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RELATED FIELDS

Volume6, Number3, September2016 pp.467–488

AN OPTIMAL MEAN-REVERSION TRADING RULE UNDER A MARKOV CHAIN MODEL

Jingzhi Tie and Qing Zhang

Department of Mathematics University of Georgia Athens, GA 30602, USA

(Communicated by Jiongmin Yong)

Abstract. This paper is concerned with a mean-reversion trading rule. In contrast to most market models treated in the literature, the underlying market is solely determined by a two-state Markov chain. The major advantage of such Markov chain model is its striking simplicity and yet its capability of capturing various market movements. The purpose of this paper is to study an optimal trading rule under such a model. The objective of the problem under consideration is to find a sequence stopping (buying and selling) times so as to maximize an expected return. Under some suitable conditions, explicit solutions to the associated HJ equations (variational inequalities) are obtained.

The optimal stopping times are given in terms of a set of threshold levels. A verification theorem is provided to justify their optimality. Finally, a numerical example is provided to illustrate the results.

1. Introduction. Major market models treated in mathematical finance in recent years can be classified into two broad categories: Cox-Ross-Rubinstein’s binomial tree model (BTM) and Brownian motion based models. See related books by Hull [15], Elliott and Kopp [8], Fouque et al. [10], Karatzas and Shreve [17], and Musiela and Rutkowski [20] among others.

The BTM is widely used in option pricing due to its simplicity and yet clear advantage when pricing American type options. However, a main drawback of the BTM is its non-Markovian nature. The lack of Markovian property makes it diffi- cult to work with mathematically, not to mention closed-form solutions. To preserve the BTM’s simplicity and enhance its mathematical tractability, a two-state Markov chain model was considered in Zhang [29]. The Markov chain model in [29] appears to be natural for not frequently traded securities such as illiquid stocks. Its capabil- ity of capturing this type of markets makes it preferable in related applications. In addition, it can be seen from Example 1 to follow that the Markov chain model is closely related to the traditional mean-reversion diffusion when the jump rates of the Markov chain are large. Related Markov chain based models can be found in Van der Hoek and Elliott [23] and Norberg [21]. Van der Hoek and Elliott introduced a stock price model based on stock dividend rates and a Markov chain noise. Norberg used a Markov chain to represent interest rate and considered a market model driven

2010Mathematics Subject Classification. Primary: 93E20, 91G80; Secondary: 49L20.

Key words and phrases. Mean-reversion Markovian asset, optimal stopping, variational inequalities.

467

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by a Markov chain. In particular, the market model in [21] resembles a GBM in which the ‘drift’ is approximated by the duration between jumps and the ‘diffusion’

is given in terms of jump times. An additional advantage of a Markov chain driven model is that its price is almost everywhere differentiable. Such differentiability is desirable in the optimal control type market analysis proposed by Barmish and Primbs [1]. In connection with dynamic programming problems, the corresponding Hamilton-Jacobi (HJ) equations are of the first order, which are easier to analyze than those under traditional Brownian motion based models.

Mean-reversion models are often used in financial markets to capture price move- ments that have the tendency to move towards an ‘equilibrium.’ There are many studies in connection with mean-reversion stock returns; see e.g., Cowles and Jones [5], Fama and French [9], and Gallagher and Taylor [11] among others. In addi- tion to equity markets, mean-reversion models are used for stochastic interest rates (Vasicek [24] and Hull [15]); stochastic volatility (Hafner and Herwartz [14]); and energy markets (Blanco and Soronow [2]. See also related results in option pricing with a mean-reversion asset by Bos, Ware and Pavlov [3].

Mathematical trading rules have been studied for many years. For example, Zhang [28] considered a selling rule determined by two threshold levels, a target price and a stop-loss limit. Such optimal threshold levels are obtained by solving a set of two-point boundary value problems. Guo and Zhang [13] studied the optimal selling rule under a model with regime switching. Using a smooth-fit technique, they obtained the optimal threshold levels by solving a set of algebraic equations.

These papers are concerned with the selling side of trading in which the underlying price models are of GBM type. Recently, Dai et al. [7] developed a trend following rule based on a conditional probability indicator. They showed that the optimal trading rule can be determined by two threshold curves which can be obtained by solving the associated HJ equations. Similar idea was developed following a confidence interval approach by Iwarere and Barmish [16]. In addition, Merhi and Zervos [18] studied an investment capacity expansion/reduction problem following a dynamic programming approach under a geometric Brownian motion market model.

In connection with mean-reversion trading, Zhang and Zhang [27] obtained a buy- low and sell-high policy by charactering the ‘low’ and ‘high’ levels in terms of two threshold levels. Further development along this line can be found in pairs trading in Song and Zhang [22]. Under a mean-reversion model, a pair of stocks can be traded simultaneously, when their price divergence exceeds a predetermined level, to bet on their eventual convergence. More details in connection with pairs trading and extensive empirical market tests can be found in Gatev et al. [12] and references therein.

