BACKWARD STOCHASTIC DIFFERENTIAL EQUATIONS WITH OBLIQUE

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BACKWARD STOCHASTIC DIFFERENTIAL EQUATIONS WITH OBLIQUE

REFLECTION AND LOCAL LIPSCHITZ DRIFT

AUGUSTE AMAN and MODESTE N’ZI URF de Math´ematiques et Informatique

22 BP 582 Abidjan 22, Cˆote d’Ivoire

(Received January, 2003; Revised September, 2003)

We consider reflected backward stochastic differential equations with time and space dependent coefficients in an orthant, and with oblique reflection. Existence and uniqueness of solution are established assuming local Lipschitz continuity of the drift, Lipschitz continuity and uniform spectral radius conditions on the reflection matrix.

Keywords:Backward Stochastic Differential Equations, Oblique Reflection, Brownian Motion.

AMS (MOS) subject classification:60H10, 60H20.

1 Introduction

It was mainly during the last decade that the theory of backward stochastic differential equations took shape as a distinct mathematical discipline. This theory has found a wide field of applications as in stochastic optimal control and stochastic games (see Hamad`ene and Lepeltier [9]), in mathematical finance via the theory of hedging and non- linear pricing theory for imperfect markets (see El Karoui et al.[6]). Backward stochastic differential equations also appear to be a powerful tool for constructing Γ−martingales on manifolds (see Darling [4]). These kind of equations provide probabilistic formulae for solutions to partial differential equations (see Pardoux and Peng [14]).

Consider the following linear backward stochastic differential equation −dYs = [Ysβs+Zsγs]ds−ZsdBs, 0≤s≤T

YT = ξ. (1.1)

As is well known, equation (1.1) was first introduced by Bismut [1, 2] when he was studying the adjoint equation associated with the stochastic maximum principle in optimal stochastic control. It is used in the context of mathematical finance as the model behind the Black and Scholes formula for the pricing and hedging option.

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The development of general backward stochastic differential equation (BSDE in short)

−dYs = f(s, Ys, Zs)ds−ZsdBs, 0≤s≤T YT = ξ

begins with the paper of Pardoux and Peng [14]. Since then, BSDEs have been inten- sively studied. For example, BSDE with reflecting barrier have been studied among others by El Karoui et al.[5], Cvitanic and Karatzas [3], Matoussi [12] and Hamad`ene et al.[10] in the one dimensional case. The higher dimensional one has been considered by Gegout-Petit and Pardoux [8] for reflection in a convex domain. The multivalued context can be found in Pardoux and Rascanu [15], N’zi and Ouknine [13], Hamad`ene and Ouknine [11] and Essaky et al [7].

These works concern the case of normal reflection at the boundary. In the last two decades, thanks to the numerous applications in queuing theory, the deterministic as well as stochastic Skorokhod problem (in a convex polyhedron with oblique reflection at the boundary) has been studied by many authors. Recently, S. Ramasubramanian [16] has considered reflected backward stochastic differential equations (RBSDE’s) in an orthant with oblique reflection at the boundary. He has established the existence and uniqueness of the solution under a uniform spectral radius condition on the reflection matrix (plus of course, a Lipschitz continuity condition on the coefficient).

The aim of this article is to weaken the Lipschitz condition on the drift to a locally Lipchitz one. The paper is organized as follows. In section 2, we introduce the under- lying assumptions and state the main result. Section 3 is devoted to the proof of the main result.

2 Assumptions and Formulation of the Main Result

Let B = {B(t) = (B1(t), ..., Bd(t)) :t ≥0}be a d− dimensional standard Brownian motion defined on a probability space (Ω,F, P) and let {F_t}be the natural filtration generated byB,withF₀ containing allP−null sets.

Let G={x∈R^d:x_i>0,1≤i≤d}denote thed−dimensional positive orthant.

We are given the following:

• T >0 is a terminal time;

• ξ is anF_T−measurable, bounded, G−valued random variable;

• b: [0;T]×Ω×R^d−→R^d, R: [0;T]×Ω×R^d−→Md(R) are both bounded measurable functions such that for every y ∈ R^d, b(., ., y) = (b1(., ., y), . . . , bd(., ., y)) and R(., ., y) = (rij(., ., y))1≤i,j≤d are F_t−predictable processes. We also assume that rii(., ., .)≡ 1. (Here Md(R) denotes the class of d×d matrices with real entries).

