# The American Put Option with A Random Time Horizon

Zhuoshu Wu111Corresponding author. Email address:
<EMAIL_ADDRESS>The University of New South Wales, Sydney
Libo Li  The University of New South Wales, Sydney

###### Abstract

This paper is intended to provide a unique valuation formula for the American
put option in a random time horizon; more precisely, such option restricts any
stopping rules to be made before the last exit time of the price of the
underlying assets at any fixed level. By exploiting the theory of enlargement
of filtrations associated with random times, this pricing problem can be
transformed into an equivalent optimal stopping problem with a semi-
continuous, time-dependent gain function whose partial derivative is singular
at certain point. The difficulties in establishing the monotonicity of the
optimal stopping boundary and the regularity of the value function lie
essentially in these somewhat unpleasant features of the gain function.
However, it turns out that a successful analysis of the continuation and
stopping sets with proper assumption can help us overcome these difficulties
and obtain the important properties that possessed by the free-boundary, which
eventually leads us to the desired free-boundary problem. After this, the
nonlinear integral equations that characterise the free-boundary and the value
function can be derived. The solution to these equations are examined in
details. ††The authors are greatly indebted to Professor Ben Goldys for very
helpful discussions.

Key Words: Option pricing, last exit time, optimal stopping, free-boundary
problem

##  1 Introduction

Consider a complete defaultable market, with only one stock and one bond whose
processes follow the stochastic differential equations:

$\displaystyle dX_{t}$ $\displaystyle=rX_{t}dt+\sigma X_{t}dW_{t},\qquad
X_{0}=x>0,$ $\displaystyle dB_{t}$
$\displaystyle=rB_{t}dt,\qquad\qquad\qquad\,\,\,\,\,\,B_{0}=1,$

where $W=(W_{t})_{t\geq 0}$ is a standard Brownian motion started at zero
under measure $P$, $r>0$ is the interest rate and $\sigma>0$ is the volatility
coefficient. This is a risk neutral model so that the discounted stock price
is a martingale under measure $P$; in other words, $P$ is the risk neutral
measure.

Then there is this option with expiration date $T$ (which can be indefinitely
long) that entitles its holders to sell one share of stock for $K$ dollars at
any time during the time frame between its purchase time and its exercise time
or the last time the stock price price at any fixed level $L$ (whichever comes
first); otherwise, it becomes worthless. How much are you willing to pay for
this option?

When a contingent claim is valued, it is conventional to assume that there is
no risk that the counter-party will default, that is, the counter party will
always be able to make the promised payment. However, this assumption is far
less defensible in the non-exchange-traded (or over-the-counter) market and
the importance of considering default risk in over-counter markets is
recognised by bank capital adequacy standards, see [17, Page 300] for a
detailed introduction of default risk. According to [11, Pages 188 and 190],
the contract given above can be viewed as the defaultable American contingent
claims, with its default time modelled by the last exit time and by assuming
the default market is complete, such contract is hedgeable.

A new factor that lies at the heart of current work is the presence of the
last exit time in the classic valuation of American contingent claim. For a
solid introduction and background of last exit time that is not stopping time,
we refer to [23, Chapter 8.2]. The most interesting feature of the last exit
time is its involvement of “the knowledge of the future”, which suggests that
this work is closely related to the optimal prediction problem, see [2, 13,
16, 18] where this problem is also known as the stopping rule problem, the
secretary problem and the optimal selection problem; and for the more recent
work, see [9, 10]. For some financial applications of this factor that are
remotely relevant to this work, we refer to the monographs [1, 11, 12].

With the aid of the classic pricing theory, the pricing problem posed above
can be formulated into a precise mathematical form:

$\displaystyle V$
$\displaystyle=\operatorname*{ess\,sup}_{\tau\in[0,T]}E_{P}\Big{(}e^{-r\tau}\left(K-X_{\tau}\right)^{+}I\\{\tau<\theta\\}\Big{)},$
(1.1)

where $\tau$ is the exercise time, $\theta=\sup\\{t\in[0,T]:X_{t}\geq L\\}$ is
the last exit time and $K>0$ is the strike price. Additionally, we assume that
$K>L$. By invoking the theory of enlargements of filtrations studied by Mansuy
and Yor [22] and Nikeghbali [23], problem (1.1) can be further reduced to an
equivalent optimal stopping problem without the presence of last exit time;
for which, the classic method proposed in [24] can be partially applied. The
resulting gain function in the finite-time formulated problem is semi-
continuous, time-dependent with its partial derivatives being singular at
certain point.

From the perspective of present work, the argument particularly consistent to
the main result presented here is that by Peskir and Shiryaev [24]. Their work
is able to provide a rather systematic methodology in a step-by-step fashion
to tackle a wide range of optimal stopping problem associated with the
American contingent claim valuation. Other useful references are [10, 9, 14].
In particular, the time-dependent reward problem was also studied in DuToit,
Peskir and Shiryaev [10, 9], Glover, Peskir and Samee [14]. A somewhat
different approach to verify the regularity the optimal stopping boundary was
developed by De Angelis [4]. A treatment of the falling-apart smooth-fit
condition was given by Qiu [28], Detemple and Kitapbayev [7].

The remainder of this paper is organised as follows: Section 2 is based on the
published paper [15] of the author and the result presented here contains
extension by considering the payoff of conventional contingent claim. This
section covers the solution for problem (1.1) as $T=\infty$. We begin by
formulating the optimal stopping problem and its associated free-boundary
problem in the form of ODE. After solving the ODE, we also proves its solution
to be the smallest superharmonic function that dominates the gain function in
Subsection 3.2. The heart of this paper is Section 3 which contains the unique
pricing formula in Theorem 3.39 for the option posed above. Furthermore, being
confronted with the singularity of the partial derivative, it becomes
challenging to justify the regularity of value function and the optimal
stopping boundary. As a result of this, one can only derive the semi-
continuity of the value function first; but fortunately, this is just enough
to confirm the existence of the optimal stopping rule in Lemma 3.7. The next
focal point of this section is once again to justify the monotonicity of the
boundary, which shall pave the way for us to address the problems caused by
the singularity and then show the regularity of the value function and the
boundary. To achieve this, the first thing we have to do is to pinpoint the
stopping set by two crucial observations in Lemma 3.11 and Lemma 3.22
respectively: (i) the relative position of the free-boundary for pricing the
American put option (which we take for granted) and the fixed level $L$ and
(ii) the maximum of the gain function for any fixed time. Based on these
observations and Assumption 3.26 concerning the choice of parameters in
problem (1.1), we are able to present the monotonicity of the boundary in
Proposition 3.17 and Proposition 3.34.

##  2 In the Infinite-time Horizon

###  2.1 Reformulation and Basics

In this subsection, we reformulate our main optimal stopping problem and prove
some basic facts. By smoothing lemma and the measurability of stopping time
$\tau$,

$\displaystyle V(x)$
$\displaystyle=\sup_{\tau}E_{x}\left(E_{x}\left(e^{-r\tau}(K-X_{\tau})^{+}I\\{\tau<\theta\\}|\mathcal{F}_{\tau}\right)\right),$
$\displaystyle=\sup_{\tau}E_{x}\left(e^{-r\tau}(K-X_{\tau})^{+}P\left(\tau<\theta|\mathcal{F}_{\tau}\right)\right).$
(2.2)

Next in line is the result of the Azéma supermartingale but first for
notational convenience, we set $Z(x)=\left(\frac{L}{x}\right)^{\alpha}\wedge
1$ and $\alpha=\frac{2r}{\sigma^{2}}-1<0$.

######  Proposition 2.1.

Let $P\left(\theta>t\big{|}\mathcal{F}_{t}\right)$ be the Azéma
supermartingales associated with the random times $\theta$. Then,

$\displaystyle P\left(\theta>t\big{|}\mathcal{F}_{t}\right)$
$\displaystyle=Z(X_{t}).$

###### Proof.

As before, the set equality suggests that

$\displaystyle\\{\theta>t\\}=\bigg{\\{}\max_{u\geq t}X_{u}\geq L\bigg{\\}},$

so that

$\displaystyle P\left(\theta>t\big{|}\mathcal{F}_{t}\right)$
$\displaystyle=P\left(\max_{u\geq 0}X_{t+u}\geq L\right)=P\left(\max_{u\geq
0}X_{t}e^{\left(r-\frac{\sigma^{2}}{2}\right)u+\sigma W_{u}}\geq L\right)$
$\displaystyle=P\left(\max_{u\geq
0}{\left(r-\frac{\sigma^{2}}{2}\right)u+\sigma
W_{u}}\geq\log{\frac{L}{X_{t}}}\right)=\left(\frac{L}{X_{t}}\right)^{\alpha},$

where the last equality follows via the general result in [30, Page 759 and
760]. ∎

Combining the above computation, the original problems (2.2) can be rewritten
as

$\displaystyle V(x)$
$\displaystyle=\sup_{\tau}E_{x}\left(e^{-r\tau}(K-X_{\tau})^{+}Z(X_{\tau})\right),$
(2.3)

and for the sake of brevity, we denote the gain function $G(x)=(K-x)^{+}Z(x)$.

Let us end this subsection by a useful property possessed by function $Z$.

######  Proposition 2.2.

The function $Z(x)$ satisfies the following ordinary differential equation,
for $x\in(0,L)\cup(L,\infty)$,

$\displaystyle
rx\frac{d}{dx}Z(x)+\frac{1}{2}\sigma^{2}x^{2}\frac{d^{2}}{dx^{2}}Z(x)=0.$
(2.4)

###### Proof.

Remember that the Doob-Meyer decomposition of the Azéma supermartingale is
given by

$\displaystyle Z(X_{t})=Z(X_{0})+M_{t}^{\theta}-A_{t}^{\theta}$ (2.5)

and the measure $dA_{t}^{\theta}$ is carried by the set $\\{t:X_{t}=L\\}$.

Now, by the local time-space formular, we have

$\displaystyle dZ(X_{t})$
$\displaystyle=\left(rX_{t}Z_{x}(X_{t})+\frac{1}{2}\sigma^{2}X_{t}^{2}Z_{xx}(X_{t})\right)I\\{X_{t}\neq
L\\}dt$ $\displaystyle+\sigma X_{t}Z_{x}(X_{t})I\\{X_{t}\neq
L\\}dW_{t}+\frac{1}{2}\left(Z_{x}(L+)-Z_{x}(L-)\right)dl_{t}^{L}(X),$

where $l_{t}^{L}$ is the local time of $X$ at the level $L$ given by

$l_{t}^{L}(X)=P-\lim_{\epsilon\to
0}\frac{1}{2\epsilon}\int_{0}^{t}I\\{|X_{s}-L|<\epsilon\\}d\langle
X,X\rangle_{s},$

which, joining with (2.5), shows that

$\displaystyle M_{t}^{\theta}$
$\displaystyle=\int_{0}^{t}\left(\sigma-\frac{2r}{\sigma}\right)\left(\frac{L}{X_{s}}\right)^{\alpha}I\\{X_{s}\neq
L\\}dW_{s}$ $\displaystyle A_{t}^{\theta}$
$\displaystyle=\int_{0}^{t}\left(\frac{1}{2}-\frac{r}{\sigma^{2}}\right)\frac{1}{L}dl_{s}^{L}(X),$

and thereby, proving that (2.4) holds true. ∎

######  Remark 2.3.

An alternative is to compute the first and second derivative of the function
$Z$ directly. Admittedly, the above proof is more clumsy than direct
computation, but latter, as we shall see in the setting of finite time, the
method of using Doob-Meyer decompostion is far more convenient.

###  2.2 The Free-boundary Problems

This subsection is devoted to formulating the free-boundary problem, whose
solution will also be provided and verified.

The point of departure is to appeal to the local time-space formula (see [26])
as $G$ is of class $C^{2}$ on $(0,\infty)\setminus\\{L,K\\}$ and obtain

$\displaystyle e^{-rt}G(X_{t})$
$\displaystyle=G(x)+\int_{0}^{t}e^{-ru}\left(-rG+rX_{u}G_{x}+\frac{1}{2}\sigma^{2}X_{u}^{2}G_{xx}\right)(X_{u})I\\{X_{u}\neq
L,X_{u}\neq K\\}du$ $\displaystyle+\int_{0}^{t}e^{-ru}\sigma
X_{s}G_{x}(X_{u})I\\{X_{u}\neq L,X_{u}\neq K\\}dW_{u}$
$\displaystyle+\frac{1}{2}\int_{0}^{t}e^{-ru}\left(G_{x}(K+)-G_{x}(K-)\right)dl_{u}^{K}(X)$
$\displaystyle+\frac{1}{2}\int_{0}^{t}e^{-ru}\left(G_{x}(L+)-G_{x}(L-)\right)dl_{u}^{L}(X)$
(2.6)

where $l_{s}^{L}$ and $l_{s}^{K}$ are the local time of $X$ at the levels $L$,
$K$ respectively, and for brevity, we denote the martingale term as

$\displaystyle M_{t}=\int_{0}^{t}e^{-ru}\sigma X_{u}G_{x}(X_{u})I\\{X_{u}\neq
L,X_{u}\neq K\\}dW_{u}.$

Now, with a little bit of effort, we can compute:

$\displaystyle\left(-rG+rxG_{x}+\frac{1}{2}\sigma^{2}x^{2}G_{xx}\right)$
$\displaystyle=\begin{cases}-rKZ-\sigma^{2}x^{2}Z_{x},&\text{for $x<L$},\\\
-rK,&\text{for $L<x<K$},\\\ 0,&\text{for $x>K$},\\\ \end{cases}$

and

$\displaystyle G_{x}(L-)=-1-\alpha(K-L)L^{-1},$ $\displaystyle G_{x}(L+)=-1,$
$\displaystyle G_{x}(K+)=0,$ $\displaystyle G_{x}(K-)=-1,$

so that (2.6) is reshuffled accordingly as

$\displaystyle e^{-rt}G(X_{t})$
$\displaystyle=G(x)+M_{t}-rK\int_{0}^{t}e^{-ru}I\\{L<X_{u}<K\\}du$
$\displaystyle\,\,\,\,\,\,\,\,-\int_{0}^{t}e^{-ru}\left(rKZ+\sigma^{2}X_{u}^{2}Z_{x}\right)(X_{u})I\\{X_{u}<L\\}du$
$\displaystyle\,\,\,\,\,\,\,\,+\int_{0}^{t}e^{-ru}\alpha(K-L)L^{-1}dl_{u}^{L}(X)+\frac{1}{2}\int_{0}^{t}e^{-ru}dl_{u}^{K}(X),$

so that

$\displaystyle E_{x}\left(e^{-rt}G(X_{t})\right)$
$\displaystyle=G(x)+E\left(-rK\int_{0}^{t}e^{-ru}I\\{L<X_{u}<K\\}du\right.$
(2.7)
$\displaystyle\qquad\qquad\qquad\left.-\int_{0}^{t}e^{-ru}\left(rK-\alpha\sigma^{2}X_{u}\right)Z(X_{u})I\\{X_{u}<L\\}du\right.$
$\displaystyle\qquad\qquad\qquad\left.+\int_{0}^{t}e^{-ru}\alpha(K-L)L^{-1}dl_{u}^{L}(X)+\frac{1}{2}\int_{0}^{t}e^{-ru}dl_{u}^{K}(X)\right).$

Immediate from (2.7), we see that the further $X$ gets away from $K$ the less
likely that the gain will increase upon continuing, and this suggests that
there exists a point $b\in(0,K)$ such that the stopping time

$\displaystyle\tau_{b}=\inf\\{t\geq 0:X_{t}\leq b\\}$ (2.8)

is optimal for problem (2.3), with the convention that the infimum of the
empty set is infinite. The optimality of $\tau_{b}$ will be verified in
section 5.1.3.

The next logical task is, of course, to determine whether or not such stopping
time $\tau_{b}$ is finite.

######  Corollary 2.4.

The stopping time $\tau_{b}$ is finite, i.e. $P_{x}(\tau_{b}<\infty)=1$.

###### Proof.

To verify $P_{x}(\tau_{b}<\infty)=1$ for all $x>0$, it suffices to show that
$P_{x}\left(\min\limits_{t\geq 0}X_{t}\leq b\right)=1$.

With appeal to general result from [30, Page 759-760], we find that

$\displaystyle P_{x}\left(\min\limits_{t\geq
0}xe^{\left(r-\frac{\sigma^{2}}{2}\right)t+\sigma W_{t}}\leq b\right)$
$\displaystyle=P_{x}\left(\min\limits_{t\geq
0}\left(\left(r-\frac{\sigma^{2}}{2}\right)t+\sigma
W_{t}\right)\leq\log{\frac{b}{x}}\right)$
$\displaystyle=P_{x}\left(-\max\limits_{t\geq
0}\left(-\left(r-\frac{\sigma^{2}}{2}\right)t-\sigma
W_{t}\right)\leq\log{\frac{b}{x}}\right)$
$\displaystyle=P_{x}\left(\max\limits_{t\geq
0}\left(-\left(r-\frac{\sigma^{2}}{2}\right)t+\sigma
W_{t}\right)\geq-\log{\frac{b}{x}}\right)$
$\displaystyle=1-P_{x}\left(\max\limits_{t\geq
0}\left(-\left(r-\frac{\sigma^{2}}{2}\right)t+\sigma
W_{t}\right)\leq-\log{\frac{b}{x}}\right),$

where the third equality holds true as $(W_{t})_{t\geq 0}\buildrel
d\over{=}(-W_{t})_{t\geq 0}$. Hence, for $x\geq b$ (i.e. $-\log\frac{b}{x}\geq
0$), by the assumption that $\alpha=\frac{2r}{\sigma^{2}}-1<0$, we have

$\displaystyle P_{x}\left(\min\limits_{t\geq 0}X_{t}\leq b\right)=1,$

which, joining with the fact that $x<b$ implies $\tau_{b}=0$, shows that
$P_{x}(\tau_{b}<\infty)=1$ for all $x>0$. ∎

With the aid of the above observations and the standard arguments based on the
strong Markov property, we are now ready to formulate the free boundary
problem to find the smallest superharmonic function $\hat{V}$ that dominates
$G$ and the unknown point $b$,

$\displaystyle\mathbb{L}_{X}\hat{V}-r\hat{V}=0\qquad\qquad\qquad\,\,\,\,\,\,\,\,\,\,\text{for
$x\in(b,\infty)$},$ (2.9)
$\displaystyle\hat{V}(x)=G(x)\qquad\qquad\qquad\qquad\,\,\,\,\text{for
$x=b$},$ (2.10)
$\displaystyle\hat{V}_{x}(x)=G_{x}(x)\qquad\qquad\qquad\,\,\,\,\,\,\,\,\,\,\,\text{for
$x=b$},$ (2.11)
$\displaystyle\hat{V}(x)=G(x)\qquad\qquad\qquad\qquad\,\,\,\,\text{for
$x\in(0,b)$},$ (2.12)
$\displaystyle\hat{V}(x)>G(x)\qquad\qquad\qquad\qquad\,\,\,\,\text{for
$x\in(b,\infty)$},$ (2.13)

where $\mathbb{L}_{X}=rx\frac{\partial}{\partial
x}+\frac{1}{2}\sigma^{2}x^{2}\frac{\partial^{2}}{\partial x^{2}}$ is the
infinitesimal generator of the strong Markov diffusion process $X$.

######  Remark 2.5.

Another message of the free-boundary problem is that the optimal stopping
level $b$ depends on the relative positions of the level $L$.

We now proceed to solve the free-boundary problem. Equation (2.9) is the well-
known Cauchy-Euler equation, whose solution gets form as follows

$\displaystyle\hat{V}(x)=x^{p},$ (2.14)

Then, by putting (2.14) into (2.9), we have the following quadratic equation

$\displaystyle\frac{1}{2}\sigma^{2}p^{2}+\left(r-\frac{\sigma^{2}}{2}\right)p-r=0,$

whose roots equal

$\displaystyle p_{1}=1,\qquad p_{2}=-\frac{2r}{\sigma^{2}},$

from which, we know that the general solution of (2.9) is

$\displaystyle\hat{V}(x)=C_{1}x+C_{2}x^{-\frac{2r}{\sigma^{2}}},$

where $C_{1}$ and $C_{2}$ are undetermined constants and by noticing that, in
fact, for all $x>0$, function $\hat{V}(x)\leq K$, implying that $C_{1}$ must
be zero.

By inspecting (2.10) and (2.11), we see that they are of different forms as
$b>L$ and $b<L$,

$\displaystyle\hat{V}(b)=C_{2}b^{-\frac{2r}{\sigma^{2}}}=\begin{cases}(K-b),&\text{for
$b>L$},\\\ (K-b)\left(\frac{L}{b}\right)^{\alpha},&\text{for
$b<L$},\end{cases}$ (2.15)

and

$\displaystyle\hat{V}_{x}(b)=-\frac{C_{2}2r}{\sigma^{2}}b^{-\frac{2r}{\sigma^{2}}-1}=\begin{cases}-1,&\text{for
$b>L$},\\\
\left(\frac{L}{b}\right)^{\alpha}\left(-1-\frac{\alpha(K-b)}{b}\right),&\text{for
$b<L$},\end{cases}$ (2.16)

and we have rediscovered Remark 2.5, from which, determining $C_{2}$ is within
easy reach:

$\displaystyle C_{2}=\begin{cases}(K-b)b^{\frac{2r}{\sigma^{2}}},&\text{for
$b>L$},\\\ (K-b)L^{\frac{2r}{\sigma^{2}}-1}b,&\text{for $b<L$}.\end{cases}$

With a little additional effort, we work out the math for $b$ by using (2.11):

$\displaystyle b=\begin{cases}\frac{2rK}{2r+\sigma^{2}},&\text{for
$\frac{2rK}{2r+\sigma^{2}}>L$},\\\ \frac{K}{2},&\text{for
$\frac{K}{2}<L$}.\end{cases}$ (2.17)

######  Remark 2.6.

Notice that, if $\frac{2r}{\sigma^{2}+2r}=\frac{1}{2}$, i.e. $\alpha=0$, then
the result of standard perpetual American put option emerges.

