VMSTA Modern Stochastics: Theory and Applications 2351-6054 2351-6046 2351-6046 VTeXMokslininkų g. 2A, 08412 Vilnius, Lithuania VMSTA90 10.15559/17-VMSTA90 Research Article Double barrier reflected BSDEs with stochastic Lipschitz coefficient MarzougueMohamedmd.marzougue@gmail.com∗ El OtmaniMohamedm.elotmani@uiz.ac.ma Laboratoire d’Analyse Mathématique et Applications (LAMA) Faculté des Sciences Agadir, Université Ibn Zohr, Maroc Corresponding author. 2017 812201744353379 2172017 14112017 14112017 © 2017 The Author(s). Published by VTeX2017 Open access article under the CC BY license.

This paper proves the existence and uniqueness of a solution to doubly reflected backward stochastic differential equations where the coefficient is stochastic Lipschitz, by means of the penalization method.

 BSDE and reflected BSDE Stochastic Lipschitz coefficient 60H20 60H30 65C30 Introduction

Backward Stochastic Differential Equations (BSDEs) were introduced (in the non-linear case) by Pardoux and Peng [21]. Precisely, given a data (ξ,f) of a square integrable random variable ξ and a progressively measurable function f, a solution to BSDE associated with data (ξ,f) is a pair of Ft -adapted processes (Y,Z) satisfying Yt=ξ+∫tTf(s,Ys,Zs)ds−∫tTZsdBs,0≤t≤T. These equations have attracted great interest due to their connections with mathematical finance [9, 10], stochastic control and stochastic games [3, 17] and partial differential equations [20, 22].

In their seminal paper [21], Pardoux and Peng generalized such equations to the Lipschitz condition and proved existence and uniqueness results in a Brownian framework. Moreover, many efforts have been made to relax the Lipschitz condition on the coefficient. In this context, Bender and Kohlmann [2] considered the so-called stochastic Lipschitz condition introduced by El Karoui and Huang [8].

Further, El Karoui et al. [11] have introduced the notion of reflected BSDEs (RBSDEs in short), which is a BSDE but the solution is forced to stay above a lower barrier. In detail, a solution to such equations is a triple of processes (Y,Z,K) satisfying Yt=ξ+∫tTf(s,Ys,Zs)ds+KT−Kt−∫tTZsdBs,Yt≥Lt0≤t≤T, where L, the so-called barrier, is a given stochastic process. The role of the continuous increasing process K is to push the state process upward with the minimal energy, in order to keep it above L; in this sense, it satisfies ∫0T(Yt−Lt)dKt=0 . The authors have proved that equation (2) has a unique solution under square integrability of the terminal condition ξ and the barrier L, and the Lipschitz property of the coefficient f.

RBSDEs have been proven to be powerful tools in mathematical finance [10], mixed game problems [6], providing a probabilistic formula for the viscosity solution to an obstacle problem for a class of parabolic partial differential equations [11].

Later, Cvitanic and Karatzas [6] studied doubly reflected BSDEs (DRBSDEs in short). A solution to such an equation related to a generator f, a terminal condition ξ and two barriers L and U is a quadruple of (Y,Z,K+,K−) which satisfies Yt=ξ+∫tTf(s,Ys,Zs)ds+(KT+−Kt+)−(KT−−Kt−)−∫tTZsdBsLt≤Yt≤Ut,∀t≤Tand∫0T(Yt−Lt)dKt+=∫0T(Ut−Yt)dKt−=0. In this case, a solution Y has to remain between the lower barrier L and upper barrier U. This is achieved by the cumulative action of two continuous, increasing reflecting processes K± . The authors proved the existence and uniqueness of the solution when f(t,ω,y,z) is Lipschitz on (y,z) uniformly in (t,ω) . At the same time, one of the barriers L or U is regular or they satisfy the so-called Mokobodski condition, which turns out into the existence of a difference of a non-negative supermartingales between L and U. In addition, many efforts have been made to relax the conditions on f, L and U [1, 15, 16, 18, 19, 27, 29] or to deal with other issues [5, 12–14, 24].

Let us have a look at the pricing problem of an American game option driven by Black–Scholes market model which is given by the following system of stochastic differential equations dSt0=r(t)St0dt,S00>0;dSt=St((r(t)+θ(t)σ(t))dt+σ(t)dBt),S0>0, 0;\\{} dS_{t}=S_{t}\big(\big(r(t)+\theta (t)\sigma (t)\big)dt+\sigma (t)dB_{t}\big),& S_{0}>0,\end{array}\right.\end{array}\]]]> where r(t) is the interest rate process, θ(t) is the risk premium process, σ(t) is the volatility process of the market. The fair price of the American game option is defined by Yt=infτ∈ℑ[0,T]supν∈ℑ[0,T]E[e−r(t)σ(t)∧θ(t)J(τ,ν)|Ft], where ℑ[0,T] is the collection of all stopping times τ with values between 0 and T, and J is a Payoff given by J(τ,ν)=Uν1{ν<τ}+Lτ1{τ≤ν}+ξ1{ν∧τ=T}. Here r(t) , σ(t) and θ(t) are stochastic, moreover they are not bounded in general. So the existence results of Cvitanic and Karatzas [6], Li and Shi [19] with completely separated barriers cannot be applied.

Motivated by the above works, the purpose of the present paper is to consider a class of DRBSDEs driven by a Brownian motion with stochastic Lipschitz coefficient. We try to get the existence and uniqueness of solutions to those DRBSDEs by means of the penalization method and the fixed point theorem. Furthermore, the comparison theorem for the solutions to DRBSDEs will be established.

The paper is organized as follows: in Section 2, we give some notations and assumptions needed in this paper. In Section 3, we establish the a priori estimates of solutions to DRBSDEs. In Section 4, we prove the existence and uniqueness of solutions to DRBSDEs via penalization method when one barrier is regular, in the first subsection, then we study the case when the barriers are completely separated, in the second subsection. In Section 5, we give the comparison theorem for the solutions to DRBSDEs. Finally, an Appendix is devoted to the special case of RBSDEs with lower barrier when the generator only depends on y; furthermore, the corresponding comparison theorem will be established under the stochastic Lipschitz coefficient.

 Notations

Let (Ω,F,(Ft)t≤T,P) be a filtered probability space. Let (Bt)t≤T be a d-dimensional Brownian motion. We assume that (Ft)t≤T is the standard filtration generated by the Brownian motion (Bt)t≤T .

We will denote by |.| the Euclidian norm on Rd .

Let’s introduce some spaces:

L2 is the space of R -valued and FT -measurable random variables ξ such that ‖ξ‖2=E[|ξ|2]<+∞.

S2 is the space of R -valued and Ft -progressively measurable processes (Kt)t≤T such that ‖K‖2=E[sup0≤t≤T|Kt|2]<+∞.

Let β>0 0$]]> and (at)t≤T be a non-negative Ft -adapted process. We define the increasing continuous process A(t)=∫0ta2(s)ds , for all t≤T , and introduce the following spaces:

L2(β,a) is the space of R -valued and FT -measurable random variables ξ such that ‖ξ‖β2=E[eβA(T)|ξ|2]<+∞.

S2(β,a) is the space of R -valued and Ft -adapted continuous processes (Yt)t≤T such that ‖Y‖β2=E[sup0≤t≤TeβA(t)|Yt|2]<+∞.

