#  $L^{p}$ optimal prediction of the last zero
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 $X$ drifting to infinity, we are
interested in finding a stopping time which minimises the $L^{p}$ distance
($p>1$) with $g$, the last time $X$ is negative. The solution is substantially
more difficult compared to the $p=1$ case for which it was shown in Baurdoux
and Pedraza, (2018) that it is optimal to stop as soon as $X$ exceeds a
constant barrier. In the case of $p>1$ treated here, we prove that solving
this optimal prediction problem is equivalent to solving an optimal stopping
problem in terms of a two-dimensional strong Markov process which incorporates
the length of the current excursion away from $0$. We show that an optimal
stopping time is now given by the first time that $X$ exceeds a non-increasing
and non-negative curve depending on the length of the current excursion away
from $0$. We also show that smooth fit holds at the boundary.

Keywords: Lévy processes, optimal prediction, optimal stopping.

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

##  1 Introduction

In recent years last passage times have received a considerable attention in
the literature. For instance, in risk theory, the capital of an insurance
company over time is studied. In the classical risk theory this is modelled by
the Cramér–Lundberg process, defined as a compound Poisson process with drift.
In more recent literature, this process has been replaced by a more general
spectrally negative Lévy process. A key quantity of interest is the moment of
ruin, which is classically defined as the first passage time below zero.
Consider instead the situation where after the moment of ruin the company may
have funds to endure a negative capital for some time. In that case, the last
passage time below zero becomes an important quantity to be studied. In this
framework, in Chiu et al., (2005) the Laplace transform of the last passage
time is derived.
Secondly, Paroissin and Rabehasaina, (2015) consider spectrally positive Lévy
processes as a degradation model. In a traditional setting, the failure time
of a device is the first time the model hits a certain critical level $b$.
However, another approach has been considered in the literature. For example,
in Barker and Newby, (2009) they considered the failure time as a last passage
time. After the last passage time the process can never go back to this level
meaning that the device is “beyond repair”.
Thirdly, Egami and Kevkhishvili, (2017) studied the last passage time of a
general time-homogeneous transient diffusion with applications to credit risk
management. They proposed the leverage process (the ratio of a company asset
process over its debt) as a geometric Brownian motion over a process that
grows at a risk free rate. It is shown there that the last passage time of the
leverage ratio is equivalent to a last passage time of a Brownian motion with
drift. In this setting the last passage represents the situation where the
company cannot recover to normal business conditions after this time has
occurred.
An important feature of last passage times is that they are random times which
are not stopping times. In the recent literature the problem of finding a
stopping time that approximates last passage times has been solved in various.
There are for example various papers in which the approximation is in $L_{1}$
sense. To mention a few: Du Toit et al., (2008) predicted the last zero of a
Brownian motion with drift in a finite horizon setting; du Toit and Peskir,
(2008) predicted the time of the ultimate maximum at time $t=1$ for a Brownian
motion with drift is attained; Glover et al., (2013) predicted the time in
which a transient difussion attains its ultimate minimum; Glover and Hulley,
(2014) predicted the last passage time of a level $z>0$ for an arbitrary
nonnegative time-homogeneous transient diffusion; Baurdoux and Van Schaik,
(2014) predicted the time at which a Lévy process attains its ultimate
supremum and Baurdoux et al., (2016) predicted when a positive self-similar
Markov process attain its pathwise global supremum or infimum before hitting
zero for the first time and Baurdoux and Pedraza, (2018) predicted the last
zero of a spectrally negative Lévy process.

In this paper we consider the problem in an $L^{p}$ sense, i.e. we are
interested in solving

$\displaystyle\inf_{\tau\in\mathcal{T}}\mathbb{E}(|\tau-g|^{p})$

where $g=\sup\\{t\geq 0:X_{t}\leq 0\\}$ is the last time a spectrally negative
Lévy process drifting to infinity is below the level zero and $p>1$. The case
when $p=1$ was solved in Baurdoux and Pedraza, (2018) for the spectrally
negative case. An optimal stopping time in this case is the first time the
process crosses above a fixed level $a^{*}>0$ which is characterised in terms
of the distribution function of the infimum of the process. The case $p>1$ is
substantially more complex, as an optimal stopping time now depends on the
length of the current excursion above the level zero. In particular, we show
that there exists a non-negative and non-increasing curve $b$ such that the
stopping time $\tau_{D}=\inf\\{t\geq 0:X_{t}\geq b(U_{t})\\}$ is optimal,
where $U_{t}=t-\sup\\{0\leq s\leq t:X_{s}\leq 0\\}$ is the length of the
current excursion above zero. Therefore, stopping is optimal when $X$ is
sufficiently large or has stayed for a sufficiently long period of time above
zero (see Figure 1).

Figure 1: Black line: $t\mapsto X_{t}$; Blue line: $t\mapsto b(U_{t})$.

The paper is organised as follows. Section 2 gives a short overview of main
results and notation on the fluctuation theory of spectrally negative Lévy
processes. In Section 3 we formulate the optimal prediction problem and we
show that it is equivalent to an optimal stopping problem which is solved in
Section 4. In particular, we show an optimal stopping time is given by the
first time $X$ exceeds a boundary $b$ which depends on the length of the
current excursion above zero. We derive various properties of $b$. For
example, in Lemma 4.10 we show that $b$ is continuous and in Theorem 4.13 we
show that smooth fit holds at the boundary. The main result, Theorem 4.14,
provides a characterisation of $b$ and the value function of the optimal
stopping problem. In Section 5 we provide two numerical examples: Firstly,
when $X$ is a Brownian motion with drift, and secondly when $X$ is a Brownian
motion with exponential jumps. Finally, some of the more technical proofs are
deferred to the Appendix.

##  2 Preliminaries

A Lévy process $X=\\{X_{t},t\geq 0\\}$ is an almost surely càdlàg process that
has independent and stationary increments such that $\mathbb{P}(X_{0}=0)=1$.
We take it to be 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 of Bichteler and Klaus,
(2002)) From the stationary and independent increments property the law of $X$
is characterised by the distribution of $X_{1}$. We hence define the
characteristic exponent of $X$, $\Psi(\theta):=-\log(\mathbb{E}(e^{i\theta
X_{1}}))$, $\theta\in\mathbb{R}$. The Lévy–Khintchine formula guarantees the
existence of constants, $\mu\in\mathbb{R}$, $\sigma\geq 0$ and a measure $\Pi$
concentrated in $\mathbb{R}\setminus\\{0\\}$ with the property that
$\int_{\mathbb{R}}(1\wedge x^{2})\Pi(\textup{d}x)<\infty$ (called the Lévy
measure) such that

$\displaystyle\Psi(\theta)=i\mu\theta+\frac{1}{2}\sigma^{2}\theta^{2}-\int_{\mathbb{R}}(e^{i\theta
y}-1-i\theta y\mathbb{I}_{\\{|y|<1\\}})\Pi(\textup{d}y).$

Moreover, from the Lévy–Itô decomposition we can write

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

where $N$ is a Poisson random measure on $\mathbb{R}^{+}\times\mathbb{R}$ with
intensity $\textup{d}t\times\Pi(\textup{d}x)$. We state now some properties
and facts about Lévy processes. The reader can refer, for example, to Bertoin,
(1998), Sato, (1999) and Kyprianou, (2014) for more details. 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 $\mathbb{P}$.

The process $X$ is a spectrally negative Lévy process if it has no negative
jumps ($\Pi(0,\infty)=0$) with no monotone paths. We state now some important
properties and fluctuation identities of spectrally negative Lévy processes
which will be of use to us later in this paper. We refer to Bertoin, (1998),
Chapter VII or Chapter 8 in Kyprianou, (2014) for details.

Due to the absence of positive jumps, we can define the Laplace transform of
$X_{1}$. We denote $\psi(\beta)$ as the Laplace exponent of the process, that
is, $\psi(\beta)=\log(\mathbb{E}(e^{\beta X_{1}}))$ for $\beta\geq 0$. For
such $\beta$ we have that

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

The function $\psi$ is infinitely often differentiable and strictly convex
function on $(0,\infty)$ with $\psi(\infty)=\infty$. In particular,
$\psi^{\prime}(0+)=\mathbb{E}(X_{1})\in[-\infty,\infty)$ determines the
behaviour of $X$ at infinity. When $\psi^{\prime}(0+)>0$ the process $X$
drifts to infinity, i.e., $\lim_{t\rightarrow\infty}X_{t}=\infty$; when
$\psi^{\prime}(0+)<0$, $X$ drifts to minus infinity and the condition
$\psi^{\prime}(0+)=0$ implies that $X$ oscillates, that is,
$\limsup_{t\rightarrow\infty}X_{t}=-\liminf_{t\rightarrow\infty}X_{t}=\infty$.
We denote by $\Phi$ the right-inverse of $\psi$, i.e.

$\displaystyle\Phi(q)=\sup\\{\beta\geq 0:\psi(\beta)=q\\},\qquad q\geq 0.$

In the particular case that $X$ drifts to infinity, we have that
$\psi^{\prime}(0+)>0$ which implies that $\psi$ is strictly increasing and
then $\Phi$ is the usual inverse with $\Phi(0)=0$. The process $X$ has paths
of finite variation if and only if $\sigma=0$ and
$\int_{(-1,0)}|x|\Pi(\textup{d}x)<\infty$, otherwise $X$ has paths of infinite
variation. Denote by $\tau_{a}^{+}$ the first passage time above the level
$a>0$,

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

The Laplace transform of $\tau_{a}^{+}$ is given by

$\displaystyle\mathbb{E}(e^{-q\tau_{a}^{+}})=e^{-\Phi(q)a}\qquad a>0.$ (1)

An important family of functions for spectrally negative Lévy processes
consists of the scale functions, usually denoted by $W^{(q)}$ and $Z^{(q)}$.
For all $q\geq 0$, the scale function
$W^{(q)}:\mathbb{R}\mapsto\mathbb{R}_{+}$ is such that $W^{(q)}(x)=0$ for all
$x<0$ and it is characterised on the interval $[0,\infty)$ as the strictly and
continuous function with Laplace transform given by

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

The function $Z^{(q)}$ is defined for all $q\geq 0$ by

$\displaystyle Z^{(q)}(x):=1+q\int_{0}^{x}W^{(q)}(y)\textup{d}y,\qquad\text{
for }x\in\mathbb{R}.$

For the case $q=0$ we simply denote $W=W^{(0)}$. When $X$ has paths of
infinite variation, $W^{(q)}$ is continuous on $\mathbb{R}$ and $W^{(q)}(0)=0$
for all $q\geq 0$, otherwise for all $q\geq 0$ $W^{(q)}(0)=1/d$, where

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

For all $q\geq 0$, $W^{(q)}$ has left and right derivatives. Moreover, when
$X$ is of infinite variation we have that $W^{(q)}\in C^{1}((0,\infty))$ with
derivative at zero given by $W^{(q)^{\prime}}(0)=2/\sigma^{2}$. When $X$ is of
finite variation $W^{(q)}\in C^{1}((0,\infty))$ when $\Pi$ has no atoms.
Henceforth, we will assume that when $X$ is of finite variation the Lévy
measure $\Pi$ has no atoms. Furthermore, for each $x\geq 0$ and $q\geq 0$,
$W^{(q)}$ has the following representation

$\displaystyle W^{(q)}(x)=\sum_{k=0}^{\infty}q^{k}W^{*(k+1)}(x),$ (2)

where $W^{*(k+1)}$ is the $(k+1)$-th convolution of $W$ with itself. Various
fluctuation identities for spectrally negative Lévy processes have been found
in terms of the scale functions. Here we list some that will be useful in
later sections. Denote by $\tau_{x}^{-}$ the first passage time below the
level $x\leq 0$, i.e.,

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

Then for any $q\geq 0$ and $x\leq a$ we have

$\displaystyle\mathbb{E}_{x}\left(e^{-q\tau_{a}^{+}}\mathbb{I}_{\\{\tau_{0}^{-}>\tau_{a}^{+}\\}}\right)=\frac{W^{(q)}(x)}{W^{(q)}(a)}.$
(3)

For any $x\in\mathbb{R}$ and $q\geq 0$,

$\displaystyle\mathbb{E}_{x}(e^{-q\tau_{0}^{-}}\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})=Z^{(q)}(x)-\frac{q}{\Phi(q)}W^{(q)}(x),$
(4)

where we understand $q/\Phi(q)$ in the limiting sense when $q=0$. Denote by
$\underline{X}_{t}=\inf_{0\leq s\leq t}X_{s}$ and
$\overline{X}_{t}=\sup_{0\leq s\leq t}X_{s}$ the running infimum and running
maximum of the process $X$ up to time $t>0$, respectively. For $q\geq 0$, let
$\textbf{e}_{q}$ be an exponential random variable with mean $1/q$ independent
of $X$, where we understand that $\textbf{e}_{q}=\infty$ almost surely when
$q=0$. Then $\overline{X}_{\textbf{e}_{q}}$ is exponentially distributed with
parameter $\Phi(q)$ and the Laplace transform of
$\underline{X}_{\textbf{e}_{q}}$ is given by

$\displaystyle\mathbb{E}(e^{\beta\underline{X}_{\textbf{e}_{q}}})=\frac{q}{\Phi(q)}\frac{\Phi(q)-\beta}{q-\psi(\beta)},\qquad\beta\geq
0.$ (5)

Denote by $\sigma_{x}^{-}$ the first time the process $X$ is below or equal to
the level $x$, i.e.

$\displaystyle\sigma_{x}^{-}=\inf\\{t>0:X_{t}\leq x\\}.$

It is easy to show that the mapping $x\mapsto\sigma_{x}^{-}$ is non-
increasing, right-continuous with left limits. The left limit is given by
$\lim_{h\downarrow 0}\sigma_{x-h}^{-}=\tau_{x}^{-}$ for all $x\in\mathbb{R}$.
Moreover, since

$\displaystyle\mathbb{E}(e^{-q\sigma_{x}^{-}}\mathbb{I}_{\\{\sigma_{x}^{-}<\infty\\}})=\mathbb{P}(\textbf{e}_{q}>\sigma_{x}^{-})=\mathbb{P}(\underline{X}_{\textbf{e}_{q}}\leq-x)$

for all $x\leq 0$ and the fact that $\underline{X}_{\textbf{e}_{q}}$ has no
atoms away from zero, we have that $\sigma_{x}^{-}$ and $\tau_{x}^{-}$ have
the same distribution for all $x<0$. When $X$ is of infinite variation, $X$
enters instantly to the set $(\infty,0)$ whilst in the finite variation case
there is a positive time before the process enters it. That implies that in
the infinite variation case $\tau_{0}^{-}=\sigma_{0}^{-}=0$ a.s. whereas in
the finite variation case, $\sigma_{0}^{-}=0$ and $\tau_{0}^{-}>0$.

Let $q>0$ and $a\in\mathbb{R}$. The $q$-potential measure of $X$ killed on
exiting $[0,a]$

$\int_{0}^{\infty}e^{-qt}\mathbb{P}_{x}(X_{t}\in\textup{d}y,t<\tau_{a}^{+}\wedge\tau_{0}^{-})\textup{d}t$

is absolutely continuous with respect to Lebesgue measure and it has a density
given by

$\displaystyle\frac{W^{(q)}(x)W^{(q)}(a-y)}{W^{(q)}(a)}-W^{(q)}(x-y)\qquad
x,y\in[0,a].$ (6)

Similarly, the $q$-potential measure of $X$ killed on exiting $(-\infty,a]$
and the $q$-potential measure of $X$ are absolutely continuous with respect to
Lebesgue measure with a density given by

$\displaystyle e^{-\Phi(q)(a-x)}W^{(q)}(a-y)-W^{(q)}(x-y),\qquad x,y\leq a,$
(7)

and

$\displaystyle\Phi^{\prime}(q)e^{-\Phi(q)(y-x)}-W^{(q)}(x-y),\qquad
x,y\in\mathbb{R},$ (8)

respectively. In the case when $X$ drifts to infinity these expression are
also valid for $q=0$.

For any $t\geq 0$ and $x\in\mathbb{R}$, we denote by $g_{t}^{(x)}$ the last
time that the process is below $x$ before time $t$, i.e.,

$\displaystyle g_{t}^{(x)}=\sup\\{0\leq s\leq t:X_{s}\leq x\\},$ (9)

with the convention $\sup\emptyset=0$. We simply denote $g_{t}:=g_{t}^{(0)}$
for all $t\geq 0$. Note that when $\mathbb{P}(X_{t}\geq 0)=\rho$ for some
$\rho\in(0,1)$, then $g_{t}/t$ follows the generalised arcsine law with
parameter $\rho$, see Theorem 13 in Bertoin, (1998). The last hitting time of
zero is of key importance in the study of Azéma’s martingale (see Azéma and
Yor, (1989)). We also define, for each $t\geq 0$, $U_{t}^{(x)}$ as the time
spent by $X$ above the level zero before time $t$ since the last visit to the
interval $(-\infty,x]$, i.e.

$\displaystyle U_{t}^{(x)}:=t-g_{t}^{(x)}\qquad t\geq 0.$

It turns out that for our optimal prediction problem

$\displaystyle\inf_{\tau\in\mathcal{T}}\mathbb{E}(|\tau-g|^{p})$

the process $U_{t}=U_{t}^{(0)}$ plays a vital role. It can be readily seen
that for all $x\in\mathbb{R}$ the process $\\{U_{t}^{(x)},t\geq 0\\}$ is not a
Markov process. We now list a number of results from Baurdoux and Pedraza,
(2019) concerning $U$. The strong Markov property holds for the two
dimensional process $\\{(U_{t},X_{t}),t\geq 0\\}$ with respect to the
filtration $\\{\mathcal{F}_{t},t\geq 0\\}$ and state space given by

$\displaystyle E=\\{(u,x):u>0\text{ and }x>0\\}\cup\\{(u,x):u=0\text{ and
}x\leq 0\\}.$

Then there exists a family of probability measures
$\\{\mathbb{P}_{u,x},(u,x)\in E\\}$ such that for any $A\in\mathbb{B}(E)$,
Borel set of $E$, we have that $\mathbb{P}_{u,x}((U_{\tau+s},X_{\tau+s})\in
A|\mathcal{F}_{\tau})=\mathbb{P}_{U_{\tau},X_{\tau}}((U_{s},X_{s})\in A)$. For
each $(u,x)\in E$, $\mathbb{P}_{u,x}$ can be written in terms of
$\mathbb{P}_{x}$ via

$\displaystyle\mathbb{E}_{u,x}(h(U_{s},X_{s})):=\mathbb{E}_{x}(h(u+s,X_{s})\mathbb{I}_{\\{\sigma_{0}^{-}>s\\}})+\mathbb{E}_{x}(h(U_{s},X_{s})\mathbb{I}_{\\{\sigma_{0}^{-}\leq
s\\}}),$ (10)

for any positive measurable function $h$. Let $F$ a $C^{1,2}(E)$ real-valued
function. In addition, in the case that $\sigma>0$ assume that
$\lim_{h\downarrow 0}F(u,h)=F(0,0)$ for all $u>0$. Then we have the following
version of Itô formula

$\displaystyle F(U_{t},$ $\displaystyle X_{t})=F(U_{0},X_{0})$
$\displaystyle\qquad+\int_{0}^{t}\frac{\partial}{\partial
u}F(U_{s},X_{s})\mathbb{I}_{\\{X_{s}\leq
0\\}}\textup{d}s+\int_{0}^{t}\frac{\partial}{\partial
x}F(U_{s-},X_{s-})\textup{d}X_{s}+\frac{1}{2}\sigma^{2}\int_{0}^{t}\frac{\partial^{2}}{\partial
x^{2}}F(U_{s},X_{s})\textup{d}s$
$\displaystyle\qquad+\int_{[0,t]}\int_{(-\infty,0)}\left(F(U_{s},X_{s^{-}}+y)-F(U_{s-},X_{s^{-}})-y\frac{\partial}{\partial
x}F(U_{s-},X_{s^{-}})\right)N(\textup{d}s\times\textup{d}y)$ (11)

Moreover, if $f$ is a $C^{1,2}(E)$ bounded function, the infinitesimal
generator $\mathcal{A}_{U,X}$ of the process $(U,X)$ is given by

$\displaystyle\mathcal{A}_{U,X}(f)(u,x)$
$\displaystyle=\frac{\partial}{\partial
u}f(u,x)\mathbb{I}_{\\{x>0\\}}-\mu\frac{\partial}{\partial
x}f(u,x)+\frac{1}{2}\sigma^{2}\frac{\partial^{2}}{\partial x^{2}}f(u,x)$
$\displaystyle\qquad+\int_{(-\infty,0)}\left(f(u,x+y)-f(u,x)-y\mathbb{I}_{\\{y>-1\\}}\frac{\partial}{\partial
x}f(u,x)\right)\mathbb{I}_{\\{x+y>0\\}}\Pi(\textup{d}y)$
$\displaystyle\qquad+\int_{(-\infty,0)}\left(f(0,x+y)-f(0,x)-y\mathbb{I}_{\\{y>-1\\}}\frac{\partial}{\partial
x}f(0,x)\right)\mathbb{I}_{\\{x\leq 0\\}}\Pi(\textup{d}y)$
$\displaystyle\qquad+\int_{(-\infty,0)}\left(f(0,x+y)-f(u,x)-y\mathbb{I}_{\\{y>-1\\}}\frac{\partial}{\partial
x}f(u,x)\right)\mathbb{I}_{\\{0<x<-y\\}}\Pi(\textup{d}y)$
$\displaystyle=\frac{\partial}{\partial
u}\widetilde{f}(u,x)-\mu\frac{\partial}{\partial
x}\widetilde{f}(u,x)+\frac{1}{2}\sigma^{2}\frac{\partial^{2}}{\partial
x^{2}}\widetilde{f}(u,x)$
$\displaystyle\qquad+\int_{(-\infty,0)}\left(\widetilde{f}(u,x+y)-\widetilde{f}(u,x)-y\mathbb{I}_{\\{y>-1\\}}\frac{\partial}{\partial
x}\widetilde{f}(u,x)\right)\Pi(\textup{d}y).$ (12)

where $\tilde{f}$ is a function that extends $f$ to the set
$\mathbb{R}_{+}\times\mathbb{R}$ given by

