# Optimization of dividend and reinsurance strategies under ruin probability
constraint

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

This paper considers nonlinear regular-singular stochastic optimal control of
large insurance company. The company controls the reinsurance rate and
dividend payout process to maximize the expected present value of the dividend
pay-outs until the time of bankruptcy. However, if the optimal dividend
barrier is too low to be acceptable, it will make the company result in
bankruptcy soon. Moreover, although risk and return should be highly
correlated, over-risking is not a good recipe for high return, the supervisors
of the company have to impose their preferred risk level and additional charge
on firm seeking services beyond or lower than the preferred risk level. These
indeed are nonlinear regular-singular stochastic optimal problems under ruin
probability constraints. This paper aims at solving this kind of the optimal
problems, that is, deriving the optimal retention ratio,dividend payout level,
optimal return function and optimal control strategy of the insurance company.
As a by-product, the paper also sets a risk-based capital standard to ensure
the capital requirement of can cover the total given risk, and the effect of
the risk level on optimal retention ratio, dividend payout level and optimal
control strategy are also presented. MSC(2000): Primary 91B30,91B70,93E20;
Secondary 60H30, 60H10. Keywords: Nonlinear regular-singular stochastic
optimal control; Ruin probability ; Optimal retention ratio; Optimal dividend
payout; Optimal return function.

##  1\. Introduction

In the present paper we consider nonlinear stochastic optimal control of
insurance company. The company controls the reinsurance rate and dividend
payout process to maximize the expected present value of the dividend pay-outs
until the time of bankruptcy. It is well known that over-risking is not a good
recipe for high return although risk and return should be highly correlated.
In fact, to reduce the risk, a risk-averse re-insurers may have their
preferred risk level and impose additional service charge on firms seeking
services beyond the target level, other re-insurers may demand additional
charges for those seeking services with risk level lower than its preferred
level as an aggressive move to gain market shares. This indeed is nonlinear
regular-singular stochastic optimal problem. The objective of the company is
to find a strategy, consisting of optimal retention ratio and dividend payment
scheme, which maximizes the expected total discounted dividend pay-outs until
the time of bankruptcy. This is a mixed regular-singular control problem on
diffusion model which has been a renewed interest recently,We refer readers to
He, Liang and et al. [12, 13, 14](2008,2009)and references therein, Højgaard
and Taksar[16, 17, 15](1999, 1998, 2001),Asmussen et all[2, 3](1997,2000),
Taksar[28](2000), Guo Xin, Liu Jun and Zhou Xunyu[11](2004), Harrison and
Taksar[19](1983), Paulsen and Gjessing [24](1997), and Radner and
Sheep[25](1996),and other authors’ works. Recent surveys can be found in
Avanzi [4].

However, we notice that the optimal dividend barrier in the nonlinear regular-
singular stochastic optimal problem may be so low that it would make the
company result in bankruptcy soon( see theorem 4.1), the company may reject
this optimal control strategy and may be prohibited to pay dividend at such a
low barrier because the insurance company is a business affected with a public
interest, and insureds and policy-holders should be protected against insurer
insolvencies (see Williams and Heins[30](1985), Riegel and Miller[26](1963),
and Welson and Taylor[29](1959)). The strategy, making the company go bankrupt
before termination of contract between insurer and policy holders or the
strategy of low solvency(1 minus ruin probability(see [5])), is not the best
way and should be prohibited even though it can win the highest profit. So the
supervisor of the company will impose some constraints on its ruin probability
and find the best equilibrium strategy between making profit and improving
security. These are turned out to be nonlinear regular-singular stochastic
optimal problems under low ruin probability constraint. This paper aims at
solving these kinds of stochastic optimal problems

Unfortunately, there are very few results concerning on these kinds of optimal
control problem with lower ruin probability and higher security. He, Hou and
Liang[13](2008) investigated the optimal control problem for linear Brownian
model, Paulsen[22](2003) and Taksar and Markussen [6](2004) studied also
similar optimal controls linear diffusion model via properties of return
function. Since the model treated in the present paper is very complicate and
different from He, Hou and Liang[13](2008) and Paulsen[22](2003), our results
can not be directly deduced from the [13, 22]. Therefore, to solve these the
problems we need to use initiated idea from the [13](2008), stochastic
analysis and PDE method to establish a complete setting for further discussing
optimal control problem of a large insurance company under lower ruin
probability constraint. This paper is the first complete presentation of the
topic, and the approach here is rather general, so we anticipate that it can
deal with other models. We aim at deriving the optimal return function, the
optimal retention rate and dividend payout level. The main result of this
paper will be presented in section 3 below. As a by-product, the paper
theoretically sets a risk-based capital standard to ensure the capital
requirement of can cover the total given risk. Moreover, based on our main
result, we also discuss how the risk affect the optimal reactions of the
insurance company by the implicit types of solutions and how the optimal
retention ratio, dividend payout level and risk-based capital standard are
affected by risk faced by the insurance company, and how the initial capital
and the premium rate impact on the company’s profit.

The paper is organized as follows: In next section, we establish nonlinear
stochastic control model of a large insurance company with ruin probability
constraint. In section 3 we present main result of this paper and its economic
and financial interpretations, and discuss how the risk affect the optimal
reactions of the insurance company by the implicit types of solutions and how
the optimal retention ratio, dividend payout level and risk-based capital
standard are affected by risk faced by the insurance company, and how the
initial capital and the premium rate impact on the company’s profit. In
section 4 we give analysis on risk of stochastic control model treated in the
present paper to explain why we study nonlinear regular-singular stochastic
optimal control of insurance company. In section 5 we give some numerical
samples to portray how the risk impacts on optimal dividend payout level and
risk-based capital based on PDE (6.25) below, and how the premium rate,
preferred reinsurance level and volatility effect on the company’s profit. The
proofs of theorems and lemmas which study properties of probability of
bankruptcy and optimal return function will be given in section 6 and
appendix.

##  2\. Nonlinear Mathematical Model

To give a mathematical formulation of the stochastic control problem treated
in this paper, let $(\Omega,\mathcal{F},\\{\mathcal{F}_{t}\\}_{t\geq
0},\mathbb{P})$ denote a filtered probability space. $\\{\mathcal{W}_{t},t\geq
0\\}$ is a standard Brownian motion on this probability space.
$\mathcal{F}_{t}$ represents the information available at time $t$ and any
decision is made based on this information. For the intuition of our diffusion
model we start from the classical Cramér-Lundberg model of a reserve(risk)
process to portray that if the insurance company shares risk with the
reinsurance and takes no dividend pay-out then its reserve process can be
approximated by the following diffusion process

$\displaystyle dR_{t}=\mu_{1}U(t)dt+\sigma U(t)d\mathcal{W}_{t},$ (2.1)

where $U(t)$ denotes retention level. In the classical Cramér-Lundberg model
claims arrive according to a Poisson process $N_{t}$ with intensity $\lambda$
on $(\Omega,\mathcal{F},\\{\mathcal{F}_{t}\\}_{t\geq 0},\mathbb{P})$. The size
of each claim is $X_{i}$. Positive 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, dividend pay-outs, 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 $X^{(U)}_{i}$, where $U$ is a
(fixed) retention level. For proportional reinsurance, $U$ 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

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

where

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

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

$\displaystyle r_{t}^{(U,\eta)}\stackrel{{\scriptstyle d}}{{\approx}}BM(\mu
Ut,\sigma^{2}U^{2}t).$

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 described by (2.1).
We refer the reader for this fact and for the specifies of the diffusion
approximations to Emanuel, Harrison and Taylor [7](1975), Grandell[8, 9,
10](1977,1978,1990), Harri-son [18](1985), Iglehart[20](1969),
Schmidli[27](1994). It is well known that over-risking is not a good recipe
for high return although risk and return are highly correlated. This leads to
question how an optimal strategy would change when the risk and return are not
linearly dependent on each other. Moreover, while a risk-averse re-insurers
may have their preferred risk level and impose additional service charge on
firms seeking services beyond the target level, other re-insurers may demand
additional charges for those seeking services with risk level lower than its
preferred level as an aggressive move to gain market shares. These make the
reserve process of the company should be the following

$\displaystyle dR(t)=[\mu_{1}U(t)-a(U(t)-p)^{2}]dt+\sigma
U(t)d\mathcal{W}_{t},\ R(0)=x,$ (2.2)

where p is the preferred reinsurance level imposed by the re-insurer and $a$
is the additional rate of charge for the deviation from the preferred level
which ensures that larger deviation is penalized heavily. If we let
$\mu=u_{1}+2ap$, $\delta=ap^{2}$, then the (2.2) becomes

$\displaystyle dR_{t}=(\mu U(t)-aU^{2}(t)-\delta)dt+\sigma
U(t)d\mathcal{W}_{t},\ \ R(0)=x,$ (2.3)

A strategy $\pi$ is a pair of non-negative càdlàg $\mathcal{F}_{t}$-adapted
processes $\\{U_{\pi}(t),L_{t}^{\pi}\\}$, where $U_{\pi}(t)\in[l,1](l>0)$
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
strategy $\pi=\\{U_{\pi}(t),L_{t}^{\pi}\\}$ is called admissible if $l\leq
U_{\pi}(t)\leq 1$ and $L_{t}^{\pi}$ is a nonnegative, non-decreasing, right-
continuous function. When $\pi$ is applied, the resulting reserve process is
denoted by $\\{R_{t}^{\pi}\\}$. We assume that the initial reserve
$R^{\pi}_{0}$ is a deterministic value $x$. In view of (2.3) the dynamics for
$R_{t}^{\pi}$ is given by

$\displaystyle dR^{\pi}_{t}=[\mu U_{\pi}(t)-a(U_{\pi}(t))^{2}-\delta]dt+\sigma
U_{\pi}(t)d\mathcal{W}_{t}-dL^{\pi}_{t},\ R^{\pi}_{0}=x.$ (2.4)

In this case, we assume the company needs to keep its reserve above $0$. The
company is considered ruin as soon as the reserves fall below $0$. We define
the time of bankruptcy by $\tau^{\pi}_{x}=\inf\\{t\geq 0:R^{\pi}_{t}\leq
0\\}$. Obviously, $\tau^{\pi}_{x}$ is an $\mathcal{F}_{t}$ -stopping time. So
the management of the insurance company should maximize the expected present
value of the dividend payout by control strategy $\pi$. Guo, Liu and Zhou[11]
proved that there exists a dividend level $b_{0}$, control strategy
$\pi^{*}_{b_{0}}$ and the time of bankruptcy $\tau_{x}^{b_{0}}$ maximizing the
expected present value of the dividend payout before bankruptcy,