S2,a(β,a) is the space of R -valued and Ft -adapted processes (Yt)t≤T such that ‖aY‖β2=E[∫0TeβA(t)|a(t)Yt|2dt]<+∞.

H2(β,a) is the space of Rd -valued and Ft -progressively measurable processes (Zt)t≤T such that ‖Z‖β2=E[∫0TeβA(t)|Zt|2dt]<+∞.

B2 is the Banach space of the processes (Y,Z)∈(S2(β,a)∩S2,a(β,a))×H2(β,a) with the norm ‖(Y,Z)‖β=‖aY‖β2+‖Z‖β2.

We consider the following conditions: (H1)

The terminal condition ξ∈L2(β,a) .

 The coefficient f:Ω×[0,T]×R×Rd⟶R satisfies (H2)

∀t∈[0,T] ∀(y,z,y′,z′)∈R×Rd×R×Rd , there are two non-negative Ft -adapted processes μ and γ such that |f(t,y,z)−f(t,y′,z′)|≤μ(t)|y−y′|+γ(t)|z−z′|.

(H3)

There exists ϵ>0 0$]]> such that a2(t):=μ(t)+γ2(t)≥ϵ .

 (H4)

For all (y,z)∈R×Rd , the process (f(t,y,z))t is progressively measurable and such that f(.,0,0)a∈H2(β,a).

The two reflecting barriers L and U are two Ft -adapted and continuous real-valued processes which satisfy (H5)

E[sup0≤t≤Te2βA(t)|Lt+|2]+E[sup0≤t≤Te2βA(t)|Ut−|2]<+∞, where L+ and U− are the positive and negative parts of L and U, respectively.

 (H6)

U is regular: i.e., there exists a sequence of (Un)n≥0 such that

∀t≤T , Utn≤Utn+1 and limn→+∞Utn=Ut P -a.s

∀n≥0 , ∀t≤T , Utn=U0n+∫0tun(s)ds+∫0tvn(s)dBs where the processes un and vn are Ft -adapted such that supn≥0sup0≤t≤T(un(t))+≤CandE[∫0T|vn(s)|2ds]12<+∞.

Let β>0 0$]]> and a be a non-negative Ft -adapted process. A solution to DRBSDE is a quadruple (Y,Z,K+,K−) satisfying (3) such that

(Y,Z)∈(S2(β,a)∩S2,a(β,a))×H2(β,a) ,

K±∈S2 are two continuous and increasing processes with K0±=0 .

A priori estimate

Let β>0 0$]]> be large enough and assume (H1)−(H6) hold. Let (Y,Z,K+,K−)∈(S2(β,a)∩S2,a(β,a))×H2(β,a)×S2×S2 be a solution to DRBSDE with data (ξ,f,L,U) . Then there exists a constant Cβ depending only on β such that E[sup0≤t≤TeβA(t)|Yt|2+∫0TeβA(t)(a2(t)|Yt|2+|Zt|2)dt+|KT+|2+|KT−|2]≤CβE[eβA(T)|ξ|2+∫0TeβA(t)|f(t,0,0)|2a2(t)dt+sup0≤t≤Te2βA(t)(|Lt+|2+|Ut−|2)].

Applying Itô’s formula and Young’s inequality, combined with the stochastic Lipschitz assumption (H2) we can write eβA(t)|Yt|2+∫tTβeβA(s)a2(s)|Ys|2ds+∫tTeβA(s)|Zs|2ds≤eβA(T)|ξ|2+β2∫tTeβA(s)a2(s)|Ys|2ds+2β∫tTeβA(s)|f(s,Ys,Zs)|2a2(s)ds+2∫tTeβA(s)YsdKs+−2∫tTeβA(s)YsdKs−−2∫tTeβA(s)YsZsdBs≤eβA(T)|ξ|2+β2∫tTeβA(s)a2(s)|Ys|2ds+6β∫tTeβA(s)a2(s)|Ys|2ds+6β∫tTeβA(s)|Zs|2ds+6β∫tTeβA(s)|f(s,0,0)|2a2(s)ds+2∫tTeβA(s)YsdKs+−2∫tTeβA(s)YsdKs−−2∫tTeβA(s)YsZsdBs. Using the fact that dKs+=1{Ys=Ls}dKs+ and dKs−=1{Ys=Us}dKs− , we have eβA(t)|Yt|2+(β2−6β)∫tTeβA(s)a2(s)|Ys|2ds+(1−6β)∫tTeβA(s)|Zs|2ds≤eβA(T)|ξ|2+6β∫tTeβA(s)|f(s,0,0)|2a2(s)ds+2∫tTeβA(s)LsdKs+−2∫tTeβA(s)UsdKs−−2∫tTeβA(s)YsZsdBs. Taking expectation on both sides above, we get E[∫0TeβA(s)a2(s)|Ys|2ds+∫0TeβA(s)|Zs|2ds]≤cβE[eβA(T)|ξ|2+∫0TeβA(s)|f(s,0,0)|2a2(s)ds+sup0≤t≤TeβA(t)|Lt+|2+|KT+|2+sup0≤t≤TeβA(t)|Ut−|2+|KT−|2] and by the Burkholder–Davis–Gundy’s inequality we obtain Esup0≤t≤TeβA(t)|Yt|2≤CβE[eβA(T)|ξ|2+∫0TeβA(s)|f(s,0,0)|2a2(s)ds+2∫tTeβA(s)LsdKs+−2∫tTeβA(s)LsdKs−] ≤CβE[eβA(T)|ξ|2+∫0TeβA(s)|f(s,0,0)|2a2(s)ds+sup0≤t≤Te2βA(t)(|Lt+|2+|Ut−|2)+|KT+|2+|KT−|2]. To conclude, we now give an estimate of KT+2 and KT−2 . From the equation KT+−KT−=Y0−ξ−∫0Tf(s,Ys,Zs)ds+∫0TZsdBs and the stochastic Lipschitz property (H2) , we have E[|KT+−KT−|2]≤4E[sup0≤t≤TeβA(t)|Yt|2+|ξ|2+(1+3β)∫0TeβA(s)|Zs|2ds+3β∫0TeβA(s)a2(s)|Ys|2ds+3β∫0TeβA(s)|f(s,0,0)|2a2(s)ds]. Combining this with (7), we derive that E|KT+|2+E|KT−|2≤CβE[eβA(T)|ξ|2+∫0TeβA(s)|f(s,0,0)|2a2(s)ds+sup0≤t≤Te2βA(t)(|Lt+|2+|Ut−|2)]+12E|KT+|2+12E|KT−|2. The desired result is obtained by estimates (6), (8) and (9).  □

 Existence and uniqueness of solution The obstacle <italic>U</italic> is regular

In this part, we apply the penalization method and the fixed point theorem to give the existence of the solution to the DRBSDE (3). We first consider the special case when the generator does not depend on (y,z) : f(t,y,z)=g(t).

Assume that ga∈H2(β,a) and (H1) – (H6) hold. Then, the doubly reflected BSDE (3) with data (ξ,g,L,U) has a unique solution (Y,Z,K+,K−) that belongs to (S2(β,a)∩S2,a(β,a))×H2(β,a)×S2×S2 .

