Explicit solutions in one-sided optimal stopping problems for one-dimensional diffusions

# Explicit solutions in one-sided optimal stopping problems for one-dimensional diffusions

Fabián Crocce and Ernesto Mordecki111Iguá 4225, Centro de Matemática, Facultad de Ciencias, Universidad de la República, Montevideo, Uruguay
###### Abstract

Consider the optimal stopping problem of a one-dimensional diffusion with positive discount. Based on Dynkin’s characterization of the value as the minimal excessive majorant of the reward and considering its Riesz representation, we give an explicit equation to find the optimal stopping threshold for problems with one-sided stopping regions, and an explicit formula for the value function of the problem. This representation also gives light on the validity of the smooth fit principle. The results are illustrated by solving some classical problems, and also through the solution of: optimal stopping of the skew Brownian motion, and optimal stopping of the sticky Brownian motion, including cases in which the smooth fit principle fails.

## 1 Introduction and problem formulation

Consider a non-terminating and regular one-dimensional (or linear) diffusion , in the sense of Itô and McKean [9] (see also Borodin and Salminen [2]). The state space of is denoted by , an interval of the real line with left endpoint and right endpoint , where . We exclude the possibility of absorbing and killing boundaries; if some of the boundaries belong to we assume it to be both entrance and exit (i.e. reflecting boundary). Denote by the probability measure associated with when starting from , and by the corresponding mathematical expectation. Denote by the set of all stopping times with respect to , the usual augmentation of the natural filtration generated by (see I.14 in [2]).

Given a non-negative lower semicontinuous reward function and a discount factor , consider the optimal stopping problem consisting in finding a function and a stopping time , such that

 Vα(x)=Ex(e−ατ∗g(Xτ∗))=supτ∈TEx(e−ατ∗g(Xτ∗)). (1)

The value function and the optimal stopping time are the solution of the problem.

The first problems in optimal stopping appeared in the framework of statistics, more precisely, in the context of sequential analysis (see the book by Wald [30]). For continuous time processes a relevant reference is the book of Shiryaev [28] that also has applications to statistics. A new impulse to these problems is related to mathematical finance, where arbitrage considerations give that in order to price an American option one has to solve an optimal stopping problem. The first results in this direction were provided by Mc Kean [15] in 1965 and Merton [16] in 1973, who respectively solved the perpetual put and call option pricing problem, by solving the corresponding optimal stopping problems in the context of the Black and Scholes model [1]. For the state of the art in the subject see the book by Peskir and Shiryaev [21] and the references therein. A new approach for solving one-dimensional optimal stopping problems for very general reward functions is provided in the recent paper [20].

When considering optimal stopping problems we typically find two classes of results. The first one consists in the explicit solution to a concrete optimal stopping problem (1). Usually in this case one has to –somehow– guess the solution and prove that this guess in fact solves the optimization problem; we call this approach “verification”. For example we can consider the papers [15], [16], [29], [26]. The second class consists of general results, for wide classes of processes and reward functions. We call this the “theoretical” approach. It typically include results about properties of the solution. In this class we mention for example [3], [8], [6]. But these two classes not always meet, as frequently in concrete problems the assumptions of the theoretical studies are not fulfilled, and, what is more important, many of these theoretical studies do not provide concrete ways to find solutions. In what concerns the first approach, a usual procedure is to apply the principle of smooth fit, that generally leads to the solution of two equations: the continuous fit equation and the smooth fit equation. Once these equations are solved, a verification procedure is needed in order to prove that the candidate is the effective solution of the problem (see chapter IV in [21]). This approach, when an explicit solution can be found, is very effective. In what concerns the second approach, maybe the most important result is Dynkin’s characterization of the solution of the value function as the least -excessive (or -superharmonic) majorant of the payoff function [3]. Other ways of classifying approaches in order to study optimal stopping problems include the martingale-Markovian dichotomy as exposed in [21].

In the present work we provide an explicit solution of a right-sided optimal stopping problem for a one-dimensional diffusion process, i.e., when the optimal stopping time has the form

 τ∗=inf{t≥0:Xt≥x∗} (2)

for some optimal threshold , under mild regularity conditions. Right-sided problems are also known as call-type optimal stopping problems. Analogous results are valid for left-sided problems.

An important byproduct of our results has to do with the smooth fit principle. Our results are independent of this principle, but they give sufficient conditions in order to guarantee it.

In section 2 some necessary definitions and preliminary results are given. Our main results are presented in section 3. In section 4 we discuss the consequences of our results related to the smooth fit principle. Finally, in section 5 we present some examples, including the optimal stopping of the skew Brownian motion and of the sticky Brownian motion (suggested by Paavo Salminen), where particular attention to the smooth fit principle is given.

