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\begin{document}

\title{When to Sell a Markov Chain Asset?\thanks{This research is supported in part by the Simons Foundation (235179).}}

\author{Q. Zhang\thanks{Department of Mathematics, University of Georgia,
Athens, GA 30602, qingz@math.uga.edu}}

\maketitle

\begin{abstract}

This paper is concerned with an optimal stock selling rule
under a Markov chain model.
The objective is to find an optimal stopping time to sell the
stock so as to maximize an expected return.
Solutions to the associated variational inequalities are obtained.
Closed-form solutions are given in terms of a set of threshold levels.
Verification theorems are provided to justify their optimality.
Finally, numerical examples are reported to illustrate the results.

\bigskip\noindent
{\bf Key words:} Markov chain asset, optimal stopping,
quasi-variational inequalities

\end{abstract}

\newpage

\baselineskip25pt
\renewcommand{\arraystretch}{1.8}

\section{Introduction}

Most market models in the literature are Brownian motion based including
geometric Brownian motion, diffusion with possible jumps and regime switching;
see related books by Duffie \cite{Duffie}, Hull \cite{Hull},
Elliott and Kopp \cite{ElliottK}, Fouque et al. \cite{FouquePR},
Karatzas and Shreve \cite{KaratzasS},
and Musiela and Rutkowski \cite{MusielaR} among others.
An alternative is the binomial tree model introduced by
Cox-Ross-Rubinstein.
The BTM is natural for financial markets because
intensive buying moves the market upwards and forceful selling
pushes it downwards.
All these transactions take place in discrete moments.
However, a main drawback of the BTM is its non-Markovian nature,
which makes it difficult to work with mathematically.
In this paper, we consider a Markov chain market model.
The main advantage of such a model is it preserves much of the flexibility of
the binomial tree structure and, in the meantime,
it is more mathematically tractable, which allows serious mathematical
analysis in related optimization problems.
Recently, several Markov chain based models are developed.
For example, van der Hoek and Elliott \cite{HoekE} introduced
a stock price model based on stock dividend rates and a Markov chain noise.
Norberg \cite{Norberg} used a Markov chain to represent interest rate and
considered a market model driven by a Markov chain. In particular,
the market model in \cite{Norberg} resembles a GBM in which
the `drift' is approximated by the duration between jumps and the
`diffusion' is given in terms of jump times.
An additional advantage of a Markov chain driven model is its price is
almost everywhere differentiable. Such differentiability is
desirable in an optimal control type
analysis proposed by Barmish and Primbs \cite{BarmishP}.
In connection with dynamic programming problems,
the corresponding Hamilton-Jacobi-Bellman equations are of first order,
which are easier to analyze than those under
traditional Brownian motion based models.
Finally, the Markov chain model is not that far apart from
a GBM because it
can be used to approximate a GBM by varying its jump rates.
In fact, it is shown in
Example 1 that a properly scaled Markov chain model converges weakly to
that of a GBM as the jump rates go to infinity.

When to sell a stock is a crucial component in stock trading.
It determines when to take profits or to cut losses.
It is probably the most emotional part for individual investors
in the trading process.
Selling rules in financial markets have been studied for many
years. For example, Zhang \cite{Zhang-trading} considered a
selling rule determined by two threshold levels: a target price
and a stop-loss limit. One makes a selling decision whenever the
price reaches either levels. Under a switching GBM,
the objective is to determine these threshold levels to maximize
an expected discounted reward function. In \cite{Zhang-trading},
such optimal threshold levels are obtained by solving a set of
two-point boundary value problems. In Guo and Zhang \cite{GuoZ},
they considered the optimal selling rule under a GBM model with
regime switching. Using a smooth-fit technique, they were able to
convert the optimal stopping problem to a set of algebraic
equations.
These algebraic equations were used to determine the optimal target levels.
In addition to these analytical results, various mathematical tools
have been developed to compute these threshold levels. For example,
a stochastic approximation technique was used in Yin, Liu and Zhang
\cite{YLZ} and a linear programming approach was developed in Helmes
\cite{Helmes}.
In addition, Merhi and Zervos \cite{MerhiZ} studied
an investment capacity expansion/reduction problem following
a dynamic programming approach under a GBM market model.
Similar problem under a more general market model was treated by
L{\o }kka and Zervos \cite{LokkaZ}.

In this paper, the stock price is assumed to follow
a Markov chain model. Under this model,
the state of the Markov chain can be estimated based on the
stock price increments. This makes the Markov chain observable.
In addition to its simplicity, the Markov chain model
is able to capture price movements of a broader range of stocks.
In this paper, under the Markov chain model,
we consider an optimal stock selling rule and obtain its
solution in terms of a set of threshold levels.
In particular, we solve the corresponding
dynamic programming problem and obtain
these threshold levels.
We point out that the standard smooth-fit method that works in
a GBM setting is not adequate in one of the
cases in this paper because of the lack
of enough equations for the unknown parameters.
To solve the problem, we need to explore other convexity
conditions to determine uniquely these parameters.
We also provide a set of sufficient conditions that guarantee their optimality.
Numerical examples are reported to illustrate these results.

This paper is organized as follows.
In \S2, we formulate the problem and make a few assumptions.
In \S3, we study properties of the value functions,
the associate HJB equations, and their solutions.
In \S4, we provide a set of sufficient conditions
that guarantee the optimality of our selling rule.
We also include three numerical examples in this section.
Some concluding remarks are given in \S5.
Some technical results are provided in an appendix.

\section{Problem Formulation}

Let $\{\alpha_t:\ t\geq0\}$ denote a two-state Markov chain with
state space ${\cal M}=\{1,2\}$ and generator
$Q=\(\begin{array}{cc}
   -\lambda_1&\lambda_1\\[-0.1in]
\lambda_2&-\lambda_2\\
\end{array}\)$, for given $\lambda_1>0$ and $\lambda_2>0$.
Let $S_t$ denote the stock price at time $t$ given by the
equation
\[
\frac{dS_t}{S_t}=f(\alpha_t)dt,\ S_0=x\geq0,\ t\geq0,
\]
where $f(1)=f_1>0$ represents the uptick return rate and
$f(2)=f_2<0$ the downtick return rate.
Let ${\cal F}_t=\{S_r:\ r\leq t\}$ denote the filtration generated by
$S_t$. Note that $\alpha_t$ is observable and
${\cal F}_t=\{\alpha_r:\ r\leq t\}$.

Let $K$ denote the fixed transaction cost.
Given $S_0=x$ and $\alpha_0=i\in{\cal M}$,
the objective of the problem is to choose an ${\cal F}_t$ stopping time
$\tau$ so as to maximize
\[
J(x,i,\tau)=E\(e^{-\rho\tau}(S_\tau-K)I_{\{\tau<\infty\}}\),
\]
where $\rho>0$ is the discount factor.

Let $V(x,i)=\sup_\tau J(x,i,\tau)$ be the value function.
Then it is easy to see that $V(x,i)\geq0$, $V(0,i)=0$, $i=1,2$.
Moreover, $V(x,i)$ is convex in $x$ for fixed $i=1,2$.

Let $\Phi(\rho)=(\rho+\lambda_1-f_1)(\rho+\lambda_2-f_2)-\lambda_1\lambda_2$.
Then the bigger root of $\Phi(\rho)=0$ is given by
\[
B_0=\frac{1}{2}\(f_1-\lambda_1+f_2-\lambda_2+
\sqrt{((f_1-\lambda_1)-(f_2-\lambda_2))^2+4\lambda_1\lambda_2}\).
\]

Note that if $\rho\leq B_0$, then following similar argument as in
Guo and Zhang \cite{GuoZ}, we can show that it is optimal not to
sell at all.
In the rest of this paper, we only consider the case when
$\rho>B_0$, which implies $\Phi(\rho)>0$.
We summarize the conditions to be imposed in the rest of this paper:

\begin{itemize}
\item[ ]{\bf (A1)} $f_1>0$ and $f_2<0$;
\item[ ]{\bf (A2)} $\Phi(\rho)>0$.
\end{itemize}

\medskip
Let
$\displaystyle (\nu_1,\nu_2)=\({\lambda_2}/{(\lambda_1+\lambda_2)},{\lambda_1}/{(\lambda_1+\lambda_2)}\)$
denote the stationary distribution of $\alpha_t$ and
let $\mu=\nu_1 f_1+\nu_2f_2$. Then, $f_2<\mu<f_1$.
Moreover, it is easy to see that
$\Phi(\mu)=(\mu-f_1)(\mu-f_2)<0$. This implies $B_0>\mu$.
Therefore, $\rho>\mu$.

Note that, for any ${\cal F}_t$ stopping time $\tau$,
\[
J(x,i,\tau)=xE\(e^{-\rho \tau}\exp\int_0^\tau f(\alpha_s)ds\)I_{\{\tau<\infty\}}
-KE e^{-\rho\tau}.
\]
In order to have finite $V(x,i)$, necessarily
\[
\sup_{\tau}E\(e^{-\rho \tau}\exp\int_0^\tau f(\alpha_s)ds\)I_{\{\tau<\infty\}}
<\infty.
\]
In view of this, $V(x,i)$ needs to be at most linear growth in $x$.
In addition, note that the stock price $S_t$ is differentiable and
the value of $\alpha_t$ can be given in terms of
the derivative of $\log(S_t)$.

