Optimal Switching of One-Dimensional Reflected BSDEs, and Associated Multi-Dimensional BSDEs with Oblique Reflection

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1 arxv: v1 [mah.pr] 17 Oc 28 Opmal Swchng of One-Dmenonal Refleced BSDE, and Aocaed Mul-Dmenonal BSDE wh Oblque Reflecon Shanjan Tang We Zhong Ocober 17, 28 Abrac In h paper, he opmal wchng problem propoed for one-dmenonal refleced backward ochac dfferenal equaon (BSDE, for hor) where he generaor, he ermnal value, and he barrer are all wched wh pove co. The value proce characerzed by a yem of mul-dmenonal refleced BSDE wh oblque reflecon, whoe exence and unquene by no mean rval and herefore carefully examned. Exence hown ung boh mehod of he Pcard eraon and penalzaon, bu under ome dfferen condon. Unquene proved by repreenaon eher a he equlbrum value proce o a ochac mxed game of wchng and oppng, or a he value proce o our opmal wchng problem for one-dmenonal refleced BSDE. Keyword. Refleced backward ochac dfferenal equaon, oblque reflecon, opmal wchng, ochac game. MR(2) Subjec Clafcaon. 93E2, 6H1, 9A15 1 Inroducon Le {W(), T } be a d-dmenonal andard Brownan moon defned on ome complee probably pace (Ω, F,P). {F, T } he naural flraon of he Brownan moon {W(), T }, augmened by all he P-null e of F. The followng noaon wll be ued n he equel. { φ : φ an {F, T }-adaped r.c.l.l. proce..e[ up φ() 2 ] < T S 2 {φ S 2 : φ connuou}, N 2 {φ S 2 : φ ncreang and φ() = }, N 2 {φ N 2 : φ connuou}, M 2 {φ : φ {F, T }-predcable and quare-negrable on [,T] Ω}. S 2 }, Deparmen of Fnance and Conrol Scence, School of Mahemacal Scence, Fudan Unvery, Shangha 2433, Chna. Th auhor uppored n par by NSFC Gran , Bac Reearch Program of Chna (973 Program) Gran 27CB81494, and Chang Jang Scholar Programme. E-mal addre: jang@fudan.edu.cn Inue of Mahemac and Deparmen of Fnance and Conrol Scence, School of Mahemacal Scence, Fudan Unvery,Shangha 2433, Chna. E-mal addre: zhongwe@fudan.edu.cn

2 Le {θ j } j= be an ncreang equence of oppng me wh value n [,T]. j, α j an F θj -meaurable random varable wh value n Λ. Aume ha a..ω, here ex an neger N(ω) < uch ha θ N = T. Then we defne a wchng raegy a: N 1 a() = α χ [θ, θ 1 ]() + α j χ (θj, θ j+1 ](). Denoe by A all he wchng raege wh nal daa α = Λ, θ =. For gven a A,ξ L 2 (Ω, F T,P;R m ), and S (S 2 ) m, conder he followng wched refleced backward ochac dfferenal equaon (abbrevaed a RBSDE): U a () = ξ a(t) + ψ(r,u a (r),v a (r),a(r))dr (L a (T) L a ()) N 1 j=1 [U a (θ j ) h αj 1, α j (θ j,u a (θ j ))]χ (,T] (θ j ) j=1 V a (r)dw(r), U a () S a() (), [,T]; [,T]; (U a () S a() ())dl a () =. (1.1) Here and n he followng χ an ndcaor funcon. The generaor ψ, he ermnal condon ξ, and he upper barrer S of RBSDE (1.1) are all wched by a. A each wchng me θ j before ermnaon, he value of U a wll jump by an amoun of U a (θ j ) h αj 1, α j (θ j,u a (θ j )) whch can be regarded a a penaly or co for he wchng. RBSDE (1.1) can be olved n a backwardly recurve way n he ubnerval [θ N 1,T] and [θ j 1,θ j ) for j = N 1,,2,1. To be prece, RBSDE (1.1) n he la ubnerval [θ N 1,T] read: U a () = ξ αn 1 + ψ(r,u a (r),v a (r),α N 1 )dr (L a (T) L a ()) V a (r)dw(r), U a () S αn 1 (), [θ N 1,T]; (U a () S αn 1 ())dl a () =. θ N 1 [θ N 1,T]; (1.2) From [3, Theorem 5.2], RBSDE (1.2) ha a unque adaped oluon on [θ N 1,T] under Hypohe 1 (ee Secon 2 below). Then we have U a (θ N 1 ) = h αn 2,α N 1 (θ N 1,U a (θ N 1 )), whch erve a he ermnal value of RBSDE (1.1) n [θ N 2,θ N 1 ). In general, n

3 he ubnerval [θ j 1,θ j ), RBSDE (1.1) read θj U a () = h αj 1, α j (θ j,u a (θ j )) + ψ(r,u a (r),v a (r),α j 1 )dr (L a (θ j ) L a ()) θj U a () S αj 1 (), [θ j 1,θ j ); θj (U a () S αj 1 ())dl a () = θ j 1 V a (r)dw(r), [θ j 1,θ j ); (1.3) for j = N 1,,2,1. Here U a (θ j ) pecfed n he nerval [θ j,θ j+1 ) and we have he followng relaon: h αj 1, α j (θ j,u a (θ j )) U a (θ j ) S αj 1 (θ j ). The exence and unquene of an adaped oluon o RBSDE (1.1) n he nerval [,T] obaned n an obvou way from he exence and unquene of an adaped oluon o RBSDE (1.1) n all he ubnerval [θ N 1,T] and [θ j 1,θ j ) for j = N 1,,2,1. In h paper, we udy he opmal wchng problem for one-dmenonal RB- SDE (1.1), where he generaor, he ermnal value, and he upper barrer are all wched wh pove co. One-dmenonal RBSDE wh fxed ngle reflecng barrer are a generalzaon of radonal opmal oppng problem, and hey were nroduced by El Karou e al.[3], who gave he fr exence and unquene reul for RBSDE. Cvanc and Karaza[1] generalzed he work of El Karou e al.[3] o he cae of fxed double reflecng barrer and lnked he oluon of one-dmenonal RBSDE wh double barrer o he well-known Dynkn game. The opmal wchng problem for RBSDE (1.1) o maxmze U a () over a A, Λ. The value proce urn ou o be decrbed by he followng yem of mul-dmenonal RBSDE wh double reflecng barrer: for Λ {1,,m}, Y () = ξ + + ψ(,y (),Z (),)d dk () dk + () Z ()dw(), [,T]; S () Y () max h,j(,y j ()), [,T]; j,j Λ ( ) Y () max h,j(,y j ()) dk + () =, j,j Λ (Y () S ()) dk () =. (1.4) The la wo equale are repecvely called he lower and he upper mnmal condon. Soluon of he above RBSDE (1.4) by no mean rval, and wll be examned carefully n h paper. The unuual feaure here ha for RBSDE (1.4), he upper barrer a fxed proce, whle he lower barrer depend on he unknown proce and herefore mplc, whch dfferen from one-dmenonal RBSDE wh fxed double barrer. In conra o RBSDE wh oblque reflecon nroduced n Hu and Tang [1], here an addonal fxed upper barrer. Th dfference wll complcae he analy of he exence and unquene for oluon o

4 RBSDE (1.4). For [,T], defne Q() {(y 1,,y m ) T R m : h,j (,y j ) y S (),,j Λ,j }. Then he ae proce Y ( ) of (1.4) forced o evolve n he me-dependen e Q( ), hank o he accumulave acon of wo ncreang procee K + and K. Exence of he oluon of RBSDE (1.4) proved by wo dfferen mehod. Aumng ha he fxed barrer uper-regular, we frly prove exence of he oluon by he mehod of a Pcard eraon, nvokng a generalzed monoonc lm heorem. A a key condon of he generalzed monoonc lm heorem, he comparon of he dfferenal of he ncreang proce neceary, whch formulaed a our Lemma 2.2. In addon, he proof of he mnmal boundary condon complcaed by he appearance of he addonal fxed barrer, and our mehod very echncal and novel. Secondly we conder he cae where he ner-conneced barrer ake a parcular form, and oban an exence reul by a penaly mehod. A pror emae play a crucal role heren. Unquene of he oluon o RBSDE (1.4) proved n Secon 5 by lnkng eher o he value proce for our opmal wchng of one-dmenonal RBSDE, or o a ochac game nvolvng boh wchng and oppng raege for onedmenonal BSDE. Recenly, Hu and Tang [1] nally formulaed and dcued he opmal wchng problem for general one-dmenonal BSDE. The value proce characerzed by he oluon of mul-dmenonal BSDE wh oblque reflecon. Laer, Hamadene and Zhang [8] generalzed he precedng work o a general form of pove co for wchng, wh a dfferen mehod of Pcard eraon. The paper a naural connuaon of he wo work. The re of he paper organzed a follow: In Secon 2, we formulae our problem, nroduce he generalzed monoonc heorem and gve ome prelmnary reul on RBSDE, whch wll be ued n ubequen argumen. In Secon 3, we prove exence of he oluon by he mehod of Pcard eraon. In Secon 4, exence of he oluon hown by he penaly mehod. Unquene of he oluon hown n Secon 5. 2 Prelmnare We make he followng aumpon on he generaor {ψ(,,,), Λ}. Hypohe 1. The generaor ψ afe he followng: () The proce ψ(,y,z,) M 2 for any (y,z,) R R d Λ. () There a conan C > uch ha for (y,y,z,z ) R R R d R d and (,) [,T] Λ, we have ψ(,y,z,) ψ(,y,z,) C( y y + z z ). We make he followng aumpon on he funcon {h,j,,j Λ}. Hypohe 2. For any (,j) Λ Λ, he funcon h,j (,y) connuou n (,y), ncreang n y, and h,j (,y) y. Hypohe 3. For any y n R and any loop { k Λ, k = 1,,n} uch ha 1 = n and k k+1 for k = 1,2,...,n 1, defne y k h k, k+1 (,y k+1 ) for k = 1,,n 1. Then y 1 < y n.