Definition 2.1: A triple Y = {Y(t) = (Y1(t), .., Yd(t)) : t ≥ 0};Z = {Z(t) = (Zij(t))1≤i,j≤d :t≥0}andK ={K(t) = (K1(t), .., Kd(t)) :t≥0}of {F_t} −progressively measurable integrable processes is said to solve RBSDE (ξ, b, R) if the following hold:

(i) (Y, Z, K) is a continuous R^d×Md(R)×R^d−valued process;

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(ii) for everyi= 1, ..., d,and 0≤t≤T, Yi(t) = ξi+

Z T t

bi(s, Y(s))ds− Xd j=1

Z T t

Zij(s)dBj+Ki(T)−Ki(t)

+X

j6=i

Z T t

rij(s, Y(s))dKj(s);

(iii) for every 0≤t≤T, Y(t)∈G;

(iv) for every 1≤i≤d, K_i(0) = 0, K_i(·) is nondecreasing and can increase only when Y_i(·) = 0,that is

Ki(t) = Z t

0

1{0}(Yi(s))dKi(s).

We make the following assumptions on the coefficientsb,R.

(A1) For every 1 ≤ i ≤ d, y 7→ bi(t, ω, y) is locally Lipschitz continuous, uniformly over (t, ω); there is a constant βi such that |bi(t, ω, y)| ≤ βi,for all (t, ω, y) ∈ [0;T]×Ω×R^d.

(A2) For 1≤i, j≤d, y7→rij(t, ω, y) is Lipschitz continuous, uniformly over (t, ω). (A3) For every i6=j there exists constant vij such that|rij(t, ω, y)| ≤vij. SetV =

(vij) with vii = 0.We assume that σ(V)< 1, where σ(V) denotes the spectral radius ofV . Therefore,

(I−V)⁻¹=I+V +V²+V³+· · · In the sequel, we putβ= (β1, ..., βd).

Remark 2.1: In view of(A3),there exists constants aj,1≤j ≤dand 0< α <1

such that X

i6=j

ai|rij(t, ω, y)| ≤X

i6=j

aivij≤αaj

for allj= 1, . . . , dand (t, ω, y)∈[0;T]×Ω×R^d.

Let Hstands for the space of all{F_t} −progressively measurable, continuous pairs of processes{Y(t) = (Y1(t), .., Yd(t)) :t≥0}and{K(t) = (K1(t), .., Kd(t)) :t≥0}such that

(i) for every 0≤t≤T, Y(t)∈G;

(ii) for every 1≤i≤d, Ki(0) = 0;Ki(·) is nondecreasing and can increase only when Yi(·) = 0;

(iii) E P_d

i=1

RT

0 e^θtai|Yi(t)|dt

<+∞;

(iv) E _d

P

i=1

RT

0 e^θtaiϕt(Ki)dt

<+∞; whereϕt(g) denotes the total variation ofgover [t, T] andθ >0 is a fixed constant which will be chosen suitably later.

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For (Y, K),(bY ,K)b ∈ H,we define the metric d((Y, K),(Y ,b K))b = E

Xd i=1

Z T 0

e^θtai|Yi(t)−Ybi|dt

!

+E Xd i=1

Z T 0

e^θtaiϕt(Ki−Kbi)dt

!

. (2.1)

It is not difficult to see that (H, d) is a complete metric space.

LetHe denote the collection of all (Y, K)∈ H such that there exists an{F_t} −progressively measurable process{D(t) = (D₁(t), ..., D_d(t)) :t≥0},with

0≤Di(t)≤((I−V)⁻¹β)i a.s. and Ki(t) = Z t

0

Di(s)ds.

SinceHe is a closed subset ofH, (H, d) is a complete metric space.e We consider the normkyk= P

ai|yi| which is equivalent to the Euclidean norm in R^d. So, we may assume that the local Lipschitz continuity in (A1) and Lipschitz continuity in(A2)are with respect to this norm.

Before stating our main result, let us remark that if (Y, K), (bY ,K)b ∈HewithDi,Dbi

being respectively the derivatives ofKi,Kbi then ϕ_t(K_i−Kb_i) =

Z T t

|D_i(s)−Db_i(s)|ds.

Therefore, using integration by parts in (2.1), we have d((Y, K),(Y ,b K))b = E

Xd i=1

Z T 0

e^θtai|Yi(t)−Ybi|dt

!