## 2 Definitions and preliminary results

Denote by the infinitesimal generator of the diffusion , and by its domain. For any stopping time and for any the following discounted version of the Dynkin’s formula holds:

 f(x)=Ex(∫τ0e−αt(α−L)f(Xt)dt)+Ex(e−ατf(Xτ)). (3)

The resolvent of the process is the operator defined by

 Rαu(x)=∫∞0e−αtEx(u(Xt))dt,

applied to a function . The range of the operator is independent of and coincides with the domain of the infinitesimal generator . Moreover, for any , , and for any , . In other terms, and are inverse operators. Denoting by and the scale function and the speed measure of the diffusion respectively, we have that, for any , the lateral derivatives with respect to the scale function exist for every . Furthermore, they satisfy

 ∂+f∂s(x)−∂−f∂s(x)=m({x})Lf(x), (4)

and the following identity holds for :

 ∂+f∂s(z)−∂+f∂s(y)=∫(y,z]Lf(x)m(dx). (5)

This last formula allows us to compute the infinitesimal generator of at as Feller’s differential operator [7]

 Lf(x)=∂∂m∂+∂sf(x). (6)

The infinitesimal generator at and (if they belong to ) can be computed as and respectively.

Given a function , and we give to the meaning given in (6) if it makes sense. We also define, if , and if , , if the limit exists.

There exist two continuous functions decreasing, and increasing, solutions of , such that any other continuous function is a solution of the differential equation if and only if , with in . Denoting by the hitting time of level , we have

 Ex(e−ατz)=⎧⎪⎨⎪⎩ψα(x)ψα(z),x≤z,φα(x)φα(z),x≥z. (7)

The functions and , though not necessarily in , also satisfy (4) for all which allow us to conclude that in case , the derivative at of both functions with respect to the scale exists. From (5) applied to , and taking into account we obtain for

 ∂+ψα∂s(z)−∂+ψα∂s(y)=∫(y,z]αψα(x)m(dx);

the right hand side is strictly positive since is positive and charge every open set. We conclude that is strictly increasing. In an analogous way it can be proven that is increasing as well. The previous consideration, together with the fact that and allow us to conclude for :

 −∞<∂−φα∂s(x)≤∂+φα∂s(x)<0<∂−ψα∂s(x)≤∂+ψα∂s(x)<∞.

The Green function of the process with discount factor is defined by

 Gα(x,y)=∫∞0e−αtp(t;x,y)dt,

where is the transition density of the diffusion with respect to the speed measure (this density always exists, see [9] or [2]). The Green function may be expressed in terms of and as follows:

 Gα(x,y)={w−1αψα(x)φα(y),x≤y,w−1αψα(y)φα(x),x≥y, (8)

where is the Wronskian, given by

 wα=∂ψ+α∂s(x)φα(x)−ψα(x)∂φ+α∂s(x).

Observe that the Wronskian is positive and independent of [9],[2]. Given , under the condition , an application of Fubini’s Theorem gives

 Rαu(x)=∫IGα(x,y)u(y)m(dy). (9)

A non-negative Borel function is called -excessive for the process if for all and , and for all . A 0-excessive function is said to be excessive.

Consider the process killed at an independent exponential time of parameter , i.e.

 Yt={Xt,t

with a random variable with exponential distribution of parameter independent of , and the cemetery state, at which any function is defined to be zero. It is easy to see that the Green function of the process coincides with ; a Borel function is excessive for if and only if it is -excessive for . In fact, the non-discounted optimal stopping problem for the process has the very same solution (value function and optimal stopping time) as the -discounted optimal stopping problem for .

For general reference on diffusions and Markov processes see [2, 9, 22, 5, 11].

## 3 Main results

Our departing point, inscribed in the Markovian approach, is Dynkin’s characterization of the optimal stopping problem solution. Dynkin’s characterization [3] states that, if the reward function is lower semi-continuous, is the value function of the non-discounted optimal stopping problem with reward if and only if is the least excessive function such that for all . Applying this result for the killed process , and taking into account the relation between and , we obtain that , the value function of the problem with discount , is characterized as the least -excessive majorant of .

The second step uses Riesz’s decomposition of an -excessive function. We recall this decomposition in our context (see [12, 13, 4]). A function is -excessive if and only if there exist a non-negative Radon measure and an -harmonic function such that

 u(x)=∫(ℓ,r)Gα(x,y)μ(dy)+(α-harmonic function). (10)

Furthermore, the previous representation is unique. The measure is called the representing measure of .

