REFLECTED BACKWARD STOCHASTIC DIFFERENTIAL EQUATION WITH JUMPS AND RANDOM OBSTACLE

: In this paper we give a solution for the one-dimensional re(cid:176)ected backward stochastic diﬁerential equation when the noise is driven by a Brownian motion and an independent Poisson point process. We prove existence and uniqueness of the solution in using penalization and the Snell envelope theory. However both methods use a contraction in order to establish the result in the general case. Finally, we highlight the connection of such re(cid:176)ected BSDEs with integro-diﬁerential mixed stochastic optimal control.


Introduction
The notion of non-linear Backward Stochastic Differential Equations (BSDE in short) was introduced by Pardoux & Peng ( [11], 1990). Their aim was to give a probabilistic interpretation of a solution of second order quasi-linear partial differential equations. Since then, these equations have gradually become an important mathematical tool which is encountered in many fields such as financial mathematics, stochastic optimal control, partial differential equations, ...
In ([6], 1997), El-Karoui et al. introduced the notion of a reflected BSDE, which is actually a backward equation but one of the components of the solution is forced to stay above a given barrier, which is an adapted continuous process.
Recently in ( [1], 1995), Barles et al. considered standard BSDEs when the noise is driven by a Brownian motion and an independent Poisson random measure. They have shown the existence and uniqueness of the solution, in addition, the link with integral-partial differential equations is studied.
In this paper our aim is to study the one-dimensional reflected BSDE (RBSDE in short) when the noise is driven by a Brownian motion and an independent Poisson random measure. This is the natural generalization of the work of El-Karoui et al.. The component (Y t ) t≤1 of the solution which is forced to stay above a given barrier is, in our frame, no longer continuous but just right continuous and left limited (rcll in short) (see equation (1) below). It has jumps which arise naturally since the noise contains a Poisson random measure part.
The problem we consider here can be studied in a more general setting, namely, multivalued backward stochastic differential equations (see [10], for the continuous case). But for the sake of simplicity, we limit ourselves to the reflected framework. Finally, two other interesting papers on BSDEs with jumps but without reflection can be mentioned, namely those of R.Situ [12], and S. Tang & X. Li [13]. This latter is motivated by control problems.
In this work we mainly show the existence and uniqueness of the solution for the reflected BSDE with jumps (i.e. whose noise includes a Poisson random measure part) for a given, -terminal value ξ which is square integrable random variable -coefficient f (t, ω, y, z, v) which is a function, uniformly Lipschitz with respect to (y, z, v) -barrier (S t ) t≤1 , which is the moving obstacle and which is a rcll process whose jumps are inaccessible.
In the proof of our result, we use two methods, the penalization and the Snell envelope theory. However in order to prove the result in the general case, both methods use a contraction (fixed point argument) since we do not have an efficient comparison theorem for solutions of standard BSDEs whose noise contains a Poisson measure part (see e.g. [1] for a counterexample). On the other hand, the fact that the jumping times of the moving barrier S are inaccessible is of crucial role. Finally we highlight the connection of our reflected BSDEs with integral-differential mixed stochastic optimal control. Nevertheless our results can be applied in mathematical finance, especially for the evaluation of American contingent claims when the dynamic of the prices contains a Poisson point process part.
This paper is divided into three sections.
In Section 1, we begin to show the uniqueness of the solution of the reflected BSDE when it exits. Then using the penalization method we show the existence of a solution when the function f does not depend on (y, z, v). Therefore we construct a contraction which has a fixed point which is the solution of our reflected BSDE with jumps. Finally we study the regularity of the non-decreasing process K which is absolutely continuous in the case when the barrier is regular.
In Section 2, using the Snell envelope theory, once again, we prove the existence of the solution if the function f does not depend on (y, z, v). Furthermore, as in Section 1, we obtain the solution in the general setting. Section 3 is devoted to the link of our reflected BSDEs with integral-differential mixed stochastic optimal control. We show that the value function of the problem is solution of an appropriate reflected BSDE with jumps. In addition an optimal control exits and its expression is given. Our result generalizes that of J.P. Lepeltier & B.Marchal ([9]) on the same subject.