+E Xd i=1

Z T 0

e^θt−1

θ ai|Di(t)−Dbi(t)|dt

!

= E

Z T 0

e^θt||Y(t)−Yb(t)||dt

! +E

Z T 0

e^θt−1

θ ||D(t)−D(t)||dtb

! .

For everyz∈Md(R),we put

|||z|||=



 Xd j=1

Xd i=1

ai|zij|²





1/2

.

Let Hdenote the space of all F_t-progressively measurable processes Z = (Zij)1≤i,j≤d

such that

E Z T

0

|||Z(t)|||²dt

!

<+∞, endowed with the norm

|Z|=

"

E Z T

0

|||Z(t)|||²dt

!#^1/2 .

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It is clear thatHis a Banach space.

Now, we state our main result:

Theorem 2.1: Assume(A1)-(A3). Letξbe a bounded,F_T−measurableG−valued random variable. Then there is a unique couple ((Y, K), Z) ∈ H×He solving RBSDE (ξ, b, R).

3 Proof of the Main Result

The proof of Theorem 2.1 needs some preliminary lemmas.

Lemma 3.1: Letb be a process satisfying assumption (A1). Then there exists a sequence of processes bⁿ such that

(i) for each n, bⁿ is Lipschitz continuous and |bⁿ_i(t, ω, y)| ≤βi , for all1≤i≤dand (t, ω, y)∈[0, T]×Ω×R^d;

(ii) for every p, ρp(bⁿ−b)→0asn→+∞, where ρ_p(f) =E

Z T 0

e^θs sup

|x|<p

||f(s, x)||ds

! .

Proof:Letψnbe a sequence of smooth functions with support in the ballB(0, n+ 1) such that supψn = 1.It not difficult to see that the sequence (bⁿ)_n≥1of truncated functions defined bybⁿ =bψn,satisfies all the properties quoted above.

In view of Ramasubramanian [16], there exists a unique couple of processes{((Yⁿ(t), Kⁿ(t)), Zⁿ(t)) :t≥0} ∈H×He solution to the RBSDE (ξ, bⁿ, R).

We formulate some uniform estimates for the processes{((Yⁿ(t), Kⁿ(t)), Zⁿ(t)) :t≥0}

in the following way.

Lemma 3.2: Assume (A1)-(A3). Then there exists a constant C, such that for every n≥1

E Z T

0

e^θt||Yⁿ(t)||dt

! +E

Z T 0

e^θt−1

θ ||Dⁿ(t)||dt

!

< C. (3.1)

Proof:Let the triple (Yⁿ, Kⁿ, Zⁿ) be the unique solution of RBSDE (ξ, bⁿ, R).We have for everyi= 1, ..., d,and 0≤t≤T

Yⁿ(t) = ξi+ Z T

t

bⁿ_i(s, Yⁿ(s))ds− Z T

t

Xd j=1

Z_ijⁿ(s)dBj+K_iⁿ(T)−K_iⁿ(t)

+X

j6=i

Z T t

rij(s, Yⁿ(s))dKj(s).

Since

ϕt(K_iⁿ) = Z T

t

|Dⁿ_i(s)|ds,

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applying Theorem 3.2 [16] and using integration by parts, we obtain

E Z T

0

θe^θt|Y_iⁿ(t)|dt

! +E

Z T 0

θe^θtϕt(K_iⁿ)dt

!

= E

Z T 0

! +E

Z T 0

e^θt−1

|Dⁿ_i(t)|dt

!

≤ E e^θT −1

|ξi| +E

Z T 0

e^θt−1

|bⁿ_i(t, Yⁿ(t))|dt

!

+E



Z T 0

e^θt−1 X

j6=i

|r_ij(t, Yⁿ(t))||Dⁿ_j(t)|dt



.

Therefore

E Z T

0

! +E

Z T 0

θe^θtϕ_t(K_iⁿ)dt

!

≤E( e^θT−1

|ξi|) +E Z T

0

e^θt−1 βidt

!

+E



Z T 0

e^θt−1 X

j6=i

υ_ij((I−V)⁻¹β)_jdt



. (3.2)

Let us note that

d((Yⁿ, Kⁿ),(0,0)) =E Z T

0

e^θtkYⁿ(t)kdt

! +E

Z T 0

e^θt−1

θ kDⁿ(t)kdt

! .