The conclusion runs as follows:

######  Theorem 2.7.

If $K<2L$, that is $b<L$, then the value function $V$ from (2.2) is given
explicitly by

$\displaystyle V(x)$ $\displaystyle=\begin{cases}G(x),&\text{for
$x\in(0,b]$},\\\
\frac{L^{\alpha}K^{2}}{4}x^{-\frac{2r}{\sigma^{2}}},&\text{for
$x\in(b,\infty)$};\end{cases}$ (2.18)

if, on the other hand, $2rK>L(\sigma^{2}+2r)$, that is $b>L$, then

$\displaystyle V(x)$ $\displaystyle=\begin{cases}G(x),&\text{for
$x\in(0,b)$},\\\
\left(K-\frac{2rK}{2r+\sigma^{2}}\right)\left(\frac{2rK}{2r+\sigma^{2}}\right)^{\frac{2r}{\sigma^{2}}}x^{-\frac{2r}{\sigma^{2}}},&\text{for
$x\in[b,\infty)$}.\end{cases}$ (2.19)

The stopping time $\tau_{b}$ from (2.8) are optimal in problem (2.2).

###### Proof of Solution (2.18).

We wish to prove that $V(x)$ in (2.2) equals $\hat{V}(x)$ in (2.15) for all
$x\in(0,\infty)$, to achieve which, we proceed via three steps (proving three
statements):

(i) $\hat{V}(x)\geq G(x)$ for all $x\in(0,\infty)$. To prove this, we set

$\displaystyle h(x)=\hat{V}(x)-G(x)=\begin{cases}0,&\text{for
$x\in(0,\frac{K}{2}]$},\\\
\frac{L^{\alpha}K^{2}}{4}x^{-\frac{2r}{\sigma^{2}}}-(K-x)\left(\frac{L}{x}\right)^{\alpha},&\text{for
$x\in[\frac{K}{2},L]$},\\\
\frac{L^{\alpha}K^{2}}{4}x^{-\frac{2r}{\sigma^{2}}}-(K-x),&\text{for
$x\in[L,\infty)$}.\end{cases}$

Knowing that $\hat{V}(x)=G(x)$ for all $x\in(0,\frac{K}{2}]$ reduces our task
to verifying that $h(x)\geq 0$ in the domain $[\frac{K}{2},\infty)$. By
rearrangement, we have

$\displaystyle
h(x)=\begin{cases}\left(\frac{L}{x}\right)^{\alpha}x^{-1}\left(\frac{K}{2}-x\right)^{2},&\text{for
$x\in[\frac{K}{2},L]$},\\\
\left(\frac{L}{x}\right)^{\alpha}\frac{K^{2}}{4}x^{-1}-K+x\geq\frac{K^{2}}{4}x^{-1}-K+x=x^{-1}\left(\frac{K}{2}-x\right)^{2},&\text{for
$x\in[L,\infty)$},\end{cases}$

which yields that $h(x)\geq 0$, and thereby proving statement (i).

(ii) ${V}(x)\leq\hat{V}(x)$ for all $x\in(0,\infty)$. By invoking the local
time-space formula, we obtain

$\displaystyle
e^{-rt}\hat{V}(X_{t})=\hat{V}(x)+\int_{0}^{t}e^{-rs}\left(-r\hat{V}+\mathbb{L}_{X}\hat{V}\right)(X_{s})I\\{X_{s}\neq
b\\}ds$ $\displaystyle\qquad\,\,\,\,\,\,\,\,\,+\int_{0}^{t}e^{-rs}\sigma
X_{s}\hat{V}_{x}(X_{s})I\\{X_{s}\neq
b\\}dW_{s}+\frac{1}{2}\int_{0}^{t}e^{-rs}\left(\hat{V}_{x}(b+)-\hat{V}_{x}(b-)\right)dl_{s}^{b}(X^{x})$
$\displaystyle\qquad=\hat{V}(x)+{M}_{t}+\int_{0}^{t}e^{-rs}\left(-r\hat{V}+\mathbb{L}_{X}\hat{V}\right)(X_{s})ds,$
(2.20)

where the second equality holds via the smooth-fit condition (2.11). Since
$\hat{V}(x)=G(x)$ for all $0<x<b$ such that

$\displaystyle-r\hat{V}+\mathbb{L}_{X}\hat{V}=-r{G}+\mathbb{L}_{X}{G}=-rKZ-\sigma^{2}x^{2}Z_{x},$

which, together with (2.9), tells us that

$\displaystyle-r\hat{V}+\mathbb{L}_{X}\hat{V}\leq 0,$ (2.21)

and then, immediate from statement (i) and (2.21) for all $x\in(0,\infty)$, we
see that

$\displaystyle e^{-rt}{G}(X_{t})$ $\displaystyle\leq
e^{-rt}\hat{V}(X_{t})\leq\hat{V}(x)+M_{t}.$ (2.22)

Let $\tau_{n}=\tau\wedge n$ for $n\geq 0$ and then we know that, for every
bounded stopping time $\tau_{n}$ of $X$, taking the expectation under measure
$P_{x}$ yields $E_{x}\left(M_{\tau_{n}}\right)=E_{x}\left(M_{n}\right)=0$ so
that

$\displaystyle E_{x}\left(e^{-r\tau_{n}}G(X_{\tau_{n}})\right)\leq\hat{V}(x).$

Next, let $n\to\infty$, by Fatou’s lemma,
$E_{x}\left(e^{-r\tau}G(X_{\tau})\right)\leq\hat{V}(x)$. After this, by taking
the supremum over all the stopping times of $X$,

$\displaystyle{V}(x)=\sup_{\tau}E_{x}\left(e^{-r\tau}G(X_{\tau})\right)\leq\hat{V}(x),$
(2.23)

and the desired statement (ii) follows.

(iii) ${V}(x)\geq\hat{V}(x)$ for all $x\in(0,\infty)$. Put $t=\tau_{b}\wedge
n$ for $n\geq 0$ in (2.20), with $\tau_{b}$ being the finite stopping time
defined in (2.8),

$\displaystyle e^{-r(\tau_{b}\wedge n)}\hat{V}\left(X_{\tau_{b}\wedge
n}\right)=\hat{V}(x)+{M}_{\tau_{b}\wedge n},$

and then by taking its expectation under measure $P_{x}$, we have
$E_{x}\left({M}_{\tau_{b}\wedge n}\right)=E_{x}\left({M}_{n}\right)=0$ so that

$\displaystyle E_{x}\left(e^{-r(\tau_{b}\wedge
n)}\hat{V}\left(X_{\tau_{b}\wedge n}\right)\right)=\hat{V}(x).$

Let $n\to\infty$, via an appeal to the dominated convergence theorem and the
fact that ${V}(X_{\tau_{b}})={G}(X_{\tau_{b}})$, we see that

$\displaystyle\hat{V}(x)$
$\displaystyle=E_{x}\left(e^{-r\tau_{b}}\hat{V}\left(X_{\tau_{b}}\right)\right)=E_{x}\left(e^{-r\tau_{b}}{G}\left(X_{\tau_{b}}\right)\right)\leq\sup_{\tau}E_{x}\left(e^{-r\tau}{G}\left(X_{\tau}\right)\right)=V(x),$
(2.24)

which yields statement (iii). Finally, combining (2.23) and (2.24) completes
the proof. ∎

######  Remark 2.8.

The above proof is the same as showing that $\hat{V}$ is the smallest
superharmonic function that dominates $G$: step (i) shows the domination, step
(ii) we have $\hat{V}$ being superharmonic function because of inequality
(2.22) and (2.24) in step (iii) tells us that $\hat{V}$ is the smallest
superharmonic function that dominates $G$. Therefore, $\tau_{b}$ is indeed
optimal given that it is finite almost surely. See [24, Page 40, Theorem 2.7]
for more details.

###### Proof of Solution (2.19).

Similar argument works for this solution, but since solution (2.19) is not
differentiable on $L$, which should require extra care in the following part
of step (ii):

$\displaystyle
e^{-rt}\hat{V}(X_{t})=\hat{V}(x)+\int_{0}^{t}e^{-rs}\left(-r\hat{V}+\mathbb{L}_{X}\hat{V}\right)(X_{s})I\\{X_{s}\neq
b,X_{s}\neq L\\}ds$ $\displaystyle\qquad+\int_{0}^{t}e^{-rs}\sigma
X_{s}\hat{V}_{x}(X_{s})I\\{X_{s}\neq b,X_{s}\neq L\\}dW_{s}$
$\displaystyle\qquad+\frac{1}{2}\int_{0}^{t}e^{-rs}\left(\hat{V}_{x}(b+)-\hat{V}_{x}(b-)\right)dl_{s}^{b}(X^{x})+\frac{1}{2}\int_{0}^{t}e^{-rs}\left(\hat{V}_{x}(L+)-\hat{V}_{x}(L-)\right)dl_{s}^{L}(X^{x})$
$\displaystyle=\hat{V}(x)+{M}_{t}+\int_{0}^{t}e^{-rs}\left(-r\hat{V}+\mathbb{L}_{X}\hat{V}\right)(X_{s})ds+\frac{1}{2}\int_{0}^{t}e^{-rs}\left(G_{x}(L+)-G_{x}(L-)\right)dl_{s}^{L}(X^{x})$
$\displaystyle\leq\hat{V}(x)+{M}_{t}+\int_{0}^{t}e^{-rs}\left(-r\hat{V}+\mathbb{L}_{X}\hat{V}\right)(X_{s})ds,$
(2.25)

where the second equality holds via the smooth-fit condition and the
inequality follows via $G_{x}(L+)-G_{x}(L-)<0$. Since $\hat{V}(x)=G(x)$ for
all $0<x<b$ such that

$\displaystyle-r\hat{V}+\mathbb{L}_{X}\hat{V}=-r{G}+\mathbb{L}_{X}{G}=-rKZ-\sigma^{2}x^{2}Z_{x},$

which, together with (2.9), tells us that $-r\hat{V}+\mathbb{L}_{X}\hat{V}\leq
0$. This, combining with statement (i) yields $e^{-rt}{G}(X_{t})\leq
e^{-rt}\hat{V}(X_{t})\leq\hat{V}(x)+M_{t}$. The rest of the proof follows
similarly as the proof of (2.18). ∎

Figure 1: This figure displays the maps $x\mapsto V(x)$ and $x\mapsto G(x)$
with chosen parameters $r=0.05$, $\sigma=0.4$, and the optimal stopping points
$b=2.6923$ for $L=2$ and $b=3.5$ for $L=4$.

######  Remark 2.9.

Figure 1 shows that the value of perpetual contract is a decreasing function
of $L$, and is bounded above by the global maximum of $G$ and below by $0$.

##  3 In the Finite-time Horizon

###  3.1 Reformulation and Basics

In this subsection, we once again reformulate problem (1.1) and collect some
elementary facts that will be used later.

By the measurability of stopping time $\tau$, we see that

$\displaystyle V$
$\displaystyle=\sup_{0\leq\tau\leq{T}}E_{0,x}\left(e^{-r\tau}\left(K-X_{\tau}\right)^{+}I\\{\theta>\tau\\}\right)$
$\displaystyle=\sup_{0\leq\tau\leq{T}}E_{0,x}\left(e^{-r\tau}\left(K-X_{\tau}\right)^{+}P(\theta>\tau|\mathcal{F}_{\tau})\right).$
(3.26)

The process $X$ is a strong Markov (diffusion) process with the infinitesimal
generator given by

$\displaystyle\mathbb{L}_{X}=rx\frac{\partial}{\partial
x}+\frac{\sigma^{2}}{2}x^{2}\frac{\partial^{2}}{\partial x^{2}}.$

In the view of (3.26), the first thing is, of course, to calculate
$P(\theta>\tau|\mathcal{F}_{\tau})$ and let us introduce the notation,

$\displaystyle
Z(t,x)=\left(\Phi\left(\frac{-\log{\frac{L}{x}}+\left(r-\frac{\sigma^{2}}{2}\right)(T-t)}{\sigma\sqrt{T-t}}\right)+\left(\frac{L}{x}\right)^{\alpha}\Phi\left(\frac{-\log{\frac{L}{x}}-\left(r-\frac{\sigma^{2}}{2}\right)(T-t)}{\sigma\sqrt{T-t}}\right)\right)\wedge
1,$

where $\Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-\frac{z^{2}}{2}}dz$
and $\alpha=\frac{2r}{\sigma^{2}}-1<0$.

######  Proposition 3.1.

Let $P\left(\theta>t|\mathcal{F}_{t}\right)$ be the Azéma supermartingale
associated with the random time $\theta$. Then, for
$r-\frac{\sigma^{2}}{2}\in\mathbb{R}$ and $\sigma>0$, for

$\displaystyle P\left(\theta>t|\mathcal{F}_{t}\right)$
$\displaystyle=Z(t,X_{t}).$

###### Proof.

First of all, suppose that

$\displaystyle{d}_{t}$ $\displaystyle=\inf\\{u\geq t:X_{u}\geq L\\}$
$\displaystyle=t+\inf\\{u\geq 0:\bar{X}_{0}=X_{t},\bar{X}_{u}\geq L\\},$

where the second equality follows from the Markov property of $X$ and for easy
reference, let

$\displaystyle\bar{h}_{L}$ $\displaystyle=\inf\\{u\geq
0:\bar{X}_{0}=X_{t},\bar{X}_{u}\geq L\\},$

in plain language, $\bar{h}_{L}$ is the first time when a geometric Brownian
motion $\bar{X}$, starting from $X_{t}$, hits the level $L$.

Then, it follows that

$\displaystyle P(\theta>t|\mathcal{F}_{t})$ $\displaystyle=1-P\left(\theta\leq
t|\mathcal{F}_{t}\right)$
$\displaystyle=1-P\left(d_{t}>T|\mathcal{F}_{t}\right)$
$\displaystyle=1-P\left(\bar{h}_{L}>T-t|\mathcal{F}_{t}\right).$

Upon observing that, in fact,

$\displaystyle\\{\bar{h}_{L}\leq T-t|\mathcal{F}_{t}\\}$
$\displaystyle=\bigg{\\{}\max_{0\leq u\leq T-t}\bar{X}_{u}\geq
L|\mathcal{F}_{t}\bigg{\\}}=\bigg{\\{}\max_{0\leq u\leq
T-t}X_{t}e^{\left(r-\frac{\sigma^{2}}{2}\right)u+\sigma W_{u}}\geq
L\bigg{\\}},$ (3.27)

after which, an application of general result from [30, Page 759] proves that,
for $\log{\frac{L}{X_{t}}}\geq 0$, $r-\frac{\sigma^{2}}{2}\in\mathbb{R}$ and
$\sigma>0$,

$\displaystyle P\left(\theta>t|\mathcal{F}_{t}\right)$
$\displaystyle=\Phi\left(\frac{-\log{\frac{L}{X_{t}}}+\left(r-\frac{\sigma^{2}}{2}\right)(T-t)}{\sigma\sqrt{T-t}}\right)$
$\displaystyle\qquad\qquad\qquad+\left(\frac{L}{X_{t}}\right)^{\alpha}\Phi\left(\frac{-\log{\frac{L}{X_{t}}}-\left(r-\frac{\sigma^{2}}{2}\right)(T-t)}{\sigma\sqrt{T-t}}\right).$

while for $\log{\frac{L}{X_{t}}}<0$, observe that event (3.27) happens almost
surely. As for $t=T$, from the set equality (3.27), it follows that

$\displaystyle P(\theta>t|\mathcal{F}_{t})=\begin{cases}1,&\text{for
$\log{\frac{L}{X_{t}}}\leq 0$},\\\ 0,&\text{for
$\log{\frac{L}{X_{t}}}>0$}.\end{cases}$

∎

Having introduced the Azéma supermartingale, the following facts are just
around the corner.

######  Proposition 3.2.

The function $Z(t,x)$ satisfies the following partial differential equation,
for $(t,x)\in[0,T]\times\\{(0,L)\cup(L,\infty)\\}$,

$\displaystyle\frac{\partial}{\partial t}Z(t,x)+rx\frac{\partial}{\partial
x}Z(t,x)+\frac{1}{2}\sigma^{2}x^{2}\frac{\partial^{2}}{\partial
x^{2}}Z(t,x)=0.$ (3.28)

###### Proof.

Notice that direct computation is certainly an option, but making use of the
Doob-Meyer decomposition shall be a much efficient way.

Let us apply the local time-space formula to obtain

$\displaystyle dZ(t,X_{t})$ $\displaystyle=\left(\frac{\partial}{\partial
t}Z+rX_{t}\frac{\partial}{\partial
x}Z+\frac{1}{2}\sigma^{2}{X_{t}}^{2}\frac{\partial^{2}}{\partial
x^{2}}Z\right)(t,X_{t})I\\{X_{t}\neq L\\}dt$
$\displaystyle\,\,\,\,\,\,\,+\sigma X_{t}\frac{\partial}{\partial
x}Z(t,X_{t})I\\{X_{t}\neq L\\}dW_{t}+\frac{1}{2}\left(\frac{\partial}{\partial
x}Z(t,L+)-\frac{\partial}{\partial x}Z(t,L-)\right)dl_{t}^{L},$ (3.29)

where $l_{t}^{L}$ is the local time of $X$ at the level $L$ given by

$l_{t}^{L}=P-\lim_{\epsilon\to
0}\frac{1}{2\epsilon}\int_{0}^{t}I\\{|X_{s}-L|<\epsilon\\}d\langle
X,X\rangle_{s},$

after which, we can lean on the Doob-Meyer decomposition
$Z(t,X_{t})=Z(0,x)+M_{t}-A_{t}$ to conclude that

$\displaystyle M_{t}=\int_{0}^{t}\sigma X_{s}\frac{\partial}{\partial
x}Z(t,X_{s})I\\{X_{s}\neq L\\}dW_{s},\qquad
A_{t}=\frac{1}{2}\int_{0}^{t}\frac{\partial}{\partial x}Z(s,L-)dl_{s}^{L},$

where $M_{t}$ is the c$\grave{a}$dl$\grave{a}$g martingale and $A_{t}$ is the
predictable projection whose measure $dA_{t}$ is carried by the set
$\\{t:X_{t}=L\\}$, and consequently, (3.28) follows. ∎

######  Corollary 3.3.

(i) The map $t\mapsto Z(t,x)$ is decreasing on $[0,T]$; (ii)The map $x\mapsto
Z(t,x)$ is increasing on $(0,\infty)$.

###### Proof.

These two statements follow directly from the computation in (A.85) and
(A.86). ∎

Summarising our findings so far, by the strong Markov property of $X$, we can
rewrite and generalise the original problem (3.26) as

$\displaystyle V(t,x)$
$\displaystyle=\sup_{0\leq\tau\leq{T-t}}E_{t,x}\left(e^{-r\tau}\left(K-X_{t+\tau}\right)^{+}Z(t+\tau,X_{t+\tau})\right),$
(3.30)

and for simplicity, let the gain function $G(t,x)=(K-x)^{+}Z(t,x)$.

###  3.2 The Free-boundary Problem

Before we turn to formulating the free-boundary problem, we squeeze the
important concepts of continuation and stopping sets into the text.

######  Definition 3.4.

The continuation set $\mathcal{C}$ and the stopping set $\mathcal{D}$ are
defined as follows:

$\displaystyle\mathcal{C}$
$\displaystyle=\\{(t,x)\in[0,T)\times(0,\infty):V(t,x)>G(t,x)\\},$ (3.31)
$\displaystyle\mathcal{D}$
$\displaystyle=\\{(t,x)\in[0,T)\times(0,\infty):V(t,x)=G(t,x)\\}\cup\\{(T,x):x\in(0,\infty)\\}.$
(3.32)

Based on the above definition, the first entry time $\tau_{\mathcal{D}}$ of
$X$ into $\mathcal{D}$ is then defined as

$\displaystyle\tau_{\mathcal{D}}=\inf\\{0\leq s\leq
T-t:(t+s,X_{t+s}^{x})\in\mathcal{D}\\}\wedge(T-t).$ (3.33)

It is then not far-fetched to ask for the optimality of $\tau_{\mathcal{D}}$,
which, according to [24, Corollary 2.9, Page 46], is the same as showing the
(semi)continuity of $V$ and $G$.

######  Remark 3.5.

It is worth mentioning that $Z_{x}(t,x)$ (see (A.86)) has a singularity at
point $(L,T)$, so establishing the continuity of the value function can only
be pursued later.

######  Lemma 3.6.

The value function $V$ is l.s.c on $[0,T]\times(0,\infty)$.

###### Proof.

As was noticed before, the supremum of an l.s.c function defined an l.s.c
function, so we only have to establish the continuity (thereby l.s.c) of
$E_{t,x}\left(e^{-r\tau}G\left(t+\tau,X_{t+\tau}\right)\right)$ on
$[0,T)\times(0,\infty)$ and the assertion will follow immediately.

To achieve this, we begin by showing that (i) $t\mapsto
E_{t,x}\left(e^{-r\tau}G\left(t+\tau,X_{t+\tau}\right)\right)$ is continuous
on $[0,T)$ for each $x\in(0,\infty)$. Take any $t_{1}<t_{2}$ in $[0,T)$ so
that

$\displaystyle 0$ $\displaystyle\leq
E_{t_{1},x}\left(e^{-r\tau}G(t_{1}+\tau,X_{t_{1}+\tau})\right)-E_{t_{2},x}\left(e^{-r\tau}G(t_{2}+\tau,X_{t_{2}+\tau})\right)$
$\displaystyle=E\left(e^{-r\tau}\left(K-X_{\tau}^{x}\right)^{+}\left(Z(t_{1}+\tau,X_{\tau}^{x})-Z(t_{2}+\tau,X_{\tau}^{x})\right)\right)$
$\displaystyle\leq
KE\left(Z(t_{1}+\tau,X_{\tau}^{x})-Z(t_{2}+\tau,X_{\tau}^{x})\right),$

where the first inequality is based on the map $t\mapsto G(t,x)$ is
decreasing, and first equality follows via the time-homogeneous property of
GBM; after which, let $t_{1}\to t_{2}$, we see that by the dominated
convergence theorem (function $Z:[0,T]\times(0,\infty)\mapsto[0,1]$),

$E_{t_{1},x}\left(e^{-r\tau}G(t_{1}+\tau,X_{t_{1}+\tau})\right)-E_{t_{2},x}\left(e^{-r\tau}G(t_{2}+\tau,X_{t_{2}+\tau})\right)\to
0,$

and assertion (i) follows.