5 In Secon 4, we hall pecalze he funcon h,j o he form: h,j (,y) = y k(,j) for ome povely valued funcon k defned on Λ Λ. We hall make he followng aumpon on k, whch andard n he leraure. Hypohe 3. The funcon k : Λ Λ R afe he followng: (),j Λ, k(,j) > for j, and k(,) =. (),j,l Λ uch ha j, j l, k(,j) + k(j,l) k(,l). Remark 2.1. Hypohe 3 mean ha here no free loop of nananeou wchng. Hypohee 2 and 3 are afed when h,j (,y) = y k(,j) for (,y) [, T] R and, j Λ wh he funcon k afyng Hypohe 3 (). Defnon 2.1. An adaped oluon of yem (1.4) a quadruple (Y,Z,K +,K ) {Y (),Z(),K + (),K (); T } (S 2 ) m (M 2 ) m (N 2 ) m (N 2 ) m, akng value n R m R m d R m R m and afyng (1.4). We recall here he generalzed monoonc lm heorem [14, Theorem 3.1], whch wll be ued n our mehod of Pcard eraon. Lemma 2.1. (generalzed monoonc heorem) We aume he followng equence of Iô procee: y () = y () + g ()d K +, () + K, () + z ()dw(), = 1,2, afy () for each, g M 2,K +, N 2,K, N 2 ; () K,j () K,j () K, () K, (), T, j; () For each [,T], {K, ()} =1 ncreangly converge o K () wh E K (T) 2 < ; (v) (g,z ) =1 converge o (g,z) weakly n M2 ; (v) For each [,T], {y ()} =1 ncreangly converge o y() wh E up y() 2 <. T Then he lm y of {y } =1 ha he followng form y() = y() + g()d K + () + K () + z()dw() where K + (rep. K ) he weak (rep. rong) lm of {K +, } =1 (rep. {K, } =1 ) n M 2 and K +,K N 2. Moreover, for any p [,2), lm E z () z() p d =. If furhermore, K + connuou, hen we have lm E z () z() 2 d =.

6 When we apply he above generalzed monoonc heorem o RBSDE (1.4), we need o compare he dfferenal of he ncreang proce K for Λ. However, uch a knd of conderaon doe no eem o be avalable n he leraure due o he appearance of he lower barrer, whch mplc and hu varyng wh he fr unknown varable. The followng lemma fll n uch a gap, whch wll be ued n Secon 3. Aume ha ξ L 2 (Ω, F T,P) and L and U are {F, T }-adaped connuou procee afyng E[ up { L() U() 2 }] <, L() U(), [,T]. T Conder he followng one-dmenonal RBSDE wh fxed double reflecng barrer: Ŷ () = ξ + L() + ψ(,ŷ (),Ẑ())d d ˆK () d ˆK + () Ŷ () U(), [,T]; (Ŷ () L())d ˆK + () =, Ẑ()dW(), [,T]; (Ŷ () U())d ˆK () =. (2.1) Defnon 2.2. A barrer U called uper-regular f here ex a equence of procee {U n } n=1 uch ha () U n () U n+1 () and lm Un () = U() for [,T]; () For n 1 and [,T], we have du n () = u n ()d + v n ()dw() where u n an {F, T }-adaped proce uch ha up up n 1 T u n () < and v n M 2. A barrer V called ub-regular f he barrer V uper-regular. Noe ha he concep of our uper-regular barrer dencal o he defnon of he regular upper barrer by Hamadène e al. [5]. Lemma 2.2. Aume ha ψ 1 and ψ 2 afy Hypohe 1 and ha he barrer U uper-regular. Aume ha ξ 1,ξ 2 L 2 (Ω, F T,P), and L 1,L 2, and U are {F, T }-adaped connuou procee afyng E[ up { L j () U() 2 }] <, L j () U(), [,T], j = 1,2. T For j = 1,2, le (Ŷ j,ẑj, ˆK +,j, ˆK,j ) be he unque adaped oluon of RBSDE (2.1) aocaed wh daa (ξ j,ψ j,l j,u). Moreover, aume ha () ξ 1 ξ 2 ; () ψ 1 (,y,z) ψ 2 (,y,z), (y,z) R R d ; () L 1 L 2. Then we have (1) Ŷ 1 () Ŷ 2 (), ˆK,1 () ˆK,2 (), T; (2) ˆK,1 (r) ˆK,1 () ˆK,2 (r) ˆK,2 (), r T.

7 Proof. For j = 1,2, and n 1, he followng penalzed RBSDE wh a ngle reflecng barrer: Ŷ j,n () = ξ j + ψ j (,Ŷ j,n (),Ẑj,n ())d n (Ŷ j,n () U()) + d + d ˆK j,n () Ŷ j,n () L j (), [,T]; (Ŷ j,n () L j ())d ˆK j,n () = Ẑ j,n ()dw(), [,T]; ha a unque adaped oluon, denoed by (Ŷ j,n,ẑj,n, ˆK j,n ). In vew of he comparon heorem [3, Theorem 4.1], we have Ŷ 1,n () Ŷ 2,n (), T. Nong ha he barrer U uper-regular, by he proof of [5, Theorem 42.2 and Remark 42.3], we have Ŷ j () = lm Ŷ j,n (), The dered reul hen follow. ˆK,j () = lm n (Ŷ j,n () U()) + d, [,T]. Remark 2.2. In a ymmerc way, aumng ha he lower barrer ub-regular and fxed, we can compare he ncremen of ˆK+ when he upper barrer vare. The followng lemma gve he connuou dependency of RBSDE wh one r.c.l.l. (rgh connuou and lef lmed) reflecng barrer, whch wll be ued o prove he lower mnmal boundary condon. Lemma 2.3. Aume ha ξ j L 2 (Ω, F T,P), ψ j afe Hypohe 1, and L j S 2 for j = 1,2. For j = 1,2, denoe by (Y j,z j,k j ) S 2 M 2 N 2 he unque adaped oluon of he followng RBSDE: Se Y j () = ξ j + T ψ j (,Y j (),Z j ())d + Z j ()dw(), Y j () L j (), [,T]; (Y j ( ) L j ( ))dk j () =. Y j () Y j () Y j ( ), [,T]; [,T] for j = 1,2. Then here a conan c > uch ha ( ) E up Y 1 () Y 2 () 2 + Z 1 () Z 2 () 2 d T +E ( K 1 (T) K 2 (T) 2 + Y 1 () Y 2 () 2) T dk j () ( ) ce ξ 1 ξ ψ 1 (,Y 1 (),Z 1 ()) ψ 2 (,Y 1 (),Z 1 ()) 2 d +c(e[ up L 1 () L 2 () 2 ]) 2Φ(T) T

8 where Φ(T) [ 2 E ξ j 2 + j=1 ] ψ j (,,) 2 d + up L j () + 2. T Remark 2.3. Lemma 2.3 can be exended o he mul-dmenonal cae where for j = 1,2, Y j,ξ j,ψ j,k j,s j, and L j are all R m -valued, Z j R m d -valued, and z race(zz T ) for z R m d. The proof mlar o [3, Propoon 3.6] and omed. 3 Exence: he mehod of Pcard eraon We have he followng exence reul for RBSDE (1.4). Theorem 3.1. Le Hypohee 1, 2 and 3 be afed. Aume ha he upper barrer S uper-regular wh S() Q() for [,T], and ha he ermnal value ξ L 2 (Ω, F T,P;R m ) ake value n Q(T). Then RBSDE (1.4) ha an adaped oluon (Y,Z,K +,K ). Proof. We ue he mehod of Pcard eraon. The whole proof dvded no he followng x ep. Sep 1. Conrucon of Pcard equence of oluon {Y n,zn,kn ; Λ} n. For Λ, he followng RBSDE wh ngle reflecng barrer: Y () = ξ + Y ψ(,y (),Z (),)d dk () Z ()dw(), () S (), [,T]; (Y () S ())dk () = [,T]; (3.1) ha a unque adaped oluon, denoed by (Y,Z,K ). For n 1 and Λ, conder he followng RBSDE: Noe ha Y n () = ξ + S () () () + ψ(,y n (),Zn (),)d dk +,n () Z n ()dw(), dk,n () [,T]; Y n() max h,j(,yj n 1 ()), [,T]; j,j Λ max h,j(,yj n 1 ()))dk +,n () =, j,j Λ S ())dk,n () =. (3.2) max h,j(,yj n 1 ()) max h,j(,s j ()) S (), j,j Λ j,j Λ [,T]