Multiplying (3.2) byai and adding leads to

θd((Yⁿ, Kⁿ),(0,0))≤E Xd i=1

e^θT−1 ai|ξi|

! +E

Xd i=1

Z T 0

(e^θt−1)aiβidt

!

+E



Z T 0

(e^θt−1) Xd i=1

X

j6=i

a_iv_ij((I−V)⁻¹β)_jdt



.

In view of the inequality

X

i6=j

υijai≤αaj,

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we have

θd((Yⁿ, Kⁿ),(0,0))≤E Xd i=1

e^θT−1 ai|ξi|

! +E

Xd i=1

Z T 0

(e^θt−1)aiβidt

!

+αE



Z T 0

(e^θt−1) Xd j=1

aj((I−V)⁻¹β)jdt





≤ e^θT−1 E||ξ||

+(||β||+α||((I−V)⁻¹β)||) Z T

0

(e^θt−1)dt

≤C.

Hence inequality (3.1) is proved.

Now, we shall prove the convergence of the sequence (Yⁿ, Kⁿ, Zⁿ)n≥1.

Theorem 3.1: Assume(A1)-(A3). Then there exists ((Y, K), Z) ∈H×He such that

n→+∞lim (

E Z T

0

e^θtkYⁿ(t)−Y(t)kdt

! +E

Z T 0

e^θt−1

θ kDⁿ(t)−D(t)kdt )

= 0 and

n→+∞lim E Z T

0

|||Zⁿ(t)−Z(t)|||²dt

!

= 0, where

Ki(t) = Z t

0

Di(s)ds, i= 1, . . . , d.

Proof: It follows from the same idea used in the proof of inequality (3.1) that E

Z T 0

θe^θt|Y_i^m(t)−Y_iⁿ(t)|dt

! +E

Z T 0

θe^θtϕt(K_i^m−K_iⁿ)dt

!

= E

Z T 0

! +E

Z T 0

e^θt−1

|D^m_i (t)−D_iⁿ(t)|dt

!

≤ E Z T

0

(e^θt−1)|b^m_i (t, Y^m(t))−bⁿ_i(t, Yⁿ(t))|dt

!

+E



Z T 0

(e^θt−1)

X

j6=i

rij(t, Y^m(t))D^m_j (t)−rij(t, Yⁿ(t)Dⁿ_j(t) dt





≤ E Z T

0

(e^θt−1)|b^m_i (t, Y^m(t))−bⁿ_i(t, Yⁿ(t))|dt

!

+E



Z T 0

(e^θt−1)X

j6=i

|r_ij(t, Y^m(t)−r_ij(t, Yⁿ(t))||D_jⁿ(t)|dt





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+E



Z T 0

(e^θt−1)X

j6=i

|rij(t, Y^m(t))||D_j^m(t)−Dⁿ_j(t)|dt



.

For an arbitrary number N > 1, let LN be the Lipschitz constant of b in the ball B(0, N). We put

A^N_m,n={ω∈Ω,||Y^m(t, ω)||+||Yⁿ(t, ω)||> N}, A^N_m,n= Ω\A^N_m,n. It follows that

E Z T

0

! +E

Z T 0

(e^θt−1)|D^m_i (t)−Dⁿ_i(t)|dt

!

≤ E Z T

0

(e^θt−1)|b^m_i (t, Y^m(t))−bⁿ_i(t, Yⁿ(t))|1_AN m,ndt

!

+E Z T

0

(e^θt−1)|b^m_i (t, Y^m(t))−bⁿ_i(t, Yⁿ(t))|1_AN m,n

dt

!

+E



Z T 0

(e^θt−1)X

j6=i

|rij(t, Y^m(t)−rij(t, Yⁿ(t))||Dⁿ_j(t)|dt





+E



Z T 0

(e^θt−1)X

j6=i

|rij(t, Y^m(t))| |D^m_j (t)−D_jⁿ(t)|dt





= I₁+I₂+I₃+I₄. (3.3)

It not difficult to check that I2 = E

Z T 0

(e^θt−1)|b^m_i (t, Y^m(t))−bⁿ_i(t, Yⁿ(t))|1_AN m,n

dt

!

≤ E Z T

0

(e^θt−1)|b^m_i (t, Y^m(t))−b_i(t, Y^m(t))|1_AN m,n

dt

+E Z T

0

(e^θt−1)|bi(t, Y^m(t))−bi(t, Yⁿ(t))|1_AN m,n

dt

!

