# Optimal cash management using impulse control

Peter Lakner and Josh Reed Leonard N. Stern School of Business New York
University 44 West 4th St. New York, NY 10012<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We consider the impulse control of Levy processes under the infinite horizon,
discounted cost criterion. Our motivating example is the cash management
problem in which a controller is charged a fixed plus proportional cost for
adding to or withdrawing from his/her reserve, plus an opportunity cost for
keeping any cash on hand. Our main result is to provide a verification theorem
for the optimality of control band policies in this scenario. We also analyze
the transient and steady-state behavior of the controlled process under
control band policies and explicitly solve for the optimal policy in the case
in which the Levy process to be controlled is the sum of a Brownian motion
with drift and a compound Poisson process with exponentially distributed jump
sizes.

Keywords: impulse control; cash management problem; Lévy processes; Brownian
motion.

AMS Subject Classification: 60G51, 60J25, 93E20.

##  1 Introduction

Impulse control problems have a long history related to applications to the
cash management problem. In the present paper, we consider the impulse control
of Levy processes. Our motivating application is the cash management problem
in which there exists a system manager who must control the amount of cash
he/she has on hand. We assume that the manager’s cash on hand fluctuates due
to randomly occurring withdrawals from and deposits to his/her account but
that the manager is charged a fixed plus proportional cost for any specific,
intentional adding to or withdrawing from his/her reserves and that there
exists an opportunity cost for keeping too little or too much cash on hand.
The manager’s objective is to minimize his/her long run opportunity cost of
keeping cash on hand plus any cost incurred from depositing or withdrawing
from the reserve. An alternative motivating application which is also
considered in the literature is a manager who wishes to control his/her
inventory level. The manager’s inventory level fluctuates randomly and he/she
may increase or decrease his/her inventory level at will by expediting or
salvaging parts, paying a fixed plus proportional cost to do so. The manager’s
objective is to minimize his/her long run inventory holding costs plus costs
of expediting and salvaging.

Our first main result in the paper is to provide a verification theorem for
the optimality of control band policies for the impulse control of Levy
processes. Our result is fairly general and holds for a wide class of
opportunity cost functions. We then explicitly calculate the Laplace transform
with respect to time and steady-state distribution of any spectrally positive
Levy process controlled under a control band policy. In Section 5, we consider
the special case of a Levy process which is comprised of the sum of a Brownain
motion and a compound Poisson process with exponentially distributed jump
sizes. In this specific case, we show how one may use the results derived in
this paper in order to characterize the value function for the associated
control problem, and characterize the band levels in the optimal control
policy as a solution of a system of equations. Moreover, we also show how one
may determine the steady-state distribution of the controlled process when the
underlying Levy process is the sum of a negative drift and a compound Poisson
process with exponentially distributed jump sizes.

The technique of impulse control was originally developed by Bensoussan and
Lions [3, 4] and extended by Richard et. al [8, 18]. Harrison, Selke and
Taylor [11] and Sulem [19] also consider the impulse control of Brownian
motion and explicitly calculate the critical parameters determining the
optimal policy. In [10], an iterative computational scheme is to provided in
order to determine the optimal policy for the impulse control of Brownain
motion. Recently, Ormeci, Dai and Vande Vate [ODV08] have considered impulse
control of Brownian motion under the average cost criterion and again show
that a control band policy is optimal. Cadenillas, Zapatero, and Sarkar [7]
and Cadenillas, Lakner, and Pinedo [6] solved the Brownian case with a mean-
reverting drift. None of the above mentioned works allow there to be jumps in
the process to be controlled. In [5], the optimality of an $(s,S)$ policy is
proven for a process which is the sum of a constant drift, a Brownian motion,
and a compound Poisson process. In related work, Bar-Ilan, Perry and Stadje
[2] have also considered the problem of impulse control of Levy processes for
the specific case in which the Levy process is a sum of a Brownian motion and
a compound Poisson process. Assuming that a control band policy is optimal,
their main results are to evaluate the cost functionals of the resulting
policy through a fundamental identity derived from the martingale originally
introduced by Kella and Whitt [13]. In [21], Yamazaki uses a scale function
approach to study one-sided impulse control problems for spectrally positive
Levy demand processes. In [9], an impulse control problem is studied for a
refracted Levy process where the ruin time is modeled by a Parisian delay.
Finally, [20] studies the impulse control of a geometric Levy process.

##  2 The Model

In this Section we provide the specifics of the model described in the
Introduction. All forthcoming processes are assumed to live on a probability
triplet equipped with a filtration ${\cal F}=\\{{\cal F}_{t},\ 0\leq
t<\infty\\}$. We begin by assuming that $Y_{t}$ is a Levy process started from
$x$ with Levy measure $\nu$ such that

$\int_{\left\\{|y|\geq 1\right\\}}|y|\nu(dy)<\infty.$ (1)

The process $Y_{t}$ will be used to represent the cash on hand process
assuming that the manager exerts no control by making no deposits to or
withdrawal from his/her fund. Let $J(\omega,dt,dy)=J(dt,dy)$ be the jump
measure of $Y$. Then, $Y_{t}$ has the Ito-Levy decomposition

$Y_{t}=x+\mu t+\sigma w_{t}+A_{t}+M_{t}$

where $M_{t}$ is the following martingale

$M_{t}=\int_{\left\\{|x|<1\right\\}}x\left\\{J((0,t],dx)-t\nu(dx)\right\\}$

and $A$ is the sum of the “large” jumps

$A_{t}=\sum_{0<s\leq t}\Delta Y_{s}1_{\left\\{|\Delta Y_{s}|\geq 1\right\\}}.$

The process $w$ is assumed to be a standard Wiener process and $\mu$ is a
constant. We do not make any assumption regarding $\sigma$, it may be zero or
non-zero. Also we allow $\nu(\Re)=0$ in which case $Y$ is continuous. The case
when both $\sigma=0$ and $\nu(\Re)=0$ is also included, although this case is
trivial ($Y$ is deterministic in this case). In general, we will use $P_{x}$
to denote the probability measure under which $Y_{t}$ is started from $x$ and
$E_{x}$ its associated expectation operator. We note that $M$ is a quadratic
pure jump local martingale ([17], Chapter II, Section 6) with quadratic
variation

$\left[M\right]_{t}=\sum_{0<s\leq t}\left(\Delta
Y_{s}\right)^{2}1_{\left\\{|\Delta Y_{s}|\leq
1\right\\}}=\int_{(0,t]\times\Re}y^{2}1_{\\{|y|<1\\}}J(ds,dy),$

which has expected value

$E\left[M\right]_{t}=t\int_{\Re}y^{2}1_{\\{|y|<1\\}}\nu(dy)<\infty.$ (2)

It follows ([17], Chapter II, Corollary 3 to Theorem 27) that $M$ is a square-
integrable martingale.

We let

$(T,\Xi)=(\tau_{1},\tau_{2},\dots,\tau_{n},\dots,\
\xi_{1},\xi_{2},\dots,\xi_{n},\dots,)$

denote the impulse control policy used by the manager where
$0\leq\tau_{1}<\tau_{2}<\tau_{3}\dots$ are stopping times and $\xi_{n}\not=0$
(a.s.) is an ${\cal F}_{\tau_{n}}$ measurable random variable for each $n\geq
1$. Positive values of $\xi_{n}$ represent deposits by the manager into
his/her fund and negative values represent withdrawal. In order to simplify
future developments we define $\tau_{0}=0$ and $\xi_{0}=0$. In principle we
also allow that with positive probability there are only finitely many
interventions. However, we introduce the following convention which
essentially changes nothing, but simplifies future formulas. Let
$\Lambda(\omega)\leq\infty$ be the number of interventions. For those
$\omega$’s such that $\Lambda(\omega)<\infty$ we define

$\tau_{\Lambda(\omega)+k}=\infty\quad\hbox{and}\quad\xi_{\Lambda(\omega)+k}=0$
(3)

for $k\geq 1$.

As described in the Introduction, the controlled cash on hand process $X_{t}$
follows the dynamics

$X_{t}=Y_{t}+\sum_{i=1}^{\infty}1_{\left\\{\tau_{i}\leq t\right\\}}\xi_{i}$

and has RCLL paths. Let $\lambda>0$ be a fixed discount factor.

######  Definition 1.

A control policy $(T,\Xi)$ is called admissible if the corresponding
discounted controlled cash on hand process is bounded, that is, there exists a
constant $K\in(0,\infty)$ such that

$e^{-\lambda t}|X_{t}|<K,\quad\hbox{a.s.},\ t\in[0,\infty).$

We emphasize that the constant $K$ is not ”universal”; for different policies
it may be different.

Let the opportunity cost function be $\phi(x)\geq 0$ for the cash on hand
being at level $x$. We shall assume that for some constants $K_{1},K_{2}>0$

$K_{1}\phi(x)+K_{2}\geq|x|,\quad x\in\Re.$ (4)

The manager’s total cost is the sum of his/her expected discounted opportunity
costs as well as impulse control costs and is given by

$I(x,T,\Xi)=E_{x}\left[\int_{0}^{\infty}e^{-\lambda
t}\phi(X_{t})dt+\sum_{n=1}^{\infty}e^{-\lambda\tau_{n}}g(\xi_{n})\right]$ (5)

where the manager’s impulse control costs are given by

$g(\xi)=\begin{cases}C+c\xi,&\textrm{if~{}}\xi>0,\\\
0,&\textrm{if~{}}\xi=0,\\\ D-d\xi,&\textrm{if~{}}\xi<0.\end{cases}$

We assume that the fixed costs $C$ and $D$ are positive and the variable costs
$c$ and $d$ are non-negative constants. The term $e^{-\lambda\tau_{n}}$ in (5)
is well defined once we set $e^{-\infty}=0.$

One of our primary objectives in this paper is to identify the optimal impulse
control $(T,\Xi)$ that minimizes the above total cost $I(x,T,\Xi)$. The value
function of this optimization problem is

$V(x)=\inf\left\\{I(x,T,\Xi),\ (T,\Xi)\ \hbox{is an admissible impulse
control}\right\\}.$

Notice that extending policies according to our convention (3) does not change
the corresponding cost.

In the section that follows we show that the impulse control takes the form of
a control band policy which arise frequently as the solution to impulse
control problems. Moreover, in Section 5 we provide an example in which we are
able to explicitly identify the parameters corresponding to this policy.

##  3 Main Results

In this Section, we provide the main result of the paper, Theorem 3.1, showing
that a solution of an ordinary differential equation that satisfies some
additional conditions, must be the value function $V$. Also included in the
statement of Theorem 3.1 is the optimal impulse control policy which turns out
to be a double bandwidth control policy. We begin first with some preliminary
results before providing the statement of Theorem 3.1.