9 due o Hypohe 2 and he aumpon ha S() Q() for [,T]. In vew of [5, Theorem 42.2 and Remark 42.3], RBSDE (3.2) ha a unque oluon,zn,k+,n,k,n ) S 2 M 2 N 2 N 2. Sep 2. Convergence of {Y n,k,n ; Λ} n 1. Snce h j (, ) ncreang, he lower barrer of RBSDE (3.1) ncreang wh n by nducon. RBSDE (3.1) can be vewed a havng he lower barrer. Then from Lemma 2.2, we have Y () Y 1 (), T. Snce h,j (,y) ncreang n y, ung Lemma 2.2 agan, we have for n 1 and Λ, Y n n+1 () Y K,n (), K,n () K,n+1 (), T; (r) K,n () K,n+1 (r) K,n+1 (), r T. Hence, he equence {(),K,n ())} n=1 ha a lm, denoed by (Y (),K ()). Snce Y () Y n () S (), T, Λ, we have upe[ up Y n () 2 ] E up ( Y () 2 + S() 2 ) <, Λ. (3.3) n 1 T T Applyng Faou lemma and he domnaed convergence heorem, we have E[ up Y () 2 ] <, lm E Y T n () Y () 2 d =, Λ. (3.4) For Λ, he followng RBSDE Ỹ () = ξ + + ψ(,ỹ(), Z (),)d d K + () Z ()dw(), S () Ỹ() max j,j Λ h,j(,s j ()), (Ỹ() max j,j Λ h,j(,s()))d K + (Ỹ() S ())d K () = () =, [,T]; d K () [,T]; (3.5) ha a unque adaped oluon, denoed by (Ỹ, Z, K +, K ) S2 M 2 N 2 N 2. By Lemma 2.2, we know ha for n 1 and Λ, K,n () K n+1 () K K,n Then, K (r) K,n upe[ up n 1 T (), T, () K,n+1 (r) K,n+1 () K (r) K (), r T. connuou for Λ. Hence, K,n () 2 ] E[ up T K () 2 ] E[ up T K () 2 ] <, Λ. (3.6)

10 From Dn heorem, we have lm up T K,n () K () 2 =, Λ. From he domnaed convergence heorem, we have lm E[ up K,n () K () 2 ] =, Λ. (3.7) T Sep 3. Unform boundedne of {ψ(,y n,zn,),zn,k+,n ; Λ} n 1 n M 2 M 2 N 2. Applyng Iô lemma o Y n() 2, we have for Λ, Y n () 2 + Z n () 2 d = ξ Y n ()ψ(,y n (),Zn (),)d + 2 Y n ()dk+,n () 2 Y n ()dk,n () 2 Y n ()Z n ()dw(). Ung he Lpchz propery of ψ, he upper mnmal boundary condon n (3.2) and he elemenary nequaly: ab 1 α a2 + αb 2, we have for any arbrary pove real number α, E Y n () 2 + E[ ( E ξ ( +E 2 ( E ξ E( α [ up Y n T Z n () 2 d] ) Y n ()(ψ(,,,) + C Y n () + C Z n () )d Y n ()dk+,n () 2 ψ(,,,) 2 d + c () 2 ] + α K +,n ) S ()dk,n () Y n () 2 d ) Z n () 2 d (T) 2 + up S () 2 + K,n (T) 2). T (3.8) Here and n he equel, c a pove conan whoe value only depend on he Lpchz coeffcen C and may change from lne o lne. From RBSDE (3.2), we know ha for Λ, K +,n (T) = Y n () ξ ψ(,y n (),Zn (),)d + K,n (T) + Z n ()dw(). Hence, ( E K +,n (T) 2 c 1+E Subung (3.9) no (3.8) and leng α = 1 3c ( Y n () 2 + Z n () 2 )d+e K,n (T) 2), Λ. (3.9) and =, we have ( E Z n () 2 d ce 1 + up Y n () 2 + Y n () 2 d + K,n (T) 2), Λ. T

11 From (3.3) and (3.6), we know Then from (3.9), we know upe n 1 Z n () 2 d <, Λ. (3.1) upe K +,n (T) 2 <, Λ. (3.11) n 1 From (3.3), (3.1) and he Lpchz propery of ψ, we know upe n 1 ψ(,y n (),) 2 d <, Λ. Therefore, whou lo of generaly, we can aume ha for Λ, {ψ(,y n,zn,)} n, {Z n} n, and {K +,n } n converge weakly n M 2 o ψ,z, and K +, repecvely. Sep 4. Verfcaon of he fr equaon of RBSDE (1.4). From he fr equaon of (3.2), we have Y n () = Y n () ψ(,y n (),Zn (),)d K+,n ()+K,n ()+ Z n ()dw(). All he aumpon of he generalzed monoonc lm heorem (ee Lemma 2.1) are hown o be afed n prevou ep. Therefore, for Λ, he lm Y r.c.l.l. and ha he form: Y () = ξ + ψ ()d and K + N 2. Moreover, for any p [,2), Hence, we have for Λ, dk () + dk + () Z ()dw(), lm E Z n () Z () p d =, Λ. lm E ψ(,y n (),Z n (),) ψ(,y (),Z (),) p d = ; Y () = ξ + ψ () = ψ(,y (),Z (),), + a.e.,a..; ψ(,y (),Z (),)d dk + () Z ()dw(). dk () (3.12) Sep 5. The mnmal boundary condon. In vew of RBSDE (3.2), we have max {h,j(,yj n 1 ())} Y n () S (), [,T], Λ. j,j Λ Pang o he lm, we have max {h,j(,y j ())} Y () S (), [,T], Λ. (3.13) j,j Λ

12 Snce T we have = Hence, () S ())dk,n () = and Y n () = Y n ( ) Y ( ) S (), () S ())dk,n () On he oher hand, for Λ, Snce we have (Y ( ) S ())dk,n (), Λ. (Y ( ) S ())dk,n () =, Λ. (S () Y ( ))d(k () K,n ()) up (S () Y ( ))[K (T) K,n (T)]. T (Y ( ) S ())dk lm K,n (T) = K () = lm (T), Λ, (Y ( ) S ())dk,n () =, Λ. We have ju proved he upper mnmal boundary condon. I reman o prove he lower mnmal boundary condon. The echnque ued n [8] found dffcul o be drecly appled o our cae nce he correpondng argumen on he malle ψ-upermarngale no rue n he cae of double barrer. We hall vew he RBSDE wh double barrer a RBSDE wh ngle lower barrer by akng he ncreang procee K,n a gven. For Λ and n 1, he followng RBSDE Ȳ n () = ξ + Ȳ n () (Ȳ n ( ) +K,n () + ψ(,ȳ n (), Z n (),)d K,n (T) d +,n K () h () max j,j Λ h,j(,y j ()), h ( ))d K +,n () = ha a unque adaped oluon, denoed by (Ȳ n, Z n Defne Then ( X n, Z n, K +,n ). Z n ()dw(), [,T], X n Ȳ n K,n, ψ n (,y,z,) ψ(,y + K,n (),z,), Λ., K +,n ) afe he followng RBSDE: X n() = (ξ K,n (T)) + +,n + d K () ψ n (, X n (), Z n (),)d Z n ()dw(), X n() h () K,n (), [,T], ( Xn ( ) h ) ( ) + K,n +,n () d K () =. (3.14) (3.15)

13 For Λ, le (Ȳ, Z, K + ) be he oluon of he followng RBSDE: Ȳ () = ξ + ψ(,ȳ(), Z (),)d K (T) + K () + d K + () Ȳ () h (), [,T]; Z ()dw(), [,T]; (3.16) Defne (Ȳ( ) h ( ))d K + () =. X Ȳ K, ψ (,y,z,) ψ(,y + K (),z,), Λ. Then ( X, Z, K + ) afe he followng RBSDE: X () = (ξ K (T)) + ψ (, X (), Z (),)d + d K + () Z ()dw(), X () h () K (), [,T], ( X ( ) h ) ( ) + K () d K + () =. Snce ψ n (,y,z,) ψ (,y,z,) (3.17) = ψ(,y + K,n (),z,) ψ(,y + K (),z,) n vew of Lemma 2.3, we have C K,n () K (), E[ up T ( c E( K +c(e[ up K T X n () X () 2 ] (T) K,n (T) 2 ) + C K () K,n () 2 ]) 1 2 (Φ n (T)) 1 2, ) () K,n () 2 d (3.18) where Snce ( ) Φ n (T) E (ξ K,n (T)) 2 + ψ K n(,,,) 2 d +E[ up (( h () K,n ()) + ) 2 ] +E T ( (ξ K (T))2 + +E[ up (( h () K ())+ ) 2 ]. T ) ψ K (,,,) 2 d ψ n (,,,) ψ(,,,) + CK,n (), ψ (,,,) ψ(,,,) + CK (),

