Trading a Mean-Reverting Asset with Regime Switching: An Asymptotic Approach

Abstract

This chapter is concerned with an optimal trading rule. The underlying asset price is governed by a mean-reverting model with regime switching. In particular, the equilibrium levels are dependent of a two-state Markov chain so that the underlying process always moves around its current equilibrium. The objective is to buy and sell the asset so as to maximize the overall return. In this chapter we consider the case in which the jump rates of the Markov chain go to infinite and study the asymptotic properties of the corresponding value functions. We establish the convergence of the value functions to those of a limiting problem, which is easier to solve. In addition, the solution of the limiting problem is used to construct a trading rule which is shown via a numerical example to be nearly optimal.

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1 Introduction

This chapter is concerned with mean-reversion trading with regime switching. It is a continuation of the study developed in Zhang and Zhang [15]. In Zhang and Zhang [15], a mean-reversion trading rule was considered. The objective was to buy and sell the asset so as to maximize an overall return. They followed the dynamic programming approach and used the associated HJB equations (quasi-variational inequalities) to characterize the value functions. They showed that the solution to the original optimal stopping problem can be obtained by solving two quasi-algebraic equations. In addition, they obtained sufficient conditions in the form of a verification theorem. Nevertheless, only the basic mean-reversion model with constant equilibrium was considered in Zhang and Zhang [15]. It is important to extend the results to account for more realistic settings. It is the purpose of this chapter to consider the mean-reversion model in which the equilibrium is subject to random jumps governed by a two-state Markov chain and to study the corresponding trading rules.

A mean-reversion model is often used in financial and energy markets to capture price movements that have the tendency to move towards an “equilibrium” level. Studies that support the mean-reversion stock returns can be traced back to the 1930s (see Cowles and Jones [3]) in empirical literature. The research was furthered by many researchers including Fama and French [6], and Gallagher and Taylor [7] among others. In addition to stock markets, mean-reversion models are also used to characterize stochastic volatility (see Hafner and Herwartz [8]) and asset prices in energy markets (see Blanco and Soronow [1] and de Jong and Huisman [4]). See also related results in option pricing with a mean-reversion asset by Bos, Ware and Pavlov [2].

Trading rules in financial markets have been studied for many years. For example, an investment capacity expansion/reduction problem was considered in Merhi and Zervos [11]. Under a geometric Brownian motion market model, the authors used the dynamic programming approach and obtained an explicit solution to the singular control problem. A more general diffusion market model was treated by Løkka and Zervos [10] in connection with an optimal investment capacity adjustment problem. More recently, Johnson and Zervos [9] studied an optimal timing of investment problem under a general diffusion market model. The objective was to maximize the expected cash flow by choosing when to enter an investment and when to exit the investment. An explicit analytic solution was obtained in Johnson and Zervos [9]. Recently, Dai et al. [5] provided a theoretical justification of trend following trading. In particular, the underlying stock price was formulated as a geometric Brownian motion with regime switching. Two regimes were considered: the up trend (bull market) and the down trend (bear market). The switching process was modeled as a two-state Markov chain which is not directly observable. The trading decisions were based on current information represented by both the stock price and historical information with the probability in the bull phase conditioning to all available historical price levels as a proxy. Assuming trading one share with a fixed percentage transaction cost, they showed that the strategy that optimizes the discounted expected return is a simple implementable trend following system. This strategy was characterized by two threshold curves for the conditional probability in a bull regime signaling buy and sell, respectively. The main advantage of this approach is that the conditional probability in a bull market can be obtained directly using actual historical stock price data through a differential equation.

In this chapter, we focus on a mean-reversion model in which its equilibrium is subject to random jumps. Such model can be applied to assets with a “staircase” price behavior. We consider trading involving both buying and selling actions. The objective is to buy and sell the underlying asset sequentially in order to maximize a discounted reward function. Slippage cost associated with each transaction is imposed. We assume that a fixed percentage slippage cost is incurred with each transaction. In general, this is a class of challenging problems because a closed-form solution is difficult to obtain. In this chapter, we consider the case in which the underlying Markov chain jumps frequently between its two states. This leads to a class of singular perturbation problems. The idea is to approximate the value functions of the original problem by the value functions of a limiting problem. The limiting problem is easier to solve. The solution of the limiting problem leads to admissible trading rules that are typically as good as the optimal ones for the original problem. There are substantial studies along the line of singular perturbations. We refer the readers to Sethi and Zhang [13] and Yin and Zhang [14] for related literature. In this chapter, we study the problem using the dynamic programming approach and establish the associated HJB equations (quasi-variational inequalities) for the value functions. Following a viscosity solution approach, we establish asymptotic properties of the value functions. Then using a numerical example, we show how the solution for the limiting problem can be used to construct a set of trading rules for the original problem.

This chapter is organized as follows. In Sect. 11.2, we formulate the problem under consideration. In Sect. 11.3, we study properties of the value functions and the associated HJB equations. In Sect. 11.4, we provide asymptotic properties of the value functions and describe the corresponding limiting problem. In Sect. 11.5, we demonstrate further related approximation schemes. A numerical example is given in Sect. 11.6 in which the closed-form solution obtained in Zhang and Zhang [15] is used to construct a trading rule for the original problem. The performance of the trading rule is provided in this example. Finally, some concluding remarks are provided in Sect. 11.7. Some technical definitions and assumption verification details are given in Appendix.

2 Problem Formulation

Let $X_{t} \in \mathbb{R}$ denote a mean-reverting diffusion with regime-switching governed by

$$\displaystyle{ \mathit{dX}_{t} = a(b(\alpha _{t}) - X_{t})\mathit{dt} +\sigma (\alpha _{t})\mathit{dW }_{t},\ X_{0} = x, }$$

(11.1)

where $a > 0$ is the rate of reversion, b(j), j = 1, 2, is the equilibrium level for each state, $\sigma (j) > 0$, j = 1, 2, is the volatility, $\alpha _{t} \in \{ 1,2\}$ is a two-state Markov chain, and W _t is a standard Brownian motion. In this chapter, we assume that $\alpha _{t}$ and W _t are independent.

Let h(x) be a smooth function. We consider the model in which the asset price is given by S _t  = h(X _t ). For example, the function h(x) = e ^x is used in Zhang and Zhang [15]. In this chapter, we consider h(x) that equals e ^x except when x is large. The main reason for specifying h(x) is to facilitate subsequent analysis without affecting much of the applicability.

Let

$$\displaystyle{ 0 \leq \phi _{1} \leq \psi _{1} \leq \phi _{2} \leq \psi _{2} \leq \cdots }$$

(11.2)

denote a sequence of stopping times. A buying decision is made at $\phi _{k}$ and a selling decision at $\psi _{k}$, k = 1, 2, ….

We consider the case that the net position at any time can be either flat (no stock holding) or long (with one share of stock holding). Let i = 0, 1 denote the initial net position. If initially the net position is long (i = 1), then one should sell the stock before acquiring a share. The corresponding sequence of stopping times is denoted by $\varLambda _{1} = (\psi _{1},\phi _{2},\psi _{2},\phi _{3},\ldots )$. Likewise, if initially the net position is flat (i = 0), then one should first buy a stock before selling a share. The corresponding sequence of stopping times is denoted by $\varLambda _{0} = (\phi _{1},\psi _{1},\phi _{2},\psi _{2},\ldots )$.

In addition, we consider the problem with at most N round trips of trading. We use the notation $\varLambda _{1}^{n} = (\psi _{1},\phi _{2},\psi _{2},\phi _{3},\ldots, \phi _{n},\psi _{n})$ and $\varLambda _{0}^{n} = (\phi _{1},\psi _{1},\phi _{2},\psi _{2},\ldots, \phi _{n},\psi _{n})$ to label the corresponding stopping times limited to n round trips for n = 0, 1, …, N.