\section{HJB Equations}
Let ${\cal A}$ denote the generator of $(S_t,\alpha_t)$, i.e., for any differentiable
functions $h(x,i)$, $i=1,2$,
\[
\left\{\begin{array}{l}
\displaystyle
{\cal A} h(x,1)=xf_1h'(x,1)+\lambda_1(h(x,2)-h(x,1)),\\[0.1in]
\displaystyle
{\cal A} h(x,2)=xf_2h'(x,2)+\lambda_2(h(x,1)-h(x,2)),\\
\end{array}\right.
\]
where $h'$ denotes the derivative of $h$ with respect to $x$.
The associated HJB equations should have the form:
\beq{HJB}
\left\{\begin{array}{l}
\min\{\rho v(x,1)-{\cal A} v(x,1),\ v(x,1)-(x-K)\}=0,\\
\min\{\rho v(x,2)-{\cal A} v(x,2),\ v(x,2)-(x-K)\}=0.
\end{array}\right.
\end{equation}

In this section, we solve these HJB equations.
First, if the price $S_t$ is small, then one should hold the
position because the price is not attractive regardless $\alpha_t=1$ or 2.
In view of this, we expect the existence of $x^*$ such that
no selling is $S_t<x^*$. The corresponding interval $(0,x^*)$ gives
a continuation region.
Note that $V(x,i)\geq0$ implies $x^*\geq K$.
On this interval, the equalities
$\rho v(x,i)-{\cal A} v(x,i)=0$, $i=1,2$, must hold.
Using the generator ${\cal A}$, we can write
\[
\left\{\begin{array}{l}
(\rho+\lambda_1)v(x,1)=xf_1 v'(x,1)+\lambda_1 v(x,2),\\
(\rho+\lambda_2)v(x,2)=xf_2 v'(x,2)+\lambda_2 v(x,1).\\
\end{array}\right.
\]
Using the first equation, we write
\[
v(x,2)=\frac{1}{\lambda_1}\((\rho+\lambda_1)v(x,1)-xf_1 v'(x,1)\).
\]
Substitute this into the second equation and simplify to obtain
\beq{joint-ode}
x^2f_1f_2v''(x,1)+x(f_1f_2-D_1)v'(x,1)+D_2 v(x,1)=0,
\end{equation}
where
\beq{D1-D2}
\left\{\begin{array}{l}
D_1=(\rho+\lambda_1)f_2+(\rho+\lambda_2)f_1,\\
D_2=(\rho+\lambda_1)(\rho+\lambda_2)-\lambda_1\lambda_2.
\end{array}\right.
\end{equation}

Let $\beta_1<0$ and $\beta_2>0$ denote the roots of
\beq{char-eqn}
f_1f_2\beta^2-D_1\beta+D_2=0.
\end{equation}
Then,
\beq{char-root}
\left\{\begin{array}{l}
\displaystyle
\beta_1=\frac{D_1+\sqrt{D_1^2-4f_1f_2 D_2}}{2f_1f_2}<0,\\ [0.1in]
\displaystyle
\beta_2=\frac{D_1-\sqrt{D_1^2-4f_1f_2 D_2}}{2f_1f_2}>0.\\
\end{array}\right.
\end{equation}
The general solution to (\ref{joint-ode}) can be given as
\[
v(x,1)=A_1 x^{\beta_1}+A_2 x^{\beta_2},
\]
for some constants $A_1$ and $A_2$.

On $(0,x^*)$, the convexity condition implies that $v(x,1)$
is bounded. Necessarily, $A_1=0$. Therefore, $v(x,1)=A_2x^{\beta_2}$.
Substitute this back into the first equation to obtain
$v(x,2)=\kappa_2 A_2 x^{\beta_2}$, where
$\kappa_2={(\rho+\lambda_1-f_1\beta_2)}/{\lambda_1}$.

Recall that $\alpha_t$ is observable. One should hold the position longer
under the condition $\alpha_t=1$ (uptick) than that under $\alpha_t=2$ (downtick).
In view of this,
we consider the HJB equations on $(x^*,x_0^*)$ for some $x_0^*>x^*$.
The idea is to sell if $(S_t,\alpha_t)\in [x^*,\infty)\times\{2\}$
and hold if $(S_t,\alpha_t)\in(0,x_0^*)\times\{1\}$
till $S_t$ reaching $x_0^*$.
Clearly, $v(x,2)=x-K$ and
$\rho v(x,1)-{\cal A} v(x,1)=0$, on $(x^*,x_0^*)$.
Using this, we solve the equation
\[
\rho v(x,1)-{\cal A} v(x,1)=0,\\
\]
which gives
\[
(\rho+\lambda_1)v(x,1)=xf_1 v'(x,1)+\lambda_1(x-K).
\]
It is easy to see a particular solution
\beq{A0-B0}
\phi_0(x)=A_0x+B_0,\mbox{ where }A_0=\frac{\lambda_1}{\rho+\lambda_1-f_1}
\mbox{ and } B_0=-\frac{\lambda_1 K}{\rho+\lambda_1}.
\end{equation}
Let $\gamma_1=(\rho+\lambda_1)/f_1$. Then, the general solution can be
given by
\[
v(x,1)=C_1x^{\gamma1}+\phi_0(x),
\]
for any constant $C_1$.

Next we consider two separate cases to continue solving the HJB equations.

\subsection*{Case I: $\rho\leq f_1$}
Assuming $\rho\leq f_1$, we first show that $x_0^*=\infty$.
If not, we must have $v(x,1)=v(x,2)=x-K$ for $x>x_0^*$.
In order to satisfy the HJB equations (\ref{HJB}), $v(x,1)$ has to
satisfy the inequality
$\rho v(x,1)-{\cal A} v(x,1)\geq0$, for $x>x_0^*$. Plugging
$v(x,1)=v(x,2)=x-K$ in this inequality, we have
\[
\rho (x-K)\geq x f_1.
\]
Therefore, $(\rho-f_1)x\geq \rho K$, for $x>x_0^*$. This
contradicts $\rho\leq f_1$. Hence $x_0^*=\infty$.

In view of these, on $(x^*,\infty)$, $v(x,1)=C_1x^{\gamma1}+\phi_0(x)$
and $v(x,2)=x-K$. This means never sell when $\alpha_t=1$.
Recall the linear growth property and nonnegativity
of $v(x,i)$. It follows that $C_1=0$ because $\gamma_1>1$.

Next, we determine the values of $x^*$ and $A_2$.
Recall that $v(x,1)$ and $v(x,2)$ are convex on $(0,\infty)$.
Necessarily, they are continuous. In particular, they
are continuous at $x=x^*$. Therefore,
\[
\left\{\begin{array}{l}
A_2(x^*)^{\beta_2}=A_0x^*+B_0,\\
\kappa_2A_2(x^*)^{\beta_2}=x^*-K.
\end{array}\right.
\]
Solving these equations, we have
\beq{x^*-Case-I}
x^*=-\frac{K+\kappa_2 B_0}{\kappa_2 A_0-1},
\end{equation}
and
\beq{A2}
A_2=\frac{A_0x^*+B_0}{(x^*)^{\beta_2}}.
\end{equation}

It is elementary to check that
\beq{x*-alternative}
x^*=\(\frac{\rho+\lambda_1-f_1}{\rho+\lambda_1}\)\(\frac{K\beta_2}{\beta_2-1}\).
\end{equation}

The solutions to the HJB equations (\ref{HJB}) should have the form:
\beq{value-fn-Case-I}
\left\{\begin{array}{l}
v(x,1)=\left\{\begin{array}{ll}
A_2 x^{\beta_2}&\mbox{ if }0\leq x\leq x^*,\\
A_0x+B_0&\mbox{ if }x> x^*,\\
\end{array}\right.\\
v(x,2)=\left\{\begin{array}{ll}
\kappa_2A_2 x^{\beta_2}&\mbox{ if }0\leq x\leq x^*,\\
x-K&\mbox{ if }x> x^*.\\
\end{array}\right.\\
\end{array}\right.
\end{equation}

\begin{thm}\hspace*{-0.08in}{\bf.}\hspace*{0.05in}\label{HJB-Case-I}
Assume $\rho\leq f_1$.
Then the functions $v(x,i)$, $i=1,2$, given above are continuous on
$(0,\infty)$ and differentiable on $(0,\infty)-\{x^*\}$.
They satisfy the HJB equations {\rm (\ref{HJB})}.
In particular, the following inequalities hold:
\beq{v-ineq}
\left\{\begin{array}{ll}
A_2 x^{\beta_2}\geq x-K&\mbox{ on }(0,x^*),\\
\kappa_2A_2 x^{\beta_2}\geq x-K&\mbox{ on }(0,x^*),\\
v(x,1)\geq x-K&\mbox{ on }(x^*,\infty),\\
\rho v(x,2)-{\cal A} v(x,2)\geq 0&\mbox{ on }(x^*,\infty).
\end{array}\right.
\end{equation}
\end{thm}