14 up (( h () K,n ()) + ) 2 up (( h ()) + ) 2 T T up (( max Y j()) + ) 2 j,j Λ T j Λ up T Y j () 2, and up () K T(( h ())+ ) 2 j Λ up T Y j () 2, we have Φ n (T) 4E(ξ 2 + ψ(,,,) 2 d) + 2 E[ up Y j () 2 ] j Λ T +2(C 2 T + 1)E( K,n (T) 2 + K (T) 2 ). From (3.3) and (3.6), we ee From (3.7), (3.18) and (3.19), we ee upφ n (T) <. (3.19) n 1 lm E[ up X n () X () 2 ] =, Λ. T So here a ubequence of { X n } n 1 convergng o X,a.e.,a.. Whou lo of generaly, aume ha Se lm X n = X, a.e.,a.., Λ. (3.2) X n Y n K,n, Λ. Then from refleced BSDE (3.2) we know ha (X n,zn,k+,n ) he oluon of he followng refleced BSDE wh ngle reflecng barrer: X n() = (ξ K,n (T)) + ψ K n(,x n (),Zn (),)d + dk +,n () Z n ()dw(), [,T]; (3.21) X n() max h,j(,yj n 1 ()) K,n (), [,T]; j,j Λ ( ) X n () max h,j(,yj n 1 ()) + K,n () dk +,n () =. j,j Λ Comparng wh refleced BSDE (3.15) and ung he comparon heorem for r.c.l.l. reflecng barrer [6, Theorem 1.5], we know ha In vew of (3.2), we have X n () X n () = Y n () K,n (), (,) [,T] Λ. X () Y () K (), (,) [,T] Λ. (3.22) Noe ha due o he appearance of he addonal fxed upper barrer, no clear wheher he lower barrer of (3.17) no le han ha of (3.21). Such a dffculy go around by comparng (3.21) and (3.15).

15 On he oher hand, from (3.17) and [11, Theorem 2.1], we know ha X ( ) he malle ψ -upermarngale wh he lower barrer { h () K (), T }. From (3.12) and (3.13), can be ealy obaned ha {Y () K (), T } a ψ -upermarngale wh he ame lower barrer and ermnal value. Hence, Togeher wh (3.22), we have Then X () Y () K X () = Y () K (), T, Λ. (), T, Λ. Ȳ () = Y (), T, Λ. From he unquene of he Doob-Meyer Decompoon, follow ha Z () = Z (), K+ () = K + (), T, Λ. Hence, for Λ, (Y,Z,K +,K ) almo afe RBSDE (1.4) excep ha boh mnmal boundary condon are replaced by ( Y ( ) max h,j(,y j ( )) )dk + () =, (Y ( ) S())dK () =. j,j Λ If we furher prove he connuy of Y, hen we know {(Y,Z,K +,K ), Λ} an adaped oluon of RBSDE (1.4). Sep 6. The connuy of Y and K + for Λ. Snce K connuou, we know Y () = K + (). Suppoe Y 1 ( ) < for ome 1 Λ and ome [,T], hen K + 1 ( ) >. From he mnmal boundary condon of (4.2), we know Y 1 ( ) = max j 1,j Λ h 1,j(,Y j ( )). The ndex aanng he maxmum denoed by 2 and 2 1. Hence, h 1, 2 (,Y 2 ( )) = Y 1 ( ) > Y 1 ( ) h 1, 2 (,Y 2 ( )). I follow ha Y 2 ( ) <. Repeang hee argumen, we can oban ha Y k ( ) = h k, k+1 (,Y k+1 ( )), k = 2,3,. Snce Λ a fne e, here mu be a loop n Λ, whou lo of generaly, we aume ha 1 = n. Then for he loop { n, n 1,, 1 }, we have Y n 1 ( ) = h n 1, n (,Y n ( )),,Y 1 ( ) = h 1, 2 (,Y 2 ( )), and Y n ( ) = Y 1 ( ). Th conradc Hypohe 3. Therefore, Y () =, T, Λ. So K +. 4 Exence: he penaly mehod In Theorem 3.1, every componen of he upper barrer S aumed o be uperregular. Le k(,j) be he wchng co from ae o ae j n he opmal wchng problem (ee Hu and Tang [1]) afyng Hypohe 3, and le he funcon h,j nroduced n he precedng econ ake he parcular form h,j (,y) = y k(,j). Then we can prove by a penaly mehod he exence of an adaped oluon o RBSDE (1.4) whou he uper-regulary aumpon on he upper barrer S.

16 4.1 Mul-dmenonal RBSDE wh fxed ngle reflecng barrer. In wha follow, we conder he mul-dmenonal RBSDE wh fxed ngle reflecng barrer, how he exence and unquene by a penaly mehod, and gve a comparon heorem. For wo m-dmenonal vecor x (x 1,,x m ) T and y (y 1,,y m ) T, we mean by x y ha x y for Λ. For a vecor x (x 1,,x m ) T, x + defned a he m-dmenonal vecor (x + 1,,x+ m )T. Conder he followng mul-dmenonal RBSDE wh fxed ngle reflecng barrer: Y () = ξ + φ(,y (),Z())d dk() Z()dW(), [,T]; (4.1) Y () S(), [,T]; (Y () S ())dk () =, Λ. We make he followng aumpon on he generaor φ, he ermnal value ξ (ξ 1,,ξ m ) T, and he barrer S (S 1,,S m ) T. Hypohe 4. () The proce φ(,,) (M 2 ) m. For Λ, ξ L 2 (Ω, F T,P) and S S 2 wh ξ S (T). () There a conan C > uch ha for any (,y,y,z,z ) [,T] (R m ) 2 (R m d ) 2, we have φ(,y,z) φ(,y,z ) C( y y + z z ), 4 y,φ(,y + + y,z) φ(,y,z ) 2 χ {y <} z z 2 + C y 2. =1 (4.2) We have Theorem 4.1. Le Hypohe 4 be afed. Then RBSDE (4.1) ha a unque adaped oluon (Y,Z,K) (S 2 ) m (M 2 ) m d (N 2 ) m. Proof. For any pove neger n, conder he followng penalzed BSDE: Y n () = ξ + φ(,y n (),Z n ())d n Z n ()dw(), [,T]. () S()) + d (4.3) From Pardoux and Peng [13], we know ha for each n, BSDE (4.3) ha a unque adaped oluon,z n ) (S 2 ) m (M 2 ) m d. Defne for (,y,z) [,T] R m R m d, φ n (,y,z) φ(,y,z) n(y S()) + and K n () n () S()) + d.

17 In vew of Hypohe 4 (), we have for all y R m, 4 y,φ n (,y + + y,z) φ n+1 (,y,z ) 4 y,φ(,y + + y,z) φ(,y,z ) + (y S()) + 4 y,φ(,y + + y,z) φ(,y,z ) 2 χ {y <} z z 2 + C y 2. =1 Applyng he comparon heorem of mul-dmenonal BSDE (ee [9, Theorem 2.1]), we deduce ha Y n+1 () Y n (), T. For [,1], he equence {Y n ()} n 1 almo urely adm a lm, whch denoed by Y () below. Applyng Iô lemma o compue Y n () 2, we have Y n () 2 + = ξ Z n () 2 d Y n (),φ(,y n (),Z n ()) d Y n (),dk n () Y n (),Z n ()dw(). Takng expecaon on boh de, n vew of he followng nequaly we have E Y n () 2 + E E ξ 2 + 2E E E ξ 2 + E E S(),dK n () Y n (),dk n () Z n () 2 d S(),dK n (), [ φ(,,) + C( Y n () + Z n () )] Y n () d φ(,,) 2 d + (3C 2 + 2C + 1)E Y n () 2 d Z n () 2 d + m E( up S () 2 ) + αe K n (T) K n () 2 α T (4.4) (4.5) for an arbrary pove real number α. From (4.3), we have K n (T) K n () = ξ Y n () + φ(,y n (),Z n ())d Z n ()dw(). (4.6) Furher, we have E K n (T) K n () 2 4E( Y n () 2 + ξ 2 ) + 4(3TC 2 + 1)E +12TE ( φ(,,) 2 + C 2 Y n () 2 )d. Z n () 2 d (4.7)

18 Seng n (4.5), n vew of (4.7), we have α = 1 12(3TC 2 + 1) 2 3 E Y n () E Z n () 2 d c(1 + E Y n () 2 )d for a conan c whch doe no dependen on n. I hen follow from Gronwall nequaly ha whch mple ha up E Y n () 2 + E Z n () 2 d c, n = 1,2, T E K n (T) 2 c, n = 1,2, (4.8) In vew of (4.4), applyng he Burkholder-Dav-Gundy nequaly, we have E up Y n () 2 + E Z n () 2 d + E K n (T) 2 c, n = 1,2, T Recallng ha Y () = lm Y n (), ung Faou lemma, we have E up Y () 2 c. T I hen follow from Lebegue domnaed convergence heorem ha lm E Y n () Y () 2 d =. For pove neger n 1 and n 2, applyng Iô lemma o compue Y n 1 () Y n 2 () 2, we have for [,T], = 2 Y n 1 () Y n 2 () Z n 1 () Z n 2 () 2 d Y n 1 () Y n 2 (),φ(,y n 1 (),Z n 1 ()) φ(,y n 2 (),Z n 2 ()) d Y n 1 () Y n 2 (),d(k n 1 () K n 2 ()) Y n 1 () Y n 2 (),(Z n 1 () Z n 2 ())dw(). Takng expecaon and leng =, we have E Y n 1 () Y n 2 () 2 + E 2C(C + 1)E +2E Z n 1 () Z n 2 () 2 d Y n 1 () Y n 2 () 2 d E Z n 1 () Z n 2 () 2 d 2 () S()) +,dk n 1 () + 2E (4.9) 1 () S()) +,dk n 2 ().