To take advantage of many new features of Markov chain models and explore related mean-reversion markets, it is the purpose of this paper to study the corre- sponding trading strategies. The objective is to buy and sell the mean-reversion security to maximize a reward function. A fixed (commission or slippage) cost will be imposed to each transaction. We study the problem following a dynamic programming approach and establish the associated HJ equations for the value functions. We show that the corresponding optimal stopping (buying and selling) times can be determined by four threshold levels depending on market states. These key levels can be obtained by solving a set of algebraic like equations. In solving the associated HJ equations, hypergeometric series arise naturally in the analysis.

Various boundary conditions and convexity conditions of hypergeometric functions

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are involved extensively. We provide a set of sufficient conditions that guarantee the optimality of the trading rules. Finally, to illustrate the results, we present a market backtesting example with the daily closing prices of Dow Jones Industrial Average from 1961 to 1980.

The main contributions of this paper include: (1) Introduction of a new mean- reversion model driven by a two-state Markov chain. Such model is very simple in structure and yet powerful enough to capture many important market features.

(2) Development of mean-reversion trading strategies that are simple to implement.

The corresponding optimal stopping problem is solved using a smooth-fit technique and hypergeometric analysis is involved in a substantial way.

This paper is organized as follows. In§2, we formulate the mean-reversion trading problem under consideration. In §3, we study preliminary properties of the value functions. In§4, we consider the associate HJ equations and their solutions. Then in§5, we discuss additional inequalities including the variational inequalities arising from the HJ equations and related convexity conditions. In§6, we give a verification theorem under suitable conditions. A numerical example is considered in§7. Several technical lemmas are postponed and provided in Appendix.

2. Problem formulation. Let{αt: t≥0}denote a two-state Markov chain with state space M={1,2} and the generatorQ=

−λ1 λ1

λ2 −λ2

. Here λ1 and λ2

are positive.

LetSt denote the stock price (or a function of the price) at timetgiven by the equation

dSt

dt =a(b−St) +f(αt), t≥0, (1) where a > 0 is the rate of reversion and b the equilibrium. Here f(1) = f1 > 0 represents the uptick return rate andf(2) =f2<0 the downtick return rate.

Example 2.1. (Convergence to a Mean-Reversion Diffusion). In this exam- ple, we present a simple Markov chain model and demonstrate how it approaches the corresponding mean-reversion diffusion. Given ε >0, we considerf1=σ/√

ε, f2 = −σ/√

ε, λ1 = λ2 = 1/ε. Using the asymptotic normality given in Yin and Zhang [26, Theorem 5.9], we can show that the solution to (1) St=Stε converges weakly to the solution ofdSt=a(b−St)dt+σdWt, whereWtis a standard Brownian motion.

In particular, taking a= 1,b= 2,σ= 0.5, andε= 0.1, we generate the Monte Carlo sample paths of (Stε, αεt). One sample is given in (a) of Figure 1. Similarly, we varyεand plot the corresponding sample paths in (b) and (c) of Figure1. It is clear from these pictures that asεgets smaller and smaller, the fluctuation ofαtis more and more rapidly and the correspondingStε approaches to a mean-reversion diffusion.

Remark 2.1. The main advantage of our model is its simplicity and flexibility in capturing various market conditions. For example, in a typical sideways (trendless) market, the price range can be precisely defined. In our model, such ‘price band’

arises naturally. A candidate for the band can be given by the interval [b+f2/a, b+ f1/a].

We consider the state processXt=St−b. ThenXtsatisfies dXt

dt =−aXt+f(αt), t≥0, (2)

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0 0.5 1 1.5 2 2.5 3

0 5 10 15 20

0 0.5 1 1.5 2 2.5 3

0 5 10 15 20

0 0.5 1 1.5 2 2.5 3

0 5 10 15 20

Figure 1. Monte Carlo Sample Paths of (Stε, αεt) withε=0.1, 0.01, 0.001, resp.

with initial vector (X0, α0) = (x, α). Let I0 = [f2/a, f1/a]. Then Xt ∈ I0 for all t≥0 ifX0∈ I0.

LetFt ={Xr : r ≤t} denote the filtration generated by Xt. Note that αt is observable andFt={αr: r≤t}. Let

0≤τ1b≤τ1s≤τ2b≤τ2s≤ · · · (3)

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denote a sequence of stopping times with respect to {Ft}. A buying decision is made atτnb and a selling decision atτns, n= 1,2, . . ..

We consider the case that the net position at any time can be either long with one share of the stock or flat with no stock position. Leti= 0,1 denote the initial net position. If the initial net position is long (i = 1), then one should sell the stock before acquiring any future shares. The corresponding sequence of stopping times is denoted by Λ1 = (τ1s, τ2b, τ2s, τ3b, . . .). Likewise, if the initial net position is flat (i= 0), then one should start to buy a share of the stock. The corresponding sequence of stopping times is denoted by Λ0= (τ1b, τ1s, τ2b, τ2s, . . .).

