International Journal of Mathematics and Mathematical SciencesVolume 2011 (2011), Article ID 264603, 28 pagesdoi:10.1155/2011/264603

Research Article

Optimal Selling Rule in a Regime Switching Lévy Market

Moustapha Pemy

Department of Mathematics, Towson University, Towson, MD 21252-0001, USA

Received 5 March 2011; Accepted 5 April 2011

Academic Editor: Giuseppe Marino

Copyright © 2011 Moustapha Pemy. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

This paper is concerned with a finite-horizon optimal selling rule problem when the underlying stock price movements are modeled by a Markov switching Lévy process. Assuming that the transaction fee of the selling operation is a function of the underlying stock price, the optimal selling rule can be obtained by solving an optimal stopping problem. The corresponding value function is shown to be the unique viscosity solution to the associated HJB variational inequalities. A numerical example is presented to illustrate the results.

1. Introduction

One of the major decision investors have to make on a daily basis is to identify the best time to sell or buy a particular stock. Usually if the right decision is not taken at the right time, this will generally result in large losses for the investor. Such decisions are mainly affected by various macro- and micro-economical parameters. One of the main factors that affect decision making in the marketplace is the trend of the stock market. In this paper, we study trading decision making when we assume that market trends are subject to change and that these fluctuations can be captured by a combination of a latent Markov chain and a jump process. In fact, we model the stock price dynamics with a regime switching Lévy process. Regime switching Lévy processes are obtained by combining a finite number of geometric Lévy processes modulated by a finite-state Markov chain. This type of processes clearly capture the main features of a wide variety of stock such as energy stock and commodities which usually display a lot of spikes and seasonality. Selling rule problems in general have been intensively studied in the literature, and most of the work have been done when the stock price follows a geometric Brownian motion or a simple Markov switching process. Among many others, we can cite the work of Zhang [1]; in this paper, a selling rule is determined by two threshold levels, and a target price and a stop-loss limit are considered. One makes a selling decision whenever the price reaches either the target price or the stop-loss limit. The objective is to choose these threshold levels to maximize an expected return function. In [1], such optimal threshold levels are obtained by solving a set of two-point boundary value problems. Recently Pemy and Zhang [2] studied a similar problem in the case where there is no jump process associated and the underlying dynamics is just a traditional Markov switching process built by coupling a set of geometric Brownian motions.

In this paper, we extend the result of Pemy and Zhang [2], we consider an optimal selling rule among the class of almost all stopping times under a regime switching Lévy model. We study the case when the stock has to be sold within a prespecified time limit. Given a transaction cost which is a function of the underlying stock price, the objective is to choose a stopping time so as to maximize an expected return. The optimal stopping problem was studied by McKean [3] back to the 1960s when there is no switching; see also Samuelson [4] in connection with derivative pricing and Øksendal [5] for optimal stopping in general. In models with regime switching, Guo and Zhang [6] considered the model with a two-state Markov chain. Using a smooth-fit technique, they were able to convert the optimal stopping problem to a set of algebraic equations under certain smoothness conditions. Closed-form solutions were obtained in these cases. However, it can be shown with extensive numerical tests that the associated algebraic equations may have no solutions. This suggests that the smoothness () assumption may not hold in these cases. Moreover, the results in [5, 6] are established on an infinite time horizon setup. However, in practice, an investor often has to sell his stock holdings by a certain date due to various nonprice-related considerations such as year-end tax deduction or the need for raising cash for major purchases. In these cases, it is necessary to consider the corresponding optimal selling with a finite horizon. It is the purpose of this paper to treat the underlying finite horizon optimization problem with possible nonsmoothness of the solutions to the associated HJB variational inequalities. We resort to the concept of viscosity solutions and show that the corresponding value function is indeed the only viscosity solution to the HJB variational inequalities. We clearly prove that the value function of this optimal stopping time problem is the unique viscosity solution to the associated HJB variational inequalities, which enables us to run some numerical schemes in order to approximate the value function and derive the both the continuation region and the stopping region. It is well known that the optimal stopping rule can be determined by the corresponding value function; see, for example, Krylov [7] and Øksendal [5] for diffusions, Pham [8] for jump diffusions, and Guo and Zhang [6] and [9] for regime switching diffusions.

The paper is organized as follows. In the next section, we formulate the problem under consideration and then present the associated HJB inequalities and their viscosity solutions. In Section 3, we obtain the continuity property of the value function and show that it is the only viscosity solution to the HJB equations. In Section 4, we give a numerical example in order to illustrate our results. To better present the results without undue technical difficulties, all proofs are moved to the appendix placed at the end of the paper.

2. Problem Formulation

Given an integer , let denote a Markov chain with an matrix generator , that is, for and for and a Lévy process . Let be the Poisson random measure of , then it is defined as follows: for any Borel set , The differential form of is denoted by . Let be the Lévy measure of ; we have for any Borel set . We define the differential form as follows: From Lévy-Khintchine formula, we have In our regime switching Lévy market model, the stock price denoted by satisfies the following Lévy stochastic differential equation where is the initial price and is a finite time. For each state , the rate of return, the volatility and the jump intensity are known and satisfied the linear growth condition. There exists a constant such that for all and all , we have is the standard Weiner process, and represents the differential form of the jump measure of . The processes , , and are defined on a probability space and are independent of each other.

