Two-sided optimal stopping for L\'evy processes

Infinite horizon optimal stopping problems for a L\'evy processes with a two-sided reward function are considered. A two-sided verification theorem is presented in terms of the overall supremum and the overall infimum of the process. A result to compute the angle of the value function at the optimal thresholds of the stopping region is given. To illustrate the results, the optimal stopping problem of a compound Poisson process with two-sided exponential jumps and a two-sided payoff function is solved. In this example, the smooth-pasting condition does not hold.


Introduction
The presence of the maximum in the solution of optimal stopping problems for Lévy processes is nowadays understood, as it was summarized in the monograph by Kyprianou [8]. After the work of Surya [15] (see also Mordecki and Mishura [12]) it became clear that one sided problems with arbitrary payoff functions could be solved with the help of an averaging function. More recently, some formulas appeared for two sided problems, where the infimum and the supremum took part in the solution. For instance, in Mordecki and Salminen [13], with the help of representation techniques, a formula for the value function through the sum two averaging functions (one for the maximum and another for the minimum) was obtained. Afterwards, a similar formula was obtained through verification techniques by Christensen et al. [6]. This second result involved the supremum of the two averaging functions.
The purpose of the present work is then to obtain a verification theorem for the optimal stopping problem of a Lévy process in the two sided case, through the sum of two averaging functions, and to provide a simple example that seems not possible to be solved with the existing techniques.
References on optimal stopping problems for Lévy processes with two sided solutions, to our knowledge, are few. Perpetual American strangle options are considered by Chang and Sheu [5], through the solution of a free boundary integro-differential problem with moving boundaries. In this case, the smooth pasting condition is a key ingredient, and a non vanishing gaussian part in the Lévy process is used to ensure this condition. These results follow previously obtained unpublished ones by Boyarchenko [3]. Similar type of techniques were applied in Buonaguidi and Muliere [4] to solve a statistical problem: the Bayesian sequential testing of two simple hypothesis. More recently, De Donno et al. [7] found disconnected continuation regions in American put options with negative discount rates in Lévy models.
The content of the rest of the paper is as follows. In section 2 we formulate the verification result for two-sided optimal stopping with the corresponding proof. In section 3 a result to compute the angle (i.e. the difference of the right and left derivatives) of the value function at a critical threshold is obtained. Section 4 contains an example: the optimal stopping problem of a compound Poission process with two-sided exponential jumps, (no gaussian component) and payoff function g(x) = |x|.

A verification result for optimal stopping
Let X = {X t : t ≥ 0} be a Lévy process defined on a stochastic basis (Ω, F, F = (F t ) t≥0 , P x ) departing from X 0 = x. The corresponding expectation is denoted by E x , and for short we denote E = E 0 and P = P 0 . The Lévy-Khintchine formula characterizes the law of the process, stating, for z ∈ iR, that E e zXt = e tψ(z) with where a ∈ R, σ ≥ 0 and Π(dy) is a non-negative measure (the jump measure) that satisfies R (1 ∧ y 2 )Π(dy) < ∞. Here h(y) = y1 {|y|<1} is a truncation function. For general references on Lévy processes see Bertoin [2] or Kyprianou [8]. The set of stopping times is the set of random variables Observe that we allow the possibility τ = ∞ as for several optimal stopping problems, the optimal stopping time is within this class. A key rôle in the solution of two-sided problems is played by the overall supremum and infimum of the process, defined respectively by where e r is an exponential random variable of parameter r > 0, independent of X.
Observe that as r > 0 both random variables M and I are proper. Given a non-negative continuous payoff function g(x), a Lévy process X, and a discount factor r > 0, the optimal stopping problem (OSP) consists in finding the value function V (x) and the optimal stopping rule τ * such that Two-sided stopping for Lévy processes Theorem 2.1. Consider a Lévy process X, a discount rate r > 0, and a continuous reward function g : R → [0, ∞). Assume that there exist two points −x 1 < 0 < x 2 and two continuous monotonous functions: Q 1 non-increasing with Q 1 (x) = 0 for −x 1 ≤ x; Q 2 non-decreasing with Q 2 (x) = 0 for x ≤ x 2 (named averaging functions); and such that Define the function Then, if the condition As usual in optimal stopping, in order to prove Theorem 2.1, we verify two statements: The proof of these two facts are stated in two corresponding lemmas. Their proofs follow, with the necessary modifications, from the respective proofs in Mordecki and Mishura [12]. For a definition of r-excessive function see Christensen et al. [6].
is r-excessive, and, in consequence, Proof. The fact that (b) follows from (a) is standard. In fact, if h is r-excessive where θ t is the shift operator, giving the super-martingale property. To prove (a) we know, from Lemma 2.2 in Christensen et al. [6], that the following two functions are excessive: The proof concludes as h = u + v, and excessivity is preserved by summation.