Let 0 < K < 1 denote the percentage of slippage (or commission) per transaction. Given the initial states X ₀ = x, $\alpha _{0} =\alpha$, and initial net position i = 0, 1, the reward functions of the decision sequences $\{\varLambda _{i}^{n},\,n = 0,1,\ldots, N\}$ are given as follows:

$$\displaystyle{ J_{i}^{n}(x,\alpha, \varLambda _{ i}^{n}) = \left \{\begin{array}{ll} E\Biggl \{\sum _{k=1}^{n}\left [{e}^{-\rho \psi _{k} }S_{\psi _{k}}(1 - K) - {e}^{-\rho \phi _{k} }S_{\phi _{k}}(1 + K)\right ]\Biggr \}, &\mbox{ if }i = 0, \\ E\Biggl \{{e}^{-\rho \psi _{1} }S_{\psi _{1}}(1 - K) & \\ \quad +\sum _{ k=2}^{n}\left [{e}^{-\rho \psi _{k} }S_{\psi _{k}}(1 - K) - {e}^{-\rho \phi _{k} }S_{\phi _{k}}(1 + K)\right ]\Biggr \},&\mbox{ if }i = 1,\\ \end{array} \right . }$$

(11.3)

where ρ > 0 is the discount factor.

For i = 0, 1 and n = 0, 1, …, N, let $V _{i}^{n}(x,\alpha )$ denote the value functions with the initial state $(X_{0},\alpha _{0}) = (x,\alpha )$ and initial net positions i = 0, 1. That is,

$$\displaystyle{ V _{i}^{n}(x,\alpha ) =\sup _{ \varLambda _{i}^{n}}J_{i}^{n}(x,\alpha, \varLambda _{ i}^{n}). }$$

(11.4)

Remark 11.1.

In (29), we allow the equalities, i.e., one is allowed to buy and sell at the same time. Nevertheless, owing to the existence of the positive slippage cost K, simultaneous buying and selling only cause negative returns and therefore are automatically ruled out by optimality conditions.

Let Q = (q _ij ) denote the generator of $\alpha _{t}$ and let $\mathcal{A}$ denote the generator of $(X_{t},\alpha _{t})$, i.e.,

$$\displaystyle{\mathcal{A}f(x,\alpha ) = a(b(\alpha ) - x)\frac{\partial f(x,\alpha )} {\partial x} + \frac{{\sigma }^{2}(\alpha )} {2} \frac{{\partial }^{2}f(x,\alpha )} {\partial {x}^{2}} + Qf(x,\cdot )(\alpha ),}$$

where $Qf(x,\cdot )(\alpha ) = q_{\alpha 1}f(x,1) + q_{\alpha 2}f(x,2)$, $\alpha = 1,2$.

In Fig. 11.1, a sample path of $(X_{t},\alpha _{t})$ is provided. The picture was generated using the Monte Carlo method with

Fig. 11.1

Mean-reversion with regime switching (a) A sample path of X _t , (b) a sample path of α _t

$$\displaystyle{a = 0.8,\,b(1) = 3,\,b(2) = 1,\,\sigma (1) = 0.7,\,\sigma (2) = 0.3,\,Q = \left (\begin{array}{cc} - 0.91& 0.91\\ 0.62 & - 0.62 \end{array} \right ),\,X_{0} = 1.}$$

It is clear from Fig. 11.1, when $\alpha _{t} = 1$, the equilibrium b(1) = 3 serves as an attractor for X _t pulling it upwards; when $\alpha _{t}$ switched to 2, the new equilibrium b(2) = 1 pulls X _t downwards and so on.

As mentioned in the introduction, a closed-form solution to the problem is difficult to obtain. In this chapter, we consider the case in which the Markov chain jumps frequently between its two states. We aim at the corresponding asymptotic properties. In particular, we consider case where the generator has the following form:

$$\displaystyle{{Q}^{\varepsilon } = \frac{1} {\varepsilon } \widetilde{Q}+\widehat{Q} = \frac{1} {\varepsilon } \left (\begin{array}{c c} -\lambda & \lambda \\ \mu &-\mu \end{array} \right )+\left (\begin{array}{c c} -\lambda _{1} & \lambda _{1} \\ \mu _{1} & -\mu _{1} \end{array} \right ),}$$

where ɛ > 0 is a small parameter, and $\lambda$, μ, $\lambda _{1}$, and μ ₁ are positive constants. We study the convergence of the problem as $\varepsilon \rightarrow 0$. For related Markov models in connection with manufacturing systems, see Sethi and Zhang [13].

Remark 11.2.

The Markov chain $\alpha _{t}$ generated by ${Q}^{\varepsilon }$ represents the regime of the underlying market. We focus on the market with frequent regime changes in $\alpha _{t}$. Such a scenario often arises in a prolonged sideways market such as Dow Jones Industrial Average during the 1960s and 1980s. Its behavior can be captured by our regime-switching model with a relatively small $\varepsilon$. In this chapter, we aim at models with a not-so-small $\varepsilon$ and construct near optimal trading rules from the optimal solution of the corresponding limiting problem as $\varepsilon \rightarrow 0$. A major advantage of our approach is that one does not have to identify the state of $\alpha _{t}$, which is difficult during the period when it is changing rapidly.

The corresponding Markov chain will be labeled as $\alpha _{t}^{\varepsilon }$. Similarly, we use $X_{t}^{\varepsilon }$ for X _t , $S_{t}^{\varepsilon }$ for S _t , $J_{i}^{n,\varepsilon }$ for $J_{i}^{n}$, and $V _{i}^{n,\varepsilon }$ for $V _{i}^{n}$ from now on to emphasize the dependence on $\varepsilon$. Using this notation, the optimal trading problem ${\mathcal{P}}^{N,\varepsilon }$ can be written as follows:

$$\displaystyle{{\mathcal{P}}^{N,\varepsilon } : \left \{\begin{array}{ll} \max &J_{i}^{n,\varepsilon }(x,\alpha, \varLambda _{i}^{n}) \\ & = \left \{\begin{array}{ll} E\Biggl \{\sum _{k=1}^{n}\left [{e}^{-\rho \psi _{k} }S_{\psi _{k}}^{\varepsilon }(1 - K) - {e}^{-\rho \phi _{k} }S_{\phi _{k}}^{\varepsilon }(1 + K)\right ]\Biggr \}, &\mbox{ if }i = 0, \\ E\Biggl \{{e}^{-\rho \psi _{1} }S_{\psi _{1}}^{\varepsilon }(1 - K) & \\ \quad +\sum _{ k=2}^{n}\left [{e}^{-\rho \psi _{k} }S_{\psi _{k}}^{\varepsilon }(1 - K) - {e}^{-\rho \phi _{k} }S_{\phi _{k}}^{\varepsilon }(1 + K)\right ]\Biggr \},&\mbox{ if }i = 1,\\ \end{array} \right . \\ \mbox{ s.t.} &\mathit{dX}_{t}^{\varepsilon } = a(b(\alpha _{ t}^{\varepsilon }) - X_{ t}^{\varepsilon })\mathit{dt} +\sigma (\alpha _{ t}^{\varepsilon })\mathit{dW }_{ t},\ X_{0}^{\varepsilon } = x, \\ \mbox{ value fn}&V _{i}^{n,\varepsilon }(x,\alpha ) =\sup _{\varLambda _{i}^{n}}J_{i}^{n,\varepsilon }(x,\alpha, \varLambda _{i}^{n}),\ n = 0,1,\ldots, N, \end{array}\right .}$$