For a function $f:\Re\mapsto\Re$ we define the operator

$Mf(x)=\inf\left\\{f(x+\eta)+g(\eta),\ \eta\in\Re\setminus\\{0\\}\right\\}.$

We shall also use the linear operator ${\cal A}$ associated with the
uncontrolled process $Y$, that is, for $f\in C^{2}(\Re)$

${\cal A}f(x)={1\over 2}\sigma^{2}f^{\prime\prime}(x)+\mu
f^{\prime}(x)+\int_{\Re}\left[f(x+y)-f(x)-f^{\prime}(x)y1_{\left\\{|y|<1\right\\}}\right]\nu(dy).$

Our assumption (1) and Taylor’s theorem implies that the integral on the
right-hand side is finite whenever $f^{\prime}$ and $f^{\prime\prime}$ are
bounded. Indeed,

$\displaystyle\left|\int_{\Re}\left[f(x+y)-f(x)-f^{\prime}(x)y1_{\left\\{|y|<1\right\\}}\right]\nu(dy)\right|$
(6) $\displaystyle\leq$
$\displaystyle\int_{\Re}\left|f(x+y)-f(x)\right|1_{\left\\{|y|\geq
1\right\\}}\nu(dy)+\int_{\Re}\left|f(x+y)-f(x)-f^{\prime}(x)y\right|1_{\left\\{|y|<1\right\\}}\nu(dy)$
$\displaystyle\leq$
$\displaystyle\mathrm{const}\times\left[\int_{\Re}|y|1_{\\{|y|\geq
1\\}}\nu(dy)+\int_{\Re}y^{2}1_{\\{|y|<1\\}}\nu(dy)\right]$ $\displaystyle<$
$\displaystyle\infty.$

In order to prove our results we shall actually need to extend the domain of
${\cal A}$ to a larger class of functions ${\cal D}$ defined below.

######  Definition 2.

Let ${\cal D}$ be the class of functions $f:\Re\mapsto\Re$ for which there
exist an integer $n\geq 0$ and a set of real numbers
$S=\\{x_{1},x_{2},\dots,x_{n}\\}$ (if $n=0$ then $S$ is the empty set) such
that the following conditions hold: (i) $f\in C^{1}(\Re)\cap
C^{2}(\Re\setminus S)$ (ii) The derivative $f^{\prime}$ is bounded on $\Re$
and the second derivative $f^{\prime\prime}$ is bounded on $\Re\setminus S$.
We shall call the points in $S$ the exceptional points.

One can see easily that (6) holds even if $f\in{\cal D}$, thus we extend the
operator ${\cal A}$ to ${\cal D}$. This way ${\cal A}f(x)$ may be undefined if
$x$ is an exceptional point of $f$, but this will not cause any problems. For
$f\in{\cal D}$ it may be shown that Ito’s rule applied to $f(X_{t})$ holds in
its usual form (see the Appendix).

We now conjecture that the optimal impulse control policy takes a double
bandwidth control policy form. In particular, we assume that there exist
constants $a<\alpha\leq\beta<b$ such that

$\tau^{*}_{n}=\inf\left\\{t\geq\tau^{*}_{n-1}:\ X_{t-}+\Delta
Y_{t}\in\Re\setminus(a,b)\right\\}$ (7)

(recall that $\tau^{*}_{0}=0$) and that for $n\geq 1$ the jump size is given
by

$\xi^{*}_{n}=\begin{cases}\beta-\left(X(\tau^{*}_{n}-)+\Delta
Y(\tau^{*}_{n})\right),&\textrm{if}~{}X(\tau^{*}_{n}-)+\Delta
Y(\tau^{*}_{n})\geq b,\\\ \alpha-\left(X(\tau^{*}_{n}-)+\Delta
Y(\tau^{*}_{n})\right),&\textrm{if}~{}X(\tau^{*}_{n}-)+\Delta
Y(\tau^{*}_{n})\leq a.\end{cases}$ (8)

Note that $\tau_{1}^{*}=0$ if and only if $x\in\Re\setminus(a,b)$. For $n>1$
it is possible that $\Delta Y(\tau_{n}^{*})=0$ in which case
$X(\tau_{n}^{*}-)$ is either equal to $a$ or $b$, $\xi_{n}^{*}$ is $\alpha-a$
or $\beta-b$ and $X(\tau_{n}^{*})$ equal to $\alpha$ or $\beta$, respectively.
However, it is also possible that $\Delta Y(\tau_{n}^{*})\not=0$ in which case
$X(\tau_{n}^{*}-)+\Delta Y(\tau_{n}^{*})$ may be either larger than $a$ or
smaller then $b$, but we still have $X(\tau_{n}^{*})$ equal to $\alpha$ or
$\beta$, respectively. This control policy is admissible since the
corresponding cash at hand process is bounded, even without discounting.

######  Proposition 3.1.

Suppose that for some $a<\alpha\leq\beta<b$ the sequence of stopping times
$\left(\tau^{*}_{n}\right)_{\\{n\geq 1\\}}$ is given in (7). If $Y$ is not
constant then $\tau^{*}_{n}<\infty$, $\tau^{*}_{n}<\tau^{*}_{n+1}$ for $n\geq
1$ almost surely, and $\lim_{n\to\infty}\tau^{*}_{n}=\infty$.

Proof: It is obvious that $\tau^{*}_{n}<\infty$ since otherwise the Levy
process $Y$ would be bounded, that is, a constant. Next we show
$\tau^{*}_{n}<\tau^{*}_{n+1}$. Let $Y^{\alpha}$ be the process $Y$ started at
$Y_{0}=\alpha$ and $Y^{\beta}$ be the process $Y$ started at $Y_{0}=\beta$.
Let $\tau_{\alpha}=\inf\left\\{s\geq 0:\ Y_{s}^{\alpha}\notin(a,b)\right\\}$
and $\tau_{\beta}=\inf\left\\{s\geq 0:\ Y_{s}^{\beta}\notin(a,b)\right\\}$. By
the right-continuity of $Y$ we have $\tau_{\alpha}>0$ and $\tau_{\beta}>0$
a.s. But conditionally on $X(\tau^{*}_{n})=\alpha$ the inter-arrival time
$\tau^{*}_{n+1}-\tau^{*}_{n}$ has the same distribution as $\tau_{\alpha}$ and
conditionally on $X(\tau^{*}_{n})=\beta$ the inter-arrival time
$\tau^{*}_{n+1}-\tau^{*}_{n}$ has the same distribution as $\tau_{\beta}$.
Hence $\tau^{*}_{n}<\tau^{*}_{n+1}$. Finally we show that
$\lim_{n\to\infty}\tau^{*}_{n}=\infty$. We first show that either
$P[X(\tau^{*}_{n})=\alpha,\ \hbox{i.o.}]=1$ or $P[X(\tau^{*}_{n})=\beta,\
\hbox{i.o.}]=1$ or both. First note that if both
$P[X(\tau^{*}_{1})=\alpha|X(0)=\alpha]=1$ and
$P[X(\tau^{*}_{1})=\beta|X(0)=\beta]=1$ (which actually is not even possible),
then by the strong Markov property of $X_{t}$, either
$P[X(\tau^{*}_{n})=\alpha,\ \hbox{i.o.}]=1$ or $P[X(\tau^{*}_{n})=\beta,\
\hbox{i.o.}]=1$, but not both. On the other hand, if
$P[X(\tau^{*}_{1})=\alpha|X(0)=\alpha]<1$ and
$P[X(\tau^{*}_{1})=\beta|X(0)=\beta]<1$, then, by the strong Markov property
of $X_{t}$, both $P[X(\tau^{*}_{n})=\alpha,\ \hbox{i.o.}]=1$ and
$P[X(\tau^{*}_{1})=\beta,\ \hbox{i.o.}]=1$. Finally, note that if
$P[X(\tau^{*}_{1})=\alpha|X(0)=\alpha]=1$ and
$P[X(\tau^{*}_{1})=\beta|X(0)=\beta]<1$, then by the strong Markov property,
$P[X(\tau^{*}_{n})=\alpha,\ \hbox{i.o.}]=1$. In the reverse case, one has that
$P[X(\tau^{*}_{n})=\beta,\ \hbox{i.o.}]=1$ Suppose now, for the sake of
argument, that $P[X(\tau^{*}_{n})=\alpha,\ \hbox{i.o.}]=1$. Then let $S_{0}=0$
and $S_{n+1}=\min\\{\tau^{*}_{m}>S_{n}:\ X(\tau^{*}_{m})=\alpha\\}$. Then
$S_{2}-S_{1},S_{3}-S_{2},\dots$ is an i.i.d. sequence of random variables.
Since $S_{n+1}-S_{n}>0$, a.s., so $E[S_{n+1}-S_{n}]>0$ which implies that
$\sum_{n=1}^{\infty}(S_{n+1}-S_{n})$ diverges, i.e. $S_{n}\to\infty$. But this
implies $\tau^{*}_{n}\to\infty$.

The following is now the main result of this Section.

######  Theorem 3.1.

Suppose that there exist constants $a<\alpha\leq\beta<b$ and a function
$f:\Re\mapsto(0,\infty)$ such that $f\in C^{1}(\Re)\cap
C^{2}\left(\Re\setminus\\{a,b\\}\right)$, the second derivative
$f^{\prime\prime}$ is bounded on $\Re\setminus\\{a,b\\}$ and the following
conditions are satisfied: (i) ${\cal A}f(x)-\lambda f(x)+\phi(x)=0$ for
$x\in(a,b)$; (ii) $f(x)\leq Mf(x)$ for $x\in(a,b)$; (iii) ${\cal
A}f(x)-\lambda f(x)+\phi(x)\geq 0$ for $x\in\Re\setminus[a,b]$; (iv)
$f(a)=Mf(a)=f(\alpha)+C+c(\alpha-a),\quad f(b)=Mf(b)=f(\beta)+D+d(b-\beta)$;
(v) $f$ is linear on $(-\infty,a]$ with slope $-c$, and also linear on
$[b,\infty)$ with slope $d$.

Then $f(x)=V(x)$. Furthermore, the control $(T^{*},\Xi^{*})$ given in (7) -
(8) with these values for $a,\alpha,\beta,b$ is optimal.

In order to provide the proof of Theorem 3.1, we first need the following.

######  Lemma 3.2.

Let $(T,\Xi)$ be a control such that $I(x,T,\Xi)<\infty$ Then

$\lim_{t\to\infty}E_{x}\left[e^{-\lambda t}|X_{t}|\right]=0$ (9)

and

$\lim_{n\to\infty}\tau_{n}(\omega)=\infty,\quad a.s.$ (10)

Proof: First we prove (9). By (4)

$\int_{0}^{\infty}E_{x}\left[e^{-\lambda
t}|X_{t}|\right]dt\leq\int_{0}^{\infty}E_{x}\left[e^{-\lambda
t}\left(K_{1}\phi(X_{t})+K_{2}\right)\right]dt,$

which is finite by our assumption that $I(x,T,\Xi)<\infty$. Next we prove
(10). Suppose the opposite, i.e., that $P(G)>0$ where $G=\\{\omega\in\Omega:\
\tau_{n}\uparrow\tau<\infty\\}$. Then

$\displaystyle\sum_{n=1}^{\infty}E_{x}\left[e^{-\lambda\tau_{n}}g(\xi_{n})\right]\geq\sum_{n=1}^{\infty}E_{x}\left[e^{-\lambda\tau_{n}}1_{G}\right]\hbox{min}\left\\{C,D\right\\}\geq\sum_{n=1}^{\infty}E_{x}\left[e^{-\lambda\tau}1_{G}\right]\hbox{min}\left\\{C,D\right\\}=\infty$

which contradicts our assumption that $I(x,T,\Xi)<\infty$. $\bullet$

By Lemma 3.2 without any loss of generality we can and shall consider only
those policies for which (9) and (10) hold.