\noindent{\it Proof.}
It is sufficient to show these four inequalities.
First, note that $\Phi(\rho)>0$ implies $\rho+\lambda_1-f_1>0$.
Under the condition $\rho\leq f_1$, we have $A_0\geq1$.
The third inequality in (\ref{v-ineq}) follows from $B_0>-K$.
In addition, the first inequality follows from the second one
because $0<\kappa_2<1$ as shown in Appendix (\lemref{kappa2}).
To show the second inequality, we claim that $A_2>0$ and
\beq{convex-2}
\beta_2\kappa_2A_2(x^*)^{\beta_2-1}<1.
\end{equation}
To see $A_2>0$, notice that (\ref{x*-alternative}) implies
\[
x^*>\frac{(\rho+\lambda_1-f_1)K}{\rho+\lambda_1},
\]
because $\beta_2>1$ (see \lemref{beta2>1} in Appendix).
Therefore $A_0 x^*+B_0$ is positive, so is $A_2$.
To show (\ref{convex-2}), use again (\ref{x*-alternative}), which yields
\[
x^*<\frac{K\beta_2}{\beta_2-1}.
\]
This is equivalent to (\ref{convex-2}) because
$\kappa_2 A_2(x^*)^{\beta_2}=x^*-K$.
Let $\phi(x)=\kappa_2A_2x^{\beta_2}-(x-K)$.
In view of the above claim and the definition of $x^*$, it follows
that, on $(0,x^*)$,
\[
\begin{array}{l}
\phi(x^*)=\kappa_2A_2(x^*)^{\beta_2}-(x^*-K)=0,\\
\phi'(x^*)=\beta_2\kappa_2A_2(x^*)^{\beta_2-1}-1< 0,\\
\phi''(x)=\beta_2(\beta_2-1)\kappa_2A_2x^{\beta_2-2}>0.
\end{array}
\]
Consequently, $\phi'(x)$ is increasing on $(0,x^*)$, which implies
$\phi'(x)<0$. Hence, $\phi(x)$ is decreasing. Therefore,
$\phi(x)>0$ on $(0,x^*)$, which implies the second inequality in
(\ref{v-ineq}).

It remains to show the last inequality $\rho v(x,2)-{\cal A} v(x,2)\geq0$
in (\ref{v-ineq}).
This is equivalent to
\[
\rho(x-K)\geq xf_2+\lambda_2(\phi_0(x)-(x-K)).
\]
It follows that
\[
(\rho+\lambda_2-f_2-\lambda_2 A_0)x\geq (\rho+\lambda_2)K+\lambda_2B_0.
\]
Using the notation $\Phi(\rho)$ and $D_2$, we have
\[
\(\frac{\Phi(\rho)}{\rho+\lambda_1-f_1}\)x\geq \frac{KD_2}{\rho+\lambda_1}.
\]
Therefore, we need
\[
x\geq\frac{K(\rho+\lambda_1-f_1)D_2}{(\rho+\lambda_1)\Phi(\rho)},
\]
for $x\geq x^*$.
It suffices to show this inequality when $x=x^*$. Using the expression in
(\ref{x*-alternative}), we only have to show
\[
\frac{\beta_2}{\beta_2-1}\geq\frac{D_2}{\Phi(\rho)}.
\]
Rewrite this to obtain
\beq{1-beta2}
\frac{1}{\beta_2}\geq 1-\frac{\Phi(\rho)}{D_2}.
\end{equation}
Now, if $1-{\Phi(\rho)}/{D_2}\leq 0$ (i.e., $D_2\leq \Phi(\rho)$), then
we are done because $\beta_2>1$. Otherwise, $D_2>\Phi(\rho)$.
Under this condition, we rewrite (\ref{1-beta2}) as
\[
\beta_2\leq\frac{D_2}{D_2-\Phi(\rho)},
\]
which is equivalent to
\beq{rho-v2}
\sqrt{D_1^2-4f_1f_2D_2}\leq D_1-\frac{2f_1f_2}{D_2-\Phi(\rho)}.
\end{equation}
Note that
\beq{Phi-D1-D2}
\Phi(\rho)=D_2-D_1+f_1f_2.
\end{equation}
Therefore, $D_2>\Phi(\rho)$ implies that $D_1>f_1f_2$.
Under this condition, it is easy to check
\[
D_1-\frac{2f_1f_2D_2}{D_2-\Phi(\rho)}>
f_1f_2-\frac{2f_1f_2D_2}{D_2-\Phi(\rho)}=
-f_1f_2\(\frac{D_2+\Phi(\rho)}{D_2-\Phi(\rho)}\)>0.
\]
Square both sides of (\ref{rho-v2}) to obtain
\[
D_1^2-4f_1f_2D_2\leq D_1^2-\frac{4f_1f_2D_1D_2}{D_2-\Phi(\rho)}
+\frac{4f_1^2f_2^2D_2^2}{(D_2-\Phi(\rho))^2}.
\]
Simplify this inequality to have
\[
D_1(D_2-\Phi(\rho))-f_1f_2D_2\geq(D_2-\Phi(\rho))^2.
\]
Furthermore, using (\ref{Phi-D1-D2}), we have $D_2-\Phi(\rho)=D_1-f_1f_2$.
Substitute this into the above inequality to obtain
\[
D_1(D_1-f_1f_2)-f_1f_2D_2\geq (D_1-f_1f_2)^2.
\]
This is equivalent to
\[
D_2-D_1\geq -f_1f_2,
\]
which leads $\Phi(\rho)\geq 0$, which holds under the
assumption $\Phi(\rho)>0$.
Therefore, $\rho v(x,2)-{\cal A} v(x,2)\geq0$ on $(x^*,\infty)$.
The proof is compete. \hfill$\Box$

\begin{rem}\hspace*{-0.08in}{\bf.}\hspace*{0.05in}
{\rm
Using (\ref{x*-alternative}),
one can show $\beta_2A_2(x^*)^{\beta_2-1}=A_0$. This implies that
$v(x,1)$ is differentiable at $x=x^*$. On the other hand,
following (\ref{convex-2}), we can see that
$v(x,2)$ is not differentiable at $x=x^*$.
}
\end{rem}

\subsection*{Case II: $\rho> f_1$}
We consider the second case when $\rho> f_1$. Note that a large $\rho$
encourages selling sooner. Naturally, we expect $x_0^*<\infty$.
The solutions to the HJB equations (\ref{HJB}) should have the form:

\beq{value-fn-Case-II}
\left\{\begin{array}{l}
v(x,1)=\left\{\begin{array}{ll}
A_2 x^{\beta_2}&\mbox{ if }0\leq x\leq x^*,\\
C_1x^{\gamma_1}+\phi_0(x)&\mbox{ if }x^*< x\leq x_0^*,\\
x-K&\mbox{ if }x> x_0^*,\\
\end{array}\right.\\
v(x,2)=\left\{\begin{array}{ll}
\kappa_2A_2 x^{\beta_2}&\mbox{ if }0\leq x\leq x^*,\\
x-K&\mbox{ if }x> x^*.\\
\end{array}\right.\\
\end{array}\right.
\end{equation}
We need to determine the values of $A_2$, $C_1$, $x_0^*$, and $x^*$.
Again, following the continuity of the value functions
at $x^*$ and $x_0^*$, we have
\beq{cont-conds-case-II}
\left\{\begin{array}{l}
A_2(x^*)^{\beta_2}=C_1(x^*)^{\gamma_1}+\phi_0(x^*),\\
\kappa_2A_2(x^*)^{\beta_2}=x^*-K,\\
C_1(x_0^*)^{\gamma_1}+\phi_0(x_0^*)=x_0^*-K.\\
\end{array}\right.
\end{equation}
Note that there are only three equations, which are not adequate to
determine uniquely the values of the four unknowns.
We need to find further conditions.
Note that
to satisfy the HJB equations (\ref{HJB}), the following inequalities
have to hold:

\beq{VI-1-Case-II}
\left\{\begin{array}{l}
A_2x^{\beta_2}\geq x-K,\\
\kappa_2A_2x^{\beta_2}\geq x-K,\\
\end{array}\right.
\mbox{    on }(0,x^*);
\end{equation}

\beq{VI-2-Case-II}
\left\{\begin{array}{l}
C_1x^{\gamma_1}+\phi_0(x)\geq x-K,\\
\rho v(x,2)-{\cal A} v(x,2)\geq0,\\
\end{array}\right.
\mbox{    on }(x^*,x_0^*);
\end{equation}

\beq{VI-3-Case-II}
\left\{\begin{array}{l}
\rho v(x,1)-{\cal A} v(x,1)\geq0,\\
\rho v(x,2)-{\cal A} v(x,2)\geq0,\\
\end{array}\right.
\mbox{    on }(x_0^*,\infty).
\end{equation}

First, we consider (\ref{VI-1-Case-II}).
Note that convexity of $v(x,2)$ at $x=x^*$ implies
\beq{case-II-x^*}
\beta_2\kappa_2A_2(x^*)^{\beta_2-1}\leq 1.
\end{equation}
Under this condition, following from similar
argument used to prove the second inequality in
(\ref{v-ineq}) with $\phi(x)=\kappa_2A_2x^{\beta_2}-(x-K)$
for possibly different $x^*$, we can show
the second inequality in (\ref{VI-1-Case-II}) holds, so does the first one.
Therefore, the inequalities in (\ref{VI-1-Case-II}) are equivalent to
(\ref{case-II-x^*}), which can be simplified and written as:
\beq{case-II-x^*-v2}
x^*\leq \frac{K\beta_2}{\beta_2-1}.
\end{equation}

Next, we consider (\ref{VI-3-Case-II}). The first inequality implies
\[
\rho(x-K)\geq xf_1, \mbox{ for }x< x_0^*.
\]
It follows that
\beq{x_0^*-cond}
x_0^*\geq \frac{\rho K}{\rho-f_1}.
\end{equation}
The second inequality in (\ref{VI-3-Case-II})
is automatically satisfied because $f_2<0$.