It remains to show that (ii) $t\mapsto
E_{t,x}\left(e^{-r\tau}G\left(t+\tau,X_{t+\tau}\right)\right)$ is continuous
on $(0,\infty)$ for each $t\in[0,T)$. We simply observe that

$\displaystyle 0$
$\displaystyle\leq\bigg{|}E_{t,x_{1}}\left(e^{-r\tau}G(t+\tau,X_{t+\tau})\right)-E_{t,x_{2}}\left(e^{-r\tau}G(t+\tau,X_{t+\tau})\right)\bigg{|}$
$\displaystyle\leq
E\bigg{|}G(t+\tau,X_{\tau}^{x_{1}})-G(t+\tau,X_{\tau}^{x_{2}})\bigg{|}$
$\displaystyle\leq
E\bigg{|}Z(t+\tau,X_{\tau}^{x_{1}})(X_{\tau}^{x_{1}}-X_{\tau}^{x_{2}})\bigg{|}+KE\bigg{|}Z(t+\tau,X_{\tau}^{x_{1}})-Z(t+\tau,X_{\tau}^{x_{2}})\bigg{|}$
$\displaystyle\leq
E\bigg{|}X_{\tau}^{x_{1}}-X_{\tau}^{x_{2}}\bigg{|}+KE\bigg{|}Z(t+\tau,X_{\tau}^{x_{1}})-Z(t+\tau,X_{\tau}^{x_{2}})\bigg{|},$

where the third inequality follows from $(K-y)^{+}-(K-z)^{+}\leq(z-y)^{+}$ for
$y,z\in\mathbb{R}$; after which, let $x_{1}\to x_{2}$, the monotone
convergence theorem (it is applicable because of Corollary 3.3) suggests

$E_{t,x_{1}}\left(e^{-r\tau}G(t+\tau,X_{t+\tau})\right)-E_{t,x_{2}}\left(e^{-r\tau}G(t+\tau,X_{t+\tau})\right)\to
0.$

By combining assertions (i) and (ii), the conclusion follows. ∎

With the aid of Lemma 3.6, we can then establish the optimality of $\tau_{D}$.
(See [24, Corollary 2.9, Page 46] for conditions of existence of the optimal
stopping time.)

######  Lemma 3.7.

The stopping time $\tau_{\mathcal{D}}$ is optimal in (3.30).

###### Proof.

Note that function $G$ is continuous (thereby u.s.c) on
$[0,T)\times(0,\infty)$, which, together with Lemma 3.6, yield that

$\bar{\tau}_{D}=\inf\\{s\in[0,T-t):(t+s,X_{t+s}^{x})\in\bar{\mathcal{D}}\\},$

is optimal (with the convention that the infimum of the empty set is infinite)
and $\bar{\mathcal{D}}=\mathcal{D}\setminus\\{(T,x):x\in(0,\infty)\\}$ in the
following problem

$\displaystyle\bar{V}(t,x)=\sup_{\tau\in[0,T-t)}E_{t,x}\left(e^{-r\tau}G\left(t+\tau,X_{t+\tau}\right)\right),$

so that
$\bar{V}(t,x)=E_{t,x}\left(e^{-r\bar{\tau}_{D}}G\left(t+\bar{\tau}_{D},X_{t+\bar{\tau}_{D}}\right)\right)$,
where its corresponding stopping set is defined as
$\bar{\mathcal{D}}=\mathcal{D}\setminus\\{(T,y):y\in[1,\infty)\\}$.
Furthermore, observe that

$\displaystyle V(t,y)$
$\displaystyle=\sup_{\tau\in[0,T-t]}{E}_{t,x}\left(e^{-r\tau}G\left(t+\tau,X_{t+\tau}\right)\right)$
$\displaystyle=\sup_{\tau\in[0,T-t]}{E}_{t,x}\left(e^{-r\tau}G\left(t+\tau,X_{t+\tau}\right)I\\{\tau<T-t\\}+e^{-r(T-t)}G\left(T,X_{T}\right)I\\{\tau=T-t\\}\right)$
$\displaystyle=\max\bigg{\\{}\sup_{\tau\in[0,T-t)}{E}_{t,x}\left(e^{-r\tau}G\left(t+\tau,X_{t+\tau}\right)\right),\widetilde{E}_{t,x}\left(e^{-r(T-t)}G\left(T,X_{T}\right)\right)\bigg{\\}}$
$\displaystyle=\max\bigg{\\{}\widetilde{E}_{t,x}\left(e^{-r\bar{\tau}_{D}}G\left(t+\bar{\tau}_{D},X_{t+\bar{\tau}_{D}}\right)\right),\widetilde{E}_{t,x}\left(e^{-r(T-t)}G\left(T,X_{T}\right)\right)\bigg{\\}},$

which, together with Corollary 3.3, implies that

$\displaystyle
V(t,y)=\begin{cases}&{E}_{t,x}\left(e^{-r\bar{\tau}_{D}}G\left(t+\bar{\tau}_{D},X_{t+\bar{\tau}_{D}}\right)\right),\,\,\bar{\tau}_{D}<T-t,\\\
&{E}_{t,x}\left(e^{-r(T-t)}G\left(T,X_{T}\right)\right),\qquad\,\,\,\bar{\tau}_{D}>T-t,\end{cases}$

which implies that $\tau_{D}=\bar{\tau}_{D}\wedge(T-t)$, and the desired claim
follows. ∎

It will be shown in the next section that the continuation and stopping sets
can be defined as

$\displaystyle\mathcal{C}=\\{(t,x)\in[0,T)\times(0,\infty):x>b(t)\\},$ (3.34)
$\displaystyle\mathcal{D}=\\{(t,x)\in[0,T)\times(0,\infty):x\leq
b(t)\\}\cup\\{(T,x):x\leq b(T)\\},$ (3.35)

where $b:[0,T]\mapsto\mathbb{R}$ is the unknown optimal stopping boundary such
that $\tau_{\mathcal{D}}$ can be rewritten as

$\displaystyle\tau_{\mathcal{D}}=\inf\\{0\leq s\leq T-t:X_{t+s}^{x}\leq
b(t+s)\\}\wedge(T-t),$ (3.36)

and by Theorem 2.4 in [24, Page 37] and Corollary 3.7, the free-boundary
problem can be formulated as:

$\displaystyle{V}_{t}+\mathbb{L}_{x}{V}-r{V}=0\qquad\qquad\qquad\qquad\,\,\,\text{for
$(t,x)\in\mathcal{C}$},$ (3.37)
$\displaystyle{V}(t,x)={G}(t,x)\qquad\qquad\qquad\qquad\,\,\,\,\,\,\,\,\,\,\text{for
$x=b(t)$, instantaneous stopping},$ (3.38)
$\displaystyle{V}_{x}(t,x)={G}_{x}(t,x)\qquad\qquad\qquad\qquad\,\,\,\,\,\text{for
$x=b(t)\neq L$, smooth-fit},$ (3.39)
$\displaystyle{V}(t,x)>{G}(t,x)\qquad\qquad\qquad\qquad\,\,\,\,\,\,\,\,\,\,\text{for
$(t,x)\in\mathcal{C}$},$ (3.40)
$\displaystyle{V}(t,x)={G}(t,x)\qquad\qquad\qquad\qquad\,\,\,\,\,\,\,\,\,\,\text{for
$(t,x)\in\mathcal{D}$}.$ (3.41)

###  3.3 The Continuation and Stopping Sets

As a further preparation for proceeding the detailed analysis, we lean on the
local time-space formula, as the function $G$ is not differentiable $K$ and
$L$, to obtain

$\displaystyle e^{-rs}{G}\left(t+s,X_{t+s}^{x}\right)$
$\displaystyle={G}(t,x)$
$\displaystyle\,\,\,+\int_{0}^{s}e^{-ru}\left(-r{G}+{G}_{t}+\mathbb{L}_{X}{G}\right)(t+u,X_{t+u}^{x})I\\{X_{t+u}\neq
K,X_{t+u}\neq L\\}du$ $\displaystyle\,\,\,+\int_{0}^{s}e^{-ru}\sigma
X_{t+u}^{x}{G}_{x}(t+u,X_{t+u}^{x})I\\{X_{t+u}\neq K,X_{t+u}\neq L\\}dW_{u}$
$\displaystyle\,\,\,+\frac{1}{2}\int_{0}^{s}e^{-ru}\left({G}_{x}(t+u,L+)-{G}_{x}(t+u,L-)\right)dl_{u}^{L}(X^{x})$
$\displaystyle\,\,\,+\frac{1}{2}\int_{0}^{s}e^{-ru}\left({G}_{x}(t+u,K+)-{G}_{x}(t+u,K-)\right)dl_{u}^{K}(X^{x}),$
(3.42)

where $l_{s}^{K}$ and $l_{s}^{L}$ are the local times of $X$ at the level $K$
and $L$ respectively, and for simplicity, we denote the martingale term as

$\displaystyle M_{s}=\int_{0}^{s}e^{-ru}\sigma
X_{t+u}^{x}{G}_{x}(t+u,X_{t+u}^{x})I\\{X_{t+u}\neq K,X_{t+u}\neq L\\}dW_{u},$

so that $E_{t,x}\left(M_{s}\right)=0$ under measure $P_{t,x}$ for each
$s\in[0,T-t]$.

A fairly easy calculation then shows that

$\displaystyle{G}_{x}(t,L+)=-1,\qquad{G}_{x}(t,L-)=-1+(K-L)Z_{x}(t,L-),$
$\displaystyle{G}_{x}(t,K+)=0,\qquad\,\,\,{G}_{x}(t,K-)=-1,$

and that, upon putting
$H(t,x)=\left(-r{G}+{G}_{t}+\mathbb{L}_{X}{G}\right)(t,x)$ and using
Proposition 3.2,

$\displaystyle
H(t,x)=\begin{cases}\left(-rKZ-\sigma^{2}x^{2}Z_{x}\right)(t,x),&\text{for
$x<L<K$},\\\ -rK,&\text{for $L\leq x<K$},\\\ 0,&\text{for $x>K$},\end{cases}$
(3.43)

after which, (3.42) is reshuffled accordingly as

$\displaystyle e^{-rs}{G}\left(t+s,X_{t+s}^{x}\right)$
$\displaystyle={G}(t,x)+M_{s}+\int_{0}^{s}e^{-ru}H\left(t+u,X_{t+u}^{x}\right)I\\{X_{t+u}\neq
L,X_{t+u}\neq K\\}du$
$\displaystyle\qquad\qquad\,+\frac{1}{2}\int_{0}^{s}e^{-ru}dl_{u}^{K}(X^{x})-\frac{1}{2}\int_{0}^{s}e^{-ru}(K-L)Z_{x}(t+u,L-)dl_{u}^{L}(X^{x}).$

######  Remark 3.8.

Note how very nicely the property of function $Z$ in Proposition 3.2 has saved
us from massive computations in the finite-time problem setup!

Using this, we find that

$\displaystyle
E_{t,x}\left(e^{-rs}{G}\left(t+s,X_{t+s}\right)\right)={G}(t,x)+E_{t,x}\left(-rK\int_{0}^{s}e^{-ru}I\\{L\leq
X_{t+u}<K\\}du\right)$
$\displaystyle\,\,\,\,+E_{t,x}\left(\int_{0}^{s}e^{-ru}\left(-rKZ-\sigma^{2}X_{u}^{2}Z_{x}\right)(t+u,X_{t+u})I\\{X_{t+u}<L\\}du\right)$
$\displaystyle\,\,\,\,-\frac{1}{2}E_{t,x}\left(\int_{0}^{s}e^{-ru}(K-L)Z_{x}(t+u,L-)dl_{u}^{L}(X)\right)+\frac{1}{2}E_{t,x}\left(\int_{0}^{s}e^{-ru}dl_{u}^{K}(X)\right),$
(3.44)

from which, what intuition would suggest is the following:

######  Lemma 3.9.

All points $(t,x)\in[0,T)\times(K,\infty)$ belong to the continuation set
$\mathcal{C}$.

###### Proof.

The proof follows, in fact, from verbal statement, since stopping at
$[0,T)\times(K,\infty)$ gives one null payoff, while waiting, there is a
positive probability to collect strictly positive payoff. ∎

######  Remark 3.10.

Another message, as we shall see in a minute, of equation (3.44) is that the
optimal stopping sets over $[0,T]$ depends on the relative positions of the
level $L$ and of the optimal stopping boundary $B(t)$ of the standard American
put option with payoff $G^{A}(x)=(K-x)^{+}$. (See [3] and [7].)

For notational convenience, we let $V^{A}(t,x)$, $G^{A}(x)=(K-x)^{+}$ and
$B(t)$ denote the value function, the gain function and the optimal stopping
boundary for the standard American put option with the same maturity and
strike price as the current contract. Then, we know from [24] that

$\displaystyle V^{A}(t,x)>G^{A}(x)\qquad\text{for $x>B(t)$},$ $\displaystyle
V^{A}(t,x)=G^{A}(x)\qquad\text{for $x\leq B(t)$},$

where $B(t)$ is unique and $B(T-)=K$; and thus, its stopping set
$\mathcal{D}^{A}$ and continuation set $\mathcal{C}^{A}$ equal:

$\displaystyle\mathcal{D}^{A}=\\{(t,x)\in[0,T]\times(0,\infty):x\leq B(t)\\},$
$\displaystyle\mathcal{C}^{A}=\\{(t,x)\in[0,T]\times(0,\infty):x>B(t)\\}.$

Also we define $t_{*}\in[0,T)$ as follows:

$\displaystyle\begin{cases}&t_{*}>0,\qquad\text{if $B(t_{*})=L$},\\\
&t_{*}=0,\qquad\text{if $B(0)>L$}.\end{cases}$

and since the map $t\mapsto B(t)$ is strictly increasing, $t_{*}$ is unique.
Moreover, let us emphasise that $t_{*}<T$, as $L<K$.

######  Lemma 3.11.

All points $(t,x)\in[t_{*},T]\times[L,B(t)]$ belongs to the stopping set
$\mathcal{D}$.

###### Proof.

Since $(t,x)\in[t_{*},T]\times[L,B(t)]$, it follows from the construction of
the stopping set for the standard American put option that

$\displaystyle V^{A}(t,x)=G^{A}(x),$

and notice that in this set, $G^{A}(x)={G}(t,x)$.

Now, knowing that

$\displaystyle{V}(t,x)$ $\displaystyle=\sup_{0\leq\tau\leq
T-t}E_{t,x}\left(e^{-r\tau}Z\left(t+\tau,X_{t+\tau}\right)(K-X_{t+\tau})^{+}\right)$
$\displaystyle\leq\sup_{0\leq\tau\leq
T-t}E_{t,x}\left(e^{-r\tau}(K-X_{t+\tau})^{+}\right)=V^{A}(t,x),$

where the inequality is due to function $Z$ being a probability and
consequently,

$\displaystyle 0=V^{A}(t,x)-G^{A}(x)\geq{V}(t,x)-{G}(t,x),$

on the set $\\{(t,x)\in[t_{*},T]\times[L,B(t)]\\}$, which in the sense of
${V}\geq{G}$ forces the equation ${V}={G}$ and thereby, proving the desired
assertion. ∎

With a little additional work, we can extend Lemma 3.11 to the following
result:

######  Lemma 3.12.

All points $(t,x)\in[t_{*},T]\times(0,B(t)]$ belongs to the stopping set
$\mathcal{D}$.

###### Proof.

The proof amounts to showing that $[t_{*},T]\times[0,L]$ also belongs to the
stopping set $\mathcal{D}$.

Lemma 3.11 tells us that $[t_{*},T]\times\\{L\\}\subset\mathcal{D}$. Suppose
that we run the process $X$ starting at point $(t,x)\in[t_{*},T]\times(0,L]$,
and we see from (3.44) that in order to compensate the negative term inside
the expectation, the process shall hit level $K$ as the local time term is
positive, but $L<K$, indicating the process can not hit level $K$ before
exercising at level $L$. Therefore, it is optimal to stop immediately on this
set and the conclusion follows. ∎

######  Remark 3.13.

Lemmas 3.11 and 3.12 show that the stopping set is not empty and that if
$t_{*}=0$, then all points $(t,x)\in[0,T]\times(0,L]$ belongs to the stopping
set $\mathcal{D}$, in the sense that

$\displaystyle[0,T]\times[L,B(t)]\subset[0,T]\times(0,B(t)],$

so that in $[0,T]\times[L,B(t)]$, the relation $V^{A}(t,x)=G^{A}(x)$ and
$G^{A}(x)={G}(t,x)$ once again entails ${V}={G}$, after which, the similar
argument as Lemma 3.12 shows $(t,x)\in[0,T]\times(0,L]$ belongs to
$\mathcal{D}$.

$0$$T$$x$$B(t)$$K$$L$

$0$$T$$x$$B(t)$$K$$L$$L+\delta$

Figure 2: Two sketches to illuminate Lemmas 3.11 and 3.12 for $t_{*}=0$. The
area coloured in yellow is the subset of $\mathcal{D}$ and that
$\mathcal{D}^{A}\subset\mathcal{D}$.

$0$$T$$x$$B(t)$$K$$L$$t_{*}$

$0$$T$$x$$B(t)$$K$$L$$L+\delta$$t_{*}$$t_{\delta}$

Figure 3: Two sketches to illuminate Lemmas 3.11 and 3.12 for $t_{*}>0$. The
area coloured in green is the subset of $\mathcal{D}$.

With Remark 3.13 in mind, we now enter the discussions of the properties for
sets $\mathcal{C}$ and $\mathcal{D}$ with $t_{*}=0$ and $t_{*}>0$.

### In the Case of $t_{*}=0$, that is $L<B(0)$.

######  Proposition 3.14.

The stopping set $\mathcal{D}$ is right-connected and the continuation set
$\mathcal{C}$ is left-connected.

###### Proof.

Let $0\leq t_{1}\leq t_{2}\leq T$ and consider the stopping time
$\tau_{L}=\inf\\{0\leq u\leq T-t_{2}:X_{t_{2}+u}^{x}\leq L\\}$ and suppose
that $\tau_{2}$ is the optimal stopping time for ${V}(t_{2},x)$ so that
$\tau_{2}<\tau_{L}$, which is a consequence of Remark 3.13. Then, by the time-
homogeneous property of GBM,

$\displaystyle{V}(t_{1},x)-{V}(t_{2},x)\geq
E\left(e^{-r\tau_{2}}{G}\left(t_{1}+\tau_{2},X_{\tau_{2}}^{x}\right)-e^{-r\tau_{2}}{G}\left(t_{2}+\tau_{2},X_{\tau_{2}}^{x}\right)\right)$
$\displaystyle\,\,\,={G}(t_{1},x)-{G}(t_{2},x)+E\left(\int_{0}^{\tau_{2}}e^{-ru}\left(H\left(t_{1}+u,X_{u}^{x}\right)-H\left(t_{2}+u,X_{u}^{x}\right)\right)I\\{L<X_{u}^{x}<K\\}du\right)$
$\displaystyle\,\,\,={G}(t_{1},x)-{G}(t_{2},x),$

where the second equality follows from
$H\left(t_{1}+u,X_{u}^{x}\right)=H\left(t_{2}+u,X_{u}^{x}\right)=-rK$. This
establishes the following relation:

$\displaystyle{V}(t_{1},x)-{G}(t_{1},x)\geq{V}(t_{2},x)-{G}(t_{2},x),$

from which, we see that $(t_{1},x)\in\mathcal{D}$ implies
$(t_{2},x)\in\mathcal{D}$ and that $(t_{2},x)\in\mathcal{C}$ implies
$(t_{1},x)\in\mathcal{C}$. ∎

######  Proposition 3.15.

The stopping set $\mathcal{D}$ is down-connected and the continuation set
$\mathcal{C}$ is up-connected.

###### Proof.

Immediate from the Proposition 3.14 and the similar argument of the proof in
Lemma 3.12. ∎

We now have sufficient mathematics at our disposal to go around the
singularity of $Z_{x}(t,x)$ for the establishment of the continuity of the
value function for $t_{*}=0$.

######  Lemma 3.16.

The value function $V$ is continuous on $[0,T)\times(0,\infty)$.

###### Proof.

The proof involves proving two assertions: (i) The map $t\mapsto V(t,x)$ is
continuous on $[0,T)$ for each $x\in(0,\infty)$ given and fixed. Take any
$t_{1}<t_{2}$ in $[0,T)$, let $\epsilon>0$ and $\tau_{1}^{\epsilon}$ be the
stopping time such that

$\displaystyle
E_{t_{1},x}\left(e^{-r\tau_{1}^{\epsilon}}G\left(t_{1}+\tau_{1}^{\epsilon},X_{t_{1}+\tau_{1}^{\epsilon}}\right)\right)\geq
V(t_{1},x)-\epsilon.$

Then, by setting $\tau_{2}^{\epsilon}=\tau_{1}^{\epsilon}\wedge(T-t_{2})$, we
have

$\displaystyle
E_{t_{2},x}\left(e^{-r\tau_{2}^{\epsilon}}G\left(t_{2}+\tau_{2}^{\epsilon},X_{t_{2}+\tau_{2}^{\epsilon}}\right)\right)\leq
V(t_{2},x).$

and that

$\displaystyle 0\leq V(t_{1},x)-V(t_{2},x)$ $\displaystyle\leq
E\left(e^{-r\tau_{1}^{\epsilon}}G\left(t_{1}+\tau_{1}^{\epsilon},X_{t_{1}+\tau_{1}^{\epsilon}}^{x}\right)\right)-E\left(e^{-r\tau_{2}^{\epsilon}}G\left(t_{2}+\tau_{2}^{\epsilon},X_{t_{2}+\tau_{2}^{\epsilon}}^{x}\right)\right)+\epsilon$
$\displaystyle\leq
E\left(e^{-r\tau_{2}^{\epsilon}}\left(G\left(t_{1}+\tau_{1}^{\epsilon},X^{x}_{\tau_{1}^{\epsilon}}\right)-G\left(t_{2}+\tau_{2}^{\epsilon},X^{x}_{\tau_{2}^{\epsilon}}\right)\right)\right)+\epsilon$
$\displaystyle\leq
E\left(G\left(t_{1}+\tau_{1}^{\epsilon},X^{x}_{\tau_{1}^{\epsilon}}\right)-G\left(t_{2}+\tau_{2}^{\epsilon},X^{x}_{\tau_{2}^{\epsilon}}\right)\right)+\epsilon$
$\displaystyle\leq
E\left(\left(X^{x}_{\tau_{2}^{\epsilon}}-X^{x}_{\tau_{1}^{\epsilon}}\right)^{+}Z\left(t_{1}+\tau_{1}^{\epsilon},X^{x}_{\tau_{1}^{\epsilon}}\right)\right.$
$\displaystyle\qquad\,\,\,\,+\left.\left(K-X^{x}_{\tau_{2}^{\epsilon}}\right)^{+}\left(Z\left(t_{1}+\tau_{1}^{\epsilon},X^{x}_{\tau_{1}^{\epsilon}}\right)-Z\left(t_{2}+\tau_{2}^{\epsilon},X^{x}_{\tau_{2}^{\epsilon}}\right)\right)\right)+\epsilon,$
(3.45)

where the first inequality is because of the map $t\mapsto V(t,x)$ being
decreasing and the last inequality holds via

$\displaystyle(K-y)^{+}-(K-z)^{+}\leq(z-y)^{+}\qquad\text{for
$y,z\in\mathbb{R}$}.$ (3.46)

By letting $t_{1}\to t_{2}$, using $\tau_{1}^{\epsilon}\to\tau_{2}^{\epsilon}$
and then $\epsilon\to 0$ on (3.45), the dominated convergence theorem allows
us to conclude that

$\displaystyle V(t_{1},x)-V(t_{2},x)\to 0,$

and assertion (i) follows.