Note that the sequence $\varLambda _{0}^{n} = (\phi _{1},\psi _{1},\ldots, \phi _{n},\psi _{n})$ can be regarded as a combination of a buy at $\phi _{1}$ and then followed by the sequence of stopping times $\varLambda _{1}^{n} = (\psi _{1},\phi _{2},\psi _{2},\ldots, \phi _{n},\psi _{n})$. In view of this, we have

$$\displaystyle{\begin{array}{rl} V _{0}^{n,\varepsilon }(x,\alpha ) \geq &J_{0}^{n,\varepsilon }(x,\alpha, \varLambda _{ 0}^{n}) \\ =&E\left \{{e}^{-\rho \psi _{1}}S_{\psi _{ 1}}^{\varepsilon }(1 - K) +\sum _{ k=2}^{n}\left [{e}^{-\rho \psi _{k}}S_{\psi _{ k}}^{\varepsilon }(1 - K) - {e}^{-\rho \phi _{k}}S_{\phi _{ k}}^{\varepsilon }(1 + K)\right ]\right \} \\ & -{\mathit{Ee}}^{-\rho \phi _{1}}S_{\phi _{ 1}}^{\varepsilon }(1 + K) \\ =&J_{1}^{n,\varepsilon }(X_{\phi _{1}}^{\varepsilon },\alpha, \varLambda _{ 1}^{n}) -{\mathit{Ee}}^{-\rho \phi _{1}}S_{\phi _{ 1}}^{\varepsilon }(1 + K).\\ \end{array} }$$

In particular, setting $\phi _{1} = 0$ (recall that $S_{t}^{\varepsilon } = h(X_{t}^{\varepsilon })$), we obtain the inequality

$$\displaystyle{ V _{0}^{n,\varepsilon }(x,\alpha ) \geq V _{ 1}^{n,\varepsilon }(x,\alpha ) - h(x)(1 + K). }$$

(11.5)

Similarly, we can show that

$$\displaystyle{ V _{1}^{n,\varepsilon }(x,\alpha ) \geq V _{ 0}^{n-1,\varepsilon }(x,\alpha ) + h(x)(1 - K). }$$

(11.6)

Formally, the associated HJB equations should have the form:

$$\displaystyle{ \begin{array}{l} \min \left \{\rho V _{0}^{n,\varepsilon }(x,\alpha ) -\mathcal{A}V _{ 0}^{n,\varepsilon }(x,\alpha ),\;V _{ 0}^{n,\varepsilon }(x,\alpha ) - V _{ 1}^{n,\varepsilon }(x,\alpha ) + h(x)(1 + K)\right \} = 0, \\ \min \left \{\rho V _{1}^{n,\varepsilon }(x,\alpha ) -\mathcal{A}V _{ 1}^{n,\varepsilon }(x,\alpha ),\;V _{ 1}^{n,\varepsilon }(x,\alpha ) - V _{ 0}^{n-1,\varepsilon }(x,\alpha ) - h(x)(1 - K)\right \} = 0,\\ \end{array} }$$

(11.7)

for $n = 1,2,\ldots, N$ and $\alpha = 1,2$. Here, we follow the convention that $V _{0}^{0,\varepsilon }(x,\alpha ) = 0$.

Next, we impose conditions on h(x).

Assumption. h(x), h ′(x), $\mathit{xh}'(x)$, and h ′(x) are bounded and Lipschitz.

Example 11.1.

An immediate example satisfying the above conditions can be given as follows. Let $h_{0}(x) = \left \{\begin{array}{ll} {e}^{x} &\mbox{ for }x \leq M, \\ {e}^{M}&\mbox{ for }x > M, \end{array} \right .$ for a fixed M. Take h(x) to be the convolution of h ₀ with the kernel $\varPsi (x) = (1/\sqrt{2\pi }){e}^{-{x}^{2}/2 }$. Validation of these conditions is provided in Appendix.

Under these assumptions, we can show, following a similar approach as in Sethi and Zhang [13, Chap. 8], that $V _{i}^{n,\varepsilon }(x,\alpha )$ are the viscosity solutions (see the definition given in Appendix) of the HJB equations (34).

In this chapter, C (and C _i ) are generic positive constants with convention $C + C = C$ and $\mathit{CC} = C$, etc.

3 Properties of the Value Functions

In this section, we consider the basic properties of the value functions. In particular, we establish the boundedness and Lipschitz continuity of these functions.

Lemma 11.1.

There exists a constant C ₀ such that

$$\displaystyle{0 \leq V _{i}^{n,\varepsilon }(x,\alpha ) \leq C_{ 0},}$$

for ɛ > 0, $x \in \mathbb{R}$, $\alpha = 1,2$, i = 0,1, and n = 0,1,…,N.

Proof.

In view of the definition of $V _{i}^{n,\varepsilon }(x,\alpha )$, it is clear that they are nonnegative. It remains to establish their upper bounds. Let

$$\displaystyle{F(x,\alpha ) = a(b(\alpha ) - x)h'(x) + \frac{{\sigma }^{2}(\alpha )} {2} h''(x) -\rho h(x).}$$

Then, using Dynkin’s formula, we have

$$\displaystyle{{ \mathit{Ee}}^{-\rho \psi _{k} }S_{\psi _{k}}^{\varepsilon } -{\mathit{Ee}}^{-\rho \phi _{k} }S_{\phi _{k}}^{\varepsilon } = E\int _{\phi _{k}}^{\psi _{k} }{e}^{-\rho s}F(X_{ s}^{\varepsilon },\alpha _{ s})\mathit{ds}. }$$

(11.8)

It is easy to see that the function $F(x,\alpha )$ is bounded above on $\mathbb{R}$ by the boundedness assumptions on h(x). Let C be an upper bound of F. It follows that

$$\displaystyle{{ \mathit{Ee}}^{-\rho \psi _{k} }S_{\psi _{k}}^{\varepsilon } -{\mathit{Ee}}^{-\rho \phi _{k} }S_{\phi _{k}}^{\varepsilon } \leq \mathit{CE}\int _{\phi _{k}}^{\psi _{k} }{e}^{-\rho t}\mathit{dt}. }$$

(11.9)

Using the definition of $J_{0}^{n,\varepsilon }(x,\alpha, \varLambda _{0}^{n})$, we have

$$\displaystyle{\begin{array}{rl} J_{0}^{n,\varepsilon }(x,\alpha, \varLambda _{0}^{n})& \leq \sum _{k=1}^{n}\left ({\mathit{Ee}}^{-\rho \psi _{k} }S_{\psi _{k}}^{\varepsilon } -{\mathit{Ee}}^{-\rho \phi _{k} }S_{\phi _{k}}^{\varepsilon }\right ) \\ & \leq \sum _{k=1}^{n}\mathit{CE}\int _{\phi _{ k}}^{\psi _{k} }{e}^{-\rho t}\mathit{dt} \\ & \leq C\int _{0}^{\infty }{e}^{-\rho t}\mathit{dt} := C_{ 0}. \end{array} }$$

This implies that $0 \leq V _{0}^{n,\varepsilon }(x,\alpha ) \leq C_{0}$.

Similarly, letting $C_{h} =\sup \vert h(x)\vert$, we have the inequalities

$$\displaystyle{J_{1}^{n,\varepsilon }(x,\alpha, \varLambda _{ 1}^{n}) \leq C_{ 0} +{ \mathit{Ee}}^{-\rho \psi _{1} }h(X_{\psi _{1}}^{\varepsilon })(1 - K) \leq C_{0} + C_{h}(1 - K) := C_{0}.}$$

Therefore, $0 \leq V _{1}^{n,\varepsilon }(x,\alpha ) \leq C_{0}$. This completes the proof.

Lemma 11.2.

