# Optimal Selling Strategies under Regime-Switching Market Environment with Finite Expiry

Authors: Jie Xing; Taoshun He
Journal: Discrete Dynamics in Nature and Society
 (2021)

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## Abstract

This paper addresses an optimal stock liquidation problem over a finite-time horizon; to that end, we model it as an optimal stopping problem in a regime-switching market. The optimal stopping time is written as a solution to a system of Volterra type integral equations. Moreover, it reveals that when the risk-free interest rate is always lower than the return rate of the stock, it is never optimal to sell the stock early; otherwise, one should sell the stock in bear market if the stock price reaches a critical value and hold the stock in bull market until the maturity date. Finally, we present a trinomial tree method for numerical implementation. The numerical results are consistent with the theoretical findings.


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## Body

## 1. Introduction

“What is the best time to sell an asset to maximize revenue?” is a fundamental problem in finance. Given a fixed transaction cost, the objective is to choose a stopping time so as to maximize an expected return. This kind of optimal stopping problem was studied by McKean firstly under Black–Scholes model [1]. It is well known that the stochastic variability in the market parameters is not reflected in the Black–Scholes model. At present, the market trend is considered in literature when dealing with the optimal stock liquidation problem. In [2], Wang et al. proposed a Markov regime-switching DCC-GARCH model and estimated the model parameters by a two-stage maximum likelihood method. Hou et al. presented a spatial-temporal neural network frame to forecast the stock price movement [3].In [4], Zhang provided a suboptimal selling rule in a regime-switching market by a two-point-boundary approach. For regime-switching exponential Gaussian diffusion model, Eloe et al. used a similar approach to derive an optimal selling rule by two threshold levels [5]. Guo and Zhang used a smooth-fit technique to analyze the optimal selling boundary under certain parameter conditions for infinite horizontal problem [6]. The finite horizontal optimal selling rule under regime-switching model was firstly studied by Pemy and Zhang using viscosity solution approach [7]. Later, Pemy studied a similar problem in the case where the underlying dynamic is a Markov switching Lévy process [8]. Bian et al. investigated the optimal liquidation problem with regime-switching until the exit time, which is essentially an optimal control problem rather than an optimal stopping problem [9]. Recently, Vaicenavicius put forward an optimal liquidation problem involved with drift uncertainty and regime-switching volatility and solved it by constructing approximating sequences of optimal stopping problem with constant volatility [10].It is known that Pemy and Zhang only discussed the value function to the related HJB equation, and up to the author’s knowledge, the optimal selling strategies under regime-switching model are still unknown so far [7]. In filling this gap, this paper is concerned with the optimal selling strategies. That is, we mainly focus on the early exercise boundary with respect to the optimal stopping problem rather than the value function. Indeed, the role of the present paper should be viewed as a supplement to [7]. On the other hand, the impacts of varying parameters on the selling decision were not discussed in [7]. We discuss it rigorously by variational inequality approach. It is important to distinguish this paper from the existing literature on optimal stopping problems with regime-switching volatility. Buffington and Elliott discussed American options in regime-switching market and obtained an approximate valuation of American option [11]. Jacka and Ocejo studied the regularity of the value function for a finite horizontal optimal stopping problem in the presence of regime-switching uncertainty [12]. The abovementioned authors mainly concentrated on the value function and did not investigate the free boundary arising from American option pricing problem. Boyarchenkp and Levendorski calculated the early exercise boundaries for American options by using a generalization of Carr’s randomization procedure for regime-switching models [13]. But they did not study the properties of the free boundary and the early exercise premium was unknown. Liu and Privault dealt with optimal prediction problem in a regime-switching model and derived Volterra type integral equations [14]. Since the optimal stopping problem involves with a system of interacting integral equations, making it more difficult to solve directly, they solved the optimal stopping problem by using a recursive algorithm later and their method did not rely on the Volterra equations [10]. Recently, Luo and Xing handled an optimal stopping problem in pricing variable annuities under regime-switching volatility, but the drift term in the model was assumed to be constant [15].In the above optimal stopping problems, there exists a unique free boundary for each state in common, whereas the existence of early exercise boundary in our problem depends on the parameters in the model. Furthermore, since there has been less focus on the development of a realistic modelling framework for analyzing the impact of various sources of risk in influencing the early exercise boundary, it is crucial to make rigorous sensitivity analysis when we make decisions. The main contribution of this paper is that we discuss the optimal selling decisions rigorously, which is not investigated in [7]. Specifically, the contributions are three-fold: properties of the early exercise boundary, regime-switching Volterra type integral equations, and sensitivity analysis of optimal selling strategy. Firstly, by using the properties of value function, we derive some basic properties of free boundary. Then, by dynamic programing principle, we convert the regime-switching optimal stopping problem into an optimal stopping problem with constant-coefficient geometric Brownian motion, and the proof is novel compared with the method used in [14]. Finally, we use a variational inequality approach inspired by [16] to derive sensitivity analysis with respect to the model parameters and the numerical results are consistent with the theorems.The remainder of this paper is organized as follows. In Section2, we formulate the optimal selling problem in a regime-switching market with finite expiry. In Section 3, we give some preliminary results which will be frequently used later. Section 4 presents the main results, namely, the properties of free boundary, integral equation for the free boundary, and sensitivity analysis of the free boundary. In Section 5, we use the trinomial tree method mentioned in [17] to calculate the optimal stopping problem and perform some numerical experiments. Finally, Section 6 concludes.