Finally, go back to (\ref{VI-2-Case-II}).
Again, the convexity of $v(x,1)$ at $x=x_0^*$ yields
\beq{1-ineq-VI-2-Case-II}
\gamma_1 C_1(x_0^*)^{\gamma_1-1}+A_0\leq 1.
\end{equation}

It follows from the third equality in (\ref{cont-conds-case-II}) that
\beq{C1}
C_1=\frac{x_0^*-K-\phi_0(x_0^*)}{(x_0^*)^{\gamma_1}}.
\end{equation}
Under the condition $x_0^*\geq \rho K/(\rho-f_1)$, it is easy to see
$C_1>0$.
Note that $0<A_0<1$ under $\rho>f_1$.
Let $\phi(x)=C_1x^{\gamma_1}+\phi_0(x)-(x-K)$. Then,
it is direct to check that $\phi(x_0^*)=0$, $\phi'(x_0^*)\leq0$,
$\phi''(x)=\gamma_1(\gamma_1-1)C_1x^{\gamma_1-2}>0$.
Therefore, $\phi'(x)$ is increasing on $(x^*,x_0^*)$.
Thus, $\phi'(x)<0$ on $(x^*,x_0^*)$, which implies
$\phi(x)$ is decreasing. Therefore, $\phi(x)\geq0$ on $(x^*,x_0^*)$.
The first inequality in (\ref{VI-2-Case-II}) follows from
(\ref{1-ineq-VI-2-Case-II}).

Use (\ref{C1}) and rewrite (\ref{1-ineq-VI-2-Case-II}) to obtain
\[
\gamma_1(x_0^*-K-\phi_0(x_0^*))\leq (1-A_0)x_0^*,
\]
which leads to
\[
(\gamma_1-1)(1-A_0)x_0^*\leq \gamma_1(K+B_0).
\]
Recall that $\gamma_1>1$ and $A_0<1$. It follows that
\[
x_0^*\leq \frac{\gamma_1(K+B_0)}{(\gamma_1-1)(1-A_0)}=\frac{\rho K}{\rho-f_1}.
\]
Combining the opposite inequality (\ref{x_0^*-cond}), we have
\[
x_0^*=\frac{\rho K}{\rho-f_1}.
\]

Next, we claim that the second inequality in (\ref{VI-2-Case-II})
follows from
\beq{2nd-ineq}
C_1(x^*)^{\gamma_1}+\phi_0(x^*)\leq \frac{(\rho+\lambda_2)(x^*-K)-x^* f_2}{\lambda_2}.
\end{equation}
To see this, let
\[
\phi(x)=C_1x^{\gamma_1}+\phi_0(x)-\frac{(\rho+\lambda_2)(x-K)-x f_2}{\lambda_2}.
\]
Then, using $v(x,1)=C_1x^{\gamma_1}+\phi_0(x)$ and $v(x,2)=x-K$,
the second inequality becomes $\phi(x)\leq0$
on $x\in(x^*,x_0^*)$.
Under (\ref{2nd-ineq}), $\phi(x^*)\leq0$. Note also that
\[
\phi'(x_0^*)=\gamma_1C_1(x_0^*)^{\gamma_1}+A_0-\frac{\rho+\lambda_2-f_2}{\lambda_2}
\leq 1-\frac{\rho+\lambda_2-f_2}{\lambda_2}<0.
\]
In addition, $\phi''(x)=\gamma_1(\gamma_1-1)C_1x^{\gamma_1-2}>0$.
In view of this, $\phi'(x)$ is increasing. Therefore, $\phi'(x)<0$ on
$(x^*,x_0^*)$, which implies $\phi(x)$ is decreasing.
So $\phi(x)\leq\phi(x^*)\leq0$ on $(x^*,x_0^*)$.

Furthermore, using (\ref{cont-conds-case-II}), we can rewrite
(\ref{2nd-ineq}) as
\[
\frac{x^*-K}{\kappa_2}\leq \frac{(\rho+\lambda_2)(x^*-K)-x^* f_2}{\lambda_2},
\]
which in turn gives
\[
(\lambda_2-\kappa_2(\rho+\lambda_2-f_2))x^*\leq (\lambda_2-\kappa_2(\rho+\lambda_2))K.
\]
This inequality is equivalent to
\[
x^*\leq \frac{(\lambda_2-\kappa_2(\rho+\lambda_2))K}
{\lambda_2-\kappa_2(\rho+\lambda_2-f_2)},
\]
because $\lambda_2-\kappa_2(\rho+\lambda_2-f_2)>0$ as shown in
\lemref{cond3-ineq} in Appendix.

In view of (\ref{case-II-x^*-v2}), $x^*$ has to be bounded above by
\[
X_0:=\min\left\{x_0^*,\, \frac{K\beta_2}{\beta_2-1},\,
\frac{(\lambda_2-\kappa_2(\rho+\lambda_2))K}
{\lambda_2-\kappa_2(\rho+\lambda_2-f_2)}\right\}.
\]
In view of \lemref{K-beta2} (Appendix), we have
\[
x_0^*=\frac{\rho K}{\rho-f_1}<\frac{K\beta_2}{\beta_2-1}
\]
Therefore, an upper bound for $x^*$
\[
X_0=\min\left\{\frac{K\beta_2}{\beta_2-1},\,
\frac{(\lambda_2-\kappa_2(\rho+\lambda_2))K}{\lambda_2-\kappa_2(\rho+\lambda_2-f_2)}\right\}.
\]

To obtain $x^*$, we only need to solve the first two equations
in (\ref{cont-conds-case-II}). Eliminating $A_2$, we obtain
\beq{x^*-Case-II}
C_1(x^*)^{\gamma_1}+\phi_0(x^*)=\frac{x^*-K}{\kappa_2},
\end{equation}
on $[K,x_0^*]$.

Let $\phi^*(x)=C_1x^{\gamma_1}+\phi_0(x)-{(x-K)}/{\kappa_2}$, Then it is easy
to check that $\phi_0(K)$ is positive, so is $\phi^*(K)$. In addition,
$\phi^*(x_0^*)=(x_0^*-K)(1-1/\kappa_2)<0$ because $0<\kappa_2<1$.
Furthermore, it can be shown that $(\phi^*)''(x)>0$,
which implies that $(\phi^*)'(x)$
increasing. Using (\ref{1-ineq-VI-2-Case-II}) to obtain $(\phi^*)'(x)<0$,
which implies $\phi^*(x)$ is decreasing.
Therefore, $\phi^*(x)$ has a unique zero $x^*$ on $[K,x_0^*]$.

Recall that
\[
x_0^*=\frac{\rho K}{\rho-f_1}
\mbox{ and }
C_1=\frac{x_0^*-K-\phi_0(x_0^*)}{(x_0^*)^{\gamma_1}}.
\]
Let $x^*$ be the solution of (\ref{x^*-Case-II}) over $[K,x_0^*]$ and let
\[
A_2=\frac{x^*-K}{\kappa_2(x^*)^{\beta_2}}.
\]

We have proved the following results.

\begin{thm}\hspace*{-0.08in}{\bf.}\hspace*{0.05in}\label{HJB-Case-II}
Assume $\rho> f_1$.
Then the functions $v(x,i)$, $i=1,2$, given in
{\rm (\ref{value-fn-Case-II})}
are continuous on $(0,\infty)$.
Moreover, $v(x,1)$ is differentiable on $(0,\infty)-\{x^*,x_0^*\}$
and $v(x,2)$ is differentiable on $(0,\infty)-\{x^*\}$.
If $x^*\leq X_0$, then they satisfy the HJB equations {\rm (\ref{HJB})}.
\end{thm}

\begin{rem}\hspace*{-0.08in}{\bf.}\hspace*{0.05in}
{\rm
Note that $X_0\in[K,x_0^*]$. Recall that $\phi^*(x)$ is decreasing on
$[K,x_0^*]$. A sufficient condition
for $x^*\leq X_0$ is $\phi^*(X_0)\leq0$.
}
\end{rem}

\section{Verification Theorems and Numerical Examples}

First, we give two verification theorems depending on
$\rho\leq f_1$ and $\rho> f_1$.
We only prove \thmref{VerificationThm1}. The proof of
\thmref{VerificationThm2} can be given in a similar way.

\begin{thm}\label{VerificationThm1}\hspace*{-0.08in}{\bf.}\hspace*{0.05in}
Assume the conditions of \thmref{HJB-Case-I}. Then,
$v(x,i)=V(x,i)$, $i=1,2$. Moreover, let
$D=(0,\infty)\times\{1\}\cup(0,x^*)\times\{2\}$
denote the continuation region. Then
\[
\tau^*=\inf\{t\geq0;\, (S_t,\alpha_t)\not\in D\}
\]
is an optimal selling time.
\end{thm}

\noindent{\it Proof.}
We only sketch the proof because it is similar to that of
Zhang and Zhang \cite[Theorem 5]{ZhangZ}.
For any given stoppting time $\tau$ and $n=1,2,\ldots$, we have
\beq{verthm1ineq}
\begin{array}{rl}
v(x.i)\geq& Ee^{-\rho(\tau\wedge n)}v(S_{\tau\wedge n},\alpha_{\tau\wedge n})\\
=&Ee^{-\rho\tau}v(S_{\tau},\alpha_{\tau})I_{\{\tau<n\}}
+Ee^{-\rho n}v(S_{n},\alpha_{n})I_{\{\tau\geq n\}}\\
\geq &Ee^{-\rho\tau}(S_\tau-K)I_{\{\tau<n\}}
+Ee^{-\rho n}v(S_{n},\alpha_{n})I_{\{\tau\geq n\}}.\\
\end{array}
\end{equation}
Recall the linear growth of $v(x,i)$ and $Ee^{-\rho t}S_t\to0$ under (A2)
as $t\to\infty$. The second term goes to 0 and $n\to\infty$
Note also that the first term converges to
$Ee^{-\rho\tau}(S_{\tau}-K)I_{\{\tau<\infty\}}=J(x,i,\tau)$.
It follows that $v(x,i)\geq J(x,i,\tau)$.