19 A a conequence, E Z n 1 () Z n 2 () 2 d 4C(C + 1)E +4E +4E Y n 1 () Y n 2 () 2 d 2 () S()) +,dk n 1 () 1 () S()) +,dk n 2 (). (4.1) Followng almo he ame argumen a n he proof of [3, Lemma 6.1], we have E up () S()) + 2 a. (4.11) T Then from (4.8) and (4.11), we know ha a n 1,n 2, E 2 () S()) +,dk n 1 () + E Togeher wh (4.1), we oban lm E Z n 1 () Z n 2 () 2 d =. n 1,n 2 1 () S()) +,dk n 2 (). In vew of (4.9), ung he Burkholder-Dav-Gundy nequaly, we conclude lm E up Y n 1 () Y n 2 () 2 =. n 1,n 2 T In vew of (4.3) and he defnon of K n, we know lm E up K n 1 () K n 2 () 2 =. n 1,n 2 T From he above convergence, we conclude ha here ex (Z,K) (M 2 ) m d (N 2 ) m, afyng lm E Z n () Z() 2 d = and lm E up K n () K() 2 =. T Pang o lm n equaon (4.3), we know ha afe he followng equaon: (Y,Z,K) (S 2 ) m (M 2 ) m d (N 2 ) m Y () = ξ + φ(,y (),Z())d dk() Z()dW(). From (4.11), ung Faou lemma, we know ha E up (Y () S()) + 2 lm E up () S()) + 2 =, T T whch mple ha Y () S(), T. (4.12)

20 Therefore, (Y () S ())dk (), Λ. (4.13) Snce,K n ) end o (Y,K) n (S 2 ) 2m, we have for Λ, Thu, (Y () S ())dk () = lm () S ())dk n () = lm n () S ()) () S ()) + d. (4.14) (Y () S ())dk () =, Λ. (4.15) We conclude ha (Y,Z,K) an adaped oluon o RBSDE (4.1). Unquene of he oluon follow from Lemma 2.3 and Remark 2.3. Remark 4.1. For he exence and unquene for mul-dmenonal RBSDE, we refer he reader o Gegou-Pe and Pardoux [4]. Remark 4.2. The comparon heorem of mul-dmenonal BSDE fr eablhed by Hu and Peng [9] under he ronger condon on he generaor φ ha φ(,y,z) connuou for any fxed (y,z) and φ(,,) (S 2 ) m. By he mehod of approxmaon, can be hown ha he comparon heorem ll hold f Hypohe 4 () afed. Thank o he above exence and unquene reul, we can prove he followng comparon heorem for mul-dmenonal RBSDE wh a fxed ngle reflecng barrer. Theorem 4.2. Aume ha (φ 1,ξ 1 ),(φ 2,ξ 2 ), and S afy Hypohe 4. Furher, aume ha () ξ 1 ξ 2 ; () There a pove conan C uch ha for (y,y ) (R m ) 2,(z,z ) (R m d ) 2, and [,T], 4 y,φ 1 (,y + + y,z) φ 2 (,y,z ) 2 χ {y <} z z 2 + C y 2. (4.16) For j = 1,2, denoe by (Y j,z j,k j ) he adaped oluon of RBSDE (4.1) aocae wh he daa (ξ j,φ j,s). Then, we have (1) Y 1 () Y 2 (), K 1 () K 2 (), T; (2) K 1 (r) K 1 () K 2 (r) K 2 (), r T. Proof. For j = 1,2 and pove neger n, he followng BSDE: Y j,n () = ξ j + =1 φ j (,Y j,n (),Z j,n ())d n Z j,n ()dw(), [,T] (Y j,n () S j ()) + d (4.17)

21 ha a unque adaped oluon, denoed by (Y j,n,z j,n,k j,n ). In vew of (4.16), we have 4 y,(φ 1 (,y + + y,z) n(y + + y S()) + ) (φ 2 (,y,z ) n(y S()) + ) 4 y,φ 1 (,y + + y,z) φ 2 (,y,z ) ny + 2 χ {y <} z z 2 + C y 2. =1 By he comparon heorem of mul-dmenonal BSDE (ee [9, Theorem 2.1]), follow ha Y 1,n () Y 2,n (), [,T],n = 1,2,. (4.18) Then from he proof of Theorem 4.1, we know ha for [,T] and j = 1,2, Y j () = K j () = lm Y j,n (), lm n (Y j,n () S()) + d. (4.19) The dered reul hen follow from (4.18) and (4.19). 4.2 Mul-dmenonal RBSDE wh oblque reflecon. Conder he followng RBSDE: for Λ, Y () = ξ + ψ(,y (),Z (),)d dk () + dk + () Z ()dw(), [,T]; S () Y () max {Y j() k(,j)}, [,T]; j,j Λ ( ) Y () max {Y j() k(,j)} dk + () =, j,j Λ (Y () S ())dk () =. (4.2) The followng heorem preen he exence of he oluon whou uperregulary aumpon on S. Theorem 4.3. Le Hypohee 1 and 3 be afed. Aume ha S (S 2 ) m wh S() Q() for [,T] and ξ L 2 (Ω, F T,P;R m ) wh ξ(ω) Q(T). Here we have defned for [,T], Q() {(y 1,,y m ) T R m : y j k(,j) y S (),,j Λ,j }. Then RBSDE (4.2) ha an adaped oluon (Y,Z,K +,K ) (S 2 ) m (M 2 ) m d (N 2 ) 2m. Proof. The proof dvded no four ep. Sep 1. The approxmang equence of penalzed RBSDE.

22 For any pove neger n, conder he followng RBSDE: Λ, Y n() = ξ + Y n +n T ψ(,y n (),Z n (),)d () Y n l () + k(,l)) d Z n ()dw(), () S (), [,T]; () S ())dk,n () =. [,T]; dk,n () They urn ou o be RBSDE well uded n he precedng ubecon. Defne for (,y,z,) [,T] R m R m d Λ, (4.21) ψ n (,y,z,) ψ(,y,z,) + n (y y l + k(,l)), ψ n (,y,z) ( ψ n (,y,z,1),, ψ n (,y,z,m)) T. Snce y,(y+ + y y+ l y l + k(,l)) (y y l + k(,l)) = y,(y y+ l y l + k(,l)) (y y l + k(,l)) (4.22) for y,y R m and,l Λ, and ψ n (,y,z,) doe no depend on z j for j, we have for (y,y ) (R m ) 2,(z,z ) (R m d ) 2, and (,l) (Λ) 2, 4 y, ψ n (,y + + y,z) ψ n (,y,z ) = 4 y,ψ(,y+ + y,z,) ψ(,y,z,) 4n =1 y,(y+ + y y+ l y l + k(,l)) (y y l + k(,l)), 4 y,ψ(,y+ + y,z,) ψ(,y,z,) 2 =1 χ {y <} z z 2 + 2C 2 y 2. =1 (4.23) I eay o check ha ψ n (,y,z) alo Lpchz connuou n (y,z). From Theorem 4.1, we know ha RBSDE (4.21) ha a unque adaped oluon,z n,k,n ) wh Y n 1,,Y n m) T (S 2 ) m, Z n (Z n 1,,Z n m) T (M 2 ) m d, and K,n (K,n 1,,K,n m ) T (N 2 ) m.

23 Moreover, we have from (4.23) ha for (y,y ) (R m ) 2,(z,z ) (R m d ) 2, and (,l) (Λ) 2, 4 y, ψ n+1 (,y + + y,z) ψ n (,y,z ) = 4 y, ψ n (,y + + y,z) ψ n (,y,z ) 4 4 y, ψ n (,y + + y,z) ψ n (,y,z ) 2 χ {y <} z z 2 + 2C 2 y 2. =1 From Theorem 4.2, we know y,(y+ + y y+ l y l + k(,l)), Y n () Y n+1 (), K,n () K,n+1 (), T. (4.24) The equence { (),K,n ())} n 1 adm a lm, whch denoed by (Y (),K ()) below wh Y () (Y 1 (),,Y m ()) T and K () (K 1 (),,K m()) T, for [,T]. Sep 2. A pror emae. The followng lemma he key o our ubequen argumen. Lemma 4.1. There a pove conan c whch ndependen of n, uch ha E up T Y n () 2 + E K,n (T) 2 + E n 2 E Z n () 2 d c, () Y n j () + k(,j)) 2 d c, Λ. I proof follow. Applyng Iô-Meyer formula [12] o compue () Y n j ()+ k(,j)) 2, we have = 2 where () Y n +2n n j () + k(,j)) 2 + () Y n j () + k(,j)) 2 d χ L () Z n () Zj n () 2 d,j,n () Y n j () + k(,j)) [ψ(,y n j (),Z n j (),j) ψ(,y n (),Z n (),)]d +2n () Y n j () + k(,j)) (Z n () Zn j ())dw() () Yj n () + k(,j)) d(k,n l, l j () Y n j () + k(,j)) j () Y n () K,n j ()) () + k(j,)) d () Y n j () + k(,j)) [ j () Y n l () + k(j,l)) () Yl n () + k(,l)) ]d L,j,n n {(,ω) : Y () Y j n () + k(,j) < }. (4.25)