In this paper, we consider the optimal selling rule with a finite horizon . We assume that the transaction cost function is the function of the stock price itself. In this case, we take into account all costs associated with the selling operation. The main objective of this selling problem is to sell the stock by time so as to maximize the quantity , where is a discount rate.

Let and let denote the set of -stopping times such that a.s. The value function can be written as follows:

Given the value function , it is typical that an optimal stopping time can be determined by the following continuation region: as follows: It can be proved that if , then Thus, is the optimal stopping time; see [9].

The process is a Markov process with generator defined as follows: where The corresponding Hamiltonian has the following form: Note that for all . Let . Formally, the value function satisfies the HJB equation In order to study the possibility of existence and uniqueness of a solution of (2.12), we use a notion of viscosity solution introduced by Crandall et al. [10].

Definition 2.1. We say that is a viscosity solution of Iffor all , and for each , is continuous in , moreover, there exist constants and such that for each , whenever such that has local maximum at , and for each , whenever such that has local minimum at . Let be a function that satisfies (2.3). It is a viscosity subsolution (resp. supersolution) if it satisfies (2.4) (resp. (2.5)).

3. Properties of Value Functions

In this section, we study the continuity of the value function, show that it satisfies the associated HJB equation as a viscosity solution, and establish the uniqueness. We first show the continuity property.

Lemma 3.1. For each , the value function is continuous in . Moreover, it has at most linear growth rate, that is, there exists a constant such that .

The continuity of the value function and its at most linear growth will be very helpful in deriving the maximum principle which itself guarantees the uniqueness of the value function. The following lemma is a simple version of the dynamic programming principle in optimal stopping. A similar result has been proven in Pemy [9]. For general dynamic programming principle, see Krylov [7] for diffusions, Pham [8] for jump diffusions, and Yin and Zhang [11] for dynamic models with regime switching.

Definition 3.2. For each , a stopping time is said to be -optimal if

Lemma 3.3. Let two stopping time such that a.s., then one has In particular for any stopping time , one has Let such that for any , where an -optimal stopping time. Then, one has With Lemma 3.3 at our hand, we proceed and show that the value function is a viscosity solution of the variational inequality (2.13).

Theorem 3.4. The value function is a viscosity solution of (2.13).

3.1. Uniqueness of the Viscosity Solution

In this subsection, we will prove that the value function defined in (2.6) is the unique viscosity solution of the HJB equation (2.13). We begin by recalling the definition of parabolic superjet and subjet.

Definition 3.5. Let . Define the parabolic superjet by and its closure is Similarly, we define the parabolic subjet and its closure

We have the following result.

Lemma 3.6. (resp. ) consist of the set of where and has a global maximum (resp. minimum) at .

A proof can be found in Fleming and Soner [12].

The following result from Crandall et al. [10] is crucial for the proof of the uniqueness.

Theorem 3.7 (Crandall et al. [10]). For , let be locally compact subsets of , and , and let be upper semicontinuous in , and the parabolic superjet of , and twice continuously differentiable in a neighborhood of . Set for , and suppose is a local maximum of relative to . Moreover, let us assume that there is an such that for every there exists a such that for Then, for each , there exists such that

We have the following maximum principle.

Theorem 3.8 (Comparison Principle). If and are continuous in and are, respectively, viscosity subsolution and supersolution of (2.13) with at most a linear growth. Then,

The following lemma is be very useful in derivation of the maximum principle.

Lemma 3.9. Let be the set of functions on which are continuous with respect to and Lipschitz continuous with respect to the variable . For fixed and , let the operator on be defined as follows: Then, there exits a constant such that for any and .

Remark 3.10. Theorem 3.8 obviously implies the uniqueness of the viscosity solution of the variational inequality (2.13). If we assume that (2.13) has two solutions and with linear growth, then they are both viscosity subsolution and supersolution of (2.13). Therefore, using the fact that is subsolution and is supersolution, Theorem 3.8 implies that for all . And conversely, we also have for all , since is subsolution and is supersolution. Consequently, we have for all , which confirms the fact the value function defined on (2.6) is the unique solution of the variational inequality (2.13).

4. Numerical Example

This example is for a stock which share price roughly around $55 in average; in this example we assume that the market has two main movements: an uptrend and a downtrend. Thus the Markov chain takes two states , where denotes the uptrend and denotes the downtrend. The transaction fee , the discount rate , the return vector is , the volatility vector is , the intensity vector is , the time (in year), and the generator of the Markov chain is

Figures 1 and 2 represent the value function computed by solving the nonlinear system of equations (2.13) and the free boundary curves in both trends. These curves divide the plane in two regions. The region below the free boundary curve is the continuation region, and the region above the free boundary curve is the stopping region. In other terms, whenever the share price at any given time within the interval is below the free boundary curve, it is optimal for the investor to hold the stock. And whenever the share price is above the free boundary, it is optimal for the investor to sell. This selling rule is obviously easy to implement. This example clearly shows that the selling rule derived by our method can be very attractive to practitioners in an automated trading setting.