$\displaystyle J(x,\pi)$ $\displaystyle=$ $\displaystyle{\bf
E}\large[(\int_{0}^{\tau}e^{-cs}dL^{\pi}_{s})],$ (2.5) $\displaystyle
V(x,b_{0})$ $\displaystyle=$
$\displaystyle\sup\limits_{\pi\in\Pi}J(x,\pi)=J(x,\pi^{*}_{b_{0}}),$ (2.6)

where $c$ denotes the discount rate, $\Pi$ is the set of all admissible
strategies. If the optimal dividend level $b_{0}$ is unacceptably low, then it
will result in the company go to bankruptcy early ( see theorem 4.1 below). To
take security and solvency into consideration and set a risk-based capital and
dividend standard to ensure the capital and dividend requirement of can cover
the total risk, we introduce our optimal control problem of nonlinear
stochastic model (2.4) as follows.Let
$\Pi_{b}=\\{\pi\in\Pi:\int_{0}^{\infty}I_{\\{s:R^{\pi}(s)<b\\}}dL_{s}^{\pi}=0\\}$
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
strategy $\pi$ we define the optimal return function $V(x)$ by

$\displaystyle J(x,\pi)$ $\displaystyle=$ $\displaystyle{\bf
E}\big{\\{}\int_{0}^{\tau^{\pi}_{x}}e^{-ct}dL_{t}^{\pi}\big{\\}},$
$\displaystyle V(x,b)$ $\displaystyle=$
$\displaystyle\sup_{\pi\in\Pi_{b}}\\{J(x,\pi)\\},$ (2.7) $\displaystyle V(x)$
$\displaystyle=$ $\displaystyle\sup_{b\in\mathfrak{B}}\\{V(x,b)\\}$ (2.8)

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

$\displaystyle J(x,\pi^{*})=V(x),$ (2.9)

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 strategy
is $\pi_{b}$. $1-\varepsilon$ is the standard of security and less than
solvency for given risk level $\varepsilon>0$.

The main purpose of this paper is to derive the optimal return function
$V(x)$, the optimal retention rate $U^{*}(t)$ and dividend payout level
$b^{*}$ as well as a risk-based capital $x(\varepsilon,b^{*})$ to ensure the
capital requirement of can cover the total risk $\varepsilon$.

##  3\.  Main result

In this section we first present main result of this paper, then give its
economic and financial interpretations .

######  Theorem 3.1.

Let level of risk $\varepsilon\in(0,1)$ 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 (6.1) below, and
$V(x)=f(b_{0},x)=J(x,\pi_{b_{o}}^{\ast})$. The optimal strategy
$\pi_{b_{o}}^{\ast}$ is
$\\{U^{\ast}(R^{\pi_{b_{o}}^{\ast}}_{t}),L^{\pi_{b_{o}}^{\ast}}_{t}\\}$, where
$\\{R^{\pi_{b_{o}}^{\ast}}_{t},L^{\pi_{b_{o}}^{\ast}}_{t}\\}$ is uniquely
determined by the following stochastic differential equation

$\displaystyle\left\\{\begin{array}[]{l l l}dR_{t}^{\pi_{b_{o}}^{\ast}}=(\mu
U^{*}_{b_{0}}(R^{\pi_{b_{o}}^{\ast}}_{t})-a{U^{*}_{b_{0}}}^{2}(R^{\pi_{b_{o}}^{\ast}}_{t})-\delta)dt+\sigma
U^{*}_{b_{0}}(R^{\pi_{b_{o}}^{\ast}}_{t})d{W}_{t}-dL_{t}^{\pi_{b_{o}}^{\ast}},\\\
R_{0}^{\pi_{b_{o}}^{\ast}}=x,\\\ 0\leq R^{\pi_{b_{o}}^{\ast}}_{t}\leq
b_{0},\\\
\int^{\infty}_{0}I_{\\{t:R^{\pi_{b_{o}}^{\ast}}_{t}<b_{0}\\}}(t)dL_{t}^{\pi_{b_{o}}^{\ast}}=0.\end{array}\right.$
(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
$\mathbb{P}[\tau_{b^{\ast}}^{\pi_{b^{*}}^{\ast}}\leq T]=\varepsilon$. The
optimal return function $V(x)$ is $g(x,b^{*})$ defined by (6.8), 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 strategy $\pi_{b^{*}}^{\ast}$ is
$\\{U^{\ast}_{b^{*}}(R^{\pi_{b^{*}}^{\ast}}_{t}),L^{\pi_{b^{*}}^{\ast}}_{t}\\}$,
where $\\{R^{\pi_{b^{*}}^{\ast}}_{t},L^{\pi_{b^{*}}^{\ast}}_{t}\\}$ is
uniquely determined by the following stochastic differential equation

$\displaystyle\left\\{\begin{array}[]{l l l}dR_{t}^{\pi_{b^{*}}^{\ast}}=(\mu
U^{*}_{b^{*}}(R^{\pi_{b^{*}}^{\ast}}_{t})-a{U^{*}_{b^{*}}}^{2}(R^{\pi_{b^{*}}^{\ast}}_{t})-\delta)dt+\sigma
U^{*}_{b^{*}}(R^{\pi_{b^{*}}^{\ast}}_{t})d{W}_{t}-dL_{t}^{\pi_{b^{*}}^{\ast}},\\\
R_{0}^{\pi_{b^{*}}^{\ast}}=x,\\\ 0\leq R^{\pi_{b^{*}}^{\ast}}_{t}\leq
b^{*},\\\
\int^{\infty}_{0}I_{\\{t:R^{\pi_{b^{*}}^{\ast}}_{t}<b^{*}\\}}(t)dL_{t}^{\pi_{b^{*}}^{\ast}}=0.\end{array}\right.$
(3.12)

The solvency of the company is $1-\varepsilon$. (3) Moreover,

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

Where $U^{*}_{b}(x)$ is defined by (6.21) and (6.29) below.

Economic and financial explanation of theorem 3.1 is as follows.(1) For a
given level of risk and time horizon, if ruin probability is less than the
level of risk, the optimal control problem of (2) and (2.8) is the traditional
(2.5) and (2.6), the company has higher solvency, so it will have good
reputation. The solvency constraints here do not work. This is a trivial case.
(2) If ruin probability is large than the level of risk $\varepsilon$, the
traditional optimal strategy will not meet the standard of security and
solvency, the company needs to find a sub-optimal strategy
$\pi_{b^{*}}^{\ast}$ to improve its solvency. The sub-optimal reserve process
$R^{\pi_{b^{*}}^{\ast}}_{t}$ is a diffusion process reflected at $b^{*}$, the
process $L^{\pi_{b^{*}}^{\ast}}_{t}$ is the process which ensures the
reflection. The sub-optimal action is to pay out everything in excess of
$b^{*}$ as dividend and pay no dividend when the reserve is below $b^{*}$, and
$U_{b^{*}}^{*}(x)$ is the sub-optimal feedback control function. The solvency
is $1-\varepsilon$. (3) On the one hand, the inequality (3.13) states that
$\pi_{b^{*}}^{\ast}$ will reduce the company’s profit, on the other hand, in
view of (3.13) and $\mathbb{P}[\tau_{b^{\ast}}^{\pi_{b^{*}}^{\ast}}\leq
T]=\varepsilon$ as well as lemma 6.4 below, the cost of improving solvency is
minimal $g(x,b_{0})-g(x,b^{*})$. Therefore the strategy $\pi_{b^{*}}^{\ast}$
is the best equilibrium action between making profit and improving solvency.
(4) The risk-based capital $x(\varepsilon,b^{*})$ to ensure the capital
requirement of can cover the total risk $\varepsilon$ can be determined by
numerical solution of $1-\phi^{b^{*}}(x,b^{*})=\varepsilon$ based on (6.25).
We see from the figure 5 that risk-based capital $x(\varepsilon,b^{*})$
decreases with risk $\varepsilon$, i.e., $x(\varepsilon,b^{*})$ increases with
solvency , so does risk-based dividend level $b^{*}(\varepsilon)$ (see the
figure 1). (5) We also see from the figures 2 and 4 below that the premium
rate will increase the company’s profit, higher risk will get higher return.
(6) We also see from the figure 3 below shows that the value function $g(x,p)$
increases with $(x,p)$, i.e., the initial capital and the premium rate will
increases the company’s profit.

##  4\. Analysis of risk on model (2.4)

The first result of this section is the following, which states that the
company has to find optimal strategy to improve its solvency.

######  Theorem 4.1.

Let $\\{R^{\pi_{b_{0}}^{\ast}}_{t},L^{\pi_{b_{0}}^{\ast}}_{t}\\}$ be defined
by the following SDE( see Lions and Sznitman [23](1984))

$\displaystyle\left\\{\begin{array}[]{l l l}dR_{t}^{\pi_{b_{o}}^{\ast}}=(\mu
U^{*}_{b_{0}}(R^{\pi_{b_{o}}^{\ast}}_{t})-a{U^{*}_{b_{0}}}^{2}(R^{\pi_{b_{o}}^{\ast}}_{t})-\delta)dt+\sigma
U^{*}_{b_{0}}(R^{\pi_{b_{o}}^{\ast}}_{t})d{W}_{t}-dL_{t}^{\pi_{b_{o}}^{\ast}},\\\
R_{0}^{\pi_{b_{o}}^{\ast}}=x,\\\ 0\leq R^{\pi_{b_{o}}^{\ast}}_{t}\leq
b_{0},\\\
\int^{\infty}_{0}I_{\\{t:R^{\pi_{b_{o}}^{\ast}}_{t}<b_{0}\\}}(t)dL_{t}^{\pi_{b_{o}}^{\ast}}=0.\end{array}\right.$
(4.5)

Then for any $x\in(0,b_{0}]$ we have

$\displaystyle{\bf P}(\tau_{x}^{b_{0}}\leq
T)\geq\varepsilon_{0}(b_{0},\sigma^{2},\mu,p,l,a)\equiv\frac{4[1-\Phi(\frac{b_{0}}{l\sigma\sqrt{T}})]^{2}}{\exp\\{\frac{(\mu-a-\delta)^{2}T}{\sigma^{2}}\\}}>0,$
(4.6)

where $\Phi(\cdot)$ is the standard normal distribution function.

