#  Optimal control of a big financial company with debt liability under
bankrupt probability constraints

Zongxia Liang
Department of Mathematical Sciences, Tsinghua University, Beijing 100084,
China. Email<EMAIL_ADDRESS>
Bin Sun
Department of Mathematical Sciences, Tsinghua University, Beijing 100084,
China. Email<EMAIL_ADDRESS>
###### Abstract.

This paper considers an optimal control of a big financial company with debt
liability under bankrupt probability constraints. The company, which faces
constant liability payments and has choices to choose various
production/business policies from an available set of control policies with
different expected profits and risks, controls the business policy and
dividend payout process to maximize the expected present value of the
dividends until the time of bankruptcy. However, if the dividend payout
barrier is too low to be acceptable, it may result in the company’s bankruptcy
soon. In order to protect the shareholders’ profits, the managements of the
company impose a reasonable and normal constraint on their dividend strategy,
that is, the bankrupt probability associated with the optimal dividend payout
barrier should be smaller than a given risk level within a fixed time horizon.
This paper aims at working out the optimal control policy as well as optimal
return function for the company under bankrupt probability constraint by
stochastic analysis, PDE methods and variational inequality approach.
Moreover, we establish a risk-based capital standard to ensure the capital
requirement of can cover the total given risk by numerical analysis and give
reasonable economic interpretation for the results. MSC(2000): Primary 91B30,
93E20, 65K10 ; Secondary 60H05, 60H10. Keywords: Regular-singular stochastic
optimal control; Stochastic differential equations with reflection; Debt
liability; Bankrupt probability constraints; Optimal dividend barrier;
Dividend payout process.

##  1\.  Introduction

In this paper, we study a model of a big financial company, which has the
possibility to choose various production/business policies with different
expected profits and associated risks. But at the same time, the company has
liability in which it has to pay out at a constant rate no matter what the
business plan is. The company controls the business policy and dividend payout
process to maximize the expected present value of the dividends until the time
of bankruptcy. Recently, there has been an upsurge of interest in dealing with
such optimal dividend control problems. We refer readers to Avanzi [4](2009)
and references therein, Højgaard and Taksar[19, 20](1999, 2001), Asmussen et
al.[2, 3](1997, 2000), He and Liang[17, 18](2008,2009). For results in the
model with debt liability, see Choulli, Taksar and Zhou [7, 8, 6, 31,
32](2008, 2004, 2003, 2001, 2000). Guo, Liu and Zhou [13](2004) is a
theoretical work on constrained nonlinear singular-regular stochastic control
problem. The optimal dividend payout for diffusions with solvency constraints
is solved in Paulsen [29](2003). According to Miller Modigliani, the
managements of the company choose the maximum of shareholders’ return as their
goals. We see from these literatures that the optimality is achieved by using
an optimal dividend payout barrier $b$, i.e. whenever the liquid reserve of
the company goes above $b$, the excess is paid out to the shareholders as
dividends. However, the optimal dividend barrier $b$ may be too low to be
acceptable because this low dividend payout barrier may result in the
company’s bankruptcy soon. Thus, the company may be prohibited to pay
dividends at such a low level in order to avoid bankruptcy. So the managements
of the company impose some constraints on its dividends payout strategy. One
reasonable and normal constraint is that the optimal dividend barrier $b$
should be such that the bankrupt probability is not larger than some
predetermined risk level $\varepsilon$ within a fixed time horizon $T$ and the
cost for the safety is minimal. Based on the new idea, He, Hou and
Liang[16](2008) studied the linear regular-singular optimal control problem of
insurance company with proportional reinsurance policy under bankrupt
probability constraints as we state above. They succeeded to find the optimal
control policy under bankrupt probability constraints by proving the bankrupt
probability is decreasing in the dividend barrier and the existence of the
dividend barrier satisfying any given bankrupt probability constraints.
Furthermore, Liang, Huang and Yao[25, 26, 24](2010) gave a exact result of
such a control problem by proving the bankrupt probability is continuous and
strictly increasing w.r.t. the dividend barrier $b$. These new results are
mainly about the insurance company with proportional reinsurance policy.
Motivated by these works, under any given bankrupt probability constraint, we
are interested in a common company, such as a big financial company, insurance
company, …, facing constant liability payments, which controls the business
policy and dividend payout process to maximize the expected present value of
the dividends until the time of bankruptcy. Based on the relationship between
those parameters that govern the company’s reserve process, we try to derive
the optimal control policy as well as optimal return function as well as a
risk-based capital standard to ensure the capital requirement of can cover the
total risk in several distinct cases of the qualitative behavior of the
company under some bankrupt probability constraints. Moreover, we also give a
robust analysis of the optimal return function and the optimal dividend
strategy w.r.t. the model parameters and the constrained risk level of
bankrupt probability. The paper is organized as follows. In section 2, we
established a rigorous stochastic control model of a big financial company
facing constant liability payments with some bankrupt probability constraints.
In section 3, we present main result of this paper and its reasonable economic
interpretation. In section 4, we give risk analysis of the model we deal with
to state the setting treated in this paper is well defined and why we study
such regular-singular stochastic optimal control of the company. In section 5,
we give some numerical examples to portray how the debt rate $\delta$, the
constrained risk level of bankrupt probability, the initial capital $x$ , the
volatility rate $\sigma^{2}$ and the profit rate $\mu$ impact on the optimal
return function and the optimal dividend strategy. We will list the properties
of the optimal return function and bankrupt probability in section 6. The
proof of main result will be given in section 7. The procedure of solving the
HJB equations and proofs of lemmas which are used to prove the main result
will be presented in the appendix.

##  2\. Mathematical Model

To give a mathematical formulation of our optimal control problem treated in
this paper, We start with a filtered probability space
$(\Omega,\mathcal{F},\\{\mathcal{F}_{t}\\}_{t\geq 0},\mathbf{P})$ and a one-
dimensional standard Brownian motion $\\{\mathcal{W}_{t}\\}_{t\geq 0}$ on it,
adapted to the filtration $\mathcal{F}_{t}$. $\mathcal{F}_{t}$ represents the
information available at time $t$ and any decision made up to time $t$ is
based on this information. For the intuition of our diffusion model we start
from the classical Cramér-Lundberg model of a reserve(risk) process. In this
model claims arrive according to a Poisson process $N_{t}$ with intensity
$\lambda$ on $(\Omega,\mathcal{F},\\{\mathcal{F}_{t}\\}_{t\geq
0},\mathbf{P})$. The size of each claim is $X_{i}$. Random variables $X_{i}$
are i.i.d. and are independent of the Poisson process $N_{t}$ with finite
first and second moments given by $\mu$ and $\sigma^{2}$ respectively. If
there is no reinsurance and dividend payout, the reserve (risk) process of
insurance company is described by

$r_{t}=r_{0}+pt-\sum^{N_{t}}_{i=1}X_{i},$

where $p$ is the premium rate. If $\eta>0$ denotes the safety loading, the $p$
can be calculated via the expected value principle as

$p=(1+\eta)\lambda\mu.$

In a case where the insurance company shares risk with the reinsurance, the
sizes of the claims held by the insurer become $X^{a}_{i}$, where $a$ is a
(fixed) retention level. For proportional reinsurance, $a$ denotes the
fraction of the claim covered by the insurance company . Consider the case of
cheap reinsurance for which the reinsuring company uses the same safety
loading as the insurance company, the reserve process of the insurance company
is given by

$r^{(a,\eta)}_{t}=u+p^{(a,\eta)}t-\sum^{N_{t}}_{i=1}X^{(a)}_{i},$

where

$\displaystyle p^{(a,\eta)}=(1+\eta)\lambda\mathbb{E}\\{X^{(a)}_{i}\\}.$

Then by center limit theorem it is well known that for large enough $\lambda$

$\displaystyle\\{r_{t}^{(a,\eta)}\\}_{t\geq 0}\stackrel{{\scriptstyle
d}}{{\approx}}BM(\mu at,\sigma^{2}a^{2}t).$

in $\mathcal{D}[0,\infty)$ (the space of right continuous functions with left
limits endowed with the skorohod topology), where $\mu=\eta\lambda E(X_{i})$,
$\sigma=\sqrt{\lambda E(X_{i}^{2})}$ and $BM(\mu,\sigma^{2})$ stands for
Brownian motion with the drift coefficient $\mu$ and diffusion coefficient
$\sigma$ on $(\Omega,\mathcal{F},\\{\mathcal{F}_{t}\\}_{t\geq 0},\mathbb{P})$.
So the passage to the limit works well in the presence of a big portfolios,
the reserve (risk) process of the insurance company can be approximated by the
following diffusion process

$\displaystyle dR_{t}=\mu a(t)dt+\sigma a(t)d\mathcal{W}_{t},$ (2.1)

where $a(t)$ denotes retention level. We refer readers for this fact and for
the specifies of the diffusion approximations to Emanuel, Harrison and Taylor
[9](1975), Grandell[10](1977), Grandell[11](1978), Grandell[12](1990),
Harrison[15](1985), Iglehart[21](1969), Schmidli[30](1994). In this paper, we
consider a common company which faces constant liability payments. In view of
the diffusion approximations for the classical Cramér-Lundberg model described
above, we assume that the reserve process of the company facing constant
liability payments is given by the following diffusion process

$\displaystyle dR_{t}=(a(t)\mu-\delta)dt+a(t)\sigma d{\mathcal{W}}_{t},\quad
R_{0}=x$ (2.2)

where $x$ is the initial reserve of the company, $\mu$ is the expected profit
per unit time (profit rate), $\sigma$ is the volatility rate of the reserve
process in the absence of any risk control, $\delta$ represents the amount of
money the company has to pay per unit time (the debt rate) irrespective of
what business activities rate it chooses. In this model, the business
activities rate that the company chooses at time $t$ are modeled by $a(t)$,
which takes values in the interval $[\alpha,\beta]$, where
$0<\alpha<\beta<+\infty$. The restriction $\alpha>0$ reflects the fact that
there are statutory reasons that its business activities rate cannot be
reduced to zero, unless the company faces bankruptcy. In our model, a policy
$\pi$ is a pair of non-negative càdlàg $\mathcal{F}_{t}$-adapted processes
$\\{a_{\pi}(t),L_{t}^{\pi}\\}$, where $a_{\pi}(t)\in[\alpha,\beta]$
corresponds to the risk exposure at time $t$ and $L_{t}^{\pi}$ corresponds to
the cumulative amount of dividend pay-outs distributed up to time $t$. A
policy $\pi=\\{a_{\pi}(t),L_{t}^{\pi}\\}$ is called admissible if $\alpha\leq
a_{\pi}(t)\leq\beta$ and $L_{t}^{\pi}$ is a nonnegative, non-decreasing,
right-continuous function. We denote $\Pi$ the set of all admissible policies.
When a admissible policy $\pi$ is applied, the resulting reserve process is
denoted by $\\{R_{t}^{\pi}\\}$. In view of (2.2) we rewrite $R_{t}^{\pi}$ as
follows

$\displaystyle dR^{\pi}_{t}=(a_{\pi}(t)\mu-\delta)dt+a_{\pi}(t)\sigma
d{\mathcal{W}}_{t}-dL^{\pi}_{t},\quad R^{\pi}_{0}=x,$ (2.3)

where $x(>0)$ is the initial (capital) reserve. In addition, we assume the
company needs to keep its reserve above 0 to avoid bankruptcy. For the given
control policy $\pi$, we define the time of bankruptcy as
$\tau^{\pi}_{x}=\inf\big{\\{}t\geq 0:R^{\pi}_{t}\leq 0\big{\\}}$.
$\tau^{\pi}_{x}$ is an $\mathcal{F}_{t}$ -stopping time. The objective of the
company is to maximize the expected present value of the dividends payout
until the time of bankrupt by choosing control policy $\pi$ from the
admissible set $\Pi$. Choulli, Taksar and Zhou[7](2003) proved that there
exists an optimal dividend barrier $b_{0}$ and the optimal policy
$\pi^{*}_{b_{0}}=\\{a_{\pi^{*}_{b_{0}}}(\cdot),L^{\pi^{*}_{b_{0}}}_{\cdot}\\}$,
which maximize the expected present value of the dividends payout before
bankruptcy, i.e., $b_{0}$ is such that

$\displaystyle
J(x,\pi)=\mathbf{E}\big{\\{}\int_{0}^{\tau^{\pi}_{x}}e^{-ct}dL_{t}^{\pi}\big{\\}},$
(2.4) $\displaystyle
V(x,b_{0})=\sup_{\pi\in\Pi}\\{J(x,\pi)\\}=J(x,\pi^{*}_{b_{0}}),$ (2.5)

where $c$ denotes the discount rate. If the optimal dividend barrier is too
low, the bankrupt probability within a fixed time will be bigger than a given
level. This is not acceptable by the management of the company. Taking the
balance of profit and risk into consideration, we impose small bankrupt
probability constraint on the company’s control policy. We describe our
optimal control problem as follows. Let
$\Pi_{b}=\big{\\{}\pi\in\Pi:\int_{0}^{\infty}{I}_{\\{s:R^{\pi}(s)<b\\}}dL_{s}^{\pi}=0\big{\\}}$
for $b\geq 0$ . Then it is easy to see that $\Pi=\Pi_{0}$ and
$b_{1}>b_{2}\Rightarrow\Pi_{b_{1}}\subset\Pi_{b_{2}}$. For a given admissible
policy $\pi$, we define the optimal return function $V(x)$ by

$\displaystyle J(x,\pi)$ $\displaystyle=$ $\displaystyle{\bf
E}\big{\\{}\int_{0}^{\tau^{\pi}_{x}}e^{-ct}dL_{t}^{\pi}\big{\\}},$
$\displaystyle V(x,b)$ $\displaystyle=$
$\displaystyle\sup_{\pi\in\Pi_{b}}\\{J(x,\pi)\\},$ (2.6) $\displaystyle V(x)$
$\displaystyle=$ $\displaystyle\sup_{b\in\mathfrak{B}}\\{V(x,b)\\}$ (2.7)

and the optimal policy $\pi^{*}$ by

$\displaystyle J(x,\pi^{*})=V(x),$ (2.8)

where

$\displaystyle\mathfrak{B}:=\big{\\{}b\ :\ \mathbb{P}[\tau_{b}^{\pi_{b}}\leq
T]\leq\varepsilon\ ,\ J(x,\pi_{b})=V(x,b)\mbox{ and}\
\pi_{b}\in\Pi_{b}\big{\\}},$

$c>0$ is a discount rate, $\tau_{b}^{\pi_{b}}$ is the time of bankruptcy
$\tau_{x}^{\pi_{b}}$ when the initial reserve $x=b$ and the control policy is
$\pi_{b}$. $\varepsilon$ is a given constrained risk level of bankrupt
probability. $1-\varepsilon$ is the standard of security and less than
solvency for a given risk level $\varepsilon>0$. $\mathfrak{B}$ is called the
risk constrained set. The main purpose of this paper is to derive the optimal
return function $V(x)$, the optimal policy $\pi^{*}$ as well as the optimal
dividend payout barrier $b^{*}$ and try to give a reasonable economic
interpretation and discuss effect of the debt rate $\delta$, the constrained
risk level $\varepsilon$ of bankrupt probability, the initial capital $x$, the
volatility rate $\sigma^{2}$ and the profit rate $\mu$ on the optimal return
function and the optimal dividend strategy $\pi^{*}$.