The third step is based on the fact that the resolvent and the infinitesimal generator of a Markov process are inverse operators. Suppose that we can write

 Vα(x)=∫IGα(x,y)(α−L)Vα(y)m(dy), (11)

where is the infinitesimal generator and is the speed measure of the diffusion. Assuming that the stopping region has the form , and taking into account that is -harmonic in the continuation region and in the stopping region we obtain as a suitable candidate to be the representing measure

 μ(dy)=⎧⎨⎩0, if yx∗, (12)

This approach was initiated by Salminen in [23] (see also [17]). According to Salminen’s approach, once the excessive function is represented as an integral with respect to the Martin kernel ,

 Vα(x)=∫IMα(x,y)κ(dy) (13)

one has to find the representing measure . Martin and Green kernels are related by , where is a reference point. Therefore, Riesz’s representation of is related with the one in (13) by considering and as the -harmonic function.

It is useful to observe that when the optimal stopping problem (1) is right-sided with optimal threshold it has a value function of the form

 Vα(x)={Ex(e−ατx∗)g(x∗),x

Furthermore, for all and, in virtue of equation (7), we have

 Vα(x)={g(x∗)ψα(x∗)ψα(x),x

The state space of the process can include or not the left endpoint and the right endpoint . In order to simplify, with a slight abuse of notation, we write , , , to denote respectively , , , .

We say that the function satisfies the right regularity condition (RRC) if there exist a point and a function (not necessarily non-negative) such that for and

 ~g(x)=∫IGα(x,y)(α−L)~g(y)m(dy)(x∈I). (15)

Proposition 3.5 gives conditions in order to verify the inversion formula (15). Informally speaking, the RRC is fulfilled by functions that satisfy all the local conditions –regularity conditions– to belong to for , and does not increase as quick as does when approaching (in the case ). Observe that if satisfies the RRC for certain it also satisfies it for any greater value; and of course, if itself satisfy (15) then it satisfies the RRC for all in . To take full advantage of the following result it is desirable to find the least such that the RRC holds.

The main result follows.

###### Theorem 3.1.

Consider a diffusion and a reward function that satisfies the RRC for some . The optimal stopping problem is right-sided with optimal threshold if and only if:

 g(x∗)≥w−1αψα(x∗)∫(x∗,r]φα(y)(α−L)g(y)m(dy), (16)
 (α−L)g(x)≥0,x∈I:x>x∗ (17)

and

 g(x∗)ψα(x∗)ψα(x)≥g(x),x∈I:x

Furthermore, in the previous situation:

• Riesz’s representation of the value function has representing measure as given in (12) with

 k=g(x∗)−w−1αψα(x∗)∫(x∗,r]φα(y)(α−L)g(y)m(dy)w−1αψα(x∗)φα(x∗),

while the -harmonic part vanishes;

• if and the inequality (16) is strict, then is the smallest number satisfying this strict inequality and (17), in particular

 g(x∗)≤w−1αψα(x∗)∫[x∗,r]φα(y)(α−L)g(y)m(dy); (19)

which implies that ;

###### Remark 3.2.

From this theorem we obtain an algorithm to solve right-sided optimal stopping problems which works in most cases: (i) Find the largest root of the equation

 g(x∗)=w−1αψα(x∗)∫(x∗,r]φα(y)(α−L)g(y)m(dy); (20)

(ii) Verify for ; (iii) Verify . If these steps are fulfilled, the problem is right-sided with optimal threshold . Observe that if , then inequalities (16) and (19) are equalities;

###### Proof.

We start by observing that if the problem is right-sided with threshold then (17) holds. In general is non-negative in the stopping region (this can be seen with the help of the Dynkin’s operator, see ex. 3.17 p. 310 in [22], see also equation (10.1.35) in [18]). Under the made assumption the value function is given by (14), which implies (18) since the value function dominates the reward. To finish the proof of the “only-if” part it remains to prove (16). Consider defined by

 Wα(x):=∫(x∗,r]Gα(x,y)(α−L)g(y)m(dy);

observe that is -excessive in virtue of (17) and Riesz’s representation. Let be defined by

 ~Vα(x):=Wα(x)+kGα(x,x∗),

where is such that , i.e. . Observe that, by (8), is the right-hand side of (16). In fact, (16) holds if and only if . By the definition of and the representation (8) of we get for

 ~Vα(x)=ψα(x)ψα(x∗)~Vα(x∗)=ψα(x)ψα(x∗)g(x∗)=Vα(x).