The economic interpretation of theorem 4.1 is the following.(1) The lower
boundary $\varepsilon_{0}(b_{0},\sigma^{2},\mu,p,l,a)$ of ruin probability for
the company is an increasing function of $(\sigma^{2},l)$, thus higher
volatility $\sigma^{2}$ and fraction of the claim covered by the company will
make the company have larger risk. (2) The lower boundary
$\varepsilon_{0}(b_{0},\sigma^{2},\mu,p,l,a)$ of ruin probability for the
company is a decreasing function of $(b_{0},\mu,p,a)$, so early making
dividend will increasing the company’s risk. The premium rate, preferred
reinsurance level and additional rate of charge for the deviation from the
preferred level will decrease the company’s risk.

###### Proof.

Let $\\{R^{(1)}_{t}\\}$ be a stochastic process satisfying

$\displaystyle\left\\{\begin{array}[]{l l l}dR^{(1)}_{t}=(\mu
U^{*}_{b_{0}}(R^{(1)}_{t})-a{U^{*}_{b_{0}}}^{2}(R^{(1)}_{t})-\delta)dt+\sigma
U^{*}_{b_{0}}(R^{(1)}_{t})d{W}_{t},\\\ R^{(1)}_{0}=b_{0}\\\
\end{array}\right.$ (4.9)

where $U^{*}_{b_{0}}(\cdot)$ is defined by (6.29). Define a measure ${\bf Q}$
on $\mathcal{F}_{T}$ by

$d{\bf Q}(\omega)=M_{1}(T)dP(\omega)$

where

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

Since $\\{M_{1}(t)\\}$ is a martingale w.r.t.$\mathcal{F}_{t}$, ${\bf
E}\big{[}M_{1}(T)\big{]}=1$. Using Girsanov theorem, ${\bf Q}$ is a
probability measure on $\mathcal{F}_{T}$ and the process $\\{R^{(1)}_{t}\\}$
satisfies the following SDE

$\displaystyle dR_{t}^{(1)}=U^{*}_{b_{0}}(R_{t}^{(1)})\sigma
d\tilde{W}_{t},R_{0}^{(1)}=b_{0}$ (4.10)

where $\tilde{W}_{t}$ is a Brownian motion on
$(\Omega,\mathcal{F},\\{\mathcal{F}_{t}\\}_{t\geq 0},{\bf Q})$. Define a time
changes $\rho(t)$ by

$\displaystyle\dot{\rho}(t)=\frac{1}{{U^{*}_{b_{0}}}^{2}(R_{t}^{(1)})\sigma^{2}},$
(4.11)

and $\hat{R}_{t}^{(1)}$ by $R_{\rho(t)}^{(1)}$. Then $\rho(t)$ is a strictly
increasing function and

$\displaystyle\hat{R}_{t}^{(1)}=b+\hat{W}_{t}$

where $\hat{W}_{t}$ is also a standard Brownian motion on
$(\Omega,\mathcal{F},\\{\mathcal{F}_{t}\\}_{t\geq 0},{\bf Q})$. Noticing that
${U^{*}_{b_{0}}}^{2}(R_{t}^{(1)})\geq l^{2}>0$, where $l$ is a positive low
boundary of optimal retention ratio $U^{*}_{b_{0}}\in[l,1]$, we have

$\displaystyle\dot{\rho}(t)\leq\frac{1}{l^{2}\sigma^{2}}.$ (4.12)

Moreover, $\rho(t)\leq\frac{1}{l^{2}\sigma^{2}}t$ and $\rho^{-1}(t)\geq
l^{2}\sigma^{2}t$. So

$\displaystyle{\bf{Q}}[\tau^{(1)}\leq T]$ $\displaystyle=$
$\displaystyle{\bf{Q}}[\inf\\{t:\hat{R}_{\rho^{-1}(t)}^{(1)}\leq 0\\}\leq T]$
$\displaystyle=$ $\displaystyle{\bf{Q}}[\inf\\{\rho(t):b_{0}+\hat{W}_{t}\leq
0\\}\leq T]$ $\displaystyle=$ $\displaystyle{\bf{Q}}[\inf\\{t:\hat{W}_{t}\leq-
b_{0}\\}\leq\rho^{-1}(T)]$ $\displaystyle\geq$
$\displaystyle{\bf{Q}}[\inf\\{t:\hat{W}_{t}\leq-b_{0}\\}\leq
l^{2}\sigma^{2}T]$ $\displaystyle=$ $\displaystyle
2[1-\Phi(\frac{b_{0}}{l\sigma\sqrt{T}})]>0,$

where $\tau^{(1)}=\inf\\{t>0;R_{t}^{(1)}\leq 0\\}$ is a stopping time. Using
comparison theorem for one-dimensional Itô process, we have
${\bf{P}}[R^{\pi_{b_{o}}^{\ast}}_{t}\leq R_{t}^{(1)}]=1$. By
${\bf{E}}^{\bf{P}}[M_{1}(T)^{2}]\leq\exp\\{\frac{(\mu-a-\delta)^{2}T}{\sigma^{2}}\\}$
and Hölder inequalities we have

$\displaystyle{\bf{P}}[\tau_{x}^{b_{0}}\leq T]$ $\displaystyle\geq$
$\displaystyle{\bf{P}}[\tau_{b_{0}}^{b_{0}}\leq T]$ $\displaystyle\geq$
$\displaystyle{\bf{P}}[\tau^{(1)}\leq T]$ $\displaystyle\geq$
$\displaystyle{\bf{Q}}[\tau^{(1)}\leq
T]^{2}/{\bf{E}}^{{\bf{P}}}[M_{1}(T)^{2}]$ $\displaystyle\geq$
$\displaystyle\frac{4[1-\Phi(\frac{b_{0}}{l\sigma\sqrt{T}})]^{2}}{\exp\\{\frac{(\mu-a-\delta)^{2}T}{\sigma^{2}}\\}}>0.$

∎

The second result of this section is the following. It sates that the
restrained set $\mathfrak{B}$ above is non-empty for any $\varepsilon>0$. So
the (2),(2.8) and (2.9) are well defined.

######  Theorem 4.2.

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

$\displaystyle\left\\{\begin{array}[]{l l l}dR_{t}^{\pi_{b}^{\ast}}=(\mu
U^{*}_{b}(R^{\pi_{b}^{\ast}}_{t})-a{U^{*}_{b}}^{2}(R^{\pi_{b}^{\ast}}_{t})-\delta)dt+\sigma
U^{*}_{b}(R^{\pi_{b}^{\ast}}_{t})d{W}_{t}-dL_{t}^{\pi_{b}^{\ast}},\\\
R_{0}^{\pi_{b}^{\ast}}=b,\\\ 0\leq R^{\pi_{b}^{\ast}}_{t}\leq b,\\\
\int^{\infty}_{0}I_{\\{t:R^{\pi_{b}^{\ast}}_{t}<b\\}}(t)dL_{t}^{\pi_{b}^{\ast}}=0,\end{array}\right.$
(4.17)

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

$\displaystyle\lim_{b\rightarrow\infty}{\bf P}[\tau_{b}^{b}\leq T]=0.$ (4.18)

###### Proof.

Let $x_{2}$ be defined as in (6.17). For $b>x_{2}$, by comparison theorem for
SDE, we have

$\displaystyle{\bf P}\\{\tau_{b}^{b}\leq T\\}\leq{\bf
P}\\{\tau_{(b+x_{2})/2}^{b}\leq T\\}.$

It is easy to see that

$\displaystyle{\bf P}\\{\tau_{(b+x_{2})/2}^{\pi^{*}_{b}}\leq T\\}$
$\displaystyle\leq$ $\displaystyle{\bf P}\\{R_{t}^{(2)}=x_{2}\ \mbox{or}\
R_{t}^{(2)}=b\ \mbox{for some $t\geq 0$ }\\}$ $\displaystyle\leq$
$\displaystyle{\bf P}\\{\sup_{0\leq t\leq T}R_{t}^{(2)}\geq b\\}+{\bf
P}\\{\inf_{0\leq t\leq T}R_{t}^{(2)}\leq x_{2}\\},$

where $\\{R_{t}^{(2)}\\}$ is the unique solution of the following SDE

$\displaystyle\left\\{\begin{array}[]{l l l}dR^{(2)}_{t}=(\mu
U^{*}_{b}(R^{(2)}_{t})-a{U^{*}_{b}}^{2}(R^{(2)}_{t})-\delta)dt+\sigma
U^{*}_{b}(R^{(2)}_{t})d{W}_{t},\\\ R^{(2)}_{0}=(b+x_{2})/2.\end{array}\right.$
(4.21)

Define a measure $\mathbb{Q}_{1}$ on $\mathcal{F}_{T}$ by

$\displaystyle d{\bf
P}(\omega)=\widetilde{M}_{T}(\omega)d\mathbb{Q}_{1}(\omega),$

where

$\displaystyle\widetilde{M}_{t}$ $\displaystyle=$
$\displaystyle\exp\big{\\{}\int_{0}^{t}\frac{(\mu
U^{*}_{b}(R^{(2)}_{t})-a{U^{*}_{b}}^{2}(R^{(2)}_{t})-\delta)}{\sigma
U_{b}^{*}(R_{t}^{(2)})}dW_{s}$ $\displaystyle+$
$\displaystyle\frac{1}{2}\int_{0}^{t}\frac{(\mu
U^{*}_{b}(R^{(2)}_{t})-a{U^{*}_{b}}^{2}(R^{(2)}_{t})-\delta)^{2}}{[\sigma
U_{b}^{*}(R_{t}^{(2)})]^{2}}ds\big{\\}}$

is a martingale. Then $\mathbb{Q}_{1}$ is a probability measure on
$\mathcal{F}_{T}$. By Girsanov theorem

$\displaystyle\hat{\mathcal{W}_{t}}:=\int_{0}^{t}\frac{(\mu
U^{*}_{b}(R^{(2)}_{t})-a{U^{*}_{b}}^{2}(R^{(2)}_{t})-\delta)}{\sigma
U_{b}^{*}(R_{s}^{(2)})}ds+\mathcal{W}_{t},t\leq T$

is a standard Brownian motion on
$(\Omega,\mathcal{F},\\{\mathcal{F}_{t}\\}_{t\geq 0},\mathbb{Q}_{1})$. So the
(4.21) becomes