$V _{i}^{n,\varepsilon }(x,\alpha )$ are Lipschitz, i.e., there exists C ₀ such that

$$\displaystyle{\vert V _{i}^{n,\varepsilon }(x_{ 1},\alpha ) - V _{i}^{n,\varepsilon }(x_{ 2},\alpha )\vert \leq C_{0}\vert x_{1} - x_{2}\vert .}$$

for $\varepsilon > 0$, $x_{1},x_{2} \in \mathbb{R}$, $\alpha = 1,2$, i = 0,1, and n = 0,1,…,N.

Proof.

Given x ₁ and x ₂, let $X_{t}^{1}$ and $X_{t}^{2}$ be solutions of (28) with $X_{0}^{1} = x_{1}$ and $X_{0}^{2} = x_{2}$, respectively. We claim that: There exists an constant $C_{0}$ such that for any stopping time τ,

$$\displaystyle{ \left \vert E\left [{e}^{-\rho \tau }(h(X_{\tau }^{1}) - h(X_{\tau }^{2})\right ]\right \vert \leq C_{ 0}\vert x_{1} - x_{2}\vert . }$$

(11.10)

Let

$$\displaystyle{G(x,y,\alpha ) = ab(\alpha )[h'(x)-h'(y)]-a[\mathit{xh}'(x)-\mathit{yh}'(y)]+\frac{{\sigma }^{2}(\alpha )} {2} [h''(x)-h''(y)]-\rho [h(x)-h(y)].}$$

Then, using the Lipschitz assumptions on h(x), we can see that

$$\displaystyle{\vert G(x,y,\alpha )\vert \leq C_{0}\vert x - y\vert,}$$

for some constant C ₀. Then, applying Dynkin’s formula, we have

$$\displaystyle{E\left [{e}^{-\rho \tau }(h(X_{ t}^{1}) - h(X_{ t}^{2}))\right ] = h(x_{ 1}) - h(x_{2}) + E\int _{0}^{\tau }{e}^{-\rho t}G(X_{ t}^{1},X_{ t}^{2},\alpha _{ t})\mathit{dt}.}$$

It follows that

$$\displaystyle{\begin{array}{rl} \left \vert E\left [{e}^{-\rho \tau }(h(X_{ t}^{1}) - h(X_{ t}^{2}))\right ]\right \vert &\leq \vert h(x_{ 1}) - h(x_{2})\vert + E\int _{0}^{\infty }{e}^{-\rho t}\vert G(X_{ t}^{1},X_{ t}^{2},\alpha _{ t})\vert \mathit{dt} \\ & \leq C_{0}\vert x_{1} - x_{2}\vert + C_{0}E\int _{0}^{\infty }{e}^{-\rho t}\vert X_{ t}^{1} - X_{ t}^{2}\vert \mathit{dt}.\\ \end{array} }$$

Note that

$$\displaystyle{X_{t}^{1} - X_{ t}^{2} = x_{ 1} - x_{2} - a\int _{0}^{t}(X_{ s}^{1} - X_{ s}^{2})\mathit{ds}.}$$

Therefore, $X_{t}^{1} - X_{t}^{2} = (x_{1} - x_{2}){e}^{-\mathit{at}}$. In view of this, we have

$$\displaystyle{\left \vert E\left [{e}^{-\rho \tau }(h(X_{ t}^{1}) - h(X_{ t}^{2}))\right ]\right \vert \leq C_{ 0}\vert x_{1}-x_{2}\vert +C_{0}E\int _{0}^{\infty }{e}^{-\rho t}\vert x_{ 1}-x_{2}\vert {e}^{-\mathit{at}}\mathit{dt} = C_{ 0}\vert x_{1}-x_{2}\vert,}$$

which proves the claim. Using this inequality, for any given $\varLambda _{i}^{n}$, it is easy to see that

$$\displaystyle{\vert J_{i}^{n,\varepsilon }(x_{ 1},\alpha, \varLambda _{i}^{n}) - J_{ i}^{n,\varepsilon }(x_{ 2},\alpha, \varLambda _{i}^{n})\vert \leq C_{ 0}\vert x_{1} - x_{2}\vert .}$$

4 Asymptotic Properties

In this section, we study the asymptotic properties of the value functions as $\varepsilon \rightarrow 0$. We first characterize the limiting problem and then establish the desired convergence.

Lemma 11.3.

For each $(x,\alpha )$, if for some subsequence of $\varepsilon$, $V _{i}^{n,\varepsilon }(x,\alpha ) \rightarrow V _{i}^{n,0}(x,\alpha )$, then $V _{i}^{n,0}(x,\alpha ) = V _{i}^{n,0}(x)$.

Proof.

Let ${\tau }^{\varepsilon }$ denote the first jump time of $\alpha _{t}^{\varepsilon }$. Then ${\tau }^{\varepsilon } \rightarrow 0$ a.s. as $\varepsilon \rightarrow 0$. Following the dynamic programming principle, we have

$$\displaystyle{V _{i}^{n,\varepsilon }(x,\alpha ) \geq {\mathit{Ee}}^{{-\rho \tau }^{\varepsilon }}V _{ i}^{n,\varepsilon }(X_{{\tau }^{\varepsilon }},\alpha _{{\tau }^{\varepsilon }}).}$$

If $\alpha = 1$, then sending $\varepsilon \rightarrow 0$, we have

$$\displaystyle{{V }^{n,0}(x,1) \geq {V }^{n,0}(x,2).}$$

Similarly,

$$\displaystyle{ {V }^{n,0}(x,2) \geq {V }^{n,0}(x,1). }$$

Therefore, ${V }^{n,0}(x,1) = {V }^{n,0}(x,2)$.

Let $(\nu _{1},\nu _{2})$ denote the equilibrium distribution corresponding to $\widetilde{Q}$, i.e.,

$$\displaystyle{(\nu _{1},\nu _{2}) = \left ( \frac{\mu } {\lambda +\mu }, \frac{\lambda } {\lambda +\mu }\right ).}$$

so that $(\nu _{1},\nu _{2})\widetilde{Q} = (0,0)$. Let $\overline{X}_{t}$ denote the corresponding mean-reversion process with mean $\overline{b} =\nu _{1}b(1) +\nu _{2}b(2)$ and volatility $\overline{\sigma } = \sqrt{\nu {_{1 } \sigma }^{2 } (1) +\nu { _{2 } \sigma }^{2 } (2)}$. The stock price driven by $\overline{X}_{t}$ is denoted by $\overline{S}_{t} = h(\overline{X}_{t})$.

Given a sequence of $\sigma \{W_{r} :\, r \leq t\}$ measurable stopping times

$$\displaystyle{0 \leq \phi _{1} \leq \psi _{1} \leq \phi _{2} \leq \psi _{2} \leq \cdots \,,}$$

one can define the set of stopping times $\varLambda _{i}^{n}$ as before for n = 0, 1, …, N and i = 0, 1. The limiting problem ${\mathcal{P}}^{N,0}$ can be defined as follows:

$$\displaystyle{{\mathcal{P}}^{N,0} : \left \{\begin{array}{ll} \max &\overline{J}_{i}^{n}(x,\varLambda _{i}^{n}) \\ & = \left \{\begin{array}{ll} E\Biggl \{\sum _{k=1}^{n}\left [{e}^{-\rho \psi _{k} }\overline{S}_{\psi _{k}}(1 - K) - {e}^{-\rho \phi _{k} }\overline{S}_{\phi _{k}}(1 + K)\right ]\Biggr \}, &\mbox{ if }i = 0, \\ E\Biggl \{{e}^{-\rho \psi _{1} }\overline{S}_{\psi _{1}}(1 - K) & \\ \quad +\sum _{ k=2}^{n}\left [{e}^{-\rho \psi _{k} }\overline{S}_{\psi _{k}}(1 - K) - {e}^{-\rho \phi _{k} }\overline{S}_{\phi _{k}}(1 + K)\right ]\Biggr \},&\mbox{ if }i = 1,\\ \end{array} \right . \\ \mbox{ s.t.} &d\overline{X}_{t} = a(\overline{b} -\overline{X}_{t})\mathit{dt} + \overline{\sigma }\mathit{dW }_{t},\ \overline{X}_{0} = x, \\ \mbox{ value fn}&\overline{V }_{i}^{n}(x) =\sup _{\varLambda _{i}^{n}}\overline{J}_{i}(x,\varLambda _{i}^{n}). \end{array} \right .}$$