Remark 2.3.
When comparing the representations of payoffs for two-sided problems in Theorem 2.7 in Christensen et al. [6] and Proposition 4.4. in Mordecki and Salminen [13], it should be noticed that in the first case, the excessive function is E x (f (M ) ∨ g(I)), and in the second, E x (f (M ) + g(I)). These two constructions give different representations of the payoff of the problem. In the present work we use the second one.
On the other side, Observe that if e r < τ * we have M < x 2 and I > −x 1 , and then sup 0≤t≤er Consider nowX = {X s = X τ * +s − X τ * : s ≥ 0} that, by the strong Markov property, is independent of F τ * and has the same distribution as X, and denote byẼ x the expectation w.r.t.X. Based on these considerations, we have where we change variables according to v = u − τ * to pass from (2.11) to (2.12), and denote s = t − τ * to pass from (2.13) to (2.14). The same relation holds with V 2 and Q 2 . Summing up these two relations, and in view of (2.8), we conclude the proof of the Lemma.
Proof of Theorem 2.1. The proof now follows, as, according to Lemma 2.2 the non- by Doob's optional sampling Theorem first and the application of condition (2.5) second. This establishes (2.6). On its turn, Lemma 2.4 gives equality (2.7), concluding the proof.

On the smooth pasting condition
Smooth pasting results for general Lévy processes were obtained for put American perpetual options for Lévy processes in Alili and Kyprianou [1], and afterwards generalized to put-type (bounded) payoffs by Surya [15]. The following result gives some natural necessary conditions for smooth pasting, that depend on the exponential moments of the process and the behaviour of the averaging function. They can be applied both to one-sided problems (as the ones considered in Mordecki and Mishura [12]) and to the two sided problems considered in the present paper.
for some A > 0 and some α > 0. Regarding the process, assume that E e αX1 < e r .
(3.6) Two-sided stopping for Lévy processes To compute the limit in (3.4), as Q(x) = 0 for x ≤ x 0 , we have To consider the term in (3.6), denote x = x 0 + y ≥ x 0 + h, We now apply Lemma 1 in Mordecki [10] to the Lévy process αX, to obtain that condition (3.1) implies that E e αM < ∞. In consequence, we have In conclussion This concludes the proof of the Theorem.

An application
To illustrate our results, we consider a compound Poisson process X = {X t : t ≥ 0} with double-sided exponential jumps, given by i ,      (2) , are independent. We consider then the OSP (2.1) for the function g(x) = |x| and the Lévy process X.

Wiener-Hopf factorization
The characteristic exponent of X is given by The equation ψ(z) = r has two roots We apply the Wiener-Hopf factorization to determine the law of M and I. We directly obtain the factors: In conclusion, due to the uniqueness of the factorization (see Thm. 5(ii) Ch. VI of Bertoin [2]), we obtain This means that the random variables M and −I have defective exponential distributions with parameters r 2 and r 1 , and atoms at zero of respective size r 2 /α 2 and r 1 /α 1 . With a slight abuse of notation, we denote these respective densities where δ 0 (x)dx denotes the Dirac mass measure at x = 0. In order to introduce our result, we also need the following notations.  Consider the Lévy process X in (4.1), the payoff function g(x) = |x|, and r > 0. Denote

6)
where u is the unique root of the equation Then, the value function x, for x 2 < x, (4.9) and stopping time defined in (2.2) conform the solution of the OSP (2.1).

