Long Jiang

Sheng-Jun Fan

Comptes Rendus Mathématique

C. R. Acad. Sci. Paris, Ser. I 348 (2010) 89–92Contents lists available at ScienceDirectC. R. Acad. Sci. Paris, Ser. Iwww.sciencedirect.comProbability TheoryUniqueness result for the BSDE whose generator is monotonic in yand uniformly continuous in z ✩Un résultat d’unicité pour une équation différentielle stochastique rétrograde dont legénérateur g est monotone en y et uniformément continue en zSheng-Jun Fan, Long JiangCollege of Sciences, China University of Mining & Technology, Xuzhou, Jiangsu 221116, PR Chinaa r t i c l e i n f o a b s t r a c tArticle history:Received 24 January 2009Accepted after revision 6 October 2009Available online 22 December 2009Presented by Paul MalliavinIn this Note, we prove that if g is continuous, monotonic and has a general growth iny, g is uniformly continuous in z, and (g(t,0,0))t∈[0,T ] is square integrable, then foreach square integrable terminal condition ξ , the one-dimensional backward stochasticdifferential equation (BSDE) with the generator g has a unique solution. This generalizessome corresponding (one-dimensional) results.© 2009 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.r é s u m éDans cette Note on démontre que si g est continue, monotone, de croissance quelconqueen y, g uniformément continue en z et (g(t,0,0))t∈[0,T ] est de carré intégrable, alorspour toute condition finale ξ de carré intégrable, en dimension un, l’équation différentiellestochastique rétrograde (BSDE) de générateur g, a une solution unique. Ce résultatgénéralise des résultats connus dans le cas de la dimension un.© 2009 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.1. IntroductionWe consider the following one-dimensional backward stochastic differential equation (BSDE for short):yt = ξ +T∫tg(s, ys, zs)ds −T∫tzs · dBs, t ∈ [0, T ], (1)where ξ is a square integral random variable termed the terminal condition, the random function g(ω, t, y, z) : Ω ×[0, T ]×R × Rd → R is progressively measurable for each (y, z), termed the generator of the BSDE (1), and B is a d-dimensionalBrownian motion. The solution (y·, z·) is a pair of square integrable, adapted processes. The triple (ξ, T , g) is called theparameters of the BSDE (1).✩ Supported by the National Natural Science Foundation of China (No. 10671205), National Basic Research Program of China (No. 2007CB814901), YouthFoundation of China University of Mining & Technology (Nos. 2006A041 and 2007A029) and Qing Lan Project.E-mail addresses: f_s_j@126.com (S.-J. Fan), jianglong365@hotmail.com (L. Jiang).1631-073X/$ – see front matter © 2009 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.doi:10.1016/j.crma.2009.10.02390 S.-J. Fan, L. Jiang / C. R. Acad. Sci. Paris, Ser. I 348 (2010) 89–92Such equations, in nonlinear case, were firstly introduced by [11], who established an existence and uniqueness resultof a solution of the BSDE (1) under the Lipschitz assumption of the generator g . Since then, many efforts have been donein relaxing the Lipschitz hypothesis on g; see, for instance [10,9,8,1–3], etc. In particular, under the conditions that gis continuous, monotonic and has a general growth in y, g is Lipschitz continuous in z, and (g(t,0,0))t∈[0,T ] is squareintegrable, [12] proved the existence and uniqueness of the solution to the BSDE (1). Furthermore, [4] proved the existenceof the solution to the BSDE (1) if the above Lipschitz continuity condition is replaced with the continuity and linear growthcondition. Recently, under the conditions that g does not depend on y, g is uniformly continuous in z, and (g(t,0))t∈[0,T ]is a bounded process, [7] obtained a uniqueness result on the solution of the BSDE (1).Enlightened by these results, this Note proves that if g is continuous, monotonic and has a general growth in y, g isuniformly continuous in z, and the process (g(t,0,0))t∈[0,T ] is