Let us compute for . In this region we have , where is the extension given by the RRC. For can use the inversion formula (15). Denoting by we have

 ~Vα(x)−g(x) =kGα(x,x∗)−∫[ℓ,x∗]Gα(x,y)ν~g(dy) =kw−1αφα(x)ψα(x∗)−w−1αφα(x)∫[ℓ,x∗]ψα(y)ν~g(dy) =φα(x)φα(x∗)(~Vα(x∗)−g(x∗))=0,

because . So far, we have proved that for all . We are ready to prove that , based on the uniqueness of Riesz’s decomposition: the -excessive function has Riesz’s representation given by its definition, and, if then

 Wα(x)=−kGα(x,x∗)+Vα(x)

would give another Riesz’s representation (the representing measure being , where is the representing measure of ). An easy way of verifying that the measures are not the same is to observe that the former does not charge , while the latter do.

To prove the “if” statement observe that, assuming (16) (17) and (18), function , already defined, is -excessive (by Riezs’s representation, bearing in mind that ) and dominates . By Dynkin’s characterization, the value function is the minimal -excessive function that dominates . Therefore . Since satisfies (14) (we have proved this in the first part of this proof), it follows that is the expected reward associated with the hitting time of the set , then

 ~Vα(x)≤Vα(x)=supτEx(e−ατg(Xτ)),

concluding that the problem is right-sided with threshold .

The consideration about Riesz’s representation of stated in the “furthermore” part are a direct consequence of the made proof.

To prove the minimality of , suppose that there exists such that satisfying the strict inequality in (16) and (17). Let us check , in contradiction with the fact that is a majorant of . Considering the extension given by the RRC and denoting we have

 Vα(x∗∗)−g(x∗∗)=−∫[ℓ,x∗]Gα(x∗∗,y)ν~g(dy)+kGα(x∗∗,x∗)=s1+s2+s3+s4,

where

 s1 =−∫[ℓ,x∗∗]Gα(x∗∗,y)ν~g(dy), s2 =−w−1αψα(x∗∗)∫(x∗∗,x∗]φα(y)ν~g(dy), s3 =ψα(x∗∗)ψα(x∗)φα(x∗)φα(x∗∗)∫[ℓ,x∗∗]Gα(x∗∗,y)ν~g(dy), s4 =ψα(x∗∗)ψα(x∗)∫(x∗∗,x∗]Gα(x∗,y)ν~g(dy).

To check that , use

 k=1Gα(x∗,x∗)∫[ℓ,x∗]Gα(x∗,y)ν~g(dy),andGα(x∗∗,x∗)Gα(x∗,x∗)=ψα(x∗∗)ψα(x∗).

Finally, observe that

 s1+s3=(ψα(x∗∗)ψα(x∗)φα(x∗)φα(x∗∗)−1)∫[ℓ,x∗∗]Gα(x∗∗,y)ν~g(dy)<0,

because the first factor is negative and the second one positive, by the assumption about , and

 s4+s2=w−1αψα(x∗∗)ψα(x∗)∫(x∗∗,x∗](ψα(y)φα(x∗)−ψα(x∗)φα(y))ν~g(dy)<0,

because the measure is positive and the integrand non-positive (it is increasing and vanishes in ). We have obtained that , concluding the proof. ∎

The previous theorem gives necessary and sufficient conditions for the problem (1) to be right-sided under the RRC. The following result gives simpler sufficient conditions for the problem to be right-sided.

###### Theorem 3.3.

Consider a diffusion and a reward function such that (15) is fulfilled for . Suppose that there exists a root of the equation , such that if and if , and that . Then the optimal stopping problem (1) is right-sided, with optimal threshold

 x∗=min{x:b(x)≥0}, (21)

where

 b(x)=∫[ℓ,x]ψα(y)(α−L)g(y)m(dy)(x∈I).
###### Proof.

The idea is to apply Theorem 3.1, with defined in (21). By the assumptions on and the fact that is strictly positive in any open set, we obtain that the function is decreasing in and increasing in . Moreover if . Since is right continuous and increasing in , the set with . Observe that, by (15) and (8) we get

 g(x)=w−1αφα(x)b(x)+∫(x,r]Gα(x,y)(α−L)g(y)m(dy),

and is equivalent to (16). Since we have for . It only remains to verify (18). By definition of there exists a signed measure whose support is contained in , and for and such that

 ∫[ℓ,x∗]ψα(y)σℓ(dy)=0.

Furthermore is a positive measure supported in . Using the inversion formula for and (8), we have for

 g(x)−ψα(x)ψα(x∗)g(x∗)=∫[ℓ,x∗]Gα(x,y)σℓ(dy)≤Gα(x,c)ψα(c)∫[ℓ,x∗]ψα(y)σℓ(dy)=0,

where the inequality follows from the following facts: if then and

 ψα(y)Gα(x,c)ψα(c)≤Gα(x,y),

while if then

 ψα(y)Gα(x,c)ψα(c)≥Gα(x,y).