$\displaystyle\widetilde{f}(u,x)=\left\\{\begin{array}[]{cc}f(u,x)&u>0\text{
and }x>0,\\\ f(0,x)&u\geq 0\text{ and }x\leq 0,\\\ f(0,0)&u=0\text{ and
}x>0.\end{array}\right.$ (16)

In addition, let $q\geq 0$ and $X$ be a spectrally negative Lévy process and
$K:E\mapsto\mathbb{R}$ a left-continuous function in each argument. Assume
that there exists a non-negative function
$C:\mathbb{R}_{+}\times\mathbb{R}\mapsto\mathbb{R}$ such that $u\mapsto
C(u,x)$ is a non-decreasing function for all $x\in\mathbb{R}$, $|K(u,s)|\leq
C(u,x)$ and
$\mathbb{E}_{x}\left(\int_{0}^{\infty}e^{-qr}C(r,X_{r})\textup{d}r\right)<\infty$
for all $x\in\mathbb{R}$. Then we have that for any $(u,x)\in E$ that

$\displaystyle\mathbb{E}_{u,x}$
$\displaystyle\left(\int_{0}^{\infty}e^{-qr}K(U_{r},X_{r})\textup{d}r\right)=K^{+}(u,x)+\lim_{\varepsilon\downarrow
0}\mathbb{E}_{x}\left(\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}}e^{-q\tau_{0}^{-}}K^{-}(X_{\tau_{0}^{-}}-\varepsilon)\right)$
$\displaystyle+e^{\Phi(q)x}\left[1-\psi^{\prime}(\Phi(q)+)e^{-\Phi(q)x}W^{(q)}(x)\right]\lim_{\varepsilon\downarrow
0}\frac{\mathbb{E}_{\varepsilon}\left(\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}}e^{-q\tau_{0}^{-}}K^{-}(X_{\tau_{0}^{-}}-\varepsilon)\right)+K^{+}(0,\varepsilon)}{\psi^{\prime}(\Phi(q)+)W^{(q)}(\varepsilon)},$
(17)

where $K^{+}$ and $K^{-}$ are given by

$\displaystyle\begin{array}[]{ll}K^{+}(u,x)=\displaystyle{\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{-}}e^{-qr}K(u+r,X_{r})\textup{d}r\right)},\\\
K^{-}(x)=\displaystyle{\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{+}}e^{-qr}K(0,X_{r})\textup{d}r\right)},\\\
\end{array}$

for all $(u,x)\in E$. We conclude this section by collecting some additional
results about the last passage time

$\displaystyle g=g_{\infty}=\sup\\{t\geq 0:X_{t}\leq 0\\}.$ (18)

The Laplace transform of $g$ was found in Chiu et al., (2005) as

$\displaystyle\mathbb{E}_{x}(e^{-qg})=e^{\Phi(q)x}\Phi^{\prime}(q)\psi^{\prime}(0+)+\psi^{\prime}(0+)(W(x)-W^{(q)}(x)),\qquad
q\geq 0.$ (19)

The distribution function of $g$ under $\mathbb{P}_{x}$ is found by observing
that

$\displaystyle\mathbb{P}_{x}(g\leq\gamma)$
$\displaystyle=\mathbb{P}_{x}(X_{u+\gamma}>0\text{ for all }u\in(0,\infty))$
$\displaystyle=\mathbb{E}_{x}(\mathbb{P}_{x}(X_{u+\gamma}>0\text{ for all
}u\in(0,\infty)|\mathcal{F}_{\gamma}))$
$\displaystyle=\mathbb{E}_{x}(\mathbb{P}_{X_{\gamma}}(\tau_{0}^{-}=\infty))$
$\displaystyle=\mathbb{E}_{x}(\psi^{\prime}(0+)W(X_{\gamma})),$ (20)

where we used the tower property of conditional expectation in the second
equality and the Markov property of Lévy processes in the third. Note that the
law of $g$ under $\mathbb{P}_{x}$ may have an atom at zero given by

$\displaystyle\mathbb{P}_{x}(g=0)=\mathbb{P}_{x}(\tau_{0}^{-}=\infty)=\psi^{\prime}(0+)W(x).$

For our optimal prediction problem we require the $p$-th moment of $g$ to be
finite.The following result is from Doney and Maller, (2004) (see Theorem 1,
Theorem 4, Theorem 5 and Remark (ii)).

######  Lemma 2.1.

Let $X$ be a spectrally negative Lévy process drifting to infinity. Then for a
fixed $p>0$ the following are equivalent.

  1. 1.

$\mathbb{E}_{x}(g^{p})<\infty$ for some (hence every) $x\leq 0$.

  2. 2.

$\int_{(-\infty,-1)}|x|^{1+p}\Pi(\textup{d}x)<\infty$.

  3. 3.

$\mathbb{E}((-\underline{X}_{\infty}^{p}))<\infty$.

  4. 4.

$\mathbb{E}_{x}((\tau_{0}^{+})^{p+1})<\infty$ for some (hence every) $x\leq
0$.

  5. 5.

$\mathbb{E}_{x}((\tau_{0}^{-})^{p}\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})<\infty$
for some (hence every) $x\geq 0$.

The next lemma states that when $\tau_{0}^{+}$ has finite $p$-th moment under
$\mathbb{P}_{x}$, then the function $\mathbb{E}_{x}((\tau_{0}^{+})^{p})$ has a
polynomial bound in $x$. It will be of use later to deduce a lower bound for
our optimal prediction problem.

######  Lemma 2.2.

Let $p>0$ and suppose $\mathbb{E}_{x}((\tau_{0}^{+})^{p+1})<\infty$ for some
$x\leq 0$. Then for each $0\leq r\leq p$ there exist non-negative constants
$A_{r}$ and $C_{r}$ such that

$\displaystyle\mathbb{E}_{x}((\tau_{0}^{+})^{r})\leq
A_{r}+C_{r}|x|^{r}\qquad\text{and}\qquad\mathbb{E}_{x}(g^{r})\leq
2^{r}[\mathbb{E}(g^{r})+A_{r}]+2^{r}C_{r}|x|^{r},\qquad x\leq 0.$

Here $\lfloor p\rfloor$ denotes the integer part of $p$.

###### Proof.

From equation (1) we know that

$\displaystyle
F(\theta,x):=\mathbb{E}_{x}(e^{-\theta\tau_{0}^{+}})=e^{\Phi(\theta)x},\qquad
x\leq 0.$

Then using the formula of Faà di Bruno (see for example Spindler, (2005)) we
have that for any $n\geq 1$,

$\displaystyle\frac{\partial^{n}}{\partial\theta^{n}}F(\theta,x)=\sum_{k=1}^{n}e^{\Phi(\theta)x}x^{k}\sum_{\begin{subarray}{c}k_{1}+\cdots+k_{n}=k,\\\
k_{1}+\cdots+nk_{n}=n\end{subarray}}\frac{n!}{k_{1}!k_{2}!\cdots
k_{n}!}\left(\frac{\Phi^{\prime}(\theta)}{1!}\right)^{k_{1}}\left(\frac{\Phi^{\prime\prime}(\theta)}{2!}\right)^{k_{2}}\cdots\left(\frac{\Phi^{(n)}(\theta)}{n!}\right)^{k_{n}}.$

Then evaluating at zero the above equation, using $\Phi(0)=0$ and the fact
that $\Phi^{(i)}(0)<\infty$ for $i=1,\ldots,\lfloor p\rfloor+1$, we can find
constants $A_{r},C_{r}\geq 0$ such that
$\mathbb{E}_{x}((\tau_{0}^{+})^{r})\leq A_{r}+C_{r}|x|^{r}$ for any
$r\in\\{1,\ldots,\lfloor p\rfloor+1\\}$. For any non integer $r<\lfloor
p\rfloor+1$ we can use Hölder’s inequality to obtain

$\displaystyle\mathbb{E}_{x}((\tau_{0}^{+})^{r})\leq[\mathbb{E}_{x}((\tau_{0}^{+})^{\lfloor
r\rfloor+1})]^{\frac{r}{\lfloor r\rfloor+1}}\leq(A_{\lfloor
r\rfloor+1}+C_{\lfloor r\rfloor+1}|x|^{\lfloor r\rfloor+1})^{\frac{r}{\lfloor
r\rfloor+1}}.$

The result follows from the inequality $(a+b)^{q}\leq 2^{q}(a^{q}+b^{q})$
which is true for any $q>0$ and $a,b>0$. Now we show that the second
inequality holds. From the strong Markov property we get that for any $x<0$

$\displaystyle\mathbb{E}_{x}(g^{r})\leq
2^{r}\mathbb{E}(g^{r})+2^{r}\mathbb{E}_{x}((\tau_{0}^{+})^{r})\leq
2^{r}[\mathbb{E}(g^{r})+A_{r}]+2^{r}C_{r}|x|^{r}.$

∎

We conclude this section with a lemma showing some properties of the function
$x\mapsto\mathbb{E}_{x}(g^{p})$.

######  Lemma 2.3.

Let $p>0$ and assume that $\int_{(-\infty,-1)}|x|^{p+1}\Pi(dx)<\infty$. Then
$x\mapsto\mathbb{E}_{x}(g^{p})$ is a non-increasing, non-negative and
continuous function. Moreover,

$\displaystyle\lim_{x\rightarrow-\infty}\mathbb{E}_{x}(g^{p})=\infty\qquad\text{and}\qquad\lim_{x\rightarrow\infty}\mathbb{E}_{x}(g^{p})=0.$

###### Proof.

It follows from the definition of $g$ that
$x\mapsto\mathbb{E}_{x}(g^{p})=\mathbb{E}(g^{(-x)})$ is non-negative and non-
increasing. In order to check continuity notice that by integration by parts
we get

$\displaystyle\mathbb{E}_{x}(g^{p})$
$\displaystyle=p\int_{0}^{\infty}s^{p-1}\mathbb{P}_{x}(g>s)\textup{d}s$
$\displaystyle=p\int_{0}^{\infty}s^{p-1}\mathbb{E}_{x}(1-\psi^{\prime}(0+)W(X_{s}))\textup{d}s$

where the last equality follows from (20). Take $x\in\mathbb{R}$ and
$\delta\in\mathbb{R}$. Then using the equation above we have that

$\displaystyle|\mathbb{E}_{x}(g^{p})-\mathbb{E}_{x+\delta}(g^{p})|\leq
p\psi^{\prime}(0+)\mathbb{E}\left(\int_{0}^{\infty}s^{p-1}|W(X_{s}+x+\delta)-W(X_{s}+x)|\textup{d}s\right).$
(21)

First, suppose that $X$ is of infinite variation and thus $W$ is continuous on
$\mathbb{R}$. From the fact that $X$ drifts to $\infty$ we know that
$W(\infty)=1/\psi^{\prime}(0+)$ and therefore it follows that
$s^{p-1}(1-\psi^{\prime}(0+)W(X_{s}))$ is integrable with respect to the
product measure $\mathbb{P}_{x}\times\lambda([0,\infty))$, where $\lambda$
denotes Lebesgue measure. We can now invoke the dominated convergence theorem
to deduce that $x\mapsto\mathbb{E}_{x}(g^{p})$ is continuous.

Next, in the case that $X$ is of finite variation we have that $W$ has a
discontinuity at zero. However, the set $\\{s\geq 0:X_{s}=x\\}$ is almost
surely countable and thus has Lebesgue measure zero. We can again use the
dominated convergence theorem to conclude that continuity also holds in this
case.

We prove now the asymptotic behaviour of $\mathbb{E}_{x}(g^{p})$. Note that
when $x$ tends to $-\infty$ the random variable $g^{(-x)}\rightarrow\infty$.
Then using Fatou’s lemma

$\displaystyle\liminf_{x\rightarrow-\infty}\mathbb{E}_{x}(g^{p})\geq\mathbb{E}(\liminf_{x\rightarrow-\infty}(g^{(-x)})^{p})=\infty$

Therefore, $\lim_{x\rightarrow-\infty}\mathbb{E}_{x}(g^{p})=\infty$. In the
other hand, note that for $x>0$

$\displaystyle\mathbb{P}_{x}(g^{p}=0)=\mathbb{P}_{x}(g=0)=\mathbb{P}_{x}(\tau_{0}^{-}=\infty)=\psi^{\prime}(0+)W(x)\xrightarrow{x\rightarrow\infty}1.$
(22)

Hence we deduce that the sequence $\\{(g^{(-n)})^{p}\\}_{n\geq 1}$ converges
in probability to $0$ (under the measure $\mathbb{P}$) when $n$ tends to
infinity. Moreover, since the sequence $\\{\mathbb{E}((g^{-n})^{p})\\}_{n\geq
1}$ is a non-increasing sequence of positive numbers we get that

$\displaystyle\sup_{n\geq
1}\mathbb{E}((g^{-n})^{p})\leq\mathbb{E}(g^{p})<\infty,$

where the last inequality holds due to Lemma 2.1 and assumption. Then
$\\{(g^{(-n)})^{p}\\}_{n\geq 1}$ is an uniformly integrable family of random
variables. The latter together with the convergence in probability allows us
to conclude that $\mathbb{E}_{x}(g^{p})\rightarrow 0$ when
$x\rightarrow\infty$ as claimed. ∎

##  3 Optimal prediction problem

Denote by $V_{*}$ the value of the optimal prediction problem, i.e.

$\displaystyle V_{*}=\inf_{\tau\in\mathcal{T}}\mathbb{E}(|\tau-g|^{p}),$ (23)

where $\mathcal{T}$ is the set of all stopping times with respect to
$\mathbb{F}$, $p>1$ and $g$ is the last zero of $X$ given in (18). Since $g$
is only $\mathcal{F}$ measurable standard techniques of optimal stopping times
are not directly applicable. However, there is an equivalence between the
optimal prediction problem (23) and an optimal stopping problem. The next
lemma, inspired in the work of Urusov, (2005), states such equivalence.

######  Lemma 3.1.

Let $p>1$ and let $X$ be a spectrally negative Lévy process drifting to
infinity such that $\int_{(-\infty,-1)}|x|^{p+1}\Pi(\textup{d}x)<\infty$.
Consider the optimal stopping problem

$\displaystyle
V=\inf_{\tau\in\mathcal{T}}\mathbb{E}\left(\int_{0}^{\tau}G(s-g_{s},X_{s})\textup{d}s\right),$
(24)

where the function $G$ is given by

$\displaystyle G(u,x)=u^{p-1}\psi^{\prime}(0+)W(x)-\mathbb{E}_{x}(g^{p-1}),$

for $u\geq 0$ and $x\in\mathbb{R}$. Then we have that
$V_{*}=pV+\mathbb{E}(g^{p})$ and a stopping time minimises (23) if and only if
it minimises (24).

###### Proof.

Let $\tau\in\mathcal{T}$. Then the following equality holds

$\displaystyle|\tau-g|^{p}$
$\displaystyle=\int_{0}^{\tau}\varrho(s-g)\textup{d}s+g^{p},$ (25)

where the function $\varrho$ is defined by

$\displaystyle\varrho(x)=p\left[\frac{(-x)^{p}}{x}\mathbb{I}_{\\{x<0\\}}+x^{p-1}\mathbb{I}_{\\{x\geq
0\\}}\right].$

Taking expectations in equation (25) and then using Fubini’s theorem and the
tower property for conditional expectation we obtain

$\displaystyle\mathbb{E}(|\tau-g|^{p})$
$\displaystyle=\int_{0}^{\infty}\mathbb{E}\left(\varrho(s-g)\mathbb{I}_{\\{s\leq\tau\\}}\textup{d}s\right)+\mathbb{E}(g^{p})$
$\displaystyle=\int_{0}^{\infty}\mathbb{E}\left[\mathbb{I}_{\\{s\leq\tau\\}}\mathbb{E}\left(\varrho(s-g)|\mathcal{F}_{s}\right)\textup{d}s\right]+\mathbb{E}(g^{p})$
$\displaystyle=\mathbb{E}\left(\int_{0}^{\tau}\mathbb{E}\left(\varrho(s-g)|\mathcal{F}_{s}\right)\textup{d}s\right)+\mathbb{E}(g^{p}).$

To evaluate the conditional expectation inside the last integral, note that
for all $t\geq 0$ we can write the the time $g$ as

$\displaystyle g=g_{t}\vee\sup\\{s\in(t,\infty):X_{s}\leq 0,\\}$

recalling that $g_{t}=g_{t}^{(0)}$ defined in (9). Hence, using the Markov
property for Lévy processes and the fact that $g_{s}$ is $\mathcal{F}_{s}$
measurable we have that

$\displaystyle\mathbb{E}(\varrho(s-g)|\mathcal{F}_{s})$
$\displaystyle=\mathbb{E}\left(\varrho\left(s-\left[g_{s}\vee\sup\\{r\in(s,\infty):X_{r}\leq
0\\}\right]\right)|\mathcal{F}_{s}\right)$
$\displaystyle=\varrho(s-g_{s})\mathbb{E}(\mathbb{I}_{\\{X_{r}>0\text{ for all
}r\in(s,\infty)\\}}|\mathcal{F}_{s})$
$\displaystyle\qquad+\mathbb{E}(\varrho(s-\sup\\{r\in(s,\infty):X_{r}\leq
0\\})\mathbb{I}_{\\{X_{r}\leq 0\text{ for some
}r\in(s,\infty)\\}}|\mathcal{F}_{s})$
$\displaystyle=\varrho(s-g_{s})\mathbb{P}_{X_{s}}(g=0)+\mathbb{E}_{X_{s}}(\varrho(-g)\mathbb{I}_{\\{g>0\\}})$
$\displaystyle=p(s-g_{s})^{p-1}\psi^{\prime}(0+)W(X_{s})-p\mathbb{E}_{X_{s}}(g^{p-1}).$

Then we have that

$\displaystyle\mathbb{E}(|\tau-g|^{p})=p\mathbb{E}\left(\int_{0}^{\tau}G(s-g_{s},X_{s})\textup{d}s\right)+\mathbb{E}(g^{p})$

∎

######  Remark 3.2.

A close inspection of the proof of Lemma 3.1 tells us that the function
$\varrho$ corresponds to the right derivative of the function $f(x)=|x|^{p}$.
Therefore, using similar arguments we can actually extend the result to any
convex function $d:\mathbb{R}_{+}\times\mathbb{R}_{+}\mapsto\mathbb{R}_{+}$.
That is, under the assumption that $\mathbb{E}(d(0,g))<\infty$, the optimal
prediction problem

$\displaystyle V_{d}=\inf_{\tau\in\mathcal{T}}\mathbb{E}(d(\tau,g))$

is equivalent to the optimal stopping problem

$\displaystyle\inf_{\tau\in\mathcal{T}}\mathbb{E}\left[\int_{0}^{\tau}G_{d}(g_{s},s,X_{s})\textup{d}s\right]$

where
$G_{d}(\gamma,t,x)=\varrho_{d}(s,\gamma)\psi^{\prime}(0+)W(x)+\mathbb{E}_{x}(\varrho_{d}(s,g+s)\mathbb{I}_{\\{g>0\\}})$
and $\varrho_{d}$ is the right derivative with respect the first argument of
$d$.

##  4 Solution to the optimal stopping problem

In order to solve the optimal stopping problem (24) using the general theory
of optimal stopping (see e.g. Peskir and Shiryaev, (2006)) we have to extend
it to an optimal stopping problem driven by a strong Markov process. For every
$(u,x)\in E$, we define the optimal stopping problem

$\displaystyle
V(u,x)=\inf_{\tau\in\mathcal{T}}\mathbb{E}_{u,x}\left[\int_{0}^{\tau}G(U_{s},X_{s})\textup{d}s\right],$
(26)

where the function $G$ is given by
$G(u,x)=u^{p-1}\psi^{\prime}(0+)W(x)-\mathbb{E}_{x}(g^{p-1})$ for any $u\geq
0$ and $x\in\mathbb{R}$. Therefore we have that
$V_{*}=pV(0,0)+\mathbb{E}(g^{p})$. Note that using the definition of
$\mathbb{E}_{u,x}$ we have that (26) takes the form

$\displaystyle
V(u,x)=\inf_{\tau\in\mathcal{T}}\mathbb{E}_{x}\left(\int_{0}^{\tau}\left\\{G(u+s,X_{s})\mathbb{I}_{\\{\sigma_{0}^{-}>s\\}}+G(U_{s},X_{s})\mathbb{I}_{\\{\sigma_{0}^{-}\leq
s\\}}\right\\}\textup{d}s\right).$ (27)

The optimal stopping problem (26) is given in terms of a function $G$ which
involves the function $x\mapsto\mathbb{E}_{x}(g^{p-1})$. Recall that for a
fixed $p>1$, the function $G$ is given by
$G(u,x)=u^{p-1}\psi^{\prime}(0+)W(x)-\mathbb{E}_{x}(g^{p-1})$ for all
$(u,x)\in E$. Then as a consequence of Lemma 2.3 we have the following
behaviour. For all $x\in\mathbb{R}$, the function $u\mapsto G(u,x)$ is non-
decreasing. In particular when $x<0$, $u\mapsto
G(u,x)=-\mathbb{E}_{x}(g^{p-1})$ is a strictly negative constant. For fixed
$u\geq 0$, $x\mapsto G(u,x)$ is a non-decreasing right-continuous function
which is continuous everywhere apart from possibly at $x=0$ (since $W$ is
discontinuous at zero when $X$ is of finite variation) such that for all
$u\geq 0$,

$\displaystyle\lim_{x\rightarrow-\infty}G(u,x)=-\infty\qquad\text{ and
}\qquad\lim_{x\rightarrow\infty}G(u,x)=u^{p-1}\geq 0.$

Moreover, we have that $\lim_{u\rightarrow\infty}G(u,x)=\infty$ and
$G(0,x)=-\mathbb{E}_{x}(g^{p-1})<0$ for all $x\geq 0$. We then define the
function

$\displaystyle h(u)$ $\displaystyle=\inf\\{x\in\mathbb{R}:G(u,x)\geq 0\\}.$
(28)

From the description of $G$ above we have that $h$ is a non-negative and non-
decreasing function such that $h(u)<\infty$ for all $u\in(0,\infty)$,
$h(0)=\infty$ and $\lim_{u\rightarrow\infty}h(u)=0$. Moreover, since $W$ is
strictly increasing on $(0,\infty)$, the function

$\displaystyle T(x)=\frac{\mathbb{E}_{x}(g^{p-1})}{\psi^{\prime}(0+)W(x)}$

is continuous and strictly decreasing on $[0,\infty)$. Then there exists an
inverse function $T^{-1}$ which is continuous and strictly decreasing on
$(0,u_{h}^{*}]$ with

$\displaystyle u_{h}^{*}:=\frac{\mathbb{E}(g^{p-1})}{\psi^{\prime}(0+)W(0)},$
(29)

where we understand $1/0=\infty$ when $X$ is of infinite variation. Hence we
can write

$\displaystyle
h(u)=\left\\{\begin{array}[]{lc}T^{-1}(u^{p-1})&u<(u_{h}^{*})^{\frac{1}{p-1}}\\\
0&u\geq(u_{h}^{*})^{\frac{1}{p-1}}\end{array}\right..$

Therefore, since $T^{-1}(u_{h}^{*}-)=0$, we conclude that $h$ is a continuous
function on $[0,\infty)$. From the definition of $h$ we clearly have that
$G(u,x)\geq 0$ if and only if $x\geq h(u)$.