2 Reflected BSDEs with respect to Brownian motion and an independent Poisson point process.
Let (Ω, F, (F t ) t≤1 )) be a stochastic basis such that F 0 contains all P -null sets of F, F t+ = >0 F t+ = F t , ∀t < 1, and suppose that the filtration is generated by the two following mutually independent processes : is a martingale for every A ∈ U satisfying λ(A) < ∞. λ is assumed to be a σ-finite measure on (U, U) satisfying On the other hand, let: -S 2 be the set of F t -adapted right continuous with left limit (rcll in short) processes (Y t ) t≤1 with values in IR and IE[sup t≤1 |Y t | 2 ] < ∞.
Let us now introduce our reflected BSDE with jumps (in short, RDBSDE; "D" for discontinuous) associated with (f, ξ, S). A solution is a quadruple (Y, Z, K, V ) := (Y t , Z t , K t , V t ) t≤1 of processes with values in IR 1+d × IR + × L 2 (U, U, λ; IR) and which satisfies : In our definition, the jumps of Y are those of its Poisson part since K is continuous To begin with, we are going to show the uniqueness of the solution of the RDBSDE (1) under the above assumptions on f , ξ and S.
P roof : Assume (Y, Z, K, V ) and (Y , Z , K , V ) are two solutions of (1). First let us underline that On the other hand, using Itô's formula with the discontinuous semi-martingale Y − Y and set . 0 Y s − Y s Z s − Z s dB s are (F t , P )-martingales, then taking the expectation in both sides yields, for any t ≤ 1, If we choose α ≥ 4k and β ≥ 4k, and denote Henceforth from Gronwall's lemma and the right continuity of (Y t − Y t ) t≤1 , we get Y = Y . Consequently (Y, Z, V, K) = (Y , Z , V , K ) whence the uniqueness of the solution of (1) We are going now to show that equation (1) has a solution in using two methods. Roughly speaking, the first one is based on the penalization and the second on the well known Snell envelope theory of processes. However, in order to obtain the result in the general frame, both methods use a contraction. The penalization, as it has been used e.g. in [6], is not workable since we do not have an efficient comparison theorem for solutions of BSDEs whose noise is driven by a Lévy process (see [1] for a counter-example). That is the reason for which, in a first time, we suppose that the map f (t, ω, y, z, v) does not depend on (y, z, v).

The penalization method.
First let us assume the map f does not depend on (y, z, v), i.e., P-a.s., f (t, ω, y, z, v) ≡ g(t, ω), for any t, y, z and v. In the following result, we establish the existence of the solution of the RDBSDE associated with (g, ξ, S).
. It exists according to Barles et al.'s result [1]. So, and From now on the proof will be divided into six steps.
Indeed, using the generalized Itô's formula with the convex function x −→ (x + ) 2 and Y n − Y n+1 implies that the following process (A n t ) t≤1 , where where for any n ≥ 0 and t ≤ 1, Through estimates (2) we deduce that s (e))μ(ds, de)} is an (F t , P )-martingale. Now taking expectation in both sides of (3) yields , P-a.s. and then Y n ≤ Y n+1 since Y n and Y n+1 are right continuous processes.
Step 2: There exists a constant C ≥ 0 such that Indeed by Itô's rule we obtain, Taking the expectation in both sides yields, is a universal non-negative real constant. But for any t ≤ 1 we have, where C is a constant. Now plugging this inequality in the previous one yields, (g(s)) 2 ds] whereC is an appropriate real constant. Now Gronwall's inequality leads to the desired result for Step 3: There exists a constant C ≥ 0 such that for any Indeed for n ≥ 0, using once again Itô's formula we obtain, On the other hand using the Burkholder-Davis-Gundy inequality ( [4],p.304) we get, and Here c 1 , c 2 , c 3 and c 4 are universal non-negative real constants. Now combining these inequalities with (5) yields, Finally choosing 2(c 3 + c 4 ) < 1 we obtain IE[sup t≤1 |Y n s | 2 ] ≤ C. Now let Y t = lim inf n→∞ Y n t , t ≤ 1. Since the sequence (Y n ) n≥0 is non-decreasing then, using Fatou's lemma, It follows that for any t ≤ 1, Y t < ∞ and then P-a.s., Y n t → Y t as n → ∞. In addition the Lebesgue's dominated convergence theorem implies that Step 4: lim n→∞ IE[sup t≤1 |(Y n t − S t ) − | 2 ] = 0. This property is the key point in the proof of our result.