##  3\. Main Result

In this section we first present main result of this paper, then, together
with numerical results in section 5 below, give its reasonable economic
interpretation. The proof of the main result will be given in section 7.

######  Theorem 3.1.

Let $\varepsilon\in(0,1)$ be the constrained risk level of bankrupt
probability and time horizon $T$ be given. (i) If
$\mathbf{P}[\tau_{b_{0}}^{\pi^{*}_{b_{0}}}\leq T]\leq\varepsilon$, then the
optimal return function $V(x)$ is $f(b_{0},x)$ defined by (6.5) below, and
$V(x)=f(b_{0},x)=J(x,\pi_{b_{o}}^{\ast})$. The optimal policy
$\pi_{b_{o}}^{\ast}$ is
$\\{a^{\ast}_{b_{o}}(R^{\pi_{b_{o}}^{\ast}}_{t}),L^{\pi_{b_{o}}^{\ast}}_{t}\\}$,
where $\\{R^{\pi_{b_{o}}^{\ast}}_{t},L^{\pi_{b_{o}}^{\ast}}_{t}\\}$ is
uniquely determined by the following stochastic differential equation

$\displaystyle\left\\{\begin{array}[]{l l
l}dR_{t}^{\pi_{b_{o}}^{\ast}}=(a^{*}_{b_{o}}(R^{\pi_{b_{o}}^{\ast}}_{t})\mu-\delta)dt+\sigma
a^{*}_{b_{o}}(R^{\pi_{b_{o}}^{\ast}}_{t})d{\mathcal{W}}_{t}-dL_{t}^{\pi_{b_{o}}^{\ast}},\\\
R_{0}^{\pi_{b_{o}}^{\ast}}=x,\\\ 0\leq R^{\pi_{b_{o}}^{\ast}}_{t}\leq
b_{0},\\\
\int^{\infty}_{0}I_{\\{t:R^{\pi_{b_{o}}^{\ast}}_{t}<b_{0}\\}}(t)dL_{t}^{\pi_{b_{o}}^{\ast}}=0.\end{array}\right.$
(3.5)

The solvency of the company is bigger than $1-\varepsilon$. (ii) If
$\mathbf{P}[\tau_{b_{0}}^{\pi^{*}_{b_{0}}}\leq T]>\varepsilon$, there is a
unique optimal dividend $b^{\ast}(\geq b_{0})$ satisfying
$\mathbf{P}[\tau_{b^{\ast}}^{\pi_{b^{*}}^{\ast}}\leq T]=\varepsilon$. The
optimal return function $V(x)$ is $g(x,b^{*})$ defined by (6.20), that is,

$\displaystyle V(x)=g(x,b^{*})=\sup_{b\in\mathfrak{B}}\\{V(x,b)\\}$ (3.6)

and

$\displaystyle
b^{*}\in\mathfrak{B}:=\big{\\{}b:\mathbb{P}[\tau_{b}^{\pi^{*}_{b}}\leq
T]\leq\varepsilon,\ J(x,\pi^{*}_{b})=V(x,b)\mbox{ and}\ \pi^{*}_{b}\in\Pi_{b}\
\big{\\}}.$ (3.7)

The optimal policy
$\pi_{b^{*}}^{\ast}=\\{a^{\ast}_{b^{*}}(R^{\pi_{b^{*}}^{\ast}}_{t}),L^{\pi_{b^{*}}^{\ast}}_{t}\\}$,
where $\\{R^{\pi_{b^{*}}^{\ast}}_{t},L^{\pi_{b^{*}}^{\ast}}_{t}\\}$ is
uniquely determined by the following stochastic differential equation

$\displaystyle\left\\{\begin{array}[]{l l l}dR_{t}^{\pi_{b^{*}}^{\ast}}=(\mu
a^{*}_{b^{*}}(R^{\pi_{b^{*}}^{\ast}}_{t})-\delta)dt+\sigma
a^{*}_{b^{*}}(R^{\pi_{b^{*}}^{\ast}}_{t})d{W}_{t}-dL_{t}^{\pi_{b^{*}}^{\ast}},\\\
R_{0}^{\pi_{b^{*}}^{\ast}}=x,\\\ 0\leq R^{\pi_{b^{*}}^{\ast}}_{t}\leq
b^{*},\\\
\int^{\infty}_{0}I_{\\{t:R^{\pi_{b^{*}}^{\ast}}_{t}<b^{*}\\}}(t)dL_{t}^{\pi_{b^{*}}^{\ast}}=0.\end{array}\right.$
(3.12)

The solvency of the company is $1-\varepsilon$. (iii) Moreover,

$\displaystyle\frac{g(x,b^{*})}{g(x,b_{0})}\leq 1,$ (3.13)

where $a^{*}_{b}(x)$ is defined by (6.25) below.

Economic interpretation of Theorem 3.1 is as follows. (1) For a given
constrained risk level $\varepsilon$ of bankrupt probability and time horizon
$T$, if the company’s bankruptcy probability is less than this given risk
constraint level $\varepsilon$, the optimal control problem of (2) and (2.7)
is the traditional problem (2.4) and (2.5). The bankrupt probability
constraints here do not work. (2) If the company’s bankruptcy probability is
larger than this given constrained risk level $\varepsilon$, the traditional
optimal control policy fails to meet the requirement of bankrupt probability
constraint. So the company has to find an optimal policy $\pi_{b^{*}}^{\ast}$
to improve its solvency ability and ensure the company operates safely. The
optimal reserve process $R^{\pi_{b^{*}}^{\ast}}_{t}$ is a diffusion process
reflected at a dividend barrier $b^{*}$, and the process
$L^{\pi_{b^{*}}^{\ast}}_{t}$ is the dividend payout process that ensures the
reflection. $a^{*}_{b^{*}}$ is the optimal feedback control function. The
optimal policy means that the company will pay out the amount of reserve in
excess of $b^{*}$ as dividend once its reserve is bigger than $b^{*}$. Under
this control policy, we can guarantee that the company’s bankrupt probability
can stay below $\varepsilon$. (3)The inequality (3.13) shows that the optimal
control policy $\pi_{b^{*}}^{\ast}$ will decrease the company’s profit-earning
capability. We can treat this decrease of the profit as the cost of keeping
the company’s risk at a low level and the cost, $g(x,b_{0})-g(x,b^{*})$, is
minimal in view of ( 3.7), Lemma 6.3 and Lemma 6.20 below. Thus
$\pi_{b^{*}}^{\ast}$ is the best equilibrium control policy between making
profit and improving solvency. (4) From the figure 1 in Example 5.1 below we
see that the optimal return function $g(x)$ is decreasing w.r.t. the debt rate
$\delta$. Figure 1 shows that the higher debt rate will lessen the company’s
profit, so the company should keep its debt rate at a appropriate level.(5)We
can see from figure 2 in Example 5.2 below that the optimal dividend barrier
$b^{*}$ is decreasing w.r.t. the constrained risk level $\varepsilon$ of
bankrupt probability. And the optimal dividend barrier $b^{*}$ is uniquely
decided by $\mathbf{P}[\tau_{b^{\ast}}^{\pi_{b^{*}}^{\ast}}\leq
T]=\varepsilon$, i.e., $1-\phi^{b^{*}}(b^{*},b^{*})=\varepsilon$ (see Lemma
6.5 below). The the optimal dividend barrier $b^{*}$ is also increasing
function of the volatility $\sigma^{2}$ (see the figure 3 in Example 5.3 below
). (6) We call $R_{0}:=x_{b^{*}}(\varepsilon)$ the risk-based capital standard
to ensure the capital requirement of can cover the total given risk
$\varepsilon$, where $x_{b^{*}}(\varepsilon)$ is determined by
$1-\phi^{b^{*}}(x,b^{*})=\varepsilon$ (see Lemma 6.5). We see from the figure
4 in Example 5.4 below that risk-based capital $x_{b^{*}}(\varepsilon)$
decreases with risk level $\varepsilon$. Since the optimal feedback control
function $a^{*}_{b^{*}}(x)$ is increasing w.r.t. $x$, in view of comparison
theorem for SDE, the constrained risk level $\varepsilon$ lessens the optimal
business activities rate $a^{*}(t)$, but improves dividend payout process
$L^{*}_{t}(\varepsilon)$. It also lessens the optimal return function(see
Example 5.7 below). (7) We can see from the figures 5 and 6 below that the
optimal return function $g(x)$ is increasing in both the profit rate $\mu$ and
the volatility rate $\sigma$.

##  4\.  Risk Analysis

In this section, we proceed a risk analysis on the model we are studying. We
first work out the lower boundary of bankrupt probability when we applied
$b_{0}$ as the dividend barrier. It proves that the risk constrained set
$\mathfrak{B}$ is not $\Re_{+}=[0,+\infty)$. So the topic of this paper is
fundamental to studying the optimal control problem under bankrupt probability
constraints. It also states that the company has to find optimal policy to
improve its solvency.

######  Theorem 4.1.

Let $\\{R^{\pi_{b_{0}}^{\ast}}_{t},L^{\pi_{b_{0}}^{\ast}}_{t}\\}$ be the
unique solution of the following SDE( see Lions and Sznitman [27](1984))

$\displaystyle\left\\{\begin{array}[]{l l l}dR_{t}^{\pi_{b_{o}}^{\ast}}=(\mu
a^{*}_{b_{o}}(R^{\pi_{b_{o}}^{\ast}}_{t})-\delta)dt+\sigma
a^{*}_{b_{o}}(R^{\pi_{b_{o}}^{\ast}}_{t})d{W}_{t}-dL_{t}^{\pi_{b_{o}}^{\ast}},\\\
R_{0}^{\pi_{b_{o}}^{\ast}}=b_{0},\\\ 0\leq R^{\pi_{b_{o}}^{\ast}}_{t}\leq
b_{0},\\\
\int^{\infty}_{0}I_{\\{t:R^{\pi_{b_{o}}^{\ast}}_{t}<b_{0}\\}}(t)dL_{t}^{\pi_{b_{o}}^{\ast}}=0.\end{array}\right.$
(4.5)

Then

$\displaystyle{\bf P}(\tau_{b_{0}}^{\pi^{*}_{b_{0}}}\leq T)$
$\displaystyle\geq$
$\displaystyle\frac{4[1-\Phi(\frac{b_{0}}{\alpha\sigma\sqrt{T}})]^{2}}{\exp\\{\frac{T}{\sigma^{2}}{\max\\{\mu-\frac{\delta}{\beta},|\mu-\frac{\delta}{\alpha}|\\}}^{2}\\}}$
(4.6) $\displaystyle\equiv$
$\displaystyle\varepsilon_{0}(b_{0},\mu,\delta,\sigma,T,\alpha,\beta)>0,$

where $\tau^{\pi^{*}_{b_{o}}}_{b_{o}}=\inf\big{\\{}t\geq
0:R^{\pi^{*}_{b_{o}}}_{t}\leq 0\big{\\}}$ and $\Phi(\cdot)$ is the standard
normal distribution function.

###### Proof.

Since $a^{*}_{b_{o}}(x)$ (defined by (6.25)) is a bounded Lipschitz continuous
function w.r.t. $x$, the following SDE

$\displaystyle\left\\{\begin{array}[]{l l l}dX_{t}=(\mu
a^{*}_{b_{0}}(X_{t})-\delta)dt+\sigma a^{*}_{b_{0}}(X_{t})d{W}_{t},\\\
X_{0}=b_{0}\\\ \end{array}\right.$ (4.9)

has a unique solution $X_{t}$. By comparison theorem for one-dimensional Itô
process, we have

$\displaystyle{\bf{P}}[R^{\pi_{b_{o}}^{\ast}}_{t}\leq X_{t}]=1$ (4.10)

Define a measure ${\bf Q}$ on $\mathcal{F}_{T}$ by

$\displaystyle d{\bf{Q}}(\omega)=M(T)d{\bf P}(\omega)$ (4.11)

where

$\displaystyle M(t)$ $\displaystyle\equiv$
$\displaystyle\exp\big{\\{}-\int_{0}^{t}\frac{(\mu
a^{*}_{b_{0}}(X_{t})-\delta)}{\sigma a^{*}_{b_{0}}(X_{t})}dW_{s}$
$\displaystyle-$ $\displaystyle\frac{1}{2}\int_{0}^{t}\frac{(\mu
a^{*}_{b_{0}}(X_{t})-\delta)^{2}}{[\sigma
a^{*}_{b_{0}}(X_{t})]^{2}}ds\big{\\}}.$

Since $\\{M(t)\\}$ is a martingale w.r.t. $\mathcal{F}_{t}$, we have ${\bf
E}\big{[}M(T)\big{]}=1$. Moreover, noticing that $a^{*}_{b_{o}}(X_{t})$
belongs to $[\alpha,\beta]$, we obtain

$\displaystyle{\bf{E}}^{\bf{P}}[M(T)^{2}]\leq\exp\\{\frac{T}{\sigma^{2}}{\max\\{\mu-\frac{\delta}{\beta},|\mu-\frac{\delta}{\alpha}|\\}}^{2}\\}$
(4.12)

Using Girsanov theorem, ${\bf Q}$ is a probability measure on
$\mathcal{F}_{T}$ and the process $\\{X_{t}\\}$ satisfies the following SDE