$\displaystyle dR_{t}^{(2)}=\sigma
U^{*}_{b}(R^{(2)}_{t})\hat{\mathcal{W}_{t}},\ R^{(2)}_{0}=(b+x_{2})/2.\
\mbox{a.e.},\ \mathbb{Q}_{1}$

Firstly, we now estimate ${\bf P}\\{\sup_{0\leq t\leq T}R_{t}^{(2)}\geq b\\}$.
By SDE (4.21), Hölder’s inequalities,Chebyshev inequalities and B-D-G
inequalities for martingales (see Ikeda and Watanabe [21](1981))

$\displaystyle{\bf P}\\{\sup_{0\leq t\leq T}R_{t}^{(2)}\geq b\\}$
$\displaystyle\leq$ $\displaystyle[{\bf
E}^{\mathbb{Q}_{1}}\\{\widetilde{M}^{2}_{T}\\}]^{\frac{1}{2}}\mathbb{Q}_{1}\\{\sup_{0\leq
t\leq T}R_{t}^{(2)}\geq b\\}^{\frac{1}{2}}$ (4.22)

and

$\displaystyle\mathbb{Q}_{1}\\{\sup_{0\leq t\leq T}R_{t}^{(2)}\geq b\\}$
$\displaystyle\leq$ $\displaystyle{\mathbb{Q}_{1}}\\{\sup_{0\leq t\leq
T}|\int_{0}^{t}\sigma
U^{*}_{b}(R^{(2)}_{s})d\hat{\mathcal{W}_{s}}|\geq\frac{b-x_{2}}{2}\\}$ (4.23)
$\displaystyle\leq$ $\displaystyle\frac{4{\bf
E}^{\mathbb{Q}_{1}}\\{\sup_{0\leq t\leq T}|\int_{0}^{t}\sigma
U^{*}_{b}(R^{(2)}_{s})d\hat{\mathcal{W}_{s}}|\\}^{2}}{(b-x_{2})^{2}}$
$\displaystyle\leq$ $\displaystyle\frac{16{\bf
E}^{\mathbb{Q}_{1}}\\{\int_{0}^{T}(\sigma
U^{*}_{b}(R^{(2)}_{s}))^{2}ds|\\}}{(b-x_{2})^{2}}$ $\displaystyle\leq$
$\displaystyle\frac{16T\sigma^{2}\widetilde{B}^{2}}{(b-x_{2})^{2}},$

where ${\bf E}^{\mathbb{Q}_{1}}$ denotes the expectation w.r.t.
$\mathbb{Q}_{1}$ .
Next we estimate $P\\{\inf_{0\leq t\leq T}R_{t}^{(1)}\leq x_{2}\\}$. Since
$U^{*}_{b}(x)=1$ for $x\geq x_{2}$,

$\displaystyle{\bf P}\\{\inf_{0\leq t\leq T}R_{t}^{(2)}\leq x_{2}\\}$
$\displaystyle=$ $\displaystyle 1-\mathbf{P}\\{\inf_{0\leq t\leq
T}R_{t}^{(2)}>x_{2}\\}$ (4.24) $\displaystyle=$ $\displaystyle 1-{\bf
P}\\{\inf_{0\leq t\leq T}\\{\mu t+\sigma W_{t}\\}>-\frac{b-x_{2}}{2}\\}$
$\displaystyle\rightarrow 1-1=0\ \mbox{as }\ b\rightarrow\infty.$

Finally, since $l\leq U^{*}_{b}\leq 1$,

$\displaystyle{\bf E}^{\mathbb{Q}_{1}}\\{\widetilde{M}^{2}_{T}\\}\leq
C(T)<\infty.$ (4.25)

So the equation (4.18) follows from (4.22)-(4.25). ∎

##  5\. Numerical examples

In this section we consider some numerical samples to demonstrate how the risk
$\varepsilon$ impacts on optimal dividend payout level $b^{*}$ and risk-based
capital $x$ based on PDE (6.25) below, and how the premium rate, preferred
reinsurance level and volatility effect on the company’s profit.

######  Example 5.1.

Let $\mu=2$, $\sigma^{2}=50$, $l=0.5$,$a=0.1$, $\delta=0.01$, $c=0.05$,
$T=500$ and solve $b(\varepsilon)$ by $1-\phi(T,b)=\varepsilon$, we get the
figure 1 below. It shows that the risk $\varepsilon$ greatly impacts on
dividend payout level $b$. The dividend payout level $b$ decreases with the
risk $\varepsilon$, so the risk $\varepsilon$ increases the company’s profit.

Figure 1.  Dividend payout level $b$ as a function of $\varepsilon$
(Parameters: $\mu=2$, $\sigma^{2}=50$, $l=0.5$,$a=0.1$, $\delta=0.01$,
$c=0.05$, $T=500$)

######  Example 5.2.

Let $b=100,\sigma^{2}=50,l=0.5,a=0.1,\delta=0.01,c=0.05,T=500$ the figure 2
below shows that the value function $g(x,\mu)$ increases with $(x,\mu)$, so
does the company’s profit.

Figure 2.  Value function $g(x,\mu)$ as a function of $(x,\mu)$ (Parameters:
$b=100,\sigma^{2}=50,l=0.5,a=0.1,\delta=0.01,c=0.05,T=500$ )

######  Example 5.3.

Let $\mu=2,b=100,\sigma^{2}=50,l=0.5,a=0.5,c=0.05,T=500$. The figure 3 below
shows that the value function $g(x,p)$ increases with $(x,p)$, so does the
company’s profit.

Figure 3.  Value function $g(x,p)$ as a function of $(x,p)$ (Parameters:
$\mu=2,b=100,\sigma^{2}=50,l=0.5,a=0.5,c=0.05,T=500$ )

######  Example 5.4.

Let $\mu=2,b=100,l=0.5,a=0.1,\delta=0.01,c=0.05,T=500$. The figure 4 below
shows that the value function $g(x,\sigma^{2})$ increases with
$(x,\sigma^{2})$, so does the company’s profit.

Figure 4.  Value function $g(x,\sigma^{2})$ as a function of $(x,\sigma^{2})$
(Parameters: $\mu=2,b=100,l=0.5,a=0.1,\delta=0.01,c=0.05,T=500$)

######  Example 5.5.

Let $\mu=2,\sigma^{2}=50,l=0.5,a=0.1,\delta=0.01,c=0.05,T=500$. The figure 5
below shows that the initial capital $x(\epsilon)$ decreases with $\epsilon$.

Figure 5.  Initial capital $x(\epsilon)$ as a function of $\epsilon$
(Parameters: $\mu=2,\sigma^{2}=50,l=0.5,a=0.1,\delta=0.01,c=0.05,T=500$)

##  6\. Properties on $V(x,b)$ and ruin probability

In this section, to prove Theorem 3.1, we list some lemmas on properties of
$V(x,b)$ and ruin probability which will be used late. The rigorous proofs of
these lemmas will be given in the appendix below. Throughout this paper we
assume that $\mu/2a>1$ and $0<l\leq U(t)\leq 1$.

######  Lemma 6.1.

There exists $b_{0}>0$ such that if $f(x)\in C^{2}$ satisfies the following
HJB equations and the boundary conditions,

$\displaystyle\max\limits_{U\in[l,1]}[\frac{1}{2}\sigma^{2}U^{2}f^{{}^{\prime\prime}}(x)+(\mu
U-aU^{2}-\delta)f^{{}^{\prime}}(x)-cf(x)]=0,$ (6.1) $\displaystyle\mbox{ for}\
\ 0\leq x\leq b_{0},$ $\displaystyle f^{{}^{\prime}}(x)=1,\ \mbox{for}\ x\geq
b_{0},$ $\displaystyle f^{{}^{\prime\prime}}(x)=0,\ \mbox{for}\ x\geq b_{0},$
$\displaystyle f(0)=0,$

then we have the following,

$\displaystyle\max\mathcal{L}f(x)\leq 0,f^{{}^{\prime}}(x)\geq 1,\ \mbox{
for}\ x\geq 0,$ $\displaystyle f(0)=0,$

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

######  Lemma 6.2.

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

$\displaystyle\max\limits_{U\in[l,1]}\big{[}\frac{1}{2}\sigma^{2}U^{2}g^{{}^{\prime\prime}}(x)+(\mu
U-aU^{2}-\delta)g^{{}^{\prime}}(x)-cg(x)\big{]}=0,$ (6.2) $\displaystyle\mbox{
for}\ \ 0\leq x\leq b,$ $\displaystyle g^{{}^{\prime}}(x)=1,\ \mbox{for}\
x\geq b,$ $\displaystyle g^{{}^{\prime\prime}}(x)=0,\ \mbox{for}\ x>b,$
$\displaystyle g(0)=0,$

then we have the following,

$\displaystyle\max\mathcal{L}g(x)\leq 0,\ \mbox{ for}\ x\geq 0,$ (6.3)
$\displaystyle g^{{}^{\prime}}(x)\geq 1,\ \mbox{ for}\ x\geq b,$
$\displaystyle g(0)=0,$

where $g^{\prime\prime}(b):=g^{\prime\prime}(b-)$, $\mathcal{L}$ is defined as
same as in Lemma 6.1. Indeed, the function $g(x)$ can be written as follows,