Let $\overline{\mathcal{A}}$ denote the generator of $\overline{X}_{t}$, i.e.,

$$\displaystyle{\overline{\mathcal{A}}f(x) = a(\overline{b} - x)\frac{\mathit{df }(x)} {\mathit{dx}} + \frac{{\overline{\sigma }}^{2}} {2} \frac{{d}^{2}f(x)} {{\mathit{dx}}^{2}} .}$$

The associated HJB equations for the limiting problem should have the form:

$$\displaystyle{ \begin{array}{l} \min \left \{\rho \overline{V }_{0}^{n}(x) -\overline{\mathcal{A}}\overline{V }_{ 0}^{n}(x),\;\overline{V }_{ 0}^{n}(x) -\overline{V }_{ 1}^{n}(x) + h(x)(1 + K)\right \} = 0, \\ \min \left \{\rho \overline{V }_{1}^{n}(x) -\overline{\mathcal{A}}\overline{V }_{ 1}^{n}(x),\;\overline{V }_{ 1}^{n}(x) -\overline{V }_{ 0}^{n-1}(x) - h(x)(1 - K)\right \} = 0,\\ \end{array} }$$

(11.11)

for $n = 1,2,\ldots, N$.

The definition of viscosity solution of the above HJB equations is also given in Appendix. We can show the following lemma, where the uniqueness can be obtained along the line of Pham [12].

Lemma 11.4.

$\overline{V }_{i}^{n}(x)$ are the unique viscosity solutions of the HJB equations ( 38 ).

Next, we give the main result of this chapter. We show that the value functions of the original problem converge to those of the limiting problem. This suggests that the optimal solution of the limiting problem can be used to construct a trading rule for the original problem. We refer the readers to Sethi and Zhang [13] for similar approach in connection with manufacturing systems.

Theorem 11.1.

As $\varepsilon \rightarrow 0$, we have

$$\displaystyle{V _{i}^{n,\varepsilon }(x,\alpha ) \rightarrow \overline{V }_{ i}^{n}(x),}$$

for $n = 0,1,\ldots, N$, i = 0,1, $x \in \mathbb{R}$, and $\alpha = 1,2$.

Proof.

Recall the Lipschitz properties of $V _{i}^{n,\varepsilon }$ in Lemma  11.2. In view of the Arzela–Ascoli Theorem, for each sequence of $\{\varepsilon \rightarrow 0\}$, there exists a further subsequence (still indexed by $\varepsilon$) such that $V _{i}^{n,\varepsilon }(x,\alpha )$ converges. Denote the limit by $V _{i}^{n,0}(x,\alpha )$. Then by Lemma  11.3, $V _{i}^{n,0}(x,\alpha ) = V _{i}^{n,0}(x)$. It suffices to show that $V _{i}^{n,0}(x)$ is a viscosity solution of (38) because Lemma  11.4 implies that $V _{i}^{n,0}(x) = \overline{V }_{i}^{n}(x)$. Following Lemma A.25 in Yin and Zhang [14], for each $i = 0,1$, take a function $\phi _{i}(x) \in {C}^{2}$ such that $V _{i}^{n,0}(x) -\phi _{i}(x)$ has a strictly local maximum at any given x ₀ in a neighborhood N(x ₀). Choose $x_{i,\alpha }^{n,\varepsilon } \in N(x_{0})$ such that

$$\displaystyle{V _{i}^{n,\varepsilon }(x_{ i,\alpha }^{n,\varepsilon },\alpha ) -\phi _{ i}(x_{i,\alpha }^{n,\varepsilon }) =\max _{ x\in N(x_{0})}\{V _{i}^{n,\varepsilon }(x,\alpha ) -\phi _{ i}(x)\}.}$$

Then, $x_{i,\alpha }^{n,\varepsilon } \rightarrow x_{0}$, as $\varepsilon \rightarrow 0$. First, fix i = 0. We are to show the following inequality:

$$\displaystyle{ \min \left \{\rho V _{0}^{n,0}(x_{ 0}) -\overline{\mathcal{A}}\phi _{0}(x_{0}),V _{0}^{n,0}(x_{ 0}) - V _{1}^{n,0}(x_{ 0}) + h(x_{0})(1 + K)\right \} \leq 0. }$$

(11.12)

If

$$\displaystyle{ V _{0}^{n,0}(x_{ 0}) - V _{1}^{n,0}(x_{ 0}) + h(x_{0})(1 + K) \leq 0, }$$

then (39) holds. Otherwise,

$$\displaystyle{V _{0}^{n,0}(x_{ 0}) - V _{1}^{n,0}(x_{ 0}) + h(x_{0})(1 + K) > 0.}$$

Then there exists $N_{0}(x_{0}) \subset N(x_{0})$ such that

$$\displaystyle{V _{0}^{n,\varepsilon }(x) - V _{ 1}^{n,\varepsilon }(x) + h(x)(1 + K) > 0}$$

on N ₀(x ₀) for $\varepsilon$ small enough. Recall that $V _{i}^{n,\varepsilon }$ is a viscosity solution to (34). $V _{0}^{n,\varepsilon }(x,\alpha )$ must satisfy (45). Necessarily,

$$\displaystyle{\rho V _{0}^{n,\varepsilon }(x_{ 0,\alpha }^{n,\varepsilon },\alpha ) -{\mathcal{A}}^{\phi _{0} }V _{0}^{n,\varepsilon }(x_{ 0,\alpha }^{n,\varepsilon },\alpha ) \leq 0,}$$

for $\alpha = 1,2$.

It follows that

$$\displaystyle{ \nu _{1}(\rho V _{0}^{n,\varepsilon }(x_{ 0,1}^{n,\varepsilon },1)-{\mathcal{A}}^{\phi _{0} }V _{0}^{n,\varepsilon }(x_{ 0,1}^{n,\varepsilon },1))+\nu _{ 2}(\rho V _{0}^{n,\varepsilon }(x_{ 0,2}^{n,\varepsilon },2)-{\mathcal{A}}^{\phi _{0} }V _{0}^{n,\varepsilon }(x_{ 0,2}^{n,\varepsilon },2)) \leq 0. }$$

(11.13)

Note that

$$\displaystyle{ \begin{array}{l} \nu _{1 } \left (\frac{\lambda } {\varepsilon }\right )(V _{0}^{n,\varepsilon }(x_{ 1}^{\varepsilon },2) - V _{ 0}^{n,\varepsilon }(x_{ 1}^{\varepsilon },1)) +\nu _{ 2}\left (\frac{\mu } {\varepsilon }\right )(V _{0}^{n,\varepsilon }(x_{ 2}^{\varepsilon },1) - V _{ 0}^{n,\varepsilon }(x_{ 2}^{\varepsilon },2)) \\ \leq \nu _{1}\left (\frac{\lambda }{\varepsilon }\right )[V _{0}^{n,\varepsilon }(x_{2}^{\varepsilon },2) -\phi (x_{2}^{\varepsilon }) - (V _{0}^{n,\varepsilon }(x_{1}^{\varepsilon },1) -\phi (x_{1}^{\varepsilon }))] \\ \quad +\nu _{2}\left (\frac{\mu } {\varepsilon }\right )[V _{0}^{n,\varepsilon }(x_{ 1}^{\varepsilon },1) -\phi (x_{ 1}^{\varepsilon }) - (V _{ 0}^{n,\varepsilon }(x_{ 2}^{\varepsilon },2) -\phi (x_{ 2}^{\varepsilon }))] = 0. \end{array} }$$

(11.14)

Using this inequality and sending $\varepsilon \rightarrow 0$ in (40) to obtain $\rho V _{0}^{n,0}(x_{0}) -\overline{\mathcal{A}}\phi _{0}(x_{0}) \leq 0$, which yields (39). Similarly, we can show

$$\displaystyle{\min \left \{\rho V _{1}^{n,0}(x_{ 0}) -\overline{\mathcal{A}}\phi _{1}(x_{0}),V _{0}^{n,0}(x_{ 0}) - V _{0}^{n-1,0}(x_{ 0}) - h(x_{0})(1 - K)\right \} \leq 0.}$$

Thus, $V _{i}^{n,0}(x)$ is a viscosity subsolution to (38).