(ii) The map $x\mapsto V(t,x)$ is continuous on $(0,\infty)$ for each
$t\in[0,T)$ given and fixed.

Note that by the up-down connectedness of $\mathcal{C}$ and $\mathcal{D}$, for
any $x_{1}<x_{2}$ in $(0,\infty)$,

$\displaystyle G(t,x_{2})-G(t,x_{1})\leq V(t,x_{2})-V(t,x_{1}),$

and let $\tau_{2}$ be optimal for $V(t,x_{2})$ such that

$\displaystyle G(t,x_{2})-G(t,x_{1})$ $\displaystyle\leq
V(t,x_{2})-V(t,x_{1})$ $\displaystyle\leq
E\left(e^{-r\tau_{2}}G(t+\tau_{2},X_{\tau_{2}}^{x_{2}})\right)-E\left(e^{-r\tau_{2}}G(t+\tau_{2},X_{\tau_{2}}^{x_{1}})\right),$
(3.47)

after which, by letting $x_{1}\to x_{2}$, the dominated convergence theorem
leads us to

$V(t,x_{2})-V(t,x_{1})\to 0$

and thereby, proving the left-continuity of $V$.

And, needless to say, if $x_{2}\leq B(t)$, Lemma 3.12 tells us that
$\tau_{2}=0$ and the continuity of $V$ follows directly from that of the gain
function $G$.

On the other hand, if $x_{2}>B(t)$, we know there exists $\delta>0$ so that
$B(t)>L+\delta>L$ (see figure 2) and that for
$0<x_{2}-x_{1}<\frac{x_{1}\delta}{2L}$, immediately from the monotonicity of
$B$, we have

$X_{\tau_{2}}^{x_{2}}\geq B(t+\tau_{2})>B(t)>L+\delta,$

which implies $X_{\tau_{2}}^{x_{1}}>L$ a.s.; more precisely, by the strong
solution of GBM and the optimality of $\tau_{2}$,
$X_{\tau_{2}}^{x_{2}}>L+\delta$ entails

$e^{\left(r-\frac{\sigma^{2}}{2}\right)\tau_{2}+\sigma
W_{\tau_{2}}}>\frac{L+\delta}{x_{2}},$

and that for $x_{2}\in\left(0,x_{1}\left(1+{\delta}/{(2L)}\right)\right)$,

$\displaystyle
X_{\tau_{2}}^{x_{1}}=x_{1}e^{\left(r-\frac{\sigma^{2}}{2}\right)\tau_{2}+\sigma
W_{\tau_{2}}}>\frac{x_{1}(L+\delta)}{x_{2}}>\frac{x_{1}(L+\delta)}{x_{1}\left(1+\delta/(2L)\right)}=L+\frac{\delta
L}{\delta+2L}>L,$

after which, it follows that as $Z(t,x)=1$ for $x>L$,

$\displaystyle G(t,x_{2})-G(t,x_{1})$ $\displaystyle\leq
V(t,x_{2})-V(t,x_{1})$ $\displaystyle\leq
E\Big{(}e^{-r\tau_{2}}G\left(t+\tau_{2},X_{\tau_{2}}^{x_{2}}\right)\Big{)}-E\Big{(}e^{-r\tau_{2}}G\left(t+\tau_{2},X_{\tau_{2}}^{x_{1}}\right)\Big{)}$
$\displaystyle\leq(x_{1}-x_{2})^{+}E\left(e^{-\frac{\sigma^{2}}{2}\tau_{2}+\sigma
W_{\tau_{2}}}\right)=(x_{1}-x_{2})^{+},$

and then by letting $x_{2}\to x_{1}$, we have $V(t,x_{2})-V(t,x_{1})\to 0$,
that is, the right-continuity of $V$. By combining the claims that $V$ is left
and right-continuous for $x\in(0,\infty)$ for each $t\in[0,T)$ given and
fixed, statement (ii) follows. ∎

By taking advantage of the above analysis , we obtain the following result:

######  Proposition 3.17 (The Free-boundary Problem).

The stopping set and the continuation set are of the forms:

$\displaystyle\mathcal{C}=\\{(t,x)\in[0,T)\times(0,\infty):x>b(t)\\},$
$\displaystyle\mathcal{D}=\\{(t,x)\in[0,T)\times(0,\infty):x\leq
b(t)\\}\cup\\{(T,x):x\leq b(T)\\},$

where the map $t\mapsto b(t)$ is increasing.

The free-boundary problem is rearranged as follows:

$\displaystyle{V}_{t}+\mathbb{L}_{x}{V}-r{V}=0\qquad\qquad\text{for
$(t,x)\in\mathcal{C}$},$ (3.48)
$\displaystyle{V}(t,x)=K-x\qquad\qquad\,\,\,\,\,\,\,\,\text{for $x=b(t)$,
instantaneous stopping},$ (3.49)
$\displaystyle{V}_{x}(t,x)=-1\qquad\qquad\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{for
$x=b(t)$, smooth-fit},$ (3.50)
$\displaystyle{V}(t,x)>{G}(t,x)\qquad\qquad\,\,\,\,\,\,\text{for
$(t,x)\in\mathcal{C}$},$ (3.51)
$\displaystyle{V}(t,x)={G}(t,x)\qquad\qquad\,\,\,\,\,\,\text{for
$(t,x)\in\mathcal{D}$}.$ (3.52)

Just to complete the picture of the free-boundary problem, we now justify
equation (3.50).

###### Proof of the Smooth-fit Condition.

Let a point $(t,x)\in(0,T)\times(0,\infty)$ lying on the boundary $b$ be
fixed, i.e. $x=b(t)$. Then, $L<x<K$ and for all $\epsilon>0$ such that
$L<x+\epsilon<K$, we have

$\displaystyle\frac{V(t,x+\epsilon)-V(t,x)}{\epsilon}$
$\displaystyle\geq\frac{G(t,x+\epsilon)-G(t,x)}{\epsilon}=\frac{(K-x-\epsilon)-(K-x)}{\epsilon}=-1,$

of which, taking the limit as $\epsilon\to 0$, we have
$\lim\limits_{\epsilon\to 0}\frac{V(t,x+\epsilon)-V(t,x)}{\epsilon}\geq-1$.

In order to prove the converse inequality, we fix a sufficiently small
$\epsilon$ so that $0<\epsilon<\frac{x-L}{2}$ and that $L<x+\epsilon<K$, and
consider the optimal stopping time $\tau_{\epsilon}$ for $V(t,x+\epsilon)$ so
that the monotonicity of $b$ tells us that
$X_{\tau_{\epsilon}}^{x+\epsilon}=b(t+\tau_{\epsilon})>b(t)\geq L$ implies
$X_{\tau_{\epsilon}}^{x}>L$ almost surely, as before, by the strong solution
of GBM,

$e^{\left(r-\frac{\sigma^{2}}{2}\right)\tau_{\epsilon}+\sigma
W_{\tau_{\epsilon}}}>\frac{b(t)}{x+\epsilon}=\frac{x}{x+\epsilon},$

so that, for $x>L$,

$\displaystyle xe^{\left(r-\frac{\sigma^{2}}{2}\right)\tau_{\epsilon}+\sigma
W_{\tau_{\epsilon}}}>\frac{x^{2}}{x+\epsilon}=\frac{x^{2}-\epsilon^{2}+\epsilon^{2}}{x+\epsilon}>x-\epsilon>x-\frac{x-L}{2}=\frac{x+L}{2}>L.$

Then, we have

$\displaystyle\frac{V(t,x+\epsilon)-V(t,x)}{\epsilon}\leq\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}G(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon})\right)-E\left(e^{-r\tau_{\epsilon}}G(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x})\right)$
$\displaystyle\qquad=\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}\left(\left(K-X_{\tau_{\epsilon}}^{x+\epsilon}\right)^{+}-\left(K-X_{\tau_{\epsilon}}^{x}\right)^{+}\right)Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon})\right.$
$\displaystyle\qquad\qquad\,\,\,+\left.e^{-r\tau_{\epsilon}}\left(K-X_{\tau_{\epsilon}}^{x}\right)^{+}\left(Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon})-Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x})\right)\right)$
$\displaystyle\qquad\leq\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}\left(X_{\tau_{\epsilon}}^{x}-X_{\tau_{\epsilon}}^{x+\epsilon}\right)Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon})I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}\right.$
$\displaystyle\qquad\qquad\,\,\,+\left.e^{-r\tau_{\epsilon}}\left(K-X_{\tau_{\epsilon}}^{x}\right)^{+}\left(Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon})-Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x})\right)\right)$
$\displaystyle\qquad=-E\left(e^{-\frac{\sigma^{2}\tau_{\epsilon}}{2}+\sigma
W_{\tau_{\epsilon}}}Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon})I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}\right)$
$\displaystyle\qquad\,\,\,\,\,\,\,+\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}\left(K-X_{\tau_{\epsilon}}^{x}\right)^{+}\left(Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon})-Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x})\right)\right)$
$\displaystyle\qquad\leq-E\left(e^{-\frac{\sigma^{2}\tau_{\epsilon}}{2}+\sigma
W_{\tau_{\epsilon}}}Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon})I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}\right)$
$\displaystyle\qquad\,\,\,\,\,\,\,+\frac{1}{\epsilon}E\left(\left(K-X_{\tau_{\epsilon}}^{x}\right)^{+}\left(Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon})-Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x})\right)\right)$
$\displaystyle\qquad\leq-E\left(e^{-\frac{\sigma^{2}\tau_{\epsilon}}{2}+\sigma
W_{\tau_{\epsilon}}}Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon})I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}\right)$
$\displaystyle\qquad\,\,\,\,\,\,\,+\frac{K}{\epsilon}E\left(I\\{X_{\tau_{\epsilon}}^{x}<L\\}\left(Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon})-Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x})\right)\right)$
$\displaystyle\qquad=-E\left(e^{-\frac{\sigma^{2}\tau_{\epsilon}}{2}+\sigma
W_{\tau_{\epsilon}}}Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon})I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}\right)$

where the second inequality follows from

$\displaystyle\left(K-X_{\tau_{\epsilon}}^{x+\epsilon}\right)^{+}-\left(K-X_{\tau_{\epsilon}}^{x}\right)^{+}$
$\displaystyle=\left(X_{\tau_{\epsilon}}^{x}-X_{\tau_{\epsilon}}^{x+\epsilon}\right)I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}-\left(K-X_{\tau_{\epsilon}}^{x}\right)^{+}I\\{X_{\tau_{\epsilon}}^{x+\epsilon}\geq
K\\}$
$\displaystyle\leq\left(X_{\tau_{\epsilon}}^{x}-X_{\tau_{\epsilon}}^{x+\epsilon}\right)I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\},$
(3.53)

and the second equality is from the strong solution of $X$ and linearity;
moreover, the fourth inequality is due to the fact that
$I\\{X_{\tau_{\epsilon}}^{x}\geq L\\}=1$ entails
$I\\{X_{\tau_{\epsilon}}^{x+\epsilon}\geq L\\}=1$ and that
$Z\left(t+\epsilon,X_{\tau_{\epsilon}}^{x}\right)=Z\left(t+\epsilon,X_{\tau_{\epsilon}}^{x+\epsilon}\right)=1$,
while we used $I\\{X_{\tau_{\epsilon}}^{x}<L\\}=0$ for chosen $\epsilon$ in
the last step. Via an appeal to the dominated convergence theorem, upon
letting $\epsilon\to 0$, we have $\tau_{\epsilon}\to 0$ (see A.1 for its
justification) so that

$\displaystyle E\left(e^{-\frac{\sigma^{2}\tau_{\epsilon}}{2}+\sigma
W_{\tau_{\epsilon}}}Z(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon})I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}\right)\to
1.$

Thus,

$\displaystyle\lim\limits_{\epsilon\to
0}\frac{V(t,x+\epsilon)-V(t,x)}{\epsilon}\leq-1,$

after which, the conclusion follows via the fact that for $(t,x)$ and
$(t,x-\epsilon)$ belong to $\mathcal{D}$,

$\frac{V(t,x)-V(t,x-\epsilon)}{\epsilon}=\frac{G(t,x)-G(t,x-\epsilon)}{\epsilon}=-1.$

∎

######  Remark 3.18.

The noteworthy feature of the value function is that it is not smooth in the
stopping set $\mathcal{D}$, to be more specific, at level $L$. This fact
implies the presence of a local time term in the contract pricing formula.
(see [7, Page 7] and [28] for similar results)

To prepare for the main theorem presented in Subsection 3.4, we further
justify the continuity of the optimal stopping boundary.

######  Lemma 3.19.

The optimal stopping boundary $b$ is continuous on $[0,T]$ and $b(T-)=K$.

###### Proof.

Th proof proceeds in three steps, which follows essentially from [24, Page
383].

(i) The boundary $b$ is right-continuous on $[0,T]$.

Let us fix $t\in[0,T)$ and take a sequence $t_{n}\downarrow t$ as
$n\to\infty$. Since $b$ is increasing, the right-limit $b(t+)$ exists.
Remember that $\mathcal{D}$ is a closed set, meaning that its limit point
$\lim_{n\to\infty}(t_{n},b(t_{n}))\to(t,b(t+))$ is contained in $\mathcal{D}$,
it then follows, together with the construction of the stopping set that
$b(t+)\leq b(t)$. However, the fact that $b$ is increasing suggests that
$b(t+)\geq b(t)$ and thereby, forcing the equation $b(t+)=b(t)$ and proving
statement (i).

(ii) The boundary $b$ is left-continuous on $[0,T)$.

Assume that, for contradiction, at some point $t\in(0,T)$, the function makes
a jump, that is, $b(t)>b(t-)$. Then, fix a point $y_{*}\in(b(t-),b(t))$. By
(3.49) and (3.50), we have

$\displaystyle
V(s,y_{*})-G(s,y_{*})=\int_{b(s)}^{y_{*}}\int_{b(s)}^{x}\left(V_{xx}-G_{xx}\right)(s,z)dzdx,$
(3.54)

for each $s\in(t-\delta,t)$, where $\delta>0$ and $t-\delta>0$. Despite the
fact that function $G$ is not differentiable at $L$, the fact that $b(t)>L$
once again comes to our rescue.

Knowing that the value function satisfies (3.48), we see that, for
$(t,x)\in\mathcal{R}$, where $\mathcal{R}$ is a curved trapezoid formed by the
vertices $(s,b(s))$, $(s,y_{*})$, $(t,b(t-))$ and $(t,y_{*})$,

$\displaystyle V_{xx}(t,x)=\frac{2}{\sigma^{2}x^{2}}\left(rV-
V_{t}-rxV_{x}\right)(t,x)>\frac{2}{\sigma^{2}x^{2}}rV(t,x)>\frac{2}{\sigma^{2}x^{2}}rG(t,x)\geq
c>0,$

where the first strict inequality follows via the fact that both the maps
$t\mapsto V(t,x)$ and $x\mapsto V(t,x)$ are decreasing on $\mathcal{R}$, and
the last one is due to the fact that $x<K$ on $\mathcal{R}$. Moreover,
$G_{xx}(t,x)=0$ on $\mathcal{R}$.

Then, from the continuity of the value function and the gain function, upon
letting $s\uparrow t$ on (3.54) and using dominated convergence theorem,

$\displaystyle V(t-,y_{*})-G(t-,y_{*})=V(t,y_{*})-G(t,y)$
$\displaystyle\geq\int_{b(t-)}^{y_{*}}\int_{b(t-)}^{x}cdzdx$
$\displaystyle=\frac{c(y_{*}-b(t-))^{2}}{2}>0,$

which is a contradiction, since $(t,y_{*})\in\mathcal{D}$, indicating that
such point cannot exist and $b(t-)=b(t)$ must hold true. Immediate from
statements (i) and (ii), the continuity of $b$ follows.

(iii) $b(T-)=K$.

In Lemma 3.9, we learned that $b(T-)\in[L,K]$.

Assume that, $b(T-)<K$ so that there exists a point $y_{*}\in(b(T-),K)$ and
let $s\in[T-\delta,T)$ with $0<\delta<T$. Then, by rerunning the proof of
statement (ii) with the above modifications and letting $s\uparrow T$, we find
that from the continuity of $V$ and $G$ on $[0,T]\times[L,\infty)$,

$\displaystyle V(T,y_{*})-G(T,y_{*})>0,$

which contradicts the fact that $\\{(T,x):x<b(T)\\}\subset\mathcal{D}$ and
thus implies that $b(T-)=K$ (because of its continuity, $b(T)=K$). ∎

### In the Case of $0<t_{*}<T$, that is $L>B(0)$.

######  Proposition 3.20.

On the state space $[0,T]\times[L,\infty)$, the stopping set $\mathcal{D}$ is
right-connected and the continuation set $\mathcal{C}$ is left-connected.

###### Proof.

Since on the state space $[0,T]\times[L,\infty)$, the map $t\mapsto V(t,x)$ is
decreasing and that $G(t,x)=(K-x)^{+}$, the map $t\mapsto V(t,x)-G(t,x)$ is
decreasing also. That is, for any $t_{1},t_{2}\in[0,T]$, if $t_{1}<t_{2}$,
then

$\displaystyle V(t_{1},x)-G(t_{1},x)\geq V(t_{2},x)-G(t_{2},x),$

such that $(t_{1},x)\in\mathcal{D}$ implies $(t_{1},x)\in\mathcal{D}$ and
$(t_{2},x)\in\mathcal{C}$ implies $(t_{1},x)\in\mathcal{C}$, which is
precisely the claim. ∎

Proposition 3.20 immediately tells us the following result, which we state
without proof:

######  Corollary 3.21.

On the state space $[0,T]\times[L,\infty)$,
$\mathcal{D}^{A}\subseteq\mathcal{D}$.

To fill the last slot of the diagram, we shall show that the stopping set
$\mathcal{D}$ is not empty on state space $[0,t_{*}]\times(0,L]$, but first we
present some important facts:

######  Lemma 3.22.

(i) The gain function $G$ is uniformly continuous on the state space
$[0,t_{*}]\times(0,L]$;

(ii) The map $t\mapsto\max\limits_{x\in(0,L]}G(t,x)$ is continuous on
$[0,t_{*}]$.

(iii) Given that for each $t\in[0,t_{*}]$, $\max\limits_{x\in(0,L]}G(t,x)$ is
reached at a unique point $x^{*}(t)$, in particular,

$\displaystyle\max\limits_{x\in(0,L]}G(t,x)=G(t,x^{*}(t)),$ (3.55)

the map $t\mapsto x^{*}(t)$ is continuous on $[0,t_{*}]$.

######  Remark 3.23.

Knowing from derivative computation that for each fixed $t$, the maximum of G
achieved in $(0,L]$ is the global maximum over the domain $(0,\infty)$, we can
write

$\max\limits_{x\in(0,\infty)}G(t,x)=\max\limits_{x\in(0,L]}G(t,x).$

As a matter of fact, $t\mapsto x^{*}(t)$ is non-decreasing, see Figure 4.

###### Proof of Lemma 3.22 .

(i) To show that $G$ is uniformly continuous on $[0,t_{*}]\times(0,L]$, is to
show that for every $\epsilon>0$, there exists a $\delta>0$ so that for all
$(t_{1},x_{1}),(t_{2},x_{2})\in[0,t_{*}]\times(0,L]$, $\lvert
t_{1}-t_{2}\rvert+\lvert x_{1}-x_{2}\rvert<\delta$ implies $\lvert
G(t_{1},x_{1})-G(t_{2},x_{2})\rvert<\epsilon$.

Notice that

$\displaystyle\lvert G(t_{1},x_{1})-G(t_{2},x_{2})\rvert$
$\displaystyle\leq\lvert x_{2}-x_{1}\rvert Z(t_{1},x_{1})+\lvert
K-x_{2}\rvert\lvert Z(t_{1},x_{1})-Z(t_{2},x_{2})\rvert$ $\displaystyle<\lvert
x_{2}-x_{1}\rvert+K\lvert Z(t_{1},x_{1})-Z(t_{2},x_{2})\rvert.$

Observe that on $[0,t_{*}]\times(0,L]$,

$\displaystyle\lvert Z(t_{1},x_{1})-Z(t_{2},x_{2})\rvert$
$\displaystyle=\lvert
Z(t_{1},x_{1})-Z(t_{2},x_{1})+Z(t_{2},x_{1})-Z(t_{2},x_{2})\rvert$
$\displaystyle\leq\lvert Z(t_{1},x_{1})-Z(t_{2},x_{1})\rvert+\lvert
Z(t_{2},x_{1})-Z(t_{2},x_{2})\rvert$ $\displaystyle\leq M\lvert
t_{1}-t_{2}\rvert+N\lvert x_{1}-x_{2}\rvert,$

where the second inequality holds true by the mean value theorem and the fact
that the partial derivatives of $Z_{x}(t,x)$ and $Z_{t}(t,x)$ are bounded, say
by constants $N$ and $M$, on the set $[0,t_{*}]\times(0,L]$ (see (A.85),
(A.86)). Hence,

$\displaystyle\lvert G(t_{1},x_{1})-G(t_{2},x_{2})\rvert<KM\lvert
t_{1}-t_{2}\rvert+(KN+1)\lvert x_{1}-x_{2}\rvert,$

and we can now choose $\delta=\frac{\epsilon}{\max\\{KM,KN+1\\}}$ and verify
that if $(t_{1},x_{1}),(t_{2},x_{2})\in[0,t_{*}]\times(0,L]$ satisfy $\lvert
t_{1}-t_{2}\rvert+\lvert x_{1}-x_{2}\rvert<\delta$, then

$\displaystyle\lvert G(t_{1},x_{1})-G(t_{2},x_{2})\rvert<\epsilon,$

thus proving statement (i) as desired.