To show the equality, first if $(S_0,\alpha_0)=(x,i)\not\in D$, then $\tau^*=0$.
This implies $v(x,i)=x-K=J(x,i,\tau^*)$.
If $(x,i)\in D$, then similarly as in (\ref{verthm1ineq}), we have
\[
\begin{array}{rl}
v(x.i)=& Ee^{-\rho(\tau^*\wedge n)}v(S_{\tau^*\wedge n},\alpha_{\tau^*\wedge n})\\
=&Ee^{-\rho\tau^*}v(S_{\tau^*},\alpha_{\tau^*})I_{\{\tau^*<n\}}
+Ee^{-\rho n}v(S_{n},\alpha_{n})I_{\{\tau^*\geq n\}}\\
=&Ee^{-\rho\tau^*}(S_{\tau^*}-K)I_{\{\tau^*<n\}}
+Ee^{-\rho n}v(S_{n},\alpha_{n})I_{\{\tau^*\geq n\}}\\
\to&Ee^{-\rho\tau^*}(S_{\tau^*}-K)I_{\{\tau^*<\infty\}}+0\\
=&J(x,i,\tau^*),
\end{array}
\]
as $n\to\infty$.
\hfill$\Box$

\begin{thm}\label{VerificationThm2}\hspace*{-0.08in}{\bf.}\hspace*{0.05in}
Assume the conditions of \thmref{HJB-Case-II}. Then,
$v(x,i)=V(x,i)$, $i=1,2$. Moreover, let
$D=(0,x_0^*)\times\{1\}\cup(0,x^*)\times\{2\}$
denote the continuation region. Then
\[
\tau^*=\inf\{t\geq0;\, (S_t,\alpha_t)\not\in D\}
\]
is an optimal selling time.
\end{thm}

\begin{cor}\hspace*{-0.08in}{\bf.}\hspace*{0.05in}
Let ${\cal T}$ denote the class of almost sure finite
${\cal F}_t$ stopping times. Then,
\[
\sup_{\tau\in{\cal T}} E\(e^{-\rho\tau}\exp\int_0^\tau f(\alpha_t)dt\)<\infty.
\]
\end{cor}

\noindent{\it Proof.}
Given $\alpha_0=i$, we have
\[
x E\(e^{-\rho\tau}\exp\int_0^\tau f(\alpha_t)dt\)I_{\{\tau<\infty\}}
\leq J(x,i,\tau)+K\leq V(x,i)+K\leq \max_i V(x,i)+K.
\]
Set $x=1$ to obtain
\[
\sup_{\tau\in{\cal T}} E\(e^{-\rho\tau}\exp\int_0^\tau f(\alpha_t)dt\)
\leq \max_i V(1,i)+K<\infty.\quad\Box
\]

\subsubsection*{Example 1 (Convergence to a Brownian motion).}

In this example, given $\varepsilon>0$, we consider
\[
f_1=\mu+\frac{\sigma}{\sqrt{\varepsilon}},\
f_2=\mu-\frac{\sigma}{\sqrt{\varepsilon}},\
\lambda_1=\lambda_2=\frac{1}{\varepsilon}.\
\]

Using the asymptotic normality given in Yin and Zhang \cite[Theorem 5.9]{YinZ},
we can show that $S_t=S_t^\varepsilon$ converges weakly to
\[
S_t^0=S_0e^{\mu t+\sigma W_t},\mbox{ as }\varepsilon\to0,
\]
where $W_t$ is a standard Brownian motion.
Such a limit is the solution to the stochastic
differential equation
\[
\frac{dS_t}{S_t}=\(\mu+\frac{\sigma^2}{2}\)dt+\sigma dW_t.
\]

It is elementary to show that, as $\varepsilon\to0$,
\[
\beta_2
=\beta_2^\varepsilon\to \beta_0=\frac{-\mu+\sqrt{\mu^2+2\rho \sigma^2}}{\sigma^2}.
\]
This implies that $x^*=x^{*,\varepsilon}$ defined in (\ref{x*-alternative})
converges to the selling threshold $x^0=K\beta_0/(\beta_0-1)$ obtained in
${\O}$ksendal \cite[Example 10.2.2]{Oksendal}.

Taking $\mu=0.2$ and $\sigma=0.3$,
we give sample paths of $\log(S_t^\varepsilon)$ and
$\alpha_t^\varepsilon$ with varying $\varepsilon$
in Figure~\ref{MonteCarlo}. It is clear from the pictures
that as $\varepsilon$ gets smaller and smaller, the fluctuation of
$\alpha_t$ is more and more rapidly and the corresponding $S_t^\varepsilon$ approaches
to a GBM.

\begin{center}
\begin{figure}[htb]
\begin{center}
\mbox{\subfigure[{\small $\varepsilon=0.1$}]{
         \psfig{figure=MC_0.1.eps,width=0.3\linewidth }} \quad
      \subfigure[{\small $\varepsilon=0.01$}]{
         \psfig{figure=MC_0.01.eps,width=0.3\linewidth }} \quad
      \subfigure[{\small $\varepsilon=0.001$}]{
         \psfig{figure=MC_0.001.eps,width=0.3\linewidth }}}
\end{center}
\caption{Monte Carlo Sample Paths: $(\log(S_t^\varepsilon),\alpha_t^\varepsilon)$}\label{MonteCarlo}
\end{figure}
\end{center}

\subsubsection*{Example 2 (Case II).}

In this example, we consider Case II with $\rho>f_1$ and
use the following parameters
\[
f_1=0.07,\,
f_2=-0.03,\,
\lambda_1=\lambda_2=1,\,
\rho=0.10,\,
K=0.01.
\]
Solving the equation (\ref{x^*-Case-II}) with $x_0^*=\rho K/(\rho-f_1)$, we
have $(x^*,x_0^*)=(0.012478,0.033333)$ and $X_0=0.013326$.
The corresponding value functions
are given in Figure~\ref{VF-CaseII}, in which
$V(x,1)$ is given by the upper curve and $V(x,2)$ the lower one.

\begin{center}
\begin{figure}
\begin{center}
\epsfysize=2.0in
\centerline{\epsfbox{case2.eps}}
\vspace{-3ex}
\noindent\caption{{\small Value Functions $v(x,1)$ and $v(x,2)$}}\label{VF-CaseII}
\end{center}
\end{figure}
\end{center}

\subsubsection*{Example 3 (Model Calibration and a Market Test).}
First we give a model calibration method. We consider
\[
f_1=\mu+\sigma_1\mbox{ and }
f_2=\mu+\sigma_2,
\]
with $\nu_1\sigma_1-\nu_2\sigma_2=0$.
Given $T$,
let $Y_t=\log(S_t/S_0)=\int_0^t f(\alpha_s)ds$, $0\leq t\leq T$.
Then, $\mu$ can be approximated by $Y_T/T$.
To estimate $\sigma_1$ and $\sigma_2$, given step size $\delta>0$, let
$n\delta=T$,
\[
\Delta Z_k=\log(S_{(k+1)\delta})-\log(S_{k\delta}),
\mbox{ for } k=0,1,2,\ldots,n,
\]
and
$\overline Z=\(\sum_{k=0}^{n-1} \Delta Z_k\)/n$.
Then, $\overline Z\approx \delta \mu$.
In addition, using Yin an Zhang \cite[Theorem 5.9]{YinZ},
we can show
\[
E\(\Delta Z_k-\overline Z\)^2\approx \delta
\(\frac{2\lambda_1\lambda_2(\sigma_1+\sigma_2)^2}{(\lambda_1+\lambda_2)^3}\).
\]
Let
\[
\sigma_0^2=\frac{\sum_{k=0}^{n-1} (\Delta Z_k-\overline Z)^2}{n-1}.
\]
Then, by the Law of Large Numbers, we have
\[
\sigma_0^2\approx \delta\(\frac{2\lambda_1\lambda_2(\sigma_1+\sigma_2)^2}
{(\lambda_1+\lambda_2)^3}\).
\]
Using $\nu_1\sigma_1=\nu_2\sigma_2$, we have
\[
\sigma_1
=\frac{\sigma_0}{\sqrt{\delta}}\sqrt{\frac{\lambda_1(\lambda_1+\lambda_2)}{2\lambda_2}}
\mbox{ and }
\sigma_2
=\frac{\sigma_0}{\sqrt{\delta}}\sqrt{\frac{\lambda_2(\lambda_1+\lambda_2)}{2\lambda_1}},
\]
Finally, we estimate $\lambda_1$ and $\lambda_2$.
Let $R=\nu_1/\nu_2$. Then, $\lambda_2=R\lambda_1$.
\[
\begin{array}{l}
R_1=\#\{k:\, \Delta Z_k<0\mbox{ and }\Delta Z_{k+1}\geq0\},\\
R_2=\#\{k:\, \Delta Z_k>0\mbox{ and }\Delta Z_{k+1}\leq0\}.\\
\end{array}
\]
Then, it follows that
\[
\frac{R_1}{\lambda_1}+\frac{R_2}{\lambda_2}=T.
\]
Therefore, the jump rates are given by
\[
\left\{\begin{array}{l}
\displaystyle
\lambda_1=\frac{1}{T}\(R_1+\frac{R_2}{R}\),\\
\displaystyle
\lambda_2=\frac{1}{T}\(R R_1+R_2\).\\
\end{array}\right.
\]

We test our selling rules using Apple Inc. (AAPL) daily closing prices
during 2009/1/2 and 2013/3/28, sse Figure~\ref{AAPL} (a).
Suppose we owned 100 AAPL shares at the beginning of 2009.
We evaluate at the end of each half year during this period
based on that half year stock prices to determine if we should sell the shares
in the near future.