LetK >0 denote the fixed transaction cost (e.g., slippage and/or commission).

Given the initial state vector (X0, α0) = (x, α), the initial net positioni= 0,1, and the decision sequences (Λ0and Λ1), we consider the corresponding reward functions

Ji(x, α,Λi) =



















 E

( X

n=1

h

e−ρτns(Xτns −K)−e−ρτnb(Xτb n+K)i

)

, ifi= 0, E

(

e−ρτ1s(Xτ1s−K) +

X

n=2

h

e−ρτns(Xτns −K)−e−ρτnb(Xτb n+K)i

)

, ifi= 1, (4) whereρ >0 is a given discount factor.

In this paper, given random variablesxn, the termE{

X

n=1

xn} is interpreted as

lim sup

N→∞

E

N

X

n=1

xn.

For i = 0,1, let Vi(x, α) denote the value functions with the initial vector (X0, α0) = (x, α) and initial net positionsi= 0,1. That is,

Vi(x, α) = sup

Λi

Ji(x, α,Λi). (5)

It follows from (2) thatXt=xe−at+e−atRt

0earf(αr)dr. Therefore,Ji(x, α,Λi) is linear inxandVi(x, α) is convex inxfor fixedi= 0,1 andα= 1,2.

In practice, the transaction costK >0 is typically small when compared to all other parameters. Here we assume it is smaller than both (f1/a) and (−f2/a). In addition, bothλ1andλ2 are large numbers relative toa. So we also assumeλ1> a andλ2> a.

We summarize conditions to be imposed in this paper:

(A1)f1>0 andf2<0;

(A2)K <min{f1/a,|f2|/a}.

(A3)λ1> aandλ2> a.

Note that in this paper the stock price Xt is differentiable and the value of αt can be given in terms of the derivative ofXtwith respect tot.

3. Properties of the value functions.

Lemma 3.1. The following inequalities hold.

(a) 0≤V0(x, α)≤

ρ+ 2a ρa

max{f1,|f2|}, for all(x, α)∈ I0× M.

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(b) −K ≤ V0(x, α)−V1(x, α) +x ≤ K, for all (x, α) ∈ I0× M. Therefore, V1(x, α)is also bounded onI0× M.

Proof. It is clear thatV0(x, α)≥0. To show it is bounded from above, we first note that

d(e−ρtXt)

dt =e−ρt(−(ρ+a)Xt+f(αt)).

Given Λ0, integrate both sides over [τnb, τns] to obtain e−ρτnsXτns−e−ρτnbXτb

n= Z τns

τnb

e−at(−(ρ+a)Xt+f(αt))dt.

It follows that, for (x, α)∈ I0× M, J0(x, α,Λ0) =E

X

n=1

h

e−ρτns(Xτns−K)−e−ρτnb(Xτb n+K)i

≤E

X

n=1

h

e−ρτnsXτns−e−ρτnbXτb n

i

≤E

X

n=1

Z τns τnb

e−at(−(ρ+a)Xt+f(αt))dt

≤E Z

0

e−at((ρ+a)|Xt|+|f(αt)|)dt

ρ+ 2a ρa

max{f1,|f2|}.

This implies (a).

To show thatV0(x, α)−V1(x, α) +x≤K, or V1(x, α)≥V0(x, α) +x−K, we note that, for a given Λ1 withτ1s= 0,

V1(x, α)≥J1(x, α,Λ1) +x−K.

Since the rest stopping times in Λ1are arbitrary, it follows thatV1(x, α)≥V0(x, α)+

x−K. Similarly, we can show the second inequality in (b).

4. HJ equations. LetAdenote the generator of (Xt, αt), i.e., for any differentiable functionsh(x, i),i= 1,2,

( Ah(x,1) = (−ax+f1)h0(x,1) +λ1(h(x,2)−h(x,1)), Ah(x,2) = (−ax+f2)f2h0(x,2) +λ2(h(x,1)−h(x,2)),

whereh0denotes the derivative ofhwith respect tox. The associated HJ equations should have the form:

min{ρv0(x, α)− Av0(x, α), v0(x, α)−v1(x, α) +x+K}= 0,

min{ρv1(x, α)− Av1(x, α), v1(x, α)−v0(x, α)−x+K}= 0, (6) forα= 1,2.

In this paper, our goal is to find the value functions by solving these HJ equations with all value function properties satisfied. Then we proceed to show that such functionsvi(x, α) are equal to the value functions.

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4.1. Solving (ρ− A)vi(x, α) = 0. To solve these HJ equations, for each fixed i= 0,1, we consider equations when ρvi(x, α)− Avi(x, α) = 0,α= 1,2. Using the generatorA, we can write

(ρ+λ1)vi(x,1) = (−ax+f1)v0i(x,1) +λ1vi(x,2),

(ρ+λ2)vi(x,2) = (−ax+f2)v0i(x,2) +λ2vi(x,1). (7) For each fixed i = 0,1 and α = 1,2, let yα(x) = vi(x, α). Then, using the first equation, we can writey2 in terms ofy1:

y2(x) = 1 λ1

[(ρ+λ1)y1(x) + (ax−f1)y01(x)].