Proof of Theorem 3.1: Using condition (v) we can extend (iv) to $(-\infty,a]$
and $[b,\infty)$:

$f(x)=Mf(x)=f(\alpha)+C+c(\alpha-x),\quad x\leq a$ (11)
$f(x)=Mf(x)=f(\beta)+D+d(x-\beta),\quad x\geq b.$ (12)

Indeed, suppose that $x<a$. Then for $\eta\in[0,a-x]$ the quantity
$f(x+\eta)+C+c\eta=f(x)+C$ does not depend on $\eta$. In addition
$f(x+\eta)+g(\eta)$ is a decreasing function of $\eta$ on $(-\infty,0)$. Hence
by (iv) and (v)

$Mf(x)=\inf\\{f(x+\eta)+C+c\eta,\ \eta\geq
a-x\\}=\inf\\{f(a+\gamma)+C+c(a+\gamma-x);\gamma\geq 0\\}$
$=Mf(a)+c(a-x)=f(a)+c(a-x)=f(x)$

Also by (iv)

$Mf(a)+c(a-x)=f(\alpha)+C+c(\alpha-a)+c(a-x)=f(\alpha)+C+c(\alpha-x).$

We deal with the $x>b$ case similarly. Let
$(T,\Xi)=\\{\tau_{1},\tau_{2},\dots,\xi_{1},\xi_{2},\dots\\}$ be an arbitrary
impulse control for which (9) and (10) hold. We define the local martingale

$U_{t}=\int_{(0,t]\times\Re}e^{-\lambda
s}\left\\{f(X_{s-}+y)-f(X_{s-})-f^{\prime}(X_{s-})y1_{\\{|y|<1\\}}\right\\}\left(J(ds,dy)-ds\nu(dy)\right)$
(13)

Since $U$ is a local martingale, there exists a sequence of stopping times
$\left(S_{m}\right)$ such that $\lim_{m\to\infty}S_{m}=\infty$, almost surely,
and $\\{U_{t\wedge S_{m}},\ t\in[0,\infty)\\}$ is a martingale for every $m$.
Fix $t>0$ $m\geq 1$. For the sake of brevity let
$T_{n}=\min\\{\tau_{n},S_{m},t\\}$. The stopping time $T_{n}$ depends on $m$
and $t$ but we suppress these dependencies in the notation. We have the
following decomposition

$\displaystyle e^{-\lambda T_{n}}f\left(X_{T_{n}}\right)-f(x)$
$\displaystyle=$ $\displaystyle\sum_{i=1}^{n}\left\\{e^{-\lambda
T_{i}}f\left(X_{T_{i}-}+\Delta Y_{T_{i}}\right)-e^{-\lambda
T_{i-1}}f\left(X_{T_{i-1}}\right)\right\\}+\sum_{i=1}^{n}e^{-\lambda
T_{i}}\left[f\left(X_{T_{i}}\right)-f\left(X_{T_{i}-}+\Delta
Y_{T_{i}}\right)\right].$

First we deal with the first sum on the right-hand side of (3). By the
generalized Ito’s rule (Proposition A.1) and integration by parts follows that

$\displaystyle e^{-\lambda T_{i}}f\left(X_{T_{i}-}+\Delta
Y_{T_{i}}\right)-e^{-\lambda T_{i-1}}f\left(X_{T_{i-1}}\right)$
$\displaystyle=$ $\displaystyle\int_{(T_{i-1},T_{i}]}e^{-\lambda
s}f^{\prime}(X_{s-})\left\\{dA_{s}+dM_{s}+\sigma dw_{s}\right\\}$
$\displaystyle+\int_{T_{i-1}}^{T_{i}}e^{-\lambda s}\left\\{{\sigma^{2}\over
2}f^{\prime\prime}(X_{s-})+\mu f^{\prime}(X_{s-})-\lambda
f(X_{s-})\right\\}ds$ $\displaystyle+\sum_{T_{i-1}<s\leq T_{i}}e^{-\lambda
s}\left\\{f(X_{s})-f(X_{s-})-f^{\prime}(X_{s-})\Delta X_{s}\right\\}.$

The right-hand side can be written as

$\displaystyle\int_{(T_{i-1},T_{i}]}e^{-\lambda
s}f^{\prime}(X_{s-})\left\\{dM_{s}+\sigma dw_{s}\right\\}$
$\displaystyle+\int_{T_{i-1}}^{T_{i}}e^{-\lambda s}\left\\{{\sigma^{2}\over
2}f^{\prime\prime}(X_{s-})+\mu f^{\prime}(X_{s-})-\lambda
f(X_{s-})\right\\}ds$ $\displaystyle+\sum_{T_{i-1}<s\leq T_{i}}e^{-\lambda
s}\left\\{f(X_{s})-f(X_{s-})-f^{\prime}(X_{s-})\Delta X_{s}1_{|\Delta
X_{s}|<1}\right\\},$

which is equal to

$\displaystyle\int_{(T_{i-1},T_{i}]}e^{-\lambda
s}f^{\prime}(X_{s-})\left\\{dM_{s}+\sigma dw_{s}\right\\}$
$\displaystyle+\int_{T_{i-1}}^{T_{i}}e^{-\lambda s}\left\\{{\sigma^{2}\over
2}f^{\prime\prime}(X_{s-})+\mu f^{\prime}(X_{s-})-\lambda
f(X_{s-})\right\\}ds+$
$\displaystyle+\int_{(T_{i-1},T_{i}]\times\Re}e^{-\lambda
s}\left\\{f(X_{s-}+y)-f(X_{s-})-f^{\prime}(X_{s-})y1_{\\{|y|<1\\}}\right\\}J(ds,dy).$

Since the left-hand side of (6) is finite, the above expression can be written
as

$\displaystyle\int_{(T_{i-1},T_{i}]}e^{-\lambda
s}f^{\prime}(X_{s-})\left\\{dM_{s}+\sigma dw_{s}\right\\}$
$\displaystyle+\int_{T_{i-1}}^{T_{i}}e^{-\lambda s}\left\\{{\sigma^{2}\over
2}f^{\prime\prime}(X_{s-})+\mu f^{\prime}(X_{s-})-\lambda
f(X_{s-})\right\\}ds$
$\displaystyle+\int_{(T_{i-1},T_{i}]\times\Re}e^{-\lambda
s}\left\\{f(X_{s-}+y)-f(X_{s-})-f^{\prime}(X_{s-})y1_{\\{|y|<1\\}}\right\\}\left(J(ds,dy)-ds\nu(dy)\right)$
$\displaystyle+\int_{(T_{i-1},T_{i}]\times\Re}e^{-\lambda
s}\left\\{f(X_{s-}+y)-f(X_{s-})-f^{\prime}(X_{s-})y1_{\\{|y|<1\\}}\right\\}ds\nu(dy)$
$\displaystyle=$ $\displaystyle\int_{(T_{i-1},T_{i}]}e^{-\lambda
s}f^{\prime}(X_{s-})\left\\{dM_{s}+\sigma dw_{s}\right\\}$
$\displaystyle+\int_{T_{i-1}}^{T_{i}}e^{-\lambda s}\left\\{{\cal
A}(X_{s-})-\lambda f(X_{s-})\right\\}ds$
$\displaystyle+\int_{(T_{i-1},T_{i}]\times\Re}e^{-\lambda
s}\left\\{f(X_{s-}+y)-f(X_{s-})-f^{\prime}(X_{s-})y1_{\\{|y|<1\\}}\right\\}\left(J(ds,dy)-ds\nu(dy)\right).$

By conditions (i) and (iii) we end up with the inequality

$\displaystyle e^{-\lambda T_{i}}f\left(X_{T_{i}-}+\Delta
Y_{T_{i}}\right)-e^{-\lambda T_{i-1}}f\left(X_{T_{i-1}}\right)$ (15)
$\displaystyle\geq$ $\displaystyle\int_{(T_{i-1},T_{i}]}e^{-\lambda
s}f^{\prime}(X_{s-})\left\\{dM_{s}+\sigma dw_{s}\right\\}$
$\displaystyle+\int_{(T_{i-1},T_{i}]\times\Re}e^{-\lambda
s}\left\\{f(X_{s-}+y)-f(X_{s-})-f^{\prime}(X_{s-})y1_{\\{|y|<1\\}}\right\\}\left(J(ds,dy)-ds\nu(dy)\right)$
$\displaystyle-\int_{T_{i-1}}^{T_{i}}e^{-\lambda s}\phi(X_{s})ds.$

There is a minor problem here since (i) and (iii) implies the inequality
${\cal A}f(x)-\lambda f(x)+\phi(x)\geq 0$ only for
$x\in\Re\setminus\\{a,b\\}$. But either $\sigma=0$, in which case the
inequality holds for all $x\in\Re$, or, $\sigma\not=0$, in which case (15)
still holds by Lemma A.3 in the Appendix.

For our candidate optimal policy $(T^{*},\Xi^{*})$ actually equality holds in
(15) by condition (i).

Next we deal with the second sum on the right-hand side of (3). By (11), (12),
and (ii) whenever $T_{i}=\tau_{j}$ for some $j\geq 1$ then

$f(X_{T_{i}})-f\left(X_{T_{i}-}+\Delta
Y_{T_{i}}\right)\geq-g\left(X_{T_{i}}-(X_{T_{i}-}+\Delta
Y_{T_{i}})\right)=-g(\xi_{j}).$ (16)

If $T_{n}=t\wedge S_{m}$ then the left-hand side of (16) is zero, hence

$\sum_{i=1}^{n}e^{-\lambda
T_{i}}\left[f\left(X_{T_{i}}\right)-f\left(X_{T_{i}-}+\Delta
Y_{T_{i}}\right)\right]\geq-\sum_{j\leq n,\tau_{j}\leq t\wedge
S_{m}}e^{-\lambda\tau_{j}}g(\xi_{j})$ (17)

By (11) and (12) this becomes an equality for $(T^{*},\Xi^{*})$.