Remark 4.2.
An application of Theorem 3.1 (or more directly in this case, the computation of the corresponding derivatives in formula (4.9)) shows that the smooth pasting condition does not hold in any of the thresholds of the problem: the averaging functions have non-vanishing derivatives at the roots, and both the maximum and the infimum have atoms at the origin.
Proof. First observe that the r.h.s of (4.7) is decreasing, as the numerator decreases and the denominator increases with u (all coefficients are positive). It decreases from when u = 0 to E 1 + E 2 , as u → ∞. Then, equation (4.7) has only one positive root. Here it was used that G 1 G 2 < 1, fact that follows directly from the definitions (4.4). We need a technical result.
have a unique common positive root.
Proof of the Lemma. Denoting for short δ = r 1 + r 2 , equation (4.10) reads Multiplying by the denominators, we obtain We first take on the l.h.s. the terms with the factor u. Afer computations in the l.h.s., many terms simplify, and in the r.h.s we take a common factor. The result of this computations is the equation (4.11) and equation (4.11) simplifies to (4.7), concluding the proof of the lemma.
We continue with the proof of Theorem 4.1. In view of the previous lemma and definitions (4.5) and (4.6), we obtain that u = x 1 + x 2 . In order to apply Theorem 2.1, introduce the functions We prove that these two functions are monotonous, non-increasing and non-decreasing respectively, and continuous. The equation: is equivalent to (4.14) We proceed to check the equivalence between (4.13) and (4.14) by using the definition of G 1 (4.4) and the equality u = x 1 + x 2 . Multiplying by the denominator in (4.13) we obtain that, after substitution of u by x 1 + x 2 and rearrengments gives, that, dividing by the the (1 − e −(r1+r2)u ) gives the first equality in (4.14). The second one derives from the definition of D 2 (4.8), concluding the equivalence between (4.13) and (4.14).
Observe now that the statement (4.14) is equivalent to Q 1 (−x 1 ) = 0, giving the continuity of Q 1 , that is clearly positive and non-increasing. Identical arguments apply to Q 2 .
To check equality (2.3), consider first x ≥ −x 1 and compute we obtain Regarding the maximum, now for x > x 2 , Summing up, To show that the second summand in the r.h.s. of the previous formula vanishes, we compute Taking into account (4.14) and the first equation in (4.15), we obtain and substitute in view of the definitions of E 1 in (4.2) and F 1 in (4.3). Similarly, for x < −x 1 , we obtain E x Q 1 (I) + Q 2 (M ) = −x. This concludes the verification of (2.3).
This gives that, for x ≥ −x 1 , we have: Applying the corresponding same arguments (or considering the the dual Lévy process In particular, this gives that D 1 > 0, and, for − so the definition (2.4) gives (4.9). To conclude with the proof, it remains to verify condition (2.5) for x ∈ [−x 1 , x 2 ]. We take 0 ≤ x ≤ x 2 . We have

Remark 4.4.
It is interesting to note that the function x − E 2 appearing in the first two summands in the r.h.s. of (4.12), in the terminology of Surya [15], is the averaging function of the one sided problem with payoff function g 2 (x) = x + (see Mordecki [11]).
So the remaining term in the r.h.s. in (4.12) is a correction due to the presence of the infimum in the two-sided problem. As x 2 is the root of Q 2 , we obtain x 2 ≥ E 2 .
Remark 4.5. The proof of Theorem 4.1 also provides bounds to find u numerically. First observe that Q 2 (x) ≤ x for x ≥ 0, so E Q 2 (M ) ≤ E 2 . Furthermore, in view of (4.14) and (4.18), we have In view of the previous remark, the conclusion is that E 1 + E 2 ≤ u ≤ E 1 (1 + F 2 ) + E 2 (1 + F 1 ).