Its derivative with respect toxis given by y02(x) = 1

λ1

[(ρ+λ1+a)y10(x) + (ax−f1)y100(x)].

Substitute these into the second equation in (7) to obtain (ax−f1)(ax−f2)y001(x) + [(ρ+λ2)(ax−f1)

+(ρ+λ1+a)(ax−f2)]y10(x) +ρ(ρ+λ12)y1(x) = 0.

To simplify this equation, let

ξ=ax−f2

f1−f2 andu(ξ) =y1(x). Then, it follows that

ax−f2= (f1−f2)ξ and ax−f1= (f1−f2)(ξ−1).

Moreover, applying the chain rule to obtain y10(x) =

a f1−f2

u0(ξ) and y100(x) = a

f1−f2 2

u00(ξ).

This reduces the equation fory1(x) into an equation foru(ξ) as

a2ξ(ξ−1)u00(ξ) +a[(ρ+λ2)(t−1) + (ρ+λ1+a)t]u0(ξ) +ρ(ρ+λ12)]u(ξ) = 0.

Next, we introduce the additional parameters γ1

a, γ2= ρ+λ12

a , γ3=ρ+λ2 a .

Then, we can write the above differential equation in terms of (γ1, γ2, γ3) as ξ(1−ξ)u00(ξ) + [γ3−(γ12+ 1)t]u0(ξ)−γ1γ2u(ξ) = 0. (8) This is exactly the classical hypergeometric equation. Related properties of hyper- geometric functions can be found in [19] and [25].

For any real numbersηi,i= 1,2,3, let F(η1, η23, ξ) =

X

n=0

1)n2)n

n!(η3)n

ξn, where

(η)0= 1 and (η)n=η(η+ 1)· · ·(η+n−1) = Γ(η+n) Γ(η) ,

for anyη =ηi. Then, for 0< ξ <1, two independent solutions of (8) can be given by

F ρ

a,ρ+λ12

a ;ρ+λ2

a , ξ

andF ρ

a,ρ+λ12

a ;ρ+λ2+a a ,1−ξ

.

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Let

F1(x) =F ρ

a,ρ+λ12

a ;ρ+λ2

a ,ax−f2

f1−f2

, Fe1(x) =F

ρ

a,ρ+λ12

a ;ρ+λ2+a

a ,1−ax−f2

f1−f2

.

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We consider these solutions over a neighborhood inI0 containing the lower end f2/a. Then any solution over this interval can be written as a linear combination of F1 and Fe1, i.e., for some constants C1 and Ce1, vi(x,1) =C1F1(x) +Ce1Fe1(x), i= 0,1. In particular, we need this solution to be bounded nearf2/a. This requires Ce1= 0 becauseFe1(x)→ ∞asx→f2/a(see Lemma 8.1in Appendix).

Therefore, on the neighborhood ofI0containingf2/awe can assume thatv1(x,1)

=C1F1(x). Using this solution, we can writev1(x,2) in terms ofF1(x):

v1(x,2) = C1 λ1

[(ρ+λ1)F1(x) + (ax−f1)F10(x)] :=C1G1(x). (10) Alternatively, we can use the second equation in (7) and writey1in terms ofy2:

y1(x) = 1

λ2[(ρ+λ2)y2(x) + (ax−f2)y02(x)]

and its derivative in terms ofy02 andy002 y01(x) = 1

λ2

[(ρ+λ2+a)y20(x) + (ax−f2)y200(x)].

Substituting these into the first equation in (7) to obtain (ax−f1)(ax−f2)y002(x) + [(ρ+λ1)(ax−f2)

+(ρ+λ2+a)(ax−f1)]y10(x) +ρ(ρ+λ12)y2(x) = 0.

Similarly, letw(ξ) =y2(x) with

ξ=f1−ax f1−f2

,

Then, we can write the above differential equation in terms ofw(ξ):

ξ(1−ξ)w00(ξ) + [γ30 −(γ12+ 1)t]w0(ξ)−γ1γ2w(ξ) = 0, withγ30 = (ρ+λ1)/a. Its two independent solutions can be given by

F ρ

a,ρ+λ12

a ;ρ+λ1

a , ξ

andF ρ

a,ρ+λ12

a ;ρ+λ1+a a ,1−ξ

. Let

F2(x) =F ρ

a,ρ+λ12

a ;ρ+λ1

a ,f1−ax f1−f2

, Fe2(x) =F

ρ

a,ρ+λ12

a ;ρ+λ2+a

a ,1−f1−ax f1−f2

.

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Similarly, we consider these solutions over a neighborhood in I0 containing the upper end f1/a. Again, any solution can be written as a linear combination of F2

andFe2, i.e., for some constantsC2 andCe2,vi(x,2) =C2F2(x) +Ce2Fe2(x),i= 0,1.