## 2. Problem Set-Up

Consider a probability spaceΩ,ℱ,ℙ. Consider also a standard Brownian motion B=Bt,t≥0. We assume that the continuous time Markov chain α=αt,t≥0 takes values in S=1,2, i.e. there are only two states: “bull” and “bear,” and the generator of αt is of the form(1)A=−λ1λ1λ2−λ2,where λ1 and λ2 are the positive constants. We denote by F=ℱt,t≥0 the ℙ−augmentation of the filtration ℱtB,α,t≥0 generated by the stochastic process B and α. Here, ℱtB,α≔σBv,αv:0≤v≤t for t∈0,+∞. Moreover, the stock price satisfies the following Markovian switching stochastic differential equation:(2)dXs=Xsμαsds+σαsdBs,s>t,Xt=x,where x is the initial stock price, μαs∈μ1,μ2 is the rate of return, and σαs∈σ1,σ2 is the volatility.LetΛt,T denote the set of all ℱt−stopping times taking value in t,T a.s. Given a transaction cost a>0, the agent aims to maximize Ee−rτ−tXτ−a by choosing a stopping time τ, where r is the risk-free interest rate. Therefore, we consider the optimal stopping problem as follows:(3)Vt,x,i=supτ∈Λt,TEe−rτ−tXτt,x−a|Xt=x,αt=i,i=1,2,where Xt,x=Xst,x,s≥t is the strong solution to the SDE (2) starting from x at time t.Denotegx=x−a. By Markov property of X, we can rewrite the optimal stopping problem as follows:(4)Vt,x,i=supτ∈Λ0,T−tEi,xe−rτgXτx,i∈S,where Xx=X0,x is the strong solution to the SDE (2) starting from x at time 0 and E⋅|X0=x,α0=i≔Ei,x⋅. For simplicity, we will always omit the superscripts in Ei,x.Besides, we assume that the following assumption holds throughout this paper.Assumption 1.
The rate of return satisfiesμ1=μ1<μ2=μ2.