We assume the risk free rate to be $\rho=0.03$ and transaction cost
$K=0.01$. We use the calibration method discussed earlier and obtain
the following results.

\begin{center}
{\small
\begin{tabular}{|c|c|c|c|c|c|}\hline
Periods&$f_1$&$f_2$&$\lambda_1$&$\lambda_2$&$\Phi(0.03)$\\ \hline
1st half of 2009 &10.45&-10.61&100.48&124.23&-336.06\\ \hline
2nd half of 2009 &3.21&-2.32&102.15&141.44&-217.41\\ \hline
1st half of 2010 &3.06&-3.15&97.98&127.02&-83.18\\ \hline
2nd half of 2010 &2.27&-1.92&103.57&134.25 &-103.09 \\ \hline
1st half of 2011 &1.80&-1.85&117.19&125.00&-5.02 \\ \hline
2nd half of 2011 &3.01&-2.72&97.98 &107.95&-60.56 \\ \hline
1st half of 2012 &5.39 &-4.80& 108.21&127.19 &-185.32 \\ \hline
2nd half of 2012&4.89 &-5.13 &135.25 &130.95 &35.79 \\ \hline
\end{tabular}

\vspace*{0.1in}
\centerline{Table 1. Parameter Values at the End of Each Period}
}
\end{center}

Note that in all periods $\rho\leq f_1$. Therefore, only Case I applies
in this example.
In Table 1, we should hold through the next half year
if $\Phi(0.03)\leq 0$ and sell (following our
selling rule) if $\Phi(0.03)>0$. Clearly, a selling decision has to be
made at the end of 2012. Using the parameter values
$f_1=4.89$, $f_2=-5.13$, $\lambda_1=135.25$, and $\lambda_2=130.95$,
we obtain $x^*=0.017213$ and the corresponding value functions
$v(x,i)=V(x,i)$, $i=1,2$, which are plotted in Figure~\ref{AAPL} (b).
Therefore, one should sell as soon as
$\alpha_t$ turns to $2$ after the new year. This occurs on
the second trading day (January 3) of 2013.
The shares should be sold at the close of that day at \$542.10/share.
As can be seen in this example, our selling rule helps to achieve
the goal of letting your profits run and cutting your losses short.

\begin{center}
\begin{figure}[htb]
\begin{center}
\mbox{\subfigure[{\small AAPL (2009/1/2--2013/3/28)}]{
         \psfig{figure=AAPL_09_2013.eps,width=0.4\linewidth }} \quad
      \subfigure[{\small Value Functions}]{
         \psfig{figure=case1.eps,width=0.4\linewidth }}}
\end{center}
\caption{Apple Inc.  Daily Closes and Value Functions $v(x,1)$ and $v(x,2)$}\label{AAPL}
\end{figure}
\end{center}

\section{Conclusion}

In this paper, we considered an optimal stock selling rule under
a Markov chain model. The model is natural for financial markets
due to its simple structure and the solutions obtained are
intuitive and easy to implement.

It would be interesting to consider more general models with
multi-scale structure as treated in Yin and Zhang \cite{YinZ}
so as to capture both long-term and short-term market movements.
Such extension and related optimization problems could be
subjects of future studies.

\section{Appendix}

\begin{lem}\label{beta2>1}\hspace*{-0.08in}{\bf.}\hspace*{0.05in}
Under the assumption $\Phi(\rho)>0$,
the bigger root of {\rm (\ref{char-eqn})} $\beta_2>1$.
\end{lem}

\noindent{\it Proof.}
Recall the definition of $D_1$ and $D_2$ given in (\ref{D1-D2}).
It is easy to check $\Phi(\rho)>0$ implies
\[
D_2>D_1-f_1f_2.
\]
This leads to
\[
\sqrt{D_1-4f_1f_2D_2}>D_1-2f_1f_2.
\]
Therefore, we have $\beta_2>1$. \hfill$\Box$

\begin{lem}\label{kappa2}\hspace*{-0.08in}{\bf.}\hspace*{0.05in}
Let $\kappa_2={(\rho+\lambda_1-f_1\beta_2)}/{\lambda_1}$,
where $\beta_2$ is given in {\rm (\ref{char-root})}.
Then, $0<\kappa_2<1$.
\end{lem}

\noindent{\it Proof.}
To see $\kappa_2>0$, it suffices to show
$\rho+\lambda_1>f_1\beta_2$. Recall that $f_2<0$. We only need to
show $2(\rho+\lambda_1)f_2<D_1-\sqrt{D_1^2-4f_1f_2D_2}$,
with $D_1$ and $D_2$ given in (\ref{D1-D2}).

This is equivalent to
\beq{kappa2-1}
\sqrt{D_1^2-4f_1f_2D_2}<D_1-2(\rho+\lambda_1)f_2.
\end{equation}
It is easy to check $D_1-2(\rho+\lambda_1)f_2=(\rho+\lambda_2)f_1-(\rho+\lambda_1)f_2>0$.
Square both sides of (\ref{kappa2-1}) to obtain
\[
D_1^2-4f_1f_2D_2<D_1^2-4(\rho+\lambda_1)f_2D_1+
4((\rho+\lambda_1)f_2)^2.
\]
Simplify this inequality to obtain
\[
D_2<(\rho+\lambda_1)(\rho+\lambda_2).
\]
This clearly holds. Therefore, $\kappa_2>0$.

Similarly, to show $\kappa_2<1$, it suffices to show
$\rho< f_1\beta_2$. This is equivalent to
\[
\sqrt{D_1^2-4f_1f_2D_2}>D_1-2\rho f_2.
\]
Square and simplify to obtain
$\rho \lambda_1 f_1>\rho\lambda_1 f_2$.
This holds because $f_1>0$ and $f_2<0$. \hfill$\Box$

\begin{lem}\label{K-beta2}\hspace*{-0.08in}{\bf.}\hspace*{0.05in}
Under $\rho>f_1$, we have
\[
\frac{K\beta_2}{\beta_2-1}<\frac{\rho K}{\rho-f_1}.
\]
\end{lem}

\noindent{\it Proof.}
It is easy to see this inequality is equivalent to $f_1\beta_2>\rho$.
Using the definition of $\beta_2$ and noting that $f_2<0$, we have
\[
\sqrt{D_1^2-4f_1f_2D_2}>D_1-2\rho f_2.
\]
If $D_1-2\rho f_2<0$, then we are done. Otherwise, square both sides to
obtain
\[
D_1^2-4f_1f_2D_2>D_1^2-4\rho f_2 D_1+4\rho^2f_2^2.
\]
Simplify this inequality to have
\[
\rho\lambda_1f_1>\rho\lambda_1 f_2,
\]
which clearly holds because $f_1>0$ and $f_2<0$.

\begin{lem}\label{cond3-ineq}\hspace*{-0.08in}{\bf.}\hspace*{0.05in}
Under $\rho>f_1$, we have
\beq{app-cond3}
\lambda_2-\kappa_2(\rho+\lambda_2-f_2)>0,\\
\end{equation}
which implies $\lambda_2-\kappa_2(\rho+\lambda_2)>0$.
\end{lem}