(ii) The uniform continuity of $G$ means that for every $\epsilon>0$, there
exists a $\delta>0$ such that for all $t_{1},t_{2}\in[0,t_{*}]$ and
$x\in(0,L]$,

$\displaystyle\text{$\lvert t_{1}-t_{2}\rvert<\delta$ implies $\lvert
G(t_{1},x)-G(t_{2},x)\rvert<\epsilon$ },$

which, after rearranging, equals

$\displaystyle G(t_{2},x)-\epsilon\leq G(t_{1},x)\leq G(t_{2},x)+\epsilon,$

so we must have,

$\displaystyle\max_{x\in(0,L]}G(t_{1},x)$
$\displaystyle\geq\max_{x\in(0,L]}G(t_{2},x)-\epsilon,$
$\displaystyle\max_{x\in(0,L]}G(t_{1},x)$
$\displaystyle\leq\max_{x\in(0,L]}G(t_{2},x)+\epsilon,$

after which, the conclusion follows from

$\displaystyle\lvert\max_{x\in(0,L]}G(t_{1},x)-\max_{x\in(0,L]}G(t_{2},x)\rvert<\epsilon.$

(iii) Assume that, for contradiction, statement (iii) is false. To say that,
$t\mapsto x^{*}(t)$ is not continuous on $[0,t_{*}]$ means that there exists a
sequence $(t_{n})\subseteq[0,t_{*}]$ where $(t_{n})\to t$ (Since the the set
is compact, the limit point $t$ is always contained on $[0,t_{*}]$.) such that
$x^{*}(t_{n})$ does not converge $x^{*}(t)$.

According to statement (ii) and the definition of $x^{*}(t)$,

$\displaystyle\lim_{t_{n}\to
t}\max_{x\in(0,L]}G(t_{n},x)\to\max_{x\in(0,L]}G(t,x)=G(t,x^{*}(t)),$

but the assumption is

$\displaystyle\lim_{t_{n}\to
t}\max_{x\in(0,L]}G(t_{n},x)\to\max_{x\in(0,L]}G(t,x)=G\left(t,\lim_{t_{n}\to
t}x^{*}(t_{n})\right)\neq G(t,x^{*}(t)),$

which is a contradiction, we may therefore conclude that statement (iii) holds
true. ∎

What Lemma 3.22 tells us is that:

######  Lemma 3.24.

All points $(t,x)\in[0,t_{*}]\times(0,x^{*}(t)]$ belongs to the stopping set
$\mathcal{D}$.

###### Proof.

According to the definition of $x^{*}$, Remark 3.23 and the fact that
$t\mapsto G(t,x)$ is decreasing,

$\displaystyle G(t,x^{*}(t))\geq\max_{x\in(0,\infty)}G(t+\tau,x)\geq
G(t+\tau,x),$

and by domination, taking expectation under measure $P_{t,x*(t)}$,

$\displaystyle G(t,x^{*}(t))\geq
E_{t,x^{*}(t)}\left(e^{-r\tau}G(t+\tau,X_{t+\tau})\right),$

so that $G(t,x^{*}(t))\geq V(t,x^{*}(t))$, but $V\geq G$ forces the equation
$G(t,x^{*}(t))=V(t,x^{*}(t))$, and thus showing that all
$(t,x^{*}(t))\in\mathcal{D}$. Furthermore, due to Lemma 3.22 and Remark 3.23,
an analogous argument as Lemma 3.12 does the rest for us. ∎

Figure 4: This figure displays the maps $t\mapsto x^{*}(t)$ with chosen
parameters $r=0.05$, $\sigma=0.4$, $L=6.5$, $K=7$, $T=5$, $t_{*}\approx 4.98$.

$0$$T$$x$$B(t)$$K$$L$$t_{*}$$L+\delta$$t_{\delta}$$x_{*}(t)$

$0$$T$$x$$B(t)$$K$$L$$t_{*}$$L+\delta$$t_{\delta}$$x_{*}(t)$

Figure 5: Two sketches to illuminate Lemmas 3.24 for $t_{*}>0$. The areas
coloured in green is $\mathcal{S}\subset\mathcal{D}$ and the blue dashed lines
are plotting function $x^{*}:[0,t_{*}]\mapsto(0,L]$. The right figure is to
illuminate the possible case stated in Remark 3.35.

######  Remark 3.25.

It has been high-lighted in multiple literatures (if not all of them, such as
[19, Page 113]) that if either the maps $t\mapsto H(t,x)$ or $x\mapsto H(t,x)$
are monotone (the later monotonicity is used to establish the up-connectedness
of the sets), we will not be able to compare two integrals with different
signs and proves most of the properties possessed by the optimal stopping
boundary, which is probably the only downfall of such analytical method but
opens the door to many interesting research possibilities; this is,
unfortunately, the story of $K\leq L$. By narrowing down the scale of the
continuation set (the idea behind this is that, in the American put option
pricing problem, the lower bound for the continuation set when $T<\infty$ is
the optimal stopping level obtained from pricing the perpetual American put
option), we can make an additional assumption to further develop some insights
towards the shape of sets $\mathcal{C}$ and $\mathcal{D}$.

######  Assumption 3.26.

Parameters $r$ and $\sigma$ can be chosen such that the map of $t\mapsto
H(t,x)$ is decreasing on the sets
$[0,t_{*}]\times[x^{*}(t),\infty)\cup[t_{*},T]\times[B(t),\infty)\cup\\{t_{*}\\}\times[x^{*}(t_{*}),L]$.

######  Remark 3.27.

Note that the above assumption holds true in the state space
$[0,t_{*}]\times[x^{*}(t),L]$ generally, see Figure 6. In addition, for $x>L$,
$H=-rK$.

Figure 6: This figure displays the maps $x\mapsto H_{t}(t,x)$ with chosen
parameters $r=0.05$, $\sigma=0.4$, $L=6.5$, $K=7$, $T=5$. Different-coloured
lines indicate different $t$ (and thus different $x^{*}(t)$).

For future reference, set

$\displaystyle\mathcal{S}=[0,t_{*}]\times(0,x^{*}(t)]\cup\\{t_{*}\\}\times[x^{*}(t_{*}),L]\cup[t_{*},T]\times(0,B(t)),$

so that $\mathcal{S}\subseteq\mathcal{D}$.

######  Proposition 3.28.

The map $t\mapsto V(t,x)-G(t,x)$ is decreasing on $[0,T]$.

###### Proof.

Let $0\leq t_{1}<t_{2}\leq T$ and consider the stopping times:

$\displaystyle\tau_{S}=\inf\\{0\leq u\leq T-t_{2}:X_{u}^{x}\in\mathcal{S}\\}$

and $\tau_{2}$ be the optimal stopping time for $V(t_{2},x)$, by the fact that
$\mathcal{S}\subseteq\mathcal{D}$, a direct result is $\tau_{2}\leq\tau_{S}$.
Then, we have

$\displaystyle V(t_{1},x)-V(t_{2},x)\geq
E\left(e^{-r\tau_{2}}G\left(t_{1}+\tau_{2},X_{\tau_{2}}^{x}\right)\right)-E\left(e^{-r\tau_{2}}G\left(t_{2}+\tau_{2},X_{\tau_{2}}^{x}\right)\right)$
$\displaystyle=G(t_{1},x)-G(t_{2},x)$
$\displaystyle\,\,\,+E\left(\int_{0}^{\tau_{2}}e^{-ru}\left(-rKZ-\sigma^{2}X_{u}^{2}Z_{x}\right)\left(t_{1}+u,X_{u}\right)I\\{X_{u}^{x}<L\\}du\right)$
$\displaystyle\,\,\,-E\left(\int_{0}^{\tau_{2}}e^{-ru}\left(-rKZ-\sigma^{2}X_{u}^{2}Z_{x}\right)\left(t_{2}+u,X_{u}\right)I\\{X_{u}^{x}<L\\}du\right)$
$\displaystyle\,\,\,+E\left(-\frac{1}{2}\int_{0}^{\tau_{2}}e^{-ru}(K-L)Z_{x}(t_{1},L-)dl_{u}^{L}(X^{x})\right)$
$\displaystyle\,\,\,-E\left(-\frac{1}{2}\int_{0}^{\tau_{2}}e^{-ru}(K-L)Z_{x}(t_{2},L-)dl_{u}^{L}(X^{x})\right)$
$\displaystyle=G(t_{1},x)-G(t_{2},x)$
$\displaystyle\,\,\,+E\left(\int_{0}^{\tau_{2}}e^{-ru}\left(H\left(t_{1}+u,X_{u}^{x}\right)-H(t_{2}+u,X_{u}^{x})\right)I\\{X_{u}^{x}<L\\}du\right)$
$\displaystyle\,\,\,+E\left(\frac{1}{2}\int_{0}^{\tau_{2}}e^{-ru}(K-L)\left(Z_{x}(t_{2},L-)-Z_{x}(t_{1},L-)\right)dl_{u}^{L}(X^{x})\right)\geq
G(t_{1},x)-G(t_{2},x),$

where the second inequality holds true via the map $t\mapsto H(t,x)$ is
decreasing and $t\mapsto Z_{x}(t,L-)$ is increasing (see A). In particular, we
have arrive at the inequality:

$\displaystyle V(t_{1},x)-G(t_{1},x)\geq V(t_{2},x)-G(t_{2},x),$

which is precisely the claim. ∎

The usefulness of Proposition 3.28 is well demonstrated by the following
results:

######  Proposition 3.29.

(i) The stopping set $\mathcal{D}$ is right-connected, i.e. increasing w.r.t
time $t$;

(ii) The continuation set $\mathcal{C}$ is left-connected, i.e. decreasing
w.r.t time $t$;

(iii) The stopping set $\mathcal{D}$ is down-connected;

(iv) The continuation set $\mathcal{C}$ is up-connected.

###### Proof.

Assertions (i) and (ii) are immediate from Proposition 3.28, from which, (iii)
and (iv) follow. ∎

######  Remark 3.30.

Once again, because the gain function is not smooth in the stopping set
$\mathcal{D}$ at level $L$, the presence of the local-time term in the pricing
formula should not come to us as a surprise; above all, the twist comes when
the optimal stopping boundary is crossing or remaining for some time at $L$
(consider the situation for $B(t)<L<K$ and $x^{*}(t)=L$ for all
$t\in[0,t_{*}]$, then $\\{(t,x)\in[0,T]\times[0,L]\\}\subset\mathcal{D}$), the
smooth-fit condition fails to hold. See [7, Page 19] for similar treatment of
such problem.

######  Remark 3.31.

The minimal conditions under which the smooth-fit condition can hold in
greater generality are the regularity of the diffusion process $X$ and the
differentiability of the gain function $G$. For further contribution and
examples, we refer to [24, Page 155].

To further construct the continuity of the value function, we need a
definition first:

######  Definition 3.32.

Let times $t_{b}$ and $t^{b}$ be defined as follows:

$\displaystyle\begin{cases}b(t)<L,&\text{for $t\in[0,t_{b})$,}\\\
b(t)=L,&\text{for $t\in[t_{b},t^{b}]$,}\\\ b(t)>L,&\text{for
$t\in(t^{b},T]$},\end{cases}$

and if $b(t)\geq L$ for all $t\in[0,T]$, we set $t_{b}=0$. Note also that if
$t_{b}=t^{b}$, then $t^{b}<T$.

As the reader has, hopefully noticed in the proof of Lemma 3.16 and
Proposition 3.17, the difficulty in the proof of continuity of $V$ and smooth-
fit condition lies in the singular point $(L,T)$ of $Z_{x}(t,x)$ (and thereby
in the applicability of the dominated convergence theorem). To overcome this,
the key idea is to keep the state space away from such a point by taking the
continuity and the monotonicity of $B$ in $[t_{*},T]$ into account (and these
properties come for free!).

######  Lemma 3.33.

The value function is continuous on $[0,T)\times(0,\infty)$.

###### Proof.

The proof once again involves two statements, but with statement (i) of Lemma
3.16 remained unchanged, we confine ourselves by proving (ii) the map $x\to
V(t,x)$ is continuous on $(0,\infty)$ for all $t\in[0,T]$. As before, the up-
down connectedness of $\mathcal{C}$ and $\mathcal{D}$ suggests that for any
$x_{1}<x_{2}$ in $(0,\infty)$,

$\displaystyle G(t,x_{2})-G(t,x_{1})\leq V(t,x_{2})-V(t,x_{1}),$

and let $\tau_{2}$ be optimal for $V(t,x_{2})$ so that

$\displaystyle G(t,x_{2})-G(t,x_{1})$ $\displaystyle\leq
V(t,x_{2})-V(t,x_{1})$ $\displaystyle\leq
E\left(e^{-r\tau_{2}}G\left(t+\tau_{2},X_{\tau_{2}}^{x_{2}}\right)\right)-E\left(e^{-r\tau_{2}}G\left(t+\tau_{2},X_{\tau_{2}}^{x_{1}}\right)\right),$
(3.56)

in which, by letting $x_{1}\to x_{2}$, the dominated convergence theorem
yields $V(t,x_{1})-V(t,x_{2})\to 0$ and thereby showing the left-continuity of
$V$ w.r.t $x$.

If $x_{2}\leq\max\\{x^{*}(t),B(t)\\}$ for $t\in[0,T]$, Lemma 3.24 (see Figure
5) suggests that $\tau_{2}=0$, which, together with the continuity of $G$,
shows that $V(t,x_{2})-V(t,x_{1})\to 0$ as $x_{2}\to x_{1}$. If, on the other
hand, $x_{2}>\max\\{x^{*}(t),B(t)\\}$ for $t\in[0,T]$, then

$\displaystyle V(t,x_{2})-V(t,x_{1})\leq
E\Big{(}e^{-r\tau_{2}}\left(G\left(t+\tau_{2},X_{\tau_{2}}^{x_{2}}\right)-G\left(t+\tau_{2},X_{\tau_{2}}^{x_{1}}\right)\right)\Big{)}$
$\displaystyle\qquad=E\left(e^{-r\tau_{2}}\left(\left(K-X_{\tau_{2}}^{x_{2}}\right)^{+}-\left(K-X_{\tau_{2}}^{x_{1}}\right)^{+}\right)Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{2}}\right)\right)$
$\displaystyle\qquad\qquad+E\left(e^{-r\tau_{2}}\left(K-X_{\tau_{2}}^{x_{1}}\right)^{+}\left(Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{2}}\right)-Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{1}}\right)\right)\right)$
$\displaystyle\qquad\leq
KE\Big{(}e^{-r\tau_{2}}\left(Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{2}}\right)-Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{1}}\right)\right)\Big{)}$
$\displaystyle\qquad\leq
KE\Big{(}e^{-r\tau_{2}}I\\{X_{\tau_{2}}^{x_{2}}>L+\delta\\}\left(Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{2}}\right)-Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{1}}\right)\right)\Big{)}$
$\displaystyle\qquad\qquad+KE\Big{(}e^{-r\tau_{2}}I\\{X_{\tau_{2}}^{x_{2}}\leq
L+\delta\\}\left(Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{2}}\right)-Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{1}}\right)\right)\Big{)},$
(3.57)

where the second inequality is because of inequality (3.46) and
$X_{\tau_{2}}^{x_{2}}>X_{\tau_{2}}^{x_{1}}$.

Before we proceed, recall that $L\geq B(t)\geq K$ for $t\in[t_{*},T]$ (and
$L<K$). By employing the continuity and monotonicity of $B$, intermediate
value theorem tells us that there exists $t_{\delta}\in(t_{*},T)$ so that
$B(t_{\delta})=L+\delta$ and because $\tau_{2}<\tau_{\mathcal{S}}$, where
$\tau_{\mathcal{S}}=\inf\\{u\in[0,T-t]\in X_{u}^{x_{2}}\in\mathcal{S}\\}$ (Set
$\mathcal{S}$ is the area coloured in green in Figure 5), we know that
$X_{\tau_{2}}^{x_{2}}\leq L+\delta$ implies $t+\tau_{2}<t_{\delta}$ a.s. (see
Figure 3), equation (3.57) therefore equals

$\displaystyle V(t,x_{2})-V(t,x_{1})\leq
KE\Big{(}e^{-r\tau_{2}}I\\{X_{\tau_{2}}^{x_{2}}>L+\delta\\}\left(Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{2}}\right)-Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{1}}\right)\right)\Big{)}$
$\displaystyle\qquad\qquad\qquad\qquad\,\,\,\,\,+KE\Big{(}e^{-r\tau_{2}}I\\{t+\tau_{2}<t_{\delta}\\}\left(Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{2}}\right)-Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{1}}\right)\right)\Big{)}$
$\displaystyle\qquad=KE\Big{(}e^{-r\tau_{2}}I\\{X_{\tau_{2}}^{x_{2}}>L+\delta\\}\big{(}Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{2}}\right)-Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{1}}\right)\big{)}\Big{)}$
$\displaystyle\qquad\qquad+K(x_{2}-x_{1})E\Big{(}e^{-\frac{\sigma^{2}}{2}\tau_{2}+\sigma
W_{\tau_{2}}}I\\{t+\tau_{2}<t_{\delta}\\}Z_{x}\left(t+\tau_{2},X_{\tau_{2}}^{x_{3}}\right)\Big{)}$
$\displaystyle\qquad=KE\Big{(}e^{-r\tau_{2}}I\\{X_{\tau_{2}}^{x_{2}}>L+\delta\\}\left(Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{2}}\right)-Z\left(t+\tau_{2},X_{\tau_{2}}^{x_{1}}\right)\right)\Big{)}+KC_{2}(x_{2}-x_{1})$
(3.58)

where the last two steps are due to the mean value theorem for
$x_{3}\in[x_{1},x_{2}]$, the strong solution of GBM and the fact that
$Z_{x}(t,x)\leq C_{2}$ in $[0,t_{\delta})\times(0,L+\delta]$ (see A.87).

Then, we can choose $\delta>0$ so that $X_{\tau_{2}}^{x_{1}}>L$ almost surely
whenever $X_{\tau_{2}}^{x_{2}}>L+\delta$ and
$x_{2}-x_{1}<\frac{x_{1}\delta}{2L}$ (the justification is the same as that in
the proof of Lemma 3.16, we omit further details), by setting such $\delta$ in
(3.58), joining with the fact that $Z(t,x)=1$ for $x\geq L$,

$\displaystyle G(t,x_{2})-G(t,x_{1})\leq V(t,x_{2})-V(t,x_{1})\leq
KC_{2}(x_{2}-x_{1}),$

in which, let $x_{2}\to x_{1}$, we have $V(t,x_{2})-V(t,x_{1})\to 0$, that is,
$V$ is right continuous w.r.t $x$. The conclusion follows as $V$ is right and
left continuous for any $x\in(0,\infty)$ for each $t\in[0,T)$ given and fixed.
∎

To this point, we have been working really hard on the discussion of the
stopping and continuation sets and now we can finally return to our main
purpose, i.e. the formulation of free-boundary problem,

######  Proposition 3.34 (The Free-boundary Problem).

The stopping set and the continuation set are of the forms:

$\displaystyle\mathcal{C}$
$\displaystyle=\\{(t,x)\in[0,T)\times(0,\infty):x>b(t)\\},$
$\displaystyle\mathcal{D}$
$\displaystyle=\\{(t,x)\in[0,T)\times(0,\infty):x\leq b(t)\\},$

where the map $t\mapsto b(t)$ is increasing. The free-boundary problem in
section 2.2 is therefore reshuffled accordingly as

$\displaystyle{V}_{t}+\mathbb{L}_{x}{V}-r{V}=0\qquad\qquad\text{for
$(t,x)\in\mathcal{C}$},$ (3.59)
$\displaystyle{V}(t,x)=G(t,x)\qquad\qquad\,\,\,\,\,\,\,\text{for $x=b(t)$
(instantaneous stopping)},$ (3.60)
$\displaystyle{V}_{x}(t,x)=G_{x}(t,x)\qquad\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{for
$x=b(t)<L$ and $x=b(t)>L$ (smooth-fit)},$ (3.61)
$\displaystyle{V}_{x}(t,L\pm)=G_{x}(t,L\pm)\qquad\,\,\,\text{for $x=b(t)=L$,
(smooth-fit breaking down),}$ (3.62)
$\displaystyle{V}(t,x)>{G}(t,x)\qquad\qquad\,\,\,\,\,\,\text{for
$(t,x)\in\mathcal{C}$},$ (3.63)
$\displaystyle{V}(t,x)={G}(t,x)\qquad\qquad\,\,\,\,\,\,\text{for
$(t,x)\in\mathcal{D}$}.$ (3.64)

###### Proof of the Smooth-fit Condition (3.61).

The proof for $b(t)>L$ is essentially the same as that of Proposition 3.17. It
thus remains to show (3.61) holds true for $b(t)<L$. Let a point
$(t,x)\in[0,T)\times(0,\infty)$ lying on the boundary $b$ be fixed, i.e.
$x=b(t)<L$. Then, let $\epsilon>0$ such that $x+\epsilon<L$, and

$\displaystyle\frac{V(t,x+\epsilon)-V(t,x)}{\epsilon}\geq\frac{G(t,x+\epsilon)-G(t,x)}{\epsilon},$

where the inequality is due to $(t,x+\epsilon)\in\mathcal{C}$ and
$(t,x)\in\mathcal{D}$; of which, we take the limit as $\epsilon\to 0$ and
obtain

$\displaystyle\frac{\partial^{+}}{\partial
x}V(t,x)\geq-Z(t,x)+(K-x)\frac{\partial}{\partial x}Z(t,x).$ (3.65)