The latter facts about give us some intuition about the optimal stopping rule
for the optimal stopping problem (26). Since we are dealing with a
minimisation problem, before stopping we want the process $(U,X)$ to be in the
set in which $G$ is negative as much as possible. Then the fact that
$G(U_{t},X_{t})$ is strictly negative when $X_{t}<h(U_{t})$ suggests that it
is never optimal to stop on this region. When $X_{t}>h(U_{t})$ we have that
$G(U_{t},X_{t})\geq 0$ but with strictly positive probability $(U,X)$ can
enter the set in which $G$ is negative. Moreover, $t\mapsto U_{t}$ is strictly
increasing when $X$ is in the positive half line and then $t\mapsto h(U_{t})$
gets closer to zero when the current excursion away from $(-\infty,0]$ is
sufficiently large and then $G(U_{t},X_{t})\geq 0$ even when $X_{t}$ is
relatively close to zero. That implies that it is optimal to stop when the
current excursion away from $(-\infty,0]$ is large and $X$ takes a
sufficiently large values. That suggest the existence of a non-negative curve
$b\geq h$ such that it is optimal to stop when $X$ crosses above $b(U_{t})$.
We will formally show in the next Lemmas the existence of such boundary.

Note that if there exists a stopping time $\tau$ for which the expectation of
the right hand side of (26) is minus infinity then $V$ would also be minus
infinity. The next Lemma provides the finiteness of a lower bound of $V$ that
will ensure that $V$ only takes finite values, its proof is included in the
Appendix.

######  Lemma 4.1.

Let $p>1$ and $X$ be a spectrally negative Lévy process drifting to infinity.
Assume that $\int_{(-\infty,-1)}|x|^{p+1}\Pi(dx)<\infty$. Then

$\displaystyle
0\leq\mathbb{E}_{x}\left(\int_{0}^{\infty}\mathbb{E}_{X_{s}}(g^{p-1})\textup{d}s\right)<\infty\qquad\text{for
all }x\in\mathbb{R}.$

We now prove the finiteness of the function $V$.

######  Lemma 4.2.

Let $p>1$. For every $(u,x)\in E$ we have that $V(u,x)\in(-\infty,0]$. In
particular $V(u,x)<0$ for $(u,x)\in U:=\\{(u,x)\in E:x<h(u)\\}$, where $h$ is
defined in (28).

###### Proof.

Taking the stopping time $\tau=0$ we deduce that for all $(u,x)\in E$,
$V(u,x)\leq 0$. In order to check that $V(u,x)>-\infty$ we use that
$G(u,x)\geq-\mathbb{E}_{x}(g^{p-1})$ to get

$\displaystyle
V(u,x)=\inf_{\tau\in\mathcal{T}}\mathbb{E}_{u,x}\left[\int_{0}^{\tau}G(U_{s},X_{s})\textup{d}s\right]\geq-\sup_{\tau\in\mathcal{T}}\mathbb{E}_{x}\left[\int_{0}^{\tau}\mathbb{E}_{X_{s}}(g^{p-1})\textup{d}s\right],$

for all $(u,x)\in E$. Hence by Lemma 4.1 we have that

$\displaystyle
V(u,x)\geq-\mathbb{E}_{x}\left[\int_{0}^{\infty}\mathbb{E}_{X_{s}}(g^{p-1})\textup{d}s\right]>-\infty,\qquad(u,x)\in
E.$

Now we prove that $V(u,x)<0$ when $(u,x)\in U$. From the definition of $h$ we
have that if $(u,x)\in U$ then $G(u,x)<0$. Take $(u,x)\in U$ and consider the
stopping time

$\displaystyle\tau_{U^{*}}:=\inf\\{t\geq 0:(U_{t},X_{t})\in E\setminus U\\}.$

Note that under the measure $\mathbb{P}_{u,x}$, $\tau_{U^{*}}>0$. Then for all
$s<\tau_{U^{*}}$, $(U_{s},X_{s})\in U$ which implies that $G(U_{s},X_{s})<0$.
Hence, by the definition of $V$, we have that

$\displaystyle
V(u,x)\leq\mathbb{E}_{u,x}\left[\int_{0}^{\tau_{U*}}G(U_{s},X_{s})\textup{d}s\right]<0.$

∎

Using the general theory of optimal stopping it can be shown that the optimal
stopping time $\tau_{D}$ is the first entrance of $(U,X)$ to a set
$D:=\\{(u,x)\in E:V(u,x)=0\\}$, i.e.

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

Then the function $V$ takes the form

$\displaystyle
V(u,x)=\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{D}}G(U_{s},X_{s})\textup{d}s\right),\qquad(u,x)\in
E.$

Now we prove some basic properties of $V$ which will be of help to deduce the
shape of the set $D$.

######  Lemma 4.3.

Let $p>1$. We have the following monotonicity property of $V$. For all
$(u,x),(v,y)\in E$ such that $u\leq v$ and $x\leq y$ we have that $V(u,x)\leq
V(v,y)$.

###### Proof.

From equation (27) we have that

$\displaystyle V(u,x)$
$\displaystyle=\inf_{\tau\in\mathcal{T}}\mathbb{E}_{x}\left(\int_{0}^{\tau}\left\\{G(u+s,X_{s})\mathbb{I}_{\\{\sigma_{0}^{-}>s\\}}+G(U_{s},X_{s})I_{\\{\sigma_{0}^{-}\leq
s\\}}\right\\}\textup{d}s\right)$
$\displaystyle=\inf_{\tau\in\mathcal{T}}\mathbb{E}\left(\int_{0}^{\tau}\left\\{G(u+s,X_{s}+x)\mathbb{I}_{\\{\sigma_{-x}^{-}>s\\}}+G(U_{s}^{(-x)},X_{s}+x)I_{\\{\sigma_{-x}^{-}\leq
s\\}}\right\\}\textup{d}s\right)$

where $\sigma_{-x}^{-}=\inf\\{t\geq 0:X_{t}\leq-x\\}$ and
$U_{s}^{(-x)}=s-\sup\\{t\geq 0:X_{t}\leq-x\\}$. Recall that for all $s\geq 0$,
$x\mapsto U_{s}^{(-x)}$ and $x\mapsto\sigma_{-x}^{-}$ are non-decreasing and
that the function $G$ is non-decreasing in each argument. Define the function

$\displaystyle
G^{\prime}(u,x):=G(u+s,X_{s}+x)\mathbb{I}_{\\{\sigma_{-x}^{-}>s\\}}+G(U_{s}^{(-x)},X_{s}+x)\mathbb{I}_{\\{\sigma_{-x}^{-}\leq
s\\}}.$

Analysing by cases it is easy to show that $G^{\prime}$ is non-decreasing in
each argument. Using the monotonicity of the integral and expectation operator
we have the desired result.

∎

Recall from Lemma 4.2 that $V(u,x)<0$ for all $(u,x)\in E$ such that $x<h(u)$.
Hence if $(u,x)\in D$ we have that $x\geq h(u)\geq 0$. We then define the
function $b:(0,\infty)\mapsto\mathbb{R}$ by

$\displaystyle b(u)=\inf\\{x>0:V(u,x)=0\\}.$

Then it directly follows that $b(u)\geq h(u)\geq 0$ for all $u>0$. Moreover,
since $h(0)=\infty$ we have that $\lim_{u\downarrow 0}b(u)=\infty$.
Furthermore, since $V$ is monotone in each argument we deduce that $u\mapsto
b(u)$ is non-increasing and $V(u,x)=0$ for all $x>b(u)$. Therefore $\tau_{D}$
takes the form

$\displaystyle\tau_{D}=\inf\\{t\geq 0:X_{t}\geq b(U_{t})\\}.$

Now we introduce some notation so that we can express $V$ in terms of the
original measure $\mathbb{P}$. Define the stopping times

$\displaystyle\tau_{b}^{u,x}$ $\displaystyle=\inf\\{t>0:X_{t}+x\geq
b(u+t)\\}\qquad u\geq 0\text{ and }x\in\mathbb{R},$ (30)
$\displaystyle\tau_{b}^{g,x}$ $\displaystyle=\inf\\{t>0:X_{t}+x\geq
b(U_{t}^{(-x)})\\}\qquad\text{ }x\in\mathbb{R}.$ (31)

Then by the strong Markov property we have that

$\displaystyle V(u,x)$
$\displaystyle=\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{D}}G(U_{s},X_{s})\textup{d}s\right)$
$\displaystyle=\mathbb{E}\left(\int_{0}^{\sigma_{-x}^{-}\wedge\tau_{b}^{u,x}}G(u+s,X_{s}+x)\textup{d}s+\mathbb{I}_{\\{\sigma_{-x}^{-}\leq\tau_{b}^{u,x}\\}}\int_{\sigma_{-x}^{-}}^{\tau_{b}^{g,x}}G(U_{s}^{(-x)},X_{s}+x)\textup{d}s\right)$
$\displaystyle=\mathbb{E}\left(\int_{0}^{\sigma_{-x}^{-}\wedge\tau_{b}^{u,x}}G(u+s,X_{s}+x)\textup{d}s+\mathbb{I}_{\\{\sigma_{-x}^{-}\leq\tau_{b}^{u,x}\\}}V(0,X_{\sigma_{-x}^{-}})\right).$
(32)

Note that in the last equation we do not longer have explicitly the process
$\\{U_{t}^{(-x)},t\geq 0\\}$. So this alternative representation of $V$ in
terms of the original measure $\mathbb{P}$ will be useful to prove further
properties of $b$ and $V$. We now state some basic properties of the function
$b$.

######  Lemma 4.4.

The function $b:\mathbb{R}_{+}\mapsto\mathbb{R}$ is non-increasing with $0\leq
h(u)\leq b(u)$. We have that $\lim_{u\downarrow 0}b(u)=\infty$ and
$b(u)<\infty$ for all $u>0$. Moreover,

$\displaystyle\lim_{u\rightarrow\infty}b(u)=\lim_{u\rightarrow\infty}h(u).$

###### Proof.

We prove that for all $u>0$, $b(u)<\infty$. Assume that there exist $u>0$ such
that $b(u)=\infty$. Define $u_{0}:=\sup\\{u>0:b(u)=\infty\\}>0$. By the
monotonicity of $b$ we have that $b(u)=\infty$ for all $u<u_{0}$. Let $x>0$
and recall that $V(u,x)$ is bounded from below by
$-\mathbb{E}(\int_{0}^{\infty}\mathbb{E}_{X_{s}}(g^{p-1}))\textup{d}s)>-\infty$.
Then from the general version of Fatou’s lemma we have that for all
$u\in(0,u_{0})$,

$\displaystyle 0$ $\displaystyle\geq\liminf_{x\rightarrow\infty}V(u,x)$
$\displaystyle\geq\mathbb{E}\left(\liminf_{x\rightarrow\infty}\int_{0}^{\sigma_{-x}^{-}\wedge\tau_{b}^{u,x}}G(u+s,X_{s}+x)\textup{d}s+\mathbb{I}_{\\{\sigma_{-x}^{-}\leq\tau_{b}^{u,x}\\}}\int_{\sigma_{-x}^{-}}^{\tau_{b}^{g,x}}G(U_{s}^{(-x)},X_{s}+x)\textup{d}s\right)$
$\displaystyle=\int_{0}^{u_{0}}(u+s)^{p-1}\textup{d}s$ $\displaystyle>0,$

where we used that $\liminf_{x\rightarrow\infty}\sigma_{-x}^{-}=\infty$,
$\liminf_{x\rightarrow\infty}\tau_{b}^{u,x}=u_{0}>0$ and that
$\lim_{x\rightarrow\infty}G(u,x)=u^{p-1}$. From this contradiction we conclude
that $b(u)<\infty$ for all $u>0$.

Next we show that

$\displaystyle\lim_{u\rightarrow\infty}h(u)=\lim_{u\rightarrow\infty}b(u).$

Note that, since $b$ is non-increasing and it is bounded from below by
$h^{*}:=\lim_{u\rightarrow\infty}h(u)\geq 0$, the limit
$b^{*}:=\lim_{u\rightarrow\infty}b(u)$ exists and $b^{*}\geq h^{*}$. We prove
by contradiction that $b^{*}=h^{*}$. Suppose $b^{*}>h^{*}$, define the
stopping time

$\displaystyle\sigma_{*}=\inf\\{t\geq 0:X_{t}\notin(h^{*},b^{*})\\}.$

Take $u>0$ and $x\in(h^{*},b^{*})$. From the fact that $b(u)\geq b^{*}$ and
that $h^{*}\geq 0$ we have that $\sigma_{*}\leq\tau_{D}\wedge\sigma_{0}^{-}$
under $\mathbb{P}_{u,x}$. Then we have that

$\displaystyle V(u,x)$
$\displaystyle=\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{D}}G(U_{s},X_{s})ds\right)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\sigma_{*}}G(u+s,X_{s})ds\right)+\mathbb{E}_{u,x}\left(V(U_{\sigma_{*}},X_{\sigma_{*}})\right)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\sigma_{*}}G(u+s,X_{s})ds\right)+\mathbb{E}_{x}\left(V(u+\sigma_{*},X_{\sigma_{*}})\mathbb{I}_{\\{X_{\sigma_{*}}>0\\}}\right)+\mathbb{E}_{x}\left(V(0,X_{\sigma_{*}})\mathbb{I}_{\\{X_{\sigma_{*}}\leq
0\\}}\right),$ (33)

where in the last equality we used the Markov property of the two dimensional
process $\\{(U_{t},X_{t}),t\geq 0\\}$. The function $u\mapsto V(u,x)$ is non-
decreasing and bounded from above by zero, thus we have that
$\lim_{u\rightarrow\infty}V(u,x)$ exists and
$-\infty<\lim_{u\rightarrow\infty}V(u,x)\leq 0$. By the dominated convergence
theorem we also conclude that
$-\infty<\lim_{u\rightarrow\infty}\mathbb{E}_{x}\left(V(u+\sigma_{*},X_{\sigma_{*}})\mathbb{I}_{\\{X_{\sigma_{*}}>0\\}}\right)\leq
0$. Moreover, using the general version of Fatou’s lemma and the fact that
$\lim_{u\rightarrow\infty}G(u,x)=\infty$ we have that

$\displaystyle\lim_{u\rightarrow\infty}\mathbb{E}_{x}\left(\int_{0}^{\sigma_{*}}G(u+s,X_{s})\textup{d}s\right)=\infty.$

Therefore, taking $u\rightarrow\infty$ in (4) we get that

$\displaystyle\lim_{u\rightarrow\infty}V(u,x)=\infty.$

Which yields the desired contradiction. Therefore we conclude that
$b^{*}=h^{*}$. ∎

Now we show the continuity of the value function $V$ and right continuity of
$b$. It turns out that $b$ is continuous, the proof of this makes use of a
variational inequality and will be proved later.

######  Lemma 4.5.

The function $V$ is continuous on $E$ and $b$ is right-continuous. Moreover,
in the case that $X$ is of infinite variation we have that

$\displaystyle\lim_{h\downarrow 0}V(u,h)=V(0,0)$

for all $u>0$.

###### Proof.

First, we show that the function $u\mapsto V(u,x)$ is continuous for all $x>0$
fixed. Take $u_{1},u_{2}>0$ and $x>0$, then since the stopping time
$\tau^{*}_{(u_{1},x)}:=\tau_{b}^{u_{1},x}\mathbb{I}_{\\{\tau_{b}^{u_{1},x}<\sigma_{-x}^{-}\\}}+\tau_{b}^{g,x}\mathbb{I}_{\\{\tau_{b}^{u_{1},x}\geq\sigma_{-x}^{-}\\}}$
is optimal for $V(u_{1},x)$ (under $\mathbb{P}$) we have that

$\displaystyle|V(u_{2},x)-V(u_{1},x)|$
$\displaystyle\leq\mathbb{E}_{x}\left(\int_{0}^{\sigma_{0}^{-}\wedge\tau_{b}^{u_{1},0}}|G(u_{2}+s,X_{s})-G(u_{1}+s,X_{s})|\textup{d}s\right)$
$\displaystyle\leq\mathbb{E}_{x}\left(\int_{0}^{\tau_{b(u_{1})}^{+}}|G(u_{2}+s,X_{s})-G(u_{1}+s,X_{s})|\textup{d}s\right)$
$\displaystyle\leq\mathbb{E}\left(\int_{0}^{\tau_{b(u_{1})}^{+}}|(u_{2}+s)^{p-1}-(u_{1}+s)^{p-1}|\textup{d}s\right)$
$\displaystyle=\frac{1}{p}|\mathbb{E}((\tau_{b(u_{1})}^{+}+u_{2})^{p})-\mathbb{E}((\tau_{b(u_{1})}^{+}+u_{1})^{p})-[u_{2}^{p}-u_{1}^{p}]|$

where $\tau_{b(u_{1})}^{+}=\inf\\{t\geq 0:X_{t}>b(u_{1})\\}$. Thus tending
$u_{2}\mapsto u_{1}$, with the dominated convergence theorem and the fact that
$\mathbb{E}((\tau_{a}^{+}+u)^{p})<\infty$ for all $u,a\geq 0$ we get that
$u\mapsto V(u,x)$ is continuous uniformly over all $x>0$.

Now we show that $x\mapsto V(u,x)$ is continuous. We start by considering the
case $x<0$ and $u=0$. Then $\tau_{0}^{+}\leq\tau_{D}$, and using the Markov
property and Fubini’s theorem we can write for all $x<0$,

$\displaystyle V(0,x)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{+}}G(0,X_{s})ds\right)+\mathbb{E}_{x}\left(\int_{\tau_{0}^{+}}^{\tau_{D}}G(U_{s},X_{s})\textup{d}s\right)$
$\displaystyle=-\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{+}}\mathbb{E}_{X_{s}}(g^{p-1})\textup{d}s\right)+V(0,0)$
$\displaystyle=-\int_{(-\infty,0)}\mathbb{E}_{z}(g^{p-1})\int_{0}^{\infty}\mathbb{P}_{x}(X_{s}\in
dz,s<\tau_{0}^{+})\textup{d}s+V(0,0)$

Using the $0$-potential measure of $X$ killed on exiting the interval
$(-\infty,0]$ (see equation (7)) we have that

$\displaystyle V(0,x)$
$\displaystyle=-\int_{0}^{\infty}\mathbb{E}_{-z}(g^{p-1})[W(z)-W(x+z)]\textup{d}z+V(0,0)$
$\displaystyle=-\int_{0}^{\infty}\mathbb{E}_{-z}(g^{p-1})\int_{(x+z,z]}W(\textup{d}u)\textup{d}z+V(0,0)$
$\displaystyle=-\int_{[0,\infty)}W(\textup{d}u)\int_{u}^{u-x}\mathbb{E}_{-z}(g^{p-1})\textup{d}z+V(0,0)$
$\displaystyle=-\int_{0}^{-x}\int_{[0,\infty)}\mathbb{E}_{-u-z}(g^{p-1})W(\textup{d}u)\textup{d}z+V(0,0).$
(34)

Therefore $x\mapsto V(0,x)$ is a continuous function in $(-\infty,0]$.

Next, suppose that $u>0$ and $x>0$. Recall from equation (32) that we can
write

$\displaystyle V(u,x)$
$\displaystyle=\mathbb{E}\left(\int_{0}^{\sigma_{-x}^{-}\wedge\tau_{b}^{u,x}}G(u+s,X_{s}+x)\textup{d}s\right)+\mathbb{E}(V(0,X_{\sigma_{-x}^{-}}+x)\mathbb{I}_{\\{\sigma_{-x}^{-}\leq\tau_{b}^{u,x}\\}}).$

Note that for all $s\leq\tau_{b}^{u,x}\wedge\sigma_{-x}^{-}$, it holds that
$0<X_{s}+x\leq b(u+s)\leq b(u)$. Moreover, it is easy to show that for any
$x>0$ we have that $\lim_{h\rightarrow 0}\sigma_{x+h}^{-}=\sigma_{x}^{-}$ a.s.
and $\lim_{h\rightarrow 0}\tau_{b}^{u,x+h}=\tau_{b}^{u,x}$ a.s. Then by the
dominated convergence theorem and the fact that $V$ is continuous on
$(-\infty,0]$ and $x\mapsto G(u,x)$ is continuous on $(0,\infty)$ we conclude
that $x\mapsto V(u,x)$ is continuous in $(0,\infty)$ for all $u>0$. Note that
when $X$ is of infinite variation, $\lim_{h\downarrow
0}\sigma_{-h}^{-}=\tau_{0}^{-}=0$ a.s. and the latter argument also tells us
that for all $u>0$,

$\displaystyle\lim_{h\downarrow 0}V(u,h)=V(0,0).$

Using the fact that $V$ is continuous on $(0,\infty)\times(0,\infty)$ and a
standard argument (see e.g. Du Toit et al., (2008)) shows that $b$ is indeed
right-continuous.