For all n∈N , let (Yn,Zn,Kn+) be the Ft -adapted process with values in (S2(β,a)∩S2,a(β,a))×H2(β,a)×S2 being a solution to the reflected BSDE with data (ξ,g(t)−n(y−Ut)+,L) . That is Ytn=ξ+∫tTg(s)ds−n∫tT(Ysn−Us)+ds+KTn+−Ktn+−∫tTZsndBsYtn≥Lt,∀t≤Tand∫0T(Ytn−Lt)dKtn+=0. We denote Ktn−:=n∫0t(Ysn−Us)+ds and gn(s,y):=g(s)−n(y−Us)+ .

We have divided the proof of Theorem 1 into sequence of lemmas.

There exists a positive constant C such that sup0≤t≤Tn(Ytn−Ut)+≤CP-a.s.

For all n,m≥0 , let (Yn,m,Zn,m) be the solution to the following BSDE Ytn,m=ξ−∫tT{g(s)+m(Ysn,m−Ls)−−n(Ysn,m−Us)+}ds−∫tTZsn,mdBs. We denote Y¯n,m=Yn,m−Um . Then we have Y¯tn,m=ξ−UTm+∫tT(g(s)+um(s))ds−n∫tT(Y¯sn,m−(Us−Usm))+ds+m∫tT(Y¯sn,m−(Ls−Usm))−ds−∫tT(Zsn,m−vn(s))dBs. For n≥0 , let Dn be the class of Ft -progressively measurable process taking values in [0,n] . For ν∈Dn and λ∈Dm we denote Rt=e−∫0t(ν(s)+λ(s))ds . Applying Itô’s formula to RtY¯tn,m and using the same arguments as on page 2042 of [6], one can show that Y¯tn,m≤ess supλ∈Dmess infν∈DnE[∫tTe−∫ts(ν(r)+λ(r))dr|um(s)|ds|Ft]. From the assumption (H6)(ii) , we can write Y¯tn,m∨0≤Cn . It follows that ∀t≤T,n(Y¯tn,m∨0) →m→+∞n(Ytn−Ut)+≤CP-a.s.  □

There exists a positive constant Cβ′ depending only on β such that for all n≥0 E[sup0≤t≤TeβA(t)|Ytn|2+∫0TeβA(t)a2(t)|Ytn|2dt+∫0TeβA(t)|Ztn|2dt+|KTn+|2]≤Cβ′E[eβA(T)|ξ|2+∫0TeβA(t)|g(t)a(t)|2dt+sup0≤t≤Te2βA(t)|Ut−|2+sup0≤t≤Te2βA(t)|Lt+|2].

Itô’s formula implies for t≤T : βE∫tTeβA(s)a2(s)|Ysn|2ds+E∫tTeβA(s)|Zsn|2ds≤EeβA(T)|ξ|2+β2E∫tTe2βA(s)a2(s)|Ysn|2ds+2βE∫tTeβA(s)|g(s)|2a2(s)ds+2E[supn≥0sup0≤t≤Tn(Ytn−Ut)+∫tTeβA(s)Us−ds]+2E[∫tTeβA(s)LsdKsn+]. Here we used the fact that −nYsn(Ysn−Us)+≤nU−(Ysn−Us)+ and dKsn+=1{Ysn=Ls}dKsn+ . We conclude, by the Burkholder–Davis–Gundy’s inequality, that Esup0≤t≤TeβA(t)|Ytn|2+E∫0TeβA(s)a2(s)|Ysn|2ds+E∫0TeβA(s)|Zsn|2ds≤cp′E[eβA(T)|ξ|2+∫0TeβA(s)|g(s)|2a2(s)ds+sup0≤t≤Te2βA(t)|Ut−|2+sup0≤t≤Te2βA(t)|Lt+|2+|KTn+|2]. In the same way as (9), we can prove that E|KTn+|2≤Cp′E[eβA(T)|ξ|2+∫0TeβA(s)|g(s)|2a2(s)ds+sup0≤t≤Te2βA(t)|Ut−|2+sup0≤t≤Te2βA(t)|Lt+|2]. We obtain the desired result.  □

There exist two Ft -adapted processes (Yt)t≤T and (Kt+)t≤T such that Yn↘Y , Kn+↗K+ and E[sup0≤t≤T|Ktn+−Kt+|2] →n→+∞0.

The comparison Theorem 5 (below) shows that Yt0≥Ytn≥Ytn+1 and Ktn+≤Kt(n+1)+ for all t≤T . Therefore, there exist processes Y and K+ such that, as n→+∞ , for all t≤T , Ytn↘Yt and Ktn+↗Kt+ . Since the process K+ is continuous, it follows by Dini’s theorem that E[sup0≤t≤T|Ktn+−Kt+|2] →n→+∞0.  □

E[sup0≤t≤TeβA(t)|(Ytn−Ut)+|2] →n→+∞0.

Since Yt≤Ytn≤Yt0 , we can replace Ut by Ut∨Y0 ; that is, we may assume that Esup0≤t≤TeβA(t)|Ut|2<+∞ .

Let (Y˜n,Z˜n,K˜n) be the solution to the following Reflected BSDE associated with (ξ,g−n(y−U),L) : Y˜tn=ξ+∫tT(g(s)−n(Y˜sn−Us))ds+K˜Tn−K˜tn−∫tTZ˜sndBsY˜tn≥Lt,∀t≤Tand∫0T(Y˜tn−Lt)dK˜tn=0. The comparison Theorem 5 shows that Yn≤Y˜n and dK˜n≤dKn+≤dK+ . Let τ≤T be a stopping time. Then we can write Y˜τn=E[e−n(T−τ)ξ+∫τTe−n(s−τ)(g(s)+nUs)ds+∫τTe−n(s−τ)dK˜sn|Fτ]. Since Esup0≤t≤TeβA(t)Ut2<+∞ , we obtain e−n(T−τ)ξ+n∫τTe−n(s−τ)Usds →n→+∞ξ1τ=T+Uτ1τ<TP-a.s.inL2 and the conditional expectation converges also in L2 . Moreover, |∫τTe−n(s−τ)g(s)ds|2≤∫τTeβA(s)|g(s)a(s)|2ds∫τTe−2n(s−τ)e−βA(s)a2(s)ds. Then ∫τTe−n(s−τ)g(s)ds →n→+∞0P-a.s. inL2. In addition, 0≤∫τTe−n(s−τ)dK˜sn≤∫τTe−n(s−τ)dKs+ →n→+∞0inL1. Consequently, Y˜τn →n→+∞ξ1τ=T+Uτ1τ<TP-a.s. inL1. Therefore, Yτ≤Uτ P -a.s. We deduce, from Theorem 86 page 220 in Dellacherie and Meyer [7], that Yt≤Ut for all t≤T P -a.s and then eβA(t)(Ytn−Ut)+↘0 for all t≤T P -a.s. By Dini’s theorem, we have sup0≤t≤TeβA(t)(Ytn−Ut)+↘0 P-a.s. and the result follows from the Lebesgue’s dominated convergence theorem.  □

There exist two processes (Zt)t≤T and (Kt−)t≤T such that E∫0TeβA(t)a2(t)|Ytn−Yt|2dt+E∫0TeβA(t)|Ztn−Zt|2dt →n→+∞0. Moreover, Esup0≤t≤TeβA(t)|Ytn−Yt|2+Esup0≤t≤T|Ktn−−Kt−|2 →n→+∞0.