## 3. Some Preliminary Results

The following notations will be used throughout this paper. Define the stopping region and continuation region as follows:(5)D≔t,x,i∈0,T×0,+∞×S:Vt,x,i=gx,C≔t,x,i∈0,T×0,+∞×S:Vt,x,i>gx,Di≔t,x∈0,T×0,+∞:Vt,x,i=gx,i∈S,Ci≔t,x∈0,T×0,+∞:Vt,x,i>gx,i∈S.We also introduce the infinitesimal generator ofXt,αt as follows:(6)Aft,x,i≔ℒft,x,i+λift,x,j−ft,x,i,j≠i,where(7)ℒft,x,i≔12x2σi2fxxt,x,i+xμifxt,x,i,where fx denotes ∂ft,x,i/∂x, and fxx is defined similarly.Lemma 1.
The value functionVt,x,i is decreasing in t, increasing in x, and convex in x.Proof.
The proof is easy; we refer to Appendix A.Next, we show thatV⋅,x,i is 1/2−Hölder continuous and Vt,⋅,i is Lipschitz continuous.Lemma 2.
For eachx,i∈ℝ+×S, the function V⋅,x,i is 1/2−Hölder continuous uniformly over a neighborhood of x; for each t,i∈0,T×S, the value function Vt,⋅,i is Lipschitz continuous in ℝ+. Moreover, it has at most linear growth rate, i.e., there exists a constant C such that Vt,x,i≤C1+x.Proof.
For the proof, we refer to Appendix B.It is clear that(8)Esup0≤s≤Te−rsXs−a≤Esup0≤s≤Te−rsXs+a<+∞,where the last inequality follows from the moment estimates (see [18] or [12]). By Corollary 2.9 in [19], the above inequality together with the fact that V and g are continuous (Lemma 2) implies that(9)τDt,x,i≔inf0≤s≤T−t:t+s,Xsx,αs∈Dis an optimal stopping time in (4).The next important lemma gives the property of discounted stock price process. The proof can be found in [6].Lemma 3.
Suppose thatη>η˜ are two roots of(10)x2−μ1−λ1+μ2−λ2x+μ1−λ1μ2−λ2−λ1λ2=0,where λ1, λ2 are defined in (1). If η>r, then e−rtXt is a submartingale. If η≤r, then e−rtXt is a supermartingale.Remark 1.
Under Assumption1, a direct computation shows(11)η=μ1−λ1+μ2−λ2+Γ2,where(12)Γ=μ1−λ1+μ2−λ22−4μ1−λ1μ2−λ2−λ1λ2.
Simple calculation gives that(13)μ2−μ1+λ1+λ22−Γ=4λ2μ2−μ1>0,Γ−μ1+λ1−μ2+λ22=−4λ1μ1−μ2>0,which implies that μ1<η<μ2. We will see that the root η affects the shape of the stopping region in the following proposition and the next section. Economically speaking, the decisions to sell the stock are closely related to η.Proposition 1.
Letη be defined by (10).Case 1.
Ifη>r, then the optimal stopping time τD=T−t, and(14)Vt,x,i=Ee−rT−tgXT−tx.Case 2.
Ifη≤r, then the function x⟼Vt,x,i−gx is decreasing. Moreover, D is upward connected: if t,x′,i∈D, then t,x,i∈D for x>x′.Proof.
Ifη>r, by Lemma 3, it is easy to see that e−rtgXt is also a submartingale. By optional sampling theorem, for any τ∈0,T−t,(15)Ee−rτgXτx≤Ee−rT−tgXT−tx.
Hence, by the definition ofV(see (4)), we conclude that τ=T−t is an optimal stopping time of problem (10) and(16)Vt,x,i=Ee−rT−tgXT−tx.
Ifη≤r, for any x′<x, let τ be the optimal stopping time for Vt,x,i, then τ is suboptimal for Vt,x′,i. By Lemma 3, e−rtXt is supermartingale. The optional sampling theorem leads to(17)Vt,x′,i−Vt,x,i≥Ee−rτXτx′−a−Ee−rτXτx−a=Ee−rτXτx′−Xτx=x′−xEe−rτexpHτ≥x′−x,where Ht is defined by(18)Ht=∫0tμαs−12σ2αsds+∫0tσαsdBs.
This implies the monotonicity of the functionx⟼Vt,x,i−gx. If t,x′,i∈D, then the above inequality leads to(19)0≤Vt,x,i−gx≤Vt,x′,i−gx′=0forx>x′.
That is,t,x,i∈D.Remark 2.
From Proposition1, we know that if the risk-free interest rate is always lower than the return rate of the stock, then it is never optimal to sell the stock before the maturity date.