To show that $V _{i}^{n,0}(x)$ is a viscosity supersolution to (38), note that

$$\displaystyle{\min \left \{\rho V _{0}^{n,\varepsilon }(x_{ 0},\alpha _{0}) -{\mathcal{A}}^{\psi }V _{ 0}^{n,\varepsilon }(x_{ 0},\alpha _{0}),\;V _{0}^{n,\varepsilon }(x_{ 0},\alpha _{0}) - V _{1}^{n,\varepsilon }(x_{ 0},\alpha _{0}) + h(x)(1 + K)\right \} \geq 0}$$

implies

$$\displaystyle{ V _{0}^{n,0}(x_{ 0}) - V _{1}^{n,0}(x_{ 0}) + h(x)(1 + K) \geq 0. }$$

Moreover, following similar argument as in (41), we can show that

$$\displaystyle{\rho V _{0}^{n,0}(x_{ 0}) -\overline{\mathcal{A}}\psi _{0}(x_{0}) \geq 0.}$$

Hence,

$$\displaystyle{\min \left \{\rho V _{0}^{n,0}(x_{ 0}) -\overline{\mathcal{A}}\psi _{0}(x_{0}),\;V _{0}^{n,0}(x_{ 0}) - V _{1}^{n,0}(x_{ 0}) + h(x)(1 + K)\right \} \geq 0.}$$

Similarly, we can show the inequality with i = 1. Therefore $V _{i}^{n,0}(x)$ is a viscosity supersolution. This completes the proof.

5 Further Approximations

In this section, we show that the value function $\overline{V }_{i}^{n}(x)$ can be further approximated by taking N to be very large and h(x) to be very close to e ^x. In this case, we can use the closed-form solution obtained in Zhang and Zhang [15] to come up with an approximate solution for the original problem.

Recall the definition of $\varLambda _{i}$ and its N-th round trip truncation $\varLambda _{i}^{N}$. Let

$$\displaystyle{\overline{J}_{i}(x,\varLambda _{i}) =\limsup _{N\rightarrow \infty }\overline{J}_{i}^{N}(x,\varLambda _{ i}^{N})}$$

and $\overline{V }_{i}(x) =\sup _{\varLambda _{i}}\overline{J}_{i}(x,\varLambda _{i})$. It is easy to see that

$$\displaystyle{\lim _{N\rightarrow \infty }\overline{V }_{i}^{N}(x) = \overline{V }_{ i}(x).}$$

In fact, for each $\delta > 0$, let $\varLambda _{i,\delta }$ be a sequence of stopping times such that $\overline{J}_{i}(x,\varLambda _{i,\delta }) \geq \overline{V }_{i}(x)-\delta$. Then, noticing that $\overline{V }_{i}^{N}(x)$ is monotonically increasing in N, we have

$$\displaystyle{\overline{V }_{i}(x)-\delta \leq \overline{J}_{i}(x,\varLambda _{i,\delta }) =\limsup _{N\rightarrow \infty }\overline{J}_{i}^{N}(x,\varLambda _{ i,\delta }^{N}) \leq \limsup _{ N\rightarrow \infty }\overline{V }_{i}^{N}(x) \leq \overline{V }_{ i}(x).}$$

Next, we consider approximating ${e}^{x}$ by particular choices of h(x). Recall Example  11.1 and the definition of h ₀(x). For each $\gamma > 0$, let $\varPsi _{\gamma }(x) = (1/\gamma )\varPsi (x/\gamma )$ and $h_{\gamma }(x)$ be the convolution of h ₀ and $\varPsi _{\gamma }$. Then, $h_{\gamma }(x) \rightarrow h_{0}(x)$ as $\gamma \rightarrow 0$ for all x. Therefore, we can approximate e ^x by $h_{\gamma }(x)$ by choosing a small enough $\gamma$ on $[-M,M]$.

In view of these, the original problem with a large N can be approximated by the limiting problem with a large N and a large M. In the next section, we study a numerical example demonstrating how these approximations work.

6 A Numerical Example

The optimal trading rule in the limiting problem with N = ∞ and $h(x) = {e}^{x}$ was treated in Zhang and Zhang [15]. The main result can be summarized as follows.

Lemma 11.5.

Let $(x_{1}^{{\ast}},x_{2}^{{\ast}})$ be a pair satisfying the following conditions:

$$\displaystyle{x_{1}^{{\ast}}\leq \frac{1} {a}\left (\frac{{\overline{\sigma }}^{2}} {2} + a\overline{b}-\rho \right ) \leq x_{2}^{{\ast}},\ x_{ 2}^{{\ast}}- x_{ 1}^{{\ast}} >\log \left (\frac{1 + K} {1 - K}\right ),}$$

and

$$\displaystyle{ \begin{array}{l} {\left (\begin{array}{cc} \int _{0}^{\infty }\eta (t){e}^{-\kappa (\overline{b}-x_{1}^{{\ast}})t }\mathit{dt} & -\int _{0}^{\infty }\eta (t){e}^{\kappa (\overline{b}-x_{1}^{{\ast}})t }\mathit{dt} \\ \int _{0}^{\infty }t\eta (t){e}^{-\kappa (\overline{b}-x_{1}^{{\ast}})t }\mathit{dt}& \int _{0}^{\infty }t\eta (t){e}^{\kappa (\overline{b}-x_{1}^{{\ast}})t }\mathit{dt}\\ \end{array} \right )}^{-1}\left (\begin{array}{c} {e}^{x_{1}^{{\ast}} }(1 + K) \\ {e}^{x_{1}^{{\ast}} }(1 + K)/\kappa \\ \end{array} \right ) \\ ={ \left (\begin{array}{cc} \int _{0}^{\infty }\eta (t){e}^{-\kappa (\overline{b}-x_{2}^{{\ast}})t }\mathit{dt} & -\int _{0}^{\infty }\eta (t){e}^{\kappa (\overline{b}-x_{2}^{{\ast}})t }\mathit{dt} \\ \int _{0}^{\infty }t\eta (t){e}^{-\kappa (\overline{b}-x_{2}^{{\ast}})t }\mathit{dt}& \int _{0}^{\infty }t\eta (t){e}^{\kappa (\overline{b}-x_{2}^{{\ast}})t }\mathit{dt}\\ \end{array} \right )}^{-1}\left (\begin{array}{c} {e}^{x_{2}^{{\ast}} }(1 - K) \\ {e}^{x_{2}^{{\ast}} }(1 - K)/\kappa \\ \end{array} \right ) \\ \end{array} }$$

(11.15)

where $\kappa = \sqrt{2a}/\overline{\sigma }$ and $\eta (t) = {t}^{(\rho /a)-1}\exp \left (-{t}^{2}/2\right )$.