We can now apply Theorem 3.1 completing the proof. ∎

### 3.1 On the right regularity condition (RRC)

In order to apply the previous results it is necessary to verify the inversion formula (15). As we have seen in the preliminaries, if we have , and if equation (9) holds for , we have (15). This is the content of the following result.

###### Lemma 3.4.

Assume , and

 ∫IGα(x,y)|(α−L)f(y)|m(dy)<∞for % all x∈I.

Then satisfies equation (15).

The conditions of the previous lemma are very restrictive in order to solve concrete problems, as reward functions typically satisfy . The following result extends the previous one to unbounded functions.

###### Proposition 3.5.

Consider the case . Suppose is such that in (6) can be defined for all . Assume

 ∫IGα(x,y)|(α−L)u(y)|m(dy)<∞, (22)

and

 limz→r−u(z)ψα(z)=0. (23)

Suppose also that for each there exist a function such that for . Then satisfies (15).

###### Proof.

By (9) we have . Consider a strictly increasing sequence and denote by the function of the hypothesis. By the continuity of the sample paths, by our assumptions on the right boundary , we have . Applying formula (3) to and the stopping time we obtain, for ,

 un(x)=Ex(∫τrn0e−αt(α−L)un(Xt)dt)+Ex(e−ατrn)un(rn),

using and for we have

 u(x)=Ex(∫τrn0e−αt(α−L)u(Xt)dt)+Ex(e−ατrn)u(rn).

Taking limits when , by (7) and (23) we have

 Ex(e−ατrn)u(rn)=ψα(x)ψα(rn)u(rn)→0.

To compute the limit of the first term above we use dominated convergence theorem and (22). The result is

 u(x) =∫∞0Ex(e−αt(α−L)u(Xt))dt=∫IGα(x,y)(α−L)u(y)m(dy)

concluding the proof. ∎

## 4 On the principle of smooth fit

The principle of smooth fit (SF) holds when condition is satisfied, being a helpful tool to find candidate solutions to optimal stopping problems. In [23] Salminen proposes an alternative version of this principle, considering derivatives with respect to the scale function. We say that there is scale smooth fit (SSF) when the value function has derivative at with respect to the scale function. Note that if also has derivative with respect to the scale function they coincide, since in . In [19] Peskir presents two interesting examples: one of them consists on the optimal stopping problem of a regular diffusion with a differentiable payoff function in which the principle of SF does not hold, but the alternative principle of SSF does; while in the other the principle of SF holds but the principle of SSF fails. Later, Samee [25] analysed the validity of the principle of smooth fit for killed diffusions and introduced other alternative principles considering derivatives of and with respect to the scale function . See also the paper by Jacka [10] for a study of the principle of smooth fit related to the Snell envelope.

We now analyse the relation between Riez’s representation of , stated in the previous section, and the principle of smooth fit. We start by proving that in (12) implies that the reward function has derivatives with respect to the function . Then we follow by stating some corollary results.

###### Theorem 4.1.

Given a diffusion and a reward function , if the value function associated with the problem (1) satisfies

 Vα(x)=∫(x∗,r]Gα(x,y)(α−L)g(y)m(dy),

for , then is differentiable at with respect to .

###### Proof.

For

 Vα(x)=w−1αψα(x)∫(x∗,r]φα(y)νg(dy),

and the left derivative of with respect to in is

 ∂V−α∂ψα(x∗)=w−1α∫(x∗,r]φα(y)νg(dy).

For

 Vα(x)=φα(x)w−1α∫(x∗,x)ψα(y)νg(dy)+ψα(x)w−1α∫[x,r]φα(y)νg(dy).

Computing the difference between and we obtain

 Vα(x)−Vα(x∗) +w−1α∫(x∗,x)(φα(x)ψα(y)−ψα(x∗)φα(y))νg(dy).

Then

 ∂V+α∂ψα(x∗) =limx→x∗+Vα(x)−Vα(x∗)ψα(x)−ψα(x∗) =w−1αlimx→x∗+∫[x,r]φα(y)νg(dy) +w−1αlimx→x∗+∫(x∗,x)φα(x)ψα(y)−ψα(x∗)φα(y)νg(dy)ψα(x)−ψα(x∗).

If the last limit vanishes, we obtain that the right derivative exists, and

 ∂V+α∂ψα(x∗)=∂V−α∂ψα(x∗)=w−1α∫(x,r]φα(y)νg(dy).

This means that we have to prove