## 4. Main Results

In this section, we shall investigate the shape of the stopping region of the primal problem (4). We start with some basic properties of the free boundary.Theorem 1.
Consider the problem (4) with μ1<η≤r.Case 1: ifμ1<η≤r≤μ2, then there exists a continuous function b1t>0 defined by(20)b1t=infx:Vt,x,1=gx,such that(21)D1=t,x:x≥b1t,andD2=∅. Moreover, b1t is nonincreasing and b1T=ra/r−μ1.Case 2: ifμ1<η≤μ2<r, then for every i∈S, there exists a continuous function bit>0 defined by(22)bit=infx:Vt,x,i=gx,such that(23)Di=t,x:x≥bit.
Moreover,bit is nonincreasing and biT=ra/r−μi for i∈S.Proof.
Firstly, we assert that(24)t,x:−rgx+ℒgt,x,i>0⊂Ci,i∈S.
By Dynkin’s formula, we obtain that(25)Ee−rsgXsx=gx+E∫0se−rv⋅−rg+ℒgt+v,Xvx,αvdv.
For anyt,x∈t,x:−rgx+ℒgt,x,i>0, we can choose an open neighborhood U⊂0,T×0,+∞ of t,x such that(26)−rgx+ℒgt,x,i>0,for any t,x∈U. Let s=τU in (25) be the first exist time of U when t+s,Xsx started at t,x at time 0. Denote Ti to be the first transition time of α with α0=i. Since −rgXvx+ℒgt+v,Xvx,αv>0 for v∈0,τU∧Ti a.s., by (4), we derive that(27)Vt,x,i≥Ee−rτU∧TigXτU∧Tix=gx+E∫0τU∧Tie−rv⋅−rg+ℒgt+v,Xvx,αvdv>gx.
This implies thatt,x∈Ci.Case 1: (24) is equivalent to(28)Di⊂t,x:−rgx+ℒgt,x,i≤0,i∈S.Under the assumption thatμ1<r≤μ2, this leads to D1⊂t,x:x≥ra/r−μ1 and D2=∅. Now, we shall follow the steps in [19] to show that the value function Vt,x,i satisfies(29)Vtt,x,i+AVt,x,i−rVt,x,i=0inC.Define a killed process(30)X˜t≔Xt,ift<T,Δ,ift≥T,whereT is a random variable exponentially distributed with parameter r>0 and independent of X, Δ is a fictitious point, and all functions are assumed to take value zero at Δ. By a direct computation, we have(31)EgX˜t=EgXt1t<T=EgXt1t<T|t<TℙT>t+EgXt1t<T|t≥TℙT≤t=Ee−rtgXt,where the second equality follows from law of total expectation and the last equality follows from the fact thatXt is independent of T. Now, we can rewrite the value function as(32)Vt,x,i=EgX˜τDx.We chooset,x,i∈C and a neighborhood U⊂C of t,x,i. Define σ=infs∈0,T−t:t+s,Xsx,αs∉U. It is clear σ≤τD. This leads to σ+τD°θσ=τD (see [19]). By strong Markov property,(33)EVσ,X˜σ,ασ=EEgX˜τD°θσ|ℱσ=EEgX˜σ+τD°θσ|ℱσ=EEgX˜τD|ℱσ=EgX˜τD=Vt,x,i.Here,θ is the shift operator. This implies the characteristic operator of t,X˜t,αt is identically zero:(34)limU⟶t,x,iEVσ,X˜σ,ασ−Vt,x,iEσ=0.By the fact that the characteristic operator coincides with the infinitesimal generator oft,X˜t,αt, we derive that (29) holds.We claim thatD1≠∅. Otherwise, by (29) and letting t⟶T−, we obtain that(35)limt⟶T−Vtt,x,1=rVT,x,1−12x2σ2VxxT,x,1−μ1xVxT,x,1−λ1VT,x,2−VT,x,1=r−μ1x−ra>0,forx>ra/r−μ1, where the second equality follows from the fact that VT,x,i=gx. As Vtt,x,1≤0 (Lemma 1), this leads to a contradiction. By Proposition 1 and D1≠∅, we obtain that b1t defined by (20) is well defined and (21) holds.We shall deduce thatb1t is nonincreasing. Otherwise, there exist t1<t2<T such that b1t1<b1t2. By the definition of b1t, we derive that(36)Vt2,b1t1,1>gb1t1=Vt1,b1t1,1.This contradicts with the fact thatV⋅,x,i is nonincreasing (Lemma 1).Next, we prove the continuity ofb1t. Otherwise, suppose that there exist some t0∈0,T such that(37)0≤x2=limt⟶t0+b1t<x1=limt⟶t0−b1t.Then, the domainN=0,t0×x2,x1×1⊂C1.For anyϵ>0, by a similar argument as (33), we have(38)EVt+ε,X˜ε,αε=EEgX˜τD°θε|ℱε=EgX˜τD+ε≤supτ∈Λt,TEgX˜τ=Vt,x,i,where the inequality follows from the fact