+t)\\}}\textup{d}r\right)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}G^{(\theta)}(r+t,X_{r})\mathbb{I}_{\\{X_{r}<c^{(\theta)}(r+t)\\}}\textup{d}r\right)$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\int_{0}^{m_{\theta}-t}e^{-\rho(X_{r}-c^{(\theta)}(r+t))}H(r+t)\mathbb{I}_{\\{X_{r}>c^{(\theta)}(r+t)\\}}\textup{d}r\right),$

where
$H(r)=\int_{(-\infty,0)}U^{(\theta)}(r,c^{(\theta)}(r)+y)\Pi(\textup{d}y)$ for
all $r\in[0,m_{\theta})$ and in the last equality we used the explicit form of
$\Pi(\textup{d}y)$. Then it suffices to show that $H(t)=\mathcal{U}(t)$ for
all $t\in[0,m_{\theta})$.

Let $t\geq 0$. For any $\delta\in(0,m_{\theta}-t)$, consider the stopping time

$\displaystyle\tau_{\delta}=\inf\\{s\in[0,\delta]:X_{s}\notin[c^{(\theta)}(s+t),c^{(\theta)}(s+t)+h]\\}.$

Note that for any $s<\tau_{\delta}$ we have that
$X_{s}\in(c^{(\theta)}(s+t),c^{(\theta)}(s+t)+h)$ and
$X_{\delta}\in(c^{(\theta)}(s+t),c^{(\theta)}(s+t)+h)$ in the event
$\\{\tau_{\delta}=\delta\\}$. Then using the strong Markov property at time
$\tau_{\delta}$, we have that for any
$x\in[c^{(\theta)}(t),c^{(\theta)}(t)+h)$,

$\displaystyle 0$
$\displaystyle=U^{(\theta)}(t,x)=-\mathbb{E}_{x}\left(\int_{0}^{\tau_{\delta}}e^{-\rho(X_{r}-c^{(\theta)}(r+t))}\mathcal{U}(r+t)\textup{d}r\right)+\mathbb{E}_{x}(U^{(\theta)}(t+\tau_{\delta},X_{\tau_{\delta}})\mathbb{I}_{\\{X_{\tau_{\delta}}<c^{(\theta)}(t+\tau_{\delta})\\}}),$

where in the last equality we used the fact that $U^{(\theta)}(t,x)=0$ for all
$x\in[c^{(\theta)}(t),c^{(\theta)}(t)+h]$, the continuity of $c^{(\theta)}$
and that $X$ can only cross above $c^{(\theta)}$ by creeping. By using the
compensation formula for Poisson Random measures, we obtain that

$\displaystyle\mathbb{E}_{x}(U^{(\theta)}(t+\tau_{\delta},X_{\tau_{\delta}})\mathbb{I}_{\\{X_{\tau_{\delta}}<c^{(\theta)}(t+\tau_{\delta})\\}})$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\tau_{\delta}}\int_{(-\infty,0)}U^{(\theta)}(t+r,X_{r}+y)\mathbb{I}_{\\{X_{r}+y<c^{(\theta)}(t+r)\\}}\textup{d}r\Pi(\textup{d}y)\right)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\tau_{\delta}}e^{-\rho(X_{r}-c^{(\theta)}(r+t))}H(r+t)\textup{d}r\right).$

Hence we conclude that for any $\delta>0$,

$\displaystyle 0$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\tau_{\delta}}e^{-\rho(X_{r}-c^{(\theta)}(r+t))}[H(r+t)-\mathcal{U}(r+t)]\textup{d}r\right)$
(34)

and hence,

$\displaystyle
0\leq\mathbb{E}_{x}\left(\int_{0}^{\tau_{\delta}}[H(r+t)-\mathcal{U}(r+t)]\textup{d}r\right).$

By continuity of $H$ and $\mathcal{U}$ we obtain that

$\displaystyle 0\leq\lim_{\delta\downarrow
0}\frac{1}{\mathbb{E}_{x}(\tau_{\delta})}\mathbb{E}_{x}\left(\int_{0}^{\tau_{\delta}}[H(r+t)-\mathcal{U}(r+t)]\textup{d}r\right)=H(t)-\mathcal{U}(t).$

Moreover, from equation (34) we conclude that $H(t)=\mathcal{U}(t)$ for all
$t\in[0,m_{\theta})$ and the conclusion holds. ∎