24 We clam ha he la hree erm of (4.25) are all equal o or le han. In fac, due o (4.21), we have and Y n j () k(,j) S j() k(,j) S () (nong S() Q()) (4.26) = () Y n () Y n j () + k(,j)) d(k,n j () + k(,j)) dk,n () () S ()) dk,n () In vew of Hypohe 3 (), we have (S () Y n ())dk,n () =. () K,n j ()) {(y 1,,y m ) T R m : y y j + k(,j) <,y j y + k(j,) < } = whch mmedaely gve () Y n j () + k(,j)) j () Y n () + k(j,)) =. From Hypohe 3 (), ung he propery ha x 1 x 2 (x 1 x 2 ), we have () Y n j () + k(,j)) [ j () Y n l () + k(j,l)) () Y n l () + k(,l)) ] () Y j n() + k(,j)) (Yj n() Y n () + k(j,l) k(,l)) () Y j n() + k(,j)) (Yj n() Y n () k(,j)) = () Y n j () + k(,j)) () Y n j () + k(,j))+ =. Takng expecaon on boh de of (4.25), we have Nong ha E () Yj n () + k(,j)) 2 + E +2nE 2E () Y n j () + k(,j)) 2 d χ L () Z n () Zj n () 2 d,j,n ( () Y n j () + k(,j)) ψ(,y n (),Z n (),) ψ(,yj n(),zn j (),j) d. (4.27) ψ(,y n(),zn (),) ψ(,y n j (),Zn j (),j) ψ(,y n(),zn (),) ψ(,y n (),Zn (),j) + ψ(,y n(),zn (),j) ψ(,y n j (),Zn j (),j) c( ψ(,,) + Y n() + Zn () + Y n () Y j n() + Zn () Zn j () ) ( c 1 + ψ(,,) + Y n() + Zn () + Y n () Y j n () + k(,j) ) + Z n() Zn j ()

25 for a pove conan c (ndependen of n and pobly varyng from lne o lne), n vew of (4.27), we have E () Yj n () + k(,j)) 2 + E +2nE () Y n j () + k(,j)) 2 d χ L () Z n () Zj n () 2 d,j,n ( n 2 + 2c)E () Y j n () + k(,j)) 2 d + 2c2 n E χ L (),j,n ( 1 + ψ(,,) 2 + Y n () 2 + Z n () 2 + Z n () Zn j () 2) d. So for uffcenly large n, n 2 E ( () Y j n () + k(,j)) 2 d c 1 + E Applyng Iô lemma o compue Y n() 2, we have ) ( Y n () 2 + Z n () 2 )d. (4.28) Y n () 2 + Z n () 2 d ( = 2 Y n () ψ(,y n (),Zn (),) + n m () Y l n () + k(,l)) ) d T +ξ 2 2 Y n ()dk,n () 2 Y n ()Zn ()dw(). (4.29) Ung he elemenary nequaly: we oban ha 2 Y n ()dk,n 2ab 1 α a2 + αb 2 for a,b,α >, () = 2 1 α up T S ()dk,n () S () 2 + α K,n (T) K,n () 2 (4.3) for an arbrary pove real number α. Then akng expecaon on boh de of (4.29), we have E Y n () 2 + E 2E Z n () 2 d m Y n () ( ψ(,y n (),Zn (),) + n () Y j n () + k(,l)) )d +E(ξ 2 ) 2E Y n ()dk,n () c ε E Y n () 2 d + ε n 2 E +εe Z n () 2 d + 1 α E up T () Y n l () + k(,l)) 2 d + E(ξ ) 2 S () 2 + αe K,n (T) K,n () 2 (4.31)

26 for arbrary pove real number ε,α, and a conan c ε dependng on ε. On he oher hand, from equaon (4.21), we have K,n (T) K,n () = ξ Y n () + +n In vew of (4.28), we have ψ(,y n (),Zn (),)d () Y n l () + k(,l)) d Z n ()dw(). E K,n (T) K,n () 2 ( c Eξ 2 + E Y n () 2 + E +c [ c ) ( Y n () 2 + Z n () 2 )d n 2 E Y n () Yl n () + k(,l)) 2 d 1 + E Y n () 2 + E ( Y n () 2 + Z n () 2 )d ]. (4.32) Furher, n vew of (4.31), we have E Y n () 2 + E c ε,α (1 + E +(ε + α)ce Z n () 2 d Y n () 2 d) + αce Y n () α E up Z n () 2 d T S () 2 for any pove number ε,α, and a conan c ε,α dependng on ε,α. Seng α = ε = 1 3c, we have from Gronwall nequaly ha From (4.28) and (4.32), we have E Y n () 2 + E Z n () 2 d c. n 2 E () Yj n () + k(,j)) 2 d c, E K +,n (T) 2 c. Moreover, n vew of (4.29), applyng he Burkholder-Dav-Gundy nequaly, we have E up Y n () 2 c. T The proof of Lemma 4.1 hen complee. Sep 3. The convergence of penalzed BSDE. In vew of Lemma 4.1, ung Faou lemma, we have E[ up ( Y () 2 + K () 2 )]. T

27 Then applyng Lebegue domnaed convergence heorem, we have E Y n () Y () 2 d + E K,n () K () 2 d a. (4.33) For pove neger n 1 and n 2, applyng Iô lemma o Y n 1 () Y n 2 () 2, we have Y n 1 () Y n 2 () 2 + Z n 1 () Z n 2 () 2 d Snce = 2 = (ψ(,y n 1 (),Z n 1 (),) ψ(,y n 2 (),Z n 2 (),)) 1 () Y n 2())d +2n 1 m 2n 2 m 2 2 T 1 () Y n 1 l () + k(,l)) 1 2 () Y n 2 l () + k(,l)) 1 1 () Y n 2 ())d(k,n 1 () K,n 2 ()) () Y n 2())d () Y n 2())d 1 () Y n 2 ())(Z n 1 () Z n 2 ())dw(), Λ. 1 () Y n 2 ())d(k,n 1 (S () Y n 2 ())dk,n 1 () + n vew of (4.34), we have for Λ, () K,n 2 ()) E Y n 1 () Y n 2 () 2 + E Z n 1 () Z n 2 () 2 d = 2E m +2n 1 E (4.34) (S () Y n 1 ())dk,n 2 ()), Λ, (4.35) (ψ(,y n 1 (),Z n 1 (),) ψ(,y n 2 (),Z n 2 (),)) 1 () Y n 2())d m 2n 2 E 2C(C + 1)E +2(E +2(E 1 () Y n 1 l () + k(,l)) 1 2 () Y n 2 l () + k(,l)) 1 Y n 1 () Y n 2() 2 d E Y n 1 () Y n 2 () 2 d) 1 2 Y n 1 () Y n 2 () 2 d) 1 2 (n 2 1 E (n 2 2E () Y n 2())d () Y n 2())d Seng =, n vew of (4.33) and Lemma 4.1, we have Z n 1 () Z n 2 () 2 d ( 1 () Y n 1 l () + k(,l)) ) 2 d) 1 2 ( 2 () Y n 2 l () + k(,l)) ) 2 d) 1 2. E Z n 1 () Z n 2 () 2 d, Λ, a n 1,n 2. (4.36)

28 So here ex Z (Z 1,,Z m ) T (M 2 ) m d uch ha lm E Z n () Z () 2 d =, Λ. In vew of (4.34), applyng he Burkholder-Dav-Gundy nequaly, we have E( up Y n 1 () Y n 2 () 2 ), Λ a n 1,n 2. (4.37) T From (4.21), we have d(k,n 1 () K,n 2 ()) = (ψ(,y n 1 (),Z n 1 (),) ψ(,y n 2 (),Z n 2 (),))d +d 1 () Y n 2 ()) + n 1 1 () Y n 1() + k(,l)) d m n 2 2 () Y n 2 l () + k(,l)) d (Z n 1 () Z n 2 ())dw(). Snce he proce {(K,n 1 () K,n 2 ()), [,T]} of fne varaon, quadrac varaon. By Iô lemma follow ha = 2 (K,n 1 () K,n 2 ()) (K,n 1 () K,n 2 ())(ψ(,y n 1 (),Z n 1 (),) ψ(,y n 2 (),Z n 2 (K,n 1 () K,n 2 ())(n 1 (K,n 1 () K,n 2 ())(n 2 l 1 () Y n 1() + k(,l)) )d l 2 () Y n 2() + k(,l)) )d (K,n 1 () K,n 2 ())d 1 (K,n 1 () K,n 2 ())(Z,n 1 () Y n 2()) () Z,n 2())dW(). Applyng he Burkholder-Dav-Gundy nequaly, we have l (),))d E up c(e T +2(E +2(E +E up {2 T K,n 1 () K,n 2 () 2 ( K,n 1 () K,n 2 () 2 + Y n 1 () Y n 2 () 2 + Z n 1 () Z n 2 K,n 1 () K,n 2 () 2 d) 1 2 K,n 1 () K,n 2 () 2 d) 1 2 () 2 )d (n 2 1 E ( 1 () Y n 1 l () + k(,l) ) 2 d) 1 2 (n 2 2 E ( 2 () Y n 2 l () + k(,l) ) 2 d) 1 2 (K,n 1 () K,n 2 ())d 1 () Y n 2())} E up K,n 1 () K,n 2 T () 2 (4.38)