We proceed to prove that $\lim_{(u,x)\rightarrow(0,0)^{+}}V(u,x)=V(0,0)$ when
$X$ is of finite variation. In this case we know that $\sigma_{0}^{-}=0$ and
$\tau_{0}^{-}>0$. Due to the strong Markov property,

$\displaystyle V(0,0)$
$\displaystyle=\mathbb{E}\left(\int_{0}^{\tau_{b}^{0,0}\wedge\tau_{0}^{-}}G(s,X_{s})\textup{d}s\right)+\mathbb{E}(\mathbb{I}_{\\{\tau_{0}^{-}<\tau_{b}^{0,0}\\}}V(0,X_{\tau_{0}^{-}})),$

where $\tau_{b}^{0,0}=\inf\\{t>0:X_{t}\geq b(s)\\}$. From the right-continuity
of $b$ and the fact that $X$ creeps upwards we can easily show that
$\lim_{(u,x)\rightarrow(0,0)^{+}}\tau_{b}^{u,x}=\tau_{b}^{0,0}$ a.s. Taking
limits in (32) we have from the dominated convergence theorem,

$\displaystyle\lim_{(u,x)\rightarrow(0,0)^{+}}V(u,x)$
$\displaystyle=\lim_{(u,x)\rightarrow(0,0)^{+}}\mathbb{E}\left(\int_{0}^{\sigma_{-x}^{-}\wedge\tau_{b}^{u,x}}G(u+s,X_{s}+x)\textup{d}s\right)+\lim_{(u,x)\rightarrow(0,0)^{+}}\mathbb{E}(V(0,X_{\sigma_{-x}^{-}}+x)\mathbb{I}_{\\{\sigma_{-x}^{-}\leq\tau_{b}^{u,x}\\}})$
$\displaystyle=\mathbb{E}\left(\int_{0}^{\tau_{b}^{0,0}\wedge\tau_{0}^{-}}G(s,X_{s})\textup{d}s\right)+\mathbb{E}(\mathbb{I}_{\\{\tau_{0}^{-}<\tau_{b}^{0,0}\\}}V(0,X_{\tau_{0}^{-}}))$
$\displaystyle=V(0,0),$

where we used that $\lim_{x\downarrow 0}\sigma_{-x}^{-}=\tau_{0}^{-}$.
Therefore $V$ is continuous on the set $E$.

∎

From the proof of the previous result we obtain the following.

######  Corollary 4.6.

For all $(u,x)\in E$ we have that there exist non-negative constants
$A^{\prime}_{p-1}$ and $C^{\prime}_{p-1}$ such that

$\displaystyle V(u,x)\geq-A^{\prime}_{p-1}-C^{\prime}_{p-1}|x|^{p}+V(0,0).$
(35)

###### Proof.

From equation (4) and the fact that $x\mapsto\mathbb{E}_{x}(g^{p-1})$ is is
non-increasing and bounded from above by a polynomial (see Lemmas 2.2 and 2.3)
we have the inequalities for $x<0$

$\displaystyle V(0,x)$ $\displaystyle\geq
x\int_{[0,\infty]}\mathbb{E}_{x-u}(g^{p-1})W(\textup{d}u)+V(0,0)$
$\displaystyle\geq\frac{1}{\psi^{\prime}(0+)}2^{p-1}[\mathbb{E}(g^{p-1})+A_{p-1}]x+\frac{1}{\psi^{\prime}(0+)}2^{p-1}C_{p-1}x\mathbb{E}(|x+\underline{X}_{\infty}|^{p-1})+V(0,0)$
$\displaystyle\geq\frac{1}{\psi^{\prime}(0+)}2^{p-1}[\mathbb{E}(g^{p-1})+A_{p-1}+2^{p-1}C_{p-1}\mathbb{E}((-\underline{X}_{\infty})^{p-1})]x-\frac{1}{\psi^{\prime}(0+)}2^{p-1}C_{p-1}|x|^{p}+V(0,0).$

The general statement holds since $V$ is non-decreasing in each argument. ∎

At this point we know from Lemma 4.5 that the function $b$ is a right-
continuous function. In order to show left continuity we make use of a
variational inequality that is satisfied by the value function $V$. The
oncoming paragraphs will be dedicated on introducing that.

It is well known that for every optimal stopping problem there is a free
boundary problem which is stated in terms of the infinitesimal generator (see
e.g. Peskir and Shiryaev, (2006) Chapter III). In particular in this case,
provided that the value function is smooth enough, we have that $V$ solves the
Dirichlet/Poisson problem. That is,

$\displaystyle\mathcal{A}_{U,X}(V)=-G\qquad\text{in }E\setminus D,$

where $\mathcal{A}_{U,X}$ corresponds to the infinitesimal generator of the
process $(U,X)$ given in (12). However, in our setting turns out to be
challenging to show that $V$ is a $C^{1,2}$ function. Lamberton and Mikou,
(2008) showed that we can state an analogous (in)equality in the sense of
distributions (see in particular section 2 for its definition). The next lemma
is an extension of Proposition 2.5 in Lamberton and Mikou, (2008). Since the
proof is similar to the aforementioned proposition is omitted.

######  Lemma 4.7.

The distribution $\mathcal{A}_{U,X}(V)+G$ is a non-negative measure on $E$.
Moreover, on $E\setminus D$ we have $\mathcal{A}_{U,X}(V)+G=0$.

Let $\widetilde{V}$ be the extension of $V$ to the set
$\mathbb{R}_{+}\times\mathbb{R}$ as in (16). We have the following remark.

######  Remark 4.8.

  * i)

In Lamberton and Mikou, (2008) the definition of the infinitesimal generator
in the sense of distributions assumes that the value function is a bounded
Borel measurable function. In our setting such condition can be relaxed by the
fact that
$(u,x)\mapsto\int_{(-\infty,-1)}|\widetilde{V}(u,x+y)|\Pi(\textup{d}y)$ is a
locally integrable function on $\mathbb{R}_{+}\times\mathbb{R}$. This can
proven as a consequence of Corollary 4.6.

  * ii)

We note from (12) that $\mathcal{A}_{U,X}(V)=\partial/\partial
u\widetilde{V}+\mathcal{A}_{X}(\widetilde{V})$ in the sense of distributions,
where $\mathcal{A}_{X}$ corresponds to the infinitesimal generator of $X$.

For the proof of left-continuity of $b$ we define an auxiliary function. For
$(u,x)\in E$,

$\displaystyle\Lambda(u,x):=\left\\{\begin{array}[]{lc}\int_{(-\infty,0)}\widetilde{V}(u,x+y)\Pi(\textup{d}y)+G(u,x)&x>b(u)\\\
0&x\leq b(u).\end{array}\right.$

The next lemma states some basic properties of the function $\Lambda$.

######  Lemma 4.9.

The function $\Lambda$ is a non-decreasing (in each argument) function such
that $0<\Lambda(u,x)<\infty$ for all $x>b(u)$. Moreover, is strictly
increasing in each argument and continuous on the set $D$ and
$\Lambda=\mathcal{A}_{U,X}+G$ almost everywhere on $E$ in the sense of
distributions.

###### Proof.

It follows from the finiteness of $\Pi$ in any set away of a neighbourhood of
zero, the fact that $V$ vanishes in $D$ and Corollary 4.6 that
$|\Lambda(u,x)|<\infty$ for all $(u,x)\in E$. The fact that $\Lambda$ is
continuous on $D$ follows from the continuity of $V$ and $G$, the dominated
convergence theorem and the fact that $\Pi$ has no atoms. Moreover, $\Lambda$
is strictly increasing in each argument on $D$ since $V$ is non-decreasing in
each argument and $G$ is strictly increasing in each argument on $D$. From
Lemma 4.7 we deduce that $\mathcal{A}_{U,X}+G=\Lambda$ in the sense of
distributions on $E$. ∎

Now we are ready to show that the function $b$ is continuous, the proof is
based on the ideas of Lamberton and Mikou, (2008) (Theorem 4.2). We include
the proof for completeness.

######  Lemma 4.10.

The function $b$ is continuous.

###### Proof.

From Lemma 4.5 we already know that $b$ is right continuous. We then show left
continuity of $b$. We proceed by contradiction. Suppose there is a point
$u_{*}>0$ such that $b(u_{*}-):=\lim_{h\downarrow 0}b(u_{*}-h)>b(u_{*})$. Then
since $b$ is non-decreasing we have for all
$(u,x)\in(0,u_{*})\times(b(u_{*}),b(u_{*}-))$ that $V(u,x)<0$. Thus,
$(0,u_{*})\times(b(u_{*}),b(u_{*}-))\subset E\setminus D$. From Lemma 4.7 we
obtain that $\mathcal{A}_{U,X}(V)+G=0$ on
$(0,u_{*})\times(b(u_{*}),b(u_{*}-))$. Hence, by Remark 4.8 ii) we obtain

$\displaystyle\mathcal{A}_{X}(\widetilde{V})+G=-\frac{\partial}{\partial
u}\widetilde{V}\leq 0\qquad\text{on }(0,u_{*})\times(b(u_{*}),b(u_{*}-)),$

where $\mathcal{A}_{X}$ represents the generator of $X$ in the sense of
distributions and the inequality follows from the fact
$u\mapsto\widetilde{V}(u,x)$ is non-decreasing. Thus, by continuity of
$\widetilde{V}$ and $G$ on $(0,\infty)\times(0,\infty)$ we have that for any
$u\in(0,u_{*})$ and any test function $\psi$ with compact support in
$(b(u_{*}),b(u_{*}-))$

$\displaystyle\int_{\mathbb{R}}\left\\{\widetilde{V}(u,x)\left[-\mu\frac{\partial}{\partial
x}\psi(x)+\frac{1}{2}\sigma^{2}\frac{\partial^{2}}{\partial
x^{2}}\psi(x)+B_{X}^{*}(\psi)(x)\right]+G(u,x)\psi(x)\right\\}\textup{d}x\leq
0,$

where
$B_{X}^{*}(\psi)(x)=\int_{(-\infty,0)}(\psi(x-y)-\psi(x)+y\mathbb{I}_{\\{|y|\leq
1\\}})\Pi(\textup{d}y)$. Taking $u\uparrow u_{*}$ and using the fact that
$\widetilde{V}(u_{*},x)=0$ for all $x\geq b(u_{*})$ we get that

$\displaystyle 0$ $\displaystyle\geq\lim_{u\uparrow
u_{*}}\int_{-\infty}^{b(u_{*})}\widetilde{V}(u,x)\int_{(-\infty,x-b(u_{*}))}\psi(x-y)\Pi(\textup{d}y)\textup{d}x+\int_{b(u_{*})}^{b(u_{*}-)}G(u_{*},x)\psi(x)\textup{d}x$
$\displaystyle=\lim_{u\uparrow
u_{*}}\int_{b(u_{*})}^{b(u_{*}-)}\psi(x)\int_{(-\infty,x-b(u_{*}))}\widetilde{V}(u,x+y)\Pi(\textup{d}y)\textup{d}x+\int_{b(u_{*})}^{b(u_{*}-)}G(u_{*},x)\psi(x)\textup{d}x$
$\displaystyle=\int_{b(u_{*})}^{b(u_{*}-)}\psi(x)\Lambda(u_{*},x)\textup{d}x$
$\displaystyle>0,$

where the strict inequality follows from the fact that $\Lambda$ is strictly
positive in each argument on $D$ (see Lemma 4.9). Therefore $b(u-)=b(u)$ for
all $u>0$ and $b$ is continuous. ∎

From Lemma 4.4 we know that $b$ and $h$ converge at the same limit when $u$
tends to infinity. Moreover, from the discussion about $h$ after Lemma 2.3 we
know that in case that $X$ is of finite variation there exists a value
$u_{h}^{*}<\infty$ for which $h(u)=0$ for all $u\geq u^{*}_{h}$. That suggests
a similar behaviour for $b$, the next lemma addresses that conjecture.

######  Lemma 4.11.

Denote as $u_{b}=\inf\\{u>0:b(u)=0\\}$. If $X$ is of infinite variation or
finite variation and infinite activity we have that $u_{b}=\infty$. Otherwise
$u_{b}=u^{*}$, where $u^{*}$ is the unique solution to

$\displaystyle G(u,0)+\int_{(-\infty,0)}V(0,y)\Pi(\textup{d}y)=0.$ (36)

###### Proof.

From the fact that $h(u)>0$ for all $u>0$ when $X$ is of infinite variation
and inequality $b(u)\geq h(u)$ we have that assertion is true for this case.
Suppose that $X$ has finite variation with infinite activity, that is
$\Pi(-\infty,0)=\infty$, and assume that $u_{b}<\infty$. Then since $b$ is
non-increasing we have that $b(u)=0$ for all $u>u_{b}$ and then $V(u,x)=0$ for
all $x>0$ and $u>u_{b}$. From Lemma 4.9 we have that

$\displaystyle G(u,x)+\int_{(-\infty,-x)}V(0,x+y)\Pi(dy)\geq 0\text{ for all
}x>0$

for all $u>u_{b}$. Taking $x\downarrow 0$ in the equation above and using the
expression for $V(0,z)$ for $z<0$ given in (4) we have that for any $u>u_{b}$

$\displaystyle 0\leq G(u,0)-\lim_{x\downarrow
0}\int_{(-\infty,0)}\int_{0}^{-x+y}\int_{[0,\infty)}\mathbb{E}_{-u-z}(g^{p-1})W(\textup{d}u)\textup{d}z\Pi(\textup{d}y)+\lim_{x\downarrow
0}V(0,0)\Pi(-\infty,-x)=-\infty$

which is a contradiction and then $u_{b}=\infty$. Now assume that $X$ has
finite variation with $\Pi(-\infty,0)<\infty$. Assume that $b(u^{*})>0$, then
$V(u^{*},x)<0$ for $x\in(0,b(u^{*}))$. Moreover, using the compensation
formula for Poisson random measures we have that for all $u>0$ and $x<b(u)$,

$\displaystyle\mathbb{E}_{u,x}$
$\displaystyle(V(0,X_{\tau_{0}^{-}})\mathbb{I}_{\\{\tau_{0}^{-}<\tau_{D}\\}})$
$\displaystyle=\mathbb{E}_{u,x}\left(\int_{[0,\infty)}\int_{(-\infty,0)}V(0,X_{s-}+y)\mathbb{I}_{\\{\underline{X}_{s-}>0\\}}\mathbb{I}_{\\{X_{s-}+y<0\\}}\mathbb{I}_{\\{X_{r}\leq
b(U_{r})\text{ for all }r<s\\}}N(\textup{d}s,\textup{d}y)\right)$
$\displaystyle=\mathbb{E}_{u,x}\left(\int_{0}^{\infty}\int_{(-\infty,0)}V(0,X_{s}+y)\mathbb{I}_{\\{\underline{X}_{s-}>0\\}}\mathbb{I}_{\\{X_{s}+y<0\\}}\mathbb{I}_{\\{X_{r}\leq
b(U_{r})\text{ for all }r<s\\}}\Pi(\textup{d}y)\textup{d}s\right)$
$\displaystyle=\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{0}^{-}\wedge\tau_{D}}\int_{(-\infty,0)}V(0,X_{s}+y)\mathbb{I}_{\\{X_{s}+y<0\\}}\Pi(\textup{d}y)\textup{d}s\right).$

Then from the Markov property we have that for all $x<b(u^{*})$

$\displaystyle V(u^{*},x)$
$\displaystyle=\mathbb{E}_{u^{*},x}\left(\int_{0}^{\tau_{D}\wedge\tau_{0}^{-}}G(u^{*}+s,X_{s})\textup{d}s\right)+\mathbb{E}_{u^{*},x}(V(0,X_{\tau_{0}^{-}})\mathbb{I}_{\\{\tau_{0}^{-}<\tau_{D}\\}})$
$\displaystyle=\mathbb{E}_{u^{*},x}\left(\int_{0}^{\tau_{D}\wedge\tau_{0}^{-}}\left[G(u^{*}+s,X_{s})+\int_{(-\infty,0)}V(0,X_{s}+y)\mathbb{I}_{\\{X_{s}+y<0\\}}\Pi(dy)\right]\textup{d}s\right)$
$\displaystyle>0,$

where the strict inequality follows from the fact that that $X$ is of finite
variation and then $\tau_{D}\wedge\tau_{0}^{-}>0$, the definition of $u^{*}$
and the fact that $G$ and $V$ are non-decreasing in each argument. Then we are
contradicting the fact that $V(u^{*},x)<0$ and we conclude that $b(u^{*})=0$
and $u_{b}\leq u^{*}$. Moreover, from Lemma 4.9 we know that for all $u>u_{b}$

$\displaystyle G(u,x)+\int_{(-\infty,-x)}V(0,x+y)\Pi(\textup{d}y)\geq 0\text{
for all }x>0.$

Taking $x\downarrow 0$ we get that for all $u\geq u_{b}$,
$G(u,0)+\int_{(-\infty,0)}V(0,y)\Pi(\textup{d}y)\geq 0$. The latter implies
that $u^{*}\leq u_{b}$ (since $u\mapsto G(u,0)$ is strictly increasing).
Therefore we conclude that $u^{*}=u_{b}$ and the proof is complete. ∎

As we mentioned before it is challenging to prove the existence of the
derivatives of $V$. However, it is possible to show that the derivative of $V$
with respect to $x$ exists and is equal to zero (smooth fit condition). In
order to prove the smooth fit we first prove a technical lemma that ensures
that the optimal stopping time has moments of order $p$.

######  Lemma 4.12.

Let $p>1$ and assume that $\int_{(-\infty,0)}|x|^{p+1}\Pi(dx)$. Then for all
$(u,x)\in E$,

$\displaystyle\mathbb{E}_{u,x}((\tau_{D})^{p})<\infty.$

Recall from Lemma 4.11 that when $X$ is of infinite variation or finite
variation with infinite activity we have that $b(u)>0$ for all $u>0$. In the
case that $X$ is of finite variation we have that $b(u)>0$ only if $u<u_{b}$
where $u_{b}$ is the solution to (36). In such cases we can guarantee that the
smooth fit condition holds which is proven in the following Theorem. Since the
proof is rather long and technical it can be found in the Appendix.

######  Theorem 4.13 (Smooth fit condition).

Suppose that $u>0$ is such that $b(u)>0$. Then the derivative of $x\mapsto
V(u,x)$ exists at the point $x=b(u)$ and

$\displaystyle\frac{\partial}{\partial x}V(u,b(u))=0.$

Recall from equation (4) that when $x<0$

$\displaystyle
V(0,x)=-\int_{0}^{-x}\int_{[0,\infty)}\mathbb{E}_{-u-z}(g^{p-1})W(\textup{d}u)\textup{d}z+V(0,0).$

Note that the first term on the right-hand side of the equation above does not
depend on the boundary $b$. Then, for $x<0$, the value function $V(0,x)$ is
characterised by the value $V(0,0)$. Moreover, from Lemma 4.11 we know that
when $X$ is of finite variation with $\Pi(-\infty,0)<\infty$, the value
$u_{b}$ is the unique solution to

$\displaystyle
G(u,0)-\int_{(-\infty,0)}\int_{0}^{-y}\int_{[0,\infty)}\mathbb{E}_{-u-z}(g^{p-1})W(\textup{d}u)\textup{d}z\Pi(\textup{d}y)+V(0,0)\Pi(-\infty,0)=0.$

Otherwise, $u_{b}=\infty$. Then if $X$ is of finite variation with finite
activty, $u_{b}$ is also characterised by the value $V(0,0)$. From the fact
that $\lim_{u\downarrow 0}b(u)=\infty$ (see Lemma 4.4) and since $V$ is non-
decreasing in each argument we deduce that $V(0,0)$ is a strictly negative
real number.

The next theorem gives a characterisation of the value function $V$ on the set
$(0,\infty)\times(0,\infty)$, the boundary $b$ and the values $V(0,0)$ and
$u_{b}$ as unique solutions of a system of non-linear integral equations. The
method of proof is deeply inspired on the ideas of Du Toit et al., (2008).
However, the presence of jumps adds an important level of difficulty. In
particular, when $\Pi\neq 0$, showing that a candidate solution
$(V_{*},b_{*})$ satisfies the inequality

$\displaystyle
G(u,x)+\int_{(-\infty,0)}\widetilde{V}_{*}(u,x+y)\Pi(\textup{d}y)>0$

for all $x>b_{*}(u)$ turns out to be challenging. This inequality in fact is
cornerstone for the submartingale property of the process
$V_{*}(U_{t},X_{t})+\int_{0}^{t}G(U_{s},X_{s})\textup{d}s$.