thatτDt+ϵ,x,i≤T−t+ϵ. By Dynkin’s formula and (30), we have(39)EVt+ε,X˜ε,αε−Vt,x,i=Ee−rεVt+ε,Xε,αε−Vt,x,i=E∫0εe−rvVt+AV−rVt+v,Xvx,αvdv≤0,where the last inequality follows from (38). Dividing the above inequality by ϵ, letting ϵ⟶0, and using the mean value theorem, we derive(40)Vt+AV−rVt,x,i≤0.We fixi=1 and let t⟶t0+. As Vt0+δ,x,1=gx for δ>0 and x∈x2,x1, by (40), we have(41)rVt0+,x,1−AVt0+,x,1≥Vtt0+,x,1=0,forx∈x2,x1.That means(42)Vt0,x,2≤1λ1r−μ1+λ1x−λ1+ra,forx∈x2,x1.On the other hand, for anyt,x,1∈N, by (29), letting t⟶t0−, we derive(43)Vt0,x,2=1λ1rVt0,x,1−Vt+ℒVt0,x,1+λ1Vt0,x,1≥1λ1r−μ1+λ1x−λ1+ra,forx∈x2,x1,where the last inequality follows fromVt0,x,1=gx for x∈x2,x1 and Vt≤0 (Lemma 1). By (41) and (43), as r>μ1, we derive that(44)Vxt0,x,2=r−μ1+λ1λ1>1,forx∈x2,x1,which is in contradiction with Proposition1. Hence, b1⋅ is continuous.Finally, we proveb1T=ra/r−μ1. As b1t is nonincreasing and D1⊂t,x:x≥ra/r−μ1, we have b1t is well defined and b1T≔limt⟶T−b1t≥ra/r−μ1≔x∗. Suppose that b1T>x∗. For any x¯∈x∗,b1T, since VT,x,i=gx, letting t⟶T− in (29), we have(45)VtT,x¯,1=rVT,x¯,1−AVT,x¯,1=r−μ1x¯−ra>0,which contradicts withVt≤0 (Lemma 1).Case 2: we notice thatDi⊂t,x:x≥ra/r−μi. Now, the proof is similar to Case 1.In the following theorem, we rewrite the optimal stopping problem (4) as a free boundary problem.Theorem 2.
Assume thatη≤r.Case 1: ifμ1<η≤r≤μ2, then the value function V is the unique solution to the following problem:(46)Vtt,x,1+AVt,x,1−rVt,x,1=0inC1=t,x:x<b1t,Vtt,x,2+AVt,x,2−rVt,x,2=0in0,T×0,+∞,Vt,b1t,1=gb1t,t∈0,T,Vxt,b1t,1=1,t∈0,T,VT,x,i=gx,x∈0,+∞,i∈S.In particular, the partial derivativesVxx, Vx, and Vt exist and are continuous in C.Case 2: ifμ1<η≤μ2<r, then the value function V defined in (4) is the unique solution to the following free boundary problem:(47)Vtt,x,i+AVt,x,i−rVt,x,i=0inC=t,x,i:x<bit,(48)Vt,bit,i=gbit,t∈0,T,i∈S,(49)Vxt,bit,i=1,t∈0,T,i∈S,(50)VT,x,i=gx,x∈0+∞,i∈S.
In particular, the partial derivativesVxx, Vx, and Vt exist and are continuous in C.Proof.
We only prove Case 2. Case 1 can be proved similarly. (47) follows directly from (29). (48) follows from the continuity of V (Lemma 2). (50) follows from (4) directly.
We shall prove the smooth pasting condition (49). We set x=bit. As t,x−ϵ∈Ci, we derive that(51)Vt,x,i−Vt,x−ϵ,iϵ≤x−a−x−ϵ−aϵ=1,where the inequality follows from (48). Let ϵ⟶0,(52)Vx−t,x,i≤1,where Vx− is the left derivative of V in x. For any ϵ>0, let τϵ′ be the optimal stopping time for Vt,x−ϵ,i and Ti be the first transition time of α with α0=i. Denote(53)τε≔infs∈0,Ti|Xsx−ε≥bit+s.
By (9) and Theorem 1, if Ti≥T−t, then τϵ′=τϵ; otherwise, we have(54)τϵ′t,x,i=infs∈0,T−t|Xsx−ϵ≥bαst+s=τϵ∧infs∈Ti,T−t|Xsx−ϵ≥bαst+s.
By the solution of SDE (2), we have(55)τϵ=infs∈0,Ti:∫0sμαv−12σ2αvdv+∫0sσαvdBv≥logbit+sx−ϵ≤infs∈0,Ti:μ1−12σ12s+σ1Bs≥logxx−ϵ≤infs∈0,Ti:−Cs+Bs≥1σ1logxx−ϵ,where the first inequality following from bi is nonincreasing and the last inequality follows from the fact that 1/σ1μ1−1/2σ12 can be bounded below by some constant −C<0.
By the law of iterated logarithm, we have for anyϵ′>0, there exists some s∈0,ϵ′ such that B˜s>Cs a.s. This implies that there exists some ϵ>0 small enough such that(56)B˜s−Cs>1σ1logxx−εa.s.
Hence, by the definition ofτϵ, we have τϵ≤ϵ′⟶0, as ϵ′⟶0. Equation (53) leads to limϵ⟶0τϵ′=0.