$\displaystyle dX_{t}=\sigma a^{*}_{b_{0}}(X_{t})d\tilde{W}_{t},\ X_{0}=b_{0}$
(4.13)

where $\tilde{W}_{t}=W_{t}+\int_{0}^{t}\frac{(\mu
a^{*}_{b_{0}}(X_{s})-\delta)}{\sigma a^{*}_{b_{0}}(X_{s})}d{s}$. It is easy to
see that $\tilde{W}_{t}$ is a Brownian motion on
$(\Omega,\mathcal{F},\\{\mathcal{F}_{t}\\}_{t\geq 0},{\bf Q})$. Define a time
changes $\rho(t)$ by

$\displaystyle\dot{\rho}(t)=\frac{1}{{a^{*}_{b_{0}}}^{2}(X_{t})\sigma^{2}},$
(4.14)

then $\rho(t)$ is a strictly increasing function. If we denote $X_{\rho(t)}$
by $\hat{X}_{t}$, then we have

$\displaystyle\hat{X}_{t}=b_{0}+\hat{W}_{t}.$

Noticing that $0<\alpha\leq{a^{*}_{b_{0}}}(R_{t})\leq\beta<+\infty$, we get

$\displaystyle\frac{1}{\beta^{2}\sigma^{2}}\leq\dot{\rho}(t)\leq\frac{1}{\alpha^{2}\sigma^{2}}.$
(4.15)

Due to the fact $\rho(t)=\int_{0}^{t}\dot{\rho(s)}ds$, we can deduce that
$\rho(t)\leq\frac{t}{\alpha^{2}\sigma^{2}}$ and
$\rho^{-1}(t)\geq\alpha^{2}\sigma^{2}t$, where $\rho^{-1}$ denotes the inverse
of $\rho$. Then we have

$\displaystyle{\bf{Q}}[\inf\\{t:X_{t}\leq 0\\}\leq T]$ $\displaystyle=$
$\displaystyle{\bf{Q}}[\inf\\{t:\hat{X}_{\rho^{-1}(t)}\leq 0\\}\leq T]$ (4.16)
$\displaystyle=$ $\displaystyle{\bf{Q}}[\inf\\{\rho(t):b_{0}+\hat{W}_{t}\leq
0\\}\leq T]$ $\displaystyle=$ $\displaystyle{\bf{Q}}[\inf\\{t:\hat{W}_{t}\leq-
b_{0}\\}\leq\rho^{-1}(T)]$ $\displaystyle\geq$
$\displaystyle{\bf{Q}}[\inf\\{t:\hat{W}_{t}\leq-
b_{0}\\}\leq\alpha^{2}\sigma^{2}T]$ $\displaystyle=$ $\displaystyle
2[1-\Phi(\frac{b_{0}}{\alpha\sigma\sqrt{T}})]>0.$

Using Hölder inequalities as well as (4.11),

$\displaystyle{\bf{Q}}[\inf\\{t:X_{t}\leq 0\\}\leq T]$ $\displaystyle=$
$\displaystyle\int_{\Omega}{\bf{1}}_{[\inf\\{t:X_{t}\leq 0\\}\leq
T]}d\bf{Q}(\omega)$ $\displaystyle=$
$\displaystyle\int_{\Omega}{\bf{1}}_{[\inf\\{t:X_{t}\leq 0\\}\leq
T]}M_{T}d{\bf{P}}(\omega)$ $\displaystyle=$
$\displaystyle{\bf{E}}^{{\bf{P}}}[M_{T}{\bf{1}}_{[\inf\\{t:X_{t}\leq 0\\}\leq
T]}]$ $\displaystyle\leq$
$\displaystyle{\bf{E}}^{{\bf{P}}}[M^{2}_{T}]^{\frac{1}{2}}{\bf{P}}[\inf\\{t:X_{t}\leq
0\\}\leq T]^{\frac{1}{2}}.$

Substituting (4.12) and (4.16) into (4), we get

$\displaystyle{\bf{P}}[\inf\\{t:X_{t}\leq 0\\}\leq T]$ $\displaystyle\geq$
$\displaystyle\frac{{\bf{Q}}[\inf\\{t:X_{t}\leq 0\\}\leq
T]^{2}}{{\bf{E}}^{\bf{P}}[M_{T}^{2}]}$ $\displaystyle\geq$
$\displaystyle\frac{4[1-\Phi(\frac{b_{0}}{\alpha\sigma\sqrt{T}})]^{2}}{\exp\\{\frac{T}{\sigma^{2}}{\max\\{\mu-\frac{\delta}{\beta},|\mu-\frac{\delta}{\alpha}|\\}}^{2}\\}}.$

By virtue of (4.10), we have

$\displaystyle{\bf P}[\tau_{b_{0}}^{\pi^{*}_{b_{0}}}\leq T]$ $\displaystyle=$
$\displaystyle{\bf{P}}[\inf\\{t:R^{\pi^{*}_{b_{o}}}_{t}\leq 0\\}\leq T]$
(4.18) $\displaystyle\geq$ $\displaystyle{\bf{P}}[\inf\\{t:X_{t}\leq 0\\}\leq
T]$ $\displaystyle\geq$
$\displaystyle\frac{4[1-\Phi(\frac{b_{0}}{\alpha\sigma\sqrt{T}})]^{2}}{\exp\\{\frac{T}{\sigma^{2}}{\max\\{\mu-\frac{\delta}{\beta},|\mu-\frac{\delta}{\alpha}|\\}}^{2}\\}}$
$\displaystyle\equiv$
$\displaystyle\varepsilon_{0}(b_{0},\mu,\delta,\sigma,T,\alpha,\beta)$
$\displaystyle>$ $\displaystyle 0.$

∎

The economic interpretation of theorem 4.1 is the following.Assume
$\frac{2\delta}{\mu}<\alpha$, then we have $\mu-\frac{\delta}{\alpha}>0$. So
the lower boundary of the bankrupt probability
$\varepsilon_{0}(b_{0},\mu,\delta,\sigma,T,\alpha,\beta)$ becomes to
$\frac{4[1-\Phi(\frac{b_{0}}{\alpha\sigma\sqrt{T}})]^{2}}{\exp\\{\frac{T}{\sigma^{2}}(\mu-\frac{\delta}{\beta})^{2}\\}}$.
Based on the assumption, we have the following explanations.(1) The lower
boundary of bankrupt probability for the company
$\varepsilon_{0}(b_{0},\mu,\delta,\sigma,\\\ T,\alpha,\beta)$ is an increasing
function of $(\sigma,\delta,\alpha)$, which means that higher volatility rate
$\sigma$ and debt rate $\delta$ will make the company face larger risk. In
addition, risk will increase as the lower boundary $\alpha$ of control
function $a(x)$ increases. (2) The lower boundary of bankrupt probability for
the company $\varepsilon_{0}(b_{0},\mu,\delta,\sigma,\\\ T,\alpha,\beta)$ is
decreasing in $(b_{0},\mu,\beta)$, which means that paying dividends at a
lower barrier will cause larger bankrupt probability. On the other hand, the
higher the profit rate is, the lower the risk is. Improving the upper boundary
$\beta$ of the control function $a(x)$ can also reduce the company’s risk. (3)
The company has a positive bankrupt probability within the time interval
$[0,T]$ if we set $b_{0}$ as the dividends barrier. In order to keep the
company’s risk at a low level, we need adjust our control policy and find the
optimal dividends barrier $b^{*}$ under lower constrained risk level of
bankrupt probability.  The second result is the following, which states that
the risk constrained set $\mathfrak{B}$ defined in section 2 is non-empty for
any $\varepsilon>0$, together with the first result, also guarantees our
problem (2), (2.7), (2.8) is well defined.

######  Theorem 4.2.

Let $(R^{\pi_{b}^{\ast}}_{t},L_{t}^{\pi_{b}^{\ast}})$ be defined by

$\displaystyle\left\\{\begin{array}[]{l l l}dR_{t}^{\pi_{b}^{\ast}}=(\mu
a^{*}_{b}(R^{\pi_{b}^{\ast}}_{t})-\delta)dt+\sigma
a^{*}_{b}(R^{\pi_{b}^{\ast}}_{t})d{W}_{t}-dL_{t}^{\pi_{b}^{\ast}},\\\
R_{0}^{\pi_{b}^{\ast}}=b,\\\ 0\leq R^{\pi_{b}^{\ast}}_{t}\leq b,\\\
\int^{\infty}_{0}I_{\\{t:R^{\pi_{b}^{\ast}}_{t}<b\\}}(t)dL_{t}^{\pi_{b}^{\ast}}=0,\end{array}\right.$
(4.23)

and $\tau_{b}^{\pi_{b}^{*}}=\inf\\{t\geq 0:R^{\pi_{b}^{\ast}}_{t}<0\\}$. Then

$\displaystyle\lim_{b\rightarrow\infty}{\bf P}[\tau_{b}^{b}\leq T]=0.$ (4.24)

###### Proof.

For any $b\geq 1$, we have $b\geq\sqrt{b}$. By comparison theorem on SDE (see
Ikeda and Watanabe [23](1981)), we have

$\displaystyle\mathbf{P}[\tau_{b}^{\pi^{*}_{b}}\leq
T]\leq\mathbf{P}[\tau_{\sqrt{b}}^{\pi^{*}_{b}}\leq T].$ (4.25)

Let $R^{(1)}_{t}$ satisfy the following SDE,

$\displaystyle\left\\{\begin{array}[]{l l
l}dR_{t}^{(1)}=(a^{*}(R_{t}^{(1)})\mu-\delta)dt+a^{*}(R_{t}^{(1)})\sigma
d\mathcal{W}_{t},\\\ R_{0}^{(1)}=\sqrt{b}.\end{array}\right.$ (4.28)

Then, we have

$\displaystyle\mathbf{P}[\tau_{\sqrt{b}}^{\pi^{*}_{b}}\leq T]$
$\displaystyle\leq$ $\displaystyle\mathbf{P}[R_{t}^{(1)}=0\ \mbox{or}\
R_{t}^{(1)}=b\ \mbox{for some $0\leq t\leq T$ }]$ (4.29) $\displaystyle\leq$
$\displaystyle\mathbf{P}[\sup_{0\leq t\leq T}R_{t}^{(1)}\geq
b]+\mathbf{P}[\inf_{0\leq t\leq T}R_{t}^{(1)}\leq 0].$

Next, we estimate $\mathbf{P}[\sup_{0\leq t\leq T}R_{t}^{(1)}\geq b]$ and
$\mathbf{P}[\inf_{0\leq t\leq T}R_{t}^{(1)}\leq 0]$, respectively. Hölder’s
inequality and $a^{*}(x)\leq\beta$ yield that

$\displaystyle\sup_{0\leq t\leq T}(R_{t}^{(1)})^{2}$ $\displaystyle\leq$
$\displaystyle 3(\sqrt{b})^{2}+3\sup_{0\leq t\leq
T}(\int_{0}^{t}(a^{*}(R_{s}^{(1)})\mu-\delta)ds)^{2}+$ (4.30) $\displaystyle
3\sup_{0\leq t\leq T}(\int_{0}^{t}a^{*}(R_{t}^{(1)})\sigma
d\mathcal{W}_{s})^{2}$ $\displaystyle\leq$ $\displaystyle
3b+3(\beta\mu-\delta)^{2}T^{2}+$ $\displaystyle 3\sup_{0\leq t\leq
T}(\int_{0}^{t}a^{*}(R_{t}^{(1)})\sigma d\mathcal{W}_{s})^{2}.$

By Markov inequality, B-D-G inequalities and (4.30), we obtain

$\displaystyle\mathbf{P}[\sup_{0\leq t\leq T}R_{t}^{(1)}\geq b]$
$\displaystyle\leq$ $\displaystyle\frac{\mathbf{E}[\sup_{0\leq t\leq
T}(R_{t}^{(1)})^{2}]}{b^{2}}$ (4.31) $\displaystyle\leq$
$\displaystyle\frac{3b+3(\beta\mu-\delta)^{2}T^{2}+12{\beta}^{2}{\sigma}^{2}T}{b^{2}}$
$\displaystyle\rightarrow$ $\displaystyle 0,\ \ \mbox{as}\
b\rightarrow\infty.$

Now we turn to estimating $\mathbf{P}[\inf_{0\leq t\leq T}R_{t}^{(1)}\leq 0]$.
Let $R_{t}^{(2)}$ satisfy the following SDE

$\displaystyle\left\\{\begin{array}[]{l l
l}dR_{t}^{(2)}=(a^{*}(R_{t}^{(1)})\mu-\delta)dt+a^{*}(R_{t}^{(1)})\sigma
d\mathcal{W}_{t},\\\ R_{0}^{(2)}=0.\end{array}\right.$ (4.34)

Thus we have

$\displaystyle R_{t}^{(1)}=\sqrt{b}+R_{t}^{(2)}.$ (4.35)

Therefore, by using the same argument as in (4.31) , we get

$\displaystyle\mathbf{P}[\inf_{0\leq t\leq T}R_{t}^{(1)}\leq 0]$
$\displaystyle=$ $\displaystyle\mathbf{P}[\inf_{0\leq t\leq
T}R_{t}^{(2)}\leq-\sqrt{b}]$ (4.36) $\displaystyle=$
$\displaystyle\mathbf{P}[\sup_{0\leq t\leq T}(-R_{t}^{(2)})\geq\sqrt{b}]$
$\displaystyle\leq$ $\displaystyle\frac{\mathbf{E}[\sup_{0\leq t\leq
T}(-R_{t}^{(2)})^{2}]}{(\sqrt{b})^{2}}$ $\displaystyle\leq$
$\displaystyle\frac{2(\beta\mu-\delta)^{2}T^{2}+8{\beta}^{2}{\sigma}^{2}T}{b}$
$\displaystyle\rightarrow$ $\displaystyle 0,\ \ \mbox{as}\
b\rightarrow\infty.$

Hence, (4.25), (4.29), (4.31) and (4.36) yield that

$\displaystyle\lim\limits_{b\rightarrow\infty}\mathbf{P}[\tau_{b}^{\pi^{*}_{b}}\leq
T]=0.$

∎

##  5\. Numerical examples

In this section, we present some numerical examples to give the readers an
intuitive impression on the relations between the results and model
parameters. Setting the parameters at suitable level, we portray how the debt
rate $\delta$, the constrained risk level of bankrupt probability, the initial
capital $x$ , the volatility rate $\sigma^{2}$ and the profit rate $\mu$
impact on the optimal return function and the optimal dividend strategy based
on the PDE (6.29) below. we also show the figures of the optimal return
function $g(x)$ and the associated optimal feedback control function
$a^{*}(x)$.