For all n≥p≥0 and t≤T , applying Itô’s formula and taking expectation yields that E[eβA(t)|Ytn−Ytp|2+β∫tTeβA(s)a2(s)|Ysn−Ysp|2ds+∫tTeβA(s)|Zsn−Zsp|2ds]≤2E[∫tTeβA(s)(Ysp−Us)+n(Ysn−Us)+ds]+2E[∫tTeβA(s)(Ysn−Us)+p(Ysp−Us)+ds]≤E[sup0≤t≤T(eβA(t)(Ytp−Ut)+)2]12E[(∫tTn(Ysn−Us)+ds)2]12+E[sup0≤t≤T(eβA(t)(Ytn−Ut)+)2]12E[(∫tTp(Ysp−Us)+ds)2]12 since (Ysn−Ysp)d(Ksn+−Ksp+)≤0 . Therefore, using Lemmas 2 and 5, we obtain E∫0TeβA(s)a2(s)|Ysn−Ysp|2ds+E∫0TeβA(s)|Zsn−Zsp|2ds →n,p→+∞0. It follows that (Zn)n≥0 is a Cauchy sequence in complete space H2(β,a) . Then there exists an Ft -progressively measurable process (Zt)t≤T such that the sequence (Zn)n≥0 tends toward Z in H2(β,a) . On the other hand, by the Burkholder–Davis–Gundy’s inequality, one can derive that Esup0≤t≤TeβA(t)|Ytn−Ytp|2≤E[sup0≤t≤T(eβA(t)(Ytp−Ut)+)2]12E[(∫tTn(Ysn−Us)+ds)2]12+E[sup0≤t≤T(eβA(t)(Ytn−Ut)+)2]12E[(∫tTp(Ysp−Us)+ds)2]12+12Esup0≤t≤TeβA(t)|Ytn−Ytp|2+2c2E∫tTeβA(s)|Zsn−Zsp|2ds where c is a universal non-negative constant. It follows that Esup0≤t≤TeβA(t)|Ytn−Ytp|2 →n,p→+∞0 and then E[sup0≤t≤TeβA(t)|Ytn−Yt|2+∫0TeβA(t)a2(t)|Ytn−Yt|2dt] →n→+∞0. Now, we set Kt−=Yt−Y0+∫0tg(s)ds+Kt+−K0+−∫0tZsdBs. One can show, at least for a subsequence (which we still index by n), that Esup0≤t≤T|Ktn−−Kt−|2 →n→+∞0. The proof is completed.  □

Obviously, the process (Yt,Zt,Kt+,Kt−)t≤T satisfies, for all t≤T , Yt=ξ+∫tTg(s)ds+(KT+−Kt+)−(KT−−Kt−)−∫tTZsdBs. Since Ytn≥Lt and from Lemma 5 we have Lt≤Yt≤Ut .

In the following, we want to show that ∫0T(Yt−Lt)dKt+=∫0T(Ut−Yt)dKt−=0P-a.s. Note that ∫0T(Yt−Lt)dKt+=∫0T(Yt−Ytn)dKt++∫0T(Ytn−Lt)(dKt+−dKtn+). Let ω∈Ω be fixed. It follows from Lemma 4 that, for any ε>0 0$]]>, there exists n(ω) such that ∀n≥n(ω) , Yt(ω)≤Ytn(ω)+ε . Hence ∫0T(Yt(ω)−Ytn(ω))dKt+(ω)≤εKT+(ω). On the other hand, since the function (Yt(ω)−Lt(ω))t≤T is continuous, then there exists a sequence of non-negative step functions (fm(ω))m≥0 which converges uniformly on [0,T] to Yt(ω)−Lt(ω) . That is |Yt(ω)−Lt(ω)−ftm(ω)|<ε. It follows that ∫0T(Yt(ω)−Lt(ω))d(Kt+(ω)−Ktn+(ω))≤ε(KT+(ω)+KTn+(ω))+∫0Tftm(ω)d(Kt+(ω)−Ktn+(ω)). Further, ε(KT+(ω)+KTn+(ω)) →n→+∞2εKT+(ω) and, since (fm(ω))m≥0 is a step function, ∫0Tftm(ω)d(Kt+(ω)−Ktn+(ω)) →m→+∞0. Therefore, we have lim supn→+∞∫0T(Ytn−Lt)d(Kt+−Ktn+)≤2εKT+(ω). From (12) we deduce that ∫0T(Yt−Lt)dKt+≤3εKT+(ω). The arbitrariness of ε and Y≥L , show that ∫0T(Yt−Lt)dKt+=0 . Further, by Lemma 4 and the result treated on p. 465 of Saisho [25] we can write ∫0T(Us−Ysn)n(Ysn−Us)ds →n→+∞∫0T(Us−Ys)dKs−. Since ∫0T(Us−Ysn)n(Ysn−Us)ds=∫0T(Us−Ysn)dKsn−≤0 for each n≥0 P -a.s. and for each n,m≥0 , n≠m , E[|∫0T(Ysn−Ysm)dKsm−|]≤E[sup0≤t≤TeβA(t)|Ytn−Ytm|KTm−] →n,m→+∞0. Then we have lim supn→+∞∫0T(Us−Ysn)dKtn−≤0P-a.s. Combining (13) and (14), we get ∫0T(Us−Ys)dKs−≤0P-a.s. Noting that Y≤U , we conclude that ∫0T(Us−Ys)dKs−=0 . Consequently, (Yt,Zt,Kt+,Kt−) is the solution to (3) associated to the data (ξ,g,L,U) .  □

We can now state the main result:

Assume (H1) – (H6) hold for a sufficient large β. Then DRBSDE (3) has a unique solution (Y,Z,K+,K−) that belongs to (S2(β,a)∩S2,a(β,a))×H2(β,a)×S2×S2 .

Given (ϕ,ψ)∈B2 , consider the following DRBSDE : Yt=ξ+∫tTf(s,ϕs,ψs)ds+(KT+−Kt+)−(KT−−Kt−)−∫tTZsdBst≤TLt≤Yt≤Ut,∀t≤Tand∫0T(Yt−Lt)dKt+=∫0T(Ut−Yt)dKt−=0. From (H2) and (H3) , we have |f(t,ϕt,ψt)|2≤3(a(t)4|ϕt|2+a(t)2|ψt|2+|f(t,0,0)|2). It follows from (H4) that fa∈H2(β,a) and then (15) has a unique solution (Y,Z,K+,K−) .