+E Z T

0

(e^θt−1)|bi(t, Yⁿ(t))−bⁿ_i(t, Yⁿ(t))|1_AN m,n

dt

! .

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Sinceb_i is L_N ai

−locally Lipschitz, we get

I2 ≤ E Z T

0

e^θt−1

|b^m_i (t, Y^m(t))−bi(t, Y^m(t))|1_AN m,n

dt

!

+E Z T

0

e^θt−1

|b_i(t, Yⁿ(t))−bⁿ_i(t, Yⁿ(t))|1_AN m,n

dt

!

+LN

ai

E Z T

0

e^θt−1

kY^m(t)−Yⁿ(t)kdt

!

. (3.4)

In view of the Lipschitz condition onR and the boundedness ofDⁿ_j(t), we obtain that there existsC1>0 such that

I₃ ≤ LE Z T

0

e^θt−1

kY^m(t)−Yⁿ(t)kD_jⁿ(t)dt

!

≤ LE



Z T 0

e^θt−1

kY^m(t)−Yⁿ(t)kX

j6=i

((I−V)⁻¹β)jdt





≤ LC1E Z T

0

e^θt−1

kY^m(t)−Yⁿ(t)kdt. (3.5)

Now, from the boundness ofR,we have

I4≤E



Z T 0

e^θt−1 X

j6=i

vij

D^m_j (t)−Dⁿ_j(t)dt



. (3.6)

By virtue of (3.3)-(3.6), we deduce that E

Z T 0

! +E

Z T 0

e^θt−1

|D_i^m(t)−Dⁿ_i(t)|dt

!

≤ E Z T

0

e^θt−1

|b^m_i (t, Y^m(t))−bⁿ_i(t, Yⁿ(t))|1_AN m,ndt

!

+E Z T

0

e^θt−1

|b^m_i (t, Y^m(t))−b_i(t, Y^m(t))|1_AN m,n

dt

!

+E Z T

0

e^θt−1

|b_i(t, Yⁿ(t))−bⁿ_i(t, Yⁿ(t))|1_AN m,n

dt

!

+ LN

a_i +LC1

E

Z T 0

e^θt−1

kY^m(t)−Yⁿ(t)kdt

!

+E



Z T 0

e^θt−1 X

j6=i

v_ijD_j^m(t)−Dⁿ_j(t)dt



. (3.7)

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Multiplying (3.7) byai, adding and usingP

i6=jaivij ≤αaj, we obtain θd((Y^m, K^m),(Yⁿ, Kⁿ))

≤ E Xd i=1

Z T 0

e^θt−1

ai|b^m_i (t, Y^m(t))−bⁿ_i(t, Yⁿ(t))|1_AN m,ndt

!

+E Xd

i=1

Z T 0

e^θt−1

a_i|b^m_i (t, Y^m(t))−b_i(t, Y^m(t))|1_AN m,n

dt

!

+E Xd

i=1

Z T 0

e^θt−1

ai|bi(t, Yⁿ(t))−bⁿ_i(t, Yⁿ(t))|1_AN m,n

dt

!

+ dLN+ Xd i=1

ai

! LC1

! E

Z T 0

e^θt−1

||Y^m(t)−Yⁿ(t)||dt

!

+αE Z T

0

e^θt−1

||D^m(t)−Dⁿ(t)||dt

! .

Choosingθ large enough such that ¹_θ

dL_N+Pd i=1a_i

LC₁

≤αleads to d((Y^m, K^m),(Yⁿ, Kⁿ)) ≤ αd((Y^m, K^m),(Yⁿ, Kⁿ))

+1

θρN(bⁿ−b) +1

θρN(b^m−b) +1

θC_m,n^N (3.8)

where

C_m,n^N = E Xd i=1

Z T 0

e^θt−1

ai|b^m_i (t, Y^m(t))−bⁿ_i(t, Yⁿ(t))|1_AN m,ndt

!

≤ 2 Xd i=1

Z T 0

e^θt−1 aiβiE

1_AN m,n

dt

≤ 2 N

Xd i=1

aiβi

Z T 0

e^θt−1

E(kYⁿ(t)k+kY^m(t)k)dt.

LetC2 be such that

Xd i=1

aiβi< C2. We have

C_m,n^N ≤2C2

N E Z T

0

e^θt−1

(kYⁿ(t)k+kY^m(t)k)dt.

By virtue of (3.1), there existsC >0 such that C_m,n^N ≤ C

N.