\noindent{\it Proof.}
Let $H_0=\rho+\lambda_2-f_2>0$.
Using the definition of $\kappa_2$, (\ref{app-cond3}) is
equivalent to
\[
H_0(\rho+\lambda_1-f_1\beta_2)<\lambda_1\lambda_2.
\]
Therefore, we have
\[
(\rho+\lambda_1)H_0-\(\frac{D_1-\sqrt{D_1^2-4f_1f_2D_2}}{2f_2}\)H_0<\lambda_1\lambda_2.
\]
Multiply both sides by $(2f_2)$ and rearrange the terms to obtain
\[
H_0\sqrt{D_1^2-4f_1f_2D_2}>H_0D_1-2f_2[(\rho+f_1)H_0-\lambda_1\lambda_2].
\]
Square both sides and simplify to have
\[
(\rho+\lambda_1-f_1)H_0-\lambda_1\lambda_2>0,
\]
which is exactly the assumption $\phi(\rho)>0$.
This completes the proof. \hfill$\Box$

redpajama_set_name: RedPajamaArXiv


NEW YORK - McDonald's says offering breakfast around the clock helped jolt its sales.
Sales at the world's biggest hamburger chain rose 5.7 percent in the U.S. for the final three months of 2015, boosted by unseasonably warm weather and the launch of an all-day breakfast menu in October.
That marked the second straight quarter of domestic growth as McDonald's fights to win back customers, and the best showing since early 2012. Globally, sales rose 5 percent at established locations.
McDonald's, which has more than 36,000 locations around the world, is working to turnaround its business under CEO Steve Easterbrook, who will mark one year on the job in March. The Oak Brook, Illinois, company has conceded that it failed to keep up with changing tastes, and that order speed and accuracy suffered as its menu grew more expansive.
Despite the encouraging finish to the year, McDonald's has a long way to go. For all of 2015, U.S. customer visits fell 3 percent at established McDonald's locations. That followed a 4.1 percent drop the previous year. Guest counts declined globally as well during the period.
The launch of all-day breakfast has been one of the company's most high-profile maneuvers under Easterbrook. The decision to make select items like the Egg McMuffin available around the clock generated a surge of national media coverage. Fans of McDonald's breakfast had long complained of being unable satisfy their cravings later in the day.
As the initial fanfare settles down, Easterbrook said McDonald's would keep momentum going by focusing on areas like improving order accuracy and a recently launched mobile app. In early January, McDonald's also introduced a "McPick 2 for $2" menu intended to draw bargain hunters.
The quarterly performance may also have gotten a boost from a culling of underperforming stores. McDonald's finished the year with 14,259 stores in the U.S., a decline of 91 stores from the previous year.
It was the first time McDonald's ended the year with fewer U.S. restaurants since at least 1970, according to an Associated Press review of regulatory filings.
Shares of McDonald's were up 1 percent at $119.45.


Below is the Jersey real estate market update for condominium sales for April 2010 in Downtown Jersey City. All information is gathered from the Hudson County Multiple Listing Service.
Since April 1st, 117 new Jersey City condos came on the market for sale with an average listing price of $509,071.
Active Downtown Jersey City Studio Condos: 14 condos for sale with an average listing price of $294,628. 85 days on the market.
Active Downtown Jersey City 1 Bedroom Condos: 147 condos for sale with an average listing price of $357,292. 101 days on the market.
Active Downtown Jersey City 2 Bedroom Condos: 175 condos for sale with an average listing price of $540,172. 88 days on the market.
Active Downtown Jersey City 3+ Bedroom Condos: 30 condos for sale with an average listing price of $971,622. 85 days on the market.
There are 108 Downtown Jersey City condos under contract waiting to close with an average listing price of $460,888 Average days on the market: 84 Days. Final sales prices of these units will not be published until they close.
Pending Downtown Jersey City Studio Condos: 3 condos under contract an average listing price of $236,633. 68 days on the market.
Pending Downtown Jersey City 1 Bedroom Condos: 41 condos under contract an average listing price of $373,551. 88 days on the market.
Pending Downtown Jersey City 2 Bedroom Condos: 53 condos under contract with an average listing price of $508,366. 90 days on the market.
Pending Downtown Jersey City 3+ Bedroom Condos: 11 condos under contract with an average listing price of $618,818. 47 days on the market.
Sold Downtown Jersey City Studio Condos: No studios sold in April 2010.
Sold Downtown Jersey City 1 Bedroom Condos: 13 condos sold an average listing price of $316,018 and an averages sales price of $305,461. 99 days on the market.
Sold Downtown Jersey City 2 Bedroom Condos: 27 condos sold with an average listing price of $539,581 and an average sales price of $511,981. 75 days on the market.
Sold Downtown Jersey City 3+ Bedroom Condos: 4 condos sold with an average listing price of $746,750 and an average sales price of $718,250. 68 days on the market.
Sold units occurred between 4/1/10 and 5/16/10.
Shore Club: 16 Active condos, 7 units under contract, 9 condos sold.
A Condominiums: 15 Active condos, 1 units under contract, 2 condos sold.
Mandalay on the Hudson: 8 Active condos, 8 units under contract, 4 condos sold.
Gulls Cove: 12 Active condos, 1 unit under contract, 4 condos sold.
Fulton's Landing: 6 Active condos, 2 units under contract, 0 condos sold.
James Monroe: 10 Active condos, 4 units under contract, 0 condos sold.
Liberty Terrace: 7 Active condos, 4 units under contract, 0 condos sold.
Liberty Harbor: 6 Active condos, 0 units under contract, 3 condos sold.
Portofino: 9 Active condos, 4 units under contract, 0 condos sold.
Pier House: 7 Active condos, 1 unit under contract, 0 condos sold.
700 Grove: 10 Active condos, 4 units under contract, 1 condo sold.
Montgomery Greene: 1 Active condo, 2 units under contract, 0 condos sold.
Clermont Cove: 6 Active condo, 2 units under contract, 1 condo sold.
Sugar House: 5 Active condos, 0 units under contract, 1 condo sold.
Waldo Lofts: 6 Active condos, 2 units under contract, 0 condos sold.
Zephyr Lofts: 10 Active condos, 1 unit under contract, 1 condo sold.
****The above data may not reflect all of the units that may be available or that have sold in downtown Jersey City. Inside sales offices in new construction downtown Jersey City condo buildings like Gulls Cove, Trump Plaza, Crystal Point, and 77 Hudson do not list and report all of their sales through the Hudson County MLS.
To see how this market update compares to last month's, please visit the March 2010 Jersey City condo sales report.
If you have questions about a certain downtown Jersey City Condo buildings that are not on this list, feel free to contact me, Walter Burns or leave a comment. I will get back to you with the information.


Apple Valley is home to many residents. These means that plumbing services are needed for both residential and commercial places. You need a plumbing company for the first-time installation. You need professional who will do a clean job that will keep your home in good condition. When it comes to repairing, you need experienced plumbers to help you fix your plumbing system. There are many plumbing companies in Apple Valley and selecting the one that suits your needs can be a challenge. Below is a detailed discussion of tips for that will help you select the best plumbing company in Apple Valley.
It is good to work with a company that is registered and licensed to offer plumbing services in Apple Valley. You can work with any company that has no legal documents of operating as it can land you to problems. Working with a non-registered company is not wise as it can make you lose money or receive low-quality services.
With many plumbing companies in Apple Valley, you need to check for a company that has a good reputation. You can check their previous works and how clients are reviewing the company. Visit their websites and find out how outstanding there are. You can also ask around people who have been served by the company. You will get clear information which will help you in making the right choice. Going to an experienced company is an advantage. Companies with over ten years of experience and with an excellent reputation are most preferred.
Plumbing services are many. It depends on what you require to be done in your home or business. It is advisable to select a plumbing company that offers a wide range of services ranging from repair to first time installation. A good company should offer maintenance and replacement services too. Installation of sinks, faucets, showers, bathtubs, pipes, and drains are some of the services a good company should offer. A good company should also install video pipe inspection in your home. When it comes to maintenance and repair the company should also offer quality services.
For a plumbing company to get to great height, it needs to have professional and experienced workers. For that reason, if you are looking for a plumbing company to hire, go for a company that has professional plumbers with vast experience in plumbing installation, repair, and maintenance services. Experienced workers do clean jobs and are always on time for plumbing tasks.
It is good to work with a company that you can reach any time. Go for a company that has a phone number that you can use to reach them like the case of the post office which has Apple Valley post office phone number. This means that you can contact them when you have any plumbing issue.
It is good to consider your budget before hiring plumbing company. Go for a company that offers quality services at affordable prices. Make sure that the company uses materials and apparatuses of high quality.
The article makes a very good point about the importance of different services. After all, it is likely that you will need your plumber to handle more than just fixing your leaky pipes. For example, if your water heater breaks down then you will want them to be able to fix that too.