We define a mapping φ:B2⟶B2(ϕ,ψ)⟼(Y,Z) Let φ(ϕ,ψ)=(Y,Z) and φ(ϕ′,ψ′)=(Y′,Z′) where (Y,Z,K+,K−) (resp. (Y′,Z′,K+′,K−′) ) is the unique solution to the DRBSDE associated with data (ξ,f(.,ϕ,ψ),L,U) (resp. (ξ,f(.,ϕ′,ψ′),L,U) ). Denote ΔΓ=Γ−Γ′ for Γ=Y,Z,K+,K−,ϕ,ψ and Δft=f(t,ϕ′t,ψ′t)−f(t,ϕt,ψt) . Applying Itô’s formula to eβA(t)|ΔYt|2 and taking expectation we have EeβA(t)|ΔYt|2+βE∫tTeβA(s)a2(s)|ΔYs|2ds+E∫tTeβA(s)|ΔZs|2ds≤2E∫tTeβA(s)ΔYsΔfsds≤αβE∫tTeβA(s)a2(s)|ΔYs|2ds+2αβE∫tTeβA(s)(a2(s)|Δϕs|2+|Δψs|2)ds. We have used the fact that ΔYsd(ΔKs+−ΔKs−)≤0 . Choosing αβ=4 and β>5 5$]]>, we can write ‖φ(ϕ,ψ)‖β2≤12‖(ϕ,ψ)‖β2. It follows that φ is a strict contraction mapping on B2 and then φ has a unique fixed point which is the solution to the DRBSDE (3).  □

If we consider U=+∞ , we obtain the BSDE with one continuous reflecting barrier L, then we proved the existence and uniqueness of the solution to RBSDE (2) by means of a penalization method. Before this work, Wen Lü [26] showed the existence and uniqueness result for this class of equations via the Snell envelope notion.

 Completely separated barriers

In this section we will prove the existence of solution to (3) when the barriers are completely separated, i.e., Lt<Ut , ∀t≤T . Then (H7)

there exists a continuous semimartingale Ht=H0+∫0thsdBs−Vt++Vt−,HT=ξ with h∈H2(0,a) and V±∈S2 ( V0±=0 ) are two nondecreasing continuous processes, such that Lt≤Ht≤Ut0≤t≤T.

We will show the existence by the general penalization method. We first consider the special case when the generator does not depend on (y,z) : f(t,y,z)=f(t). Let (Yn,Zn)∈(S2(β,a)∩S2,a(β,a))×H2(β,a) be solution to the following BSDE Ytn=ξ+∫tTf(s)ds−n∫tT(Ysn−Us)+ds+n∫tT(Ysn−Ls)−ds−∫tTZsndBs. We denote Ktn+:=n∫0t(Ysn−Ls)−ds , Ktn−:=n∫0t(Ysn−Us)+ds , Ktn=Ktn+−Ktn− and fn(s,y)=f(s)−n(y−Us)++n(y−Ls)− .

Now let us derive the uniform a priori estimates of (Yn,Zn,Kn+,Kn−) .

There exists a positive constant κ independent of n such that, ∀n≥0 , E[sup0≤t≤TeβA(t)|Ytn|2+∫0TeβA(t)a2(t)|Ytn|2dt+∫0TeβA(t)|Ztn|2dt+|KTn+|2+|KTn−|2]≤κ.

Consider the RBSDE with data (ξ,f,L) . That is, Y‾t=ξ+∫tTf(s)ds+K‾T−K‾t−∫tTZ‾sdBsY‾t≥Lt,∀t≤Tand∫0T(Y‾t−Lt)dK‾t=0. From Appendix A there exists a unique triplet of processes (Y‾,Z‾,K‾)∈(S2(β,a)∩S2,a(β,a))×H2(β,a)×S2 being the solution to RBSDE (18). We consider the penalization equation associated with the RBSDE (18), for n∈N , Y‾tn=ξ+∫tTf(s)ds+n∫tT(Y‾sn−Ls)−ds−∫tTZ‾sndBs. The Remark 2 implies that Y‾t0≤Y‾tn≤Y‾n+1 and Ytn≤Y‾tn for all t≤T . Therefore, as n⟶+∞ for all t≤T , Y‾tn↗Y‾t . Hence Ytn≤Y‾t .

Similarly, we consider the RBSDE with data (ξ,f,U) . There exists a unique triplet of processes (Y_,Z_,K_)∈(S2(β,a)∩S2,a(β,a))×H2(β,a)×S2 , which satisfies Y_t=ξ+∫tTf(s)ds−(K_T−K_t)−∫tTZ_sdBsY_t≤Ut,∀t≤Tand∫0T(Ut−Y_t)dK_t=0. By the penalization equation associated with the RBSDE (19) Y_tn=ξ+∫tTf(s)ds−n∫tT(Y_sn−Us)+ds−∫tTZ_sndBs and the Remark 2, we deduce that Ytn≥Y_t for all t≤T . Then we can write Esup0≤t≤TeβA(t)|Ytn|2≤max{Esup0≤t≤TeβA(t)|Y‾t|2,Esup0≤t≤TeβA(t)|Y_t|2}≤κ. On the other hand, using Itô’s formula and taking expectation implies for t≤T : βE∫tTeβA(s)a2(s)|Ysn|2ds+E∫tTeβA(s)|Zsn|2ds≤EeβA(T)|ξ|2+2E∫tTeβA(s)Ysnf(s)ds−2nE∫tTeβA(s)Ysn(Ysn−Us)+ds+2nE∫tTeβA(s)Ysn(Ysn−Ls)−ds≤EeβA(T)|ξ|2+β2E∫tTeβA(s)a2(s)|Ysn|2ds+2βE∫tTeβA(s)|f(s)|2a2(s)ds+2nE∫tTeβA(s)Us−(Ysn−Us)+ds+2nE∫tTeβA(s)Ls+(Ysn−Ls)−ds. Hence β2E∫tTeβA(s)a2(s)|Ysn|2ds+E∫tTeβA(s)|Zsn|2ds≤EeβA(T)|ξ|2+2βE∫tTeβA(s)|f(s)a(s)|2ds+1αEsup0≤t≤Te2βA(t)(|Lt+|2+|Ut−|2)+αE[∫tTn(Ysn−Us)+ds]2+αE[∫tTn(Ysn−Ls)−ds]2. Now we need to estimate E[∫tTn(Ysn−Us)+ds]2+E[∫tTn(Ysn−Ls)−ds]2 . For this, let us consider the following stopping times τ0=0,τ2l+1=inf{t>τ2l|Ytn≤Lt}∧T,l≥0τ2l+2=inf{t>τ2l+1|Ytn≥Ut}∧T,l≥0. \tau _{2l}\hspace{0.2778em}\hspace{0.2778em}|\hspace{0.2778em}\hspace{0.2778em}{Y_{t}^{n}}\le L_{t}\big\}\wedge T,& l\ge 0\\{} \tau _{2l+2}=\inf \big\{t>\tau _{2l+1}\hspace{0.2778em}\hspace{0.2778em}|\hspace{0.2778em}\hspace{0.2778em}{Y_{t}^{n}}\ge U_{t}\big\}\wedge T,& l\ge 0.\end{array}\right.\]]]> Since Y, L and U are continuous processes and L<U , τl<τl+1 on the set {τl+1<T} . In addition the sequence (τl)l≥0 is of stationary type (i.e. ∀ω∈Ω , there exists l0(ω) such that τl0(ω)=T ). Indeed, let us set G={ω∈Ω,τl(ω)<T,l≥0} , and we will show that P(G)=0 . We assume that P(G)>0 0$]]>, therefore for ω∈G , we have Yτ2l+1≤Lτ2l+1 and Yτ2l≥Uτ2l . Since (τl)l≥0 is nondecreasing sequence then τl↗τ , hence Uτ≤Yτ≤Lτ which is contradiction since L<U . We deduce that P(G)=0 . Obviously Yn≥L on the interval [τ2l,τ2l+1] , then the BSDE (17) becomes Yτ2ln=Yτ2l+1n+∫τ2lτ2l+1f(s)ds−n∫τ2lτ2l+1(Ysn−Us)+ds−∫τ2lτ2l+1ZsndBs. On the other hand, using the assumption (H7) , we get Yτ2ln≥Hτ2lon{τ2l<T}andYτ2ln=Hτ2l=ξon{τ2l=T},Yτ2l+1n≤Hτ2l+1on{τ2l+1<T}andYτ2l+1n=Hτ2l+1=ξon{τ2l+1=T}. From (22) and the definition of process H we obtain n∫τ2lτ2l+1(Ysn−Us)+ds≤Hτ2l+1−Hτ2l+∫τ2lτ2l+1f(s)ds−∫τ2lτ2l+1ZsndBs≤∫τ2lτ2l+1(hs−Zsn)dBs+∫τ2lτ2l+1|f(s)|ds+Vτ2l+1−−Vτ2l−. By summing in l, using the fact that Yn≤U on the interval [τ2l+1,τ2l+2] , we can write for t≤T E[n∫tT(Ysn−Us)+ds]2≤4(E∫tT|hs|2ds+E∫tTeβAs|Zsn|2ds+TβE∫tTeβAs|f(s)|2a2(s)ds+E|VT−|2). In the same way, we obtain E[n∫tT(Ysn−Ls)−ds]2≤4(E∫tT|hs|2ds+E∫tTeβAs|Zsn|2ds+TβE∫tTeβAs|f(s)|2a2(s)ds+E|VT+|2). Combining (23), (24) with (21), we obtain the desired result.  □