$\displaystyle g(x,b)=\left\\{\begin{array}[]{l l
l}A\big{(}e^{\alpha_{1}x}-e^{\beta_{1}x}\big{)},&0\leq x\leq x_{1},\\\
\big{(}Be^{\alpha_{2}x_{2}}+Ce^{\beta_{2}x_{2}}\big{)}\exp\big{\\{}-\int_{x}^{x_{2}}\frac{c}{\frac{1}{2}\mu\eta(y)-\delta}dy\big{\\}},&x_{1}<x<x_{2},\\\
Be^{\alpha_{2}x}+Ce^{\beta_{2}x},&x_{2}\leq x\leq b,\\\
x-b+Be^{\alpha_{2}b}+Ce^{\beta_{2}b},&x>b,\end{array}\right.$ (6.8)

where

$\displaystyle\alpha_{1}=\frac{-(\mu l-al^{2}-\delta)+\sqrt{(\mu
l-al^{2}-\delta)^{2}+2c\sigma^{2}l^{2}}}{\sigma^{2}l^{2}},$ (6.9)
$\displaystyle\beta_{1}=\frac{-(\mu l-al^{2}-\delta)-\sqrt{(\mu
l-al^{2}-\delta)^{2}+2c\sigma^{2}l^{2}}}{\sigma^{2}l^{2}},$ (6.10)
$\displaystyle\alpha_{2}=\frac{-(\mu-a-\delta)+\sqrt{(\mu-a-\delta)^{2}+2c\sigma^{2}}}{\sigma^{2}},$
(6.11)
$\displaystyle\beta_{2}=\frac{-(\mu-a-\delta)-\sqrt{(\mu-a-\delta)^{2}+2c\sigma^{2}}}{\sigma^{2}}$
(6.12) $\displaystyle
A=\frac{Be^{\alpha_{2}x_{2}}+Ce^{\beta_{2}x_{2}}}{e^{\alpha_{1}x_{1}}-e^{\beta_{1}x_{1}}}\exp\big{\\{}-\int_{x_{1}}^{x_{2}}\frac{c}{\frac{1}{2}\mu\eta(y)-\delta}dy\big{\\}},$
(6.13) $\displaystyle B=\frac{[-c+(\frac{1}{2}\mu
1-\delta)\beta_{2}]e^{\beta_{2}x_{2}}}{[-c+(\frac{1}{2}\mu
1-\delta)\beta_{2}]e^{\beta_{2}x_{2}+\alpha_{2}b}\alpha_{2}-[-c+(\frac{1}{2}\mu
1-\delta)\alpha_{2}]e^{\alpha_{2}x_{2}+\beta_{2}b}\beta_{2}},$ $\displaystyle
C=\frac{[-c+(\frac{1}{2}\mu
1-\delta)\alpha_{2}]e^{\alpha_{2}x_{2}}}{-[-c+(\frac{1}{2}\mu
1-\delta)\beta_{2}]e^{\beta_{2}x_{2}+\alpha_{2}b}\alpha_{2}+[-c+(\frac{1}{2}\mu
1-\delta)\alpha_{2}]e^{\alpha_{2}x_{2}+\beta_{2}b}\beta_{2}},$ (6.15)
$\displaystyle
x_{1}=\frac{1}{\alpha_{1}-\beta_{1}}\log\big{[}\frac{c-\beta_{1}(\frac{1}{2}\mu
l-\delta)}{c-\alpha_{1}(\frac{1}{2}\mu l-\delta)}\big{]}>0,$ (6.16)
$\displaystyle
x_{2}=x_{1}+\frac{\sigma^{2}}{2a}\big{[}\frac{G}{G-H}\log(\frac{G-l}{G-1})-\frac{H}{G-H}\log(\frac{l-H}{1-H})\big{]},$
(6.17) $\displaystyle G$ $\displaystyle=$
$\displaystyle\frac{2c\sigma^{2}+\mu^{2}+4a\delta+\sqrt{(2c\sigma^{2}+\mu^{2}+4a\delta)^{2}-16a\mu^{2}\delta}}{4a\mu},$
(6.18) $\displaystyle H$ $\displaystyle=$
$\displaystyle\frac{2c\sigma^{2}+\mu^{2}+4a\delta-\sqrt{(2c\sigma^{2}+\mu^{2}+4a\delta)^{2}-16a\mu^{2}\delta}}{4a\mu},$
(6.19) $\displaystyle K$ $\displaystyle=$
$\displaystyle(G-l)^{G/(G-H)}(l-H)^{-H/(G-H)},$ (6.20)

and $\eta(x)$ is uniquely determined by

$\displaystyle[G-\eta(x)]^{G/(G-H)}[\eta(x)-H]^{-H/(G-H)}=K\exp\big{[}-\frac{2a}{\sigma^{2}}(x-x_{1})\big{]}.$
(6.21)

######  Lemma 6.3.

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

######  Lemma 6.4.

The ruin probability $\mathbf{P}[\tau_{b}^{b}\leq T]$ is strictly increasing
w.r.t. $b$ on $[x_{2},b_{K})$, where
$b_{K}:=\inf\\{b:\mathbf{P}[\tau_{b}^{b}\leq T]=0\\}$, and $x_{2}$ is defined
by (6.17), $\tau^{b}_{b}:=\tau_{b}^{\pi_{b}^{*}}$.

######  Lemma 6.5.

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,

$\displaystyle\left\\{\begin{array}[]{l l
l}\phi_{t}^{b}(t,y)=\frac{1}{2}[U_{b}^{*}(y)]^{2}\sigma^{2}\phi_{yy}^{b}(t,y)+(\mu
U_{b}^{*}(y)-a[U_{b}^{*}(y)]^{2}-\delta)\phi_{y}^{b}(t,y),\\\
\phi^{b}(0,y)=1,\ \mbox{for}\ \ 0<y\leq b,\\\
\phi^{b}(t,0)=0,\phi_{y}^{b}(t,y)=0,\ \mbox{for}\ t>0.\end{array}\right.$
(6.25)

Then $\phi^{b}(T,y)=1-\psi^{b}(T,y)$, where $\psi^{b}(T,y):={\bf
P}\big{\\{}\tau_{y}^{b}<T\big{\\}}$, and $U^{*}(x)$ is defined by

$\displaystyle U^{*}_{b}(x)=\left\\{\begin{array}[]{l l l}l,&0\leq x\leq
x_{1},\\\ \eta(x)&x_{1}<x<x_{2},\\\ 1,&x_{2}\leq x.\end{array}\right.$ (6.29)

######  Lemma 6.6.

Let the function $\phi^{b}(t,x)$ solve the equation(6.25) and
$u(b)\equiv\phi^{b}(T,b)$. Then $u(b)$ is a continuous function of $b$ on
$[b_{0},+\infty)$.

##  7\. Proof of Main Result

In this section we will give the proof of Theorem 3.1. Before this proof we
first prove the following.

######  Theorem 7.1.

Let $f(x)$, $g(x,b)$ and $U^{*}_{b}(x)$ be as the same as in lemma 6.1, lemma
6.2 and lemma 6.5,respectively. Then
(i) If $b\leq b_{0}$ we have $V(x,b)=V(x,b_{0})=V(x)=f(x)$, the optimal
strategy associated with $V(x)$ is
$\pi_{b_{o}}^{\ast}=\\{U^{*}_{b_{0}}(R^{\pi_{b_{o}}^{\ast}}_{\cdot}),L^{\pi_{b_{o}}^{\ast}}_{\cdot}\\}$,
where the process
$\\{R^{\pi_{b_{o}}^{\ast}}_{t},L^{\pi_{b_{o}}^{\ast}}_{t}\\}$ is uniquely
determined by the following SDE,

$\displaystyle\left\\{\begin{array}[]{l l l}dR_{t}^{\pi_{b_{o}}^{\ast}}=(\mu
U^{*}_{b_{0}}(R^{\pi_{b_{o}}^{\ast}}_{t})-a{U^{*}_{b_{0}}}^{2}(R^{\pi_{b_{o}}^{\ast}}_{t})-\delta)dt+\sigma
U^{*}_{b_{0}}(R^{\pi_{b_{o}}^{\ast}}_{t})d{W}_{t}-dL_{t}^{\pi_{b_{o}}^{\ast}},\\\
R_{0}^{\pi_{b_{o}}^{\ast}}=x,\\\ 0\leq R^{\pi_{b_{o}}^{\ast}}_{t}\leq
b_{0},\\\
\int^{\infty}_{0}I_{\\{t:R^{\pi_{b_{o}}^{\ast}}_{t}<b_{0}\\}}(t)dL_{t}^{\pi_{b_{o}}^{\ast}}=0.\end{array}\right.$
(7.5)

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

$\displaystyle\left\\{\begin{array}[]{l l l}dR_{t}^{\pi_{b^{*}}^{\ast}}=(\mu
U^{*}_{b^{*}}(R^{\pi_{b^{*}}^{\ast}}_{t})-a{U^{*}_{b^{*}}}^{2}(R^{\pi_{b^{*}}^{\ast}}_{t})-\delta)dt+\sigma
U^{*}_{b^{*}}(R^{\pi_{b^{*}}^{\ast}}_{t})d{W}_{t}-dL_{t}^{\pi_{b^{*}}^{\ast}},\\\
R_{0}^{\pi_{b^{*}}^{\ast}}=x,\\\ 0\leq R^{\pi_{b^{*}}^{\ast}}_{t}\leq
b^{*},\\\
\int^{\infty}_{0}I_{\\{t:R^{\pi_{b^{*}}^{\ast}}_{t}<b^{*}\\}}(t)dL_{t}^{\pi_{b^{*}}^{\ast}}=0.\end{array}\right.$
(7.10)

###### Proof.

(i) If $b\leq b_{0}$ then 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 $V(x)\leq
f(x)=V(x,b_{0})$. For a admissible strategy $\pi=\\{a_{\pi},L^{\pi}\\}$ we
assume that $(R^{\pi}_{t},L^{\pi}_{t})$ is the process defined by (2.4). Set
$\Lambda=\\{s:L_{s-}^{\pi}\neq L_{s}^{\pi}\\}$ and let
$\hat{L}=\sum_{s\in\Lambda,s\leq t}(L_{s}^{\pi}-L_{s-}^{\pi})$ and
$\tilde{L}_{t}^{\pi}=L_{t}^{\pi}-\hat{L}_{t}^{\pi}$ denote the discontinuous
part and continuous part of $L_{s}^{\pi}$,respectively. Let
$\tau^{\varepsilon}=\inf\\{t\geq 0:R^{\pi}_{t}\leq\varepsilon\\}$. Applying
Itô formula to stochastic process $R^{\pi}_{t}$ and $f(x)$, we have

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

where

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

By lemma 6.1 the second term in the right-hand side of last equation is
nonpositive. Since $f^{{}^{\prime}}(R_{s\wedge\tau^{\varepsilon}}^{\pi})\leq
f^{{}^{\prime}}(\varepsilon)$, the third term is a square integrable
martingale. Taking expectations on both sides of Eq.(7.11) and then letting
$\varepsilon\rightarrow 0$ one has

