1. Introduction
In [1], Shiryaev and Peskir have considered the problem:
(1.1)
where 
is standard Brownian process and they found the optimal stopping time as:
(1.2)
where 
is the solution of the equation
and
.
The simulation for
, boundary 
and 
are given in Figures 1-3.
In [2], Albert Shiryaev, Zouquan Xu and Xun Yu Zhou solve the following problem: There is an investor holding a stock, and he needs to decide when to sell it for the last time with given time to sell T. It is obvious that he wants to sell at a time of highest price on the interval from 0 to T. Assume that the discounted share price 
complies with the following dynamic equation:
on a filtered probability space 
where a is the growth rate of the price and 
is the volatility, r is interest rate, 
is the standard Brownian motion with 
under the measure P. Here, 
is Pincreasing filter generated by
. Then, 
Figure 2. A simulation for the stopping time in (1.2). In this case 
is less than 0 but is not
.
Figure 3. A simulation for the stopping time in (1.2). In this case 
is very large but is not
.
where
. We define: 
and
.
Then, the following cases:
• If
, 
is an unique optimal selling time.
• If
,
or 
are optimal selling times.
• If
, 
is an unique optimal selling time.
In this paper, we will find the optimal time to sell a stock when the appreciation rate is the random variable taking one of two given values 
and
.
2. The Problem of Finding the Optimal Selling Time
Assume that the asset price process Xt follows a geometric Brownian motion with its drift is a random variable taking one of two given values 
or
, the volatility 
is constant, i.e.
(2.1)
where 
is a standard Brownian motion independent with 
of the probability space
. Assume 
is a complete probability space with nondecrease 
-field. Suppose 
and 
satisfy al < r < ah, where r is the interest rate and it is constant and the initial value of assets 
is a positive constant.
Investors holding assets need to decide when to sell it for the last time with given time to sell them is T. Knowing that at the initial time distribution of α as
At time 
we put
, where 
is the completion of the filtration generated by X.
The problem is finding 
-stopping time τ, 0 ≤ τ ≤ T such that
(2.2)
where supremum is taken in 
-stopping time τ, 0 ≤ τ ≤ T.
The price process 
and posterior probability process 
satisfying the equations
(2.3)
where 
is a P-Brownian motion defined by:
(see [3], theorem 9.1)
Define the process 
by:
and a new measure 
satisfying:
where
.
According to the Girsanov theorem, 
is a 
Brownian motion. Let
, we have
Then, price process 
and process 
satisfy the equations
(2.4)
So, X and 
are geometric Brownian motions under measure
. Moreover, 
-field generated by 
coincides with the one generated by X.
We define the likelihood process
We know that 
is a 
-Brownian motion, so 
is an 
-martingale under measure
.
We have:
Denote that 
is an expectation operator with respect to measure 
and let 
is an 
-stopping time. Then, by the property 
-martingale under measure 
of 
(see [4], theorem 11), we have:
(2.5)
Lemma 1. 
can be written as:
(2.6)
where: 
and 
Proof.
We have:
and
We consider an optimal stopping problem as:
(2.7)
where:
where supremum is taken in 
-stopping time 
with respect to filtration generated by 
It can be seen that the optimal stopping time in (2.7) can be turned to the optimal stopping time in the problem (2.2).
Now, we study the optimal stopping problem (2.7). We will prove that existing an increasing and continuous monotone function:
such that the stopping time
is an optimal stopping time for the problem (2.7).
satisfies the equation:
On the other hand, we can write 
where
With this notation, we have:
Give 
in (2.7), we have
Define the continuation region C:
and the stopping region D:
According to general theory about optimal stopping problems, the stopping time
is optimal stopping time problem (2.7). Thus, determining the optimal stopping time is sufficient to defining the stopping region D.
Theorem 1. There exists a right continuous and nondecreasing function
such that
Furthermore, supremum in (7) is achieved with the stopping time
Proof.
We know that
with every fix 
and
, assume that
, there exists a stopping time 
such that:
So:
And process 
is a submartingale, so we have:
Therefore,
. This proves that remaining a function 
such that: 
We have 
to be a submartingale for
, so all points in region
belong to the continuation region.
Therefore,
. The monotonicity of b follows from monotonicity of function
.
The right continuity of 
follows from the continuation region C is an open set.
Theorem 2. Assume that b is the function described above whose existence is proved in Lemma 1. Define stopping time:
(2.8)
Then, 
attains the supremum in (2.2).
Proof. Deduce directly from Theorem 1 and Lemma 1 by replacing 
by
.