######  Example 5.1.

Let $\mu=2$, $\sigma^{2}=50$, $c=0.05$, $\alpha=0.5$, $\beta=8$, $T=300$ and
$b=100$. Figure 1 shows that the optimal return function $g_{\delta}(x)$
decreases with the debt rate $\delta$.

Figure 1.  The optimal return function $g_{\delta}(x)$ as a function of
$\delta$. (Parameters: $\mu=2,\sigma=50,c=0.05,\alpha=0.5,\beta=8,T=300,b=100$
)

######  Example 5.2.

Let $\mu=2$, $\sigma^{2}=50$, $\delta=0.2$, $c=0.05$, $\alpha=0.5$, $\beta=8$,
$T=300$. Let $b(\varepsilon)$ be the solution of $1-\phi(T,b)=\varepsilon$,
where $\phi(T,b)$ is defined in Lemma 6.5. Thus given a constrained risk level
$\varepsilon$ of bankrupt probability, $b(\varepsilon)$ is the associated
dividends barrier. Figure 2 shows that the dividends barrier $b(\varepsilon)$
decreases with the constrained risk level $\varepsilon$.

Figure 2.  Dividends barrier $b(\varepsilon)$ as a function of $\varepsilon$.
(Parameters: $\mu=2,\sigma^{2}=50,\delta=0.2,c=0.05,\alpha=0.5,\beta=8,T=300$
)

######  Example 5.3.

Let $\mu=2$, $\delta=0.2$, $c=0.05$, $\alpha=0.5$, $\beta=8$, $T=300$. Let
$b_{\sigma}(\varepsilon)$ be the solution of $1-\phi(T,b)=\varepsilon$, where
$\phi(T,b)$ is defined in Lemma 6.5. We see from Figure 3 that at the same
constrained risk level, the bigger the volatility rate $\sigma$ is, the higher
the dividends barrier $b_{\sigma}(\varepsilon)$ is.

Figure 3.  Dividends barrier $b_{\sigma}(\varepsilon)$ as a function of
$\sigma^{2}$. (Parameters: $\mu=2,\delta=0.2,c=0.05,\alpha=0.5,\beta=8,T=300$
)

######  Example 5.4.

Let $\mu=2$, $\sigma^{2}=50$ $\delta=0.2$, $c=0.05$, $\alpha=0.5$, $\beta=8$,
$T=300$. Let $b_{\sigma}(\varepsilon)$ be the solution of
$1-\phi(T,b)=\varepsilon$, where $\phi(T,b)$ is defined in Lemma 6.5.
$R_{0}=x$ is the initial reserve and $\varepsilon$ is the constrained risk
level of bankrupt probability. We see from figure 4 that the lower the initial
reserve $x$ is, the higher the constrained risk level $\varepsilon$ is.

Figure 4.  Initial reserve $x(\varepsilon)$ as a function of the risk
restrained level $\varepsilon$. (Parameters:
$\mu=2,\sigma^{2}=50,\delta=0.2,c=0.05,\alpha=0.5,\beta=8,T=300$ )

######  Example 5.5.

Let $\sigma^{2}=50$, $\delta=0.2$, $c=0.05$, $\alpha=0.5$, $\beta=8$, $T=300$
and $b=100$. Figure 5 shows that the optimal return function $g_{\mu}(x)$
increases with the profit rate $\mu$.

Figure 5.  The optimal return function $g_{\mu}(x)$ as a function of $\mu$.
(Parameters: $\sigma^{2}=50,\delta=0.2,c=0.05,\alpha=0.5,\beta=8,T=300,b=100$
)

######  Example 5.6.

Let $\mu=2$, $\delta=0.2$, $c=0.05$, $\alpha=0.5$, $\beta=8$, $T=300$ and
$b=100$. Figure 6 shows that the optimal return function $g_{\sigma}(x)$
increases with the volatility rate $\sigma^{2}$.

Figure 6.  The optimal return function $g_{\sigma}(x)$ as a function of
$\sigma^{2}$. (Parameters:
$\mu=2,\delta=0.2,c=0.05,\alpha=0.5,\beta=8,T=300,b=100$ )

######  Example 5.7.

Let $\mu=2$, $\sigma^{2}=50$, $\delta=0.2$, $c=0.05$, $\alpha=0.5$, $\beta=8$,
$T=300$ and $b=100$. Set $x_{\alpha}=4.72$, $x_{\beta}=94.79$, the images of
the optimal return function $g(x)$ as well as the optimal feedback control
function $a^{*}(x)$ are as follows (see Figure 7 and Figure 8).

Figure 7.  The optimal return function $g(x)$. (Parameters:
$\mu=2,\sigma^{2}=50,\delta=0.2,c=0.05,\alpha=0.5,\beta=8,T=300,b=100$ )
Figure 8.  The optimal feedback control function $a(x)$. (Parameters:
$\mu=2,\sigma^{2}=50,\delta=0.2,c=0.05,\alpha=0.5,\beta=8,T=300,b=100$ )

##  6\. Properties of $V(x,b)$ and Bankrupt Probability

In this section, we will discuss some important properties of the optimal
return function $V(x,b)$ and bankrupt probability, which are used to prove the
main result of this paper. The rigorous proofs of these properties will be
given in the appendix. In view of Lemma 8.1 in the appendix, different value
of $\frac{2\delta}{\mu}$ can lead to three different cases. When
$\frac{2\delta}{\mu}<\alpha$, this case is the most complicated. We select
this case as the basis of our discussion throughout the paper, and the results
of the other two cases are almost same.

######  Lemma 6.1.

If $f(x)\in C^{2}$ and satisfies the following HJB equation and boundary
conditions,

$\displaystyle\left\\{\begin{array}[]{l l
l}\max\limits_{a\in[\alpha,\beta]}[\frac{1}{2}\sigma^{2}a^{2}f^{{}^{\prime\prime}}(x)+(\mu
a-\delta)f^{{}^{\prime}}(x)-cf(x)]=0,\ \mbox{for}\ 0\leq x\leq b_{0},\\\
f^{{}^{\prime}}(x)=1,\ \mbox{ for}\ x\geq b_{0},\\\
f^{{}^{\prime\prime}}(x)=0,\ \mbox{ for}\ x\geq b_{0},\\\
f(0)=0,\end{array}\right.$ (6.5)

then we have

$\displaystyle b_{0}=\inf\\{x\geq 0:f^{\prime\prime}(x)=0\\}$

and

$\displaystyle\left\\{\begin{array}[]{l l l}\max\mathcal{L}f(x)\leq 0\
\mbox{and }\ f^{{}^{\prime}}(x)\geq 1\ \mbox{for}\ x\geq 0,\\\
f(0)=0,\end{array}\right.$

where $\mathcal{L}=\frac{1}{2}\sigma^{2}a^{2}\frac{d^{2}}{dx^{2}}+(\mu
a-\delta)\frac{d}{dx}-c$.

######  Lemma 6.2.

Let $b>b_{0}$ be a predetermined variable. If $g\in C^{1}(R_{+})$, $g\in
C^{2}(R_{+}\setminus\\{b\\})$ and satisfies the following HJB equation and
boundary conditions,

$\displaystyle\left\\{\begin{array}[]{l l
l}\max\limits_{a\in[\alpha,\beta]}[\frac{1}{2}\sigma^{2}a^{2}g^{{}^{\prime\prime}}(x)+(\mu
a-\delta)g^{{}^{\prime}}(x)-cg(x)]=0,\ \mbox{for}\ 0\leq x\leq b,\\\
g^{{}^{\prime}}(x)=1,\ \ \mbox{ for}\ \ x\geq b,\\\
g^{{}^{\prime\prime}}(x)=0,\ \ \mbox{ for}\ \ x>b,\\\
g(0)=0,\end{array}\right.$ (6.11)

then we have

$\displaystyle\left\\{\begin{array}[]{l l l}\max\mathcal{L}g(x)\leq 0,\
\mbox{for}\ \ x\geq 0,\\\ g^{{}^{\prime}}(x)\geq 1,\ \mbox{for}\ \ x\geq b,\\\
g(0)=0,\end{array}\right.$ (6.15)

where $b_{0}$ and $\mathcal{L}$ are the same as in Lemma 6.1,
$g^{\prime\prime}(b):=g^{\prime\prime}(b-)$. The expression of $g(x)$ can be
written as

$\displaystyle g(x,b)=\left\\{\begin{array}[]{l l
l}k_{1}(e^{r_{+}(\alpha)}x-e^{r_{-}(\alpha)x}),\ 0\leq x<x_{\alpha},\\\
k_{2}[\frac{\alpha\mu-2\delta}{2c}+\int_{x_{\alpha}}^{x}exp(-\frac{\mu}{\sigma^{2}}\int_{x_{\alpha}}^{y}\frac{dv}{a(v)})dy],\
x_{\alpha}\leq x<x_{\beta},\\\
k_{3}e^{r_{+}(\beta)(x-b_{0})}+k_{4}e^{r_{-}(\beta)(x-b_{0})},\ x_{\beta}\leq
x<b,\\\ x-b+g_{3}(b),\ x\geq b,\end{array}\right.$ (6.20)

where $r_{\pm}(x),x_{\alpha},x_{\beta},k_{1},k_{2},k_{3}$ and $k_{4}$ are
given by (8.9), (8.10), (8.17), (8.21), (8.37) and (8.36), respectively.

######  Lemma 6.3.

Let $g(x,b)$ be as the same as in Lemma 6.2. Then $\frac{\partial}{\partial
b}g(x,b)\leq 0$ holds for $b\geq b_{0}$.

######  Lemma 6.4.

The bankrupt probability $\mathbf{P}[\tau_{b}^{\pi^{*}_{b}}\leq T]$ is a
strictly decreasing function w.r.t. the dividends barrier $b$ on
$[x_{\beta},D)$, $D:=\inf\\{b:\mathbf{P}[\tau_{b}^{\pi^{*}_{b}}\leq T]=0\\}$,
and $x_{\beta}$ is defined by (8.17).

From the proof of Lemma 6.2, for each $x\leq b$, if we define

$\displaystyle a^{*}(x):=arg\
\max\limits_{a\in[\alpha,\beta]}[\frac{1}{2}\sigma^{2}a^{2}g^{{}^{\prime\prime}}(x)+(\mu
a-\delta)g^{{}^{\prime}}(x)-cg(x)],$ (6.21)

then it follows that $a^{*}(x)$ can be represented as

$\displaystyle a^{*}(x)=\left\\{\begin{array}[]{l l l}\alpha,\ \ 0\leq
x<x_{\alpha},\\\ a(x),\ \ x_{\alpha}\leq x<x_{\beta},\\\ \beta,\ \ x\geq
x_{\beta},\end{array}\right.$ (6.25)

where $a(x)$ and $x_{\alpha},x_{\beta}$ are specified by (8.16), (8.10),
(8.17), respectively. We now have the following lemma.

######  Lemma 6.5.

Let $a^{*}(x)$ be defined by (6.25), and define
$\psi^{b}(T,x):={\bf{P}}[\tau^{\pi^{*}_{b}}_{x}\leq T]$, i.e., $\psi^{b}(T,x)$
is the bankrupt probability when the initial reserve of
$\\{R^{\pi^{*}_{b}}_{t}\\}_{t\geq 0}$ is $x$ and dividends barrier is $b$. Let
$\phi^{b}(t,y)\in C^{1}(0,\infty)\cap C^{2}(0,b)$ and satisfy the following
partial differential equation and the boundary conditions,

$\displaystyle\left\\{\begin{array}[]{l l
l}\phi_{t}^{b}(t,x)=\frac{1}{2}[a^{*}(x)]^{2}\sigma^{2}\phi_{xx}^{b}(t,x)+(a^{*}(x)\mu-\delta)\phi_{x}^{b}(t,x),\\\
\phi^{b}(0,x)=1,\ \mbox{for}\ \ 0<x\leq b,\\\
\phi^{b}(t,0)=0,\phi_{x}^{b}(t,b)=0,\ \mbox{for}\ t>0.\end{array}\right.$
(6.29)

Then $\phi^{b}(T,x)=1-\psi^{b}(T,x)$, i.e., $\phi^{b}(T,x)$ is probability
that the company will survive on $[0,T]$.

######  Lemma 6.6.

Let $\phi^{b}(t,x)$ solve the equation(6.29). Then $\phi^{b}(T,b)$ is
continuous with respect to the dividends barrier $b$ on $[b_{0},+\infty)$.

##  7\. Proof of Main Result

In this section, we prove the main result of this paper, which is described in
Theorem 3.1. In order to do this, we first need the following.