square integrable, then for each square integrable terminalcondition ξ , the BSDE (1) has a unique solution, which generalizes the corresponding (one-dimensional) results in [11,12,7].It is worth mentioning that we use a different method from that used in [7], and our result does not need the conditionthat (g(t,0,0))t∈[0,T ] is a bounded process.2. Main resultLet (Ω, F , P ) be a probability space carrying a standard d-dimensional Brownian motion (Bt)t0. Fix a terminal timeT > 0, let (Ft)t0 be the natural σ -algebra generated by (Bt)t0 and assume FT = F . For every positive integer n, we use| · | to denote norm of Euclidean space Rn . For t ∈ [0, T ], let L2(Ω, Ft , P ) denote the set of all Ft -measurable random variableξ such that E|ξ |2 < +∞. Let L2F (0, T ;Rn) denote the set of Ft -progressively measurable, Rn-valued process {Xt, t ∈ [0, T ]}such that‖X‖2 =̂(ET∫0|Xt |2 dt)1/2< +∞.Now, let ξ ∈ L2(Ω, FT , P ) be a terminal condition, g be the Ft -progressively measurable generator of the BSDE (1).A solution of the BSDE (1) is a pair of processes (y·, z·) in L2F (0, T ;R1+d) which satisfies BSDE (1) and y· is a continuousprocess. In this Note, we further assume that g satisfies some of the following assumptions:(H1) The process (g(t,0,0))t∈[0,T ] ∈ L2F (0, T ;R1).(H2) dP × dt − a.s., (y, z) → f (ω, t, y, z) is continuous.(H3) g is monotonic in y, i.e., there exists a constant μ  0, such that, dP × dt − a.s.,∀y1, y2, z,(g(ω, t, y1, z) − g(ω, t, y2, z))(y1 − y2)  μ|y1 − y2|2.(H4) g has a general growth with respect to y, i.e., dP × dt − a.s.,∀y, ∣∣g(ω, t, y,0)∣∣  ∣∣g(ω, t,0,0)∣∣ + ϕ(|y|),where ϕ : R+ → R+ is an increasing continuous function.(H5) g is uniformly continuous in z and uniform with respect to (ω, t, y), i.e., there exists a continuous, nondecreasingfunction φ(·) from R+ to itself with at most linear growth and φ(0) = 0 such that dP × dt − a.s.,∀y, z1, z2,∣∣g(ω, t, y, z1) − g(ω, t, y, z2)∣∣  φ(|z1 − z2|).Here and henceforth we denote the constant of linear growth for φ by A, i.e., 0  φ(x)  A(x + 1) for all x ∈ R+ (see[5] for details).(H5′) g is Lipschitz continuous in z and uniform with respect to (ω, t, y), i.e., there exists a constant C  0 such thatdP × dt − a.s.,∀y, z1, z2,∣∣g(ω, t, y, z1) − g(ω, t, y, z2)∣∣  C |z1 − z2|.Remark 1. Under the conditions of (H1)–(H4) and (H5′), [12] established the existence and uniqueness of the solution to theBSDE with the generator g . This Note aims at establishing the existence and uniqueness under the conditions of (H1)–(H5).Obviously, (H5′) can imply (H5).In the following, we will put forward and prove our main result that if g is continuous, monotonic and has a generalgrowth in y, g is uniformly continuous in z, and the process (g(t,0,0))t∈[0,T ] is square integrable, then the BSDE with thegenerator g has a unique solution, which generalizes the corresponding (one-dimensional) results in [11,12,7]. Rigorously,we have:S.-J. Fan, L. Jiang / C. R. Acad. Sci. Paris, Ser. I 348 (2010) 89–92 91Theorem 1. Assume that g satisfies (H1)–(H5). Then for each ξ ∈ L2(Ω, FT , P ), the BSDE with parameters (ξ, T , g) has a uniquesolution (y·, z·).Proof. Existence: Since g satisfies (H4) and (H5), then dP × dt − a.s., for each (y, z) ∈ R1+d we have∣∣g(ω, t, y, z)∣∣  ∣∣g(ω, t, y, z) − g(ω, t, y,0)∣∣ + ∣∣g(ω, t, y,0)∣∣ φ(|z|) + ∣∣g(ω, t,0,0)∣∣ + ϕ(|y|)∣∣g(ω, t,0,0)∣∣ + A + ϕ(|y|) + A|z|.Thus the existence of the solution to the BSDE with parameters (ξ, T , g) follows from Theorem 4.1 in [4].Uniqueness: Assume that (y·, z·) and (y′·, z′·) be two solutions to the BSDE with parameters (ξ, T , g) in L2F (0, T ;R1+d).Let ŷ· = y· − y′·, ẑ· = z· − z′· then we haveŷt =T∫t[g(s, ys, zs) − g(s, y′s, z′s)]ds −T∫tẑs · dBs, t ∈ [0, T ].Using the Tanaka–Meyer formula (see [6]), one gets that for each t ∈ [0, T ],| ŷt | =T∫tŷs| ŷs|1 ŷs 	=0[g(s, ys, zs) − g(s, y′s, z′s)]ds − (L0T − L0t ) −T∫tŷs| ŷs|1 