Theorem 3. The optimal stopping boundary b(t) satisfies the integral equation (see Equation (2.9) below):
where 
Proof.
Fix 
and
. Then,
where
.
But:
Consider:
We have:
where
.
Let 
and
then 
In a similar way, we have
where
.
Put 
we attain (2.9).
3. The Numerical Solution of the Integral Equation and Simulation Results
Below we follow [5], devided 
by the points
, 
which
, then Equation (2.8) can be discretized as:
For
, we have equation
Due to
, from the above equation we determine
, continue to
, we obtain the following equation for determining
:
Just do so until
, it has been determined
. Thus, we obtain a sequence of values 
of 
and approximate the optimal boundaries for the asset liquidation process.
We have the approximate solution of Equation (2.9) by a computer program written in Matlab software, then set the boundaries for the process 
is
. Then the optimal time to sell is the first time the line describes the process 
lies below the line describes the process
. These figures (Figures 4-19) and tables below (Tables 1-3) illustrates the stopping time in (2.8) and the solutions of Equation (2.9) in some cases.
Figure 4. The line describes b (t) with al = 0.1; ah = 0.2; r = 0.15; σ = 0.2.
Figure 5. A simulation for the stopping time in (2.8) with parameters X0 = 3; al = 0.1; ah = 0.2; r = 0.15; σ = 0.2; π0 = 0.5.
Figure 6. A simulation for the stopping time in (2.8) with parameters X0 = 3; al = 0.1; ah = 0.2; r = 0.15; σ = 0.2; π0 = 0.2. In this case π0 = 0.2 is small so is τ*.
Figure 7. A simulation for the stopping time in (2.8) with parameters X0 = 3; al = 0.1; ah = 0.2; r = 0.15; σ = 0.2; π0 = 0.4.
Figure 8. The line describes b (t) with al = 0.09; ah = 0.15; r = 0.11; σ = 0.1.
Figure 9. A simulation for the stopping time in (2.8) with parameters X0 = 3; al = 0.09; ah = 0.15; r = 0.11; σ = 0.1; π0 = 0.5.
Figure 10. A simulation for the stopping time in (2.8) with parameters X0 = 3; al = 0.09; ah = 0.15; r = 0.11; σ = 0.1; π0 = 0.4.
Figure 11. A simulation for the stopping time in (2.8) with parameters X0 = 3; al = 0.09; ah = 0.15; r = 0.11; σ = 0.1; π0 = 0.3.
Figure 12. A simulation for the stopping time in (2.8) with parameters X0 = 3; al = 0.09; ah = 0.15; r = 0.11; σ = 0.1; π0 = 0.2.
Figure 13. The line describes b (t) with al = 0.09; ah = 0.15; r = 0.13; σ = 0.1.
Figure 14. A simulation for the stopping time in (2.8) with parameters X0 = 3; al = 0.09; ah = 0.15; r = 0.13; σ = 0.1; π0 = 0.2.
Figure 15. A simulation for the stopping time in (2.8) with parameters X0 = 3; al = 0.09; ah = 0.15; r = 0.13; σ = 0.1; π0 = 0.4.
Figure 16. A simulation for the stopping time in (2.8) with parameters X0 = 3; al = 0.09; ah = 0.15; r = 0.13; σ = 0.1; π0 = 0.5.
Figure 17. A simulation for the stopping time in (2.8) with parameters X0 = 3; al = 0.09; ah = 0.15; r = 0.13; σ = 0.1; π0 = 0.5.
Figure 18. A simulation for the stopping time in (2.8) with parameters X0 = 3; al = –0.3; ah = 0.5; r = 0.1; σ = 0.2; π0 = 0.4.
Figure 19. A simulation for the stopping time in (2.9) with parameters X0 = 3; al = –0.3; ah = 0.5; r = 0.1; σ = 0.2; π0 = 0.6.
Table 1. The numerical solutions of (2.9) with al = 0.1; ah = 0.2; r = 0.15; σ = 0.2.
Table 2. The numerical solutions of (2.8) with al = 0.09; ah = 0.15; r = 0.11; σ = 0.1.
Table 3. The numerical solutions of (2.9) with al = 0.09; ah = 0.15; r = 0.13; σ = 0.1.
4. Conclusion
This paper solves the problem to find the optimal stopping time for the holding asset and make a decision when to sell assets with discounted price reaching the greatest expected value. The optimal stopping time is the first time the price of the asset hit the boundary or be at the time T. In next study, we will study the distributions and characteristics of the optimal stopping time.