######  Theorem 7.1.

Let $a^{*}_{b}(x)$ be defined by (6.25), and $f(x)$, $g(x,b)$ be as the same
as in Lemma 6.1 and Lemma 6.2, respectively. Then
(i) If $b\leq b_{0}$, we have $V(x,b)=V(x,b_{0})=V(x)=f(x)$, the optimal
policy associated with $V(x)$ is
$\pi_{b_{o}}^{\ast}=\\{a^{*}_{b_{0}}(R^{\pi_{b_{o}}^{\ast}}_{\cdot}),L^{\pi_{b_{o}}^{\ast}}_{\cdot}\\}$,
where the process
$\\{R^{\pi_{b_{o}}^{\ast}}_{t},L^{\pi_{b_{o}}^{\ast}}_{t}\\}$ is uniquely
determined by the following SDE,

$\displaystyle\left\\{\begin{array}[]{l l l}dR_{t}^{\pi_{b_{o}}^{\ast}}=(\mu
a^{*}_{b_{0}}(R^{\pi_{b_{o}}^{\ast}}_{t})-\delta)dt+\sigma
a^{*}_{b_{0}}(R^{\pi_{b_{o}}^{\ast}}_{t})d{W}_{t}-dL_{t}^{\pi_{b_{o}}^{\ast}},\\\
R_{0}^{\pi_{b_{o}}^{\ast}}=x,\\\ 0\leq R^{\pi_{b_{o}}^{\ast}}_{t}\leq
b_{0},\\\
\int^{\infty}_{0}I_{\\{t:R^{\pi_{b_{o}}^{\ast}}_{t}<b_{0}\\}}(t)dL_{t}^{\pi_{b_{o}}^{\ast}}=0.\end{array}\right.$
(7.5)

(ii) If $b>b_{0}$, we have $V(x,b)=g(x,b)$ and the optimal policy
$\pi_{b}^{\ast}$ is
$\\{a^{\ast}_{b}(R^{\pi_{b}^{\ast}}_{t}),L^{\pi_{b}^{\ast}}_{t}\\}$, where
$\\{R^{\pi_{b}^{\ast}}_{t},L^{\pi_{b}^{\ast}}_{t}\\}$ is uniquely determined
by the following SDE

$\displaystyle\left\\{\begin{array}[]{l l l}dR_{t}^{\pi_{b}^{\ast}}=(\mu
a^{*}_{b}(R^{\pi_{b}^{\ast}}_{t})-\delta)dt+\sigma
a^{*}_{b}(R^{\pi_{b}^{\ast}}_{t})d{W}_{t}-dL_{t}^{\pi_{b}^{\ast}},\\\
R_{0}^{\pi_{b}^{\ast}}=x,\\\ 0\leq R^{\pi_{b}^{\ast}}_{t}\leq b,\\\
\int^{\infty}_{0}I_{\\{t:R^{\pi_{b}^{\ast}}_{t}<b\\}}(t)dL_{t}^{\pi_{b}^{\ast}}=0.\end{array}\right.$
(7.10)

###### Proof.

(i) If $b\leq b_{0}$, since $\pi^{*}_{b_{0}}\in\Pi_{b_{0}}\subset\Pi_{b}$, we
have $V(x,b_{0})\leq V(x,b)\leq V(x)$. It suffices to show $V(x)\leq
f(x)=V(x,b_{0})$. Since its proof is similar to [7], we omit it here. (ii) If
$b\geq b_{0}$, denote $g(x,b)$ by $g(x)$ for simplicity, for any admissible
policy $\pi=\\{a_{\pi},L^{\pi}\\}$, we assume that $(R^{\pi}_{t},L^{\pi}_{t})$
is the process (2.3) associated with $\pi$. Let $\Lambda=\\{s:L_{s-}^{\pi}\neq
L_{s}^{\pi}\\}$, $\hat{L}=\sum_{s\in\Lambda,s\leq
t}(L_{s}^{\pi}-L_{s-}^{\pi})$ be the discontinuous part of $L_{s}^{\pi}$ and
$\tilde{L}_{t}^{\pi}=L_{t}^{\pi}-\hat{L}_{t}^{\pi}$ be the continuous part of
$L_{s}^{\pi}$. Applying generalized Itô formula to
$e^{-c(t\wedge\tau^{\pi}_{x})}g(R_{t\wedge\tau^{\pi}_{x}}^{\pi})$, we have

$\displaystyle
e^{-c(t\wedge\tau^{\pi}_{x})}g(R_{t\wedge\tau^{\pi}_{x}}^{\pi})$
$\displaystyle=$ $\displaystyle
g(x)+\int_{0}^{t\wedge\tau^{\pi}_{x}}e^{-cs}\mathcal{L}g(R_{s}^{\pi})ds$
(7.11) $\displaystyle+\int_{0}^{t\wedge\tau^{\pi}_{x}}a_{\pi}\sigma
e^{-cs}g^{{}^{\prime}}(R_{s}^{\pi})d\mathcal{W}_{s}\mathcal{-}\int_{0}^{t\wedge\tau^{\pi}_{x}}e^{-cs}g^{{}^{\prime}}(R_{s}^{\pi})dL_{s}^{\pi}$
$\displaystyle+\sum\limits_{s\in\Lambda,s\leq
t\wedge\tau^{\pi}_{x}}e^{-cs}[g(R_{s}^{\pi})-g(R_{s-}^{\pi})$
$\displaystyle-g^{{}^{\prime}}(R_{s-}^{\pi})(R_{s}^{\pi}-R_{s-}^{\pi})]$
$\displaystyle=$ $\displaystyle
g(x)+\int_{0}^{t\wedge\tau^{\pi}_{x}}e^{-cs}\mathcal{L}g(R_{s}^{\pi})ds$
$\displaystyle+\int_{0}^{t\wedge\tau^{\pi}_{x}}a_{\pi}\sigma
e^{-cs}g^{{}^{\prime}}(R_{s}^{\pi})d\mathcal{W}_{s}\mathcal{-}\int_{0}^{t\wedge\tau^{\pi}_{x}}e^{-cs}g^{{}^{\prime}}(R_{s}^{\pi})d\tilde{L}_{s}^{\pi}$
$\displaystyle+\sum\limits_{s\in\Lambda,s\leq
t\wedge\tau^{\pi}_{x}}e^{-cs}[g(R_{s}^{\pi})-g(R_{s-}^{\pi}))],$

where

$\displaystyle\mathcal{L}=\frac{1}{2}a^{2}\sigma^{2}\frac{d^{2}}{dx^{2}}+(\mu
a-\delta)\frac{d}{dx}-c.$

In view of the HJB equation (6.11), $\mathcal{L}g(R_{s}^{\pi})$ is always non-
positive, so is the second term on the right hand side of(7.11). By taking
mathematical expectations at both sides of (7.11), we get

$\displaystyle{\bf
E}\big{[}e^{-c(t\wedge\tau^{\pi}_{x})}g(R_{t\wedge\tau^{\pi}_{x}}^{\pi})\big{]}$
$\displaystyle\leq$ $\displaystyle g(x)-{\bf
E}\big{[}\int_{0}^{t\wedge\tau^{\pi}_{x}}e^{-cs}g^{{}^{\prime}}(R_{s}^{\pi})d\tilde{L}_{s}^{\pi}\big{]}$
$\displaystyle+{\bf E}\big{[}\sum\limits_{s\in\Lambda,s\leq
t\wedge\tau^{\pi}_{x}}e^{-cs}[g(R_{s}^{\pi})-g(R_{s-}^{\pi})]\big{]}.$

Since $g^{{}^{\prime}}(x)\geq 1$, for $x\geq b$,

$\displaystyle g(R_{s}^{\pi})-g(R_{s-}^{\pi})\leq-(L_{s}^{\pi}-L_{s-}^{\pi}),$
(7.13)

which, together with (7), implies that

$\displaystyle{\bf
E}\big{[}e^{-c(t\wedge\tau^{\pi}_{x})}g(R_{t\wedge\tau^{\pi}_{x}}^{\pi})\big{]}$
$\displaystyle+$ $\displaystyle{\bf
E}\big{[}\int_{0}^{t\wedge\tau^{\pi}_{x}}e^{-cs}dL_{s}^{\pi}\big{]}\leq g(x).$
(7.14)

By the definition of $\tau^{\pi}_{x}$ and $g(0)=0$, letting
$t\rightarrow\infty$ in (7.14), we get

$\displaystyle\liminf\limits_{t\rightarrow\infty}e^{-c(t\wedge\tau^{\pi}_{x})}g(R_{t\wedge\tau^{\pi}_{x}}^{\pi})$
$\displaystyle=$ $\displaystyle e^{-c\tau}g(0)I_{\\{\tau^{\pi}_{x}<\infty\\}}$
(7.15)
$\displaystyle+\liminf\limits_{t\rightarrow\infty}e^{-ct}g(R_{t})I_{\\{\tau^{\pi}_{x}=\infty\\}}$
$\displaystyle\geq$ $\displaystyle 0.$

We deduce from (7.14) and (7.15) that

$\displaystyle J(x,\pi)={\bf
E}\big{[}\int_{0}^{\tau^{\pi}_{x}}e^{-cs}dL_{s}^{\pi}\big{]}\leq g(x).$

So

$\displaystyle V(x,b)\leq g(x).$ (7.16)

If we choose the control policy
$\pi_{b}^{\ast}=\\{a^{*}_{b}(R^{\pi_{b}^{\ast}}_{\cdot}),L^{\pi_{b}^{\ast}}_{\cdot}\\}$,
which is uniquely determined by SDE (7.10), then all the inequalities above
become equalities. Hence

$\displaystyle V(x,b)=g(x).$

So we have

$\displaystyle V(x,b)=g(x,b).$ (7.17)

∎

Now we prove the main result of this paper. Proof of Theorem 3.1. If
$\mathbf{P}[\tau_{b_{0}}^{\pi^{*}_{b_{0}}}\leq T]\leq\varepsilon$, the
bankrupt probability constraint does not work and it turns to a usual optimal
control problem, thus the conclusion is obvious. If
$\mathbf{P}[\tau_{b_{0}}^{\pi^{*}_{b_{0}}}\leq T]>\varepsilon$, then by lemmas
6.4-6.6 there exists a unique $b^{\ast}$ solving the equation
$\mathbf{P}\\{\tau_{b}^{\pi^{*}_{b}}\leq T\\}=\varepsilon$. Moreover,
$b^{\ast}=\inf\\{b:b\in\mathfrak{B}\\}>b_{0}$. By Theorem 7.1 and Lemma 6.3,
$V(x,b)=g(x,b)$ for $b>b_{0}$ and $V(x,b)$ is decreasing w.r.t. $b$.
Therefore, we know that $b^{\ast}$ meets (3.6) and (3.7). So the optimal
policy associated with the optimal return function $V(x,b^{*})$ is
$\\{a^{\ast}_{b^{*}}(R^{\pi_{b^{*}}^{\ast}}_{t}),L^{\pi_{b^{*}}^{\ast}}_{t}\\}$,
which is uniquely determined by SDE (3.12). Due to the fact $b^{*}>b_{0}$, the
inequity (3.13) is a direct consequence of Lemma 6.3. Thus we complete the
proof. $\Box$

##  8\. Appendix

In this section, we first discuss some useful arguments, then we give the
proofs of the lemmas used in the previous sections. Due to the mathematical
model presented by (2.2), $a(t)$ is required to take values in the interval
$\alpha,\beta$, where $0<\alpha<\beta<+\infty$. Thus, $a(t)\mu-\delta$ may be
negative because $\delta>0$. If $\beta\mu\leq\delta$, there exists a trivial
solution to the corresponding HJB equations, which has been proved by Choulli,
Taksar, and Zhou[7]. In the next section, we always assume that
$\beta\mu\geq\delta$. Then the following statements are valid.

######  Lemma 8.1.

Let $\beta\mu>0$. Then
(i) $\frac{2\delta}{\mu}<\alpha$ if and only if $a(0)<\alpha$. In this case,

$\displaystyle a(0)=\frac{\mu\alpha^{2}}{2(\mu\alpha-\delta)}.$ (8.1)

(ii) $\alpha\leq\frac{2\delta}{\mu}<\beta$ if and only if $\alpha\leq
a(0)<\beta$. In this case,

$\displaystyle a(0)=\frac{2\delta}{\mu}.$ (8.2)

(iii) $\beta\leq\frac{2\delta}{\mu}$ if and only if $a(0)\geq\beta$. In this
case,

$\displaystyle a(0)=\frac{\mu\beta^{2}}{2(\mu\beta-\delta)}.$ (8.3)

###### Proof.

See Choulli, Taksar and Zhou[7] for details. ∎

Due to Lemma 8.1, there are three different cases to investigate. Since the
proof of each case is similar, we only give sketch proofs of lemmas in case
(i). Thus we suppose $\frac{2\delta}{\mu}<\alpha$ throughout the following
procedure to prove these lemmas. Proof of lemma 6.1. The complete proof is
given in Choulli, Taksar and Zhou[7](2003). $\Box$

Proof of lemma 6.2. Step 1. For each $x\geq 0$ and $a\geq 0$, define

$\displaystyle h(x,a)=\frac{1}{2}\sigma^{2}a^{2}g^{{}^{\prime\prime}}(x)+(\mu
a-\delta)g^{{}^{\prime}}(x)-cg(x).$ (8.4)

Then, by differentiating $h(x,a)$ w.r.t. $a$, we get the maximizing function
of $h(x,a)$

$\displaystyle a(x)=-\frac{\mu
g^{{}^{\prime}}(x)}{\sigma^{2}g^{{}^{\prime\prime}}(x)},\ \ x\geq 0.$ (8.5)

In view of Lemma 8.1 (i), $a(x)\leq\alpha$ for all $x$ in the right
neighborhood of 0. Substituting $a=\alpha$ into (6.11), and solving the
resulting second-order linear ODE, we get

$\displaystyle g(x)=k_{1}(e^{r_{+}(\alpha)}x-e^{r_{-}(\alpha)x}),\ \ 0\leq
x<x_{\alpha},$ (8.6)

where $k_{1}$ and $x_{\alpha}$ are to be determined and for $x>0$

$\displaystyle\left\\{\begin{array}[]{l l l}r_{+}(x)=\frac{-(\mu
x-\delta)+\sqrt{(\mu x-\delta)^{2}+2\sigma^{2}cx^{2}}}{\sigma^{2}x^{2}}\\\
r_{-}(x)=\frac{-(\mu x-\delta)-\sqrt{(\mu
x-\delta)^{2}+2\sigma^{2}cx^{2}}}{\sigma^{2}x^{2}}.\end{array}\right.$ (8.9)

Due to (8.5) and (8.6), for $x>0$

$\displaystyle a^{{}^{\prime}}(x)=\frac{-\mu
r_{+}(\alpha)r_{-}(\alpha)(r_{+}(\alpha)-r_{-}(\alpha))^{2}e^{(r_{+}(\alpha)+r_{-}(\alpha))x}}{\sigma^{2}(g^{{}^{\prime\prime}}(x))^{2}}>0.$

Therefore $a(x)$ increases to $\alpha$ at the point $x_{\alpha}$ given by

$\displaystyle
x_{\alpha}=\frac{1}{r_{+}(\alpha)-r_{-}(\alpha)}log(\frac{r_{-}(\alpha)(\mu+\alpha\sigma^{2}r_{-}(\alpha))}{r_{+}(\alpha)(\mu+\alpha\sigma^{2}r_{+}(\alpha))})>0.$
(8.10)

Step 2. In view of Proposition 8 in [7], $\alpha\leq a(x)\leq\beta$ in the
right neighborhood of $x_{\alpha}$. From (8.5), we get