Adding up (15) and (17) and considering (13), (3), we get

$\displaystyle e^{-\lambda T_{n}}f\left(X_{T_{n}}\right)-f(x)$
$\displaystyle\geq$ $\displaystyle\int_{(0,T_{n}]}e^{-\lambda
s}f^{\prime}(X_{s-})\left\\{dM_{s}+\sigma
dw_{s}\right\\}+U_{T_{n}}-\int_{0}^{T_{n}}e^{-\lambda
s}\phi(X_{s})ds-\sum_{j\leq n,\tau_{j}\leq t\wedge
S_{m}}e^{-\lambda\tau_{j}}g(\xi_{j}).$

After taking expectations on both sides, all martingale terms will disappear.
Indeed, the local martingale

$Z_{t}=\int_{(0,t]}e^{-\lambda s}f^{\prime}(X_{s-})dM_{s}$

has quadratic variation

$\left[Z\right]_{t}=\int_{(0,t]}e^{-2\lambda
s}\left(f^{\prime}(s)\right)^{2}d\left[M\right]_{s}$

([17], Chapter II, Theorem 29), and the boundedness of $f^{\prime}$ and (2)
imply that $Z$ is a square-integrable martingale (see also [17], Chapter II,
Corollary 3 to Theorem 27). Then by the Optional sampling Theorem for bounded
stopping times we have $E[Z_{T_{n}}]=0$. Also, since $\left\\{U_{s\wedge
S_{m}},s<\infty\right\\}$ is a martingale, another application of the Optional
sampling Theorem implies that $E[U_{\tau_{n}\wedge t\wedge
S_{m}}]=E[U_{T_{n}}]=0$. Hence after taking expectations on both sides we end
up with

$-E_{x}\left[e^{-\lambda T_{n}}f\left(X_{T_{n}}\right)\right]+f(x)\leq
E_{x}\left[\int_{0}^{T_{n}}e^{-\lambda s}\phi(X_{s})ds+\sum_{j:\tau_{j}\leq
T_{n}}e^{-\lambda\tau_{j}}g(\xi_{j})\right].$

Now let $n\to\infty$. By the continuity of $f$ and condition (v) $f$ satisfies
a linear growth condition $f(x)\leq K_{3}|x|+K_{4}$ for some positive
constants $K_{3},K_{4}$, hence $0\leq e^{-\lambda
T_{n}}f\left(X_{T_{n}}\right)\leq e^{-\lambda
T_{n}}(K_{3}\left|X_{T_{n}}\right|+K_{4})$, which is bounded by the
admissibility of $(T,\Xi)$. We apply the Dominated Convergence Theorem on the
left-hand side and the Monotone Convergence Theorem on the right-hand side,
and get the inequality

$-E_{x}\left[e^{-\lambda(t\wedge S_{m})}f\left(X_{t\wedge
S_{m}}\right)\right]+f(x)\leq E_{x}\left[\int_{0}^{t\wedge S_{m}}e^{-\lambda
s}\phi(X_{s})ds+\sum_{j:\tau_{j}\leq t\wedge
S_{m}}e^{-\lambda\tau_{j}}g(\xi_{j})\right].$

We apply exactly the same method as $m\to\infty$ and get

$-E_{x}\left[e^{-\lambda t}f\left(X_{t}\right)\right]+f(x)\leq
E_{x}\left[\int_{0}^{t}e^{-\lambda s}\phi(X_{s})ds+\sum_{j:\tau_{j}\leq
t}e^{-\lambda\tau_{j}}g(\xi_{j})\right].$

The linear growth of $f$ and (9) imply that $E\left[e^{-\lambda
t}f\left(X_{t}\right)\right]\to\infty$ as $t\to\infty$. So by letting
$t\to\infty$ another application of the Monotone Convergence Theorem on the
right-hand side gives

$f(x)\leq E_{x}\left[\int_{0}^{\infty}e^{-\lambda
s}\phi(X_{s})ds+\sum_{j=1}^{\infty}e^{-\lambda\tau_{j}}g(\xi_{j})\right]$

Equality holds $(T^{*},\Xi^{*})$ hence the proof is complete. $\bullet$

##  4 Analysis of the Optimal Control $(T^{*},\Xi^{*})$

We now set out to determine the transient and steady-state behavior of the
controlled cash on hand process $X_{t}$ under the optimal control
$(T^{*},\Xi^{*})$ assuming that $Y_{t}$ is a spectrally positive Levy process.
In other words,

$\displaystyle\int_{\Re_{-}}\nu(dy)=0$

and $Y_{t}$ is not a subordinator. The case of a spectrally negative Levy
process may be treated similarly. Our main result will be to determine the
Laplace transform (with respect to time) of the transition probabilities of
$X_{t}$ and also to determine the limiting distribution of $X_{t}$.

We begin by determining the Laplace transform of the transition probabilities
of $X_{t}$. Let $\mathcal{A}\in\mathcal{B}(\Re)$ be a Borel set of $\Re$ and
let $e_{q}$ be an exponential random variable with rate $q$ independent of
$X$. Now consider $P_{x}[X_{e_{q}}\in\mathcal{A}]$ for $x\in(a,b)$. It then
follows conditioning on the value of $e_{q}$ relative to the stopping time
$\tau^{*}_{1}$ that

$P_{x}[X_{e_{q}}\in{\cal A}]=E_{x}[1\\{X_{e_{q}}\in{\cal
A}\\}1\\{e_{q}<\tau^{*}_{1}\\}]+E_{x}[1\\{X_{e_{q}}\in{\cal
A}\\}1\\{e_{q}\geq\tau^{*}_{1}\\}].$ (18)

However, since $X_{t}=Y_{t}$ for $0\leq t<\tau^{*}_{1}$, it follows by the
memoryless property of the exponential distribution and the strong Markov
property that $E_{x}[1\\{X_{e_{q}}\in{\cal A}\\}1\\{e_{q}\geq\tau^{*}_{1}\\}]$
is equal to

$\displaystyle E_{\alpha}[1\\{X_{e_{q}}\in{\cal
A}\\}]P_{x}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\leq
a\\}]+E_{\beta}[1\\{X_{e_{q}}\in{\cal
A}\\}]P_{x}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq b\\}].$

Substituting the above into (18), we have

$\displaystyle E_{x}[1\\{X_{e_{q}}\in{\cal A}\\}]$ $\displaystyle=$
$\displaystyle E_{x}[1\\{X_{e_{q}}\in{\cal A}\\}1\\{e_{q}<\tau^{*}_{1}\\}]$
$\displaystyle+E_{\alpha}[1\\{X_{e_{q}}\in{\cal
A}\\}]P_{x}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\leq a\\}]$
$\displaystyle+E_{\beta}[1\\{X_{e_{q}}\in{\cal
A}\\}]P_{x}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq b\\}].$

It therefore remains to determine expressions for the three quantities on the
righthand side of (4). We proceed term by term.

First note that $\tau^{*}_{1}$ is equal to the fist time the Levy process
$Y_{t}$ exits the open interval $(a,b)$. We then have that in general one may
write

$\displaystyle E_{x}[1\\{X_{e_{q}}\in{\cal A}\\}1\\{e_{q}<\tau^{*}_{1}\\}]$
$\displaystyle=$ $\displaystyle E_{x}[1\\{Y_{e_{q}}\in{\cal
A}\\}1\\{e_{q}<\tau^{*}_{1}\\}]$ $\displaystyle=$ $\displaystyle
q\int_{0}^{\infty}e^{-qt}E_{x}[1\\{Y_{t}\in{\cal
A}\\}1\\{\tau^{*}_{1}>t\\}]dt$ $\displaystyle=$ $\displaystyle q\int_{{\cal
A}}\int_{0}^{\infty}e^{-qt}P_{x}[Y(t)\in dy,\tau^{*}_{1}>t]dt$
$\displaystyle=$ $\displaystyle q\int_{{\cal A}}U^{(q)}(x,dy),$

where $U^{{(q)}}$ is the $q$-potential measure of $Y_{t}$. This then provides
an expression for the first term on the righthand side of (4). Also note, by
Theorem 8.7 of [15], if $Y_{t}$ is spectrally positive then its $q$-potential
measure $U^{(q)}(x,dy)$ has a density $u^{(q)}(x,y)$ given by

$\displaystyle u^{(q)}(x,y)$ $\displaystyle=$ $\displaystyle
W^{(q)}(b-x)\frac{W^{(q)}(y-a)}{W^{(q)}(b-a)}-W^{(q)}(y-x),$ (21)

where

$\displaystyle\int_{0}^{\infty}e^{-sy}W^{(q)}(y)dy$ $\displaystyle=$
$\displaystyle\frac{1}{\psi(-s)-q},$ (22)

whenever $-s$ is large enough so that $\psi(-s)>q$ and $\psi(s)=\log
E_{0}[e^{sY_{1}}]$ is the Laplace exponent of $Y_{t}$. Note that
$\psi(s)<\infty$ for all $s\leq 0$ by the spectral positivity of $Y$.

Next note that setting $x=\alpha$ in (4), we obtain

$\displaystyle E_{\alpha}[1\\{X_{e_{q}}\in{\cal
A}\\}](1-P_{\alpha}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\leq
a\\}))$ $\displaystyle-E_{\beta}[1\\{X_{e_{q}}\in{\cal
A}\\}]P_{\alpha}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq b\\}]$
$\displaystyle=$ $\displaystyle E_{\alpha}[1\\{X_{e_{q}}\in{\cal
A}\\}1\\{e_{q}<\tau^{*}_{1}\\}],$

and similarly, setting $x=\beta$, we have

$\displaystyle E_{\beta}[1\\{X_{e_{q}}\in{\cal
A}\\}](1-P_{\beta}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq
b\\}])$ $\displaystyle-E_{\alpha}[1\\{X_{e_{q}}\in{\cal
A}\\}]P_{\beta}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\leq a\\}]$
$\displaystyle=$ $\displaystyle E_{\beta}[1\\{X_{e_{q}}\in{\cal
A}\\}1\\{e_{q}<\tau^{*}_{1}\\}].$

(4) and (4) constitute a set of linear equations for
$E_{\alpha}[1\\{X_{e_{q}}\in{\cal A}\\}]$ and $E_{\beta}[1\\{X_{e_{q}}\in{\cal
A}\\}]$. Moreover, so long as $q>0$, we have that

$\displaystyle(1-P_{\beta}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq
b\\}])(1-P_{\alpha}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y_{\tau^{*}_{1}}\leq
a\\}])$ $\displaystyle>$ $\displaystyle
P_{\alpha}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq
b\\}]P_{\beta}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y_{\tau^{*}_{1}}\leq a\\}]$

and so the determinant associated with (4) and (4) is non-zero and hence a
solution exists. Solving for $E_{\alpha}[1\\{X_{e_{q}}\in{\cal A}\\}]$ and
$E_{\beta}[1\\{X_{e_{q}}\in{\cal A}\\}]$ then yields that
$E_{\alpha}[1\\{X_{e_{q}}\in{\cal A}\\}]$ is given by

$\displaystyle{1\over
C_{a,\alpha,\beta,b}}\times\left(\begin{array}[]{c}E_{\alpha}[1\\{X_{e_{q}}\in{\cal
A}\\}1\\{e_{q}<\tau^{*}_{1}\\}](1-P_{\beta}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq
b\\}]\\\ \ +E_{\beta}[1\\{X_{e_{q}}\in{\cal
A}\\}1\\{e_{q}<\tau^{*}_{1}\\}]P_{\alpha}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq
b\\}]\end{array}\right)$ (27)

and $E_{\beta}[1\\{X_{e_{q}}\in{\cal A}\\}]$ is given by

$\displaystyle{1\over
C_{a,\alpha,\beta,b}}\times\left(\begin{array}[]{c}E_{\alpha}[1\\{X_{e_{q}}\in{\cal
A}\\}1\\{e_{q}<\tau^{*}_{1}\\}]P_{\beta}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y_{\tau^{*}_{1}}\leq
a\\}]\\\ +E_{\beta}[1\\{X_{e_{q}}\in{\cal
A}\\}1\\{e_{q}<\tau^{*}_{1}\\}](1-P_{\alpha}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y_{\tau^{*}_{1}}\leq
a\\}])\end{array}\right),$ (30)

where

$\displaystyle C_{a,\alpha,\beta,b}$ $\displaystyle=$
$\displaystyle(1-P_{\beta}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq
b\\}])(1-P_{\alpha}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y_{\tau^{*}_{1}}\leq
a\\}])$ $\displaystyle-
P_{\alpha}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq
b\\}]P_{\beta}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y_{\tau^{*}_{1}}\leq a\\}].$

We now proceed to compute the terms appearing on the right-hand sides of (27)
and (30).