In order to prove the converse inequality, we first note that (the exact same
argument is presented in the proof of Lemma 3.33, so we omit some details
here) there exists $\delta>0$ s.t. $B(t_{\delta})=L+\delta$ for
$t_{\delta}\in(t_{*},T)$. Then, let $\tau_{\epsilon}$ be optimal for
$V(t,x+\epsilon)$, where we can choose
$\epsilon\in\left(0,\frac{x\delta}{2L}\right)$ so that
$X_{\tau_{\epsilon}}^{x}>L$ almost surely whenever
$X_{\tau_{\epsilon}}^{x+\epsilon}>L+\delta$ (the justification follows the
same pattern as that in the proof of Lemma 3.16),

$\displaystyle\frac{V(t,x+\epsilon)-V(t,x)}{\epsilon}\leq\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}\left(G\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)-G\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)\right)\right)$
$\displaystyle\,\,\,\leq-E\left(e^{-\frac{\sigma^{2}}{2}\tau_{\epsilon}+\sigma
W_{\tau_{\epsilon}}}I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)\right)$
$\displaystyle\,\,\,\,\,\,+\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}\left(K-X_{\tau_{\epsilon}}^{x}\right)^{+}I\\{X_{\tau_{\epsilon}}^{x+\epsilon}\leq
L+\delta\\}\left(Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)-Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)\right)\right)$
$\displaystyle\,\,\,\,\,\,+\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}\left(K-X_{\tau_{\epsilon}}^{x}\right)^{+}I\\{X_{\tau_{\epsilon}}^{x+\epsilon}>L+\delta\\}\left(Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)-Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)\right)\right)$
$\displaystyle\,\,\,=-E\left(e^{-\frac{\sigma^{2}}{2}\tau_{\epsilon}+\sigma
W_{\tau_{\epsilon}}}I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)\right)$
$\displaystyle\,\,\,\,\,\,+\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}\left(K-X_{\tau_{\epsilon}}^{x}\right)^{+}I\\{X_{\tau_{\epsilon}}^{x+\epsilon}\leq
L+\delta\\}\left(Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)-Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)\right)\right)$
$\displaystyle\,\,\,=-E\left(e^{-\frac{\sigma^{2}}{2}\tau_{\epsilon}+\sigma
W_{\tau_{\epsilon}}}I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)\right)$
$\displaystyle\,\,\,\,\,\,+\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}\left(K-X_{\tau_{\epsilon}}^{x}\right)^{+}I\\{t+\tau_{\epsilon}\leq
t_{\delta}\\}\left(X_{\tau_{\epsilon}}^{x+\epsilon}-X_{\tau_{\epsilon}}^{x}\right)Z_{x}\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{y}\right)\right)$
$\displaystyle\,\,\,=-E\left(e^{-\frac{\sigma^{2}}{2}\tau_{\epsilon}+\sigma
W_{\tau_{\epsilon}}}I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)\right)$
$\displaystyle\,\,\,\,\,\,+E\left(e^{\frac{-\sigma^{2}}{2}\tau_{\epsilon}+\sigma
W_{\tau_{\epsilon}}}\left(K-X_{\tau_{\epsilon}}^{x}\right)^{+}I\\{t+\tau_{\epsilon}\leq
t_{\delta}\\}Z_{x}\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{y}\right)\right)$

where the second inequality holds true because of (3.53) and the first
equality follows from the facts that for $\epsilon$ of our choice,
$I\\{X_{\tau_{\epsilon}}^{x+\epsilon}>L+\delta\\}=1$ entails
$I\\{X_{\tau_{\epsilon}}^{x}>L\\}=1$, the second equality is derived from the
mean value theorem for $y\in[x,x+\epsilon]$. Recalling that the random
variable $X$ is dominated by a positive integrable random variable, letting
$\epsilon\to 0$, $\tau_{\epsilon}\to 0$ a.s. and exploiting the dominated
convergence theorem (the boundedness of $Z_{x}(t,x)$ for
$(t,x)\in[0,t_{\delta})\times(0,L+\delta]$ is verified in A.87) yield

$\displaystyle\frac{\partial^{+}}{\partial
x}V(t,x)\leq-Z(t,x)+(K-x)\frac{\partial}{\partial x}Z(t,x),$ (3.66)

which, together with (3.65) and the fact that
$(t,x),(t,x-\epsilon)\in\mathcal{D}$ so that

$\frac{V(t,x)-V(t,x-\epsilon)}{\epsilon}=\frac{G(t,x)-G(t,x-\epsilon)}{\epsilon},$

and this allows us to obtain (3.61). ∎

###### Proof of (3.62).

(i) Let $x=b(t)=L$ such that there exists $\epsilon>0$, $x-\epsilon<L$ and
$(t,x-\epsilon)\in\mathcal{D}$. The first part of (3.62) is a consequence of

$\displaystyle\frac{V(t,x)-V(t,x-\epsilon)}{\epsilon}$
$\displaystyle=\frac{G(t,x)-G(t,x-\epsilon)}{\epsilon}$
$\displaystyle=\frac{(K-x)Z(t,x)-(K-x+\epsilon)Z(t,x-\epsilon)}{\epsilon}$
$\displaystyle=-Z(t,x)+\frac{(K-x+\epsilon)\left[Z(t,x)-Z(t,x-\epsilon)\right]}{\epsilon},$

and via letting $\epsilon\to 0$, we see $\lim\limits_{\epsilon\to
0}\frac{V(t,x)-V(t,x-\epsilon)}{\epsilon}=-1+(K-L)Z_{x}(t,L-)$

(ii) Again, let $x=b(t)=L$ so that there exists
$0<\epsilon<\frac{x\delta}{2L}$ (where $\delta>0$ is defined as so that
$B(t_{\delta})=L+\delta$ for $t_{\delta}\in(t_{*},T)$), $L<x+\epsilon<K$ and
$(t,x+\epsilon)\in\mathcal{C}$ and that

$\displaystyle\frac{V(t,x+\epsilon)-V(t,x)}{\epsilon}\geq\frac{G(t,x+\epsilon)-G(t,x)}{\epsilon}=-1.$

Furthermore, let $\tau_{\epsilon}$ be the optimal stopping time for
$V(t,x+\epsilon)$ so that $X_{\tau_{\epsilon}}^{x+\epsilon}\geq L$ and that

$\displaystyle\frac{V(t,x+\epsilon)-V(t,x)}{\epsilon}\leq\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}\left(G\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)-G\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)\right)\right)$
$\displaystyle\,\,\,\leq-E\Big{(}e^{-\frac{\sigma^{2}}{2}\tau_{\epsilon}-\sigma
W_{\tau_{\epsilon}}}Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}\Big{)}$
$\displaystyle\,\,\,\,\,\,+\frac{1}{\epsilon}E\Big{(}e^{-r\tau_{\epsilon}}I\\{X_{\tau_{\epsilon}}^{x+\epsilon}=L\\}\left(K-X_{\tau_{\epsilon}}^{x+\epsilon}\right)^{+}\left(Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)-Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)\right)\Big{)}$
$\displaystyle\,\,\,\,\,\,+\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}I\\{X_{\tau_{\epsilon}}^{x+\epsilon}>L+\delta\\}\left(K-X_{\tau_{\epsilon}}^{x+\epsilon}\right)^{+}\left(Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)-Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)\right)\right)$
$\displaystyle\,\,\,\,\,\,+\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}I\\{L<X_{\tau_{\epsilon}}^{x+\epsilon}<L+\delta\\}\left(K-X_{\tau_{\epsilon}}^{x+\epsilon}\right)^{+}\left(Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)-Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)\right)\right)$
$\displaystyle\,\,\,=-E\Big{(}e^{-\frac{\sigma^{2}}{2}\tau_{\epsilon}-\sigma
W_{\tau_{\epsilon}}}Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}\Big{)}$
$\displaystyle\,\,\,\,\,\,+\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}I\\{t_{b}<t+\tau_{\epsilon}\leq
t_{\delta}\\}\left(K-X_{\tau_{\epsilon}}^{x+\epsilon}\right)^{+}\left(Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)-Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)\right)\right)$
$\displaystyle\,\,\,\,\,\,+\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}I\\{X_{\tau_{\epsilon}}^{x+\epsilon}>L+\delta\\}\left(K-X_{\tau_{\epsilon}}^{x+\epsilon}\right)^{+}\left(Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)-Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)\right)\right)$
$\displaystyle\,\,\,=-E\Big{(}e^{-\frac{\sigma^{2}}{2}\tau_{\epsilon}-\sigma
W_{\tau_{\epsilon}}}Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}\Big{)}$
$\displaystyle\,\,\,\,\,\,+E\Big{(}e^{-\frac{\sigma^{2}}{2}\tau_{\epsilon}-\sigma
W_{\tau_{\epsilon}}}I\\{t_{b}<t+\tau_{\epsilon}\leq
t_{\delta}\\}\left(K-X_{\tau_{\epsilon}}^{x+\epsilon}\right)^{+}Z_{x}\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{y}\right)\Big{)}$
$\displaystyle\,\,\,\,\,\,+\frac{1}{\epsilon}E\left(e^{-r\tau_{\epsilon}}I\\{X_{\tau_{\epsilon}}^{x+\epsilon}>L+\delta\\}\left(K-X_{\tau_{\epsilon}}^{x+\epsilon}\right)^{+}\left(Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x+\epsilon}\right)-Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)\right)\right)$
$\displaystyle\,\,\,=-E\Big{(}e^{-\frac{\sigma^{2}}{2}\tau_{\epsilon}-\sigma
W_{\tau_{\epsilon}}}Z\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{x}\right)I\\{X_{\tau_{\epsilon}}^{x+\epsilon}<K\\}\Big{)}$
$\displaystyle\,\,\,\,\,\,+E\Big{(}e^{-\frac{\sigma^{2}}{2}\tau_{\epsilon}-\sigma
W_{\tau_{\epsilon}}}I\\{t_{b}<t+\tau_{\epsilon}\leq
t_{\delta}\\}\left(K-X_{\tau_{\epsilon}}^{x+\epsilon}\right)^{+}Z_{x}\left(t+\tau_{\epsilon},X_{\tau_{\epsilon}}^{y}\right)\Big{)}$

where the second inequality is due to (3.53) and in the first equality, we use
the fact that

$\\{L\leq
X_{\tau_{\epsilon}}^{x+\epsilon}<L+\delta\\}=\\{t_{b}<t+\tau_{\epsilon}<t_{\delta}\\},$

and in the third equality, we apply mean value theorem for
$y\in[x,x+\epsilon]$ and with our choice of $\epsilon$,
$X_{\tau_{\epsilon}}^{x+\epsilon}>L+\delta$ entails
$X_{\tau_{\epsilon}}^{x}>L$ a.s.; from which, we know that as $\epsilon\to 0$,
$\tau_{\epsilon}\to 0$ almost surely so that
$I\\{t+\tau_{\epsilon}>t^{b}\\}\to 0$ and hence, equations in (3.62) emerge. ∎

Before we bring this section to a close, we establish the continuity of the
boundary.

######  Remark 3.35.

The proof of the continuity of $b$ on $[t^{b},T]$ can be done by following the
exact same pattern as that in Lemma 3.19, and if $b$ remains at level L in
$[t_{b},t^{b}]$, we know from the left-connectedness of $\mathcal{D}$ that
during such amount of time, $b$ must be continuous. Our task has therefore
been reduced to prove that $b$ is continuous on $[0,t_{b}]$.

######  Lemma 3.36.

The optimal stopping boundary $b$ is continuous on $[0,T]$ and $b(T-)=K$.

######  Remark 3.37.

The essential tool to prove Lemma 3.19 is the smooth-fit condition, which,
may, at times, either be difficult to verify or hold true. Following is a
rather convenient technique to prove the continuity based on the regularity of
functions $V$ and $G$ and the monotonicity of $b$.

###### Proof.

To avoid using smooth-fit condition, we use the argument provided in [4, Page
173-175].

(i) The boundary $b$ is right-continuous on $[0,t_{b}]$. The conclusion is
immediate from the fact that the stopping set is closed and $b$ is increasing
on $[0,t_{b}]$.

(ii) The boundary $b$ is left-continuous on $[0,t_{b}]$. We assume that, for
contradiction, there exists $t_{0}\in(0,t_{b})$ so that a discontinuity of $b$
occurs; that is, at $t_{0}$, one has $b(t_{0}-)<b(t_{0})$, where $b(t_{0}-)$
denotes the left limit of the boundary at $t_{0}$, which always exists as
$b(t)$ is monotonically increasing in $(0,t_{b})$.

Let $x_{1},x_{2}$ be fixed such that $b(t_{0}-)<x_{1}<x_{2}<b(t_{0})$. Then,
for fixed $t^{\prime}\in(0,t_{0})$, we define an open bounded domain
$\mathcal{R}\subset\mathcal{C}$ with
$\mathcal{R}=\\{(t^{\prime},t_{0})\times(x_{1},x_{2})\\}$. Hence, by the
structure of $\mathcal{C}$, we know the value function $V$, which is $C^{1,2}$
in $\mathcal{R}$, solves the Cauchy problem

$\displaystyle V_{t}+\mathbb{L}_{X}V-rV=0.$ (3.67)

Then, we denote $C^{\infty}_{c}([x_{1},x_{2}])$ as the set of function with
infinitely many continuous derivatives and compact support in $[x_{1},x_{2}]$,
in which, we define $\psi\geq 0$ such that $\int_{x_{1}}^{x_{2}}\psi(y)dy=1$.
By further multiplying (3.67) by function $\psi$, and integrating the product
over $(t,t_{0})\times(x_{1},x_{2})$ for some $t\in(t^{\prime},t_{0})$,

$\displaystyle\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}\left(V_{t}+\mathbb{L}_{X}V\right)(s,y)\psi(y)dyds+\left(-\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}rV(s,y)\psi(y)dyds\right)=0,$
(3.68)

where the second double integral on the left-hand side is strictly negative,
as $V>G\geq 0$ on $\mathcal{C}$. Therefore, we want to investigate the sign of
first double integral by estimating its upper bound, which if is negative, the
proof will be completed by reaching a contradiction. To do so, let us first
observe that

$\displaystyle\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}V_{t}(s,y)\psi(y)dyds=\int_{x_{1}}^{x_{2}}\bigl{(}V(t_{0},y)-V(t,y)\bigr{)}\psi(y)dy$
$\displaystyle\qquad\leq\int_{x_{1}}^{x_{2}}\bigl{(}G(t_{0},y)-G(t,y)\bigr{)}\psi(y)dy=\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}G_{t}(s,y)\psi(y)dyds,$
(3.69)

where the inequality is the consequence of $(t_{0},y)\in\mathcal{D}$ and
$(t,y)\in\mathcal{C}$.

Next in line is partial integrations (three times),

$\displaystyle\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}\mathbb{L}_{X}V(s,y)\psi(y)dyds=\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}V(s,y)\mathbb{L}_{X}^{*}\psi(y)dyds,$
(3.70)

where, if we also impose the condition
$\psi(x_{1})=\psi(x_{2})=\psi^{\prime}(x_{1})=\psi^{\prime}(x_{2})=0$ for the
boundary terms disappear, $\mathbb{L}_{X}^{*}$ is the formal adjoint of
$\mathbb{L}_{X}$ given as

$\displaystyle\mathbb{L}_{X}^{*}\psi(x)=\frac{1}{2}\frac{\partial^{2}}{\partial
x^{2}}(\sigma^{2}x^{2}\psi(x))-\frac{\partial}{\partial x}(rx\psi(x)).$

Since $G\in C^{1,2}({\mathcal{R}})$,

$\displaystyle\lvert V(s,y)-G(s,y)\rvert$ $\displaystyle=\lvert
V(s,y)-V(t_{0},y)+G(t_{0},y)-G(s,y)\rvert$ $\displaystyle\leq\lvert
V(s,y)-V(t_{0},y)\rvert+\lvert G(t_{0},y)-G(s,y)\rvert$ $\displaystyle\leq
C_{1}(t_{0}-s)^{\frac{\beta}{2}}+C_{2}(t_{0}-s),$ (3.71)

where $C_{1},C_{2}$ are constants and $\beta>0$, the first equality follows
via again $(t_{0},y)\in\mathcal{D}$ and the second inequality is due to the
fulfilment of the conditions proposed on [4, Page 172, (C.3)].

In addition, we consider $\lvert t_{0}-t\rvert<1$ and set $C_{3}=C_{1}+C_{2}$,
upon using $V\geq G$,

$\displaystyle 0\leq G(s,y)\leq V(s,y)\leq
G(s,y)+C_{3}(t_{0}-s)^{\frac{\beta}{2}},$

of which, as a result, for any $s\in(t,t_{0})$, in the view of (3.70),

$\displaystyle\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}V(s,y)\mathbb{L}_{X}^{*}\psi(y)dyds=\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}{I}\\{\mathbb{L}_{X}^{*}\psi(y)\geq
0\\}V(s,y)\mathbb{L}_{X}^{*}\psi(y)dyds$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}{I}\\{\mathbb{L}_{X}^{*}\psi(y)<0\\}V(s,y)\mathbb{L}_{X}^{*}\psi(y)dyds$
$\displaystyle\qquad\leq\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}{I}\\{\mathbb{L}_{X}^{*}\psi(y)\geq
0\\}\left(G(s,y)+C_{3}(t_{0}-s)^{\frac{\beta}{2}}\right)\mathbb{L}_{X}^{*}\psi(y)dyds$
$\displaystyle\qquad\qquad+\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}{I}\\{\mathbb{L}_{X}^{*}\psi(y)<0\\}G(s,y)\mathbb{L}_{X}^{*}\psi(y)dyds$
$\displaystyle\qquad=\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}G(s,y)\mathbb{L}_{X}^{*}\psi(y)dyds$
$\displaystyle\qquad\qquad+\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}{I}\\{\mathbb{L}_{X}^{*}\psi(y)\geq
0\\}C_{3}(t_{0}-s)^{\frac{\beta}{2}}\mathbb{L}_{X}^{*}\psi(y)dyds$
$\displaystyle\qquad=\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}\mathbb{L}_{X}G(s,y)\psi(y)dyds$
$\displaystyle\qquad\qquad+\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}{I}\\{\mathbb{L}_{X}^{*}\psi(y)\geq
0\\}C_{3}(t_{0}-s)^{\frac{\beta}{2}}\mathbb{L}_{X}^{*}\psi(y)dyds,$ (3.72)

where the last equality follows by partial integrations again.

Joining (3.69) with (3.72), from the view of equation (3.68),

$\displaystyle 0$
$\displaystyle\leq\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}\left(G_{t}+\mathbb{L}_{X}G-rG\right)(s,y)\psi(y)dyds$
$\displaystyle\qquad+C_{3}\int_{t}^{t_{0}}(t_{0}-s)^{\frac{\beta}{2}}\int_{x_{1}}^{x_{2}}I\\{\mathbb{L}_{X}^{*}\psi(y)\geq
0\\}\mathbb{L}_{X}^{*}\psi(y)dyds$
$\displaystyle=\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}\left(G_{t}+\mathbb{L}_{X}G-rG\right)(s,y)\psi(y)dyds$
$\displaystyle\qquad+C_{3}\int_{t}^{t_{0}}(t_{0}-s)^{\frac{\beta}{2}}\int_{x_{1}}^{x_{2}}I\\{\mathbb{L}_{X}^{*}\psi(y)>0\\}\mathbb{L}_{X}^{*}\psi(y)dyds$
(3.73)
$\displaystyle=\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}\left(G_{t}+\mathbb{L}_{X}G-rG\right)(s,y)\psi(y)dyds+\frac{2\gamma}{2+\beta}(t_{0}-t)^{1+\frac{\beta}{2}}$
$\displaystyle\leq\int_{t}^{t_{0}}\int_{x_{1}}^{x_{2}}(-l_{\epsilon})\psi(y)dyds+\frac{2\gamma}{2+\beta}(t_{0}-t)^{1+\frac{\beta}{2}}$
$\displaystyle=-l_{\epsilon}(t_{0}-t)+\frac{2\gamma}{2+\beta}(t_{0}-t)^{1+\frac{\beta}{2}},$
(3.74)

where the second inequality follows in the sense that in such interval
$[0,t^{*}]\times[x_{1},x_{2}]$, there exists some constant $l_{\epsilon}$ so
that

$\displaystyle(G_{t}+\mathbb{L}_{X}G-rG)(t,x)=-(rKZ+\sigma^{2}x^{2}Z_{x})(t,x)\leq-
l_{\epsilon}<0$

and that $\int_{x_{1}}^{x_{2}}\psi(y)dy=1$; since the second term in (3.73) is
strictly positive, we can set

$\displaystyle\gamma=\gamma(\psi;x_{1},x_{2}):=C_{3}\int_{x_{1}}^{x_{2}}I\\{\mathbb{L}_{X}^{*}\psi(y)>0\\}\mathbb{L}_{X}^{*}\psi(y)dy>0,$

from which, the second equality holds.

Taking the limit $t_{0}\to t$ thus leads to a contradiction in the sense that
the second term on the last equality of (3.74) is vanishing more rapidly than
its first term (that is, $0\leq-l_{\epsilon}(t_{0}-t)<0$), hence the jump may
not occur. ∎

###  3.4 The Optimal Stopping Rule

In order to prepare for the main result, we shall first verify the conditions
for the local time-space formula to be applicable.

######  Lemma 3.38.

Let the function $F(t,x)=e^{-rt}V(t,x)$ defined on $[0,T)\times(0,\infty)$,

$\displaystyle\mathcal{D}_{1}=\\{(t,x)\in[0,t_{b}]\times(0,b(t))\\},$
$\displaystyle\mathcal{D}_{2}=\\{(t,x)\in[t_{b},t^{b}]\times(0,L)\\},$
$\displaystyle\mathcal{D}_{3}=\\{(t,x)\in[t^{b},T]\times(0,L)\cup(L,b(t))\\}.$

Then, the local time-space formula is applicable to $F$ if the following
conditions are met:

(a) $F(t,x)$ is $C^{1,2}$ on
$\mathcal{C}\cup\mathcal{D}_{1}\cup\mathcal{D}_{2}\cup\mathcal{D}_{3}$;

(b) $F_{t}+\mathbb{L}_{X}F$ is locally bounded;

(c) $t\mapsto F_{x}(t,b(t)\pm)$ is continuous;

(d) $t\mapsto F_{x}(t,L\pm)$ is continuous;

(e) $F_{xx}=F_{1}+F_{2}$ on
$\mathcal{C}\cup\mathcal{D}_{1}\cup\mathcal{D}_{2}\cup\mathcal{D}_{3}$, where
$F_{1}$ is non-negative and $F_{2}$ is continuous.