######  Theorem 4.14.

Let $p>1$ and $X$ be a spectrally negative Lévy process such that
$\int_{(-\infty,-1)}|x|^{p+1}\Pi(\textup{d}x)<\infty$. For all $u>0$ and
$x>0$, the function $V$ can be written as

$\displaystyle V(u,x)$
$\displaystyle=V(0,0)\frac{\sigma^{2}}{2}W^{\prime}(x)-\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{-}}\int_{(-\infty,0)}V(u+s,X_{s}+y)\mathbb{I}_{\\{X_{s}+y>0\\}}\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>b(u+s)\\}}\textup{d}s\right)$
$\displaystyle\qquad+\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{-}}\left[G(u+s,X_{s})+\int_{(-\infty,0)}V(0,X_{s}+y)\mathbb{I}_{\\{X_{s}+y<0\\}}\Pi(\textup{d}y)\right]\mathbb{I}_{\\{X_{s}<b(u+s)\\}}\textup{d}s\right).$
(37)

The curve $b$ satisfies the equation

$\displaystyle 0$
$\displaystyle=V(0,0)\frac{\sigma^{2}}{2}W^{\prime}(b(u))-\mathbb{E}_{b(u)}\left(\int_{0}^{\tau_{0}^{-}}\int_{(-\infty,0)}V(u+s,X_{s}+y)\mathbb{I}_{\\{X_{s}+y>0\\}}\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>b(u+s)\\}}\textup{d}s\right)$
$\displaystyle\qquad+\mathbb{E}_{b(u)}\left(\int_{0}^{\tau_{0}^{-}}\left[G(u+s,X_{s})+\int_{(-\infty,0)}V(0,X_{s}+y)\mathbb{I}_{\\{X_{s}+y<0\\}}\Pi(\textup{d}y)\right]\mathbb{I}_{\\{X_{s}<b(u+s)\\}}\textup{d}s\right)$
(38)

for all $u<u_{b}$ and $b(u)=0$ for all $u\geq u_{b}$, where $u_{b}=\infty$ in
the case $X$ is of infinite variation or finite variation with
$\Pi(-\infty,0)=\infty$. Otherwise $u_{b}$ is the unique solution to

$\displaystyle
G(u,0)-\int_{(-\infty,0)}\int_{0}^{-y}\int_{[0,\infty)}\mathbb{E}_{-u-z}(g^{p-1})W(\textup{d}u)\textup{d}z\Pi(\textup{d}y)+V(0,0)\Pi(-\infty,0)=0.$
(39)

Moreover, when $X$ is of finite variation $V(0,0)$ also satisfies (37) when
$u=0$ and $x=0$. When $X$ is of infinite variation we have in addition that

$\displaystyle\frac{\partial}{\partial x}V_{+}(0,0)=\frac{\partial}{\partial
x}V_{-}(0,0),$ (40)

where $\frac{\partial}{\partial x}V_{+}(0,0)$ and $\frac{\partial}{\partial
x}V_{-}(0,0)$ are the right and left derivative of $x\mapsto\lim_{u\downarrow
0}V(u,x)$ and $x\mapsto V(0,x)$ at zero, respectively. Furthermore, the
quadruplet $(V,b,V(0,0),u_{b})$ is uniquely characterised by the equations
above, where $V$ and $b$ are considered in the class of continuous functions
such that $V(u,x)\leq 0$ for all $u,x>0$ and $b$ is non-increasing with $b\geq
h$ whereas $-\frac{1}{p}\mathbb{E}(g^{p})\leq V(0,0)<0$.

Since the proof of Theorem 4.14 is rather long we break it in a series of
Lemmas. Next subsection is entirely dedicated to that purpose.

###  4.1 Proof of Theorem 4.14

First, we show that $V$ satisfies the alternative representation mentioned.

######  Lemma 4.15.

The quadruplet $(V,b,V(0,0),u_{b})$ satisfy equations (37)-(40).

###### Proof.

We follow an analogous argument as Lamberton and Mikou, (2013) (see Theorem
3.2). Let $\rho$ be a positive $C^{\infty}$ function on $\mathbb{R}_{+}^{2}$
with support in $[0,1]\times[0,1]$ and
$\int_{0}^{\infty}\int_{0}^{\infty}\rho(v,y)\textup{d}v\textup{d}y=1$. For
$n\geq 1$ define $\rho_{n}(v,y)=n^{2}\rho(nv,ny)$. Then $\rho_{n}$ is
$C^{\infty}$ and has compact support in $[0,1/n]\times[0,1/n]$ and
$(\widetilde{V}*\rho_{n})(u,x)=\int_{0}^{\infty}\int_{0}^{\infty}\widetilde{V}(u+v,x+y)\rho_{n}(v,y)\textup{d}v\textup{d}y$
is a $C^{1,2}(E)$ function (hence with bounded derivatives in compact sets).
Moreover, from Corollary 4.6 we have that

$\displaystyle\int_{(-\infty,-1)}(\widetilde{V}*\rho_{n})(u,x+y)\Pi(\textup{d}y)\geq\int_{(-\infty,-1)}\widetilde{V}(u,x+y)\Pi(\textup{d}y)>-\infty$

for all $(u,x)\in E$. It can be shown that when $X$ is of infinite variation,
$\widetilde{V}*\rho_{n}\downarrow V$ on $E$ when $n\rightarrow\infty$ and that
(see Lamberton and Mikou, (2008)),

$\displaystyle\mathcal{A}_{U,X}(\widetilde{V}*\rho_{n})(u,x)=(\widetilde{\lambda}*\rho_{n})(u,x)\qquad\text{for
all }(u,x)\in E,$ (41)

where $\mathcal{A}_{U,X}$ is the infinitesimal generator of the process
$(U,X)$ given in (12) and $\lambda:E\mapsto\mathbb{R}$ is given by

$\displaystyle\lambda(u,x)=\left\\{\begin{array}[]{lc}\int_{(-\infty,0)}V(u,x+y)\Pi(\textup{d}y)&x\geq
b(u)\\\ -G(u,x)&x<b(u)\end{array}\right..$

When $X$ is of finite variation we have that
$(\widetilde{V}*\rho_{n})(u,x)\downarrow V(u,x)$ when $n\rightarrow\infty$ and
(41) holds for $u>0$ and $x>0$. Then when $X$ is of infinite variation,
applying Itô formula (see (11)), we have that for any $(u,x)\in E$ and
$m,t\geq 0$,

$\displaystyle\mathbb{E}_{u,x}\left((\widetilde{V}*\rho_{n})(U_{t\wedge\tau_{-m}^{-}},X_{t\wedge\tau_{-m}^{-}})\right)$
$\displaystyle=(\widetilde{V}*\rho_{n})(u,x)+\mathbb{E}_{u,x}\left(\int_{0}^{t\wedge\tau_{-m}^{-}}\mathcal{A}_{U,X}(\widetilde{V}_{n})(U_{s},X_{s})\textup{d}s\right)$
$\displaystyle=(\widetilde{V}*\rho_{n})(u,x)-\mathbb{E}_{u,x}\left(\int_{0}^{t\wedge\tau_{-m}^{-}}(\widetilde{G}*\rho_{n})(U_{s},X_{s})\mathbb{I}_{\\{X_{s}<b(U_{s})\\}}\textup{d}s\right)$
$\displaystyle\qquad+\mathbb{E}_{u,x}\left(\int_{0}^{t\wedge\tau_{-m}^{-}}\int_{(-\infty,0)}(\widetilde{V}*\rho_{n})(U_{s},X_{s}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>b(U_{s})\\}}\textup{d}s\right),$

where we used the fact that $U_{s}$ has a linear behaviour when $X_{s}>0$ and
that $\mathbb{P}_{x}(X_{s}=b(u+s))=0$ for all $s>0$ and $x\in\mathbb{R}$ when
$X$ is of infinite variation (see Sato, (1999)). Similarly, when $X$ is finite
variation with $\Pi(-\infty,0)=\infty$ we also have that
$\mathbb{P}_{x}(X_{s}=b(u+s))=0$ for all $s>0$ and $x\in\mathbb{R}$ so that
the equation above holds for $u,x>0$ and $m=0$. An analogous argument shows
that when $X$ is of finite variation with $\Pi(-\infty,0)<\infty$,

$\displaystyle\mathbb{E}_{u,x}\left((\widetilde{V}*\rho_{n})(U_{t\wedge\tau_{0}^{-}},X_{t\wedge\tau_{0}^{-}})\right)$
$\displaystyle=(\widetilde{V}*\rho_{n})(u,x)-\mathbb{E}_{u,x}\left(\int_{0}^{t\wedge\tau_{0}^{-}}(\widetilde{G}*\rho_{n})(U_{s},X_{s})\mathbb{I}_{\\{X_{s}<b(U_{s})\\}}\textup{d}s\right)$
$\displaystyle\qquad+\mathbb{E}_{u,x}\left(\int_{0}^{t\wedge\tau_{0}^{-}}\int_{(-\infty,0)}(\widetilde{V}*\rho_{n})(U_{s},X_{s}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}\geq
b(U_{s})\\}}\textup{d}s\right)$

for all $u,x>0$. Note that, since $X_{t}\geq\underline{X}_{\infty}$ for all
$t>0$ and $V$ is non-decreasing in each argument we have, for all $n\geq 1$
and $(u,x)\in E$, the inequality

$\displaystyle
0\geq\mathbb{E}_{u,x}\left((\widetilde{V}*\rho_{n})(U_{t\wedge\tau_{-m}^{-}},X_{t\wedge\tau_{-m}^{-}})\right)\geq\mathbb{E}_{x}(V(0,\underline{X}_{\infty}))\geq-A^{\prime}_{p-1}-C^{\prime}_{p-1}\mathbb{E}_{x}((-\underline{X}_{\infty})^{p})+V(0,0)>-\infty,$

where the third inequality follows from equation (35) and the last quantity is
finite by Lemma 2.1. On the other hand we have for all $(u,x)\in E$ that

$\displaystyle\left|\mathbb{E}_{u,x}\left(\int_{0}^{\infty}G(U_{s},X_{s})\mathbb{I}_{\\{X_{s}<b(U_{s})\\}}\textup{d}s\right)\right|$
$\displaystyle\leq\mathbb{E}_{u,x}\left(\int_{0}^{\infty}|G(U_{s},X_{s})|\mathbb{I}_{\\{X_{s}<b(U_{s})\\}}\textup{d}s\right)$
$\displaystyle\leq\mathbb{E}_{u,x}\left(\int_{0}^{\infty}[U_{s}^{p-1}+\mathbb{E}_{X_{s}}(g^{p-1})]\mathbb{I}_{\\{X_{s}<b(U_{s})\\}}\textup{d}s\right)$
$\displaystyle\leq\mathbb{E}_{u,x}\left(\int_{0}^{\infty}s^{p-1}\mathbb{I}_{\\{X_{s}<b(U_{s})\\}}\textup{d}s\right)+\mathbb{E}_{x}\left(\int_{0}^{\infty}\mathbb{E}_{X_{s}}(g^{p-1})\textup{d}s\right).$

From Lemma 4.1 we know that the second integral above is finite. Now we check
that the first integral above is also finite. Consider $\delta>0$ and consider
$g^{(b(\delta))}$, the last time $X$ is below the level $b(\delta)$, then
$g^{(b(\delta))}\geq g$ and $X_{s+g^{(b(\delta))}+\delta}\geq b(\delta)$ for
all $s\geq 0$. Hence, since $b$ is non-increasing we get

$\displaystyle\mathbb{E}_{u,x}\left(\int_{0}^{\infty}s^{p-1}\mathbb{I}_{\\{X_{s}<b(U_{s})\\}}\textup{d}s\right)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{g^{(b(\delta))}+\delta}s^{p-1}\mathbb{I}_{\\{X_{s}<b(U_{s})\\}}\textup{d}s\right)\leq\mathbb{E}_{x}((g^{(b(\delta))}+\delta)^{p})<\infty,$

where the last expectation is finite by Lemma 2.1. Moreover, since
$V\leq(\widetilde{V}*\rho_{n})\leq 0$ and $V*\rho_{n}\geq V*\rho_{n+1}$ for
all $n\geq 1$ we have that the mappings

$\displaystyle
m,n,t\mapsto-\int_{0}^{t\wedge\tau_{-m}^{-}}\int_{(-\infty,0)}(\widetilde{V}*\rho_{n})(U_{s},X_{s}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>b(U_{s})\\}}\textup{d}s$

are nonnegative and nondecreasing. Therefore by the dominated and monotone
convergence theorem we have that letting $n,t\rightarrow\infty$,

$\displaystyle\mathbb{E}_{u,x}\left(V(U_{\tau_{-m}^{-}},X_{\tau_{-m}^{-}})\right)$
$\displaystyle=V(u,x)-\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{-m}^{-}}G(U_{s},X_{s})\mathbb{I}_{\\{X_{s}<b(U_{s})\\}}\textup{d}s\right)$
$\displaystyle\qquad+\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{-m}^{-}}\int_{(-\infty,0)}\widetilde{V}(U_{s},X_{s}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}\geq
b(U_{s})\\}}\textup{d}s\right)$ (42)

for all $(u,x)\in E$ and $m\geq 0$ when $X$ is of infinite variation and for
$u,x>0$ and $m=0$ when $X$ is of finite variation. Then equations (37) and
(38) follow from the compensation formula and the fact that $V(u,b(u))=0$ for
all $u<u_{b}$. From the fact that $\tau_{0}^{-}>0$ a.s. when $X$ is of finite
variation we have that, taking $(u,x)\downarrow(0,0)$, (37) also holds when
$u=x=0$. Moreover, when $X$ is of infinite variation we have that tending
$m\rightarrow\infty$ and taking $u=x=0$ in (4.1),

$\displaystyle V(0,0)$
$\displaystyle=\mathbb{E}\left(\int_{0}^{\infty}G(U_{s},X_{s})\mathbb{I}_{\\{X_{s}<b(U_{s})\\}}\textup{d}s\right)-\mathbb{E}\left(\int_{0}^{\infty}\int_{(-\infty,0)}\widetilde{V}(U_{s},X_{s}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>b(U_{s})\\}}\textup{d}s\right)$
$\displaystyle=\lim_{\delta\downarrow
0}\left\\{\mathbb{E}\left(\int_{0}^{\infty}K_{1}(U_{s}+\delta,X_{s})\textup{d}s\right)-\mathbb{E}\left(\int_{0}^{\infty}K_{2}(U_{s}+\delta,X_{s})\textup{d}s\right)\right\\},$

where $K_{1}(u,x):=G(u,x)\mathbb{I}_{\\{x<b(u)\\}}$ and
$K_{2}(u,x):=\int_{(-\infty,0)}\widetilde{V}(u,x+y)\Pi(\textup{d}y)\mathbb{I}_{\\{x>b(u)\\}}$
for all $(u,x)\in E$. Note since $b$ is non-increasing we have that $K_{2}$ is
non-decreasing in each argument and for each $\delta>0$ and $s>0$,
$|K_{1}(U_{s}+\delta,X_{s})|\leq(s+\delta)^{p-1}\mathbb{I}_{\\{X_{s}<b(\delta)\\}}+\mathbb{E}_{X_{s}}(g^{p-1})$
and

$\displaystyle\mathbb{E}$
$\displaystyle\left(\int_{0}^{\infty}[(s+\delta)^{p-1}\mathbb{I}_{\\{X_{s}<b(\delta)\\}}+\mathbb{E}_{X_{s}}(g^{p-1})]\textup{d}s\right)$
$\displaystyle=\mathbb{E}\left(\int_{0}^{g^{(b(\delta))}}(s+\delta)^{p-1}\mathbb{I}_{\\{X_{s}<b(\delta)\\}}\textup{d}s\right)+\mathbb{E}\left(\int_{0}^{\infty}\mathbb{E}_{X_{s}}(g^{p-1})]\textup{d}s\right)$
$\displaystyle\leq\mathbb{E}((g^{(b(\delta))}+\delta)^{p})+\mathbb{E}\left(\int_{0}^{\infty}\mathbb{E}_{X_{s}}(g^{p-1})]\textup{d}s\right)$
$\displaystyle<\infty,$

where the last inequality follows from Lemmas 2.1 and 4.1. Hence applying (2)
to the functions $K_{1}$ and $K_{2}$ we get that

$\displaystyle V(0,0)=\lim_{\delta\downarrow 0}\lim_{\varepsilon\downarrow
0}\frac{\mathbb{E}_{\varepsilon}\left(\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}}K^{-}(X_{\tau_{0}^{-}}-\varepsilon)\right)+K^{+}(\delta,\varepsilon)}{\psi^{\prime}(0+)W(\varepsilon)},$

where for all $x\leq 0$,

$\displaystyle
K^{-}(x)=\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{+}}[K_{1}(\delta,X_{r})-K_{2}(\delta,X_{r})]\textup{d}r\right)=\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{+}}G(\delta,X_{s})\textup{d}s\right)=V(0,x)-V(0,0),$

and for all $u,x>0$,

$\displaystyle
K^{+}(u,x)=\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{-}}[K_{1}(u+s,X_{r})-K_{2}(u+s,X_{r})]\textup{d}r\right).$

Using the fact that $b$ is nonnegative and $W(x)=0$ for all $x<0$ (and then
$G(\delta,x)=G(0,x)$ for all $x<0$) we have that for all $x<0$

$\displaystyle
K^{-}(x)=\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{+}}G(\delta,X_{s})\textup{d}s\right)=V(0,x)-V(0,0),$

where the last equality follows from the expression of $V$ in terms of the
stopping time $\tau_{D}$. Moreover for all $u>0$ and $x>0$ we have that from
equation (4.1) (taking $m=0$),

$\displaystyle V_{2}(u,x)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{-}}G(u+s,X_{s})\mathbb{I}_{\\{X_{s}<b(u+s)\\}}\textup{d}s\right)-\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{-}}\int_{(-\infty,0)}\widetilde{V}(u+s,X_{s}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>b(u+s)\\}}\textup{d}s\right)$
$\displaystyle=V(u,x)-\mathbb{E}_{x}(V(0,X_{\tau_{0}^{-}})\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}}).$

Hence, rearranging the terms and by dominated convergence theorem we have that

$\displaystyle V(0,0)$ $\displaystyle=\lim_{\delta\downarrow
0}\lim_{\varepsilon\downarrow
0}\frac{\mathbb{E}(\mathbb{I}_{\\{\tau_{-\varepsilon}^{-}<\infty\\}}V(0,X_{\tau_{-\varepsilon}^{-}}))-\mathbb{E}(V(0,X_{\tau_{-\varepsilon}^{-}}+\varepsilon)\mathbb{I}_{\\{\tau_{-\varepsilon}^{-}<\infty\\}})+V(\delta,\varepsilon)-V(0,0)\mathbb{P}_{\varepsilon}(\tau_{0}^{-}<\infty)}{\psi^{\prime}(0+)W(\varepsilon)}$
$\displaystyle=\frac{\sigma^{2}}{2\psi^{\prime}(0+)}\left[-\frac{\partial}{\partial
x}V_{-}(0,0)+\frac{\partial}{\partial x}V_{+}(0,0)\right]+V(0,0),$

where in the last equality we used that
$\mathbb{P}_{\varepsilon}(\tau_{0}^{-}<\infty)=1-\psi^{\prime}(0+)W(\varepsilon)$
(see equation (4)) and the fact that $W^{\prime}(0)=2/\sigma^{2}$. Therefore
we conclude that (40) holds. The lower bound for $V(0,0)$ follows from Lemma
3.1. ∎

Now we show that $(V,b,V(0,0),u_{b})$ is the only quadruplet that satisfy
equations (37)-(40). Suppose that there exist continuous functions $H$ and $c$
on $E$ and $\mathbb{R}_{+}$, respectively, and real numbers $H_{0}<0$ and
$u_{H}>0$ such that the conclusions of the theorem hold. Following a similar
proof than du Toit and Peskir, (2008) we are going to show that $c=b$ which
will imply that $H=V$, $H_{0}=V(0,0)$ and $u_{H}=u_{b}$.

Specifically, let $H$ a non-positive continuous real valued function on $E$, a
non-increasing continuous real valued function $c$ on $(0,\infty)$ such that
$c\geq h\geq 0$, a strictly negative value $H_{0}$ and a non-negative value
$u_{H}$ such that $u_{H}=\infty$ when $X$ is of infinite variation or $X$ is
of finite variation with infinite activity. Otherwise, let $u_{H}$ be the
solution of

$\displaystyle
G(u,0)-\int_{(-\infty,0)}\int_{0}^{-y}\int_{[0,\infty]}\mathbb{E}_{-u-z}(g^{p-1})W(\textup{d}u)\textup{d}z\Pi(\textup{d}y)+H_{0}\Pi(-\infty,0)=0.$
(43)

The function $c$ is such that $c(u)>0$ for all $u<u_{H}$ and $c(u)=0$ for all
$u\geq u_{H}$ and the value of $H$ on the set $\\{(0,x):x\leq 0\\}$ is given
by

$\displaystyle
H(0,x)=-\int_{0}^{-x}\int_{[0,\infty]}\mathbb{E}_{-u-z}(g^{p-1})W(\textup{d}u)\textup{d}z+H_{0}.$
(44)

In addition suppose that $H$ and $c$ are solutions to the equations

$\displaystyle H(u,x)$
$\displaystyle=\mathbb{E}_{x}(H(0,X_{\tau_{0}^{-}})\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})+\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{-}}G(u+s,X_{s})\mathbb{I}_{\\{X_{s}<c(u+s)\\}}\textup{d}s\right)$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{-}}\int_{(-\infty,0)}\widetilde{H}(u+s,X_{s}+y)\mathbb{I}_{\\{X_{s}+y<c(u+s)\\}}\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>c(u+s)\\}}\textup{d}s\right)$
(45)

for $u>0$ and $x>0$, where $\widetilde{H}$ is the extension of $H$ to the set
$\mathbb{R}_{+}\times\mathbb{R}$ as in (16), and

$\displaystyle\mathbb{E}_{c(u)}$
$\displaystyle(H(0,X_{\tau_{0}^{-}})\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})+\mathbb{E}_{c(u)}\left(\int_{0}^{\tau_{0}^{-}}G(u+s,X_{s})\mathbb{I}_{\\{X_{s}<c(u+s)\\}}\textup{d}s\right)$
$\displaystyle=\mathbb{E}_{c(u)}\left(\int_{0}^{\tau_{0}^{-}}\int_{(-\infty,0)}\widetilde{H}(u+s,X_{s}+y)\mathbb{I}_{\\{X_{s}+y<c(u+s)\\}}\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>c(u+s)\\}}\textup{d}s\right),$
(46)

for $u<u_{H}$. Assume that when $X$ is of finite variation (4.1) also holds
for $u=x=0$ and if $X$ is of infinite variation then $H$ satisfies

$\displaystyle\frac{\partial}{\partial x}H_{+}(0,0)=\frac{\partial}{\partial
x}H_{-}(0,0),$ (47)

where $\frac{\partial}{\partial x}H_{+}(0,0)$ and $\frac{\partial}{\partial
x}H_{-}(0,0)$ are the right and left derivative of $x\mapsto\lim_{u\downarrow
0}H(u,x)$ and $x\mapsto H(0,x)$ at zero, respectively.

First, we show that $H$ has an alternative representation.