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

Since $f^{{}^{\prime}}(x)\geq 1$ for $x\geq 0$,

$\displaystyle f(R_{s}^{\pi})-f(R_{s-}^{\pi})\leq-(L_{s}^{\pi}-L_{s-}^{\pi}).$
(7.13)

So the inequalities (7) and (7.13) yield

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

By the definition of $\tau^{\pi}_{x}$ ,$f(0)=0$ and $f^{\prime}(x)\geq 1$, it
easily follows that

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

So we deduce from the inequalities ( 7.14 ) and (7.15 ) that

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

Therefore

$\displaystyle V(x)\leq f(x).$

If we choose the control strategy
$\pi_{b_{o}}^{\ast}=\\{U^{*}_{b_{0}}(R^{\pi_{b_{o}}^{\ast}}_{\cdot}),L^{\pi_{b_{o}}^{\ast}}_{\cdot}\\}$
and stochastic process
$(R^{\pi_{b_{o}}^{\ast}}_{t},L^{\pi_{b_{o}}^{\ast}}_{t})$ as in SDE (7.5, the
inequalities above become equalities, so

$\displaystyle V(x)\leq f(x)=V(x,b_{0}).$

(i) thus follows. (ii) Assume $b\geq b_{0}$. Let $(R^{\pi}_{t},L^{\pi}_{t})$
be the process as in (2.4) for $\pi\in\Pi_{b}$. Then

$\displaystyle\left\\{\begin{array}[]{l l l}\mathbf{P}\\{R_{s-}^{\pi}\geq
R_{s}^{\pi}\geq b\\}+\mathbf{P}\\{b\geq R_{s-}^{\pi}\geq R_{s}^{\pi}\\}=1,\
\forall s\geq 0,\\\ \mathcal{L}g(R_{s}^{\pi})\leq 0\ \mbox{
$s\leq\tau^{\pi}_{x}=\inf\\{t\geq 0:R^{\pi}_{s}\leq 0\\}$},\\\
g^{\prime}(x)=1\ \mbox{ $x\geq b$}.\end{array}\right.$ (7.19)

Replacing $f$ in proof of (i) above with $g$, then using (7.19) and the same
argument as in (i) we can get

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

Similarly, letting
$\pi_{b}^{\ast}=\\{U^{*}_{b}(R^{\pi_{b}^{\ast}}_{\cdot}),L^{\pi_{b}^{\ast}}_{\cdot}\\}$
we derive $V(x,b)=g(x)$. Therefore (ii) follows. ∎

Now we turn to proof of Theorem 3.1. Proof of Theorem 3.1. If
$\mathbf{P}[\tau_{b_{0}}^{\pi^{*}_{b_{0}}}\leq T]\leq\varepsilon$, then the
conclusion is obvious because the constraints does not work and the proof
reduces to the usual optimal control problem. 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 equation
$\mathbf{P}\\{\tau_{b}^{\pi^{*}_{b}}\leq T\\}=\varepsilon$ and
$x_{2}<b^{\ast}=\inf\\{b:b\in\mathfrak{B}\\}>b_{0}$. By theorem 7.1 we know
that $b^{\ast}$ meets (3.6) and (3.7) because $V(x,b)=g(x,b)$ is a decreasing
function of $b(\geq b_{0})\in[x_{2},b_{K})$ due to lemma 6.3. So the optimal
strategy associated with the optimal return function $V(x,b^{*})=g(x,b^{*})$
is
$\\{U^{\ast}_{b^{*}}(R^{\pi_{b^{*}}^{\ast}}_{t}),L^{\pi_{b^{*}}^{\ast}}_{t}\\}$
and $\\{R^{\pi_{b^{*}}^{\ast}}_{t},L^{\pi_{b^{*}}^{\ast}}_{t}\\}$ is uniquely
determined by SDE (3.12). Thus we complete the proof. $\Box$

##  8\. Appendix

In this section we will give the proofs of lemmas we concerned with throughout
this paper. Proof of lemma 6.1. Since the proof is complete similar to that of
Guo Xin, Liu Jun and Zhou Xunyu[11](2004), we omit it here. Proof of lemma
6.2. Since the proof is somewhat similar to that of He, Hou and Liang
[13](2008) and Guo, Liu and Zhou [11](2004), we only give the sketch of the
proof as follows.
If the $\max$ in (6.2) is attained in the interior of the control region,
then, by differentiating w.r.t. $U$, we can find the maximizing function
$U^{*}_{b}(x)$ can be defined by (6.29) above, and

$\displaystyle\eta(x)=\frac{\mu
g^{\prime}(x)}{2ag^{\prime}(x)-\sigma^{2}g^{\prime\prime}(x)}=\frac{2cg(x)}{\mu
g^{\prime}(x)}+\frac{2\delta}{\mu},0\leq x\leq b.$ (8.1)

Letting $x\rightarrow 0+$ we have $\eta(x)\rightarrow 2\delta/\mu<l$, whereas
taking $x\rightarrow b_{0}$ and noticing that $g^{\prime\prime}(b_{0})=0$, we
also have $\eta(x)\rightarrow\mu/2a>1$. So by (6.29) we find $0\leq
x_{1}<x_{2}\leq b_{0}<b$ such that

$\displaystyle U^{*}(x)=\left\\{\begin{array}[]{l l l}l,&0\leq x\leq x_{1},\\\
\eta(x)&x_{1}<x<x_{2},\\\ 1,&x_{2}\leq x\leq b.\end{array}\right.$ (8.5)

Putting this expression into (6.2) we have (6.8). Then by smooth fit principle
we can determine parameters $\alpha_{1}$, $\beta_{1}$,$\alpha_{2}$,
$\beta_{2}$, $x_{1}$, $x_{2}$, $A$, $B$, $C$, $G$,$H$ and $K$ by (6.9)-(6.18).
Now it remains to prove the solution $g$ defined by (6.8) satisfies (6.3). We
only need to prove

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

For $x\geq b$, we first prove $g^{\prime\prime}(b-)\geq 0$. Noticing that

$\displaystyle
g^{\prime\prime}(b-)=\frac{\alpha_{2}^{2}v(\beta_{2})e^{\beta_{2}x_{2}+\alpha_{2}b}-\beta_{2}^{2}v(\alpha_{2})e^{\alpha_{2}x_{2}+\beta_{2}b}}{v(\beta_{2})e^{\beta_{2}x_{2}+\alpha_{2}b}\alpha_{2}-v(\alpha_{2})e^{\alpha_{2}x_{2}+\beta_{2}b}\beta_{2}},$

where $v(t)\equiv-c+(\frac{1}{2}\mu-\delta)t$. Since
$|\alpha_{2}|\leq|\beta_{2}|$, $v(\beta_{2})<v(\alpha_{2})<0$,
$\alpha_{2}>0,\beta_{2}<0$ and $g^{\prime\prime}(b_{0})=0$, the numerator and
denominator of (8) are strictly negative, so $g^{\prime\prime}(b-)\geq 0$ .
Then by (6.2)

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

Thus we complete the proof.

Proof of lemma 6.3. If $x\geq b$, then using (6.8) and $g^{\prime}(x)=1$ one
has

$\displaystyle\frac{\partial}{\partial b}g(b,x)$ $\displaystyle=$
$\displaystyle-1+\alpha_{2}Be^{\alpha_{2}b}+\beta_{2}Ce^{\beta_{2}b}+B^{\prime}e^{\alpha_{2}b}+C^{\prime}e^{\beta_{2}b}$
(8.7) $\displaystyle=$ $\displaystyle
B^{\prime}e^{\alpha_{2}b}+C^{\prime}e^{\beta_{2}b}$

where $B^{\prime}$ and $C^{\prime}$ denote derivatives w.r.t $b$ of $B$ and
$C$,respectively. By the first three expressions in (6.8) the proof reduces to
showing that for $x_{2}\leq x\leq b$

$\displaystyle B^{\prime}e^{\alpha_{2}x}+C^{\prime}e^{\beta_{2}x}\leq 0.$
(8.8)

By (6.18)-(6.21) one has

$\displaystyle B^{\prime}e^{\alpha_{2}x}+C^{\prime}e^{\beta_{2}x}$
$\displaystyle=$
$\displaystyle\big{\\{}\frac{[v(\alpha_{2})e^{\alpha_{2}x_{2}+\beta_{2}x}-v(\beta_{2})e^{\beta_{2}x_{2}+\alpha_{2}x}]}{[\alpha_{2}v(\beta_{2})e^{\beta_{2}x_{2}+\alpha_{2}b}-\beta_{2}v(\alpha_{2})e^{\alpha_{2}x_{2}+\beta_{2}b}]^{2}}\big{\\}}$
$\displaystyle\times\big{[}\alpha_{2}^{2}v(\beta_{2})e^{\beta_{2}x_{2}+\alpha_{2}b}-\beta_{2}^{2}v(\alpha_{2})e^{\alpha_{2}x_{2}+\beta_{2}b}\big{]}$
$\displaystyle=$ $\displaystyle\frac{K_{1}(x)}{K_{2}^{2}(x_{2})}\times
K_{3}(b).$

where

$\displaystyle K_{1}(x)$ $\displaystyle=$ $\displaystyle
v(\alpha_{2})e^{\alpha_{2}x_{2}+\beta_{2}x}-v(\beta_{2})e^{\beta_{2}x_{2}+\alpha_{2}x},$
$\displaystyle K_{2}(x)$ $\displaystyle=$
$\displaystyle\alpha_{2}v(\beta_{2})e^{\beta_{2}x+\alpha_{2}b}-\beta_{2}v(\alpha_{2})e^{\alpha_{2}x+\beta_{2}b},$
$\displaystyle K_{3}(x)$ $\displaystyle=$
$\displaystyle\alpha_{2}^{2}v(\beta_{2})e^{\beta_{2}x_{2}+\alpha_{2}x}-\beta_{2}^{2}v(\alpha_{2})e^{\alpha_{2}x_{2}+\beta_{2}x}.$

Since $K_{1}(x)$ is an increasing function of $x$ and $K_{1}(x_{2})>0$,
$K_{1}(x)>0$ for $x\in[x_{2},b]$. Noting that $K_{3}(b)\leq 0$, we know that
(8.8) is true. $\Box$