Let Ȳ n t ,Z n t ,V n t t≤1 be the solution of the following BSDE : By comparison we have P-a.s., ∀t ≤ 1, Y n t ≥Ȳ n t , for any n ≥ 0 (the proof is similar to the one we have done to prove Y n ≤ Y n+1 in Step 1). Now let τ be an F t -stopping time such that τ ≤ 1. Then, Since S is a right continuous then ] as n → ∞ P−a.s. and in L 2 (Ω, P ). Henceforth we have also the convergence of the conditional expectation in L 2 (Ω, P ). In addition s. From that and the section theorem ( [4], p.220), we deduce that Y t ≥ S t , ∀t ≤ 1, P − a.s. and then (Y n t − S t ) − 0, ∀t ≤ 1, P-a.s. Now since Y n Y then, if we denote by p X the predictable projection of any process X, p Y n p Y and p Y ≥ p S. But for any n the jumping times of the process ( t 0 UV n s (e)μ (ds, de)) 0≤t≤1 are inaccessible since µ is a Poisson random measure. It follows that the jumping times of Y n are also inaccessible. Then for any predictable stopping time δ we have Y n δ = Y n δ− , henceforth the predictable projection of Y n is Y n − , i.e., p Y n = Y n − . In the same way we have p S = S − since we have supposed the jumping times of S inaccessible.
s. as n → ∞. Consequently, from a weak version of the Dini's theorem ( [4], p.202), we deduce that sup t≤1 (Y n t − S t ) − 0 P-a.s. as n → ∞. Therefore the dominated convergence theorem implies Step 5: lim n→∞ IE[sup t≤1 |Y n t − Y t | 2 ] = 0 and there exist F t -adapted processes Z = (Z t ) t≤1 , K = (K t ) t≤1 (K non-decreasing and K 0 = 0) and V = (V t ) t≤1 such that Indeed using Itô's formula we have for any p ≥ n ≥ 0 and t ≤ 1, So since p ≥ n, then It follows that (Z n ) n≥0 and (V n ) n≥0 are Cauchy sequences in complete spaces then there exist processes Z and V , respectively F t -progressively measurable and P ⊗ U-measurable such that the sequences (Z n ) n≥0 and (V n ) n≥0 converge respectively toward Z and V in L 2 (dP ⊗ dt) and L 2 (dP ⊗ dtλ(de)) respectively.
Step 6: The limiting process (Y, Z, K, V ) = (Y t , Z t , K t , V t ) t≤1 is the solution of the reflected DBSDE associated with (g, ξ, S).
Obviously the process (Y, Z, K, V ) satisfies On the other hand since lim n→∞ IE[sup t≤1 ((Y n t − S t ) − ) 2 ] = 0 then P-a.s., ∀t ≤ 1, Y t ≥ S t . Finally let First there exists a subsequence of (K n ) n≥0 which we still denote (K n ) n≥0 such that P−a.s.
On the other hand the result stated in Step 5 implies, for any > 0, there exists n 0 (ω) such that for any n ≥ n 0 (ω), ∀t ≤ 1, Y t (ω) − S t (ω) ≤ Y n t (ω) − S t (ω) + and K n 1 (ω) ≤ K 1 (ω) + . Therefore for n ≥ n 0 (ω) we have But the right-hand side converge to 2 K 1 (ω), as n → ∞, since f m (ω) is a step function and then Finally from (7), (8) and (9) we deduce that As is whatever and Y ≥ S then The other properties are satisfied by construction of the quadruple of processes (Y, Z, K, V ) and the proof is complete .