text:
"DNR" to restore "historic justice": terrorists make new statement on the fate of Donbas
Representatives of the self-proclaimed "DPR" authorities are actively making aggressive statements to change the situation in Donbas for the better. The President of the Democratic People's Republic of Russia, Vladimir Bidevka, made a very lively statement on the Day of Russia.
Flag of the "DPR". Photo – msk.ru
This is reported by a number of separatist media, reports "MFN".
Bidevka is confident that the "DPR" will soon be able to make a breakthrough in terms of "restoring historic justice" thanks to its loyalty to the Russian Federation.
"We made our choice in 2014 and clearly adhere to it despite the difficulties. The DPR is part of the Russian world, our homeland is Russia, and every day we get closer to our roots. These six years of struggle have shown the world that the Donbass was and is part of Russia. Soon new great achievements await us, "said the representative of the terrorists.
He also expressed his personal gratitude to Vladimir Putin and congratulated all the Russians on this occasion.
Recall, as previously stated, popular Ukrainian politicians congratulated the Russians on the occasion of the holiday of Russia: details of the scandalous statement.
Redmi Note 9 immediately available on Amazon
Huawei P40 Pro + officially arrives in Italy, and the price is even higher than expected

redpajama_set_name: RedPajamaCommonCrawl


text:
Edward Rasmusen
February 27, 1940 - November 5, 2019
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Edward L. Rasmusen, age 79 of Le Sueur, died Tuesday, November 5, 2019 at Regency Hospital in Minneapolis with his loving family by his side. Memorial service will be 11:00 a.m. on Friday, November 15 at First Lutheran Church in Le Sueur with Pastor Carl Bruihler officiating. Visitation will be from 3-7 p.m. on Thursday, November 14 at the Kolden Funeral Home in Le Sueur, and also one hour prior to service time on Friday at the church. Interment will be at a later date in Minnewaska Lutheran Church Cemetery in Starbuck, MN.
Ed was born on February 27, 1940 to Carl and Olga (Strum) Rasmusen in Starbuck. He graduated from Starbuck High School in 1958. He attended Moorhead State University receiving his BS degree in Mathematics in 1963, and in 1974 he received his MS degree in Education from Mankato State University.
Ed married Geraldine Barsness on October 19, 1963 in Starbuck. They eventually made their home in Le Sueur where Ed taught high school math from 1967 to 1997. Geri passed away in 2009. On April 20, 2013 Ed was united in marriage to Annette Kinkeade.
Ed was actively engaged in the community, serving as a volunteer ambulance driver, city council member, mayor, hospital board member and church council member. He enjoyed golf, travel and time with family, and he took great pride in his reputation for telling corny jokes and having a meticulously manicured lawn.
He is survived by his wife, Annette; son, Jason (Sarah) Rasmusen of Madison, WI; grandchildren, Caroline and Nathaniel Rasmusen; sister, Elnora Bryant of Vadnais Heights; step-children, Stephanie McElroy of Orono, Jennifer (Matthew) Johnson of Brooklyn Park, Brad (Sheryl) Boren of Maple Grove, Josh (Heather) Kinkeade of Jordan, and John (Sarah) Kinkeade of Andover and 11 step-grandchildren.
Ed is preceded in death by his parents; wife, Geraldine; brother, Howard (Lydia) Rasmusen; sister, Carol (Percy) Samborski; and brother-in-law, Harry Bryant.
To plant a tree in memory of Edward Rasmusen, please visit Tribute Store.
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Tina Martin
Simply Elegant Spathiphyllum was purchased for the family of Edward Rasmusen by Tina Martin. Send flowers
Tony Nagel
My sincere condolences to the family. I had the pleasure of having one class with Ed my senior year at LSH wayyy back in 1990-91. Ed always kept it light in class and incorporated his students into his jokes daily. The world won't be the same without ol' Mr. Green Jeans. May you rest in peace Ed.


Reading, Berkshire : Garnet, ©1995.
attributed to al-Jahiz ; translated and introduced by M.A.S. Abdel Haleem.
Did the world come about by chance or through creation? This book, attributed to the ninth-century Arab author al-Jahiz, discusses the evidence of God's design in the universe. Looking at the world around us and mankind's place in it, he argues that the design, harmony and wisdom in the natural world show that the world was created consciously. In answer to those who say that all things came about by chance, he argues that the ecosystem is playing out a divine plan. In his innovative approach he was centuries ahead of most other thinkers, and the modern reader will appreciate his holistic approach.
Add tags for "Chance or creation? : God's design in the universe". Be the first.


Home /Knowledge Base /For Buyers/Purchasing/Why do you need my VAT number?
Home /Knowledge Base /For Buyers /Why do you need my VAT number?
If you're a VAT-registered business, in most cases you can reclaim the VAT you pay when you buy goods and services for your business. In order to reclaim this money you need to have records to support your claim. By collecting your VAT number, Iconfinder can correctly prepare an invoice for your records. If you're not a VAT-registered business or organisation then you cannot reclaim the VAT you pay when you purchase goods and services.
If you do not have a VAT number or the one you provided cannot be validated, you will be charged the standard VAT rate for your billing region.


Minimum wage is set provincially, but the cost-of-living ranges from region to region. The Canadian Centre for Policy Alternatives (CCPA) calculates that the Living Wage is $20.62 in Vancouver. It's time to build a broad-based campaign to pressure the provincial government to tie the minimum wage in Vancouver to a living wage of $20.
We will build support with small businesses for commercial rent-control, and call for a four-year Rent Freeze for local small businesses in Vancouver.
COPE will call on the federal and provincial governments to increase the provision of public pensions in Canada, improving both the funding and benefit structures, with the aim of dramatically decreasing the importance of real estate appreciation and investment in Canadians' retirement income.


text:
Kedarnath Director Abhishek Kapoor Appeals to Uttarakhand Govt to Lift Ban on His Film
Kedarnath director Abhishek Kapoor has urged the Uttarakhand government to lift a ban on the movie.
The poster of Kedarnath.
"I plead with the Uttarakhand government to please lift the ban on my film Kedarnath. It is an attempt to bring peace, harmony and healing to the people of this country. Please don't deprive us of this opportunity," Kapoor tweeted.
I plead with the uttarakhand govt to pls lift the ban on my film #kedarnath. Its an attempt to bring peace, harmony and healing to the people of this country. Pls dont deprive us of this opportunity. @tsrawatbjp @RonnieScrewvala
— Abhishek Kapoor (@Abhishekapoor) December 8, 2018
While there has been no official ban on the film by the state government, it wrote to all district magistrates about the film and the controversies surrounding it. The decision followed a meeting chaired by Chief Minister Trivendra Singh Rawat that reviewed a report submitted by a committee headed by Tourism Minister Satpal Maharaj.
The government asked the district administration to take a call on their own and left it to their discretion whether or not the film should be released in their respective jurisdictions.
Following which Dehradun, Haridwar, Pauri, Tehri, Nainital, Almora and Udhamsinghnagar banned the film.
The film, starring Sushant Singh Rajput and debutante Sara Ali Khan, was shot extensively in Triyuginarayan, Gaurikund, Soneprayag, Rambada, Kedarnath and Chopta, and it had to face protests even then.
The Uttarakhand High Court on Thursday refused to ban the film and asked the petitioners to approach the District Magistrate with their complaints.
On its opening day, the movie garnered Rs 7.25 crore in India.
abhishek kapoor
Kedarnath ban
Kedarnath movie
Drunk Man in Jabalpur Smashes Toddler Daughter's Head on Floor for 'Disturbing' Him, Dumps Body in Drain
Let Democracy Not Turn Into Dictatorship: Robert Vadra Calls Priyanka Gandhi's Arrest 'Unconstitutional'


72 more remain behind bars.
On March 16 Azerbaijani President Ilham Aliyev signed a decree pardoning more than 400 prisoners who had their sentences reduced or were released the following day. Of these, 51 are recognized as political prisoners.
The decree came after pervasive pressure from human rights groups and international campaigns, demanding the release of the many political prisoners in Azerbaijan.
Among these are the Norwegian Helsinki Committee's "No Political Prisoners"-campaign which urges for the immediate and unconditional release of all political prisoners, and for an end to the practice of politically motivated arrests in Azerbaijan.
"We are happy for these unusually good news coming from Azerbaijan about the release of so many political prisoners. Norwegian Helsinki Committee is also very grateful to those who supported the No Political Prisoners campaign", says Olga Shamshur Flydal, senior adviser at NHC.
The decision was taken after appeals from the prisoners, their families, and many members of the Azerbaijani civil society. In addition, the prisoners' health and family situation as well as their behaviour and qualities were also included mentioned among the reasons behind the pardoning.
Rashad Ramazanov, scholar, critic of the government on social media was released after 2138 days of unlawful imprisonment.
Ali Insanov, former Health Minister, later government critic has been released from prison after more than 13 and half years of imprisonment. The European Court of Human Rights ruled in 2013 that Insanov's right to free trial was violated by the authorities.
Giyas İbrahimov, youth activist, member of NIDA Civic Movement, arrested and sentenced to 10 years with spurious drug charges for painting political graffiti on the monument of the late President of Azerbaijan. Giyas was released after 1041 days of unlawful imprisonment.
Nijat Aliyev, journalist, editor-in-chief of Azadxeber news agency has been released after 2492 days of unlawful imprisonment.
Among the pardoned also includes at least nine men that were arrested in 2015, during the violent police raid in Nardaran, a Baku suburb known for its Shia religious conservatism and criticism of government policies.
Several of those who just got their freedom back still say that they will not let their recent imprisonment stop them from continuing their activism and their work.
"Recent mass-release of political prisoners is without a doubt a step in the right direction. We urge the Azerbaijan authorities to release the remaining 72 and stop the practice of political prosecution and imprisonment of oppositional and critical voices in the country", Shamshur Flydal says.
Those still behind bars include journalists Afgan Mukharli, Araz Guliyev, Elchin Ismayilli and opposition activists Fuad Ahmadli, Saleh Rustamli, Ali Maharramov, Babak Hasanov. In addition, Said Dadashbayli, religious scholar Tale Bagirzade and his fellows who were charged of unlawful terrorist actions also remains in prison.
In March 2017, a Baku court sentenced Azerbaijani journalist Mehman Huseynov to two years in prison on charges of defaming an entire police station, after he gave a statement in front of the courthouse in which he described the abuses he had suffered at that police station. Just two months before release he was facing new charges that could add seven years to his sentence, if convicted. Luckily the charges were dropped. His case has become a symbol of the government's harsh crackdown on human rights defenders, political activists, and journalists.
Read article "New charges against jailed blogger dropped"
Read article "Corruption Scandal in PACE Discussed"