 

  Esup0≤t≤TeβA(t)|(Ytn−Ut)+|2 →n→+∞0 .

  Esup0≤t≤TeβA(t)|(Ytn−Lt)−|2 →n→+∞0 .

Consider the following BSDE for each n∈N Yˆtn=ξ+∫tTf(s)ds+n∫tT(Ls−Yˆsn)ds−∫tTZˆsndBs=ξ+∫tTf(s)ds+n∫tT(Yˆsn−Ls)−ds−n∫tT(Ls−Yˆsn)−ds−∫tTZˆsndBs. By the Remark 2, we have Ytn≥Yˆtn for all t≤T . Let ν be a stopping time such that ν≤T . Then Yˆνn=E[e−n(T−ν)ξ+∫νTe−n(s−ν)f(s)ds+n∫νTe−n(s−ν)Lsds|Fν]. It is easily seen that e−n(T−ν)ξ+n∫νTe−n(s−ν)Lsds →n→+∞ξ1ν=T+Lν1ν<TP-a.s. inL2. Moreover, the conditional expectation converges also in L2 . In addition, by the Hölder inequality, we have |∫νTe−n(s−ν)f(s)ds|2≤(∫νTeβA(s)|f(s)a(s)|2ds)(∫νTe−2n(s−ν)−βA(s)a2(s)ds) →n→+∞0. Thus ∫νTe−n(s−ν)f(s)ds →n→+∞0 P-a.s. inL2 .

Now, we denote yˆtn:=e−n(T−t)ξ+∫tTe−n(s−t)(f(s)+nLs)ds,y˜tn:=e−n(T−t)LT+∫tTe−n(s−t)(f(s)+nLs)ds and Xtn:=e−n(T−t)LT+n∫tTe−n(s−t)Lsds−Lt. By the fact that L is uniformly continuous on [0,T] , it can be shown that the sequence (Xtn)n≥1 uniformly converges in t, and the same for (Xtn−)n≥1 . Lebesgue’s dominated convergence theorem implies that limn→+∞Esup0≤t≤TeβA(t)|(yˆtn−Lt)−|2=limn→+∞Esup0≤t≤TeβA(t)|(y˜tn−Lt)−|2≤2limn→+∞E[sup0≤t≤TeβA(t)|Xtn−|2+sup0≤t≤TeβA(t)|∫tTe−n(s−t)f(s)ds|2]=0. So, from (25), Jensen’s inequality and Doob’s maximal quadratic inequality (see Theorem 20, p. 11 in [23]), we have Esup0≤t≤TeβA(t)|(Yˆtn−Lt)−|2≤Esup0≤t≤TeβA(t)|E[(yˆtn−Lt)−|Ft]|2≤4Esup0≤t≤TeβA(t)|(yˆtn−Lt)−|2 →n→+∞0. From the fact that Ytn≥Yˆtn for all t≤T we deduce that limn→+∞Esup0≤t≤TeβA(t)|(Ytn−Lt)−|2=0. Similarly to proof of the Lemma 5, we can obtain limn→+∞Esup0≤t≤TeβA(t)|(Ytn−Ut)+|2=0.  □

For each n≥p≥0 , we have E[sup0≤t≤TeβA(t)|Ytn−Ytp|2+∫0TeβA(t)a2(t)|Ytn−Ytp|2dt+∫0TeβA(t)|Ztn−Ztp|2dt+sup0≤t≤T|Ktn−Ktp|2] →n,p→+∞0.

Itô’s formula implies that EeβA(t)|Ytn−Ytp|2+E∫tTeβA(s)(βa2(s)|Ysn−Ysp|2+|Zsn−Zsp|2)ds≤2E∫tTeβA(s)(Ysn−Ysp)(dKsn+−dKsp+)−2E∫tTeβA(s)(Ysn−Ysp)(dKsn−−dKsp−)≤2E∫tTeβA(s)(Ysn−Ls)−dKsp++2E∫tTeβA(s)(Ysp−Ls)−dKsn++2E∫tTeβA(s)(Ysn−Us)+dKsp−+2E∫tTeβA(s)(Ysp−Us)+dKsn−. Hence βE∫tTeβA(s)a2(s)|Ysn−Ysp|2ds+E∫tTeβA(s)|Zsn−Zsp|2ds≤2Esup0≤t≤TeβA(t)(Ytn−Lt)−KTp++2Esup0≤t≤TeβA(t)(Ytp−Lt)−KTn++2Esup0≤t≤TeβA(t)(Ytn−Ut)+KTp−+2Esup0≤t≤TeβA(t)(Ytp−Ut)+KTn−. Lemma 8 implies that E∫tTeβA(s)a2(s)|Ysn−Ysp|2ds+E∫tTeβA(s)|Zsn−Zsp|2)ds →n,p→+∞0. On the other hand, by the Burkholder–Davis–Gundy’s inequality, we get Esup0≤t≤TeβA(t)|Ytn−Ytp|2 →n,p→+∞0. From the equation Ktn=Y0n−Ytn−∫0tf(s)ds+∫0tZsndBs0≤t≤T, we can conclude that Esup0≤t≤T|Ktn−Ktp|2 →n,p→+∞0. The proof is completed.  □

The main result of this section is the following:

Assume that L<U . Then the DRBSDE (3) has a unique solution (Y,Z,K+,K−) that belongs to (S2(β,a)∩S2,a(β,a))×H2(β,a)×S2×S2 .