Firs note that the expressions of the form $E_{x}[1\\{X_{e_{q}}\in{\cal
A}\\}1\\{e_{q}<\tau^{*}_{1}\\}]$ appearing in (27) and (30) have already been
determined by (4). It remains to determine expressions for the probabilities
appearing in (27) and (30). However, note that

$\displaystyle P_{x}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y_{\tau^{*}_{1}}\leq
a\\}]$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}qe^{-qt}P_{x}[\\{t\geq\tau^{*}_{1}\\}\cap\\{Y_{\tau^{*}_{1}}\leq
a\\}]dt$ $\displaystyle=$ $\displaystyle
E_{x}\left[\int_{\tau^{*}_{1}}^{\infty}qe^{-qt}dt1\\{Y_{\tau^{*}_{1}}\leq
a\\}\right]$ $\displaystyle=$ $\displaystyle
E_{x}\left[e^{-q\tau^{*}_{1}}1\\{Y_{\tau^{*}_{1}}\leq a\\}\right]$
$\displaystyle=$ $\displaystyle
Z^{(q)}(b-x)-Z^{(q)}(b-a)\frac{W^{(q)}(b-x)}{W^{(q)}(b-a)},$

where the final equality follows from Theorem 8.1 of [15] and we have the
relationship

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

In a similar fashion, using Theorem 8.1 of [15], one may compute

$\displaystyle P_{x}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq
b\\}]$ $\displaystyle=$ $\displaystyle\frac{W^{(q)}(b-x)}{W^{(q)}(b-a)}.$

Substituting the above results into (27) and (30), and subsequently into (4),
one obtains an expression for the Laplace transform of the transition
probabilities of $X_{t}$.

We now proceed towards obtaining an expression for the limiting distribution
of $X_{t}$ as $t\rightarrow\infty$. Our main result in this regard will be the
following. Recall the definition of $U^{(q)}$ as the $q$-potential measure
associated with $Y$.

######  Proposition 4.1.

If $Y_{t}$ is spectrally positive, then under a double bandwidth control
policy $(a,\alpha,\beta,b)$, the limiting distribution of $X_{t}$ is given by

$\displaystyle\pi(\mathcal{A})$ $\displaystyle=$
$\displaystyle\frac{1}{K_{a,\alpha,\beta,b}}\left(\left(1-\frac{W^{(0)}(b-\beta)}{W^{(0)}(b-a)}\right)\int_{\mathcal{A}}U^{(0)}(\alpha,dy)+\left(\frac{W^{(0)}(b-\alpha)}{W^{(0)}(b-a)}\right)\int_{\mathcal{A}}U^{(0)}(\beta,dy)\right),$

for each $\mathcal{A}\in\mathcal{B}$, where $K_{a,\alpha,\beta,b}$ is an
appropriate normalizing constant, as given by (34).

We prove Proposition 4.1 in a series of lemmas. First note that by the strong
Markov property, $X_{t}$ is a regenerative process with possible regeneration
points either $\alpha$ or $\beta$. Let us consider the point $\alpha$ and
define $n^{\star}_{\alpha}=\inf\\{n\geq 1:X(\tau^{\star}_{n})=\alpha\\}$. By
the standard theory of regenerative processes, see for instance Theorem 1.2 of
Chapter VI of [1], if we may show that
$E_{\alpha}[\tau_{n^{*}_{\alpha}}]<\infty$ and that $\tau_{n^{*}_{\alpha}}$ is
nonlattice, then $\lim_{t\to\infty}P_{x}(X_{t}\in\cal{A})=\pi({\cal A})$
exists for all ${\cal A}\in{\cal B}(\Re)$ and $x\in\Re$ and is given by

$\displaystyle\pi({\cal A})$ $\displaystyle=$
$\displaystyle\frac{E_{\alpha}[\int_{0}^{\tau_{n^{*}_{\alpha}}}1\\{X_{s}\in{\cal
A}\\}ds]}{E_{\alpha}[\tau_{n^{*}_{\alpha}}]}.$

The following lemma now shows that $E_{\alpha}[\tau_{n^{*}_{\alpha}}]<\infty$.

######  Lemma 4.2.

If the Levy process $Y_{t}$ is spectrally positive, then
$E_{\alpha}[\tau_{n^{*}_{\alpha}}]<\infty$.

Proof: Note first that

$\displaystyle\tau_{n^{\star}_{\alpha}}$ $\displaystyle=$
$\displaystyle\tau^{\star}_{1}1\\{Y(\tau^{*}_{1})\leq
a\\}+\tau_{n^{\star}_{\alpha}}1\\{Y(\tau^{*}_{1})\geq b\\}$ $\displaystyle=$
$\displaystyle\tau^{\star}_{1}+(\tau_{n^{\star}_{\alpha}}-\tau^{*}_{1})1\\{Y(\tau^{*}_{1})\geq
b\\}.$

Hence, by the strong Markov property,

$\displaystyle E_{\alpha}[\tau_{n^{\star}_{\alpha}}]$ $\displaystyle=$
$\displaystyle
E_{\alpha}[\tau^{\star}_{1}]+E_{\alpha}[(\tau_{n^{\star}_{\alpha}}-\tau^{*}_{1})1\\{Y(\tau^{*}_{1})\geq
b\\}]$ $\displaystyle=$ $\displaystyle
E_{\alpha}[\tau^{\star}_{1}]+E_{\beta}[\tau_{n^{\star}_{\alpha}}]P_{\alpha}[Y(\tau^{*}_{1})\geq
b].$

Similarly, we may show

$\displaystyle E_{\beta}[\tau_{n^{\star}_{\alpha}}]$ $\displaystyle=$
$\displaystyle
E_{\beta}[\tau^{\star}_{1}]+E_{\beta}[\tau_{n^{\star}_{\alpha}}]P_{\beta}[Y(\tau^{*}_{1})\geq
b],$

from which we obtain

$\displaystyle E_{\beta}[\tau_{n^{\star}_{\alpha}}]$ $\displaystyle=$
$\displaystyle\frac{E_{\beta}[\tau^{\star}_{1}]}{1-P_{\beta}[Y(\tau^{*}_{1})\geq
b]}.$

Now note that since $Y_{t}$ is spectrally positive, we have by Theorem 8.1 of
[15] that

$\displaystyle P_{\beta}[Y(\tau^{*}_{1})\geq b]$ $\displaystyle=$
$\displaystyle\frac{W^{0}(b-\beta)}{W^{0}(b-a)}~{}<~{}1$

and so it suffices from the above to show
$E_{\alpha}[\tau^{\star}_{1}],E_{\beta}[\tau^{\star}_{1}]<\infty$.

We now show that in general for $x\in(a,b)$, $E_{x}[\tau^{\star}_{1}]<\infty$.
Recall by [15], the potential measure of $Y_{t}$ upon exiting $[a,b]$ is given
by

$\displaystyle U(x,dy)=\int_{0}^{\infty}P_{x}[Y_{t}\in
dy,\tau^{\star}_{1}>t]dt.$

Integrating over $[a,b]$, we obtain that

$\displaystyle\int_{[a,b]}U(x,dy)$ $\displaystyle=$
$\displaystyle\int_{[a,b]}\int_{0}^{\infty}P_{x}[Y_{t}\in
dy,\tau^{\star}_{1}>t]dt$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\int_{[a,b]}P_{x}[Y_{t}\in
dy,\tau^{\star}_{1}>t]dt$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}P_{x}[\tau^{\star}_{1}>t]dt$ $\displaystyle=$
$\displaystyle E_{x}[\tau^{\star}_{1}].$

However, by Theorem 8.7 of [15], since $Y_{t}$ is spectrally positive,
$U(x,dy)$ has a density given by

$\displaystyle u(x,y)$ $\displaystyle=$ $\displaystyle
W(b-x)\frac{W(y-a)}{W(b-a)}-W(y-x).$

Integrating over $[a,b]$, we therefore find that

$\displaystyle\int_{[a,b]}U(x,dy)$ $\displaystyle=$
$\displaystyle\int_{[a,b]}\left(W(b-x)\frac{W(y-a)}{W(b-a)}-W(y-x)\right)dy$
$\displaystyle<$ $\displaystyle\infty,$

where the inequality follows since $W$ is bounded on compact sets. By the
above, this completes the proof. $\bullet$

The following lemma allows us to take the limit as $q\rightarrow 0$ in (27)
and (30) in order to obtain the limiting distribution of $X_{t}$.

######  Lemma 4.3.

For each $x\in\Re$,

$\lim_{q\to 0}P_{x}[X(e_{q})\in\cal{A}]=\pi({\cal A})$ (31)

Proof: Select $T>0$ large enough so that
$\left|P_{x}[X_{t}\in\cal{A}]-\pi({\cal A})\right|<\epsilon$. Then

$\left|P_{x}[X(e_{q})\in\cal{A}]-\pi({\cal
A})\right|=\left|\int_{0}^{\infty}\left\\{P_{x}[X_{t}\in\cal{A}]-\pi({\cal
A})\right\\}qe^{-qt}dt\right|$
$\leq\left|\int_{0}^{T}\left\\{P_{x}[X_{t}\in\cal{A}]-\pi({\cal
A})\right\\}qe^{-qt}dt\right|+\left|\int_{T}^{\infty}\left\\{P_{x}[X_{t}\in\cal{A}]-\pi({\cal
A})\right\\}qe^{-qt}dt\right|.$

The second term in the above expression is bounded by $\epsilon$ and the first
term converges to zero as $q\to 0$ by the Dominated Convergence Theorem, which
completes the proof. $\bullet$

We now provide the proof of Proposition 4.1.

Proof of Proposition 4.1: Using Lemma 4.3, we now wish to take limits
$q\rightarrow 0$ in (27) in order to determine the limiting distribution
$\delta$. However, both the numerator and denominator in (27) converge to 0 as
$q\rightarrow 0$ and so we must apply L’Hoptial’s rule. Before doing so,
however, we first must verify that both the numerator and denominator in (27)
are differentiable.