######  Lemma 4.16.

For all $(u,x)\in E$ we have that

$\displaystyle H(u,x)$
$\displaystyle=\mathbb{E}_{u,x}\left(\int_{0}^{\infty}G(U_{s},X_{s})\mathbb{I}_{\\{X_{s}<c(U_{s})\\}}\textup{d}s\right)$
$\displaystyle\qquad-\mathbb{E}_{u,x}\left(\int_{0}^{\infty}\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\mathbb{I}_{\\{X_{s}+y<c(U_{s})\\}}\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>c(U_{s})\\}}\textup{d}s\right)$
(48)

###### Proof.

First notice that, in a similar way than Lemma 4.15, from (2) and (47) we have
that

$\displaystyle H(0,0)$
$\displaystyle=\mathbb{E}\left(\int_{0}^{\infty}G(U_{s},X_{s})\mathbb{I}_{\\{X_{s}<c(U_{s})\\}}\textup{d}s\right)$
$\displaystyle\qquad-\mathbb{E}\left(\int_{0}^{\infty}\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\mathbb{I}_{\\{X_{s}+y<c(U_{s})\\}}\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>c(U_{s})\\}}\textup{d}s\right).$
(49)

Next, define for all $(u,x)\in E$ the function

$\displaystyle K(u,x)$
$\displaystyle=\mathbb{E}_{u,x}\left(\int_{0}^{\infty}G(U_{s},X_{s})\mathbb{I}_{\\{X_{s}<c(U_{s})\\}}\textup{d}s\right)$
$\displaystyle\qquad-\mathbb{E}_{u,x}\left(\int_{0}^{\infty}\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\mathbb{I}_{\\{X_{s}+y<c(U_{s})\\}}\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>c(U_{s})\\}}\textup{d}s\right).$

Take $u>0$ and $x>0$, then by the strong Markov property and equation (4.1),

$\displaystyle K(u,x)$
$\displaystyle=\mathbb{E}_{x}(K(0,X_{\tau_{0}^{-}})\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})+\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{-}}G(u+s,X_{s})\mathbb{I}_{\\{X_{s}<c(u+s)\\}}\textup{d}s\right)$
$\displaystyle\qquad-\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{-}}\int_{(-\infty,0)}\widetilde{H}(u+s,X_{s}+y)\mathbb{I}_{\\{X_{s}+y<c(u+s)\\}}\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>c(u+s)\\}}\textup{d}s\right)$
$\displaystyle=\mathbb{E}_{x}(K(0,X_{\tau_{0}^{-}})\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})+H(u,x)-\mathbb{E}_{x}(H(0,X_{\tau_{0}^{-}})\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}}).$
(50)

Note that when $X$ is of finite variation the equality above also holds for
$u=0$ and $x=0$. On the other hand, for $u=0$ and $x<0$ we have that again by
the Markov property, the fact that $X$ creeps upwards, $c$ is a nonnegative
curve and the definition of $H(0,x)$ for $x<0$ (see (44)) that

$\displaystyle K(0,x)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\tau_{0}^{+}}G(U_{s},X_{s})\textup{d}s\right)+K(0,0)=H(0,x)-H_{0}+K(0,0).$
(51)

Then when $X$ is of infinite variation we have that $H(0,x)=K(0,x)$ for all
$x<0$ (since by (49) we have that $K(0,0)=H_{0}$ in the infinite variation
case). Hence substituting (51) in (4.1), we get that for any $(u,x)$ such that
$u>0$ and $x>0$

$\displaystyle
K(u,x)-H(u,x)=[K(0,0)-H_{0}]\mathbb{P}_{x}(\tau_{0}^{-}<\infty).$

Hence we conclude that when $X$ is of infinite variation $K(u,x)=H(u,x)$ for
all $u>0$ and $x>0$. Note that when $X$ is of finite variation the equation
above also holds when $x=0$ and $u=0$ and hence we conclude that
$K(0,0)=H_{0}$ (since $\mathbb{P}_{x}(\tau_{0}^{-}<\infty)>0$). The conclusion
of the Lemma holds from equations (4.1) and (51). ∎

Define the set $D_{c}=\\{(u,x)\in E:x\geq c(u)\\}$. We show in the following
lemma that $H$ vanishes in $D_{c}$ so that $D_{c}$ corresponds to the
“stopping set” of $H$.

######  Lemma 4.17.

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

###### Proof.

Note that from equations (4.1) and (4.1) we know that $H(u,c(u))=0$ for all
$u\in(0,u_{H})$. Let $(u,x)\in D_{c}$ such that $x>c(u)$ and define
$\sigma_{c}$ as the first time that $(U,X)$ exits $D_{c}$, i.e.

$\displaystyle\sigma_{c}=\inf\\{s\geq 0:X_{s}<c(U_{s})\\}.$

From the fact that $X_{r}\geq c(U_{r})$ for all $r<\sigma_{c}$ we have that
from the Markov property and representation (48) of $H$,

$\displaystyle H(u,x)$
$\displaystyle=\mathbb{E}_{u,x}(H(U_{\sigma_{c}},X_{\sigma_{c}})\mathbb{I}_{\\{\sigma_{c}<\infty\\}})+\mathbb{E}_{u,x}\left(\int_{0}^{\sigma_{c}}G(U_{s},X_{s})\mathbb{I}_{\\{X_{s}<c(U_{s})\\}}\textup{d}s\right)$
$\displaystyle\qquad-\mathbb{E}_{u,x}\left(\int_{0}^{\sigma_{c}}\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\mathbb{I}_{\\{X_{s}+y<c(U_{s})\\}}\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>c(U_{s})\\}}\textup{d}s\right)$
$\displaystyle=\mathbb{E}_{u,x}(H(U_{\sigma_{c}},X_{\sigma_{c}})\mathbb{I}_{\\{\sigma_{c}<\infty\\}}\mathbb{I}_{\\{X_{\sigma_{c}}<c(U_{s})\\}})-\mathbb{E}_{u,x}\left(\int_{0}^{\sigma_{c}}\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\mathbb{I}_{\\{X_{s}+y<c(U_{s})\\}}\Pi(\textup{d}y)\textup{d}s\right),$

where the last equality follows from the fact that
$\mathbb{P}_{x}(X_{\sigma_{c}}=c(u+\sigma_{c}))>0$ only when $\sigma>0$ and
then $U(u,c(u))=0$ for all $u>0$ (since $u_{H}=\infty$). Then, applying the
compensation formula for Poisson random measures and the fact that
$\sigma_{c}\leq\tau_{0}$ (since $c(u)\geq 0$ for all $u>0$) we get

$\displaystyle\mathbb{E}_{u,x}$
$\displaystyle(H(U_{\sigma_{c}},X_{\sigma_{c}})\mathbb{I}_{\\{\sigma_{c}<\infty\\}}\mathbb{I}_{\\{X_{\sigma_{c}}<c(U_{s})\\}})$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\infty}\int_{(-\infty,0)}\mathbb{I}_{\\{X_{r}\geq
c(u+s)\text{ for all
}r<s\\}}\mathbb{I}_{\\{X_{s-}+y<c(u+s)\\}}\widetilde{H}(u+s,X_{s-}+y)N(\textup{d}s,\textup{d}y)\right)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\infty}\int_{(-\infty,0)}\mathbb{I}_{\\{X_{r}\geq
c(u+s)\text{ for all
}r<s\\}}\mathbb{I}_{\\{X_{s-}+y<c(u+s)\\}}\widetilde{H}(u+s,X_{s-}+y)\Pi(\textup{d}y)\textup{d}s\right)$
$\displaystyle=\mathbb{E}_{u,x}\left(\int_{0}^{\sigma_{c}}\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\mathbb{I}_{\\{X_{s}<c(U_{s})\\}}\Pi(\textup{d}y)\textup{d}s\right).$

Hence we have that $H(u,x)=0$ for all $(u,x)\in D_{c}$ as claimed. ∎

The following Lemma states that $H$ dominates the value function $V$. That
suggest that $H$ is the largest function with $H\leq 0$ that makes the process
$\\{H(U_{t},X_{t})+\int_{0}^{t}G(U_{s},X_{s})\textup{d}s,t\geq 0\\}$ a
$\mathbb{P}_{u,x}$-submartingale. The latter assertion will be shown
indirectly on the upcoming lemmas.

######  Lemma 4.18.

We have that $H(u,x)\geq V(u,x)$ for all $(u,x)\in E$.

###### Proof.

If $(u,x)\in D_{c}$ we have the inequality

$\displaystyle H(u,x)=0\geq V(u,x).$

Now we show that the inequality also holds in $E\setminus D_{c}$. Consider the
stopping time

$\displaystyle\tau_{c}=\inf\\{s\geq 0:X_{s}\geq c(U_{s})\\}.$

Then using the Markov property and equation (48) we get that for all $(u,x)\in
E$ with $x<c(u)$ (here we take $c(0):=\lim_{u\downarrow
0}c(u)\geq\lim_{u\downarrow 0}h(u)=\infty$),

$\displaystyle H(u,x)$
$\displaystyle=\mathbb{E}_{u,x}\left(H(U_{\tau_{c}},X_{\tau_{c}})\right)+\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{c}}G(U_{s},X_{s})\mathbb{I}_{\\{X_{s}<c(U_{s})\\}}\textup{d}s\right)$
$\displaystyle\qquad-\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{c}}\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>c(U_{s})\\}}\textup{d}s\right)$
$\displaystyle=\mathbb{E}_{u,x}\left(H(U_{\tau_{c}},c(U_{\tau_{c}}))\right)+\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{c}}G(U_{s},X_{s})\textup{d}s\right),$

where in the second equality we used the fact $X$ creeps upwards and
$\tau_{c}<\infty$. Note that since $X_{t}>0$ if and only if $U_{t}>0$ for all
$t>0$ and that $c(u)>0$ for all $u$ sufficiently small we have that
$c(U_{\tau_{c}})>0$ and hence $H(U_{\tau_{c}},c(U_{\tau_{c}}))=0$. Therefore

$\displaystyle
H(u,x)=\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{c}}G(U_{s},X_{s})\textup{d}s\right)\geq
V(u,x),$

where the inequality follows from the definition of $V$ as per (26). ∎

It turns out that the fact that $H$ dominates $V$ implies that $b$ dominates
the curve $c$. This fact is shown in the following Lemma.

######  Lemma 4.19.

We have that $b(u)\geq c(u)$ for all $u>0$.

###### Proof.

Note that in the case that $X$ is of finite variation with
$\Pi(-\infty,0)<\infty$ we have that $c(u)=0\leq b(u)$ for all $u>u_{H}$. We
proceed by contradiction. Suppose that there exists $u_{0}>0$ such that
$b(u_{0})<c(u_{0})$. Then in the case that $X$ is of finite variation with
$\Pi(-\infty,0)<\infty$, it holds that $u_{0}<u_{H}$. Take $x>c(u_{0})$ and
consider the stopping time

$\displaystyle\sigma_{b}$ $\displaystyle=\inf\\{s>0:X_{s}<b(U_{s})\\}.$

Then from the Markov property and the representation of $H$ given in (48) we
have that

$\displaystyle H(u_{0},x)$
$\displaystyle=\mathbb{E}_{u_{0},x}\left(H(U_{\sigma_{b}^{-}},X_{\sigma_{b}^{-}})\right)+\mathbb{E}_{u_{0},x}\left(\int_{0}^{\sigma_{b}^{-}}G(U_{s},X_{s})\mathbb{I}_{\\{X_{s}<c(U_{s})\\}}\textup{d}s\right)$
$\displaystyle\qquad-\mathbb{E}_{u_{0},x}\left(\int_{0}^{\sigma_{b}^{-}}\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>c(U_{s})\\}}\textup{d}s\right).$

Moreover, since $V(u,b(u))=0$ for $u>0$ such that $b(u)>0$ and $0\geq H\geq V$
we have that

$\displaystyle\mathbb{E}_{u_{0},x}\left(H(U_{\sigma_{b}^{-}},X_{\sigma_{b}^{-}})\right)$
$\displaystyle=\mathbb{E}_{u_{0},x}\left(H(U_{\sigma_{b}^{-}},X_{\sigma_{b}^{-}})\mathbb{I}_{\\{X_{\sigma_{b}^{-}}<b(U_{\sigma_{b}^{-}})\\}}\right)$
$\displaystyle=\mathbb{E}_{u_{0},x}\left(\int_{0}^{\sigma_{b}^{-}}\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\mathbb{I}_{\\{X_{s}+y\leq
b(U_{s})\\}}\Pi(\textup{d}y)\textup{d}s\right)$
$\displaystyle\geq\mathbb{E}_{u_{0},x}\left(\int_{0}^{\sigma_{b}^{-}}\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\Pi(\textup{d}y)\textup{d}s\right),$

where the second equality follows from the compensation formula for Poisson
random measures and the inequality since $H\leq 0$ by assumption. Hence,
combining the two inequalities above and from the fact that $x>c(u_{0})$ and
then $H(u_{0},x)=0$ we get

$\displaystyle
0\geq\mathbb{E}_{u_{0},x}\left(\int_{0}^{\sigma_{b}^{-}}\left[G(U_{s},X_{s})+\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\Pi(\textup{d}y)\right]\mathbb{I}_{\\{X_{s}<c(U_{s})\\}}\textup{d}s\right).$

Due to the continuity of $b$ and $c$ we have that there exists a value $u_{1}$
sufficiently small such that $c(v)>b(v)$ for all $v\in[u_{0},u_{1})$. Thus,
from Lemma 4.9, the fact that $u\mapsto G(u,x)$ is strictly increasing when
$x>0$ and the inequality $U\geq V$ (see Lemma 4.18) we have that for all $u>0$
and $x>b(u)$,

$\displaystyle
G(u,x)+\int_{(-\infty,0)}\widetilde{H}(u,x+y)\Pi(\textup{d}y)\geq
G(u,x)+\int_{(-\infty,0)}\widetilde{V}(u,x+y)\Pi(\textup{d}y)>0,$

where the strict inequality follows from Lemma 4.9. Note that taking $x$
sufficiently big we have that, under the measure $\mathbb{P}_{u_{0},x}$, $X$
spends a positive amount of time between the curves $b(u)$ and $c(u)$ for
$u\in[u_{0},u_{1}]$ with positive probability. Thus, since
$\sigma_{c}<\tau_{0}^{-}$ the above facts imply that

$\displaystyle
0\geq\mathbb{E}_{u_{0},x}\left(\int_{0}^{\sigma_{b}^{-}}\left[G(U_{s},X_{s})+\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\Pi(\textup{d}y)\right]\mathbb{I}_{\\{X_{s}<c(U_{s})\\}}\textup{d}s\right)>0,$

which is a contradiction and then we have that $c(u)\leq b(u)$ for all $u>0$.
∎

The next lemma shows that the function $\widetilde{H}$ is non-decreasing in
the second coordinate.

######  Lemma 4.20.

For all $u\geq 0$ we have that $\widetilde{H}(u,x)\leq\widetilde{H}(u,y)$ for
all $x\leq y$.

###### Proof.

From the definition of $H(0,x)$ when $x<0$ and Lemma 4.17 it is clear that the
assertion is true on the set $(-\infty,0]\cup[c(u),\infty)$ for each $u\geq
0$. Now, take $u>0$ and $0<x\leq y<c(u)$. Define, for all $z\in\mathbb{R}$,

$\displaystyle\tau_{c}(z)=\inf\\{t>0:X_{t}+z\geq c(U_{t}^{(-z)})\\},$

where $U_{t}^{(-z)}=t-\sup\\{0\leq s\leq t:X_{s}\leq-z\\}$. Then since $c$ is
non-increasing we have that $\tau_{c}(y)\leq\tau_{c}(x)$. Thus, from Lemma
4.16 we have that

$\displaystyle H(u,y)-H(u,x)$
$\displaystyle=\mathbb{E}_{u,0}\left(\int_{0}^{\tau_{c}(y)}G(U_{s}^{(-y)},X_{s}+y)\textup{d}s\right)-\mathbb{E}_{u,0}\left(\int_{0}^{\tau_{c}(x)}G(U_{s}^{(-x)},X_{s}+x)\textup{d}s\right)$
$\displaystyle=\mathbb{E}_{u,0}\left(\int_{0}^{\tau_{c}(y)}[G(U_{s}^{(-y)},X_{s}+y)-G(U_{s}^{(-x)},X_{s}+x)]\textup{d}s\right)$
$\displaystyle\qquad-\mathbb{E}_{u,0}\left(\int_{\tau_{c}(y)}^{\tau_{c}(x)}G(U_{s}^{(-x)},X_{s}+x)\textup{d}s\right)$
$\displaystyle=\mathbb{E}_{u,0}\left(\int_{0}^{\tau_{c}(y)}[G(U_{s}^{(-y)},X_{s}+y)-G(U_{s}^{(-x)},X_{s}+x)]\textup{d}s\right)$
$\displaystyle\qquad-\mathbb{E}_{u,x}\left(H(U_{\tau_{c}(y-x)},X_{\tau_{c}(y-x)})\right)$
$\displaystyle\geq 0,$

where the third equality follows from the Markov property and the last
inequality follows since $G$ is non-decreasing in each argument and $H\leq 0$.
Lastly, by continuity of $H$ we have that for all $u>0$ and $0<x<c(u)$,
$H(u,c(u))\geq H(u,x)\geq H(0,0)$ also holds. The proof is now complete. ∎

From Lemma 4.9 we know that
$G(u,x)+\int_{(-\infty,0)}\widetilde{V}(u,x+y)\Pi(\textup{d}y)>0$ for all
$x>b(u)$. It turns out that an analogous inequality follows for $H$ and $c$.
Moreover, it can be shown that such inequality guarantees that the process
$\\{H(U_{t},X_{t})+\int_{0}^{t}G(U_{s},X_{s})\textup{d}s,t\geq 0\\}$ is a
$\mathbb{P}_{u,x}$-submartingale for all $(u,x)\in E$.

######  Lemma 4.21.

We have that for all $(u,x)\in E$ such that $x>c(u)$

$\displaystyle
G(u,x)+\int_{(-\infty,0)}\widetilde{H}(u,x+y)\Pi(\textup{d}y)>0.$

###### Proof.

First we show that there exists $N>0$ such that for all $u>N$,

$\displaystyle
G(u,x)+\int_{(-\infty,0)}\widetilde{H}(u,x+y)\Pi(\textup{d}y)>0\qquad\text{for
all }x>c(u).$

If $X$ is such that $X$ is of finite variation with $\Pi(-\infty,0)<\infty$ we
already know from the definition of $u_{H}$ (see (43)) and since $x\mapsto
H(0,x)$ is non-decreasing that the assertion holds. Now suppose that $X$ is of
infinite variation or $X$ is of finite variation with $\Pi(-\infty,0)=\infty$
(then $c(u)>0$ for all $u>0$). By continuity of $H$ and $c$ and the fact that
$0\leq\lim_{u\rightarrow\infty}c(u)\leq\lim_{u\rightarrow\infty}b(u)=0$ we
have that for any $x>0$,

$\displaystyle\lim_{u\rightarrow\infty}H(u,x)=\lim_{u\rightarrow\infty}V(u,x)=0.$

From the dominated convergence theorem (see Corollary 4.6 and Lemma 2.1) we
deduce that

$\displaystyle\liminf_{u\rightarrow\infty}$
$\displaystyle\left[G(u,c(u)+1/u)+\int_{(-\infty,0)}\widetilde{H}(u,c(u)+1/u+y)\Pi(\textup{d}y)\right]$
$\displaystyle=\liminf_{u\rightarrow\infty}\left[G(u,b(u)+1/u)+\int_{(-\infty,0)}\widetilde{V}(u,b(u)+1/u+y)\Pi(\textup{d}y)\right]$
$\displaystyle\in[0,\infty],$

where the inequality follows from Lemma 4.9. This implies by monotonicity of
$G$ and Lemma 4.20 that there exists some $N_{1}>0$ such that for all
$0<x<1/N_{1}$ and $u>N_{1}$,

$\displaystyle
G(u,c(u)+x)+\int_{(-\infty,0)}\widetilde{H}(u,c(u)+x+y)\Pi(\textup{d}y)\geq
0.$

Moreover, since $\lim_{u\rightarrow\infty}G(u,x)=\infty$ for all $x>0$, we
have that

$\displaystyle\lim_{u\rightarrow\infty}\left[G(u,1/N_{1})+\int_{(-\infty,0)}\widetilde{H}(u,1/N_{1}+y)\Pi(\textup{d}y)\right]=\infty.$

Hence, since for all $u\geq 0$, $x\mapsto\widetilde{H}(u,x)$ is non-decreasing
we have that there exists $N_{2}$ (sufficiently large) such that for $u>N_{2}$
and all $x>1/N_{1}>c(u)$,

$\displaystyle
G(u,x)+\int_{(-\infty,0)}\widetilde{H}(u,x+y)\Pi(\textup{d}y)>0.$

Then we define $N:=\max\\{N_{1},N_{2}\\}$ and the assertion holds. Now we
proceed to prove the main claim by contradiction. Assume that there exists $u$
such that

$\displaystyle
G(u,x)+\int_{(-\infty,0)}\widetilde{H}(u,x+y)\Pi(\textup{d}y)\leq 0\text{ for
}x\in(c(u),c(u)+\varepsilon)$

for some $\varepsilon>0$. Define

$\displaystyle u^{*}=\sup\\{u>0:\text{there exists }\varepsilon>0\text{ such
that }G(u,x)+\int_{(-\infty,0)}\widetilde{H}(u,x+y)\Pi(dy)\leq 0\text{ for
}x\in(c(u),c(u)+\varepsilon)\\}$

Hence, $u^{*}<N$ and there exists $\varepsilon^{*}>0$ such that

$\displaystyle
G(u^{*},x)+\int_{(-\infty,0)}\widetilde{H}(u^{*},x+y)\Pi(\textup{d}y)\leq
0\qquad\text{for }x\in(c(u^{*}),c(u^{*})+\varepsilon^{*}).$

On the other hand, we have that for any $h>0$

$\displaystyle
G(u^{*}+h,x)+\int_{(-\infty,0)}\widetilde{H}(u^{*}+h,x+y)\Pi(\textup{d}y)>0\qquad\text{for
}x>c(u^{*}+h).$

In particular, the strict inequality holds for any $x>c(u^{*})$. Hence taking
$h\downarrow 0$ we conclude that

$\displaystyle
G(u^{*},x)+\int_{(-\infty,0)}\widetilde{H}(u^{*},x+y)\Pi(dy)=0\qquad\text{for
}x\in(c(u^{*}),c(u^{*})+\varepsilon^{*}).$

But the equality above contradicts the fact that $x\mapsto G(u^{*},x)$ is
strictly increasing on $(0,\infty)$ and $x\mapsto H(u^{*},x)$ is non-
decreasing (see Lemma 4.20). Therefore we conclude that for all $u>0$,

$\displaystyle
G(u,x)+\int_{(-\infty,0)}\widetilde{H}(u,x+y)\Pi(dy)>0\qquad\text{for all
}x>c(u)$

and the conclusion of the lemma follows. ∎

We finish the proof showing that indeed $c$ corresponds to $b$.