From Lemma 9, we obtain that there exists an adapted process (Y,Z,K)∈(S2(β,a)∩S2,a(β,a))×H2(β,a)×S2 such that E[sup0≤t≤TeβA(t)|Ytn−Yt|2+∫0TeβA(t)a2(t)|Ytn−Yt|2dt+∫0TeβA(t)|Ztn−Zt|2dt+sup0≤t≤T|Ktn−Kt|2] →n→+∞0. Then, passing to the limit as n→+∞ in the equation Ytn=ξ+∫tTf(s)ds+KTn−Ktn−∫tTZsndBs, we obtain Yt=ξ+∫tTf(s)ds+KT−Kt−∫tTZsdBs. Let τ≤T be a stopping time, by Lemma 7 we obtain that the sequences Kτn± are bounded in L2 , consequently, there exist Fτ -measurable random variables Kτ± in L2 , such that there exist the subsequences of Kτn± weakly converging in Kτ± .

Now we set Kτ=Kτ+−Kτ− . By [28] (Mazu’s Lemma, p. 120), there exists, for every n∈N , an integer N≥n and a convex combination ∑j=nNζjτ,n(Kτ±)j with ζjτ,n≥0 and ∑j=nNζjτ,n=1 such that Kτn±:= ∑j=nNζjτ,n(Kτ±)j →n→+∞Kτ±. Denoting Kτn=Kτn+−Kτn− , it follows that E|Kτn−Kτ|2 →n→+∞0. Thanks to (30), we have ‖Kτn−Kτ‖L2<ε for all ε>0 0$]]>. Therefore ‖Kτn−Kτ‖L2=‖ ∑j=nNζjτ,n((Kτ±)j−Kτ)‖L2≤ ∑j=nNζjτ,n‖(Kτ±)j−Kτ‖L2<ε. Hence E|Kτn−Kτ|2 →n→+∞0. Combining (32) and (33), we obtain Kτ=Kτ a.s. Therefore, from Theorem 86, p. 220 in [7] we have Kt=Kt for all t≤T . On the other hand, (31) implies that, for τ=T , there exists a subsequence of KTn+:=∑j=nNζjT,n(KT+)j (resp. KTn−:=∑j=nNζjT,n(KT−)j ) converging a.s. to KT+ (resp. KT− ). Then for P -a.s. ω∈Ω , the sequence KTn+(ω) (resp. KTn−(ω) ) is bounded. Using Theorem 4.3.3, p. 88 in [4], there exists a subsequence of Ktn+(ω) (resp. Ktn−(ω) ) tending to Kt+(ω) (resp. Kt−(ω) ), weakly.

On the other hand, by the definition of stopping times (τl)l≥0 , we have Ytn>Lt,on[τ2l,τ2l+1[;Ytn<Ut,on[τ2l+1,τ2l+2[. L_{t},& on[\tau _{2l},\tau _{2l+1}[;\\{} {Y_{t}^{n}} Then Lt1[τ2i,τ2i+1](t)≤Ytn≤Ut1[τ2i+1,τ2i+2](t). By summing in i, i=0,…,l and passing to limit in n, we obtain Lt≤Yt≤Ut . Now, we would have to show the Skorokhod’s conditions. Indeed, since Ktn+(ω) tends to Kt+(ω) , using the result treated in p. 465 of [25] we can write ∫0T(Ytn(ω)−Lt(ω))dKtn+(ω) →n→+∞∫0T(Yt(ω)−Lt(ω))dKt+(ω). Since ∫0T(Ytn−Lt)dKtn+≤0 , ∀n≥0 a.s., and ∀n,m≥0 , n≠m , E[|∫0T(Ytn−Ytm)dKtm+|]≤E[sup0≤t≤TeβA(t)|Ytn−Ytm|KTm+] →n,m→+∞0, then by ∫0T(Ytn−Lt)dKtm+=∫0T(Ytn−Ytm)dKtm++∫0T(Ytm−Lt)dKtm+ we have lim supn→+∞∫0T(Ytn−Lt)dKtn+≤0P-a.s. Combining (34) and (35), we get ∫0T(Yt−Lt)dKt+≤0P-a.s. Noting that Y≥L , we conclude that ∫0T(Yt−Lt)dKt+=0 . By a similar consideration, we can prove ∫0T(Ut−Yt)dKt−=0 .

Finally, using the fixed point theorem we construct a strict contraction mapping φ on B2 and conclude that (Yt,Zt,Kt+,Kt−) is the unique solution to DRBSDE (3) associated with data (ξ,f,L,U) .  □

 Comparison theorem

In this section we prove a comparison theorem for the DRBSDE under the stochastic Lipschitz assumptions on generators.

Let (Y1,Z1,K1+,K1−) and (Y2,Z2,K2+,K2−) be respectively the solutions to the DRBSDE with data (ξ1,f1,L1,U1) and (ξ2,f2,L2,U2) . Assume in addition the following:

ξ1≤ξ2 a.s.

f1(t,Y2,Z2)≤f2(t,Y2,Z2)   ∀t∈[0,T] a.s.

Lt1≤Lt2 and Ut1≤Ut2   ∀t∈[0,T] a.s.

 Then ∀t≤T,Yt1≤Yt2a.s.

Let ℜ¯=ℜ1−ℜ2 for ℜ=Y,Z,K+,K+,ξ and

ζt=1{Y¯t≠0}f1(t,Yt1,Zt1)−f1(t,Yt2,Zt1)Y¯t ;

ηt=1{Z¯t≠0}f1(t,Yt2,Zt1)−f1(t,Yt2,Zt2)Z¯t ;

δt=f1(t,Yt2,Zt2)−f2(t,Yt2,Zt2) .

 Applying the Meyer–Itô formula (Theorem 66, p. 210 in [23]), there exists a continuous nondecreasing process (At)t≤T such that |Y¯t+|2=2∫tTY¯s+(ζsY¯s+ηsZ¯s+δs)ds−2∫tTY¯s+Z¯sdBs+2∫tTY¯s+dK¯s+−2∫tTY¯s+dK¯s−−(AT−At). Suppose in addition that E∫0Tμtdt<+∞andE∫0T|γt|2dt<+∞. Let {Γt,s,0≤t≤s≤T} be the process defined as Γt,s=exp{∫ts(ζu−12|ηu|2)du+∫tsηudBu}>0 0\]]]> being a solution to the linear stochastic differential equation Γt,s=1+∫tsζuΓt,udu+∫tsηuΓt,udBu. Applying the integration by parts and taking expectation yield E[eβA(t)|Y¯t+|2]+βE∫0TeβA(s)Γt,sa2(s)|Y¯s+|2ds≤E[∫tTeβA(s)Γt,sζs|Y¯s+|2ds]+2E[∫tTeβA(s)Γt,sδsY¯s+ds]+2E∫tTeβA(s)Γt,sY¯s+dKs+−2E∫tTeβA(s)Γt,sY¯s+dKs−. Remark that Y¯s+dK¯s+=(Ls1−Ys2)1Ys1>Ys2dKs1+−(Ys1−Ls2)1Ys1>Ys2dKs2+≤0 {Y_{s}^{2}}}d{K_{s}^{1+}}-\big({Y_{s}^{1}}-{L_{s}^{2}}\big)\mathbb{1}_{{Y_{s}^{1}}>{Y_{s}^{2}}}d{K_{s}^{2+}}\le 0\]]]> and Y¯s+dK¯s−=(Ys1−Us2)1Ys1>Ys2dKs2−−(Us1−Ys2)1Ys1>Ys2dKs1−≤0. {Y_{s}^{2}}}d{K_{s}^{2-}}-\big({U_{s}^{1}}-{Y_{s}^{2}}\big)\mathbb{1}_{{Y_{s}^{1}}>{Y_{s}^{2}}}d{K_{s}^{1-}}\le 0.\]]]> Since δs≤0 and |ζs|≤a2(s) , one can derive that E[eβA(t)|Y¯t+|2]≤0. It follows that Y¯t+=0 , i.e Yt1≤Yt2 for all t≤T a.s.  □

If Ui=+∞ for i=1,2 , then dKi−=0 and the comparison holds also for the reflected BSDE (2).