Now, we prepare to proveVx−t,x,i≥1. Since(57)Vt,x,i−Vt,x−ϵ,iϵ≥Ee−rτϵ′Xτϵ′x−a−Ee−rτϵ′Xτϵ′x−ϵ−aϵ=Ee−rτϵ′Xτϵ′x−Xτϵ′x−ϵϵ=EexpHτϵ′−rτϵ′,where Ht is defined in (18). Letting ϵ⟶0, by Fatou’s lemma, we conclude that Vx−t,x,i=1. As Vt,x,i=gx for x>bit, it is easy to see that Vx+t,x,i=1. Here, Vx+ is the right derivative of V in x. We conclude that the smooth pasting condition (49) holds.
For uniqueness, letft,x,i defined on C¯ be a solution of the free boundary problem (47). We define Ms≔e−rsfs+t,Xs,αs and set τD=infs∈0,T−t:Xsx≥bαss+t. By It ô’s lemma,(58)Ms∧τD=ft,x,i+∫0s∧τDe−rvft+Af−rft+v,Xv,αvdv+Ns∧τD,where Ns∧τD,s≥0 is a martingale. Hence, by (47), Ms∧τD,s≥0 is a martingale. We know from the optional sampling theorem that(59)ft,x,i=M0=EMτD=Ee−rτDfτD+t,XτD,ατD=Ee−rτDgXτD=Vt,x,i,where the second equality follows from that Ms∧τD,s≥0 is a bounded martingale and dominated convergent theorem, the fourth equality follows from (48) and definition of τD, and the last equality follows from that τD is an optimal stopping time of Vt,x,i. This implies the uniqueness.
By Lemma2 and regularity of parabolic equation (see [20]), it is obvious that Vt, Vx, Vxx exist and are continuous in C.In the following, we decompose the value function into two components and derive a Volterra type integral equation. Firstly, we introduce the following notations:(60)Jit,x≔Ee−r+λiT−tgXT−ti,(61)Kit,x,u,biu+t≔Ee−r+λiuΦiXui1Xui>biu+t−λiVt+u,Xui,j1Xui<biu+t,j≠i,where(62)Φix=r+λi−μix+r+λia.Xti satisfies(63)dXti=Xtiμidt+σidBt,with initial condition X0i=x.Theorem 3.
Assume thatη≤r.Case 1: ifμ1<η≤r≤μ2, then the value function defined in (4) is given by(64)Vt,x,1=J1t,x−∫0T−tK1t,x,u,b1u+tdu,Vt,x,2=J2t,x+λ2E∫0T−te−r+λ2uVt+u,Xu2,1du.Furthermore, the free boundaryb1t satisfies the following nonlinear Volterra type integral equation:(65)gb1t=J1t,b1t−∫0T−tK1t,b1t,u,b1u+tdu.Case 2: ifμ1<η≤μ2<r, then the value function defined in (4) is given by(66)Vt,x,i=Jit,x−∫0T−tKit,x,u,biu+tdu,i∈S.Furthermore, the free boundarybit satisfies the following nonlinear Volterra integral equation:(67)gbit=Jit,bit−∫0T−tKit,bit,u,biu+tdu,i∈S.Proof.
We only prove Case 2 and Case 1 can be proved by a similar method. Firstly, we consider the following optimal stopping problem:(68)Vit,x=sup0≤τ≤T−tEe−r+λiτgXτi+λi∫0τe−r+λiuVt+u,Xui,j,j≠i,where Vt,x,i is defined in (4) and Xti satisfies (63). Fixing i in Case 2 of Theorem 2, by a similar Markovian argument, we conclude that Vit,x is a solution to the free boundary problem (47)–(50). The uniqueness implies that Vit,x=Vt,x,i for t,x,i∈0,T×0,+∞×S.
By the regularity ofV in Case 2 of Theorem 2, we can use the local time-space formula given in [19] to obtain that(69)e−r+λisVis+t,Xsi=Vit,x+∫0se−r+λisVit+12σi2x2Vixx+μixVix−r+λiVit+u,Xui1Xui≠biu+tdu+∫0sσiXuiVixt+u,Xui1Xui≠biu+tdBu+12∫0sVixt+u,Xui+−Vixt+u,Xui−⋅1Xui=biu+tdℓubiXi,where ℓubiXi is the local time of Xi at the curve bi given by(70)ℓubiXi≔limϵ⟶012ϵ∫0u1biv−ϵ<Xvi<biv+ϵd<Xi>v,and dℓubiXi refers to integration with respect to the continuous increasing function u↦ℓubiXi.
We notice thatVit,x=gx for x>bit and Vi satisfies (47) for x<bit. Hence,(71)Vit+12σi2x2Vixx+μixVix−r+λiVit+u,Xui=ΦiXui,ifXui>biu+t,λiVu,Xui,j,ifXui<biu+t.
Here,j≠i. Setting s=T−t together with (49) and (50), taking expectation in (69), and using the notation (60) and (61), we derive that(72)Vit,x=Ee−r+λiT−tgXT−t−E∫0T−te−r+λiuΦiXui1Xui>biu+t−λiVt+u,Xui,j1Xui<biu+tdu=Jit,x−∫0T−tKit,x,u,biu+tdu=Vt,x,i.