By Lemma 8.3 and Corollary 8.5 in [15] we have that for each $x>0$, both
$W^{q}(x)$ and $Z^{(q)}(x)$ are differentiable in $q$. Moreover, since

$\displaystyle\frac{W^{(q)}(b-x)}{W^{(q)}(b-a)}$ $\displaystyle=$
$\displaystyle E_{x}[e^{-q\tau^{*}_{1}}1\\{Y_{\tau^{*}_{1}}\geq b\\}],$

for $a\leq x<b$, it follows that

$\displaystyle\frac{d}{dq}\frac{W^{(q)}(b-x)}{W^{(q)}(b-a)}$ $\displaystyle=$
$\displaystyle-E_{x}[\tau^{*}_{1}e^{-q\tau^{*}_{1}}1\\{Y_{\tau^{*}_{1}}\geq
b\\}]$ $\displaystyle<$ $\displaystyle\infty,$

where the inequality follows as in the proof of Lemma 4.2. Finally, since for
each $a\leq x\leq b$, $U^{q}(x)$ has a density $u^{(q)}(x,y)$ given by (21) it
follows that for each $\mathcal{A}\in\mathcal{B}(\Re)$,

$\displaystyle\frac{d}{dq}\int_{\mathcal{A}}U^{(q)}(x,dy)$ $\displaystyle=$
$\displaystyle\int_{\mathcal{A}}\frac{d}{dq}u^{(q)}(x,y)dy.$

Thus, noting that

$\displaystyle(E_{\alpha}[1\\{X_{e_{q}}\in{\cal
A}\\}1\\{e_{q}<\tau^{*}_{1}\\}](1-P_{\beta}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq
b\\}])$ $\displaystyle+E_{\beta}[1\\{X_{e_{q}}\in{\cal
A}\\}1\\{e_{q}<\tau^{*}_{1}\\}]P_{\alpha}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq
b\\}])$ $\displaystyle=$ $\displaystyle
q\int_{\mathcal{A}}U^{(q)}(\alpha,dy)\left(1-\frac{W^{(q)}(b-\beta)}{W^{(q)}(b-a)}\right)+q\int_{\mathcal{A}}U^{(q)}(\beta,dy)\left(\frac{W^{(q)}(b-\alpha)}{W^{(q)}(b-a)}\right)$

and

$\displaystyle((1-P_{\beta}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq
b\\}])(1-P_{\alpha}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y_{\tau^{*}_{1}}\leq
a\\}])$ $\displaystyle-
P_{\alpha}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y(\tau^{*}_{1})\geq
b\\}]P_{\beta}[\\{e_{q}\geq\tau^{*}_{1}\\}\cap\\{Y_{\tau^{*}_{1}}\leq a\\}])$
$\displaystyle=$
$\displaystyle\left(1-\frac{W^{(q)}(b-\beta)}{W^{(q)}(b-a)}\right)\left(1-\left(Z^{(q)}(b-\alpha)-Z^{(q)}(b-a)\frac{W^{q}(b-\alpha)}{W^{q}(b-a)}\right)\right)$
$\displaystyle-\left(Z^{(q)}(b-\beta)-Z^{(q)}(b-a)\frac{W^{q}(b-\beta)}{W^{q}(b-a)}\right)\frac{W^{(q)}(b-\alpha)}{W^{(q)}(b-a)},$

we see that both the numerator and denominator in (27) are differentiable.

Let us now take derivatives on the righthand sides of (4) and (4).

Taking the derivative of the right hand side of (4) and evaluating at $q=0$ we
obtain

$\displaystyle\int_{\mathcal{A}}U^{(0)}(\alpha,dy)\left(1-\frac{W^{(0)}(b-\beta)}{W^{(0)}(b-a)}\right)+\int_{\mathcal{A}}U^{(0)}(\beta,dy)\left(\frac{W^{(0)}(b-\alpha)}{W^{(0)}(b-a)}\right).$

Next, recalling that

$\displaystyle Z^{q}(x)$ $\displaystyle=$ $\displaystyle
1+q\int_{0}^{x}W^{(q)}(y)dy,$

it follows upon taking the derivative of the righthand side of (4) and
evaluating at $q=0$ that we obtain

$\displaystyle\frac{d}{dq}\left(\left(1-\frac{W^{(q)}(b-\beta)}{W^{(q)}(b-a)}\right)\left(1-\left(Z^{(q)}(b-\alpha)-Z^{(q)}(b-a)\frac{W^{q}(b-\alpha)}{W^{q}(b-a)}\right)\right)\right)$
(34)
$\displaystyle\left.-\left(\left(Z^{(q)}(b-\beta)-Z^{(q)}(b-a)\frac{W^{q}(b-\beta)}{W^{q}(b-a)}\right)\frac{W^{(q)}(b-\alpha)}{W^{(q)}(b-a)}\right)\right)$
$\displaystyle=$
$\displaystyle\frac{W^{(0)}(b-\beta)}{W^{(0)}(b-a)}\int_{0}^{b-\alpha}W^{0}(x)dx+\frac{W^{(0)}(b-\alpha)}{W^{(0)}(b-a)}\int_{b-\beta}^{b-a}W^{0}(x)dx-\int_{0}^{b-\alpha}W^{0}(x)dx$
$\displaystyle=$ $\displaystyle K_{a,\alpha,\beta,b}.$

Thus, by (27) and Lemma 4.2 we obtain the desired result. $\bullet$

##  5 An Example

We suppose in this section that

$Y_{t}=x+\sigma w_{t}+N_{t},$

where $N$ is a compound Poisson process independent of $w$ such that the rate
of jump arrivals is equal to 1 and the Levy measure $\nu$ of $N$ is

$\nu(dy)=\theta e^{-\theta y}dy,\quad y\geq 0$

for some $\theta>0$, and $\nu((-\infty,0])=0$. In this case, we can write the
linear operator ${\cal A}$ in the form

${\cal A}f(x)={\sigma^{2}\over
2}f^{\prime\prime}(x)+\int_{0}^{\infty}\left[f(x+y)-f(x)\right]\nu(dy).$ (35)

We also specify the opportunity cost function as

$\phi(x)=(x-\rho)^{2}$

where $\rho$ is a fixed target value. Suppose now that $x\in(a,b)$. Using
(35), the equation in (i) in Theorem 3.1 may be written as

${1\over 2}\sigma^{2}f^{\prime\prime}(x)+(x-\rho)^{2}-\lambda
f(x)+\int_{0}^{\infty}\left[f(x+y)-f(x)\right]\theta e^{-\theta y}dy=0.$

This becomes

${1\over 2}\sigma^{2}f^{\prime\prime}(x)+(x-\rho)^{2}-(1+\lambda)f(x)+\theta
e^{\theta x}\int_{x}^{b}f(z)e^{-\theta z}dz+$ $e^{\theta
x}\int_{b}^{\infty}\left[f(b)+d(z-b)\right]\theta e^{-\theta z}dz=0,$ (36)

and also

${1\over 2}\sigma^{2}e^{-\theta x}f^{\prime\prime}(x)+e^{-\theta
x}(x-\rho)^{2}-(1+\lambda)e^{-\theta x}f(x)+\theta\int_{x}^{b}f(z)e^{-\theta
z}dz+\zeta=0,$ (37)

where

$\zeta=\int_{b}^{\infty}\left[f(b)+d(z-b)\right]\theta e^{-\theta z}dz.$

Let us now introduce $e^{-\theta x}f(x)=g(x)$. We then obtain the following
equation from (37):

$({1\over 2}\sigma^{2}\theta^{2}-\lambda-1)g(x)+\sigma^{2}\theta
g^{\prime}(x)+{1\over 2}\sigma^{2}g^{\prime\prime}(x)+e^{-\theta
x}(x-\rho)^{2}+\theta\int_{x}^{b}g(z)dz+\zeta=0.$ (38)

Differentiating the above with respect to $x$ we get the following
inhomogeneous linear ordinary differential equation of the third order:

${1\over 2}\sigma^{2}g^{\prime\prime\prime}+\sigma^{2}\theta
g^{\prime\prime}+\left({1\over
2}\sigma^{2}\theta^{2}-\lambda-1\right)g^{\prime}-\theta g+2e^{-\theta
x}(x-\rho)-\theta e^{-\theta x}(x-\rho)^{2}=0.$ (39)

A particular solution for the inhomogeneous equation, denoted by $g_{p}$, is
given by

$g_{p}(x)=e^{-\theta x}\left[K_{1}(x-\rho)^{2}+K_{2}(x-\rho)+K_{3}\right],$

where

$K_{1}={1\over\lambda}\quad,~{}K_{2}={2\over\theta\lambda^{2}}\quad,~{}K_{3}={1\over\lambda^{3}\theta^{2}}\left[2\lambda+2+\theta^{2}\lambda\sigma^{2}\right].$

The general solution of the homogeneous equation is given by $g_{h}$, that is,

$g_{h}(x)=L_{1}e^{c_{1}x}+L_{2}e^{c_{2}x}+L_{3}e^{c_{3}x}$

where $c_{1},c_{2},c_{3}$ are the roots of the equation

$P(x)={1\over 2}\sigma^{2}x^{3}+\sigma^{2}\theta x^{2}+\left({1\over
2}\sigma^{2}\theta^{2}-\lambda-1\right)x-\theta=0$

and $L_{1},L_{2},L_{3}$ are “free parameters”. Notice that $P(0)=-\theta<0$,
$P(-\theta)=\theta\lambda>0$, and $\lim_{x\to-\infty}P(x)=-\infty$,
$\lim_{x\to\infty}P(x)=\infty$ thus $P(x)$ has three roots, say
$c_{1}<-\theta$, $-\theta<c_{2}<0$ and $c_{3}>0$.

We have now arrived at the following family of candidate solutions:

$g(x;L_{1},L_{2},L_{3},b)=e^{-\theta
x}\left[K_{1}(x-\rho)^{2}+K_{2}(x-\rho)+K_{3}\right]+L_{1}e^{c_{1}x}+L_{2}e^{c_{2}x}+L_{3}e^{c_{3}x}.$

This gives

$f(x;L_{1},L_{2},L_{3},b)=K_{1}(x-\rho)^{2}+K_{2}(x-\rho)+K_{3}+L_{1}e^{(\theta+c_{1})x}+L_{2}e^{(\theta+c_{2})x}+L_{3}e^{(\theta+c_{3})x}.$
(40)

For simplicity we shall use the notation $f(x;L_{1},L_{2},L_{3},b)=f(x)$.

We now have 7 unknown parameters $a,\alpha,\beta,b,L_{1},L_{2},L_{3}$. From
the conditions of Theorem 3.1, we may derive the following 6 equations for
these constants:

$f^{\prime}(a)=-c$ (41) $f^{\prime}(\alpha)=-c$ (42) $f^{\prime}(b)=d$ (43)
$f^{\prime}(\beta)=d$ (44) $f(a)=f(\alpha)+C+c(\alpha-a)$ (45)
$f(b)=f(\beta)+D+d(b-\beta).$ (46)

In addition, if we trace back our derivation in the above, then we see that we
must have (36) hold for at least for one particular $x$ since in going from
(38) to (39) we took a derivative. Select $x=b$. This then gives us our $7$th
equation

${1\over
2}\sigma^{2}f^{\prime\prime}(b)+(b-\rho)^{2}-(1+\lambda)f(b)+e^{\theta
b}\zeta=0.$ (47)

We now have the following.