If Ui=+∞ and Li=−∞ for i=1,2 , then dKi±=0 and the comparison holds also for the BSDE (1).

Appendix

In this section, we study a special case of the reflected BSDE when the generator depends only on y.

We consider the following reflected BSDE Yt=ξ+∫tTf(s,Ys)ds+KT−Kt−∫tTZsdBsYt≥Lt∀t≤Tand∫0T(Yt−Lt)dKt=0 where (ξ,f,L) satisfies the following assumptions:

ξ∈S2(β,a) ;

f is Lipschitz, i.e. there exists a positive constant μ such that ∀(t,y,y′)∈[0,T]×R×R |f(t,y)−f(t,y′)|≤μ|y−y′|;

f(t,0)a∈H2(β,a) ;

E[sup0≤t≤Te2βA(t)|Lt+|2]<+∞ .

 As in [11], we prove the existence and uniqueness of a solution to (36) by means of the penalization method. Indeed, for each n∈N , we consider the following BSDE: Ytn=ξ+∫tTf(s,Ysn)ds+n∫tT(Ysn−Ls)−ds−∫tTZsndBs. We denote Ktn:=n∫0t(Ysn−Ls)−ds and fn(t,y)=f(t,y)+n(y−Lt)− . Remark that fn is Lipschitz and E|ξ|2+E∫0T|fn(t,0)|2dt≤E[eβA(T)|ξ|2]+2βE[∫0TeβA(t)|f(t,0)a(t)|2dt]+2n2TE[sup0≤t≤Te2βA(t)|Lt+|2]. From [21], there exists a unique process (Yn,Zn) being a solution to the BSDE (37). The sequence (Yn,Zn,Kn)n satisfies the uniform estimate Esup0≤t≤TeβA(t)|Ytn|2+E[∫0TeβA(s)a2(s)|Ysn|2ds+E∫0TeβA(s)|Zsn|2ds]≤CE[eβA(T)|ξ|2+∫0TeβA(s)|f(s,0)|2a2(s)ds+sup0≤t≤Te2βA(s)|Ls+|2]. where C is a positive constant depending only on β, μ and ϵ.

Now we establish the convergence of sequence (Yn,Zn,Kn) to the solution to (36). Obviously fn(t,y)≤fn+1(t,y) for each n∈N , and it follows from Remark 2 that Yn≤Yn+1 . Hence there exists a process Y such that Ytn↗Yt 0≤t≤T a.s. From the a priori estimates and Fatou’s lemma, we have E[sup0≤t≤TeβA(t)|Yt|2]≤lim infn→+∞E[sup0≤t≤TeβA(t)|Ytn|2]≤C. Then by the dominated convergence, one can derive that E[∫0TeβA(s)|Ysn−Ys|2ds] →n→+∞0. On the other hand, for all n≥p≥0 and t≤T , we have EeβA(t)|Ytn−Ytp|2+(β−2μϵ)E∫tTeβA(s)a2(s)|Ysn−Ysp|2ds+E∫tTeβA(s)|Zsn−Zsp|2ds≤2E∫tTeβA(s)(Ysn−Ls)−dKsp+E∫tTeβA(s)(Ysp−Ls)−dKsn. Similarly to Lemma 8, we can easily prove that Esup0≤t≤TeβA(t)|(Ytn−Lt)−|2 →n→+∞0. By the above result an the a priori estimates, one can derive that E[∫tTeβA(s)(Ysn−Ls)−dKsp+∫tTeβA(s)(Ysp−Ls)−dKsn] →n,p→+∞0. Thus E[∫tTeβA(s)a2(s)|Ysn−Ysp|2ds+∫tTeβA(s)|Zsn−Zsp|2ds] →n,p→+∞0. Moreover, by the Burkholder–Davis–Gundy’s inequality, one can derive that E[sup0≤t≤TeβA(t)|Ytn−Ytp|2] →n,p→+∞0. Further, from the equation (37), we have also E[sup0≤t≤T|Ktn−Ktp|2] →n,p→+∞0. Consequently there exists a pair of progressively measurable processes (Z,K) such that E∫0TeβA(t)|Ztn−Zt|2dt+Esup0≤t≤T|Ktn−Kt|2 →n→+∞0. Obviously the triplet (Y,Z,K) satisfies (36). It remains to check the Skorokhod condition. We have just seen that the sequence (Yn,Kn) tends to (Y,K) uniformly in t in probability. Then the measure dKn tends to dK weakly in probability, hence ∫0T(Ytn−Lt)dKtn →n→+∞P∫0T(Yt−Lt)dKt. We deduce from the equation (38) that ∫0T(Ytn−Lt)dKtn≤0 , n∈N , which implies that ∫0T(Yt−Lt)dKt≤0 . On the other hand, since Yt≥Lt then ∫0T(Yt−Lt)dKt≥0 . Hence ∫0T(Yt−Lt)dKt=0 .

(Special cases).

The coefficients gn(s,y)=g(s)−n(y−Us)+ and g˜n(s,y)=g(s)−n(y−Us) are Lipschitz and satisfy E∫0TeβA(s)|gn(s,0)a(s)|2ds+E∫0TeβA(s)|g˜n(s,0)a(s)|2ds≤4E∫0TeβA(s)|g(s)a(s)|2ds+4n2TϵE[supn≥0e2βA(t)|Ut−|2]<+∞. Then the Reflected BSDEs (10) and (11) have a unique solution.

 (Comparison theorem).

Let (Y1,Z1,K1) and (Y2,Z2,K2) be solutions to the Reflected BSDE (36) with data (ξ1,f1,L) and (ξ2,f2,L) respectively. If we have

f1(t,y)≤f2(t,y) a.s. ∀(t,y) ,

ξ1≤ξ2 a.s.,

 then Yt1≤Yt2 and Kt1≥Kt2 ∀t∈[0,T] a.s.

We consider the penalized equations relative to the Reflected BSDE with data (ξi,fi,L) for i=1,2 and n∈N , as follows Ytn,i=ξi+∫tTfi(s,Ysn,i)ds+n∫tT(Ysn,i−Ls)−−∫tTZsn,idBs. Let fni(t,y):=fi(t,y)+n(y−Ls)− . So, by the comparison theorem, we have Ytn,1≤Ytn,2 for t≤T . Since Ktn,i=n∫0t(Ysn,i−Ls)−ds for i=1,2 , we deduce that Ktn,1≥Ktn,2 for t≤T . But Ytn,i↗Yti and Ktn,i⟶Kti as n⟶+∞ for i=1,2 , and it follows that Yt1≤Yt2 and Kt1≥Kt2 for t≤T .  □

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