We now need this solution to be bounded nearf1/a, which impliesCe2= 0 because Fe2(x)→ ∞asx→f1/a.

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-

f1 a f2

a

b2 b1

v0(1) =v1(1)−x−K

v0(1) =v1(1)−x−K ρv0(1) =Av0(1)

v0(2) =v1(2)−x−K ρv0(2) =Av0(2) ρv0(2) =Av0(2)

-

f1 a f2

a

s1 s2

ρv1(1) =Av1(1) ρv1(1) =Av1(1)

ρv1(2) =Av1(2) v1(2) =v0(2)+x−K v1(2) =v0(2)+x−K v1(1) =v0(1)+x−K

Figure 2. Buying Thresholds: b1 andb2; Selling Thresholds: s1 ands2. Therefore, on the neighborhood ofI0containingf1/awe can assume thatv0(x,2)

=C2F2(x). Using this solution, we have v0(x,1) = C2

λ2

[(ρ+λ2)F2(x) + (ax−f2)F20(x)] :=C2G2(x). (12) Lemma 4.1. The solutions of (ρ− A)vi(x, α) = 0 can be given as

v0(x,1) =C2G1(x), v0(x,2) =C2F2(x), v1(x,1) =C1F1(x), v1(x,2) =C1G1(x), whereF1,F2,G1, andG2 are defined in (9), (11), (10), and (12), respectively.

4.2. Order of threshold levels. Ifαt= 1, letb1 denote the buying threshold (to buy whenXt< b1) ands1the selling threshold (to sell whenXt> s1), respectively.

Likewise, if αt = 2, let b2 and s2 denote the corresponding buying and selling thresholds, respectively.

In the purely selling case, it is shown in [29] thats1> s2. Intuitively,b1 should be greater thanb2 as well.

In addition, in this paper, we consider the case whenb1< s2.

The opposite inequalityb1≥s2means to buy whenαt= 1 andXt< b1and sell as soon as αt jumps to 2 andXt> s2. This case rarely arises in practice because the jump rates λ1 and λ2 are typically large so thatαt switches between 1 and 2 frequently.

In view of these, we consider the threshold levels having the following order f2

a < b2< b1< s2< s1<f1

a.

The corresponding equalities in the HJ equations in (6) are marked in Figure2.

Using convexity of the value functions and variational inequalities in (6), it is shown in Lemma8.2(Appendix) that

b2= f2−ρK

ρ+a ands1=f1+ρK ρ+a .

It is easy to check, under Assumption (A2), thatb2 and s1 given above indeed satisfy the inequalitiesb2> f2/aands1< f1/a.

4.3. Solving (ρ− A)v1(x,1) = 0 with v1(x,2) = v0(x,2) +x−K and (ρ− A)v0(x,2) = 0with v0(x,1) =v1(x,1)−x−K. Forx∈ I0, let

g1(x) =

as1−f1

ax−f1 ρ+λa1

andg2(x) =

ab2−f2

ax−f2 ρ+λa2

.

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Next, we solve the HJ equations on [s2, s1]. Note that on this interval,v0(x,1) = C2G2(x),v0(x,2) =C2F2(x), andv1(x,2) =v0(x,2) +x−K=C2F2(x) +x−K.

To solveρv1(x,1) =Av1(x,1) on this interval, we write it as (ax−f1)v01(x,1) + (ρ+λ1)v1(x,1) =λ1v1(x,2).

Differentiate (1/g1(x))v1(x,1) and then integrate froms1 toxto obtain v1(x,1) =v1(s1,1)g1(x)−λ1(ax−f1)ρ+λa1

Z x s1

(f1−as)ρ+λa1−1v1(s,2)ds.

Recall thatv1(x,2) =C2F2(x) +x−K. We write

v1(x,1) =g1(x)A+k1(x)C2+k2(x), whereA=v1(s1,1) and





k1(x) =−λ1(f1−ax)ρ+λa1 Z x

s1

(f1−as)ρ+λa1−1F2(s)ds, k2(x) =−λ1(f1−ax)ρ+λa1

Z x s1

(f1−as)ρ+λa1−1(s−K)ds.

Similarly, we solve the HJ equations on [b2, b1]. Note that on this interval, v1(x,1) = C1F1(x), v1(x,2) = C1G1(x), and v0(x,1) = v1(x,1)−x−K = C1F1(x)−x−K. To solve ρv0(x,2) = Av0(x,2) on this interval, we first write it as

(ax−f2)v00(x,2) + (ρ+λ2)v0(x,2) =λ2v0(x,1).

Differentiate (1/g2(x))v0(x,2) and then integrate over [b2, x] to obtain v0(x,2) =v0(b2,2)g2(x) +λ2(ax−f2)ρ+λa2

Z x b2

(as−f2)ρ+λa2−1v0(s,1)ds.