ŷs 	=0 ẑs · dBs=T∫t[μ| ŷs| + 1 ŷs 	=0φ(|ẑs|)] ds + (V T − Vt) −T∫tŷs| ŷs|1 ŷs 	=0 ẑs · dBs, (2)where L0t is the local time of ŷt at 0 andVt = −t∫0[(μ| ŷs| + 1 ŷs 	=0φ(|ẑs|)) − ŷs| ŷs|1 ŷs 	=0(g(s, ys, zs) − g(s, y′s, z′s))]ds − L0t .Thanks to (H3) and (H5), we know thatŷs[g(s, ys, zs) − g(s, y′s, z′s)] = ŷs[g(s, ys, zs) − g(s, y′s, zs)] + ŷs[g(s, y′s, zs) − g(s, y′s, z′s)] μ| ŷs|2 + | ŷs|φ(|ẑs|).This inequality combining that L0t is a continuous increasing process yields that (Vt)t∈[0,T ] is a continuous decreasing processwith V 0 = 0. Moreover, from (2) one also knows thatV T = ŷ0 −T∫0[μ| ŷs| + 1 ŷs 	=0φ(|ẑs|)]ds +T∫0ŷs| ŷs|1 ŷs 	=0 ẑs · dBs,then recalling that φ(·) increases at most linearly, from Hölder inequality one hasE sup0tT|Vt |2 = E|V T |2  4| ŷ0|2 + 8T ET∫0[μ2| ŷs|2 +(A|ẑs| + A)2]ds + 2ET∫0|ẑs|2 ds < +∞. (3)In the following, for each n  1, from [11], one knows that the following BSDE has a unique solution (Y n· , Zn· ) ∈L2F (0, T ;R1+d):Y nt =T∫t[μY ns + (n + 2A)∣∣Zns ∣∣ + φ(2An + 2A)]ds −T∫tZns · dBs, t ∈ [0, T ]. (4)Recalling that φ(·) is a nondecreasing function from R+ to itself with at most linear growth, one can prove that for eachn ∈ N,φ(x)  (n + 2A)x + φ(2A)(5)n + 2A92 S.-J. Fan, L. Jiang / C. R. Acad. Sci. Paris, Ser. I 348 (2010) 89–92holds true for each x ∈ R+ . In fact, if 0  x  2An+2A , the conclusion is obvious considering φ(·) is nondecreasing. And, if2An+2A < x < 1, we have (n + 2A)x > 2A = A + A > Ax + A  φ(x). Finally, in the case of x  1, we also have (n + 2A)x >2Ax = Ax + Ax  Ax + A  φ(x). Therefore, for each n  1, from (5) we haveg′(s) := μ| ŷs| + 1 ŷs 	=0φ(|ẑs|)  μ| ŷs| + 1 ŷs 	=0(n + 2A)|ẑs| + φ(2An + 2A)= μ(| ŷs|) + (n + 2A)∣∣∣∣ ŷs| ŷs|1 ŷs 	=0 ẑs∣∣∣∣ + φ(2An + 2A). (6)Obviously, g′(·) ∈ L2F (0, T ;R1). Thus, considering inequalities (3) and (6), and the fact that (0 − Vt)t∈[0,T ] is a continuousincreasing process, by using Comparison Theorem 1.3 in [13] to compare the solution of the BSDE (2) with the one of theBSDE (4), we know that for each n  1 and t ∈ [0, T ], | ŷt |  Y nt , dP − a.s.On the other hand, one may verify directly that for each t ∈ [0, T ],Y nt =1μ(eμ(T −t) − 1)φ( 2An + 2A)and Znt ≡ 0.Thus since φ(·) is a continuous function and φ(0) = 0, we have | ŷt |  limn→∞ Y nt = 0, dP − a.s.Hence, for each t ∈ [0, T ] we have, yt = y′t , dP − a.s. That is to say, the solution to the BSDE with parameters (ξ, T , g) isunique. The proof of Theorem 1 is complete. Remark 2. From the proof of Theorem 1, one can see that we need only the monotonicity condition in y (see (H3)) anduniform continuity condition in z (see (H5)) to ensure the uniqueness of the solution of the BSDE:From Theorem 1 one can easily obtain the following Corollaries which can be regarded as the extensions of the corre-sponding (one-dimensional) results in [7,11].Corollary 1. Assume that g satisfies (H1) and (H5). Moreover, let g be independent of y. Then for each ξ ∈ L2(Ω, FT , P ), the BSDEwith parameters (ξ, T , g) has a unique solution.Corollary 2. Assume that g satisfies (H1) and (H5). Moreover, let g be Lipschitz continuous in y. Then for each ξ ∈ L2(Ω, FT , P ), theBSDE with parameters (ξ, T , g) has a unique solution.AcknowledgementThe authors would like to thank the referee for his/her careful reading and helpful suggestions.References[1] K. Bahlali, Backward stochastic differential equations with locally Lipschitz coefficient, C. R. Acad. Sci. Paris, Ser. 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Uniqueness result for the BSDE whose generator is monotonic in <i>y</i> and uniformly continuous in <i>z</i>

Fan, Sheng-Jun

Jiang, Long

Numérisation de Documents Anciens Mathématiques

http://www.numdam.org/item/10.1016/j.crma.2009.10.023.pdf

Uniqueness result for the BSDE whose generator is monotonic in <i>y</i> and uniformly continuous in <i>z</i>

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