$\displaystyle g^{{}^{\prime\prime}}(x)=-\frac{\mu
g^{{}^{\prime}}(x)}{\sigma^{2}a(x)}.$ (8.11)

Substituting (8.11) into (6.11), differentiating the resulting equation, and
using (8.11) again, we obtain

$\displaystyle
a^{{}^{\prime}}(x)=\frac{\mu^{2}+2c\sigma^{2}}{\mu\sigma^{2}}(1-\frac{u}{a(x)}),$
(8.12)

with

$\displaystyle u\equiv\frac{2\delta\mu}{\mu^{2}+2c^{2}\sigma^{2}}.$ (8.13)

Integrating (8.12), we get

$\displaystyle
G(a(x))=\frac{\mu^{2}+2c\sigma^{2}}{\mu\sigma^{2}}(x-x_{\alpha})+G(\alpha),$
(8.14)

where

$\displaystyle G(z)=z+ulog(z-u).$ (8.15)

Therefore

$\displaystyle
a(x)=G^{-1}(\frac{\mu^{2}+2c\sigma^{2}}{\mu\sigma^{2}}(x-x_{\alpha})+G(\alpha)).$
(8.16)

Obviously, $a(x)$ is increasing. Let $a(x_{\beta})=\beta$, we get

$\displaystyle x_{\beta}$ $\displaystyle=$
$\displaystyle\frac{\mu\sigma^{2}}{\mu^{2}+2c\sigma^{2}}[G(\beta)-G(\alpha)]+x_{\alpha}$
(8.17) $\displaystyle=$
$\displaystyle\frac{\mu\sigma^{2}}{\mu^{2}+2c\sigma^{2}}(\beta-\alpha)+\frac{\mu\sigma^{2}u}{\mu^{2}+2c\sigma^{2}}log(\frac{\beta-u}{\alpha-u})+x_{\alpha}.$

Solving (8.11),(8.15) and (8.16), we obtain

$\displaystyle
g(x)=g(x_{\alpha})+g^{{}^{\prime}}(x_{\alpha})\int_{x_{\alpha}}^{x}exp(-\frac{\mu}{\sigma^{2}}\int_{x_{\alpha}}^{y}\frac{dv}{a(v)})dy,\
\ x_{\alpha}\leq x<x_{\beta},$

where $g(x_{\alpha})$ and $g^{{}^{\prime}}(x_{\alpha})$ are free constants to
be determined. From (8.6) and (8.10), we deduce

$\displaystyle
g(x_{\alpha})=\frac{\alpha\mu-2\delta}{2c}g^{{}^{\prime}}(x_{\alpha})$ (8.19)

Let

$\displaystyle k_{2}\equiv g^{{}^{\prime}}(x_{\alpha}),$ (8.20)

Then (8.6) and (8.19) imply

$\displaystyle
k_{1}=\frac{\alpha\mu-2\delta}{2c(e^{r_{+}(\alpha)x_{\alpha}}-e^{r_{-}(\alpha)x_{\alpha}})}k_{2}$
(8.21)

Substituting (8.19) and (8.20) into (8), we get

$\displaystyle
g(x)=k_{2}[\frac{\alpha\mu-2\delta}{2c}+\int_{x_{\alpha}}^{x}exp(-\frac{\mu}{\sigma^{2}}\int_{x_{\alpha}}^{y}\frac{dv}{a(v)})dy],\
x_{\alpha}\leq x<x_{\beta}.$ (8.22)

Step 3. In view of Proposition 9 in [7], $a(x)\geq\beta$ holds for $x\geq
x_{\beta}$. Substituting $a=\beta$ into (6.11), and solving it, we get the
following solution

$\displaystyle
g(x)=k_{3}e^{r_{+}(\beta)(x-b_{0})}+k_{4}e^{r_{-}(\beta)(x-b_{0})},\
x_{\beta}\leq x<b,$ (8.23)

where $k_{3},k_{4}$ are free constants to be determined and $r_{\pm}(\beta)$
are given by (8.9). For $x\geq b$, the solution has the following form

$\displaystyle
g(x)=x-b+k_{3}e^{r_{+}(\beta)(b-b_{0})}+k_{4}e^{r_{-}(\beta)(b-b_{0})},\ x\geq
b.$ (8.24)

Next we apply the principle of smooth fit to determine the unknown constants
above. Note that

$\displaystyle\left\\{\begin{array}[]{l l l}g(x_{\beta}-)=g(x_{\beta}+)\\\
g^{{}^{\prime}}(x_{\beta}-)=g^{{}^{\prime}}(x_{\beta}+),\end{array}\right.$
(8.27)

we arrive at

$\displaystyle\left\\{\begin{array}[]{l l
l}k_{2}\xi=k_{3}e^{r_{+}(\beta)(x_{\beta}-b_{0})}+k_{4}e^{r_{-}(\beta)(x_{\beta}-b_{0})}\\\
k_{2}\eta=k_{3}r_{+}(\beta)e^{r_{+}(\beta)(x_{\beta}-b_{0})}+k_{4}r_{-}(\beta)e^{r_{-}(\beta)(x_{\beta}-b_{0})},\end{array}\right.$
(8.30)

where

$\displaystyle\left\\{\begin{array}[]{l l
l}\xi=\frac{\alpha\mu-2\delta}{2c}+\int_{x_{\alpha}}^{x_{\beta}}exp(-\frac{\mu}{\sigma^{2}}\int_{x_{\alpha}}^{y}\frac{dv}{a(v)})dy\\\
\eta=exp(-\frac{\mu}{\sigma^{2}}\int_{x_{\alpha}}^{x_{\beta}}\frac{dv}{a(v)}).\end{array}\right.$
(8.33)

Solving (8.30) for $k_{3}$ and $k_{4}$, we get

$\displaystyle\left\\{\begin{array}[]{l l l}k_{3}=\frac{\eta-\xi
r_{-}(\beta)}{(r_{+}(\beta)-r_{-}(\beta))e^{r_{+}(\beta)(x_{\beta}-b_{0})}}k_{2}\equiv
Ak_{2}\\\ k_{4}=\frac{\xi
r_{+}(\beta)-\eta}{(r_{+}(\beta)-r_{-}(\beta))e^{r_{-}(\beta)(x_{\beta}-b_{0})}}k_{2}\equiv
Bk_{2}.\end{array}\right.$ (8.36)

Substituting (8.36) into (8.23) and using $g^{{}^{\prime}}(b-)=1$, we obtain

$\displaystyle
k_{2}=\frac{1}{Ar_{+}(\beta)e^{r_{+}(\beta)(b-b_{0})}+Br_{-}(\beta)e^{r_{-}(\beta)(b-b_{0})}}.$
(8.37)

Thus, $k_{1},k_{2},k_{3},k_{4}$ are determined by (8.21),(8.36) and (8.37). We
claim that

$\displaystyle g^{{}^{\prime\prime}}(b_{-})\geq 0.$ (8.38)

In order to prove this statement, we consider $f(x)$ in Lemma 6.1 and notice
that $A$ and $B$ in (8.36) have the same expression both in $f(x)$ and $g(x)$.
Since $f^{{}^{\prime}}(b_{0})=1$ and $f^{{}^{\prime\prime}}(b_{0})=0$,

$\displaystyle\left\\{\begin{array}[]{l l
l}k^{f}_{2}(Ar_{+}(\beta)+Br_{-}(\beta))=1\\\
k^{f}_{2}(Ar^{2}_{+}(\beta)+Br^{2}_{-}(\beta))=0,\end{array}\right.$ (8.41)

where $k^{f}_{2}$ is the corresponding constant in Lemma 6.1. From (8.41), we
know that $A<0,B>0$ due to $r_{+}(\beta)>0,r_{-}(\beta)<0)$ and $k^{f}_{2}>0$
in $f(x)$. In addition, if we let

$\displaystyle l(b)\equiv
g^{{}^{\prime\prime}}_{3}(b-)=\frac{Ar^{2}_{+}(\beta)e^{r_{+}(\beta)(b-b_{0})}+Br^{2}_{-}(\beta)e^{r_{-}(\beta)(b-b_{0})}}{Ar_{+}(\beta)e^{r_{+}(\beta)(b-b_{0})}+Br_{-}(\beta)e^{r_{-}(\beta)(b-b_{0})}},$
(8.42)

then,

$\displaystyle\frac{\partial l}{\partial b}$ $\displaystyle=$
$\displaystyle\frac{AB(r^{3}_{+}(\beta)r_{-}(\beta)+r_{+}(\beta)r^{3}_{-}(\beta)-2r^{2}_{+}(\beta)r^{2}_{-}(\beta))e^{(r_{+}(\beta)+r_{(}\beta))(b-b_{0})}}{(Ar_{+}(\beta)e^{r_{+}(\beta)(b-b_{0})}+Br_{-}(\beta)e^{r_{-}(\beta)(b-b_{0})})^{2}}$
$\displaystyle>$ $\displaystyle 0$

holds for $A>0,B<0,r_{+}(\beta)>0$ and $r_{-}(\beta)<0$. Since $b>b_{0}$, we
conclude that

$\displaystyle
g^{{}^{\prime\prime}}(b-)=l(b)>l(b_{0})=\tilde{g}^{{}^{\prime\prime}}(b_{0})=0,$
(8.44)

where $\tilde{g}(x)$ is the solution of (6.11) with replacing $b$ by $b_{0}$.
Step 4. Now we only need to prove the solution $g(x)$ satisfies (6.11). It
suffices to prove the following conditions,

$\displaystyle\max\limits_{a\in[\alpha,\beta]}[\frac{1}{2}\sigma^{2}a^{2}g^{{}^{\prime\prime}}(x)+(\mu
a-\delta)g^{{}^{\prime}}(x)-cg(x)]=0,\ \mbox{for}\ x\geq b.$

By (6.11), (8.44) and noticing that $g^{{}^{\prime}}(b-)=1$, we get

$\displaystyle\max\mathcal{L}g(x)$ $\displaystyle=$ $\displaystyle\mu
a-\delta-c(x-b+g(b))$ (8.45) $\displaystyle\leq$ $\displaystyle\mu a-\delta-
cg(b)$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}\sigma^{2}a^{2}g^{{}^{\prime\prime}}(b-)+(\mu
a-\delta)g^{{}^{\prime}}_{3}(b-)-cg(b)$ $\displaystyle\leq$ $\displaystyle 0.$

Thus, we complete the proof. We summarize the solution as follows. For
$\frac{2\delta}{\mu}<\alpha$,

$\displaystyle g(x)=\left\\{\begin{array}[]{l l
l}k_{1}(e^{r_{+}(\alpha)x}-e^{r_{-}(\alpha)x}),\ \ 0\leq x<x_{\alpha},\\\
k_{2}[\frac{\alpha\mu-2\delta}{2c}+\int_{x_{\alpha}}^{x}\exp(-\frac{\mu}{\sigma^{2}}\int_{x_{\alpha}}^{y}\frac{dv}{a(v)})dy],\
\ x_{\alpha}\leq x<x_{\beta},\\\
k_{3}e^{r_{+}(\beta)(x-b_{0})}+k_{4}e^{r_{-}(\beta)(x-b_{0})},\ \
x_{\beta}\leq x<b,\\\
x-b+k_{3}e^{r_{+}(\beta)(b-b_{0})}+k_{4}e^{r_{-}(\beta)(b-b_{0})},\ \ x\geq
b,\end{array}\right.$ (8.50)

where $r_{\pm}(x),x_{\alpha},x_{\beta},k_{1},k_{2},k_{3}$ and $k_{4}$ are
defined by (8.9), (8.10), (8.17), (8.21), (8.37) and (8.36), respectively.
$\Box$ Proof of lemma 6.3. For $b\geq b_{0},A<0,B>0$, together with (8.42) and
(8), we have

$\displaystyle\frac{\partial}{\partial b}g(b,x)$ $\displaystyle=$
$\displaystyle-\frac{(\alpha\mu-2\delta)}{2c(e^{r_{+}(\alpha)x_{\alpha}-r_{-}(\alpha)x_{\alpha}})}$
$\displaystyle\times\frac{Ar^{2}_{+}(\beta)e^{r_{+}(\beta)(b-b_{0})}+Br^{2}_{-}(\beta)e^{r_{-}(\beta)(b-b_{0})}}{(Ar_{+}(\beta)e^{r_{+}(\beta)(b-b_{0})}+Br_{-}(\beta)e^{r_{-}(\beta)(b-b_{0})})^{2}}$
$\displaystyle\leq$ $\displaystyle 0,\ \ \ 0\leq x<x_{\alpha};$
$\displaystyle\frac{\partial}{\partial b}g(b,x)$ $\displaystyle=$
$\displaystyle-(\frac{\alpha\mu-2\delta}{2c}+\int_{x_{\alpha}}^{x}exp(-\frac{\mu}{\sigma^{2}}\int_{x_{\alpha}}^{y}\frac{dv}{a(v)})dy)$
$\displaystyle\times\frac{Ar^{2}_{+}(\beta)e^{r_{+}(\beta)(b-b_{0})}+Br^{2}_{-}(\beta)e^{r_{-}(\beta)(b-b_{0})}}{(Ar_{+}(\beta)e^{r_{+}(\beta)(b-b_{0})}+Br_{-}(\beta)e^{r_{-}(\beta)(b-b_{0})})^{2}}$
$\displaystyle\leq$ $\displaystyle 0,\ \ \ x_{\alpha}\leq x<x_{\beta};$
$\displaystyle\frac{\partial}{\partial b}g(b,x)$ $\displaystyle=$
$\displaystyle(Ae^{r_{+}(\beta)(b-b_{0})}+Be^{r_{-}(\beta)(b-b_{0})})$
$\displaystyle\times\frac{(Ar^{2}_{+}(\beta)e^{r_{+}(\beta)(b-b_{0})}+Br^{2}_{-}(\beta)e^{r_{-}(\beta)(b-b_{0})})}{(Ar_{+}(\beta)e^{r_{+}(\beta)(b-b_{0})}+Br_{-}(\beta)e^{r_{-}(\beta)(b-b_{0})})^{2}}$
$\displaystyle\leq$ $\displaystyle 0,\ \ \ x_{\beta}\leq x<b;$
$\displaystyle\frac{\partial}{\partial b}g(b,x)$ $\displaystyle=$
$\displaystyle(Ae^{r_{+}(\beta)(b-b_{0})}+Be^{r_{-}(\beta)(b-b_{0})})$
$\displaystyle\times\frac{(Ar^{2}_{+}(\beta)e^{r_{+}(\beta)(b-b_{0})}+Br^{2}_{-}(\beta)e^{r_{-}(\beta)(b-b_{0})})}{(Ar_{+}(\beta)e^{r_{+}(\beta)(b-b_{0})}+Br_{-}(\beta)e^{r_{-}(\beta)(b-b_{0})})^{2}}$
$\displaystyle\leq$ $\displaystyle 0,\ \ \ x\geq b.$