######  Lemma 4.22.

We have that then $c(u)=b(u)$ for all $u>0$ and $V(u,x)=H(u,x)$ for all
$(u,x)\in E$.

###### Proof.

Suppose that there exists $u>0$ such that $c(u)<b(u)$ and take
$x\in(c(u),b(u))$. Then we have by the Markov property and representation (48)
that

$\displaystyle H(u,x)$
$\displaystyle=\mathbb{E}_{u,x}\left(H(U_{\tau_{D}},X_{\tau_{D}})\right)+\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{D}}G(U_{s},X_{s})\mathbb{I}_{\\{X_{s}<c(U_{s})\\}}\textup{d}s\right)$
$\displaystyle\qquad-\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{D}}\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\Pi(\textup{d}y)\mathbb{I}_{\\{X_{s}>c(U_{s})\\}}\textup{d}s\right),$

where $\tau_{D}=\inf\\{t>0:X_{t}\geq b(U_{t})\\}$. On the other hand, we have
that

$\displaystyle
V(u,x)=\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{D}}G(U_{s},X_{s})\textup{d}s\right).$

Hence, since $X_{\tau_{D}}=b(U_{\tau_{D}})\geq c(U_{\tau_{D}})$ and Lemma 4.17
we have that $H(U_{\tau_{D}},X_{\tau_{D}})=0$. Moreover, using the inequality
$H\geq V$ (see Lemma 4.18) we obtain that

$\displaystyle
0\geq\mathbb{E}_{u,x}\left(\int_{0}^{\tau_{D}}\left[G(U_{s},X_{s})+\int_{(-\infty,0)}\widetilde{H}(U_{s},X_{s}+y)\Pi(\textup{d}y)\right]\mathbb{I}_{\\{X_{s}>c(U_{s})\\}}\textup{d}s\right)>0.$

where the strict inequality follows from Lemma 4.21 and continuity of $b$ and
$c$. This contradiction allows us to conclude that $c(u)=b(u)$ for all $u>0$
and $H(u,x)=V(u,x)$ for all $(u,x)\in E$. ∎

######  Remark 4.23.

A close inspection of the proof tells us that the assumptions that $H\leq 0$
and that $c$ is non-increasing can be dropped when $\Pi\equiv 0$.

##  5 Examples

###  5.1 Brownian Motion with drift example

Suppose that $X_{t}$ is given by

$\displaystyle X_{t}=\mu t+\sigma B_{t},$

where $\mu>0$, $\sigma>0$ and $B=\\{B_{t},t\geq 0\\}$ is a standard Brownian
motion. Here, we consider the case $p=2$. Then

$\displaystyle G(u,x)=u\psi^{\prime}(0+)W(x)-\mathbb{E}_{x}(g).$

It is well known that for $\beta\geq 0$ and $q\geq 0$,

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

Thus, $\psi^{\prime}(0+)=\mu$, $\Phi^{\prime}(0)=\frac{1}{\mu}$ and
$\Phi^{\prime\prime}(0)=-\frac{\sigma^{2}}{\mu^{3}}$. The scale function is
given by

$\displaystyle W(x)=\frac{1}{\mu}(1-\exp(-2\mu/\sigma^{2}x)),\qquad x\geq 0.$

An easy calculation shows that $W^{*(2)}$ is given by

$\displaystyle
W^{*(2)}(x)=\frac{1}{\mu^{2}}x[1+\exp(-2\mu/\sigma^{2}x)]-\frac{\sigma^{2}}{\mu^{2}}\frac{1}{\mu}(1-\exp(-2\mu/\sigma^{2}x)),\qquad
x\geq 0.$

For all $x\in\mathbb{R}$, the value $\mathbb{E}_{x}(g)$ can be calculated from
(19) via differentiation to have

$\displaystyle\mathbb{E}_{x}(g)$
$\displaystyle=-\psi^{\prime}(0+)[\Phi^{\prime\prime}(0+)+x\Phi^{\prime}(0)^{2}]+\psi^{\prime}(0+)W^{*2}(x)$
$\displaystyle=\left\\{\begin{array}[]{lc}\frac{\sigma^{2}}{\mu^{2}}-\frac{x}{\mu}&x<0\\\
\frac{\sigma^{2}}{\mu^{2}}\exp(-2\mu/\sigma^{2}x)+\frac{x}{\mu}\exp(-2\mu/\sigma^{2}x)&x\geq
0.\end{array}\right.$

Moreover, we know that $X_{r}\sim N(\mu r,\sigma^{2}r)$ and for any $x\geq 0$
we have that

$\displaystyle\mathbb{P}_{x}(X_{r}\in\textup{d}z,\underline{X}_{r}\geq
0)=\left[\frac{1}{\sqrt{\sigma^{2}r}}\phi\left(\frac{z-x-\mu
r}{\sqrt{\sigma^{2}r}}\right)-e^{-\frac{2\mu}{\sigma^{2}}x}\frac{1}{\sqrt{\sigma^{2}r}}\phi\left(\frac{z+x-\mu
r}{\sqrt{\sigma^{2}r}}\right)\right]\textup{d}z,$

where $\phi$ is the density of a standard normal distribution. Then we
calculate for any $u>0$

$\displaystyle\mathbb{E}_{x}$
$\displaystyle\left(\int_{0}^{\tau_{0}^{-}}[(r+u)\psi^{\prime}(0+)W(X_{r})-\mathbb{E}_{X_{r}}(g)]\mathbb{I}_{\\{X_{r}<b(r+u)\\}}\textup{d}r\right)$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{b(r+u)}[(r+u)\psi^{\prime}(0+)W(z)-\mathbb{E}_{z}(g)]\mathbb{P}_{x}(X_{r}\in\textup{d}z,\underline{X}_{r}\geq
0)\textup{d}r$
$\displaystyle=\int_{0}^{\infty}\left\\{H(r,u,x,b(r+u))-e^{-2\mu/\sigma^{2}x}H(r,u,-x,b(r+u))\right\\}\textup{d}r$

where

$\displaystyle H(r,t,x,b)$
$\displaystyle=\int_{0}^{b}[(r+t)\psi^{\prime}(0+)W(z)-\mathbb{E}_{z}(g)]\frac{1}{\sqrt{\sigma^{2}r}}\phi\left(\frac{z-x-\mu
r}{\sqrt{\sigma^{2}r}}\right)dz$
$\displaystyle=(r+t)\left[\Psi\left(\frac{b-x-\mu
r}{\sigma\sqrt{r}}\right)-\Psi\left(\frac{-x-\mu
r}{\sigma\sqrt{r}}\right)\right]$
$\displaystyle\qquad-\left[\frac{x}{\mu}+t+\frac{\sigma^{2}}{\mu^{2}}\right]e^{-2\mu/\sigma^{2}x}\left[\Psi\left(\frac{b-x+\mu
r}{\sigma\sqrt{r}}\right)-\Psi\left(\frac{-x+\mu
r}{\sigma\sqrt{r}}\right)\right]$
$\displaystyle\qquad+\frac{\sigma\sqrt{r}}{\mu}e^{-2\mu/\sigma^{2}x}\left[\phi\left(\frac{b-x+\mu
r}{\sigma\sqrt{r}}\right)-\phi\left(\frac{-x+\mu
r}{\sigma\sqrt{r}}\right)\right].$

From formula (4) we know that

$\displaystyle V(0,x)$
$\displaystyle=-\int_{0}^{-x}\int_{[0,\infty)}\mathbb{E}_{-u-z}(g)W(\textup{d}u)\textup{d}z+V(0,0)$
$\displaystyle=\frac{3\sigma^{2}}{2\mu^{3}}x-\frac{1}{2\mu^{2}}x^{2}+V(0,0).$

Then,

$\displaystyle\frac{\partial}{\partial
x}V_{-}(0,0)=\frac{3\sigma^{2}}{2\mu^{3}}.$

From Theorem 4.14 we have that

$\displaystyle V(u,x)$
$\displaystyle=\left\\{\begin{array}[]{ll}V(0,0)[1-\psi^{\prime}(0+)W(x)]+\int_{0}^{\infty}\left\\{H(r,u,x,b(r+u))-e^{-2\mu/\sigma^{2}x}H(r,u,-x,b(r+u))\right\\}\textup{d}r&x>0\\\
\frac{3\sigma^{2}}{2\mu^{3}}x-\frac{1}{2\mu^{2}}x^{2}+V(0,0)&x\leq
0\end{array}\right..$

The curve $b(u)$ and $V(0,0)$ satisfy the equations

$\displaystyle\int_{0}^{\infty}H(r,u,b(u),b(r+u))\textup{d}r+V(0,0)[1-\psi^{\prime}(0+)W(b(u))]=0,$
$\displaystyle\frac{3\sigma^{2}}{2\mu^{3}}-\frac{\partial}{\partial
x}V_{+}(0,0)=0,$

for all $u>0$. Note that $\frac{\partial}{\partial x}V_{+}(0,0)$ can be
estimated via $[V(0,h)-V(0,0)]/h$ for $h$ sufficiently small. We show in
Figure 2 a numerical calculation of the optimal boundary and the value
function using the equations above. The case considered is when $\mu=1/2$ and
$\sigma=1$.

Figure 2: Numeric calculation of the optimal boundary and value function $V$
for the Brownian motion with drift case.

###  5.2 Brownian motion with exponential jumps example

Consider the case in which $p=2$ and $X$ a Brownian motion with drift and
exponential jumps and the case, this is, $X=\\{X_{t},t\geq 0\\}$ with

$\displaystyle X_{t}=\mu t+\sigma B_{t}-\sum_{i=1}^{N_{t}}Y_{i},\qquad t\geq
0,$

where $\sigma>0$, $\mu>0$, $B=\\{B_{t},t\geq 0\\}$ is a standard Brownian
motion, $N=\\{N_{t},t\geq 0\\}$ is an independent Poisson process with rate
$\lambda>0$ and $\\{Y_{i},i\geq 1\\}$ is a sequence of independent exponential
distributed random variables with parameter $\rho>0$ independent of $B$ and
$N$. We further assume that $\mu\rho>\lambda$ so $X$ drifts to infinity. The
Laplace exponent is given by for $\beta\geq 0$ by

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

where $\mu$ is a positive constant. In this case the Lévy measure is given by
$\Pi(\textup{d}x)=\lambda\rho e^{\rho x}\textup{d}x$ for all $x<0$. An easy
calculation leads to $\psi^{\prime}(0+)=\mu-\lambda/\rho$,

$\displaystyle\Phi^{\prime}(0+)=\frac{\rho}{\mu\rho-\lambda}\qquad\text{and}\qquad\Phi^{\prime\prime}(0+)=-\frac{\sigma^{2}\rho^{3}+2\lambda\rho}{\left[\mu\rho-\lambda\right]^{3}}.$

The scale function is given by

$\displaystyle
W(x)=\frac{1}{\psi^{\prime}(0+)}+\frac{e^{\zeta_{1}x}}{\psi^{\prime}(\zeta_{1}(0))}+\frac{e^{\zeta_{2}x}}{\psi^{\prime}(\zeta_{2}(0))}$

for $x\geq 0$, where

$\displaystyle\zeta_{1}=\frac{-\left(\frac{\sigma^{2}}{2}\rho+\mu\right)+\sqrt{\left(\frac{\sigma^{2}}{2}\rho-\mu\right)^{2}+2\sigma^{2}\lambda}}{\sigma^{2}}\qquad\text{and}\qquad\zeta_{2}=\frac{-\left(\frac{\sigma^{2}}{2}\rho+\mu\right)-\sqrt{\left(\frac{\sigma^{2}}{2}\rho-\mu\right)^{2}+2\sigma^{2}\lambda}}{\sigma^{2}}.$

Then we have that

$\displaystyle\mathbb{E}_{x}(g)$
$\displaystyle=-\psi^{\prime}(0+)[\Phi^{\prime\prime}(0+)+x\Phi^{\prime}(0)^{2}]+\psi^{\prime}(0+)W^{*2}(x)$
$\displaystyle=\left\\{\begin{array}[]{lc}\frac{\sigma^{2}\rho^{2}+2\lambda}{\left[\mu\rho-\lambda\right]^{2}}-\frac{\rho}{\mu\rho-\lambda}x&x<0\\\
\frac{\sigma^{2}\rho^{2}+2\lambda}{\left[\mu\rho-\lambda\right]^{2}}-\frac{\rho}{\mu\rho-\lambda}x+(\mu-\lambda/\rho)W^{*2}(x)&x\geq
0\end{array}\right..$

Therefore, since we have exponential jumps, we have that $V$, $b$ and $V(0,0)$
satisfy the equations

$\displaystyle V(u,x)$
$\displaystyle=V(0,0)\frac{\sigma^{2}}{2}W^{\prime}(x)+\int_{0}^{\infty}F_{1}(b(u+s),s,u,x)\textup{d}s-\int_{0}^{\infty}\mathcal{V}(u+s,b(u+s))F_{2}(b(u+s),s,x)\textup{d}s,$
$\displaystyle 0$
$\displaystyle=V(0,0)\frac{\sigma^{2}}{2}W^{\prime}(b(u))+\int_{0}^{\infty}F_{1}(b(u+s),s,u,b(u))\textup{d}s-\int_{0}^{\infty}\mathcal{V}(u+s,b(u+s))F_{2}(b(u+s),s,b(u))\textup{d}s,$
$\displaystyle 0$
$\displaystyle=\frac{3}{2}\Phi^{\prime\prime}(0)+\frac{\partial}{\partial
x}V_{+}(0,0)$

for all $u,x>0$, where for any $b,s,u>0$ and $x\in\mathbb{R}$

$\displaystyle F_{1}(b,s,u,x)$
$\displaystyle=\mathbb{E}\left(\left[G(u+s,X_{s}+x)+\int_{(-\infty,0)}V(0,X_{s}+x+y)\mathbb{I}_{\\{X_{s}+x+y<0\\}}\Pi(\textup{d}y)\right]\mathbb{I}_{\\{X_{s}+x<b,\underline{X}_{s}+x\geq
0\\}}\right),$ $\displaystyle F_{2}(b,s,x)$ $\displaystyle=e^{-\rho
x}\mathbb{E}\left(e^{-\rho
X_{s}}\mathbb{I}_{\\{X_{s}+x>b,\underline{X}_{s}+x\geq 0\\}}\right),$
$\displaystyle\mathcal{V}(u,b)$ $\displaystyle=\int_{0}^{b}V(u+s,y)\lambda\rho
e^{\rho y}\textup{d}y.$

We show in Figure 3 a numerical calculation of the optimal boundary and the
value function using the parametrisation $\mu=3$, $\sigma=1$, $\lambda=1$ and
$\rho=1$. The functions $F_{1}$ and $F_{2}$ above were estimated using Monte
Carlo simulations accordingly to the algorithm given in Kuznetsov et al.,
(2011).

Figure 3: Numeric calculation of the optimal boundary and value function $V$
for the Brownian motion with exponential jumps case.

##  6 Appendix

###### Proof of Lemma 4.1.

First, notice that due to the spatial homogeneity of Lévy processes and that
$x\mapsto\mathbb{E}_{x}(g^{p-1})$ is non-increasing it suffices to prove the
assertion for $x\leq 0$. Using Fubini’s theorem we have that for all $x\leq
0$,

$\displaystyle\mathbb{E}_{x}\left(\int_{0}^{\infty}\mathbb{E}_{X_{s}}(g^{p-1})\textup{d}s\right)$
$\displaystyle=\int_{(-\infty,\infty)}\mathbb{E}_{z}(g^{p-1})\int_{0}^{\infty}\mathbb{P}_{x}(X_{s}\in\textup{d}z).$

Since $X$ drifts to infinity we can use the density for the $0$-potential
measure of $X$ without killing (see equation (8)) to obtain

$\displaystyle\mathbb{E}_{x}\left(\int_{0}^{\infty}\mathbb{E}_{X_{s}}(g^{p-1})\textup{d}s\right)$
$\displaystyle=\int_{-\infty}^{\infty}\mathbb{E}_{z}(g^{p-1})\left[\frac{1}{\psi^{\prime}(0+)}-W(x-z)\right]\textup{d}z$
$\displaystyle=\frac{1}{\psi^{\prime}(0+)}\int_{-\infty}^{x}\mathbb{E}_{z}(g^{p-1})\left[1-\psi^{\prime}(0+)W(x-z)\right]\textup{d}z+\frac{1}{\psi^{\prime}(0+)}\int_{x}^{\infty}\mathbb{E}_{z}(g^{p-1})\textup{d}z.$
(52)

Now we prove that the above two integrals are finite for all $x\leq 0$. From
the fact that $z\mapsto\mathbb{E}_{z}(g^{p-1})$ is continuous on $\mathbb{R}$
and $W$ is continuous on $(0,\infty)$ we can assume without of loss of
generality that $x=0$.

First, we show that the first integral on the right hand side of (6) is
finite. From Lemma 2.2 we have that

$\displaystyle\int_{0}^{\infty}$
$\displaystyle\mathbb{E}_{-z}(g^{p-1})\left[1-\psi^{\prime}(0+)W(z)\right]\textup{d}z\leq
2^{p-1}\mathbb{E}(-\underline{X}_{\infty})[\mathbb{E}(g^{p-1})+A_{p-1}]+\frac{2^{p-1}}{p}C_{p-1}\mathbb{E}((-\underline{X}_{\infty})^{p}),$

where $A_{p-1}$ and $C_{p-1}$ are non-negative constants. In the equality
above we relied on the fact that $z\mapsto\psi(0+)W(z)$ corresponds to the
distribution function of the random variable $-\underline{X}_{\infty}$. We
conclude from Lemma 2.1 that

$\displaystyle\int_{0}^{\infty}\mathbb{E}_{-z}(g^{p-1})\left[1-\psi^{\prime}(0+)W(z)\right]\textup{d}z<\infty.$

Now we proceed to check the finiteness of the second integral in (6) when
$x=0$. Using the strong Markov property we have that

$\displaystyle\int_{0}^{\infty}\mathbb{E}_{z}(g^{p-1})\textup{d}z$
$\displaystyle=\int_{0}^{\infty}\mathbb{E}_{z}(g^{p-1}\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})\textup{d}z$
$\displaystyle\leq
2^{p-1}\int_{0}^{\infty}\mathbb{E}_{z}((\tau_{0}^{-})^{p-1}\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})dz+2^{p-1}\int_{0}^{\infty}\mathbb{E}_{z}(\mathbb{E}_{X_{\tau_{0}^{-}}}(g^{p-1})\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})\textup{d}z$
$\displaystyle\leq
2^{p-1}\int_{0}^{\infty}\mathbb{E}_{z}((\tau_{0}^{-})^{p-1}\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})\textup{d}z+2^{p-1}\int_{0}^{\infty}\mathbb{E}_{z}(\mathbb{E}_{\underline{X}_{\infty}}(g^{p-1})\mathbb{I}_{\\{\underline{X}_{\infty}<0\\}})\textup{d}z$

where in the last inequality we used the fact that $\underline{X}_{\infty}\leq
X_{\tau_{0}^{-}}$ and that $x\mapsto\mathbb{E}_{x}(g^{p-1})$ is a non-
increasing function. Using Fubini’s theorem we have that

$\displaystyle\int_{0}^{\infty}\mathbb{E}_{z}(\mathbb{E}_{\underline{X}_{\infty}}(g^{p-1})\mathbb{I}_{\\{\underline{X}_{\infty}<0\\}})\textup{d}z$
$\displaystyle=\int_{0}^{\infty}\int_{(-\infty,0)}\mathbb{E}_{y}(g^{p-1})\mathbb{P}_{z}(\underline{X}_{\infty}\in\textup{d}y)\textup{d}z$
$\displaystyle=\int_{(-\infty,0)}\mathbb{E}_{y}(g^{p-1})\int_{0}^{\infty}\mathbb{P}_{z}(\underline{X}_{\infty}\in\textup{d}y)\textup{d}z$
$\displaystyle=\int_{0}^{\infty}\mathbb{E}_{-y}(g^{p-1})[1-\psi^{\prime}(0+)W(y)]\textup{d}y$
$\displaystyle<\infty.$

It thus only remains to show that

$\displaystyle\int_{0}^{\infty}\mathbb{E}_{z}((\tau_{0}^{-})^{p-1}\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})\textup{d}z<\infty.$

For this, define the function
$F_{1}(q):=\int_{0}^{\infty}\mathbb{E}_{z}(e^{-q\tau_{0}^{-}}\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})\textup{d}z$.
Differentiating with respect to $\beta$ the equation (5) and evaluating at
zero we obtain that

$\displaystyle
F_{1}(q)=\int_{0}^{\infty}\mathbb{P}(-\underline{X}_{\textbf{e}_{q}}>z)\textup{d}z=\mathbb{E}(-\underline{X}_{\textbf{e}_{q}})=\frac{1}{\Phi(q)}-\frac{\psi^{\prime}(0+)}{q},$

where $\textbf{e}_{q}$ is an independent exponential random variable with
parameter $q>0$. On the other hand, define the function
$F_{2}(q)=\int_{0}^{\infty}\mathbb{E}_{-z}(e^{-q\tau_{0}^{+}})[1-\psi^{\prime}(0+)W(z)]\textup{d}z$.
Using the expression for the Laplace transform of $\tau_{0}^{+}$ the
definition of $W$, we have that

$\displaystyle
F_{2}(q)=\int_{0}^{\infty}e^{-\Phi(q)z}[1-\psi^{\prime}(0+)W(z)]\textup{d}z=\frac{1}{\Phi(q)}-\frac{\psi^{\prime}(0+)}{q}=F_{1}(q).$

The fact that $F_{2}=F_{1}$ implies that, when $\alpha$ is a natural number,
we can take derivatives of order $\alpha$ (with the help of the dominated
convergence theorem), at $q=0$ and conclude that