Recall thatv0(x,1) =C1F1(x)−x−K. We write

v0(x,2) =g2(x)B+h1(x)C1+h2(x), whereB=v0(b2,2) and





h1(x) =λ2(ax−f2)ρ+λa2 Z x

b2

(as−f2)ρ+λa2−1F1(s)ds, h2(x) =−λ2(ax−f2)ρ+λa2

Z x b2

(as−f2)ρ+λa2−1(s+K)ds.

Using the power series form ofF1(x) =

X

n=0

pn

ax−f2

f1−f2

n

, we can write

h1(x) =λ2

X

n=0

pn

ρ+λ2+an

ax−f2

f1−f2

n

−g2(x)

ab2−f2

f1−f2

n . Similarly, we have

h2(x) =−λ2 a

1

ρ+λ2+a((ax−f2)−(ab2−f2)g2(x)) +f2+aK ρ+λ2

(1−g2(x))

.

Likewise, writeF2(x) =

X

n=0

qn

f1−ax f1−f2

n

to obtain

k1(x) =−λ1

X

n=0

qn ρ+λ1+an

g1(x)

f1−as1 f1−f2

n

f1−ax f1−f2

n

(11)

and

k2(x) =λ1

a

− 1

ρ+λ1+a((f1−ax)−(f1−as1)g1(x)) +f1−aK

ρ+λ1 (1−g1(x))

. 4.4. Smooth-fit conditions. Note that the convexity ofvi(x, α) inxonI0implies that they are continuous in the interior ofI0. The continuity of these functions at the threshold levels leads to the following conditions:













x=b2 : C1G1(b2)−b2−K=g2(b2)B+h1(b2)C1+h2(b2);

x=b1 :

C1F1(b1)−b1−K=C2G2(b1),

g2(b1)B+h1(b1)C1+h2(b1) =C2F2(b1), x=s2 :

C1F1(s2) =g1(s2)A+k1(s2)C2+k2(s2), C1G1(s2) =C2F2(s2) +s2−K,

x=s1 : C2G2(s1) +s1−K=g1(s1)A+k1(s1)C2+k2(s1).

(13)

Recall thatb2= (f2−ρK)/(ρ+a) ands1= (f1+ρK)/(ρ+a). Using the first and last equations in (13), we can write





 B=

G1(b2)−h1(b2) g2(b2)

C1−b2+K+h2(b2)

g2(b2) =:B1C1+B2

A=

G2(s1)−k1(s1) g1(s1)

C2+s1−K−k2(s1)

g1(s1) =:A1C2+A2.

Then, we can write the second and the third equations in terms ofC1andC2: F1(b1) −G2(b1)

g2(b1)B1+h1(b1) −F2(b1)

C1

C2

=

b1+K

−g2(b1)B2−h2(b1)

. Similarly, we can write the fourth and the fifth equations as follows:

G1(s2) −F2(s2) F1(s2) −[g1(s2)A1+k1(s2)]

C1 C2

=

s2−K g1(s2)A2+k2(s2)

. If the above 2×2 matrices are invertible, then we can eliminate C1 and C2 and obtain

F1(b1) −G2(b1) g2(b1)B1+h1(b1) −F2(b1)

−1

b1+K

−g2(b1)B2−h2(b1)

=

G1(s2) −F2(s2) F1(s2) −[g1(s2)A1+k1(s2)]

−1

s2−K g1(s2)A2+k2(s2)

.

(14)

The solutions to the HJ equations (6) should have the form:





























v0(x,1) =

C1F1(x)−x−K iff2/a≤x≤b1, C2G2(x) ifb1< x≤f1/a, v0(x,2) =

C1G1(x)−x−K iff2/a≤x≤b2, g2(x)B+h1(x)C1+h2(x) ifb2< x≤b1, C2F2(x) ifb1< x≤f1/a, v1(x,1) =

C1F1(x) iff2/a≤x≤s2, g1(x)A+k1(x)C2+k2(x) ifs2< x≤s1, C2G2(x) +x−K ifs1< x≤f1/a, v1(x,2) =

C1G1(x) iff2/a≤x≤s2, C2F2(x) +x−K ifs2< x≤f1/a,

(15)

In Figure 3, these functions are so labelled in their corresponding intervals in Figure2.

(12)

-

f1 a f2

a

b2 b1

v0(1) =C1F1−x−K

v0(1) =C1F1−x−K v0(1) =C2G2

v0(2) =C1G1−x−K v0(2)=g2B+h1C1+h2 v0(2) =C2F2

-

f1 a f2

a

s2 s1

v1(1) =C1F1 v1(1)=g1A+k1C2+k2

v1(2) =C1G1 v1(2) =C2F2+x−K v1(2) =C2F2+x−K v1(1) =C2G2+x−K

Figure 3. HJ equation solutions.

5. Variational inequalities and convexity. In this section, we consider the variational inequalities in (6) and related convexity inequalities.

5.1. Variational inequalities. Note that with the functions given in (15), all variational inequalities in (6) have to be satisfied, i.e., for (x, α)∈ I0× M,





ρv0(x, α)− Av0(x, α)≥0, v0(x, α)−v1(x, α) +x+K≥0, ρv1(x, α)− Av1(x, α)≥0, v1(x, α)−v0(x, α)−x+K≥0.