29 for a pove conan c ndependen of n 1 and n 2. Idencal o he proof of (4.35), we have Hence by Iô lemma, 1 () Y n 2 ())d(k,n 1 () K,n 2 ()). 2 (K,n 1 () K,n 2 ())d 1 () Y n 2 ()) = 2((K,n 1 () K,n 2 ()) 1 () Y n 2 ())) 2 Then a a conequence, { E up 2 T ( 2E up 1 () Y n 2 ())d(k,n 1 () K,n 2()) 2 K,n 1 () K,n 2 () Y n 1 () Y n 2(). T 1 3 E up K,n 1 T (K,n 1 () K,n 2 ())d 1 K,n 1 () K,n 2 () Y n 1 () K,n 2 () Y n 2 ) () Y n 2() Togeher wh (4.33), (4.38), and Lemma 4.1, we have E( up ce T +ce up T From (4.36) and (4.37), we have Se and } ()) () 2 + 3E up Y n 1 () Y n 2 () 2. T K,n 1 () K,n 2 () 2 ) ( Y n 1 () Y n 2 () 2 + Z n 1 () Z n 2 () 2 )d Y n 1 () Y n 2 () 2. E up K,n 1 () K,n 2 () 2, Λ, a n 1,n 2. T K +,n () n K + () Y () Y () We have K +,n () = Y n () Y n () and =. () Y n l () + k(,l)) d ψ(,y (),Z (),)d + ψ(,y n (),Z n (),)d+ (4.39) dk () + Z ()dw(). dk,n ()+ lm E( up K +,n () K + () 2 ) T ( lm ce up { Y n () Y () 2 + K,n () K () 2 } T ) + Z n () Z () 2 d Z n ()dw()

30 Hence, K (K 1 +,,K+ m) (N 2 ) m, and (Y,Z,K +,K ) afe he fr equaon of (4.2). By Lemma 4.1 and Faou lemma, we oban ha E lm E lm whch mple mmedaely ha (Y () Y l () + k(,l)) 2 d c n 2 =, () Y n l () + k(,l)) 2 d Y () Y l () + k(,l), T. (4.4) Snce Y n () S() for any n and [,T], we have Hence, Y () S(), T. Y () Q(), T. Sep 4. The lower mnmal boundary condon. For,j,l Λ, j,l =, obvou ha () Y n j () + k(,j)) + () Y n l () + k(,l)) =. For,j,l Λ, j,l, we have Hence, = n. =. mn j () Y n j () + k(,j))+ () Y n l () + k(,l)) () Y n l () + k(,l)) + () Y n l () + k(,l)) () max j {Y n j () k(,j)}) + dk +,n () mn {(Y n () Yj n () + k(,j)) + () Yl n () + k(,j)) }d j On he oher hand, nce K +,n ( ) ncreang, we have Therefore, () max {Y j n () j k(,j)})+ dk +,n (). () max j {Y n j () k(,j)}) + dk +,n () =. Followng he ame argumen a n he proof of (4.14) and applyng [4, Lemma 5.8], we have In vew of (4.4), we have (Y () max j {Y j() k(,j)}) + dk + (Y () max j {Y j() k(,j)})dk + () =. () =.

31 5 Unquene The unquene of oluon defned n he followng ene: f (Y,Z,K +,K ) anoher oluon, hen Y () Y (), Z () Z(), K + () K () K + () K (), T,a.. The followng Sronger aumpon on h,j needed n our proof of he unquene reul. Hypohe 5.,j,l Λ uch ha j, j l, y R, h,j (,h j,l (,y)) < h,l (,y). Remark 5.1. I eay o check ha Hypohe 5 mple Hypohe 3. If h,j (,y) y k(,j), hen Hypohe 5 reduce o he nequaly: k(,j) + k(j,l) > k(,l) for j and j l. Le {θ j } j= be an ncreang equence of oppng me wh value n [,T]. j, α j an F θj -meaurable random varable wh value n Λ. Aume ha a..ω, here ex an neger N(ω) < uch ha θ N = T. Then we defne a wchng raegy a: N 1 a() = α χ [θ, θ 1 ]() + α j χ (θj, θ j+1 ](). We denoe by A all he wchng raege wh nal daa (α,θ ) = (,) Λ [,T]. For gven a A, conder he followng RBSDE: U a () = ξ a(t) + ψ(r,u a (r),v a (r),a(r))dr (L a (T) L a ()) N 1 j=1 [U a (θ j ) h αj 1, α j (θ j,u a (θ j ))]χ (,T] (θ j ) j=1 V a (r)dw(r), U a () S a() (), [,T]; [,T]; (5.1) (U a () S a() ())dl a () =. The generaor ψ of RBSDE (5.1) depend on he conrol a and a each wchng me θ j before ermnaon, he value of U a wll jump by an amoun of U a (θ j ) h αj 1, α j (θ j,u a (θ j )) whch can be regarded a a penaly or co for he wchng. In each ubnerval dvded by he wchng me, RBSDE (5.1) evolve a a andard RBSDE wh ngle barrer, whch can be olved n a backwardly nducve way. The opmal conrol problem for RBSDE (5.1) o maxmze U a () over a A. The oluon of RBSDE (1.4) cloely conneced wh h conrol problem. Bede, alo conneced o he ochac game conruced below. Le τ be a oppng me wh value n [,T]. For gven a A and oppng

32 me τ, conder he followng BSDE: U a,τ () = S a(τ) (τ)χ {τ<t } + ξ a(τ) χ {τ=t } + N 1 τ [U a,τ (θ j ) h αj 1, α j (θ j,u a,τ (θ j ))]χ (,τ] (θ j ) j=1 τ V a,τ (r)dw(r), [,τ]. ψ(r,u a,τ (r),v a,τ (r),a(r))dr (5.2) The ermnal value S a(τ) (τ)χ {τ<t } + ξ a(τ) χ {τ=t } of BSDE (5.2) depend on he ermnal me τ. And he value U a,τ () depend on boh he wchng raegy a and ermnal me τ. Ung he ame argumen a n RBSDE (5.1), we know ha BSDE (5.2) well defned and ha a unque oluon. Baed on ha, we conruc a zero-um ochac game a follow. Suppoe here are wo player A and B whoe benef are anagonc. The payoff U a,τ () whch afe BSDE (5.2) a reward for player A and a co for player B. Player A chooe a wchng raegy a A o a o maxmze he reward U a,τ (). Player B chooe he me τ o ermnae he game and re o mnmze he co U a,τ (). The followng heorem reveal he connecon among he oluon of he yem (1.4), he above conrol problem and ochac game. Theorem 5.1. Le Hypohee 1 and 5 be afed. Aume ha (U a,v a,l a ) he unque adaped oluon of RBSDE (5.1) for a A and (U a,τ,v a,τ ) he unque oluon of BSDE (5.2) for a A and oppng me τ. Then f (Y,Z,K +,K ) an adaped oluon of RBSDE (1.4), we have Y () = eup a A U a () = eupenf U a,τ () = enf a A τ τ eupu a,τ (), a A [,T]. Le ˆθ = and ˆα =. Defne he equence {ˆθ j, ˆα j } j=1 follow: n an nducve way a ˆθ j = nf{ ˆθ j 1 : Yˆαj 1 () = max k ˆα j 1,k Λ hˆα j 1,k(,Y k ())} T. And f ˆθ j < T, e ˆα j be he malle ndex n Λ uch ha Oherwe, e ˆα j be an arbrary ndex. Defne Yˆαj 1 (ˆθ j ) = hˆαj 1,ˆα j (ˆθ j,yˆαj (ˆθ j )). (5.3) ˆN 1 â() ˆα χ [ˆθ, ˆθ 1 ] () + j=1 ˆα j χ (ˆθj, ˆθ j+1 ] () and τ nf{ [,T) : Uâ() = Sâ() ()} T, wh he convenon ha nf +. Then, we have â A and Y () = Uâ() and Y () = Uâ,τ (), [,T]. (5.4)