$\displaystyle\int_{0}^{\infty}\mathbb{E}_{z}((\tau_{0}^{-})^{\alpha}\mathbb{I}_{\\{\tau_{0}^{-}<\infty\\}})\textup{d}z<\infty\qquad\text{
if and only if
}\qquad\int_{0}^{\infty}\mathbb{E}_{-z}((\tau_{0}^{+})^{\alpha})[1-\psi^{\prime}(0+)W(z)]\textup{d}z<\infty.$

Furthermore, if $\alpha=k+\lambda$, with $k$ a positive integer and
$0<\lambda<1$, we can draw the same conclusion using the Marchaud derivative
(see e.g. Laue, (1980)). Using Lemma 2.2 we have that

$\displaystyle\int_{0}^{\infty}\mathbb{E}_{-z}((\tau_{0}^{+})^{p-1})[1-\psi^{\prime}(0+)W(z)]\textup{d}z<\infty.$

and the proof is complete. ∎

###### Proof of Lemma 4.12.

Let $x\leq 0$ and take $\delta>0$. Then

$\displaystyle\mathbb{E}_{0,x}((\tau_{D})^{p})=\mathbb{E}_{x}((\tau_{b}^{g,0})^{p})$
$\displaystyle\leq\mathbb{E}_{x}((\tau_{b}^{g,0})^{p}\mathbb{I}_{\\{g+\delta<\tau_{b}^{g,0}\\}})+\mathbb{E}_{x}((g+\delta)^{p}\mathbb{I}_{\\{g+\delta>\tau_{b}^{g,0}\\}}).$

Note that on the event $\\{g+\delta<\tau_{b}^{g,0}\\}$ we have that

$\displaystyle\tau_{b}^{g,0}$ $\displaystyle=\inf\\{t>g+\delta:X_{t}\geq
b(U_{t})\\}$ $\displaystyle=\inf\\{t>0:X_{t+g+\delta}\geq
b(t+\delta)\\}+g+\delta$ $\displaystyle\leq\inf\\{t>0:X_{t+g+\delta}\geq
b(\delta)\\}+g+\delta,$

where the second equality follows from the fact that after $g$, the process
$X$ never goes back below zero and the last inequality holds since $b$ is non-
increasing. We have that the law of the process $\\{X_{t+g},t\geq 0\\}$ is the
same as that of $\mathbb{P}^{\uparrow}$ where
$\mathbb{P}^{\uparrow}=\mathbb{P}^{\uparrow}_{0}$ and
$\mathbb{P}^{\uparrow}_{x}$ corresponds to the law of $X$ starting at $x$
conditioned to stay positive (see Bertoin, (1998) Corollary VII.4.19). Using
the Markov property and equation VII.3.(6) in Bertoin, (1998) we get

$\displaystyle\mathbb{E}_{x}((\tau_{b}^{g,0})^{p})$ $\displaystyle\leq
2^{p}\mathbb{E}^{\uparrow}(\mathbb{E}^{\uparrow}_{X_{\delta}}[(\tau_{b(\delta)}^{+})^{p}])+(2^{p}+1)\mathbb{E}_{x}((g+\delta)^{p})$
$\displaystyle=2^{p}\mathbb{E}^{\uparrow}\left(\frac{W(b(\delta))}{W(X_{\delta})}\mathbb{E}_{X_{\delta}}[(\tau_{b(\delta)}^{+})^{p}\mathbb{I}_{\\{\tau_{0}^{-}>\tau_{b(\delta)}^{+}\\}}]\right)+(2^{p}+1)\mathbb{E}_{x}((g+\delta)^{p})$
$\displaystyle\leq
2^{p}\mathbb{E}[(\tau_{b(\delta)}^{+})^{p}]\mathbb{E}^{\uparrow}\left(\frac{W(b(\delta))}{W(X_{\delta})}\right)+(2^{p}+1)\mathbb{E}_{x}((g+\delta)^{p})$
$\displaystyle=2^{p}\mathbb{E}[(\tau_{b(\delta)}^{+})^{p}]\int_{(0,\infty)}\frac{W(b(\delta))}{W(z)}\mathbb{P}^{\uparrow}(X_{\delta}\in\textup{d}z)+(2^{p}+1)\mathbb{E}_{x}((g+\delta)^{p}),$

where the second inequality follows from the fact that
$\mathbb{E}_{x}[(\tau_{a}^{+})^{p}]\leq\mathbb{E}[(\tau_{a}^{+})^{p}]$ for all
$0\leq x\leq a$ and $X_{\delta}>0$ under $\mathbb{P}^{\uparrow}$. Thus, using
that
$\mathbb{P}^{\uparrow}(X_{\delta}\in\textup{d}z)=[zW(z)/\delta]\mathbb{P}(X_{\delta}\in\textup{d}z)$
(see e.g. Corollary VII.3.16 in Bertoin, (1998)) we have that

$\displaystyle\mathbb{E}_{x}((\tau_{b}^{g,0})^{p})$ $\displaystyle\leq
2^{p}\mathbb{E}[(\tau_{b(\delta)}^{+})^{p}]\int_{(0,\infty)}\frac{W(b(\delta))}{W(z)}\mathbb{P}^{\uparrow}(X_{\delta}\in\textup{d}z)+(2^{p}+1)\mathbb{E}_{x}((g+\delta)^{p})$
$\displaystyle=2^{p}\mathbb{E}[(\tau_{b(\delta)}^{+})^{p}]\frac{W(b(\delta))}{\delta}\mathbb{E}(X_{\delta}^{+})+2^{p}(2^{p}+1)\delta^{p}+2^{p}(2^{p}+1)\mathbb{E}_{x}((g)^{p}),$
(53)

where $X_{\delta}^{+}$ is the positive part of $X_{\delta}$. Thus from Lemma
2.1 we have that $\mathbb{E}_{x}((\tau_{b}^{g,0})^{p})$ is finite.

Next, we show that $\mathbb{E}_{u,x}((\tau_{D})^{p})<\infty$ when $u,x>0$.
From the Markov property of Lévy processes we have that

$\displaystyle\mathbb{E}_{u,x}((\tau_{D})^{p})$
$\displaystyle=\mathbb{E}_{x}((\tau_{b}^{u,0})^{p}\mathbb{I}_{\\{\tau_{b}^{u,0}<\sigma_{0}^{-}\\}})+\mathbb{E}_{x}((\tau_{b}^{g,0})^{p}\mathbb{I}_{\\{\tau_{b}^{u,0}>\sigma_{0}^{-}\\}})$
$\displaystyle\leq\mathbb{E}_{x}((\tau_{b(u)}^{+})^{p})+2^{p}\mathbb{E}_{x}((\sigma_{0}^{-})^{p}\mathbb{I}_{\\{\sigma_{0}^{-}<\infty\\}})+\mathbb{E}_{x}(\mathbb{I}_{\\{\sigma_{0}^{-}<\infty\\}}\mathbb{E}_{X_{\sigma_{0}^{-}}}[(\tau_{b}^{g,0})^{p}]).$

Using (53), the inequality $|X_{\sigma_{0}^{-}}|\leq|\underline{X}_{\infty}|$
under the event $\\{\sigma_{0}^{-}<\infty\\}$ and Lemmas 2.1 and 2.2 we deduce
that $\mathbb{E}_{u,x}((\tau_{D})^{p})<\infty$ and the proof is complete. ∎

Before proving the smooth fit condition, we first consider a technical lemma
involving the derivative of the potential measure. More specifically, for
fixed $a>0$, $x\in(0,a)$ and $r\in\mathbb{N}\cup\\{0\\}$ denote by
$U_{r}(a,x,\textup{d}y)$ as the measure

$\displaystyle
U_{r}(a,x,\textup{d}y)=\int_{0}^{\infty}t^{r}\mathbb{P}_{x}(X_{t}\in\textup{d}y,t<\sigma_{0}^{-}\wedge\tau_{a}^{+})\textup{d}t.$

######  Lemma 6.1.

Let $q\in\mathbb{N}\cup\\{0\\}$ such that
$\int_{(-\infty,-1)}|x|^{q}\Pi(\textup{d}x)<\infty$. Fix $a>0$ and $0\leq
x\leq a$. We have that for all $r\in\\{0,1,\ldots,q\\}$ the measure
$U_{r}(a,x,\textup{d}y)$ is absolutely continuous with respect to the Lebesgue
measure. It has a density $u_{r}(a,x,y)$ given by

$\displaystyle u_{r}(a,x,y)=\lim_{q\downarrow
0}(-1)^{r}\frac{\partial^{r}}{\partial
q^{r}}\left[\frac{W^{(q)}(x)W^{(q)}(a-y)}{W^{(a)}(a)}-W^{(q)}(x-y)\right],$

for $y\in(0,a]$. Moreover, for a fixed $a>0$ the functions
$x\mapsto\mathbb{E}_{x}((\tau_{a}^{+})^{r}\mathbb{I}_{\\{\sigma_{0}^{-}<\tau_{a}^{+}\\}})$
and $x\mapsto u_{r}(a,x,y)$ are differentiable on $(0,a)$ and have finite left
derivative at $x=a$ for all $y\in(0,a)$ and $r\in\\{0,1,\ldots,q\\}$.

###### Proof.

Let $a>0$ and $x\in(0,a)$. First we show that for all $r\in\\{0,1,\ldots,q\\}$
the measure $U_{r}(a,x,\textup{d}y)$ is absolutely continuous with respect to
the Lebesgue measure. Take any measurable set $A\subset(0,a)$, thus by
Fubini’s theorem

$\displaystyle\int_{A}U_{r}(a,x,\textup{d}y)$
$\displaystyle=\int_{0}^{\infty}t^{r}\mathbb{P}_{x}(X_{t}\in
A,t<\sigma_{0}^{-}\wedge\tau_{a}^{+})\textup{d}t$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\tau_{a}^{+}\wedge\sigma_{0}^{-}}t^{r}\mathbb{I}_{\\{X_{t}\in
A\\}}\textup{d}t\right).$

From Lemma 2.1 we know that $\mathbb{E}_{x}((\tau_{a}^{+})^{r})<\infty$ for
all $r\in\\{0,1,\ldots,q\\}$. Then by dominated convergence theorem we have
that

$\displaystyle\int_{A}U_{r}(a,x,\textup{d}y)$ $\displaystyle=\lim_{q\downarrow
0}\mathbb{E}_{x}\left(\int_{0}^{\tau_{a}^{+}\wedge\sigma_{0}^{-}}t^{r}e^{-qt}\mathbb{I}_{\\{X_{t}\in
A\\}}\textup{d}t\right)$ $\displaystyle=\int_{A}\lim_{q\downarrow
0}(-1)^{r}\frac{\partial^{r}}{\partial
q^{r}}\left[\frac{W^{(q)}(x)W^{(q)}(a-y)}{W^{(a)}(a)}-W^{(q)}(x-y)\right]\textup{d}y,$

where the last equality follows from (6). From the convolution representation
of $W^{(q)}$ (see equation (2)) the derivatives in the last equation above
exist and indeed $u_{r}(a,x,y)$ is a density of $U_{r}(a,x,\textup{d}y)$ for
all $y\in(0,a)$. Now we proceed to show the differentiation statements. Note
that from equations (1) and (3) we have that

$\displaystyle
f_{x}(q):=\mathbb{E}_{x}(e^{-q\tau_{a}^{+}}\mathbb{I}_{\\{\sigma_{0}^{-}<\tau_{a}^{+}\\}})=e^{\Phi(q)(x-a)}-\frac{W^{(q)}(x)}{W^{(q)}(a)},$

for any $x\in(0,a)$. Since $W$ is differentiable, the proof follows by
induction and implicit differentiation. A similar argument works for the
function $x\mapsto u_{r}(a,x,y)$. ∎

###### Proof of Theorem 4.13.

Let $0<\varepsilon<1$ and $u,x>0$ such that $b(u)>0$. Define the stopping time
$\tau_{*}:=\tau_{b}^{u,-\varepsilon}\mathbb{I}_{\\{\sigma_{\varepsilon}^{-}>\tau_{b}^{u,-\varepsilon}\\}}+\tau_{b}^{g,-\varepsilon}\mathbb{I}_{\\{\sigma_{\varepsilon}^{-}<\tau_{b}^{u,-\varepsilon}\\}}$.
It turns out that $\tau_{*}$ is an optimal stopping time for
$V(u,x-\varepsilon)$ under $\mathbb{P}_{x}$. That is,

$\displaystyle V(u,x-\varepsilon)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\tau_{b}^{u,-\varepsilon}\wedge\sigma_{\varepsilon}^{-}}G(u+s,X_{s}-\varepsilon)ds\right)+\mathbb{E}_{x}\left(\mathbb{I}_{\\{\sigma_{\varepsilon}^{-}<\tau_{b}^{u,-\varepsilon}\\}}\int_{\sigma_{\varepsilon}^{-}}^{\tau_{b}^{g,-\varepsilon}}G(U_{s}^{(\varepsilon)},X_{s}-\varepsilon)\textup{d}s\right)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\tau_{b}^{u,-\varepsilon}\wedge\sigma_{\varepsilon}^{-}}G(u+s,X_{s}-\varepsilon)ds\right)+\mathbb{E}_{x}\left(\mathbb{I}_{\\{\sigma_{\varepsilon}^{-}<\tau_{b}^{u,-\varepsilon}\\}}\int_{\sigma_{\varepsilon}^{-}}^{\tau_{b}^{g,-\varepsilon}\wedge\sigma_{0}^{-}}G(U_{s}^{(\varepsilon)},X_{s}-\varepsilon)\textup{d}s\right)$
$\displaystyle\qquad+\mathbb{E}_{x}\left(\mathbb{I}_{\\{\sigma_{0}^{-}<\tau_{b}^{g,-\varepsilon}\\}}\mathbb{I}_{\\{\sigma_{\varepsilon}^{-}<\tau_{b}^{u,-\varepsilon}\\}}\int_{\sigma_{0}^{-}}^{\tau_{b}^{g,-\varepsilon}}G(U_{s}^{(\varepsilon)},X_{s}-\varepsilon)\textup{d}s\right).$

On the other hand, from the definition of $V$ we have that

$\displaystyle V(u,x)$
$\displaystyle\leq\mathbb{E}_{x}\left(\int_{0}^{\tau_{*}\wedge\sigma_{0}^{-}}G(u+s,X_{s})\textup{d}s\right)+\mathbb{E}_{x}\left(\mathbb{I}_{\\{\sigma_{0}^{-}<\tau_{*}\\}}\int_{\sigma_{0}^{-}}^{\tau_{*}}G(U_{s},X_{s})\textup{d}s\right)$
$\displaystyle=\mathbb{E}_{x}\left(\int_{0}^{\tau_{b}^{u,-\varepsilon}\wedge\sigma_{\varepsilon}^{-}}G(u+s,X_{s})\textup{d}s\right)+\mathbb{E}_{x}\left(\mathbb{I}_{\\{\sigma_{\varepsilon}^{-}<\tau_{b}^{u,-\varepsilon}\\}}\int_{\sigma_{\varepsilon}^{-}}^{\tau_{b}^{g,-\varepsilon}\wedge\sigma_{0}^{-}}G(u+s,X_{s})\textup{d}s\right)$
$\displaystyle\qquad+\mathbb{E}_{x}\left(\mathbb{I}_{\\{\sigma_{0}^{-}<\tau_{b}^{g,-\varepsilon}\\}}\mathbb{I}_{\\{\sigma_{\varepsilon}^{-}<\tau_{b}^{u,-\varepsilon}\\}}\int_{\sigma_{0}^{-}}^{\tau_{b}^{g,-\varepsilon}}G(U_{s},X_{s})\textup{d}s\right).$

Hence for all $0<x\leq b(u)$,

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

where

$\displaystyle R_{1}^{(\varepsilon)}(u,x)$
$\displaystyle:=\frac{1}{\varepsilon}\mathbb{E}_{x}\left(\int_{0}^{\tau_{b}^{u,-\varepsilon}\wedge\sigma_{\varepsilon}^{-}}[G(u+s,X_{s})-G(u+s,X_{s}-\varepsilon)]\textup{d}s\right)\geq
0,$ $\displaystyle R_{2}^{(\varepsilon)}(u,x)$
$\displaystyle:=\frac{1}{\varepsilon}\mathbb{E}_{x}\left(\mathbb{I}_{\\{\sigma_{\varepsilon}^{-}<\tau_{b}^{u,-\varepsilon}\\}}\int_{\sigma_{\varepsilon}^{-}}^{\tau_{b}^{g,-\varepsilon}\wedge\sigma_{0}^{-}}[G(u+s,X_{s})-G(U_{s}^{(\varepsilon)},X_{s}-\varepsilon)]\textup{d}s\right)\geq
0,$ $\displaystyle R_{3}^{(\varepsilon)}(u,x)$
$\displaystyle:=\frac{1}{\varepsilon}\mathbb{E}_{x}\left(\mathbb{I}_{\\{\sigma_{0}^{-}<\tau_{b}^{g,-\varepsilon}\\}}\mathbb{I}_{\\{\sigma_{\varepsilon}^{-}<\tau_{b}^{u,-\varepsilon}\\}}\int_{\sigma_{0}^{-}}^{\tau_{b}^{g,-\varepsilon}}[G(U_{s},X_{s})-G(U_{s}^{(\varepsilon)},X_{s}-\varepsilon)]\textup{d}s\right)\geq
0.$

We will show that $\lim_{\varepsilon\downarrow 0}R_{i}^{(\varepsilon)}(u,x)=0$
for $i=1,2,3$. From the fact that $b$ is non-increasing we have that
$\tau_{b}^{u,-\varepsilon}\leq\tau_{b(u)+\varepsilon}^{+}$ and then for all
$u>0$ and $x=b(u)$ we have that

$\displaystyle R_{1}^{(\varepsilon)}(u,b(u))$
$\displaystyle\leq\frac{1}{\varepsilon}\mathbb{E}_{b(u)}\left(\int_{0}^{\tau_{b(u)+\varepsilon}^{+}\wedge\sigma_{\varepsilon}^{-}}(u+s)^{p-1}\psi^{\prime}(0+)[W(X_{s})-W(X_{s}-\varepsilon)]\textup{d}s\right)$
$\displaystyle\qquad-\frac{1}{\varepsilon}\mathbb{E}_{b(u)-\varepsilon}\left(\int_{0}^{\tau_{b(u)}^{+}\wedge\sigma_{0}^{-}}[\mathbb{E}_{X_{s}+\varepsilon}(g^{p-1})-\mathbb{E}_{X_{s}}(g^{p-1})]\textup{d}s\right)$
$\displaystyle=\frac{1}{\varepsilon}\mathbb{E}_{b(u)}\left(\int_{0}^{\tau_{b(u)+\varepsilon}^{+}\wedge\sigma_{\varepsilon}^{-}}(u+s)^{p-1}\psi^{\prime}(0+)[W(X_{s})-W(X_{s}-\varepsilon)]\textup{d}s\right)$
$\displaystyle\qquad-\frac{1}{\varepsilon}\int_{(0,b(u))}[\mathbb{E}_{z+\varepsilon}(g^{p-1})-\mathbb{E}_{z}(g^{p-1})]\int_{0}^{\infty}\mathbb{P}_{b(u)-\varepsilon}(X_{s}\in\textup{d}z,t<\tau_{b(u)}^{+}\wedge\sigma_{0}^{-})\textup{d}s$

Using the density of the $0$-potential measure of $X$ exiting the interval
$[0,b(u)]$ given in equation (6) we obtain that

$\displaystyle R_{1}^{(\varepsilon)}(u,b(u))$
$\displaystyle\leq\mathbb{E}_{b(u)}\left(\int_{0}^{\tau_{b(u)+\varepsilon}^{+}\wedge\sigma_{\varepsilon}^{-}}(u+s)^{p-1}\psi^{\prime}(0+)\frac{W(X_{s})-W(X_{s}-\varepsilon)}{\varepsilon}\textup{d}s\right)$
$\displaystyle\qquad-\int_{0}^{b(u)-\varepsilon}[\mathbb{E}_{z+\varepsilon}(g^{p-1})-\mathbb{E}_{z}(g^{p-1})]\frac{1}{\varepsilon}\left[\frac{W(b(u)-\varepsilon)W(b(u)-z)}{W(b(u))}-W(b(u)-\varepsilon-z)\right]\textup{d}z$
$\displaystyle\qquad-\frac{1}{\varepsilon}\int_{b(u)-\varepsilon}^{b(u)}[\mathbb{E}_{z+\varepsilon}(g^{p-1})-\mathbb{E}_{z}(g^{p-1})]\left[\frac{W(b(u)-\varepsilon)W(b(u)-z)}{W(b(u))}\right]\textup{d}z.$

Note that for all
$s<\tau_{b(u)+\varepsilon}^{+}\wedge\sigma_{\varepsilon}^{-}$, we have
$X_{s}\in(\varepsilon,b(u)+\varepsilon)$. Then using the fact that $W\in
C^{1}((0,\infty)$), the function $z\mapsto\mathbb{E}_{z}(g^{p-1})$ is
continuous, $\lim_{\varepsilon\downarrow
0}\tau_{b(u)+\varepsilon}^{+}\wedge\sigma_{\varepsilon}^{-}=\tau_{b(u)}^{+}\wedge\sigma_{0}^{-}=0$
a.s. under $\mathbb{P}_{b(u)}$ and the dominated convergence theorem we
conclude that

$\displaystyle\lim_{\varepsilon\downarrow 0}R_{1}^{(\varepsilon)}(u,b(u))=0.$

Now we show that $\lim_{\varepsilon\downarrow
0}R_{2}^{(\varepsilon)}(u,b(u))=0$. Take $0<x\leq b(u)$. Then using the
inequality $G(u,x)\leq u^{p-1}$, the fact that for $s<\sigma_{0}^{-}$,
$X_{s}>0$ (then $-\mathbb{E}_{-1}(g^{p-1})=G(0,-1)\leq
G(U_{s}^{(\varepsilon)},X_{s}-\varepsilon)$) and the strong Markov property at
time $\sigma_{\varepsilon}^{-}$ we get that