(a)

(b)

Figure 1: Value functions for the uptrend and for the downtrend.

(a)

(b)

Figure 2: Free boundary curves.

Appendix

A. Proofs of Results

A.1. Proof of Lemma 3.3

We have

This proves (3.2).

Now let us prove (3.4). Since for any , using (3.2), we have So using (3.1) and (3.3), we obtain for any . Thus, letting , we obtain (3.4).

A.2. Proof of Lemma 3.1

Given and , let and be two solutions of (2.4) with and , respectively. For each , we have Using the Itô-Lévy isometry, we have Taking into account (2.5), we can find such that . The inequality (A.5) becomes Let , then (A.7) becomes Applying Gronwall's inequality, we have This implies, in view of Cauchy-Schwarz inequality, that Using this inequality, we have This implies the (uniform) continuity of with respect to .

We next show the continuity of with respect to . Let be the solution of (2.4) that starts at with and . Let , we define It is easy to show that

Given , let . Then, and as .

Let . Then, . It is easy to show that for some constant .

We define We have where as . It follows that

Therefore, we have This gives the continuity of with respect to .

The joint continuity of follows from (A.10) and (A.17). Finally, the linear growth inequality follows from (A.10) and This completes the proof.

A.3. Proof of Theorem 3.4

First we prove that is a viscosity supersolution of (2.13). Given , let such that has local minimum at in a neighborhood . We define a function Let be the first jump time of from the initial state , and let be such that starts at and stays in for . Moreover, , for . Using Dynkin's formula, we have Recall that is the minimum of in . For , we have Using (A.19) and (A.21), we have Moreover, we have Combining (A.22) and (A.23), we have It follows from Lemma 3.3 that Dividing both sides by and sending lead to

By definition, . The supersolution inequality follows from this inequality and (A.27).

Now, let us prove the subsolution inequality, namely, that let and has a local maximum at , then we can assume without loss of generality that .

Define Let be the first jump time of from the state , and let be such that starts at and stays in for . Since we have , for , and let be the optimal stopping time, and for , we have from Lemma 3.3 (The appendix) Moreover, since and attains its maximum at in , then Thus, we also have This implies, using Dynkin's formula, that Note that Using (A.27) and (A.32), we obtain Recall that, by assumption. From (A.28), we deduce Dividing the last inequality by and sending give This gives the subsolution inequality. Therefore, is a viscosity solution of (2.13).

A.4. Proof of Lemma 3.9

Let and , then we have Note that from the Lévy-Khintchine inequality (2.3), one can prove ; therefore, there exists a constant such that This proves (3.14).

Now, let us proof the theorem.

A.5. Proof of Theorem 3.8

For any and , we define Note that and satisfy the linear growth. Then, we have for each and since is a continuous in , therefore it has a global maximum at a point . Observe that It implies Then, Consequently, we have By the linear growth condition, we know that there exist such that and . Therefore, there exists such that we have So, We also have and . This leads to It comes that therefore there exists such that The inequality (A.48) implies the sets and are bounded by independent of , so we can extract convergent subsequences that we also denote , , and . Moreover, from the inequality (A.45), it comes that the exists such that Using (A.43) and the previous limit, we obtain achieves its maximum at , so by the Theorem 3.7 for each , there exist , , , and such that But, we know that Therefore, we obtain Equation (A.51) implies by the definition of the viscosity solution that Consequently, we have two cases; either or First of all, we assume that . And similarly, (A.54) implies by the definition of the viscosity solution that Therefore, we have It comes that Letting , we obtain Note that the function reaches its maximum at . It follows that for all , , and , we have Again letting and using (A.61), we obtain so, we have Second of all, let assume that and, moreover, we have Let us use the operator defined on the Lemma 3.9; thus, Using the operator , thus (A.65) becomes and (A.66) becomes Combining the last two inequalities, we obtain From Lemma 3.9, there exists a constant such that Recall that , so we have Since attains its maximum at , we have ; therefore Thus, we have Consequently, from (A.74) it comes that (A.72) implies Note that from (3.11), we have Therefore, we have Similarly, for any , we have consequently, we derive that Therefore, using (2.3), we obtain for some constant . Taking into account (A.75) and (A.80), thus (A.77) becomes We know from the Maximum principle that Moreover, Note that Letting , we obtain Take , this leads to Using (A.50), we obtain Letting in (A.81), we have and taking the as goes to zero and using (A.87), we obtain Recall that is maximum of . Then, for all , , and for all , we have It comes that Letting , we obtain

Using (A.89), we have Therefore, using (A.92), we conclude that Letting in (A.64) and the previous inequality, we have This completes the proof of the theorem.

Acknowledgment

The author is grateful to the editor and the referee for their helpful suggestions and comments.

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