Let

$$\displaystyle{ \left \{\begin{array}{l} \overline{V }_{0}(x) = \left \{\begin{array}{ll} C_{2}^{{\ast}}\int _{ 0}^{\infty }\eta (t){e}^{\kappa (\overline{b}-x)t}\mathit{dt} &\mbox{ if }x \geq x_{ 1}^{{\ast}}, \\ C_{1}^{{\ast}}\int _{ 0}^{\infty }\eta (t){e}^{-\kappa (\overline{b}-x)t}\mathit{dt} - {e}^{x}(1 + K)&\mbox{ if }x < x_{ 1}^{{\ast}},\\ \end{array} \right . \\ \overline{V }_{1}(x) = \left \{\begin{array}{ll} C_{1}^{{\ast}}\int _{ 0}^{\infty }\eta (t){e}^{-\kappa (\overline{b}-x)t}\mathit{dt} &\mbox{ if }x < x_{ 2}^{{\ast}}, \\ C_{2}^{{\ast}}\int _{ 0}^{\infty }\eta (t){e}^{\kappa (\overline{b}-x)t}\mathit{dt} + {e}^{x}(1 - K)&\mbox{ if }x \geq x_{ 2}^{{\ast}} \\ \end{array} \right . \\ \end{array} \right . }$$

(11.16)

with

$$\displaystyle{\left (\begin{array}{c} C_{1}^{{\ast}} \\ C_{2}^{{\ast}}\\ \end{array} \right ) ={ \left (\begin{array}{cc} \int _{0}^{\infty }\eta (t){e}^{-\kappa (\overline{b}-x_{1}^{{\ast}})t }\mathit{dt} & -\int _{0}^{\infty }\eta (t){e}^{\kappa (\overline{b}-x_{1}^{{\ast}})t }\mathit{dt} \\ \int _{0}^{\infty }t\eta (t){e}^{-\kappa (\overline{b}-x_{1}^{{\ast}})t }\mathit{dt}& \int _{0}^{\infty }t\eta (t){e}^{\kappa (\overline{b}-x_{1}^{{\ast}})t }\mathit{dt}\\ \end{array} \right )}^{-1}\left (\begin{array}{c} {e}^{x_{1}^{{\ast}} }(1 + K) \\ {e}^{x_{1}^{{\ast}} }(1 + K)/\kappa \\ \end{array} \right ).}$$

If, on the interval $(x_{1}^{{\ast}},x_{2}^{{\ast}})$, the following inequalities hold

$$\displaystyle{{e}^{x}(1 - K) \leq \overline{V }_{ 1}(x) -\overline{V }_{0}(x) \leq {e}^{x}(1 + K),}$$

then buying when $x \leq x_{1}^{{\ast}}$ and selling when $x \geq x_{2}^{{\ast}}$ is optimal.

Example 11.2.

In this example, we take

$$\displaystyle{a = 0.8,\,b(1) = 3,\,b(2) = 1,\,\sigma (1) = 0.7,\,\sigma (2) = 0.3,\,}$$

$$\displaystyle{\lambda = 0.09,\,\mu = 0.06,\,\lambda _{1} = 0.01,\,\mu _{1} = 0.02,\,\rho = 0.5,\,\mbox{ and }K = 0.01.}$$

Then, $(\nu _{1},\nu _{2}) = (2/5,3/5)$, $\overline{b} = 9/5$, and $\overline{\sigma } = 0.5$. We solve (42) to obtain $(x_{1}^{{\ast}},x_{2}^{{\ast}}) = (1.115,1.455)$ and the value functions $\overline{V }_{i}(x)$. These functions are plotted in Fig. 11.2. In addition, we vary $\varepsilon = 0.1$, 0. 01, and 0. 001 and solve the corresponding HJB equations in (34) (using the explicit finite difference method) with N = ∞. The value functions $V _{i}^{\varepsilon }$ are also presented in Fig. 11.2. It is clear in this example that $V _{i}^{\varepsilon }$ can be approximated by $\overline{V }_{i}$ when $\varepsilon$ is small enough.

Fig. 11.2

Value function approximation. (a) $V _{1}^{\varepsilon }(x,1)$ and $\overline{V }_{1}(x)$, (b) $V _{1}^{\varepsilon }(x,2)$ and $\overline{V }_{1}(x)$, (c) $V _{0}^{\varepsilon }(x,1)$ and $\overline{V }_{0}(x)$, (d) $V _{0}^{\varepsilon }(x,2)$ and $\overline{V }_{0}(x)$

Next, we use $(x_{1}^{{\ast}},x_{2}^{{\ast}}) = (1.115,1.455)$ to construct the following trading rules for the original problem:

$$\displaystyle{ \left \{\begin{array}{ll} \mbox{ Buy: }&\mbox{ if }X_{t}^{\varepsilon } \leq x_{1}^{{\ast}}, \\ \mbox{ Sell: } &\mbox{ if }X_{t}^{\varepsilon } \geq x_{2}^{{\ast}}.\\ \end{array} \right . }$$

(11.17)

Using these trading rules, we generate the corresponding reward functions with a varying $\varepsilon$ and N = ∞. In particular, we use Monte Carlo simulations based on (28) and generate 10 K sample paths. The corresponding reward functions with ɛ = 1, 0. 01, and 0. 0001 are plotted in Fig. 11.3. Combining these two figures, one can see how the constructed trading rules in (44) work for the original problem.

Fig. 11.3

Reward functions under trading rules constructed from that of the limiting problem.

In general, the control policy obtained via a singular perturbation approach not only work when $\varepsilon$ is small but also work for the problem with not-so-small $\varepsilon$. The performance with $\varepsilon = 1$ can be seen in Fig. 11.3 in which the corresponding reward functions are fairly close to the value functions of the limiting problem, and therefore, in view of Fig. 11.2, close to those of the original problem.

7 Concluding Remarks

In this chapter, we studied the asymptotic properties of the mean-reverting trading problem. We established the convergence of the value functions and demonstrated how the optimal trading rule for the limiting problem can be used to construct a trading rule for the original problem.

In general, to use an optimal trading rule for the original problem, one needs to determine the mode (or the state of $\alpha _{t}^{\varepsilon }$). This typically involves nonlinear filtering as in Dai et al. [5]. Nevertheless, in this chapter, we showed that this is not necessary when the jump rates of $\alpha _{t}^{\varepsilon }$ is large because the constructed trading rule does not require the state information of $\alpha _{t}^{\varepsilon }$.

8 Appendix

In this appendix, we provide the definitions of viscosity solutions of the HJB equations (34) and (38). First, we consider (34). For each $f(x,\alpha )$ and $\phi (x) \in {C}^{2}$, let

$$\displaystyle{{\mathcal{A}}^{\phi }f(x,\alpha ) = a(b(\alpha ) - x)\frac{d\phi (x)} {\mathit{dx}} + \frac{{\sigma }^{2}(\alpha )} {2} \frac{{d}^{2}\phi (x)} {{\mathit{dx}}^{2}} + Qf(x,\cdot )(\alpha ).}$$

Definition 11.1.