######  Theorem 5.1.

Suppose that there exist seven constants $L_{1}\leq 0,L_{2}\leq 0,L_{3}\leq
0,a<\alpha\leq\beta<b$ satisfying the seven equations (41)-(47). We define $h$
by

$h(x)=\begin{cases}f(a)-c(x-a),&\textrm{if~{}}x\leq a,\\\
f(x),&\textrm{if~{}}a\leq x\leq b\\\ f(b)+d(x-b),&\textrm{if~{}}x\geq
b.\end{cases}$

Then $h(x)=V(x)$, i.e., $h(x)$ is the value function of the optimization
problem. Furthermore, the policy $(T^{*},\Xi^{*})$ described in (7) and (8)
with this choice of $a,\alpha,\beta,b$ is optimal.

In order to prove this theorem, we need the following lemma.

######  Lemma 5.1.

Assume the conditions of Theorem 5.1. Then there exists a constant
$\xi\in(\alpha,\beta)$ such that $h^{\prime}$ is convex on $[a,\xi]$, concave
on $[\xi,b]$. Furthermore $h^{\prime}(x)\leq-c$ if $x\in[a,\alpha]$,
$h^{\prime}(x)\geq d$ if $x\in[\beta,b]$, and $-c\leq h^{\prime}(x)\leq d$ if
$x\in[\alpha,\beta]$.

Proof: From the condition that $L_{1},L_{2},L_{3}\leq 0$ it follows that
$h^{\prime\prime\prime}(x)$ is decreasing on $(a,b)$. Therefore
$h^{\prime\prime}(x)$ has at most two zero points, which implies that
$h^{\prime}(x)$ has at most two local extreme values in $(a,b)$. By (41)-(44)
the derivative function $h^{\prime}$ must be first decreasing then increasing
then again decreasing on $[a,b]$. Since $h^{\prime\prime}$ is concave, it must
be either increasing, or decreasing, or first increasing then decreasing on
$[a,b]$. But the first two possibilities are not possible, since $h^{\prime}$
can not be neither convex nor concave on $[a,b]$. Therefore it must be first
convex then concave.

Proof of Theorem 5.1: We need to prove that the conditions of Theorem 3.1 are
satisfied. Condition (i) and the required smoothness of $h$ follows from our
construction. Next we prove (ii) and (iv). From conditions (42), (44) and
Lemma 5.1 it follows that

$Mh(x)=\begin{cases}h(\alpha)+C+c(\alpha-x),&\textrm{if~{}}a\leq
x\leq\alpha,\\\ h(x)+\hbox{min}\\{C,D\\},&\textrm{if~{}}\alpha<x<\beta,\\\
h(\beta)+D+d(x-\beta),&\textrm{if~{}}\beta\leq x\leq b.\end{cases}$

Condition (ii) follows from (45) and (46). We show condition (iv) for the case
of $x\in[a,\alpha]$, the case of $x\in[\beta,b]$ is similar and the case of
$x\in(\alpha,\beta)$ is obvious. We need to show that $0\leq
h(\alpha)-h(x)+C+c(\alpha-x)$. For $x=a$ we have equality by (45), and the
derivative of the right-hand side of the inequality with respect to $x$ is
$-h^{\prime}(x)-c$, which is non-negative on $[a,\alpha]$ by Lemma 5.1. Next
we show condition (iii). First we look at the case of $x>b$. Let

$K(x)={\cal A}h(x)-\lambda h(x)+(x-\rho)^{2};\quad x\in\Re\setminus\\{a,b\\}.$

Then

$K(x)={\sigma^{2}\over
2}h^{\prime\prime}(x)+\int_{0}^{\infty}\left[h(x+y)-h(x)\right]\nu(dy)-\lambda
h(x)+(x-\rho)^{2}=0,\quad x\in(a,b)$ (48)

and

$K(x)=\int_{0}^{\infty}\left[h(x+y)-h(x)\right]\nu(dy)-\lambda
h(x)+(x-\rho)^{2},\quad x\in(-\infty,a)\cup(b,\infty).$ (49)

Therefore $0=K(b-)={\sigma^{2}\over 2}h^{\prime\prime}(b-)+K(b+)$, and
${\sigma^{2}\over 2}h^{\prime\prime}(b-)\leq 0$ implies $K(b+)\geq 0$. On the
other hand for $x>b$ we have $K^{\prime}(x)=-\lambda d+2(x-\rho)$ and
$K^{\prime}(b+)=-\lambda d+2(b-\rho)$. We also have by (48)

$0=K^{\prime}(x)={\sigma^{2}\over
2}h^{\prime\prime\prime}(x)+\int_{0}^{\infty}\left[h^{\prime}(x+y)-h^{\prime}(x)\right]\nu(dy)-\lambda
h^{\prime}(x)+2(x-\rho),\quad x\in(a,b)$ (50)

hence

$0=K^{\prime}(b-)={\sigma^{2}\over 2}h^{\prime\prime\prime}(b-)-\lambda
d+2(b-\rho)={\sigma^{2}\over 2}h^{\prime\prime\prime}(b-)+K^{\prime}(b+).$

Since $h^{\prime\prime\prime}(b-)\leq 0$ so we must have $K^{\prime}(b+)\geq
0$. But $K^{\prime}$ is increasing on $(b,\infty)$, thus $K^{\prime}(x)\geq 0$
whenever $x\in(b,\infty)$. This in turn implies that $K$ is increasing on
$(b,\infty)$, thus $K(x)\geq 0$ for $x\in(b,\infty)$.

Next we show that $K(x)\geq 0$ for $x<a$. By (48) and (49)
$0=K(a+)=K(a-)+{\sigma^{2}\over 2}h^{\prime\prime}(a+)$ which implies that
$K(a-)\geq 0$. Hence all we need to show is that $K^{\prime}(x)\leq 0$ for
$x\leq a$. Differentiating (49) we get

$K^{\prime}(x)=\int_{0}^{\infty}\left[h^{\prime}(x+y)-h^{\prime}(x)\right]\nu(dy)-\lambda
h^{\prime}(x)+2(x-\rho),\quad x<a,$

and with a change of variable in the integral one can see that

$K^{\prime}(x)=e^{\theta(x-a)}\int_{0}^{\infty}[h^{\prime}(a+z)+c]\nu(dz)+\lambda
c+2(x-\rho),\quad x<a$ (51)

and

$K^{\prime\prime}(x)=\theta
e^{\theta(x-a)}\int_{0}^{\infty}[h^{\prime}(a+z)+c]\nu(dz)+2,\quad x<a.$ (52)

Thus $K^{\prime\prime}$ is either increasing or decreasing on $(-\infty,a)$
depending on the sign of $\int_{0}^{\infty}[h^{\prime}(a+z)+c]\nu(dz)$ which
makes $K^{\prime}$ either convex or concave on $(-\infty,a)$. However, the
fact that $\lim_{x\to-\infty}K^{\prime}(x)=-\infty$ implies that $K^{\prime}$
must be concave and $K^{\prime\prime}$ decreasing on $(-\infty,a)$. In
addition, the integral in (52) and in (51) is negative. By (50) and (51)

$0=K^{\prime}(a+)={\sigma^{2}\over
2}h^{\prime\prime\prime}(a+)+K^{\prime}(a-)$

thus $K^{\prime}(a-)\leq 0$ follows from $h^{\prime\prime\prime}(a+)\geq 0$.
Formula (51) imply that $K^{\prime}$ has at most one zero-point on
$(-\infty,a)$. These facts about $K^{\prime}$ imply that indeed
$K^{\prime}(x)\leq 0$. $\bullet$

We conclude this section by providing an example showing how the limiting
distribution $\pi$ of Proposition 4.1 may be calculated for an arbitrary
double bandwidth control policy $(a,\alpha,\beta,b)$. We will assume that
$Y_{t}=\vartheta t+N_{t}$, where $\vartheta<0$ and $N_{t}$ is a compound
Poisson process which has jumps at rate one and jump sizes which are
exponentially distributed with rate $\theta$. Note that by (21) and
Proposition 4.1, it suffices to determine the function
$W^{(0)}=\lim_{q\rightarrow 0}W^{(q)}$. By (22), the Laplace transform of
$W^{(0)}$ is given by $1/\psi(-s)$ where $\psi(s)$ is the Levy exponent of
$Y_{t}$. Morever, by (8.1) in [15] it then follows that

$\displaystyle\psi(s)$ $\displaystyle=$ $\displaystyle\vartheta
s-\int_{0}^{\infty}(1-e^{xs})\theta e^{-\theta x}dx,$

for $s<\theta$, which reduces to $\psi(s)=\vartheta s+s(\theta-s)^{-1}$. One
may now proceed to verify that

$\displaystyle\frac{1}{\psi(-s)}$ $\displaystyle=$
$\displaystyle\frac{-\theta}{s(\vartheta
s+\vartheta\theta+1)}-\frac{1}{\vartheta s+\vartheta\theta+1}.$

In the case in which $\theta\vartheta\neq-1$ inverting each of the terms in
the above, one obtains that the function $W^{(0)}$ is given by

$\displaystyle W^{(0)}(x)$ $\displaystyle=$
$\displaystyle\frac{-\theta}{\theta\vartheta+1}-\left(\frac{1}{\vartheta}-\frac{\theta}{\theta\vartheta+1}\right)\exp\left(-\frac{\theta\vartheta+1}{\vartheta}x\right).$

For the case in which $\theta\vartheta=1$, one has that
$W^{(0)}(x)=\vartheta^{-1}(\theta x+1)$. Substituting into the formula of
Proposition 4.1, one may now obtain the density of $\pi$.

##  6 Acknowledgements

The authors would like to thank Bert Zwart for his help with Section 4.

##  Appendix A Appenidx

In the Appendix, we provide a proof of the fact that Ito’s rule applies to
test functions $f\in{\cal D}$. For $f\in{\cal D}$ the second derivative
$f^{\prime\prime}(x)$ may not exist in points $S=\\{x_{1},\dots,x_{n}\\}$. We
shall call $S$ the set of exceptional points. We extend $f^{\prime\prime}$ to
the entire of $\Re$ by assuming an arbitrary value for
$f^{\prime\prime}(x_{i})$. This convention will be used in the rest of this
section. The following is then the main result.

######  Proposition A.1.

If $f\in{\cal D}$ and $X$ is a controlled cash on hand process with an
arbitrary impulse control $(T,\Xi)=(\tau_{1},\tau_{2},\dots,\tau_{n},\dots,\
\xi_{1},\xi_{2},\dots,\xi_{n},\dots,)$ then Ito’s rule holds in its usual
form:

$f(X_{t})-f(X_{0})=\int_{(0,t]}f^{\prime}(X_{s-})dX_{s}+{\sigma^{2}\over
2}\int_{(0,t]}f^{\prime\prime}(X_{s})ds+$ $\sum_{0<s\leq
t}\left\\{f(X_{s})-f(X_{s-})-f^{\prime}(X_{s-})\Delta X_{s}\right\\}.$ (53)

In order to prove Proposition A.1, we need the following two lemmas.