Thus, the proof is completed. $\Box$ Proof of lemma 6.4. We can prove that
$\mathbf{P}[\tau_{b}^{\pi^{*}_{b}}\leq T]$ is decreasing in $b$ along the
lines of Theorem 3.1 in [17](2008). Here we only need to prove that
$\mathbf{P}[\tau_{b}^{\pi^{*}_{b}}\leq T]$ is strictly decreasing in $b$ on
$[x_{\beta},D)$. We denote $\mathbf{P}[\tau_{b}^{\pi^{*}_{b}}\leq T]$ by
$\mathbf{P}[\tau_{b}^{b}\leq T]$ for simplicity. For any $b_{1},b_{2}$
satisfying $D\geq b_{2}\geq b_{1}\geq x_{\beta}$, we need to prove that

$\displaystyle\mathbf{P}[\tau_{b_{1}}^{b_{1}}\leq
T]-\mathbf{P}[\tau_{b_{2}}^{b_{2}}\leq T]>0$

By comparison theorem, we have

$\mathbf{P}[\tau_{b_{1}}^{b_{1}}\leq T]-\mathbf{P}[\tau_{b_{2}}^{b_{2}}\leq
T]\geq\mathbf{P}[\tau_{b_{1}}^{b_{2}}\leq
T]-\mathbf{P}[\tau_{b_{2}}^{b_{2}}\leq T].$

So we the proof can be reduced to proving that

$\displaystyle\mathbf{P}[\tau_{b_{1}}^{b_{2}}\leq
T]-\mathbf{P}[\tau_{b_{2}}^{b_{2}}\leq T]>0.$ (8.52)

Define stochastic processes $R_{t}^{(1)}$, $R_{t}^{(2)}$, $R_{t}^{(3)}$,
$R_{t}^{(4)}$ by the following SDEs:

$\displaystyle dR_{t}^{(1)}$ $\displaystyle=$ $\displaystyle[\mu
a_{b_{2}}^{*}(R_{t}^{(1)})-\delta]dt+a_{b_{2}}^{*}(R_{t}^{(1)})\sigma
d\mathcal{W}_{t}-dL_{t}^{b_{2}},R_{0}^{(1)}=b_{1};$ $\displaystyle
dR_{t}^{(2)}$ $\displaystyle=$ $\displaystyle[\mu
a_{b_{2}}^{*}(R_{t}^{(2)})-\delta]dt+a_{b_{2}}^{*}(R_{t}^{(2)})\sigma
d\mathcal{W}_{t}-dL_{t}^{b_{2}},R_{0}^{(2)}=b_{2};$ $\displaystyle
dR_{t}^{(3)}$ $\displaystyle=$ $\displaystyle[\mu
a_{b_{2}}^{*}(R_{t}^{(3)})-\delta]dt+a_{b_{2}}^{*}(R_{t}^{(3)})\sigma
d\mathcal{W}_{t}-dL_{t}^{b_{2}},R_{0}^{(3)}=\frac{b_{1}+b_{2}}{2};$
$\displaystyle dR_{t}^{(4)}$ $\displaystyle=$ $\displaystyle[\mu
a_{b_{2}}^{*}(R_{t}^{[4]})-\delta]dt+a_{b_{2}}^{*}(R_{t}^{(4)})\sigma
d\mathcal{W}_{t},R_{0}^{(4)}=\frac{b_{1}+b_{2}}{2},$

respectively, where $D\geq b_{2}\geq b_{1}\geq x_{\beta}$ and $a^{*}(\cdot)$
is defined by (6.25). First, let $\tau^{b_{1}}=\inf\limits_{t\geq
0}\\{t:R_{t}^{(2)}=b_{1}\\}$, $A=\\{\tau^{b_{1}}\leq T\\}$, $B=\big{\\{}\sl
R_{t}^{(2)}$ will go to bankruptcy in a time interval
$[\tau^{b_{1}},\tau^{b_{1}}+T]$ and $\tau^{b_{1}}\leq T\big{\\}}$,
$D=\\{\inf\limits_{0\leq t\leq T}R_{t}^{(3)}>b_{1}\\}$ and
$E=\\{\inf\limits_{0\leq t\leq T}R_{t}^{(4)}>b_{1},\sup\limits_{0\leq t\leq
T}R_{t}^{(4)}<b_{2}\\}$. Then

$\displaystyle\\{\tau_{b_{2}}^{b_{2}}\leq T\\}\subset B\subset A.$ (8.53)

Moreover, by using strong Markov property of $R_{t}^{[2]}$, we have

$\displaystyle\mathbf{P}[\tau_{b_{1}}^{b_{2}}\leq T]=\mathbf{P}[B|A].$ (8.54)

By comparison theorem on SDE, we have

$\displaystyle\mathbf{P}(A^{c})\geq\mathbf{P}(D)\geq\mathbf{P}(E).$ (8.55)

Since $a_{b_{2}}^{*}(x)=\beta$ we have

$\displaystyle
R_{t}^{(4)}=\frac{b_{1}+b_{2}}{2}+[\mu\beta-\delta]t+\sigma\mathcal{W}_{t}\
\mbox{ on $E$}.$ (8.56)

We deduce from (8.56) and properties of Brownian motion with drift (cf.
Andrei,Borodin,Paavo,Salminen [1] (2002)) that

$\displaystyle\mathbf{P}(E)=\frac{e^{-\mu^{\prime 2}T/2}}{\sqrt{2\pi
T}}\sum_{k=-\infty}^{\infty}\int_{b_{1}/\sigma\beta}^{b_{2}/\sigma\beta}e^{\mu^{\prime}(z-x)}[(e^{-(z-x+\frac{2k(b_{2}-b_{1})}{\sigma\beta})^{2}/2T})$
$\displaystyle-(e^{-(z+x-\frac{2b_{1}-2k(b_{2}-b_{1})}{\sigma\beta})^{2}/2T})]dz>0,$

where $\mu^{\prime}=(\beta\mu-\delta)/\sigma$ and
$x=\frac{b_{1}+b_{2}}{2\sigma\beta}$. Thus we get

$\displaystyle\mathbf{P}(A^{c})>0.$ (8.57)

We also know from Theorem 4.1 that $\mathbf{P}[\tau_{b_{1}}^{b_{2}}\leq
T]\geq\mathbf{P}[\tau_{b_{2}}^{b_{2}}\leq T]>0$, which together with (8.53),
(8.54) and (8.57), implies that

$\displaystyle\mathbf{P}[\tau_{b_{1}}^{b_{2}}\leq
T]-\mathbf{P}[\tau_{b_{2}}^{b_{2}}\leq T]$ $\displaystyle\geq$
$\displaystyle\mathbf{P}[\tau_{b_{1}}^{b_{2}}\leq T]-\mathbf{P}(B)$
$\displaystyle=$ $\displaystyle\mathbf{P}[\tau_{b_{1}}^{b_{2}}\leq
T]-\mathbf{P}(A)\mathbf{P}(B|A)$ $\displaystyle=$
$\displaystyle\mathbf{P}[\tau_{b_{1}}^{b_{2}}\leq T](1-\mathbf{P}(A))$
$\displaystyle=$ $\displaystyle\mathbf{P}[\tau_{b_{1}}^{b_{2}}\leq
T]\mathbf{P}(A^{c})$ $\displaystyle>$ $\displaystyle 0.$

Thus the proof is completed. $\Box$

Proof of lemma 6.5. Let $\phi(t,x)\equiv\phi^{b}(t,x)$. Setting
$\tau_{x}^{b}:=\tau_{x}^{\pi^{*}_{b}}$ and applying the generalized Itô
formula to $(R^{\pi_{b}^{\ast},x}_{t},L_{t}^{\pi_{b}^{\ast}})$ and $\phi(t,x)$
we have for $0<x\leq b$,

$\displaystyle\phi(T-(t\wedge\tau_{x}^{b}),R^{\pi_{b}^{\ast},x}_{t\wedge\tau_{x}^{b}})$
$\displaystyle=$ $\displaystyle\phi(T,x)$ $\displaystyle+$
$\displaystyle\int_{0}^{t\wedge\tau_{x}^{b}}(\frac{1}{2}a^{*2}(R^{\pi_{b}^{\ast},x}_{s})\sigma^{2}\phi_{yy}(T-s,R^{\pi_{b}^{\ast},x}_{s})$
$\displaystyle+$
$\displaystyle(a^{*}_{b}(R^{\pi_{b}^{\ast},x}_{s})\mu-\delta)\phi_{x}(T-s,R^{\pi_{b}^{\ast},x}_{s})$
$\displaystyle-$
$\displaystyle\phi_{t}(T-s,R^{\pi_{b}^{\ast},x}_{s}))ds-\int_{0}^{t\wedge\tau_{x}^{b}}\phi_{y}(T-s,R^{\pi_{b}^{\ast},x}_{s})dL_{s}^{b}$
$\displaystyle+$
$\displaystyle\int_{0}^{t\wedge\tau_{x}^{b}}a^{*}(R^{\pi_{b}^{\ast},x}_{s})\sigma\phi_{x}(T-s,R^{\pi_{b}^{\ast},x}_{s})dW_{s}.$

Setting $t=T$ and taking mathematical expectation at both sides of (8) yield
that

$\displaystyle\phi(T,x)$ $\displaystyle=$
$\displaystyle\mathbf{E}[\phi(T-(T\wedge\tau_{x}^{b}),R^{\pi_{b}^{\ast},x}_{T\wedge\tau_{x}^{b}})]$
$\displaystyle=$
$\displaystyle\mathbf{E}[\phi(0,R^{\pi_{b}^{\ast},x}_{T})1_{T<\tau_{x}^{b}}]+\mathbf{E}[\phi(T-\tau_{x}^{b},0)1_{T\geq\tau_{x}^{b}})]$
$\displaystyle=$ $\displaystyle\mathbf{E}[1_{T<\tau_{x}^{b}}]$
$\displaystyle=$ $\displaystyle 1-\psi(T,x).$

Thus we complete the proof. $\Box$

Define $u(x):=\frac{1}{2}a^{*2}(x)\sigma^{2}$, $v(x):=a^{*2}(x)\mu-\delta$,
the equation (4.13) becomes

$\displaystyle\phi_{t}^{b}(t,x)=u(x)\phi_{xx}^{b}(t,x)+v(x)\phi_{x}^{b}(t,x).$
(8.59)

Obviously, $u(x)$ and $v(x)$ are continuous in $[0,b]$ due to the fact that
$a^{*}(x)$ is continuous w.r.t $x$. Thus there exists a unique solution in
$C^{1}(0,\infty)\cap C^{2}(0,b)$ for (6.29). Moreover, $u^{{}^{\prime}}(x)$,
$v^{{}^{\prime}}(x)$, $u^{{}^{\prime\prime}}(x)$ are bounded in
$(0,x_{\alpha})$, $(x_{\alpha},x_{\beta})$, $(x_{\beta},b)$, respectively. Now
we are ready to prove that the bankrupt probability $\psi^{b}(T,b)$ is
continuous with respect to the dividends barrier $b(b\geq b_{0})$. Proof of
lemma 6.6. It suffices to prove that $\phi^{b}(t,x)$ is continuous in b. Let
$x=by$ and $\theta^{b}(t,y)=\phi^{b}(t,by)$, the equation (6.29) becomes

$\displaystyle\left\\{\begin{array}[]{l l
l}\theta_{t}^{b}(t,y)=[u(by)/b^{2}]\theta_{yy}^{b}(t,z)+[v(by)/b]\theta_{y}^{b}(t,y),\\\
\theta^{b}(0,y)=1,\ \mbox{for}\ 0<y\leq 1,\\\
\theta^{b}(t,0)=0,\theta_{y}^{b}(t,1)=0,\ \mbox{for}\ t>0.\end{array}\right.$
(8.63)

So the proof of Lemma 6.6 can be reduced to proving
$\lim\limits_{b_{2}\rightarrow b_{1}}\theta^{b_{2}}(t,1)=\theta^{b_{1}}(t,1)$
for fixed $b_{1}>b_{0}$. Setting
$w(t,y)=\theta^{b_{2}}(t,y)-\theta^{b_{1}}(t,y)$ and noticing that
$\theta^{b}(t,y)$ is continuous at $y=1$ for any $b\geq b_{0}$, it suffices to
show that

$\displaystyle\int_{0}^{t}\int_{0}^{1}w^{2}(s,y)dyds\rightarrow 0,\ \mbox{as}\
b_{2}\rightarrow b_{1}.$ (8.64)

Thus we have

$\displaystyle\left\\{\begin{array}[]{l l
l}w_{t}(t,y)&=&[u(b_{2}y)/b_{2}^{2}]w_{yy}(t,y)+[v(b_{2}y)/b_{2}]w_{y}(t,y)\\\
&+&\\{u(b_{2}y)/b_{2}^{2}-u(b_{1}z)/b_{1}^{2}\\}\theta_{yy}^{b_{1}}(t,y)\\\
&+&\\{u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}\\}\theta_{y}^{b_{1}}(t,y),\\\
w(0,y)&=&0,\ \mbox{for}\ 0<y\leq 1,\\\ w(t,0)&=&0,\ w_{y}(t,1)=0,\ \mbox{for}\
t>0.\end{array}\right.$ (8.70)

Multiplying the first equation in (8.70) by $w(t,y)$ and then integrating on
$[0,1]\times[0,t]$,

$\displaystyle\int_{0}^{t}\int_{0}^{1}w(s,y)w_{t}(s,y)dyds$ (8.71)
$\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}\big{\\{}[u(b_{2}y)/b_{2}^{2}]w(s,y)w_{yy}(s,y)$
$\displaystyle+\int_{0}^{t}\int_{0}^{1}[v(b_{2}y)/b_{2}]w(s,y)w_{y}(s,y)$
$\displaystyle+\int_{0}^{t}\int_{0}^{1}[u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}]w(s,y)\theta_{yy}^{b_{1}}(t,y)$
$\displaystyle+\int_{0}^{t}\int_{0}^{1}w(s,y)[v(b_{2}y)/b_{2}-v(b_{1}y)/b_{1}]w(s,y)\theta_{y}^{b_{1}}(t,y)\big{\\}}dyds$
$\displaystyle\equiv$ $\displaystyle E_{1}+E_{2}+E_{3}+E_{4}.$

We now estimate each terms at both sides of (8.71) as follows.