Hence, for any $(t,x)\in[0,m_{\theta})\times\mathbb{R}$, we can write

$\displaystyle V^{(\theta)}(t,x)$
$\displaystyle=\int_{0}^{m_{\theta}-t}K_{1}(t,x,r,b^{(\theta)}(r+t))\textup{d}r-\int_{0}^{m_{\theta}-t}\mathcal{V}(r+t)K_{2}(t,x,r,b^{(\theta)}(r+t))\textup{d}r,$

where for any $r,s\in[0,m_{\theta})$, $b\geq 0$ and $x\in\mathbb{R}$,

$\displaystyle\mathcal{V}(t)$
$\displaystyle=\int_{(-\infty,0)}V^{(\theta)}(t,b^{(\theta)}(t)+y)\Pi(\textup{d}y)$
$\displaystyle K_{1}(t,x,s,b)$
$\displaystyle=\mathbb{E}\left(G^{(\theta)}(s+t,X_{s}+x)\mathbb{I}_{\\{X_{s}<b-x\\}}\right)$
$\displaystyle K_{2}(t,x,s,b)$
$\displaystyle=\mathbb{E}\left(e^{-\rho(X_{s}+x-b)}\mathbb{I}_{\\{X_{s}>b-x\\}}\right).$

Take a value $h_{0}>0$ sufficiently small. Hence the functions $b^{(\theta)}$
and $\mathcal{V}$ satisfy the integral equations

$\displaystyle\int_{0}^{m_{\theta}-t}$ $\displaystyle
K_{1}(t,b^{(\theta)}(t),r,b^{(\theta)}(r+t))\textup{d}r-\int_{0}^{m_{\theta}-t}\mathcal{V}(r+t,b^{(\theta)}(r+t))K_{2}(t,b^{(\theta)}(t),r,b^{(\theta)}(r+t))\textup{d}r=0$
$\displaystyle\int_{0}^{m_{\theta}-t}$ $\displaystyle
K_{1}(t,b^{(\theta)}(t)+h_{0},r,b^{(\theta)}(r+t))\textup{d}r-\int_{0}^{m_{\theta}-t}\mathcal{V}(r+t,b^{(\theta)}(r+t))K_{2}(t,b^{(\theta)}(t)+h_{0},r,b^{(\theta)}(r+t))\textup{d}r=0$

for all $t\in[0,m_{\theta}]$. We can approximate the integrals above by
Riemann sums so a numerical approximation can be implement. Indeed, take
$n\in\mathbb{Z}_{+}$ sufficiently large and define $h=m_{\theta}/n$. For each
$k\in\\{0,1,2,\ldots,n\\}$, we define $t_{k}=kh$. Then the sequence of times
$\\{t_{k},k=0,1,\ldots,n\\}$ is a partition of the interval $[0,m_{\theta}]$.
Then, for any $x\in\mathbb{R}$ and $t\in[t_{k},t_{k+1})$, we approximate
$V^{(\theta)}(t,x)$ by

$\displaystyle
V^{(\theta)}_{h}(t_{k},x)=\sum_{i=k}^{n-1}[K_{1}(t_{k},x,t_{i-k+1},b_{i})-\mathcal{V}_{i}K_{2}(t_{k},x,t_{i-k+1},b_{i})]h,$

where the sequence $\\{(b_{k},\mathcal{V}_{k}),k=0,1,\ldots,n-1\\}$ is a
solution to

$\displaystyle\sum_{i=k}^{n-1}[K_{1}(t_{k},b_{k},t_{i-k+1},b_{i})-\mathcal{V}_{i}K_{2}(t_{k},b_{k},t_{i-k+1},b_{i})]=0$
$\displaystyle\sum_{i=k}^{n-1}[K_{1}(t_{k},b_{k}+h_{0},t_{i-k+1},b_{i})-\mathcal{V}_{i}K_{2}(t_{k},b_{k}+h_{0},t_{i-k+1},b_{i})]=0$

for each $k\in\\{0,1,\ldots,n-1\\}$. Note that, for $n$ sufficiently large,
the sequence $\\{(b_{k},\mathcal{V}_{k}),k=0,1,\ldots,n\\}$ is a numerical
approximation to the sequence
$\\{(b^{(\theta)}(t_{k}),\mathcal{V}(t_{k})),k=0,1,\ldots,n\\}$ (provided that
$V^{(\theta)}_{h}\leq 0$) and can be calculated by using backwards induction.
The functions $K_{1}$ and $K_{2}$ can be estimated using simulation methods.
In the figures below we include a plot of the numerical calculation of
$b^{(\theta)}$ and $V^{(\theta)}(0,x)$ using the parameters
$\theta=\log(2)/10$, $\mu=3$, $\sigma=\lambda=\rho=1$.

Figure 3: Brownian motion with drift perturbed by a compound Poisson process
with exponential sized jumps with $\mu=3$ and $\sigma=\lambda=\rho=1$. Left
hand side: Optimal boundary; Right hand side: Value function fixing $t=0$.

## References

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