33 Proof. For [,T],a A, and [,θ N ], defne Y a () Z a () K +,a () K,a () N Y αj 1 ()χ [θj 1, θ j )() + ξ αn 1 χ {=T }, j=1 N Z αj 1 ()χ [θj 1, θ j )(), j=1 N (K α + j 1 (θ j ) K α + j 1 (θ j 1 )), j=1 N (Kα j 1 (θ j ) Kα j 1 (θ j 1 )). j=1 (5.5) In vew of he jump Y αj (θ j ) Y αj 1 (θ j ) = Y a (θ j ) Y a (θ j ) a each oppng me θ j, j = 1,,N 1, we know ha (Y a,z a,k +,a,k,a ) afe he followng RBSDE: Y a () = ξ a(t) + Y a () N 1 ψ(r,y a (r),z a (r),a(r))dr (Y a (θ j ) Y a (θ j ))χ (,T] (θ j ) + K +,a (T) K +,a () j=1 (K,a (T) K,a ()) S a() () Z a (r)dw(r), (Y a () S a() ())dk,a () =, [,T]. (5.6) And for any oppng me τ, (Y a,z a,k +,a,k,a ) alo afe he followng BSDE: and Y a () = Y a (τ) + N 1 τ ψ(r,y a (r),z a (r),a(r))dr (Y a (θ j ) Y a (θ j ))χ (,τ] (θ j ) + K +,a (τ) K +,a () j=1 (K,a (τ) K,a ()) τ Z a (r)dw(r), Comparng RBSDE (5.1) wh (5.6), n vew of he fac ha [,τ]. Y a (θ j ) = Y αj 1 (θ j ) h αj 1,α j (θ j,y αj (θ j )) = h αj 1,α j (θ j,y a (θ j )) K +,a (T) K +,a (), we deduce from he comparon heorem [3, Theorem 4.1] ha From he defnon n (5.5), Hence, Y a () U a (), T. Y a () = Y (). (5.7) Y () U a (), a A. (5.8)

34 For he equence {ˆθ j } j=1, we clam ha for a..ω, here ex an neger ˆN(ω) < uch ha ˆθ ˆN = T. Oherwe, defne B j=1 {ω : ˆθ j (ω) < T } F T, hen P(B) >. For j = 1,2,, we have Yˆαj 1 (ˆθ j ) = hˆαj 1,ˆα j (ˆθ j,yˆαj (ˆθ j )), Yˆαj (ˆθ j+1 ) = hˆαj,ˆα j+1 (ˆθ j+1,yˆαj+1 (ˆθ j+1 )), on B. (5.9) Snce he equence {(ˆα j 1, ˆα j, ˆα j+1 )} j=1 ake value n Λ3, whch a fne e, here are a rple ( 1, 2, 3 ) and a ubequence j k uch ha for k = 1,2, (ˆα jk 1, ˆα jk, ˆα jk +1) = ( 1, 2, 3 ). Snce he equence {ˆθ j } j=1 ncreang and bounded by T, here a lm ˆθ. Pang o he lm n (5.9) for he ubequence {j k }, we have Y 1 (ˆθ ) = h 1, 2 (ˆθ,Y 2 (ˆθ )), Y 2 (ˆθ ) = h 2, 3 (ˆθ,Y 3 (ˆθ )), on B. (5.1) From Hypohe 5, we have Th conradc o he fac ha Th how â A. I eay o ee ha Y 1 (ˆθ ) = h 1, 2 (ˆθ,h 2, 3 (ˆθ,Y 3 (ˆθ ))) < h 1, 3 (ˆθ,Y 3 (ˆθ )) on B. Y 1 (ˆθ ) max h 1,j(ˆθ,Y j (ˆθ )). (5.11) j 1,j Λ Y â(ˆθ j ) = hˆαj 1,ˆα j (ˆθ j,y â(ˆθ j )), K +,â (T) K +,â () =. (5.12) Then (Y â,zâ,k,â ) afe RBSDE (5.1). By he unquene of he oluon of RBSDE (5.1), we have Y â() = Uâ(), T. Hence, Nong (5.8), we know Y () = Uâ(). Y () = eupu (), [,T]. (5.13) a A In vew of (5.7), follow from (5.12) ha Y â() = Y â(τ) + ˆN 1 τ ψ(r,y â(r),zâ(r),â(r))dr (Y â(ˆθ j ) hˆαj 1,ˆα j (ˆθ j,y â(ˆθ j )))χ (,τ] (ˆθ j ) j=1 (K,â (τ) K,â ()) τ Zâ(r)dW(r). (5.14)

35 For any wchng raegy a, defne ˆτ nf{ [,T) : Y a () = S a() ()} T, wh he convenon ha nf +. From he upper mnmal boundary condon n (1.4), we know ha K,a (ˆτ) K,a () =, Y a (ˆτ) = S a(ˆτ) (ˆτ)χ {ˆτ<T } + ξ a(ˆτ) χ {ˆτ=T }. (5.15) In vew of (5.7) and (5.14), we have from (5.15) ha and ˆτ Y a () = S a(ˆτ) (ˆτ)χ {ˆτ<T } + ξ a(ˆτ) χ {ˆτ=T } + N 1 ψ(r,y a (r),z a (r),a(r))dr (Y a (θ j ) Y a (θ j ))χ (,ˆτ] (θ j ) + K +,a (ˆτ) K +,a () j=1 ˆτ Z a (r)dw(r), [, ˆτ];a A. Y â() = Sâ(ˆτ) (ˆτ)χ {ˆτ<T } + ξâ(ˆτ) χ {ˆτ=T } + ˆN j=1 ˆτ ˆτ (Y â(ˆθ j ) hˆαj 1,ˆα j (ˆθ j,y â(ˆθ j )))χ (,ˆτ] (ˆθ j ) Zâ(r)dW(r), [, ˆτ]. ψ(r,y â(r),zâ(r),â(r))dr (5.16) (5.17) Comparng BSDE (5.2) wh (5.16), (5.14) and (5.17), repecvely, n vew of he fac ha Y a (θ j ) = Y αj 1 (θ j ) h αj 1,α j (θ j,y αj (θ j )) = h αj 1,α j (θ j,y a (θ j )), Y â(τ) Sâ(τ) (τ)χ {τ<t } + ξâ(τ) χ {τ=t }, and K +,a and K,â are ncreang procee, we have Y a () U a,ˆτ (), Y â() Uâ,τ (), Y â() = Uâ,ˆτ (), [,T]. Nong by defnon ha (5.18) Y a () = Y â() = Y (), [,T], we have Y () = Uâ,ˆτ (), U a,ˆτ () Uâ,ˆτ () Uâ,τ () [,T]. (5.19) Th how ha (â, ˆτ) a addle pon for he funconal U a,τ () a a funconal of (a,τ) A T, wh T beng he oaly of oppng me whch ake value n [,T]; or equvalenly, Y () = Uâ,ˆτ () = eupenf U a,τ () = enf a A τ τ eupu a,τ (), [,T]. (5.2) a A

36 Theorem 5.1 gve he unquene of Y. The unquene of oher componen (Z,K +,K ) of he oluon (Y,Z,K +,K ) a conequence of Doob-Meyer decompoon of Y. We conclude he followng reul. Theorem 5.2. Le Hypohee 1, 2 and 5 be afed. Aume ha he upper barrer S uper-regular wh S() Q() for [,T], and ha he ermnal value ξ L 2 (Ω, F T,P;R m ) ake value n Q(T). Then RBSDE (1.4) ha a unque adaped oluon (Y,Z,K +,K ). Theorem 5.3. Le Hypohee 1 and 3 () be afed. Aume ha he upper barrer S (S 2 ) m wh S() Q() for [,T], ha he ermnal value ξ L 2 (Ω, F T,P;R m ) ake value n Q(T), and ha k(,j) + k(j,l) > k(,l) for j, j l,,j,l Λ. Then RBSDE (4.2) ha a unque adaped oluon. Reference [1] Cvanc, J. and I. Karaza (1996): Backward SDE wh reflecon and Dynkn game. Annal of Probably 24 (4), [2] Dellachere, C. and P. A. Meyer (198): Probablé e Poenel, I-IV. Hermann, Par. [3] El Karou, N., C. Kapoudjan, E. Pardoux, S. Peng and M. C. Quenez (1997): Refleced oluon of backward SDE and relaed obacle problem for PDE. Annal of Probably 25 (2), [4] Gegou-Pe, A. and E. Pardoux (1996): Equaon dfférenelle ochaque rérograde réfléche dan un convexe. Sochac Rep. 57, [5] Hamadene, S., J.-P. Lepeler and A. Maou (1997): Double barrer backward SDE wh connuou coeffcen. In El Karou, N. and L. Mazlak (Ed) Backward ochac dfferenal equaon, Pman Reearch Noe n Mahemac Sere 364, [6] Hamadene, S. (22): Refleced BSDE wh dconnuou barrer and applcaon. Sochac Sochac Rep. 74(3-4), [7] Hamadene, S. and M. Haan (25): BSDE wh wo reflecng barrer: he general reul. Probab. Theory Rela. Feld 132, [8] Hamadene, S. and J. Zhang (27): The Sarng and Soppng Problem under Knghan Uncerany and Relaed Syem of Refleced BSDE. arxv:mah.pr/ [9] Hu, Y. and S. Peng (26): On he comparon heorem for mul-dmenonal BSDE. C.R.Mah Acad.Sc.Par 343, [1] Hu, Y. and S. Tang (27): Mul-dmenonal BSDE wh oblque reflecon and opmal wchng. arxv:mah.pr/ [11] Lepeler, J.-P. and M. Xu (25): Penalzaon mehod for refleced backward ochac dfferenal equaon wh one r.c.l.l. barrer. Sac Probably Leer 75, [12] Meyer, P.A. (1976): Un cour ur le négrale ochaque, Sémnare de Probablé X. Lecure Noe n. Mah. 511, Sprnger, Berln. [13] Pardoux, E. and S. Peng (199): Adaped oluon of a backward ochac dfferenal equaon. Syem Conrol Le. 14,

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