Finally, by condition (48), we derive that(73)Jit,bit−∫0T−tKit,bit,u,biu+tdu=gbit,i∈S.The next result compares the optimal selling strategies in different states.Theorem 4.
Suppose Assumption1 holds, σ1<σ2, and r>μ2. Let Xti be defined as (63). Then,(74)b10t≤b1t≤b2t≤b20t,0<t<T,where bi0i∈S is the free boundary of the following optimal stopping problem(75)Vi0t,x=sup0≤τ≤T−tEe−rτgXτi,i∈S.Proof.
We setVt,x,i≔Vit,x for i∈S and N≔t,x:V2t,x<V1t,x. We shall prove N=∅. Otherwise, in N,(76)V1t,x>V2t,x≥gx,V1t+12σ21x2V1xx+μ1xV1x−rV1=−λ1V2−V1≥0,V2t+12σ22x2V2xx+μ2xV2x−rV2≤−λ2V1−V2≤0,where the last two inequalities follow from (29) and (40). By Lemma 1, it follows that(77)V2t+12σ12x2V2xx+μ1xV2x−rV2=V2t+12σ22x2V2xx+μ2xV2x−rV2+12x2σ12−σ22V2xx+μ1−μ2xV2x≤0.
Furthermore,V1T,x=V2T,x and V1=V2 on the boundary of N. Hence, the comparison principle of variational inequality implies that V2≥V1 in N, which is a contradiction with definition of N. This leads to V2≥V1 for t,x∈0,T×0,+∞. That means C1⊂C2, i.e., b1t≤b2t.
By Markovian argument,Vi0 satisfies(78)min−Vit0−12σi2x2Vixx0−μixVix0+rVi0,Vi0−gx=0,Vi0T,x=gx.
The above inequality implies that(79)−V2t0−12σ22x2V2xx0−μ2xV2x0+rV20−λ1V1−V2≥−λ2V1−V2≥0,and by (40),(80)−V1t−12σ12x2V1xx−μ1xV1x+rV1≥λ1V2−V1≥0.
Applying comparison principle of variational inequality, we deduce thatV10≤V1 and V2≤V20, i.e., b10t≤b1t and b2t≤b20t.In the end of this section, we perform some sensitivity analyses of the optimal selling strategy.Theorem 5.
Assume thatr≥η. The free boundary bit of problem (4) is increasing with respect to μi, σii∈S, and a and decreasing with respect to r.Proof.
Assume thatV^1,V^2 is the solution to variational inequality(81)minrV−Vt−AVt,x,i,Vt,x,i−gx=0,for any t,x,i∈0,T×0,+∞×S, where σi is σ^i, V¯1,V¯2 is the solution to variational inequality (81), where σi is σ¯i, and σ^i>σ¯i for each i∈S. By Lemma 1, it follows that(82)V^it+12σ¯i2x2V^ixx+μixV^ix−rV^i+λiV^j−V^i=12x2σ¯i2−σ^i2V^ixx≤0,if V^it,x>gx, and(83)V^it+12σ¯i2x2V^ixx+μixV^ix−rV^i+λiV^j−V^i≤12x2σ¯i2−σ^i2V^ixx≤0,if V^it,x=gx. Employing comparison principle of variational system (see [16]), we conclude that V^i≥V¯i, i.e., b¯it≤b^it, for i∈S. Hence, b1t,b2t is increasing with respect to σi for each i∈S.
Next, assume thatV^1,V^2 is the solution to variational inequality (81), where a is a1, V¯1,V¯2 is the solution to variational inequality (81), where a is a2, and a1<a2. Then, we have(84)V¯it,x=sup0≤τ≤T−tEe−rτXτ−a2≥a1−a2+sup0≤τ≤T−tEe−rτXτ−a1=a1−a2+V^it,x.
This implies that ifV^it,x>x−a1, then V¯it,x≥V^it,x+a1−a2>x−a2. If we denote C^i to be the stopping region with respect to a1 and C¯i the stopping region with respect to a2, then C^i⊂C¯i, i.e., b¯i≥b^i. Hence, b1t,b2t is increasing with respect to a.
In the following, assume thatV^1,V^2 is the solution to variational inequality (81), where μi is μ^i, V¯1,V¯2 is the solution to variational inequality (81), where μi is μ¯i, and μ^i>μ¯i for each i∈S. By Lemma 1, it follows that(85)V^it+12σi2x2V^ixx+μ¯ixV^ix−rV^i+λiV^j−V^i=μ¯i−μ^ixV^ix≤0,if V^it,x>gx, and(86)V^it+12σ¯i2x2V^ixx+μ¯ixV^ix−rV^i+λiV^j−V^i≤μ¯i−μ^ixV^ix≤0,if V^it,x=gx. Employing comparison principle of variational system (see [16]), we conclude that V^i≥V¯i, i.e., b¯it≤b^it, for i∈S.
If we set(87)Vt,x,i;r=sup0≤τ≤T−tEe−rτgXτ,then for any r1<r2, it is easy to see that(88)Vt,x,i;r1=sup0≤τ≤T−tEe−r1τgXτ≥sup0≤τ≤T−tEe−r2τgXτ=Vt,x,i;r2,which implies that bit is decreasing with respect to r.