We are now ready to give the main result of this section.
P roof : It remains to show the existence which will be obtained via a fixed point of the contraction of the function Φ defined as follows: is the solution of the reflected DBSDE associated with (f (t, Y t , Z t , V t ), ξ, S). Let (Y , Z , V ) be another triple of D and Φ(Y , Z , V ) = (Ỹ ,Z ,Ṽ ), then using Itô's formula we obtain, for any t ≤ 1, It implies that Now let α great enough and such that k < < α−1 k , then Φ is a contraction on D, henceforth there exists a triple (Y, Z, V ) such that Φ(Y, Z, V ) = (Y, Z, V ) which, with K, is the unique solution of the reflected DBSDE associated with (f, ξ, S)

Regularity of the process K.
We now focus on the regularity of the process K. We are going to show that the process K = (K t ) t≤1 is absolutely continuous if the barrier S = (S t ) t≤1 is regular. Precisely we have : In addition (S m 1 ) m≥0 converges to S 1 in L 2 (Ω, dP ). Then the process K of the solution of the reflected DBSDE associated with (f, ξ, S) is absolutely continuous with respect to the Lebesgue measure dt. P roof : Let (Y, Z, K, V ) be the solution of the reflected DBSDE associated with (f, ξ, S) and for n ≥ 0, let (Y n , Z n , V n ) be the solution of the following standard BSDE : Since the solution of the reflected DBSDE associated with (f, ξ, S) is unique then, as it has been shown in Thm.1.2.a, the sequence ((Y n , Z n , Now using the generalized Itô's formula with the convex function x −→ x − 2 and the process Y n − S m implies that A n,m t , defined below, is non-decreasing in t ; Then for any t ≤ 1 we have A n,m Taking the expectation in both sides above yields, for any t ≤ 1, Then It follows that Now since the sequence of processes (S m ) m≥0 converges to S in H 2,1 and S 1 = L 2 − lim m→∞ S m 1 then taking the limit in the previous inequality as m → ∞ yields, But the inequality (10) can be written as is the usual Sobolev space consisting of all absolutely continuous functions with derivative in L 2 (0, 1). Hence the sequence (K n ) n is bounded in the Hilbert space L 2 (Ω; H 1 (0, 1; IR d )) and then there exists a subsequence of (K n ) n which converges weakly. The limiting process, which is actually K, belongs to L 2 (Ω; H 1 (0, 1; IR d )) and then P-a.s., K . (ω) ∈ H 1 (0, 1; IR d ) i.e. K is absolutely continuous with respect to Lebesgue measure dt 2.4 The Snell envelope method.
The aim of this part is to give another proof of the existence and uniqueness result using the so called Snell envelope of processes (see El Karoui et al. [6] for the continuous case). However as it is pointed out in the beginning of Section 1.2, first we assume the function f does not depend on (y, z, v), i.e. f (t, y, z, v) = g(t), then we have the following result.
1.4.a. Proposition: There exists a process (Y t , Z t , V t , K t ) t≤1 solution of the reflected DBSDE associated with (g, ξ, S). P roof : Let η := (η t ) t≤1 be the process defined as follows: then η is rcll and its jumping times τ before 1 are the same as the ones of S and then they are inaccessible since those of this latter process are so. Moreover sup 0≤t≤1 |η t | ∈ L 2 (Ω) .
The Snell envelope of η is the smallest rcll supermartingale which dominates the process η, it is given by where T t is the set of stopping time ν such that t ≤ ν ≤ 1 a.s. Now due to (11), we have IE[sup t≤1 |S t | 2 ] < ∞ and then (S t (η)) t≤1 is of class [D], i.e. the set of random variables {S τ (η), τ ∈ T 0 } is uniformly integrable. Henceforth it has the following Doob-Meyer decomposition where (K(t)) t≤1 is an (F t ) t≤1 -adapted rcll non-decreasing process such that K(0) = 0. Furthermore we have IE[K(1) 2 ] < ∞ since IE[sup t≤1 |S t | 2 ] < ∞ (see e.g. [4], p.221). It follows that IE[sup t≤1 |IE[ξ + K(1)|F t ]| 2 ] < ∞ and then, through the representation theorem of martingales with respect to (F t ) t≤1 (see [8]), there exist two processes Z = (Z t ) t≤1 and V = (V t ) t≤1 which belong respectively to H 2,d and L 2 such that, Now let us show that the process K is continuous.