###### Proof.

By the definition of $F$,

$\displaystyle F(t,x)=\begin{cases}e^{-rt}V(t,x),&(t,x)\in\mathcal{C},\\\
e^{-rt}G(t,x),&(t,x)\in\mathcal{D}_{1}\cup\mathcal{D}_{2}\cup\mathcal{D}_{3},\end{cases}$

statement (a) is immediate.

To prove statement (b) is to show that $F_{t}+\mathbb{L}_{X}F$ is locally
bounded on
$(\mathcal{C}\cup\mathcal{D}_{1}\cup\mathcal{D}_{2}\cup\mathcal{D}_{3})\cap\mathcal{K}$
for each compact set $\mathcal{K}$. Since

$\displaystyle F_{t}+\mathbb{L}_{X}F=\begin{cases}0,&(t,x)\in\mathcal{C},\\\
-rK,&(t,x)\in[t^{b},T]\times(L,b(t)),\\\
(-rKZ-\sigma^{2}x^{2}Z_{x})(t,x),&(t,x)\in\mathcal{D}_{1}\cup\mathcal{D}_{2}\cup[t^{b},T]\times(0,L),\end{cases}$

where the last case is locally bounded as it is a continuous function on
$\\{\mathcal{D}_{1}\cup\mathcal{D}_{2}\cup[t^{b},T]\times(0,L)\\}\cap\mathcal{K}$.

The maps $t\mapsto F_{x}(t,b(t)+)$ for $t\in[0,T)$ and $t\mapsto
F_{x}(t,L\pm)$ for $t\in[0,t_{b}]$ are continuous as the result of $V$ is
$C^{1,2}$ on the continuation set $\mathcal{C}$, see [7, Page 8]. Moreover,
the map $t\mapsto F_{x}(t,b(t)-)$ is continuous as
$F_{x}(t,b(t)-)=e^{-rt}G_{x}(t,b(t)-)$ and the map $t\mapsto b(t)$ is
continuous, which, joining with $F_{x}(t,L\pm)=e^{-rt}G(t,L\pm)$ for
$t\in[t^{b},T]$, justifies statements (c) and (d).

Finally, statement (e) follows via

$\displaystyle
F_{xx}(t,x)=\begin{cases}\frac{2e^{-rt}}{\sigma^{2}x^{2}}\left(rV-
rxV_{x}+V_{t}\right)(t,x),&(t,x)\in\mathcal{C},\\\
0,&(t,x)\in[t^{b},T]\times(L,b(t)),\\\ \
\frac{2e^{-rt}}{\sigma^{2}x^{2}}\left(-G_{t}+rxG_{x}-\sigma^{2}x^{2}Z_{x}-rxZ\right)(t,x),&(t,x)\in\mathcal{D}_{1}\cup\mathcal{D}_{2}\cup[t^{b},T]\times(0,L)\end{cases}$

where the first and second case indicates that $F$ is convex on $\mathcal{C}$,
which (both convexity and concavity, see [26, Page 526]) is a rather stronger
condition than (e) (remember that the optimal stopping boundary is lower
bounded by $x^{*}$ so that $V_{x}<0$) and in the third case, $G_{t}(t,x)\leq
0$ and the rest terms are continuous. ∎

######  Theorem 3.39.

The optimal stopping boundary $b:[0,T]\mapsto\mathbb{R}$ in problem (3.30) is
given as $b(t)=b_{1}(t)\vee b_{2}(t)$ where $b_{1}(t)=x^{*}(t)$ for
$t\in[0,t_{*}]$ with $x^{*}:[0,t_{*}]\times(0,L]$ being continuous non-
decreasing function defined in Lemma 3.22, $b_{2}:[0,T]\mapsto\mathbb{R}$ is
in the class of continuous non-decreasing functions $c:[0,T]\mapsto\mathbb{R}$
satisfying $0<c(t)<K$ for all $t\in[0,T)$ and $b_{2}$ can be characterised as
the unique solution of the following nonlinear integral equation:

$\displaystyle G\bigl{(}t,b_{2}(t)\bigr{)}$
$\displaystyle=E_{t,b_{2}(t)}\left(e^{-r(T-t)}G(T,X_{T})-\int_{0}^{s}e^{-ru}H\left(t+u,X_{t+u}\right){I}\\{X_{t+u}<b_{2}(t+u)\\}du\right)$
$\displaystyle\qquad+\frac{K-L}{2}\int_{0}^{s}e^{-ru}{I}\\{b_{2}(t+u)>L\\}Z_{x}\left(t+u,L-\right)dE_{t,x}\left(l_{u}^{L}(X)\right)$
$\displaystyle\qquad+\frac{K-L}{2}\int_{0}^{s}e^{-ru}{I}\\{b_{2}(t+u)=L\\}Z_{x}\left(t+u,L-\right)dE_{t,x}\left(l_{u}^{L}(X)\right),$
(3.75)

satisfying $b(T-)=K$. The value of the current contract admits the following
“early exercise premium” representation:

$\displaystyle V(t,x)$
$\displaystyle=E_{t,x}\left(e^{-r(T-t)}G\left(T,X_{T}\right)-\int_{0}^{T-t}e^{-ru}H\left(t+u,X_{{t+u}}\right){I}\\{X_{{t+u}}<b(t+u)\\}du\right)$
$\displaystyle\qquad+\frac{K-L}{2}\int_{0}^{T-t}e^{-ru}{I}\\{b(t+u)>L\\}Z_{x}\left(t+u,L-\right)dE_{t,x}\left(l_{u}^{L}(X)\right)$
$\displaystyle\qquad-\frac{K-L}{2}\int_{0}^{T-t}e^{-ru}{I}\\{b(t+u)=L\\}Z_{x}\left(t+u,L-\right)dE_{t,x}\left(l_{u}^{L}(X)\right)$
(3.76)

for all $(t,x)\in[0,T]\times(0,\infty)$.

The proof follows essentially from [24, Page 385-392].

###### Proof of formulas (3.75) and (3.76).

Since the function $F$ fulfils the conditions on Lemma 3.38, an application of
the local time-space formula yields

$\displaystyle e^{-rs}V\left(t+s,X_{t+s}^{x}\right)=V(t,x)+M_{s}$
$\displaystyle\qquad\qquad\,\,\,+\int_{0}^{s}e^{-ru}\left(-rV+V_{t}+\mathbb{L}_{X}V\right)\left(t+u,X_{t+u}^{x}\right){I}\\{X_{t+u}^{x}\neq
b(t+u),X_{t+u}^{x}\neq L\\}du$
$\displaystyle\qquad\qquad\,\,\,+\frac{1}{2}\int_{0}^{s}e^{-ru}\bigl{(}V_{x}\left(t+u,L+\right)-V_{x}\left(t+u,L-\right)\bigr{)}dl_{u}^{L}(X^{x})$
$\displaystyle\qquad\qquad\,\,\,+\frac{1}{2}\int_{0}^{s}e^{-ru}\bigl{(}V_{x}\left(t+u,b(t+u)+\right)-V_{x}\left(t+u,b(t+u)-\right)\bigr{)}dl_{u}^{b}(X^{x})$
$\displaystyle\qquad=V(t,x)+M_{s}$
$\displaystyle\qquad\qquad\,\,\,+\int_{0}^{s}e^{-ru}\left(-rG+G_{t}+\mathbb{L}_{X}G\right)\left(t+u,X_{t+u}^{x}\right){I}\\{X_{t+u}^{x}<b(t+u),X_{t+u}^{x}\neq
L\\}du$
$\displaystyle\qquad\qquad\,\,\,+\frac{1}{2}\int_{0}^{s}e^{-ru}{I}\\{b(t+u)>L\\}\bigl{(}G_{x}\left(t+u,L+\right)-G_{x}\left(t+u,L-\right)\bigr{)}dl_{u}^{L}(X^{x})$
$\displaystyle\qquad\qquad\,\,\,+\frac{1}{2}\int_{0}^{s}e^{-ru}{I}\\{b(t+u)=L\\}\bigl{(}V_{x}\left(t+u,L+\right)-V_{x}\left(t+u,L-\right)\bigr{)}dl_{u}^{L}(X^{x}),$
(3.77)

where $M_{s}=\int_{0}^{s}e^{-ru}\sigma
X_{t+u}V_{x}\left(t+u,X_{t+u}\right)\mathbb{I}\\{X_{t+u}^{x}\neq
b(t+u),X_{t+u}^{x}\neq L\\}dW_{u}$ is a martingale under measure $P_{t,x}$ for
each $s\in[0,T-t]$, and the second equality is due to the fact that the
smooth-fit condition fails as $b(t)=L$ and the gain function is not smooth as
$b(t)>L$ for $t\in[0,T]$.

Then, upon taking the expectation under measure $P_{t,x}$ of (3.77) and
invoking the optional sampling theorem, we obtain

$\displaystyle
E_{t,x}\left(e^{-rs}V\left(t+s,X_{t+s}\right)\right)=V(t,x)+E_{t,x}\left(\int_{0}^{s}e^{-ru}H\left(t+u,X_{t+u}\right){I}\\{X_{t+u}<b(t+u)\\}du\right)$
$\displaystyle\qquad+\frac{1}{2}\int_{0}^{s}e^{-ru}{I}\\{b(t+u)>L\\}\bigl{(}G_{x}\left(t+u,L+\right)-G_{x}\left(t+u,L-\right)\bigr{)}dE_{t,x}\left(l_{u}^{L}(X)\right)$
$\displaystyle\qquad+\frac{1}{2}\int_{0}^{s}e^{-ru}{I}\\{b(t+u)=L\\}\bigl{(}V_{x}\left(t+u,L+\right)-V_{x}\left(t+u,L-\right)\bigr{)}dE_{t,x}\left(l_{u}^{L}(X)\right),$
(3.78)

where

$\displaystyle
H(t,x)=(-rG+G_{t}+\mathbb{L}_{X}G)(t,x)=\begin{cases}-rK,&L<x<b(t),\\\
(-rKZ-\sigma^{2}x^{2}Z_{x})(t,x),&x<L.\end{cases}$

Next, let $s=T-t$ so that $V(T,X_{T})=G(T,X_{T})$ and that

$\displaystyle
E_{t,x}\left(e^{-r(T-t)}V\left(T,X_{T}\right)\right)=E_{t,x}\left(e^{-r(T-t)}\left(K-X_{T}\right)I\\{L<X_{T}<K\\}\right),$

which, together with (3.78) shows that

$\displaystyle V(t,x)$
$\displaystyle=E_{t,x}\left(e^{-r(T-t)}G\left(T,X_{T}\right)-\int_{0}^{T-t}e^{-ru}H\left(t+u,X_{{t+u}}\right){I}\\{X_{{t+u}}<b(t+u)\\}du\right)$
$\displaystyle\qquad-\frac{1}{2}\int_{0}^{T-t}e^{-ru}{I}\\{b(t+u)>L\\}\bigl{(}G_{x}\left(t+u,L+\right)-G_{x}\left(t+u,L-\right)\bigr{)}dE_{t,x}\left(l_{u}^{L}(X)\right)$
$\displaystyle\qquad-\frac{1}{2}\int_{0}^{T-t}e^{-ru}{I}\\{b(t+u)=L\\}\bigl{(}G_{x}\left(t+u,L+\right)-G_{x}\left(t+u,L-\right)\bigr{)}dE_{t,x}\left(l_{u}^{L}(X)\right)$

after which, we set $x=b(t)$ and because $V(t,b(t))=G(t,b(t))$, (3.75)
follows. ∎

######  Remark 3.40.

To avoid the danger of confusion in notations, in what follows, we simply
assume, without loss of generality, that $b_{2}(t)\geq b_{1}(t)$ for all
$t\in[0,T]$ (see Figure 9) such that the optimal stopping boundary is just the
unique solution for (3.75), that is $b=b_{2}$.

###### Proof of uniqueness of the solution in (3.75).

The key idea is to show that if there exists a continuous increasing function
$c:[0,T]\mapsto(0,\infty)$ satisfying $c(t)<K$ and solving (3.75), then such
$c$ must coincide with the optimal stopping boundary $b$. We proceeds via
steps:

(i) From the view of (3.76), we introduce the following functions:

$\displaystyle U(t,x)$
$\displaystyle=E_{t,x}\left(e^{-r(T-t)}G\left(T,X_{T}\right)-\int_{0}^{T-t}e^{-ru}H\left(t+u,X_{{t+u}}\right){I}\\{X_{{t+u}}<c(t+u)\\}du\right)$
$\displaystyle\,\,\,\qquad+\frac{K-L}{2}\int_{0}^{T-t}e^{-ru}{I}\\{c(t+u)>L\\}Z_{x}\left(t+u,L-\right)dE_{t,x}\left(l_{u}^{L}(X)\right)$
$\displaystyle\,\,\,\qquad+\frac{K-L}{2}\int_{0}^{T-t}e^{-ru}{I}\\{c(t+u)=L\\}Z_{x}\left(t+u,L-\right)dE_{t,x}\left(l_{u}^{L}(X)\right),$
(3.79)

and set

$\displaystyle V^{c}(t,x)=\begin{cases}U(t,x),\qquad x>c(t),\\\ G(t,x),\qquad
x\leq c(t),\end{cases}$

which is continuous on $[0,T)\times(0,\infty)$ in the sense that $c$ solves
(3.75), implying $U(t,c(t))=G(t,c(t))$. In addition, we reconstruct the
continuation and stopping sets by the means of $c$:

$\displaystyle\mathcal{C}_{c}=\\{(t,x)\in[0,T)\times(0,\infty):x>c(t)\\},$
$\displaystyle\mathcal{D}_{c}=\\{t,x)\in[0,T]\times(0,\infty):x\leq c(t)\\}.$

From the definition, $V^{c}$ has the following properties, which we state
without proof222By expressing the expectation using Law of the unconscious
statistician, this shall be clear:

(a) $V^{c}=U$ is $C^{1,2}$ and satisfies the equation
$V_{t}^{c}+\mathbb{L}_{X}V^{c}=rV^{c}$ on $\mathcal{C}_{c}$;

(b) $V^{c}=G$ is $C^{1,2}$ on
$\mathcal{D}_{c}\setminus\\{(t,x)\in[0,T]\times\\{L,c(t)\\}\\}$.

(ii) Following exactly the same pattern on Lemma 3.38, we can verify that the
local time-space formula is applicable to $e^{-rt}V^{c}(t,X_{t})$ so that

$\displaystyle e^{-rs}V^{c}\left(t+s,X_{t+s}^{x}\right)=V^{c}(t,x)+M_{s}^{c}$
$\displaystyle\,\,\,\,\,\,\,\,\,\,+\int_{0}^{s}e^{-ru}\left(-rV^{c}+V_{t}^{c}+\mathbb{L}_{X}V^{c}\right)\left(t+u,X_{t+u}^{x}\right)I\\{X_{t+u}^{x}\neq
c(t+u),X_{t+u}^{x}\neq L\\}du$
$\displaystyle\,\,\,\,\,\,\,\,\,\,+\frac{1}{2}\int_{0}^{s}e^{-ru}\bigl{(}V_{x}^{c}\left(t+u,L+\right)-V_{x}^{c}\left(t+u,L-\right)\bigr{)}dl_{u}^{L}(X^{x})$
$\displaystyle\,\,\,\,\,\,\,\,\,\,+\frac{1}{2}\int_{0}^{s}e^{-ru}\bigl{(}V_{x}^{c}\left(t+u,c(t+u)+\right)-V_{x}^{c}\left(t+u,c(t+u)-\right)\bigr{)}dl_{u}^{c}(X^{x})$
$\displaystyle=V^{c}(t,x)+M_{s}^{c}$
$\displaystyle\,\,\,\,\,\,\,\,\,\,+\int_{0}^{s}e^{-ru}\left(-rG+G_{t}+\mathbb{L}_{X}G\right)\left(t+u,X_{t+u}^{x}\right)I\\{X_{t+u}^{x}<c(t+u)\\}du$
$\displaystyle\,\,\,\,\,\,\,\,\,\,+\frac{1}{2}\int_{0}^{s}e^{-ru}I\\{c(t+u>L)\\}\bigl{(}G_{x}\left(t+u,L+\right)-G_{x}\left(t+u,L-\right)\bigr{)}dl_{u}^{L}(X^{x})$
$\displaystyle\,\,\,\,\,\,\,\,\,\,+\frac{1}{2}\int_{0}^{s}e^{-ru}\bigl{(}V_{x}^{c}\left(t+u,c(t+u)+\right)-V_{x}^{c}\left(t+u,c(t+u)-\right)\bigr{)}dl_{u}^{c}(X^{x}),$

where $M_{s}^{c}=\int_{0}^{s}e^{-ru}\sigma
X_{t+u}V_{x}^{c}(t+u,X_{t+u})I\\{X_{t+u}\neq c(t+u),X_{t+u}\neq L\\}du$ and it
is a martingale under measure $P_{t,x}$ so that
$E_{t,x}\left(M_{s}^{c}\right)=0$ for each $0\leq s\leq T-t$. Let us take the
expectation under measure $P_{t,x}$, upon recalling the definition of $V^{c}$,

$\displaystyle
E_{t,x}\left(e^{-rs}V^{c}\left(t+s,X_{t+s}\right)\right)=V^{c}(t,x)$
$\displaystyle\qquad+E_{t,x}\left(\int_{0}^{s}e^{-ru}\left(-rG+G_{t}+\mathbb{L}_{X}G\right)\left(t+u,X_{t+u}\right)I\\{X_{t+u}<c(t+u)\\}du\right)$
$\displaystyle\qquad+E_{t,x}\left(\frac{1}{2}\int_{0}^{s}e^{-ru}I\\{c(t+u)>L\\}\bigl{(}G_{x}\left(t+u,L+\right)-G_{x}\left(t+u,L-\right)\bigr{)}dl_{u}^{L}(X)\right)$
$\displaystyle\qquad+E_{t,x}\left(\frac{1}{2}\int_{0}^{s}e^{-ru}\bigl{(}V_{x}^{c}\left(t+u,c(t+u)+\right)-V_{x}^{c}\left(t+u,c(t+u)-\right)\bigr{)}dl_{u}^{c}(X)\right).$
(3.80)

(iii) The smooth-fit condition then makes it reasonable to guess that
$x\mapsto V^{c}(t,x)$ is $C^{1}$ at $c(t)$ for $0\leq t<T$ whenever $c(t)\neq
L$. This will not only be a guess if we could somehow show that
$U(t,x)=G(t,x)$. The crucial observation is that, from the strong Markov
property and time-homogeneous property of $X$ and smoothing lemma (see
(A.88)),

$\displaystyle E_{t,x}\left(e^{-rs}U\left(t+s,X_{t+s}\right)\right)-U(t,x)$
$\displaystyle=E_{t,x}\left(\int_{0}^{s}e^{-ru}H\left(t+u,X_{t+u}\right){I}\\{X_{t+u}<c(t+u)\\}du\right.$
$\displaystyle\qquad\qquad-\left.\frac{K-L}{2}\int_{0}^{s}e^{-ru}{I}\\{c(t+u)\geq
L\\}Z_{x}(t+u,L-)dl_{u}^{L}(X)\right).$ (3.81)

We are now have made the preparation needed to show that $U(t,x)=G(t,x)$ for
$(t,x)\in\mathcal{D}_{c}$. Let us consider the stopping time,

$\displaystyle\sigma_{c}=\inf\\{0\leq s\leq T-t:X_{t+s}\geq c(t+s)\\},$

so that, from the fact that $c(t)$ solves (3.75),
$U\left(t+\sigma_{c},X_{t+\sigma_{c}}\right)=G\left(t+\sigma_{c},X_{t+\sigma_{c}}\right)$.

Then, let $s=\sigma_{c}$ in equation (3.81) so that for $x\in(0,c(t)]$,

$\displaystyle U(t,x)$
$\displaystyle=E_{t,x}\left(e^{-r\sigma_{c}}U\left(t+\sigma_{c},X_{t+\sigma_{c}}\right)-\int_{0}^{\sigma_{c}}e^{-ru}H\left(t+u,X_{t+u}\right){I}\\{X_{t+u}<c(t+u)\\}du\right.$
$\displaystyle\qquad\qquad\,\,\,-\left.\frac{1}{2}\int_{0}^{\sigma_{c}}e^{-ru}{I}\\{c(t+u)\geq
L\\}\left(G_{x}\left(t+u,L+\right)-G_{x}\left(t+u,L-\right)\right)dl_{u}^{L}(X)\right)$
$\displaystyle=E_{t,x}\left(e^{-r\sigma_{c}}G\left(t+\sigma_{c},X_{t+\sigma_{c}}\right)-\int_{0}^{\sigma_{c}}e^{-ru}H\left(t+u,X_{t+u}\right)du\right.$
$\displaystyle\qquad\qquad-\left.\frac{1}{2}\int_{0}^{\sigma_{c}}e^{-ru}{I}\\{c(t+u)>L\\}\left(G_{x}\left(t+u,L+\right)-G_{x}\left(t+u,L-\right)\right)dl_{u}^{L}(X)\right)$
$\displaystyle=G(t,x)+E_{t,x}\left(\int_{0}^{\sigma_{c}}e^{-ru}H\left(t+u,X_{t+u}\right)du\right)$
$\displaystyle\qquad\qquad-
E_{t,x}\left(\frac{K-L}{2}\int_{0}^{\sigma_{c}}e^{-ru}Z_{x}(t+u,L-)dl_{u}^{L}(X)\right)$
$\displaystyle\qquad\qquad-
E_{t,x}\left(\int_{0}^{\sigma_{c}}e^{-ru}H\left(t+u,X_{t+u}\right)du\right)$
$\displaystyle\qquad\qquad+E_{t,x}\left(\frac{K-L}{2}\int_{0}^{\sigma_{c}}e^{-ru}{I}\\{c(t+u)>L\\}Z_{x}(t+u,L-)dl_{u}^{L}(X)\right)$
$\displaystyle=G(t,x)-E_{t,x}\left(\frac{K-L}{2}\int_{0}^{\sigma_{c}}e^{-ru}{I}\\{c(t+u)\leq
L\\}Z_{x}(t+u,L-)dl_{u}^{L}(X)\right)=G(t,x),$

where the first and last equality is due to the definition of $\sigma_{c}$, in
particular, for $c\leq L$, $X$ will either never reach $L$ before the process
stops or reach $L$ but stop simultaneously; and the third equality holds true
via (3.44).