(16)

In what follows, we consider these inequalities on each intervals: [f2/a, b2], [b2, b1], [b1, s2], [s2, s1], and [s1, f1/a].

First, we show that these inequalities on [f2/a, b2] are automatically satisfied.

Note that on [f2/a, b2], we have





v0(x,1) =v1(x,1)−x−K, v0(x,2) =v1(x,2)−x−K, ρv1(x,1) =Av1(x,1), ρv1(x,2) =Av1(x,2).

(17)

We need to show





ρv0(x,1)≥ Av0(x,1), ρv0(x,2)≥ Av0(x,2), v1(x,1)≥v0(x,1) +x−K, v1(x,2)≥v0(x,2) +x−K.

(18)

It is easy to see that the last two inequalities follow directly from the first two equalities in (17). To see the first inequalityρv0(x,1)≥ Av0(x,1), note that it can be written in terms ofv1(x,1)andv1(x,2):

ρ(v1(x,1)−x−K)≥(f1−ax)(v01(x,1)−1) +λ1(v1(x,2)−v1(x,1)), which is equivalent to

−ρ(x+K)≥ −(f1−ax), forx < b2,

because ρv1(x,1) = Av1(x,1). The above inequality is clearly satisfied because b2= (f2−ρK)/(ρ+a) andf2< f1.

Similarly, on [s1, f1/a], we can uses1= (f1+ρK)/(ρ+a) and show that all VIs in (16) hold.

(13)

Next, we consider the VIs on [b2, b1]. We have





v0(x,1) =v1(x,1)−x−K, ρv0(x,2) =Av0(x,2), ρv1(x,1) =Av1(x,1), ρv1(x,2) =Av1(x,2).

(19)

We need the following inequalities to hold:





ρv0(x,1)≥ Av0(x,1), v0(x,2)≥v1(x,2)−x−K, v1(x,1)≥v0(x,1) +x−K, v1(x,2)≥v0(x,2) +x−K.

(20)

The third inequality is automatically satisfied. The second and forth ones can be written as:

−K≤v0(x,2)−v1(x,2) +x≤K.

Finally, the first inequality in (20) can be given as v0(x,2)−v1(x,2) +x≤ 1

λ1

[f1−ρK−(ρ+a)x]−K.

We can work out all needed inequalities this way on all other intervals and sum- marize these inequalities as follows:









On [b2, b1] : −K≤W2(x)≤min

K, 1 λ1

[f1−ρK−(ρ+a)x]−K

, On [b1, s2] : −K≤W1(x)≤K and −K≤W2(x)≤K,

On [s2, s1] : max

−K, 1 λ2

[f2+ρK−(ρ+a)x] +K

≤W1(x)≤K, (21)

whereW1(x) =v0(x,1)−v1(x,1) +xandW2(x) =v0(x,2)−v1(x,2) +x.

5.2. Convexity. Recall that the value functions are convex onI0. We expect the solutions to the HJ equations in (6) to be also convex.

Recall also that

Gi(x) = 1 λi

((ρ+λi)Fi(x) + (ax−fi)F10(x)), fori= 1,2.

Lemma 5.1. F1,F2,G1,G2 are convex on[f2/a, f1/a].

Proof. Clearly, both F1 and F2 are convex because their second derivatives are positive.

To show the convexities ofG1 andG2, we first consider g(ξ) = (γ12−γ3)u(ξ)−(1−ξ)du

dξ(ξ),

with u(ξ) = F(γ1, γ23, ξ). Multiply its both sides by (−(1−ξ)γ12−γ3−1) to obtain

−(1−ξ)γ12−γ3−1g(ξ) =−(γ12−γ3)(1−ξ)γ12−γ3−1u(ξ)+(1−ξ)γ12−γ3du dξ(ξ).

The right hand side of the above equation is equal to d

(1−ξ)γ12−γ3F(γ1, γ23, ξ) .

(14)

We next apply the formula dn

n

(1−ξ)γ12−γ3F(γ1, γ23, ξ)

=(γ3−γ1)n3−γ2)n3)n

(1−ξ)γ12−γ3−nF(γ1, γ23+n, ξ), withn= 1 and get

−(1−ξ)γ12−γ3−1g(ξ) =(γ3−γ1)(γ3−γ2)

γ3 (1−ξ)γ12−γ3−1F(γ1, γ23+ 1, ξ).

Hence we conclude that

g(ξ) =−(γ3−γ1)(γ3−γ2) γ3

F(γ1, γ23+ 1, ξ).

Foru(ξ) =Fρ

a,ρ+λ1a2;ρ+λa 2, ξ

, the correspondingg(ξ) is given by g(ξ) = λ2λ1

ρ+λ2F ρ

a,ρ+λ12

a ;ρ+λ2

a + 1, ξ

. Then the convexity ofFρ

a,ρ+λ1a2;ρ+λa2 + 1, ξ

implies thatg(ξ) is convex. It is easy to check with change of variablet= (ax−f2)/(f1−f2) thatG1(x) = (λ1/a)g(ξ).