######  Lemma A.2.

Let $f\in{\cal D}$ with set of exceptional points $S$. Then there exists a
sequence $\left(f_{n}\right)_{n\geq 1}\subset C^{2}(\Re)$ such that the
following hold; (i) $f_{n}(x)\to f(x)$ and $f^{\prime}_{n}(x)\to
f^{\prime}(x)$ for every $x\in\Re$ as $n\to\infty$; (ii)
$f^{\prime\prime}_{n}(x)\to f^{\prime\prime}(x)$ for every $x\in\Re\setminus
S$ as $n\to\infty$; (iii) $f_{n}^{\prime}$ and $f_{n}^{\prime\prime}$ are
bounded uniformly in $n$, i.e., $|f_{n}^{\prime}(x)|\leq C_{1}$ and
$|f_{n}^{\prime\prime}(x)|\leq C_{1}$ for some constant $C_{1}$ and all $n$
and $x\in\Re$.

The proof of this lemma can be based on the proof of a similar lemma in
Økesendal [16], Appendix D with some obvious modifications.

######  Lemma A.3.

If $\sigma\not=0$ then for every $x\in\Re$

$\int_{0}^{\infty}1_{\\{x\\}}(X_{s})ds=0$

In other words, the Lebesgue measure of the time the controlled cash on hand
process spends at level $x$ is zero.

Proof:

$\displaystyle E\left[\int_{0}^{\infty}1_{\\{x\\}}(X_{s})ds\right]$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}P[X_{s}=x]ds$
$\displaystyle\leq$ $\displaystyle\int_{0}^{\infty}P[\tau_{i}=s\ \hbox{for
some}\ i]ds+\int_{0}^{\infty}P[X_{s}=x,\ \tau_{i}\not=s\ \hbox{for all}\
i]ds.$

We deal with these last two integrals separately.

$\int_{0}^{\infty}P[\tau_{i}=s\ \hbox{for some}\
i]ds\leq\int_{0}^{\infty}\sum_{i=1}^{\infty}P[\tau_{i}=s]ds=\sum_{i=1}^{\infty}\int_{0}^{\infty}P[\tau_{i}=s]ds$

and this last expression is zero because the set $\left\\{s\geq 0:\
P[\tau_{i}=s]>0\right\\}$ is either countable or finite. For the second
integral we have

$\int_{0}^{\infty}P[X_{s}=x,\ \tau_{i}\not=s\ \hbox{for all}\
i]ds=\sum_{i=0}^{\infty}\int_{0}^{\infty}P\left[X_{s}=x,\
\tau_{i}<s<\tau_{i+1}\right]ds.$

Now it suffices to show that the probability in the right-hand side is zero.
Indeed,

$\displaystyle P\left[X_{s}=x,\ \tau_{i}<s<\tau_{i+1}\right]$ $\displaystyle=$
$\displaystyle\int_{[0,s]\times\Re}P\left[X_{s}=x,\ s<\tau_{i+1}\ |\
\tau_{i}=u,X_{u}=y\right]P\left[\tau_{i}\in du,X_{u}\in dy\right]$
$\displaystyle=$ $\displaystyle\int_{[0,s]\times\Re}P\left[X_{s}-X_{u}=x-y,\
s<\tau_{i+1}\ |\ \tau_{i}=u,X_{u}=y\right]P\left[\tau_{i}\in du,X_{u}\in
dy\right]$ $\displaystyle=$
$\displaystyle\int_{[0,s]\times\Re}P\left[Y_{s}-Y_{u}=x-y,\ s<\tau_{i+1}\ |\
\tau_{i}=u,X_{u}=y\right]P\left[\tau_{i}\in du,X_{u}\in dy\right]$
$\displaystyle\leq$
$\displaystyle\int_{[0,s]\times\Re}P\left[Y_{s}-Y_{u}=x-y,\ |\
\tau_{i}=u,X_{u}=y\right]P\left[\tau_{i}\in du,X_{u}\in dy\right]$
$\displaystyle=$
$\displaystyle\int_{[0,s]\times\Re}P\left[Y_{s}-Y_{u}=x-y\right]P\left[\tau_{i}\in
du,X_{u}\in dy\right]$ $\displaystyle=$
$\displaystyle\int_{[0,s]\times\Re}P\left[Y_{s}=x\ |\
Y_{u}=y\right]P\left[\tau_{i}\in du,X_{u}\in dy\right]$

and $P\left[Y_{s}=x\ |\ Y_{u}=y\right]=0$ follows from our assumption
$\sigma\not=0$ and Sato [14], Theorem 27.4.

We now provide the proof of Proposition A.1.

Proof of Proposition A.1: Let $\left(f_{n}\right)_{n\geq 1}$ be the sequence
approximating $f$ in the sense of Lemma A.2. Ito’s rule holds for each
$f_{n}$, i.e.,

$\displaystyle f_{n}(X_{t})-f_{n}(X_{0})$ $\displaystyle=$
$\displaystyle\int_{(0,t]}f_{n}^{\prime}(X_{s-})dX_{s}+{\sigma^{2}\over
2}\int_{(0,t]}f_{n}^{\prime\prime}(X_{s})ds$ $\displaystyle+\sum_{0<s\leq
t}\left\\{f_{n}(X_{s})-f_{n}(X_{s-})-f_{n}^{\prime}(X_{s-})\Delta
X_{s}\right\\}.$

All we need to show that all three terms in the right-hand side of (A)
converge to the corresponding terms in the right-hand side of (53) as
$n\to\infty$. We can write $X=X_{0}+M_{1}(t)+A_{1}(t)$ where $M_{1}$ is a
local martingale with bounded jumps (thus also locally square-integrable) and
$A_{1}$ is a finite variation process (Jacod & Shiryaev [12], Proposition
4.17). We then have

$\int_{(0,t]}f_{n}^{\prime}(X_{s-})dM_{1}(s)\to\int_{(0,t]}f^{\prime}(X_{s-})dM_{1}(s)\
\hbox{in probability as}\ n\to\infty$

by Theorem 4.40 iii in Jacod & Shiryaev [12]. Also

$\int_{(0,t]}f_{n}^{\prime}(X_{s-})dA_{1}(s)\to\int_{(0,t]}f^{\prime}(X_{s-})dA_{1}(s)\
\hbox{a.s, as}\ n\to\infty$

by the Dominated Convergence Theorem. Therefore, the first integral in the
right-hand side of (A) indeed converges to the corresponding integral in (53).
The convergence

$\sum_{0<s\leq
t}\left\\{f_{n}(X_{s})-f_{n}(X_{s-})-f_{n}^{\prime}(X_{s-})\Delta
X_{s}\right\\}\to\sum_{0<s\leq
t}\left\\{f(X_{s})-f(X_{s-})-f^{\prime}(X_{s-})\Delta X_{s}\right\\}$

follows from the discrete time version of the Dominated Convergence Theorem
since $|f_{n}(X_{s})-f_{n}(X_{s-})-f_{n}^{\prime}(X_{s-})\Delta X_{s}|$ is is
bounded by ${C_{1}\over 2}(\Delta X_{s})^{2}$ and $\sum_{0<s\leq t}(\Delta
X_{s})^{2}<\infty$. Finally we need to show that

${\sigma^{2}\over
2}\int_{(0,t]}f_{n}^{\prime\prime}(X_{s})ds\to{\sigma^{2}\over
2}\int_{(0,t]}f^{\prime\prime}(X_{s})ds$

as $n\to\infty$. If $\sigma=0$ then there is nothing to prove and if
$\sigma\not=0$ then this follows from Lemma A.3 and the Dominated Convergence
Theorem.

## References

  * [1] Søren Asmussen, Soren Asmussen, and Sren Asmussen.  Applied probability and queues, volume 2.  Springer, 2003.
  * [2] Avner Bar-Ilan, David Perry, and Wolfgang Stadje.  A generalized impulse control model of cash management.  Journal of Economic Dynamics and Control, 28(6):1013–1033, 2004\.
  * [3] Alain Bensoussan, LIONS JL, et al.  Nouvelle formulation de problèmes de contrôle impulsionnel et applications.  1973\.
  * [4] ALAIN Bensoussan and JACQUES-LOUIS Lions.  Nouvelles méthodes en contrôle impulsionnel.  Applied Mathematics and Optimization, 1(4):289–312, 1975.
  * [5] Alain Bensoussan, RH Liu, and Suresh P Sethi.  Optimality of an (s,s) policy with compound poisson and diffusion demands: A quasi-variational inequalities approach.  SIAM journal on control and optimization, 44(5):1650–1676, 2005\.
  * [6] Abel Cadenillas, Peter Lakner, and Michael Pinedo.  Optimal control of a mean-reverting inventory.  Operations research, 58(6):1697–1710, 2010.
  * [7] Abel Cadenillas, Sudipto Sarkar, and Fernando Zapatero.  Optimal dividend policy with mean-reverting cash reservoir.  Mathematical Finance, 17(1):81–109, 2007.
  * [8] George M Constantinides and Scott F Richard.  Existence of optimal simple policies for discounted-cost inventory and cash management in continuous time.  Operations research, 26(4):620–636, 1978.
  * [9] Irmina Czarna and Adam Kaszubowski.  Optimality of impulse control problem in refracted lévy model with parisian ruin and transaction costs.  Journal of Optimization Theory and Applications, 185(3):982–1007, 2020.
  * [10] Haolin Feng and Kumar Muthuraman.  A computational method for stochastic impulse control problems.  Mathematics of Operations Research, 35(4):830–850, 2010.
  * [11] J Michael Harrison, Thomas M Sellke, and Allison J Taylor.  Impulse control of brownian motion.  Mathematics of Operations Research, 8(3):454–466, 1983.
  * [12] Jean Jacod and Albert Shiryaev.  Limit theorems for stochastic processes, volume 288.  Springer Science & Business Media, 2013.
  * [13] Offer Kella and Ward Whitt.  Useful martingales for stochastic storage processes with lévy input.  Journal of Applied Probability, 29(2):396–403, 1992.
  * [14] Sato Ken-Iti.  Lévy processes and infinitely divisible distributions.  Cambridge university press, 1999.
  * [15] Andreas E Kyprianou.  Introductory lectures on fluctuations of Lévy processes with applications.  Springer Science & Business Media, 2006.
  * [16] Bernt Oksendal.  Stochastic differential equations: an introduction with applications.  Springer Science & Business Media, 2013.
  * [17] P Protter.  Stochastic integration and stochastic differential equations, 2004.
  * [18] Scott F Richard.  Optimal impulse control of a diffusion process with both fixed and proportional costs of control.  SIAM Journal on Control and Optimization, 15(1):79–91, 1977.
  * [19] Agnès Sulem.  A solvable one-dimensional model of a diffusion inventory system.  Mathematics of Operations Research, 11(1):125–133, 1986.
  * [20] Jinbiao Wu.  Optimal exchange rates management using stochastic impulse control for geometric lévy processes.  Mathematical Methods of Operations Research, 89(2):257–280, 2019\.
  * [21] Kazutoshi Yamazaki.  Inventory control for spectrally positive lévy demand processes.  Mathematics of Operations Research, 42(1):212–237, 2017.