Proof of lemma 6.4. Because the proof of decreasing property is complete
similar to that of theorem 3.1 in [13](2008), we only need to prove that the
probability of bankruptcy is strictly decreasing on $[x_{2},b_{K}]$, that is,

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

for any $b_{2}>b_{1}\geq x_{2}$. By comparison theorem,

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

The proof can be reduced to proving that

$\displaystyle\mathbf{P}[\tau_{b_{1}}^{b_{2}}\leq
T]-\mathbf{P}[\tau_{b_{2}}^{b_{2}}\leq T]>0.$ (8.9)

To prove the inequality (8.9) we define stochastic processes $R_{t}^{[1]}$ and
$R_{t}^{[2]}$ by the following SDEs:

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

respectively, where $U_{b}^{*}(\cdot)$ is as in (6.29). Let
$\tau^{b_{1}}=\inf\limits_{t\geq 0}\\{t:R_{t}^{[2]}=b_{1}\\}$,
$A=\\{\tau^{b_{1}}\leq T\\}$ and $B=\big{\\{}\sl R_{t}^{[2]}$ will go to
bankruptcy in a time interval $[\tau^{b_{1}},\tau^{b_{1}}+T]$ and
$\tau^{b_{1}}\leq T\big{\\}}$. Then $\\{\tau_{b_{2}}^{b_{2}}\leq T\\}\subset
B\subset A$. Moreover, by using strong Markov property of $R_{t}^{[2]}$, we
have

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

So

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

By theorem 4.1, $\mathbf{P}[\tau_{b_{1}}^{b_{2}}\leq
T]\geq\mathbf{P}[\tau_{b_{1}}^{b_{1}}\leq T]>0$. So we only need to prove
$\mathbf{P}(A^{c})>0$. For doing this we define stochastic processes
$R_{t}^{[3]}$ and $R_{t}^{[4]}$ by the following SDEs

$\displaystyle\left\\{\begin{array}[]{l l l}dR_{t}^{[3]}=[\mu
U_{b_{2}}^{*}(R_{t}^{[3]})-a{U_{b_{2}}^{*}}^{2}(R_{t}^{[3]})-\delta]dt+U_{b_{2}}^{*}(R_{t}^{[3]})\sigma
d\mathcal{W}_{t}-dL_{t}^{b_{2}},\\\
R_{0}^{[3]}=\frac{b_{1}+b_{2}}{2}\end{array}\right.$

and

$\displaystyle\left\\{\begin{array}[]{l l l}dR_{t}^{[4]}=[\mu
U_{b_{2}}^{*}(R_{t}^{[4]})-a{U_{b_{2}}^{*}}^{2}(R_{t}^{[4]})-\delta]dt+U_{b_{2}}^{*}(R_{t}^{[4]})\sigma
d\mathcal{W}_{t},\\\ R_{0}^{[4]}=\frac{b_{1}+b_{2}}{2}.\end{array}\right.$

Setting $D=\\{\inf\limits_{0\leq t\leq T}R_{t}^{[3]}>b_{1}\\}$ and
$E=\\{\inf\limits_{0\leq t\leq T}R_{t}^{[4]}>b_{1},\sup\limits_{0\leq t\leq
T}R_{t}^{[4]}<b_{2}\\}$, by comparison theorem on SDE, we have
$\mathbf{P}(A^{c})\geq\mathbf{P}(D)\geq\mathbf{P}(E)$. Since
$U_{b_{2}}^{*}(x)=1$ we have

$\displaystyle
R_{t}^{[4]}=\frac{b_{1}+b_{2}}{2}+[\mu-a-\delta]t+\sigma\mathcal{W}_{t}\
\mbox{ on $E$}.$ (8.12)

We deduce from (8.12) and properties of Brownian motion with drift (cf.
Borodin and Salminen [1] (2002)) that

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

where $\mu^{\prime}=(\mu-a-\delta)/\sigma$ and
$x=\frac{b_{1}+b_{2}}{2\sigma}$. Thus the proof follows. $\Box$ Proof of lemma
6.5. Let $\phi(t,y)\equiv\phi^{b}(t,y)$. Since the stochastic process
$(R^{\pi_{b}^{\ast},y}_{t\wedge\tau_{y}^{b}},L_{t\wedge\tau_{y}^{b}}^{\pi_{b}^{\ast}})$
is continuous, by applying the generalized Itô formula to
$(R^{\pi_{b}^{\ast},y}_{t\wedge\tau_{y}^{b}},L_{t\wedge\tau_{y}^{b}}^{\pi_{b}^{\ast}})$
and $\phi(t,y)$, we have for $0<y\leq b$

$\displaystyle\phi(T-(t\wedge\tau_{y}^{b}),Y_{t\wedge\tau_{y}^{b}}^{b})$
$\displaystyle=$ $\displaystyle\phi(T,y)$ (8.13) $\displaystyle+$
$\displaystyle\int_{0}^{t\wedge\tau_{y}^{b}}(\frac{1}{2}U^{*2}(Y_{s}^{b})\sigma^{2}\phi_{yy}(T-s,Y_{s}^{b})$
$\displaystyle+$ $\displaystyle(\mu
U^{*}(Y_{s}^{b})-a[U^{*}((Y_{s}^{b}))]^{2}-\delta)\phi_{y}(T-s,Y_{s}^{b})$
$\displaystyle-$
$\displaystyle\phi_{t}(T-s,Y_{s}^{b}))ds-\int_{0}^{t\wedge\tau_{y}^{b}}\phi_{y}(T-s,Y_{s}^{b})dL_{s}^{b}$
$\displaystyle+$
$\displaystyle\int_{0}^{t\wedge\tau_{y}^{b}}a(Y_{s}^{b})\sigma\phi_{y}(T-s,Y_{s}^{b})dW_{s}^{b}.$

where
$\tau_{y}^{b}\equiv\tau^{\pi_{b}^{\ast}}_{y}=\inf\\{t:R^{\pi_{b}^{\ast},y}_{t}=0\\}$.
Letting $t=T$ and taking mathematical expectation at both sides of (8.13)
yield that

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

$\Box$ Proof of lemma 6.6. Let $a(y):=\frac{1}{2}[U_{b}^{*}(y)]^{2}\sigma^{2}$
$\mu(y):=\mu U_{b}^{*}(y)-a[U_{b}^{*}(y)]^{2}-\delta$. Then the equation
(6.25) becomes

$\displaystyle\phi_{t}^{b}(t,y)=a(y)\phi_{yy}^{b}(t,y)+\mu(y)\phi_{y}^{b}(t,y).$
(8.14)

By the properties of $U_{b}^{*}(y)$, we can easily show that $a(y)$ and
$\mu(y)$ are continuous in $[0,b]$. So there exists a unique solution in
$C^{1}(0,\infty)\cap C^{2}(0,b)$ for (6.25). Moreover, $a^{{}^{\prime}}(y)$,
$\mu^{{}^{\prime}}(y)$ and $a^{{}^{\prime\prime}}(y)$ are bounded in
$(0,x_{1})$,$(x_{1},x_{2})$ and $(x_{2},b)$. So we only need to prove that
$\phi^{b}(t,x)$ is continuous in $b$. Let $y=bz$ and
$\theta^{b}(t,z)=\phi^{b}(t,by)$, the equation (6.25) becomes

$\displaystyle\left\\{\begin{array}[]{l l
l}\theta_{t}^{b}(t,z)=[a(bz)/b^{2}]\theta_{zz}^{b}(t,z)+[\mu(bz)/b]\theta_{z}^{b}(t,z),\\\
\theta^{b}(0,z)=1,\ \mbox{for}\ 0<z\leq 1,\\\
\theta^{b}(t,0)=0,\theta_{z}^{b}(t,1)=0,\ \mbox{for}\ t>0.\end{array}\right.$
(8.18)

So the proof of Lemma 6.6 reduces to proving $\lim\limits_{b_{2}\rightarrow
b_{1}}\theta^{b_{2}}(t,z)=\theta^{b_{1}}(t,z)$ for fixed $b_{1}>b_{0}$.
Setting $w(t,z)=\theta^{b_{2}}(t,z)-\theta^{b_{1}}(t,z)$, we have

$\displaystyle\left\\{\begin{array}[]{l l
l}w_{t}(t,z)&=&[a(b_{2}z)/b_{2}^{2}]w_{zz}(t,z)+[\mu(b_{2}z)/b_{2}]w_{z}(t,z)\\\
&+&\\{a(b_{2}z)/b_{2}^{2}-a(b_{1}z)/b_{1}^{2}\\}\theta_{zz}^{b_{1}}(t,z)\\\
&+&\\{a(b_{2}z)/b_{2}^{2}-a(b_{1}z)/b_{1}^{2}\\}\theta_{z}^{b_{1}}(t,z),\\\
w(0,z)&=&0,\ \mbox{for}\ 0<z\leq 1,\\\ w(t,0)&=&0,\ w_{x}(t,1)=0,\ \mbox{for}\
t>0.\end{array}\right.$ (8.24)

By multiplying the first equation in (8.24) by $w(t,z)$ and then integrating
on $[0,t]\times[0,1]$,

$\displaystyle\int_{0}^{t}\int_{0}^{1}w(s,x)w_{t}(s,x)dxds$ (8.25)
$\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}\big{\\{}[a(b_{2}x)/b_{2}^{2}]w(s,x)w_{xx}(s,x)$
$\displaystyle+$ $\displaystyle[\mu(b_{2}x)/b_{2}]w(s,x)w_{x}(s,x)$
$\displaystyle+$
$\displaystyle[a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}]w(s,x)\theta_{xx}^{b_{1}}(t,x)$
$\displaystyle+$ $\displaystyle
w(s,x)[\mu(b_{2}x)/b_{2}-\mu(b_{1}x)/b_{1}]w(s,x)\theta_{x}^{b_{1}}(t,x)\big{\\}}dxds$
$\displaystyle\equiv$ $\displaystyle E_{1}+E_{2}+E_{3}+E_{4}.$

We now estimate terms $E_{i}$, $i=1,\cdots,4$, as follows. By definitions of
$a(x)$ and $\mu(x)$, there exit positive constants $D_{1}$, $D_{2}$ and
$D_{3}$ such that $[\mu(b_{2}z)/b_{2}]^{2}\leq D_{1}$,
$[a(bx)/b^{2}]^{\prime}\geq 0$, $[a(b_{2}x)/b_{2}^{2}]\geq D_{2}$ and
$[a(b_{2}x)/b_{2}^{2}]^{\prime}\leq D_{3}$, so by Young’s inequality, we have
for any $\lambda_{1}>0$ and $\lambda_{2}>0$