This proves that $U(t,x)=G(t,x)$ for $(t,x)\in\mathcal{D}^{c}$ as desired;
correspondingly, from the construction of $V^{c}(t,x)$, the conclusion that
$V^{c}(t,x)=U(t,x)$ on $[0,T]\times(0,\infty)$ can be drawn so that

$\displaystyle
E_{t,x}\left(e^{-rs}V^{c}\left(t+s,X_{t+s}\right)\right)-V^{c}(t,x)$
$\displaystyle=E_{t,x}\left(\int_{0}^{s}e^{-ru}H\left(t+u,X_{t+u}\right){I}\\{X_{t+u}<c(t+u)\\}du\right.$
$\displaystyle\qquad\qquad-\left.\frac{K-L}{2}\int_{0}^{s}e^{-ru}{I}\\{c(t+u)\geq
L\\}Z_{x}(t+u,L-)dl_{u}^{L}(X)\right).$ (3.82)

(iv) Before moving on, we pause for another direct consequence of (3.81):
$V^{c}(t,x)\leq V(t,x)$. To see this relation, we begin by considering the
stopping time

$\displaystyle\tau_{c}=\inf\\{0\leq s\leq T-t:X_{t+s}\leq c(t+s)\\}.$

With $s$ replaced by $\tau_{c}$ in (3.82), we once again arrives at the
relation $V^{c}\leq V$. Namely,

$\displaystyle V^{c}(t,x)$
$\displaystyle=E_{t,x}\left(e^{-r\tau_{c}}V^{c}\left(t+\tau_{c},X_{t+\tau_{c}}\right)-\int_{0}^{\tau_{c}}e^{-ru}H\left(t+u,X_{t+u}\right){I}\\{X_{t+u}<c(t+u)\\}du\right.$
$\displaystyle\qquad\qquad+\left.\frac{K-L}{2}\int_{0}^{s}e^{-ru}{I}\\{c(t+u)\geq
L\\}Z_{x}(t+u,L-)dl_{u}^{L}(X)\right)$
$\displaystyle=E_{t,x}\left(e^{-r\tau_{c}}V^{c}\left(t+\tau_{c},X_{t+\tau_{c}}\right)\right)$
$\displaystyle=E_{t,x}\left(e^{-r\tau_{c}}G\left(t+\tau_{c},X_{t+\tau_{c}}\right)\right)\leq
V(t,x),$

where the second equality is based on the definition of $\tau_{c}$: if
$c(t)>L$, then $X$ will never hit $L$ before stopping; on the other hand, if
$c(t)=L$, it will spend zero time on level $L$.

(v) Analogously, our task in this step is to prove the relation $c\geq b$
holds true on $[0,T]$.

Let us fix $(t,x)\in(0,T)\times(0,\infty)$ such that $x<b(t)\wedge c(t)$ and
consider the stopping time

$\displaystyle\sigma_{b}=\inf\\{0\leq s\leq T-t:X_{t+s}\geq b(t+s)\\}.$

Replacing $s$ by $\sigma_{b}$ in (3.82) and in (3.78) yields,

$\displaystyle
E_{t,x}\left(e^{-r\sigma_{b}}V^{c}\left(t+\sigma_{b},X_{t+\sigma_{b}}\right)\right)-V^{c}(t,x)$
$\displaystyle\qquad=E_{t,x}\left(\int_{0}^{\sigma_{b}}e^{-ru}H\left(t+u,X_{t+u}\right){I}\\{X_{t+u}<c(t+u)\\}du\right.$
$\displaystyle\qquad\qquad\qquad-\left.\frac{K-L}{2}\int_{0}^{\sigma_{b}}e^{-ru}{I}\\{c(t+u)\geq
L\\}Z_{x}(t+u,L-)dl_{u}^{L}(X)\right)$

and

$\displaystyle
E_{t,x}\left(e^{-r\sigma_{b}}V\left(t+\sigma_{b},X_{t+\sigma_{b}}\right)\right)-V(t,x)$
$\displaystyle\qquad=E_{t,x}\left(\int_{0}^{\sigma_{b}}e^{-ru}H\left(t+u,X_{t+u}\right){I}\\{X_{t+u}<b(t+u)\\}du\right.$
$\displaystyle\qquad\qquad\qquad-\left.\frac{K-L}{2}\int_{0}^{\sigma_{b}}e^{-ru}{I}\\{b(t+u)\geq
L\\}Z_{x}(t+u,L-)dl_{u}^{L}(X)\right)$

From early step (iv), we know that

$\displaystyle
E_{t,x}\left(e^{-r\sigma_{b}}V^{c}\left(t+\sigma_{b},X_{t+\sigma_{b}}\right)\right]\leq
E_{t,x}\left[e^{-r\sigma_{b}}V\left(t+\sigma_{b},X_{t+\sigma_{b}}\right)\right),$

and that $V^{c}(t,x)=V(t,x)=G(t,x)$, it then follows that

$\displaystyle
E_{t,x}\left(\int_{0}^{\sigma_{b}}e^{-ru}H\left(t+u,X_{t+u}\right){I}\\{X_{t+u}\geq
c(t+u)\\}du\right.$ (3.83)
$\displaystyle\qquad-\left.\frac{K-L}{2}\int_{0}^{\sigma_{b}}e^{-ru}\left({I}\\{b(t+u)\geq
L\\}-I\\{c(t+u)\geq L\\}\right)Z_{x}(t+u,L-)dl_{u}^{L}(X)\right)\geq 0.$

The desired claim is that $c\geq b$, so let us assume the contrary that $c<b$.
It then follows that $I\\{c(t+u)\geq L)\\}=1$ entails $I\\{b(t+u)\geq L\\}=1$.
In addition, note that $H<0$, leading to a contradiction and thereby proving
$c\geq b$.

(vi) In this final step, we wish to show that $c=b$ on $[0,T]$. To achieve
this, suppose that $c=b$ is violated such that $c>b$ (from the previous step).
Then, by choosing $x\in(b(t),c(t))$and considering the stopping time

$\displaystyle\tau_{b}=\inf\\{0\leq s\leq T-t:X_{t+s}\leq b(t+s)\\}$

and with $s$ in (3.82) and in (3.78) replaced by $\tau_{b}$, noticing that
$\left(t+\tau_{b},X_{t+\tau_{b}}\right)\in\mathcal{D}\subset\mathcal{D}_{c}$,
we arrive at

$\displaystyle
E_{t,x}\left(e^{-r\tau_{b}}G\left(t+\tau_{b},X_{t+\tau_{b}}\right)\right)=V(t,x),$
$\displaystyle
E_{t,x}\left(e^{-r\tau_{b}}G\left(t+\tau_{b},X_{t+\tau_{b}}\right)\right)=V^{c}(t,x)+E_{t,x}\left(\int_{0}^{\sigma_{b}}e^{-ru}H\left(t+u,X_{t+u}\right){I}\\{X_{t+u}<c(t+u)\\}du\right)$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\,\,\,\,\,\,-E_{t,x}\left(\frac{K-L}{2}\int_{0}^{\sigma_{b}}e^{-ru}{I}\\{c(t+u)\geq
L\\}Z_{x}(t+u,L-)dl_{u}^{L}(X)\right).$

The relation $V^{c}\leq V$ therefore suggests that

$\displaystyle
E_{t,x}\left(\int_{0}^{\sigma_{b}}e^{-ru}H\left(t+u,X_{t+u}\right){I}\\{X_{t+u}<c(t+u)\\}du\right.$
$\displaystyle\qquad\qquad-\left.\frac{K-L}{2}\int_{0}^{\sigma_{b}}e^{-ru}{I}\\{c(t+u)\geq
L\\}Z_{x}(t+u,L-)dl_{u}^{L}(X)\right)\geq 0.$ (3.84)

which leads to a contradiction since $\eqref{bmcic}$ is strictly negative, in
particular, $H<0$ so are the local time terms. This establishes the uniqueness
of the solution in (3.75).

The proof of Theorem 3.39 is finally complete. ∎

Figure 7: This figure displays the maps $t\mapsto b(t)$ and $t\mapsto B(t)$ on
$[0,T]$ with the same chosen parameters $r=0.05$, $\sigma=0.4$, $L=2$, $K=7$,
$T=10$, where functions $b$ and $B$ are defined as the optimal stopping
boundaries for problem (3.26) in the case of $L<B(0)$ and the standard
American put option (in page 82) respectively.

Figure 8: This figure displays the maps $t\mapsto b(t)$ and $t\mapsto B(t)$ on
$[0,T]$ with the same chosen parameters $r=0.05$, $\sigma=0.4$, $L=6.5$,
$K=7$, $T=10$, where functions $b$ and $B$ are defined as the optimal stopping
boundaries for problem (3.26) in the case of $B(0)<L$ and the standard
American put option (in page 82) respectively.

Figure 9: This figure displays the maps $t\mapsto b(t)$, $t\mapsto B(t)$ on
$[0,T]$ and $t\mapsto x^{*}(t)$ on $[0,t_{*}]$ with the same chosen parameters
$r=0.05$, $\sigma=0.4$, $L=5$, $K=7$, $T=10$, where functions $b$, $B$ are
defined as that in Figure 9 and the detailed definition of function $x^{*}$ is
given in Lemma 3.22.

## References

  * [1] Biagini, F. and Øksendal, B., 2005. A general stochastic calculus approach to insider trading. Applied Mathematics and Optimization, 52(2), pp.167-181.
  * [2] Boyce, W.M., 1970. Stopping rules for selling bonds. The Bell Journal of Economics and Management Science, pp.27-53.
  * [3] Broadie, M. and Detemple, J., 1995. American capped call options on dividend-paying assets. The Review of Financial Studies, 8(1), pp.161-191.
  * [4] De Angelis, T., 2015. A note on the continuity of free-boundaries in finite-horizon optimal stopping problems for one-dimensional diffusions. SIAM Journal on Control and Optimization, 53(1), pp.167-184.
  * [5] De Angelis, T. and Peskir, G., 2020. Global $C^{1}$ regularity of the value function in optimal stopping problems. The Annals of Applied Probability, 30(3), pp.1007-1031.
  * [6] De Angelis, T. and Stabile, G., 2019. On the free boundary of an annuity purchase. Finance and Stochastics, 23(1), pp.97-137.
  * [7] Detemple, J. and Kitapbayev, Y., 2018. American options with discontinuous two-level caps. SIAM Journal on Financial Mathematics, 9(1), pp.219-250.
  * [8] Detemple, J. and Kitapbayev, Y., 2021. Callable barrier reverse convertible securities. Quantitative Finance, pp.1-14.
  * [9] Du Toit, J. and Peskir, G., 2009. Selling a stock at the ultimate maximum. The Annals of Applied Probability, 19(3), pp.983-1014
  * [10] Du Toit, J., Peskir, G. and Shiryaev, A.N., 2008. Predicting the last zero of Brownian motion with drift. Stochastics: An International Journal of Probability and Stochastics Processes, 80(2-3), pp.229-245.
  * [11] Elliott, R.J., Jeanblanc, M. and Yor, M., 2000. On models of default risk. Mathematical Finance, 10(2), pp.179-195.
  * [12] Fontana, C., Jeanblanc, M. and Song, S., 2014. On arbitrages arising with honest times. Finance and Stochastics, 18(3), pp.515-543.
  * [13] Gilbert, J.P. and Mosteller, F., 2006. Recognizing the maximum of a sequence. Selected Papers of Frederick Mosteller, pp.355-398.
  * [14] Glover, K., Peskir, G. and Samee, F., 2011. The British Russian Option. Stochastics An International Journal of Probability and Stochastic Processes, 83(4-6), pp.315-332.
  * [15] Gapeev, P.V., Li, L. and Wu, Z., 2020. Perpetual American cancellable standard options in models with last passage times. Algorithms, 14(1), p.3.
  * [16] Griffeath, D. and Snell, J.L., 1974. Optimal stopping in the stock market. The Annals of Probability, pp.1-13.
  * [17] Hull, J. and White, A., 1995. The impact of default risk on the prices of options and other derivative securities. Journal of Banking and Finance, 19(2), pp.299-322.
  * [18] Karlin, S., 1962. Stochastic models and optimal policy for selling an asset. Studies in applied probability and management science.
  * [19] Kitapbayev, Y., 2014. Optimal stopping problems with applications to mathematical finance (Doctoral dissertation, University of Manchester).
  * [20] Moerbeke, P., 1974. An optimal stopping problem with linear reward. Acta Mathematica, 132, pp.111-151.
  * [21] Myneni, R., 1992. The pricing of the American option. The Annals of Applied Probability, pp.1-23.
  * [22] Mansuy, R. and Yor, M., 2006. Random times and enlargements of filtrations in a Brownian setting (Vol. 1873). Berlin: Springer.
  * [23] Nikeghbali, A., 2006. An essay on the general theory of stochastic processes. Probability Surveys, 3, pp.345-412.
  * [24] Peskir, G. and Shiryaev, A., 2006. Optimal stopping and free-boundary problems (pp. 123-142). Birkhäuser Basel.
  * [25] Peskir, G., 2005. On the American option problem. Mathematical Finance: An International Journal of Mathematics, Statistics and Financial Economics, 15(1), pp.169-181.
  * [26] Peskir, G., 2005. A change-of-variable formula with local time on curves. Journal of Theoretical Probability, 18(3), pp.499-535.
  * [27] Peskir, G., 2007. A change-of-variable formula with local time on surfaces. In Séminaire de probabilités XL (pp. 70-96). Springer, Berlin, Heidelberg.
  * [28] Qiu, S., 2016. American eagle options. Research Report, School of Mathematics, The University of Manchester.
  * [29] SD, J., 1992. Finite-horizon optimal stopping, obstacle problems and the shape of the continuation region. Stochastics: An International Journal of Probability and Stochastic Processes, 39(1), pp.25-42.
  * [30] Shiryaev, A.N., 1999. Essentials of stochastic finance: facts, models, theory (Vol. 3). World scientific.

##  Appendix A Some Additional Mathematics

#### 1\. The setting:

$\displaystyle\alpha$ $\displaystyle=\frac{2r}{\sigma^{2}}-1$ $\displaystyle
d_{1}$
$\displaystyle=\frac{-\log{\frac{L}{x}}+\left(r-\frac{\sigma^{2}}{2}\right)(T-t)}{\sigma\sqrt{T-t}}$
$\displaystyle d_{2}$
$\displaystyle=\frac{-\log{\frac{L}{x}}-\left(r-\frac{\sigma^{2}}{2}\right)(T-t)}{\sigma\sqrt{T-t}}$

#### 2\. The relation between $d_{1}$ and $d_{2}$:

$\displaystyle
e^{-\frac{d_{1}^{2}}{2}}=\left(\frac{L}{x}\right)^{\alpha}e^{-\frac{d_{2}^{2}}{2}}.$

#### 3\. The derivatives with respect to time $t$:

$\displaystyle\frac{\partial}{\partial t}d_{1}$
$\displaystyle=-\frac{1}{2}\sigma^{-1}(T-t)^{-\frac{3}{2}}\log{\frac{L}{x}}-\frac{1}{2}\left(\frac{r}{\sigma}-\frac{\sigma}{2}\right)(T-t)^{-\frac{1}{2}},$
$\displaystyle\frac{\partial}{\partial t}d_{2}$
$\displaystyle=-\frac{1}{2}\sigma^{-1}(T-t)^{-\frac{3}{2}}\log{\frac{L}{x}}+\frac{1}{2}\left(\frac{r}{\sigma}-\frac{\sigma}{2}\right)(T-t)^{-\frac{1}{2}},$

so that

$\displaystyle\frac{\partial}{\partial
t}Z(t,x)=\begin{cases}&-\sqrt{\frac{1}{2\pi}}\log{\left(\frac{L}{x}\right)}e^{-\frac{d_{1}^{2}}{2}}\sigma^{-1}(T-t)^{-\frac{3}{2}},\qquad\qquad\,\,t<T,\\\
&0,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad t=T.\end{cases}$
(A.85)

#### 4\. The derivatives with respect to time $x$:

$\displaystyle\frac{\partial}{\partial x}d_{1}=\frac{\partial}{\partial
x}d_{2}=x^{-1}\sigma^{-1}(T-t)^{-\frac{1}{2}}.$

so that

$\displaystyle\frac{\partial}{\partial
x}Z(t,x)=\begin{cases}\sqrt{\frac{2}{\pi}}e^{-\frac{d_{1}^{2}}{2}}x^{-1}\sigma^{-1}(T-t)^{-\frac{1}{2}}-\alpha\left(\frac{L}{x}\right)^{\alpha}x^{-1}\Phi(d_{2}),&x<L\\\
0,&x\geq L\,\,\text{or}\,\,t=T,\end{cases}$ (A.86)

and thus, for $(t,x)\in[0,T)\times(0,L)$

$\displaystyle\frac{\partial^{2}}{\partial x\partial
t}Z(t,x)=\sqrt{\frac{1}{2\pi}}x^{-1}\sigma^{-1}(T-t)^{-\frac{3}{2}}e^{-\frac{d_{1}^{2}}{2}}\left(1-\left(\log{\frac{L}{x}}\right)^{2}(T-t)^{-1}\sigma^{-2}+\log{\frac{L}{x}}\sigma^{-1}\left(\frac{r}{\sigma}-\frac{\sigma}{2}\right)\right)$

#### 5\. The bounded partial derivative w.r.t $x$ for
$(t,x)\in[0,t_{\delta})\times(0,\infty)$:

$\displaystyle 0\leq\frac{\partial}{\partial x}Z(t,x)$
$\displaystyle=\sqrt{\frac{2}{\pi}}e^{-\frac{d_{1}^{2}}{2}}x^{-1}\sigma^{-1}(T-t)^{-\frac{1}{2}}-\alpha\left(\frac{L}{x}\right)^{\alpha}x^{-1}\Phi(d_{2})$
(A.87)
$\displaystyle<\sqrt{\frac{2}{\pi}}\sigma^{-1}(T-t_{\delta})^{-\frac{1}{2}}x^{-1}e^{-\frac{\left(\frac{-\log{\frac{L}{x}}-\left(r-\frac{\sigma^{2}}{2}\right)T}{\sigma\sqrt{T-t_{\delta}}}\right)^{2}}{2}}-\alpha
L^{\alpha}x^{-\frac{2r}{\sigma^{2}}}\Phi(d_{2})=C_{2}<\infty.$

Note that by L’Hôpital’s rule (and that $d_{2}$ is a function of $x$ and
$\frac{2r}{\sigma^{2}}<1$, $\lim\limits_{x\infty}\frac{x}{e^{x}}\to 0$):

$\displaystyle\lim_{x\to
0}\frac{\Phi(d_{2})}{x^{\frac{2r}{\sigma^{2}}}}=\lim_{x\to
0}\frac{\sqrt{\frac{1}{{2\pi}}}e^{-\frac{d_{2}^{2}}{2}}x^{-1}\sigma^{-1}(T-t)^{-\frac{1}{2}}}{\frac{2r}{\sigma^{2}}x^{\frac{2r}{\sigma^{2}}-1}}=\sqrt{\frac{1}{2\pi}}\frac{\sigma}{2r}(T-t)^{-\frac{1}{2}}\lim_{x\to
0}\frac{e^{-\frac{d_{2}^{2}}{2}}}{x^{\frac{2r}{\sigma^{2}}}}\to 0$

#### 6\. The derivative w.r.t time $t$ for $x<L,t<T$:

$\displaystyle-
rKZ_{t}-\sigma^{2}x^{2}Z_{xt}=\sqrt{\frac{1}{2\pi}}e^{-\frac{d_{1}^{2}}{2}}(T-t)^{-\frac{3}{2}}\left(\frac{r(K-x)\log{\frac{L}{x}}}{\sigma}-\sigma
x+\sigma
x\left(\frac{\log{\frac{L}{x}}}{\sigma\sqrt{T-t}}\right)^{2}+\frac{\sigma
x}{2}\log{\frac{L}{x}}\right)$

and note that the assumption $-rKZ_{t}-\sigma^{2}x^{2}Z_{xt}\leq 0$ on set
$\mathcal{S}$, together with $Z_{t}\leq 0$, suggests that
$Z_{xt}\geq\frac{-rKZ_{t}}{\sigma^{2}x^{2}}\geq 0$.

#### 7\. The expectation of the local time:

$\displaystyle\lim_{\epsilon\to
0}\frac{1}{\epsilon}P\left(\log{\frac{L-\epsilon}{y}}<\left(r-\frac{\sigma^{2}}{2}\right)u+\sigma
W_{u}<\log{\frac{L+\epsilon}{y}}\right)=\frac{1}{\sigma\sqrt{u}L}\phi\left(\frac{\log{\frac{L}{y}-\left(r-\frac{\sigma^{2}}{2}\right)u}}{\sigma\sqrt{u}}\right).$
$\displaystyle dE_{t,x}\left[l_{u}^{L}\right]$
$\displaystyle=\sigma^{2}L^{2}\lim_{\epsilon\to
0}\frac{1}{\epsilon}P\left(\log{\frac{L-\epsilon}{x}}<\left(r-\frac{\sigma^{2}}{2}\right)u+\sigma
W_{u}<\log{\frac{L+\epsilon}{x}}\right)du$ $\displaystyle=\frac{\sigma
L}{\sqrt{u}}\phi\left(\frac{\log{\frac{L}{x}-\left(r-\frac{\sigma^{2}}{2}\right)u}}{\sigma\sqrt{u}}\right)du,$

where $\phi(z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{z^{2}}{2}}$ is the probability
density function of the standard normal law.

#### 8\. Useful result for proving smooth-fit condition:

######  Lemma A.1.

As $\epsilon\to 0$, $\tau_{\epsilon}\to 0$ almost surely.

###### Proof.