## 5. Numerical Examples

In this section, we present some numerical results obtained from trinomial tree method, which was put forward in [17] for pricing American option. Firstly, we briefly explain how to use the trinomial tree method to solve our problem.LetΔt=T/n be the time step-size. We take the jump ratios of the lattice as(89)u=eσΔt,d=e−σΔt,where(90)σ=maxi∈Sσi+1.5−1σ˜,where σ˜ is the arithmetic mean or geometric mean of σi, i∈S. For regime i, let πui, πmi, and πdi be the risk neutral probabilities corresponding to the stock price increases, remaining the same, and decreases, respectively. By moment matching technique, the values of the probabilities are given by(91)πmi=1−σi2σ2,πui=eμiΔt−e−σΔt−1−σ2/σi21−e−σΔteσΔt−e−σΔt,πdi=eσΔt−eμiΔt−1−σ2/σi2eσΔt−1eσΔt−e−σΔt.Letxj+1=uxj and xj−1=dxj and Vkxj,i be the trinomial approximation of the primal value function for regime i at asset price xj and time tk=kΔt. Then, the trinomial formula of the primal value function is given by, for k=0,1,…,n−1,(92)Vkxj,i=maxe−rΔt∑l=12pilπuiVk+1xj+1,l+πmiVk+1xj,l+πdiVk+1xj−1,l,gxj,with Vnxj,i=gxj for j=0,±1,…,±n, in which pil is the transition probability from state i to state l for the time interval with length Δt. It is given by(93)pili,l∈S=eAΔt=I+∑l=1∞ΔtlAll!,where I is the identity matrix and A is the generator matrix of the Markov chain process given by (1).Example 1.
In this example, the numerical experiments shed light on the optimal selling strategy of problem (4) with fixed parameters for two different cases.
The generator of Markov chainαt is taken as(94)A=−0.40.40.5−0.5.
The returns of the stock areμ1=0.04 and μ2=0.05 and the volatilities σ1=0.25 and σ2=0.45. The maturity date is assumed to be T=1. The transaction cost is assumed to be a=0.4. The interest rate is taken as r=0.08 for Figure 1(a) so that r>μ2 and r=0.045 for Figure 1(b) so that μ1<η<r<μ2. The number of time steps for both methods is N=10000. In Figure 1, we simulate the sample path of the stock price and plot the early exercise boundary of problem (4). Intuitively, from Figure 1(a), we see that if the risk-free interest rate is always higher than the return rate of stock, the stockholder will have motivation to sell the stock when the stock price hits the free boundary at time t0. Hence, in this case, there exists a unique free boundary for each state. Moreover, the larger return rate and volatility mean that the value function is larger, and the early exercise boundary will be higher. Intuitively, from Figure 1(b), it is clear that if the risk-free interest rate is between the return rate of bear market and return rate of the bull market, then the stockholder will sell the stock early in bear market and hold the stock until the maturity date in bull market. Hence, in this case, there exists a unique free boundary when the investor is in bear market, and the free boundary does not exist when the investor is in bull market.Figure 1
Sample path of Markov chain and stock price with (A.)α0=1, S0=0.5, and early exercise boundary when r>μ2 and (B.) α0=2, S0=1, and early exercise boundary when η≤r≤μ2.
(a)(b)Example 2.
In this example, we assess the impact of varying model parameters on the early exercise boundary.
Figure2 displays the sensitivity with respect to transaction fee. We vary the transaction fee between 1 and 4 and compute the early exercise boundary. We observe that the early exercise boundary is higher for a higher transaction fee. The higher transaction fee means that the stockholder will sell the stock if stock price is higher. That is, the critical price will be higher.
In Figure3, we show the sensitivity of the early exercise boundary with respect to interest rate. In our case, the interest rate takes values from 0.055 to 0.07. The figure shows that with a higher interest rate, the early exercise boundary is lower. Intuitively, the higher interest rate motivates the stockholder to sell the stock although the stock price is lower.
Figure4 illustrates the sensitivity with respect to the return rate of stock. For μ1,μ2=0.01,0.02,0.02,0.03,0.03,0.04,0.04,0.05, we compute the early exercise boundary. This figure shows that the free boundary is higher with a higher return rate. Intuitively, this result can be explained by the fact that the higher return rate makes the investor unwilling to sell the stock for a low stock price. Therefore, the critical price to sell stock is higher.
We analyze the sensitivity of the early exercise boundary with respect to the volatility in Figure5. For σ1,σ2=0.10,0.15,0.15,0.20,0.20,0.25,0.25,0.30, we compute the early exercise boundary. We observe that the higher the volatility, the higher is the early exercise boundary. Intuitively, if the volatility is higher, the stock price fluctuates more heavily and the investor will be more willing to hold the stock until the stock price goes up.Figure 2
The impact of varying transaction fee on the early exercise boundary for (A.) regime 1 and (B.) regime 2.
(a)(b)Figure 3
The impact of varying interest rate on the early exercise boundary for (A.) regime 1 and (B.) regime 2.
(a)(b)Figure 4
The impact of varying return rate of stock on the early exercise boundary for (A.) regime 1 and (B.) regime 2.
(a)(b)Figure 5
The impact of varying volatility on the early exercise boundary for (A.) regime 1 and (B.) regime 2.
(a)(b)


## 6. Conclusions

In this paper, we discuss the optimal selling rule under regime-switching models. To explore the optimal stopping rule, we analyze the optimal stopping problem and convert the problem into a free boundary problem. The properties of the free boundary and the involved Volterra type integral equations are established. Finally, we calculate the free boundary by trinomial tree method. Once the free boundary can be calculated, the optimal stopping rule is determined. We make a conclusion that, when the return of the stock is always higher than the interest rate, the agent should hold the stock until the maturity date; otherwise, he/she should sell the stock in the bear market when the stock price reaches a critical level and hold the stock until the maturity date in the bull market, which is consistent with intuition. Since the Volterra equations in this paper are interacting, this makes it difficult to solve the free boundary in each state independently. Therefore, the method of solving the Volterra equation without regime-switching cannot be extended to solve our problem directly. We will discuss this kind of problem in the future.



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