$\displaystyle E_{1}$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}[a(b_{2}x)/b_{2}^{2}]w(s,x)w_{xx}(s,x)dxds$
(8.26) $\displaystyle=$
$\displaystyle-\int_{0}^{t}\int_{0}^{1}[a(b_{2}x)/b_{2}^{2}]w_{x}^{2}(s,x)dxds$
$\displaystyle-\int_{0}^{t}\int_{0}^{m/b_{2}}[a(b_{2}x)/b_{2}^{2}]^{{}^{\prime}}w_{x}(s,x)w(s,x)dxds$
$\displaystyle\leq$ $\displaystyle-
D_{2}\int_{0}^{t}\int_{0}^{1}w_{x}^{2}(s,x)dxds$
$\displaystyle+D_{3}\int_{0}^{t}\int_{0}^{1}[\lambda_{1}w_{x}^{2}(s,x)+\frac{1}{4\lambda_{1}}w^{2}(s,x)]dxds$

and

$\displaystyle E_{2}$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}[\mu(b_{2}x)/b_{2}]w(s,x)w_{x}(s,x)dxds$
(8.27) $\displaystyle\leq$
$\displaystyle\lambda_{2}\int_{0}^{t}\int_{0}^{1}w_{x}^{2}(s,x)dxds$
$\displaystyle+\frac{D_{1}}{4\lambda_{2}}\int_{0}^{t}\int_{0}^{1}w^{2}(s,x)dxds.$

We decompose $E_{3}$ as follows.

$\displaystyle E_{3}$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}w(s,x)\theta_{xx}^{b_{1}}(s,x)dxds$
(8.28) $\displaystyle=$
$\displaystyle-\int_{0}^{t}\int_{0}^{1}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}w_{x}(s,x)\theta_{x}^{b_{1}}(s,x)dxds$
$\displaystyle-\int_{0}^{t}\\{\int_{0}^{x_{1}/b_{2}}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}^{\prime}w(s,x)\theta_{x}^{b_{1}}(s,x)dx$
$\displaystyle-\int_{x_{1}/b_{2}}^{x_{1}/b_{1}}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}^{\prime}w(s,x)\theta_{x}^{b_{1}}(s,x)dx$
$\displaystyle-\int_{x_{1}/b_{1}}^{x_{2}/b_{2}}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}^{\prime}w(s,x)\theta_{x}^{b_{1}}(s,x)dx$
$\displaystyle-\int_{x_{2}/b_{2}}^{x_{2}/b_{1}}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}^{\prime}w(s,x)\theta_{x}^{b_{1}}(s,x)dx$
$\displaystyle-\int_{x_{2}/b_{1}}^{1}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}w(s,x)^{\prime}\theta_{x}^{b_{1}}(s,x)dx\\}ds$
$\displaystyle=$
$\displaystyle-\int_{0}^{t}\int_{0}^{1}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}w_{x}(s,x)\theta_{x}^{b_{1}}(s,x)dxds$
$\displaystyle-\int_{0}^{t}\\{\int_{x_{1}/b_{2}}^{x_{1}/b_{1}}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}^{\prime}w(s,x)\theta_{x}^{b_{1}}(s,x)dx$
$\displaystyle-\int_{x_{1}/b_{1}}^{x_{2}/b_{2}}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}^{\prime}w(s,x)\theta_{x}^{b_{1}}(s,x)dx$
$\displaystyle-\int_{x_{2}/b_{2}}^{x_{2}/b_{1}}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}^{\prime}w(s,x)\theta_{x}^{b_{1}}(s,x)dx\\}ds$
$\displaystyle:=$ $\displaystyle E_{30}+E_{31}+E_{32}+E_{33}.$

It easily follows that

$\displaystyle\lim_{b_{2}\rightarrow b_{1}}\\{|E_{31}|+|E_{33}|\\}=0.$

Since there exists an $L>0$ such that for or all
$x\in(x_{1}/b_{2},x_{2}/b_{1})$

$\displaystyle\left\\{\begin{array}[]{l l
l}|[a(b_{2}x)/b_{2}^{2}]-[a(b_{1}x)/b_{1}^{2}]|\leq L|b_{2}-b_{1}|,\\\
|[a(b_{2}x)/b_{2}^{2}]^{\prime}-[a(b_{1}x)/b_{1}^{2}]^{\prime}|\leq
L|b_{2}-b_{1}|,\\\ |[\mu(b_{2}x)/b_{2}]-[\mu(b_{1}x)/b_{1}]|\leq
L|b_{2}-b_{1}|,\end{array}\right.$ (8.32)

we have for any $\lambda_{3}>0$

$\displaystyle E_{30}+E_{32}$ $\displaystyle=$
$\displaystyle-\int_{0}^{t}\int_{0}^{1}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}w_{x}(s,x)\theta_{x}^{b_{1}}(s,x)dxds$
$\displaystyle-$
$\displaystyle\int_{0}^{t}\int_{x_{1}/b_{1}}^{x_{2}/b_{2}}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}^{\prime}w(s,x)\theta_{x}^{b_{1}}(s,x)dxds$
$\displaystyle\leq$
$\displaystyle\frac{L^{2}(b_{2}-b_{1})^{2}}{4\lambda_{3}}\int_{0}^{t}\int_{0}^{1}[\theta_{x}^{b_{1}}(s,x)]^{2}dxds$
$\displaystyle+\lambda_{3}\int_{0}^{t}\int_{0}^{1}w_{x}^{2}(s,x)+w^{2}(s,x)dxds.$

By the boundary conditions, we estimate
$\int_{0}^{t}\int_{0}^{1}[\theta_{x}^{b}(s,x)]^{2}dxds$ for
$b\in[b_{1},b_{2}]$ as follows:

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}\theta_{t}^{b}(s,x)\theta^{b}(s,x)$
$\displaystyle-[a(bx)/b^{2}]\theta_{xx}^{b}(s,x)\theta^{b}(s,x)-[\mu(bx)/b]\theta_{x}^{b}(s,x)\theta^{b}(s,x)dxds$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{0}^{1}[\theta^{b}(s,x)]^{2}dx+\int_{0}^{t}\int_{0}^{1}[a(bx)/b^{2}][\theta_{x}^{b}(s,x)]^{2}dxds$
$\displaystyle+\int_{0}^{t}\int_{0}^{1}[a(bx)/b^{2}]^{\prime}[\theta_{x}^{b}(s,x)][\theta^{b}(s,x)]dxds$
$\displaystyle-\int_{0}^{t}\int_{0}^{1}[\mu(bx)/b][\theta_{x}^{b}(s,x)][\theta^{b}(s,x)]dxds$
$\displaystyle\geq$
$\displaystyle\lambda\int_{0}^{t}\int_{0}^{1}[\theta_{x}^{b}(s,x)]^{2}dxds-\frac{\lambda}{2}\int_{0}^{t}\int_{0}^{1}[\theta_{x}^{b}(s,x)]^{2}dxds$
$\displaystyle-\frac{D_{4}}{2\lambda}\int_{0}^{t}\int_{0}^{1}[\theta^{b}(s,x)]^{2}dxds$
$\displaystyle\geq$
$\displaystyle\frac{\lambda}{2}\int_{0}^{t}\int_{0}^{1}[\theta_{x}^{b}(s,x)]^{2}dxds-\frac{D_{4}t}{2\lambda},$

where $|[a(bz)/b^{2}]^{\prime}-[\mu(bx)/b]|^{2}<D_{4}$ and
$\lambda=\frac{l^{2}}{8b_{1}^{2}}$, so

$\displaystyle\int_{0}^{t}\int_{0}^{1}[\theta_{x}^{b}(s,x)]^{2}dxds\leq\frac{D_{4}t}{\lambda^{2}}.$

Therefore there exists a function $B^{b_{1}}(b_{2})$ satisfying

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

such that for $0\leq t\leq T$

$\displaystyle E_{3}$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}\\{a(b_{2}x)/b_{2}^{2}-a(b_{1}x)/b_{1}^{2}\\}w(s,x)\theta_{xx}^{b_{1}}(t,x)dxds$
(8.33) $\displaystyle\leq$ $\displaystyle
B^{b_{1}}(b_{2})+\lambda_{3}\int_{0}^{t}\int_{0}^{1}w_{x}^{2}(s,x)+w^{2}(s,x)dxds.$

Similarly,

$\displaystyle E_{4}$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}\\{\mu(b_{2}x)/b_{2}-\mu(b_{1}x)/b_{1}\\}w(s,x)\theta_{x}^{b_{1}}(t,x)dxds$
(8.34) $\displaystyle\leq$ $\displaystyle
B_{1}^{b_{1}}(b_{2})+\int_{0}^{t}\int_{0}^{1}w^{2}(s,x)dxds.$

with

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

Choosing $\lambda_{1}$, $\lambda_{2}$ and $\lambda_{3}$, by (8.25)-(8.27),
(8.33)-(8.34), and

$\displaystyle\int_{0}^{t}\int_{0}^{1}w(s,x)w_{t}(s,x)dxds$ $\displaystyle=$
$\displaystyle\int_{0}^{1}\frac{1}{2}w^{2}(t,x)dx,$

there exist constants $C_{1}$ and $C_{2}$ such that

$\displaystyle\int_{0}^{1}w^{2}(t,x)dx\leq
C_{1}\int_{0}^{t}\int_{0}^{1}w^{2}(s,x)dxds+C_{2}[B_{1}^{b_{1}}(b_{2})+B^{b_{1}}(b_{2})].$

By setting $F(t)=\int_{0}^{t}\int_{0}^{1}w^{2}(s,x)dxds$ and using the
Gronwall inequality, we get

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

so

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

Thus the proof is complete. $\Box$ Acknowledgements. This work is supported by
Projects 10771114 and 11071136 of NSFC, Project 20060003001 of SRFDP, and SRF
for ROCS, SEM, and the Korea Foundation for Advanced Studies. We would like to
thank the institutions for the generous financial support. We also express our
deep thanks to Professor Huaiyu Jian, Wen’an Yong and Dr Zhijie Chen for their
valuable comments and suggestions. 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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