$\displaystyle\int_{0}^{t}\int_{0}^{1}w(s,y)w_{t}(s,y)dyds$ $\displaystyle=$
$\displaystyle\int_{0}^{1}\frac{1}{2}w^{2}(t,y)dy.$ (8.72)

By property of $a^{*}(x)$ and the definition of $u(x)$ and $v(x)$ there exit
positive constants $D_{1}$, $D_{2}$ and $D_{3}$ such that
$[v(b_{2}y)/b_{2}]^{2}\leq D_{1}$, $[u(by)/b^{2}]^{\prime}\geq 0$,
$[u(b_{2}y)/b_{2}^{2}]\geq D_{2}$ and $[a(b_{2}y)/b_{2}^{2}]^{\prime}\leq
D_{3}$. For any $\lambda_{1}>0$ and $\lambda_{2}>0$, by these facts and
Young’s inequality, we estimate $E_{1}$ and $E_{2}$ as follows,

$\displaystyle E_{1}$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}[u(b_{2}y)/b_{2}^{2}]w(s,y)w_{yy}(s,y)dyds$
(8.73) $\displaystyle=$
$\displaystyle-\int_{0}^{t}\int_{0}^{1}[u(b_{2}y)/b_{2}^{2}]w_{y}^{2}(s,y)dyds$
$\displaystyle-\int_{0}^{t}\int_{x_{\alpha}/b_{2}}^{x_{\beta}/b_{2}}[u(b_{2}y)/b_{2}^{2}]^{{}^{\prime}}w_{y}(s,y)w(s,y)dyds$
$\displaystyle\leq$ $\displaystyle-
D_{2}\int_{0}^{t}\int_{0}^{1}w_{y}^{2}(s,y)dyds$
$\displaystyle+D_{3}\int_{0}^{t}\int_{0}^{1}[\lambda_{1}w_{y}^{2}(s,y)+\frac{1}{4\lambda_{1}}w^{2}(s,y)]dyds,$
$\displaystyle E_{2}$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}[v(b_{2}y)/b_{2}]w(s,y)w_{y}(s,y)dyds$
(8.74) $\displaystyle\leq$
$\displaystyle\lambda_{2}\int_{0}^{t}\int_{0}^{1}w_{y}^{2}(s,y)dyds$
$\displaystyle+\frac{D_{1}}{4\lambda_{2}}\int_{0}^{t}\int_{0}^{1}w^{2}(s,y)dyds.$

It is easy to see that $[u(by)/b^{2}]$, $[u(by)/b^{2}]^{\prime}$and
$[v(by)/b]$ are Lipschitz continuous for all
$y\in(x_{\alpha}/b_{2},x_{\beta}/b_{1})$, that is, there exists $L>0$ such
that

$\displaystyle|[u(b_{2}y)/b_{2}^{2}]-[u(b_{1}y)/b_{1}^{2}]|\leq
L|b_{2}-b_{1}|,$
$\displaystyle|[u(b_{2}y)/b_{2}^{2}]^{\prime}-[u(b_{1}y)/b_{1}^{2}]^{\prime}|\leq
L|b_{2}-b_{1}|,$ $\displaystyle|[v(b_{2}y)/b_{2}]-[v(b_{1}y)/b_{1}]|\leq
L|b_{2}-b_{1}|.$ (8.75)

$E_{3}$ has the following expressions,

$\displaystyle E_{3}$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}\\{u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}\\}w(s,y)\theta_{yy}^{b_{1}}(s,y)dyds$
(8.76) $\displaystyle=$
$\displaystyle-\int_{0}^{t}\int_{0}^{1}[u(b_{2}y)/b_{2}^{2}u(b_{1}y)/b_{1}^{2}]w_{y}(s,y)\theta_{y}^{b_{1}}(s,y)dyds$
$\displaystyle-\int_{0}^{t}\int_{0}^{x_{\alpha}/b_{2}}[u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}]^{\prime}w(s,y)\theta_{y}^{b_{1}}(s,y)dyds$
$\displaystyle-\int_{0}^{t}\int_{x_{\alpha}/b_{2}}^{x_{\alpha}/b_{1}}[u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}]^{\prime}w(s,y)\theta_{y}^{b_{1}}(s,y)dyds$
$\displaystyle-\int_{0}^{t}\int_{x_{\alpha}/b_{1}}^{x_{\beta}/b_{2}}[u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}]^{\prime}w(s,y)\theta_{y}^{b_{1}}(s,y)dyds$
$\displaystyle-\int_{0}^{t}\int_{x_{\beta}/b_{2}}^{x_{\beta}/b_{1}}[u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}]^{\prime}w(s,y)\theta_{y}^{b_{1}}(s,y)dyds$
$\displaystyle-\int_{0}^{t}\int_{x_{\beta}/b_{1}}^{1}[u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}]^{\prime}w(s,y)\theta_{y}^{b_{1}}(s,y)dyds$
$\displaystyle=$
$\displaystyle-\int_{0}^{t}\int_{0}^{1}[u(b_{2}y)/b_{2}^{2}u(b_{1}y)/b_{1}^{2}]w_{y}(s,y)\theta_{y}^{b_{1}}(s,y)dyds$
$\displaystyle-\int_{0}^{t}\int_{x_{\alpha}/b_{2}}^{x_{\alpha}/b_{1}}[u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}]^{\prime}w(s,y)\theta_{y}^{b_{1}}(s,y)dyds$
$\displaystyle-\int_{0}^{t}\int_{x_{\alpha}/b_{1}}^{x_{\beta}/b_{2}}[u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}]^{\prime}w(s,y)\theta_{y}^{b_{1}}(s,y)dyds$
$\displaystyle-\int_{0}^{t}\int_{x_{\beta}/b_{2}}^{x_{\beta}/b_{1}}[u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}]^{\prime}w(s,y)\theta_{y}^{b_{1}}(s,y)dyds$
$\displaystyle=$ $\displaystyle E_{31}+E_{32}+E_{33}+E_{34}.$

By (8) and (8.76)and Young’s inequality for any $\lambda_{3}>0$ and
$\lambda_{4}>0$,

$\displaystyle E_{31}$ $\displaystyle=$
$\displaystyle-\int_{0}^{t}\int_{0}^{1}[u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}]w(s,y)\theta_{y}^{b_{1}}(s,y)dyds$
(8.77) $\displaystyle\leq$
$\displaystyle\frac{L^{2}(b_{2}-b_{1})^{2}}{4\lambda_{3}}\int_{0}^{t}\int_{0}^{1}[\theta_{y}^{b_{1}}(s,y)]^{2}dyds$
$\displaystyle+\lambda_{3}\int_{0}^{t}\int_{0}^{1}[w_{y}^{2}(s,y)+w^{2}(s,y)]dyds$

and

$\displaystyle E_{33}$ $\displaystyle=$
$\displaystyle-\int_{0}^{t}\int_{0}^{m/b_{2}}[u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}]^{\prime}w(s,y)\theta_{y}^{b_{1}}(s,y)dyds$
(8.78) $\displaystyle\leq$
$\displaystyle\frac{L^{2}(b_{2}-b_{1})^{2}}{4\lambda_{4}}\int_{0}^{t}\int_{0}^{1}[\theta_{y}^{b_{1}}(s,y)]^{2}dyds$
$\displaystyle+\lambda_{4}\int_{0}^{t}\int_{0}^{1}[w_{y}^{2}(s,y)+w^{2}(s,y)]dyds.$

There exists a constant $D_{4}>0$ such that
$|[u(by)/b^{2}]^{\prime}-[v(by)/b]|\leq D_{4}$ and
$\lambda_{5}=\inf\limits_{b_{1}\leq b\leq b_{2}}\\{u(by)/b^{2}\\}>0$. Then by
the boundary conditions we estimate
$\int_{0}^{t}\int_{0}^{1}[\theta_{y}^{b}(s,y)]^{2}dyds$ for
$b\in[b_{1},b_{2}]$ as follows,

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}\theta_{t}^{b}(s,y)\theta^{b}(s,y)$
(8.79)
$\displaystyle-[u(by)/b^{2}]\theta_{yy}^{b}(s,y)\theta^{b}(s,y)-[v(by)/b]\theta_{y}^{b}(s,y)\theta^{b}(s,y)dyds$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{0}^{1}[\theta^{b}(s,y)]^{2}dy+\int_{0}^{t}\int_{0}^{1}[u(by)/b^{2}][\theta_{y}^{b}(s,y)]^{2}dyds$
$\displaystyle+\int_{0}^{t}\int_{0}^{1}[u(by)/b^{2}]^{\prime}\theta_{y}^{b}(s,y)\theta^{b}(s,y)dyds$
$\displaystyle-\int_{0}^{t}\int_{0}^{1}[v(by)/b]\theta_{y}^{b}(s,y)\theta^{b}(s,y)dyds$
$\displaystyle\geq$
$\displaystyle\lambda_{5}\int_{0}^{t}\int_{0}^{1}[\theta_{y}^{b}(s,y)]^{2}dyds-\frac{\lambda_{5}}{2}\int_{0}^{t}\int_{0}^{1}[\theta_{y}^{b}(s,y)]^{2}dyds$
$\displaystyle-\frac{1}{2\lambda_{5}}\int_{0}^{t}\int_{0}^{1}[\theta^{b}(s,y)]^{2}dyds$
$\displaystyle\geq$
$\displaystyle\frac{\lambda_{5}}{2}\int_{0}^{t}\int_{0}^{1}[\theta_{y}^{b}(s,y)]^{2}dyds-\frac{D_{4}}{2\lambda_{5}}$

from which we deduce that

$\displaystyle\int_{0}^{t}\int_{0}^{1}[\theta_{y}^{b}(s,y)]^{2}dyds\leq\frac{D_{4}}{\lambda_{5}^{2}}.$
(8.80)

Therefore we conclude that
$\int_{0}^{t}\int_{0}^{1}[\theta_{y}^{b}(s,y)]^{2}dyds$ is bounded. Noticing
that $w(s,y)\leq 2$ and

$\displaystyle\lim\limits_{b_{2}\rightarrow b_{1}}\\{|E_{32}|+|E_{34}|\\}=0,$
(8.81)

as well as using the equalities (8.76)-(8.81), there exists a positive
function $B_{1}^{b_{1}}(b_{2})$ such that

$\displaystyle\lim\limits_{b_{2}\rightarrow b_{1}}B_{1}^{b_{1}}(b_{2})=0$

and for $0\leq t\leq T$

$\displaystyle E_{3}$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}[u(b_{2}y)/b_{2}^{2}-u(b_{1}y)/b_{1}^{2}]w(s,y)\theta_{yy}^{b_{1}}(t,y)dyds$
$\displaystyle\leq$ $\displaystyle
B_{1}^{b_{1}}(b_{2})+(\lambda_{3}+\lambda_{4})\int_{0}^{t}\int_{0}^{1}[w_{y}^{2}(s,y)+w^{2}(s,y)]dyds.$

By the same way as in estimating $E_{3}$ we also find a positive function
$B_{2}^{b_{1}}(b_{2})$ such that

$\displaystyle\lim\limits_{b_{2}\rightarrow b_{1}}B_{2}^{b_{1}}(b_{2})=0$

and for any $\lambda_{6}>0$

$\displaystyle E_{4}$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}[v(b_{2}y)/b_{2}-v(b_{1}y)/b_{1}]w(s,y)\theta_{y}^{b_{1}}(s,y)dyds$
(8.83) $\displaystyle\leq$
$\displaystyle\frac{L^{2}(b_{2}-b_{1})^{2}}{4\lambda_{6}}\int_{0}^{t}\int_{0}^{1}[\theta_{y}^{b_{1}}(s,y)]^{2}dyds+\lambda_{6}\int_{0}^{t}\int_{0}^{1}w^{2}(s,y)dyds$
$\displaystyle\leq$ $\displaystyle
B_{2}^{b_{1}}(b_{2})+\lambda_{6}\int_{0}^{t}\int_{0}^{1}w^{2}(s,y)dyds.$

Choosing $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$ and $\lambda_{4}$ small
enough such that $\lambda_{1}D_{3}+\lambda_{2}+\lambda_{3}+\lambda_{4}\leq
D_{2}$, we deduce from (8.71)-(8.74), (8) and (8.83) that there exist positive
constants $C_{1}$ and $C_{2}$ such that

$\displaystyle\int_{0}^{1}w^{2}(t,y)dy\leq
C_{1}\int_{0}^{t}\int_{0}^{1}w^{2}(s,y)dyds+C_{2}[B_{1}^{b_{1}}(b_{2})+B_{2}^{b_{1}}(b_{2})].$

Setting $F(t)=\int_{0}^{t}\int_{0}^{1}w^{2}(s,y)dyds$ and using the Gronwall
inequality,

$\displaystyle F(t)\leq
C_{2}[B_{1}^{b_{1}}(b_{2})+B_{2}^{b_{1}}(b_{2})]\exp\\{C_{1}t\\}.$

So

$\displaystyle\lim\limits_{b_{2}\rightarrow
b_{1}}\int_{0}^{t}\int_{0}^{1}[\theta^{b_{2}}(s,y)-\theta^{b_{1}}(s,y)]^{2}dyds=0.$

Thus we complete the proof. $\Box$ Acknowledgements. This work is supported by
Project 10771114 of NSFC, Project 20060003001 of SRFDP, the SRF for ROCS, SEM
and the Korea Foundation for Advanced Studies. We would like to thank the
institutions for the generous financial support. Special thanks also go to the
participants of the seminar stochastic analysis and finance at Tsinghua
University for their feedbacks and useful conversations.

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