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Therefore

(1−α)d((Yⁿ, Kⁿ),(Y^m, K^m))≤ 1 θ

C N

+1

θρN(bⁿ−b) +1

θρN(b^m−b). (3.9) Passing to the limit onn, mandN in (3.9 ), we deduce that (Yⁿ, Kⁿ)_n∈Nis a Cauchy sequence inH. Sincee He is a Banach space, we set

n→+∞lim Yⁿ=Y, and lim

n→+∞Kⁿ=K.

If we return to the equation satisfied by the triple (Yⁿ, Kⁿ, Zⁿ)_n∈Nand use Itˆo’s formula, we have

E(|Y_i^m(t)−Y_iⁿ(t)|²) +E



Z T 0

Xd j=1

|Z_ij^m(s)−Z_ijⁿ(s)|²ds





= 2E Z T

0

|Y_i^m(s)−Y_iⁿ(s)| |b^m_i (s, Y^m(s))−bⁿ_i(s, Yⁿ(s))|ds

!

+2E Z T

0

|Y_i^m(s)−Y_iⁿ(s)| |D^m_i (s)−Dⁿ_i(s)|ds

!

+2E



Z T 0

|Y_i^m(s)−Y_iⁿ(s)|

X

j6=i

rij(s, Y^m(s))D^m_j (s)−rij(s, Yⁿ(s))Dⁿ_j(s) ds





≤ 4β_iE Z T

0

|Y_i^m(s)−Y_iⁿ(s)|ds

! + 4

(I−V)⁻¹β

i

E Z T

0

!

+4X

j6=i

vij

(I−V)⁻¹β

j

E Z T

0

!

. (3.10)

Multiplying (3.10) byai and adding leads to the existence ofC >0 such that

E Xd i=1

ai|Y_i^m(t)−Y_iⁿ(t)|²

! +E



Z T 0

Xd i=1

Xd j=1

ai

Z_ij^m(s)−Z_ijⁿ(s)²ds





≤ 4E Z T

0

Xd i=1

aiβi|Y_i^m(s)−Y_iⁿ(s)|ds

!

+4E Z T

0

Xd i=1

(I−V)⁻¹β

i

ai|Y_i^m(s)−Y_iⁿ(s)|ds

!

+4E



Z T 0

Xd i=1

X

j6=i

aivij((I−V)⁻¹β)j|Y_i^m(s)−Y_iⁿ(s)|ds



.

≤ CE Z T

0

Xd i=1

a_i|Y_i^m(s)−Y_iⁿ(s)|ds

!

. (3.11)

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Passing to the limit on m, n, we deduce that (Zⁿ)_n≥1 is a Cauchy sequence in the BanachH. SinceHis a Banach space, we put

Z= lim

n→+∞Zⁿ.

Lemma 3.3:Let(Yⁿ, Kⁿ, Zⁿ)_n≥1 be the unique solution of the RBSDE (ξ, bⁿ, R). Then

bⁿ(., Yⁿ) converges tob(., Y) in (L¹₊(Ω×[0, T], dP×e^θtdt)).

Proof:Set

A^N_n ={ω∈Ω,||Yⁿ(t, ω)||+||Y(t, ω)||> N}, A^N_n = Ω\A.

We have

E Z T

0

e^θt|bⁿ_i(t, Yⁿ(t))−bi(t, Y(t))|dt

!

≤E Z T

0

e^θt|bⁿ_i(t, Yⁿ(t))−bi(t, Y(t))|1_AN ndt

!

+E Z T

0

e^θt|bⁿ_i(t, Yⁿ(t))−b_i(t, Yⁿ(t))|1_AN n

dt

!

+E Z T

0

e^θt|bi(t, Yⁿ(t))−bi(t, Y(t))|1_AN n

dt

!

≤2βi

N E Z T

0

e^θt(kYⁿ(t)k+kY(t)k)dt

!

+E Z T

0

e^θt|bⁿ_i(t, Yⁿ(t))−bi(t, Yⁿ(t))|1_AN n

dt

!

LN

ai

E Z T

0

!

. (3.12)

Multiplying (3.12) byai and adding, we get E

Z T 0

e^θtkbⁿ(t, Yⁿ(t))−b(t, Y(t))kdt

!

≤ρ_N(bⁿ−b) + 2 N

Xd i

β_ia_iE Z T

0

e^θt(kYⁿ(t)k+kY(t)k)dt

!