SoG1is convex. Similarly, we can show the convexity ofG2. Lemma 5.2. h1(x)andk1(x)are convex.

Proof. Here, we only show the convexity ofh1(x). The convexity fork1(x) can be given in a similar way. Recall that

h1(x) =λ2(ax−f2)ρ+λa2 Z x

b2

(as−f2)ρ+λa2−1F1(s)ds, where

F1(x) =F

γ1, γ23,ax−f2

f1−f2

withγ1= ρ

a, γ2=ρ+λ12

a andγ3= ρ+λ2

a . Let

w1(ξ) =h1(x) =h1

(f1−f2)t+f2 a

, witht= ax−f2 f1−f2

. It follows that

w1(ξ) = λ2−γ3

Z ξ eb2

rγ3−1F(γ1, γ23, r)dr, whereeb2= (ab2−f2)/(f1−f2). Note that

Z

F(α, β;γ, ξ)ξγ−1dt= ξγ

γ F(α, β;γ+ 1, ξ) +k.

Apply the above formula, we obtain w1(ξ) = λ2

3

h

F(γ1, γ23+ 1, ξ)−F(γ1, γ23+ 1,eb2)i . Recall the derivative formula for hypergeometric function:

dm

mF(α, β;γ, ξ) =(α)m(β)m (γ)m

F(α+m, β+m;γ+m, ξ).

(15)

It follows that

w001(ξ) =λ2γ11+ 1)γ22+ 1)

33+ 1)(γ3+ 2) F(γ1+ 2, γ2+ 2;γ3+ 3, ξ).

In view of this and the convexity of the hypergeometric functions with positive parameters, the convexity ofw1(ξ) follows on [0,1]. So is the convexity ofh1(x).

Lemma 5.3. g2B+h1C1+h2 andg1A+k1C2+k2 are convex provided that C1≥0, C2≥0, A+A0≥0, B+B0≥0, (22) where

A0= λ1

a

f1−as1

ρ+λ1+a−f1−aK ρ+λ1

andB0= λ2

a

ab2−f2

ρ+λ2+a+f2+aK ρ+λ2

. Proof. It is easy to see that bothg1(x) andg2(x) are convex. In addition, note that g1A+k2 can be written as g1(A+A0)+ a linear function. SoA+A0>0 implies that g1A+k2 is convex. Then in view of the convexity ofk1(x) in Lemma5.2, it follows that g1A+k1C2+k2 is convex. Similarly, we can show the convexity of

g2B+h1C1+h2.

To guarantee the convexity of the functionsvi(x, α), it is sufficient to require the convexity ofv0(x,1) atx=b1and v1(x,2) atx=s2.

The convexity ofv0(x,1) atx=b1 is equivalent to (v0(b1,1))0≤(v0(b1,1))0+. Therefore, we need

C1F10(b1)−1≤C2G02(b1). (23) Similarly, The convexity ofv1(x,2) atx=s2requires

C1G01(s2)≤C2F20(s2) + 1. (24) Theorem 5.1. Assume that there exist b1 and s2 such that (a) the equalities in (14) hold with invertible matrices; (b) the additional VI inequalities in (21) hold;

(c)the inequalities in(22)hold; (d)the convexity inequalities in(23)and(24)hold.

Then, vi(x, α) defined in(15)satisfy the HJ equations (6).

6. A verification theorem.

Theorem 6.1. Assume the conditions of Theorem 5.1. Then, (a)v(x, i) =V(x, i),i= 1,2.

(b)Moreover, let

Db = (b1, f1/a)× {1} ∪(b2, f1/a)× {2}, Ds= (f2/a, s1)× {1} ∪(f2/a, s2)× {2}

denote the continuation regions. The set of stopping times in Λ0 and Λ1 can be defined as the first exit times from these continuation regions in sequence. Let Λ0 = (τ1b∗, τ1s∗, τ2b∗, τ2s∗, . . .), where τ1b∗ = inf{t : (Xt, αt) 6∈ Db}, τ1s∗ = inf{t ≥ τ1b∗ : (Xt, αt) 6∈ Ds},.., τn+1b∗ = inf{t ≥ τns∗ : (Xt, αt) 6∈ Db}, τn+1s∗ = inf{t ≥ τn+1b∗ : (Xt, αt)6∈Ds},..., and let Λ1 = (τ1s∗, τ2b∗, τ2s∗, . . .), where τ1s∗= inf{t≥0 : (Xt, αt) 6∈ Ds},.., τn+1b∗ = inf{t ≥ τns∗ : (Xt, αt) 6∈ Db}, τn+1s∗ = inf{t ≥ τn+1b∗ : (Xt, αt)6∈Ds},... Ifv0(x)≥0 andτns∗→ ∞, a.s., then, Λ0 andΛ1 are optimal.

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