First let us underline that the jumping times of K are included in the set S − (η) = η where η t = lim sup s t η s = η t− since the process η is rcll (see e.g. [EK], pp.131).
Now let τ be a predictable stopping time, then The second inequality is obtained through the fact that the process η has inaccessible jumping times, and may have a positive jump at t = 1. On the other hand, The second equality stems from the fact that τ is predictable, then M τ − = M τ since the jumping times of (M t ) t≤1 are those of its Poisson part and those latter are inaccessible. Now combining (12) and (13) yields IE[S τ − (η)] ≤ IE[S τ (η)] and then, since S(η) is a supermartingale, IE[S τ − (η)] = IE[S τ (η)] for any predictable stopping time τ . Henceforth the supermartingale (S t (η)) t≤1 is regular, i.e. p S(η) = S − (η), and then the process K is continuous (see [3], p.119).

Now let us set
and then for any t ≤ 1 we have, Finally it remains to show that 1 0 (Y t −S t )dK(t) = 0. The Snell envelope process (S t (η)) t≤1 is regular i.e. S − (η) = p S(η). Now let t ≤ 1 and δ t := inf{s ≥ t, K(s) > K(t)} ∧ 1. As S(η) is regular then δ t is the largest optimal stopping time after t (see e.g. [ The process (Y, Z, K, V ) is then solution of the reflected BSDE associated with (g, ξ, S) Now we argue as in Thm. 1.2.b. to obtain the existence and uniqueness of the solution of the reflected discontinuous BSDE associated with coefficients f which depend on (y, z, v) and which are uniformly Lipschitz with respect to those variables. Therefore we have, for the reflected backward stochastic differential equation (1) with jumps associated with (f, ξ, S) 2.5 Application of Reflected DBSDEs in mixed stochastic control.
Now we are going to highlight the link between mixed stochastic optimal control, when the noise is of gaussian and Poisson types, and RDBSDEs.
We can think of J(δ, τ ) as the profit that makes an agent who intervenes on a system whose evolution is described by a stochastic process, say (X t ) t≤1 . An intervention strategy for the agent is a pair (δ, τ ), δ is his control action and τ is the time he chooses to stop controlling. A strategy (δ * , τ * ) which maximizes J(δ, τ ), if it exists, is called optimal for the agent.
In the expression of J(δ, τ ), the term which is absolutely continuous with respect to dt is the instantaneous reward and the other is the reward at stopping for the agent.
This problem is called of mixed control type because it combines optimal control and stopping.
Assume the state evolution (X t ) t≤1 of the non-controlled system is described by a stochastic differential equation of the following type: eμ(ds, de), t ≤ 1.
The control action of the agent consists in choosing a probability P δ under which the system will evolve. So under P δ the state evolution of the controlled system is described by : e(e g(s,δ 2 (s),e) − 1)λ(de)ds, t ≤ 1 It means that the agent control action generates a drift for the dynamic of the system and a reward which is equal to J(δ, τ ), therefore he looks for optimal strategies We now go back to our general mixed control problem. Let H 1 and H 2 be the hamiltonian functions associated with this control problem, defined on [0, 1]×Ω×IR d ×D 1 and [0, 1]×Ω×L 2 (U, U, λ; IR)×D 2 respectively, as follows: H 1 (t, ω, p, d 1 ) = pf (t, ω, d 1 ) + c(t, ω, d 1 ) and H 2 (t, ω, v, d 2 ) = U v(e)(e g(t,ω,d 2 ,e) − 1)λ(de) + U h(t, ω, d 2 , e)e g(t,ω,d 2 ,e) λ(de).