+dL_NE Z T

0

! .

By virtue of (3.1), we deduce that there existsC >0 such that E

Z T 0

e^θtkbⁿ(t, Yⁿ(t))−b(t, Y(t))kdt

!

≤ρN(bⁿ−b) + C

N +dLNE Z T

0

! .

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Passing to the limit onn, N, completes the proof of Lemma 3.5.

Proof of Theorem 2.1:

Existence: Combining Lemmas (3.2)-(3.5) and passing to the limit in the RBSDE (ξ, bⁿ, R), we deduce that the triple{(Y(t), K(t), Z(t)),0≤t≤T}is a solution of our RBSDE (ξ, b, R).

Uniqueness: Let{(Y(t), K(t), Z(t)),0≤t≤T}andn

(Y⁰(t), K⁰(t), Z⁰(t)),0≤t≤To be two solutions of our RBSDE. For everyt≥0,define

(∆Y(t),∆K(t),∆Z(t),∆D(t)) = (Y(t)−Y⁰(t), K(t)−K⁰(t), Z(t)−Z⁰(t), D(t)−D⁰(t)).

We have

E Z T

0

θe^θt|∆Y_i(t)|dt+E Z T

0

e^θt−1

|∆D_i(t)|dt

!

≤ E Z T

0

e^θt−1

|bi(t, Y(t))−bi(t, Y⁰(t))|dt

!

+E



Z T 0

e^θt−1 X

j6=i

|r_ij(t, Y(t)−r_ij(t, Y⁰(t))| |D_j(t)|dt





+E



Z T 0

e^θt−1 X

j6=i

|rij(t, Y⁰(t))| |∆Dj(t)|dt



. (3.13)

For an arbitrary numberN >1, let LN be Lipschitz constant ofb in the ballB(0, N).

We put

A^N =n

ω∈Ω,kY(t, ω)k+||Y⁰(t, ω)||> No

, A^N= Ω\A^N.

By virtue of (3.13) and the Lipschitz continuity ofR, we deduce that there existsC₁>0 such that

E Z T

0

θe^θt|∆Yi(t)|dt

! +E

Z T 0

e^θt−1

|∆Di(t)|dt

!

≤ E Z T

0

e^θt−1

|b_i(t, Y(t))−b_i(t, Y⁰(t))|1_ANdt

!

+E Z T

0

e^θt−1

|b_i(t, Y(t))−b_i(t, Y⁰(t))|1_ANdt

!

+LC1E Z T

0

e^θt−1

k∆Y(t)kdt

!

+E



Z T 0

e^θt−1 X

j6=i

υ_ij|∆D_j(t)|dt



.

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From the boundeness condition on the coefficientb,we get E

Z T 0

θe^θt|∆Yi(t)|dt

! +E

Z T 0

e^θt−1

|∆Di(t)|dt

!

≤ 2βi

N E Z T

0

e^θt−1

kY(t)k+||Y⁰(t)||dt

!

+

LC1+LN

ai

E

Z T 0

(e^θt−1||∆Z(t)||dt

!

+E Z T

0

e^θt−1 X

j6=i

υij|∆Dj(t)|dt. (3.14)

Multiplying (3.14) byai, adding and using (3.1) and the inequalityP

i6=jaiυij ≤αaj , we get the existence ofC >0 such that

θd((Y, Z),(Y⁰, Z⁰)) ≤ C N

+ dL_N+ Xd

i=1

a_i

! LC₁

! E

Z T 0

e^θt−1

k∆Y(t)kdt

+αE Z T

0

e^θt−1

k∆D(t)kdt

! .

Choosingθ large enough such that ¹_θ

dLN+Pd i=1ai

LC1

≤α,we get d((Y, K),(Y⁰, K⁰))≤ C

θN +αd((Y, K),(Y⁰, K⁰)).

Finally

(1−α)d((Y, K),(Y⁰, K⁰))≤ C θN, which leads to

Y =Y⁰ andK=K⁰, by lettingN going to +∞.

By the same calculations as in (3.10) and (3.11), we obtain the existence ofC >0 such that

E Xd i=1

ai|∆Yi(t)|²

! +E



Z T 0

Xd i=1

Xd j=1

ai|∆Zij(s)|²ds





≤ CE Z T

0

Xd i=1

ai|∆Yi(s)|ds

! .

Therefore

Z=Z⁰.

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