$v_{i}^{n,\varepsilon }(x,\alpha )$ is a viscosity solution of (34) if the following hold:

• (a) $v_{i}^{n,\varepsilon }(x,\alpha )$ is uniformly continuous in $x$;

• (b) for any $\alpha _{0} \in \{ 1,2\}$ and $x_{0}$,

  $$\displaystyle{ \begin{array}{l} \min \Big\{\rho v_{0}^{n,\varepsilon }(x_{ 0},\alpha _{0}) -{\mathcal{A}}^{\phi _{0} }v_{0}^{n,\varepsilon }(x_{ 0},\alpha _{0}), \\ v_{0}^{n,\varepsilon }(x_{0},\alpha _{0}) - v_{1}^{n,\varepsilon }(x_{0},\alpha _{0}) + h(x_{0})(1 + K)\Big\} \leq 0, \\ \min \Big\{\rho v_{1}^{n,\varepsilon }(x_{ 0},\alpha _{0}) -{\mathcal{A}}^{\phi _{1} }v_{1}^{n,\varepsilon }(x_{ 0},\alpha _{0}), \\ v_{1}^{n,\varepsilon }(x_{0},\alpha _{0}) - v_{0}^{n-1,\varepsilon }(x_{0},\alpha _{0}) - h(x_{0})(1 - K)\Big\} \leq 0,\end{array} }$$

  (11.18)

  for $n = 0,1,\ldots, N$, whenever $\phi _{i}(x) \in {C}^{2}$ and $v_{i}^{n,\varepsilon }(x,\alpha _{0}) -\phi _{i}(x)$ has a local maximum at x = x ₀; and

• (c) for any $\alpha _{0} \in \{ 1,2\}$ and x ₀,

  $$\displaystyle{ \begin{array}{l} \min \Big\{\rho v_{0}^{n,\varepsilon }(x_{ 0},\alpha _{0}) -{\mathcal{A}}^{\psi _{0} }v_{0}^{n,\varepsilon }(x_{ 0},\alpha _{0}), \\ v_{0}^{n,\varepsilon }(x_{0},\alpha _{0}) - v_{1}^{n,\varepsilon }(x_{0},\alpha _{0}) + h(x_{0})(1 + K)\Big\} \geq 0, \\ \min \Big\{\rho v_{1}^{n,\varepsilon }(x_{ 0},\alpha _{0}) -{\mathcal{A}}^{\psi _{1} }v_{1}^{n,\varepsilon }(x_{ 0},\alpha _{0}), \\ v_{1}^{n,\varepsilon }(x_{0},\alpha _{0}) - v_{0}^{n-1,\varepsilon }(x_{0},\alpha _{0}) - h(x_{0})(1 - K)\Big\} \geq 0,\end{array} }$$

  (11.19)

  for $n = 0,1,\ldots, N$, whenever $\psi _{i}(x) \in {C}^{2}$ and $v_{i}^{n,\varepsilon }(x,\alpha _{0}) -\psi _{i}(x)$ has a local minimum at x = x ₀.

If (a) and (b) (resp. (a) and (c)) hold, we say that v is a viscosity subsolution (resp. viscosity supersolution).

Finally, we give the definition of viscosity solution of (38). Recall that

$$\displaystyle{\overline{\mathcal{A}}f(x) = a(\overline{b} - x)\frac{\mathit{df }(x)} {\mathit{dx}} + \frac{{\overline{\sigma }}^{2}} {2} \frac{{d}^{2}f(x)} {{\mathit{dx}}^{2}} .}$$

Definition 11.2.

$v_{i}^{n}(x)$ is a viscosity solution of (38) if the following hold:

• (a) $v_{i}^{n}(x)$ is uniformly continuous in x;

• (b) for any x ₀,

  $$\displaystyle{ \begin{array}{l} \min \left \{\rho v_{0}^{n}(x_{ 0}) -\overline{\mathcal{A}}\phi _{0}(x_{0}),\;v_{0}^{n}(x_{ 0}) - v_{1}^{n}(x_{ 0}) + h(x_{0})(1 + K)\right \} \leq 0, \\ \min \left \{\rho v_{1}^{n}(x_{ 0}) -\overline{\mathcal{A}}\phi _{1}(x_{0}),\;v_{1}^{n}(x_{ 0}) - v_{0}^{n-1}(x_{ 0}) - h(x_{0})(1 - K)\right \} \leq 0,\end{array} }$$

  (11.20)

  for $n = 0,1,\ldots, N$, whenever ϕ _i (x) ∈ C ² and $v_{i}^{n}(x) -\phi _{i}(x)$ has a local maximum at x = x ₀; and

• (c) for any x ₀,

  $$\displaystyle{ \begin{array}{l} \min \left \{\rho v_{0}^{n}(x_{ 0}) -\overline{\mathcal{A}}\psi _{0}(x_{0}),\;v_{0}^{n}(x_{ 0}) - v_{1}^{n}(x_{ 0}) + h(x_{0})(1 + K)\right \} \geq 0, \\ \min \left \{\rho v_{1}^{n}(x_{ 0}) -\overline{\mathcal{A}}\psi _{1}(x_{0}),\;v_{1}^{n}(x_{ 0}) - v_{0}^{n-1}(x_{ 0}) - h(x_{0})(1 - K)\right \} \geq 0,\end{array} }$$

  (11.21)

  for n = 0, 1, …, N, whenever ψ _i (x) ∈ C ² and $v_{i}^{n}(x) -\psi _{i}(x)$ has a local minimum at x = x ₀.

If (a) and (b) (resp. (a) and (c)) hold, we say that v is a viscosity subsolution (resp. viscosity supersolution).

Next, we give a sketch verifying the conditions in Example  11.1, i.e., we show that h(x), h ′(x), xh ′(x), and h ′(x) are bounded and Lipschitz.

First note that h ₀(x) is bounded and Lipschitz. The boundedness and Lipschitz properties of h, h ′, and h ′ follow from the equalities:

$$\displaystyle{\begin{array}{l} h(x) =\int _{ -\infty }^{\infty }h_{ 0}(x - u)\varPsi (u)\mathit{du}, \\ h'(x) =\int _{ -\infty }^{\infty }h_{ 0}(x - u)\varPsi '(u)\mathit{du}, \\ h''(x) =\int _{ -\infty }^{\infty }h_{ 0}(x - u)\varPsi ''(u)\mathit{du}.\\ \end{array} }$$

Next, we show that xh ′(x) is bounded. Note that

$$\displaystyle{ \begin{array}{ll} xh'(x)& = x\int _{-\infty }^{\infty }h_{ 0}'(u)\varPsi (x - u)\mathit{du}, \\ & = x\int _{-\infty }^{M}{e}^{u}\varPsi (x - u)\mathit{du}, \\ & = x\int _{x-M}^{\infty }{e}^{x-y}\varPsi (y)\mathit{dy},\,(\mbox{ with }y = x - u) \\ & = \frac{{\mathit{xe}}^{x}} {\sqrt{2\pi }} \int _{x-M}^{\infty }{e}^{-y}{e}^{-{y}^{2}/2 }\mathit{dy} \\ & \leq \frac{{\mathit{xe}}^{x}} {\sqrt{2\pi }} \int _{x-M}^{\infty }{e}^{-{y}^{2}/2 }\mathit{dy}. \end{array} }$$

(11.22)

Clearly, it is bounded on ( − ∞, M]. To see it is also bounded on (M, ∞), note also that

$$\displaystyle{ \frac{{\mathit{xe}}^{x}} {\sqrt{2\pi }} \int _{x-M}^{\infty }{e}^{-{y}^{2}/2 }\mathit{dy} \leq \frac{{\mathit{xe}}^{x}} {\sqrt{2\pi }} \left (\frac{\exp \left (-{(x - M)}^{2}/2\right )} {x - M} \right ). }$$

(11.23)

The boundedness follows.

Finally, to see the Lipschitz property of xh ′(x), in view of the Mean Value Theorem, it suffices to show that xh ′(x) is bounded. This can be done similarly as in (49) and (50) by noticing

$$\displaystyle{\mathit{xh}''(x) = x\int _{-\infty }^{\infty }h_{ 0}'(u)\varPsi '(x - u)\mathit{du}.}$$