Journal of Applied Mathematics and Stochastic Analysis, 16:4 (2003), 295-309.
c
Printed in the USA 2003
by North Atlantic Science Publishing Company
BACKWARD STOCHASTIC DIFFERENTIAL
EQUATIONS WITH OBLIQUE
REFLECTION AND LOCAL
LIPSCHITZ DRIFT
AUGUSTE AMAN and MODESTE N’ZI
URF de Mathématiques et Informatique
22 BP 582 Abidjan 22, Côte 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
Hamadène and Lepeltier [9]), in mathematical finance via the theory of hedging and nonlinear 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 = [Ys βs + Zs γs ]ds − Zs dBs , 0 ≤ s ≤ T
(1.1)
YT = ξ.
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.
295
296
A. AMAN and M. N’ZI
The development of general backward stochastic differential equation (BSDE in
short)
−dYs = f (s, Ys , Zs )ds − Zs dBs , 0 ≤ s ≤ T
YT = ξ
begins with the paper of Pardoux and Peng [14]. Since then, BSDEs have been intensively 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 Hamadène
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], Hamadène
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 underlying 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 {Ft } be the natural filtration
generated by B, with F0 containing all P −null sets.
Let G = {x ∈ Rd : xi > 0, 1 ≤ i ≤ d} denote the d−dimensional positive orthant.
We are given the following:
• T > 0 is a terminal time;
• ξ is an FT −measurable, bounded, G−valued random variable;
• b : [0; T ] × Ω × Rd −→ Rd , R : [0; T ] × Ω × Rd −→ Md (R) are both bounded measurable functions such that for every y ∈ Rd , b(., ., y) = (b1 (., ., y), . . . , bd (., ., y))
and R(., ., y) = (rij (., ., y))1≤i,j≤d are Ft −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} and K = {K(t) = (K1 (t), .., Kd (t)) : t ≥ 0} of {Ft } −progressively measurable integrable processes is said to solve RBSDE (ξ, b, R) if the following
hold:
(i) (Y, Z, K) is a continuous Rd × Md (R) × Rd −valued process;
Backward Stochastic Differential Equations
297
(ii) for every i = 1, ..., d, and 0 ≤ t ≤ T,
Yi (t)
= ξi +
Z
T
bi (s, Y (s))ds −
t
+
d Z
X
j=1
XZ
T
Zij (s)dBj + Ki (T ) − Ki (t)
t
T
rij (s, Y (s))dKj (s);
t
j6=i
(iii) for every 0 ≤ t ≤ T, Y (t) ∈ G;
(iv) for every 1 ≤ i ≤ d, Ki (0) = 0, Ki (·) is nondecreasing and can increase only when
Yi (·) = 0, that is
Z t
1{0} (Yi (s))dKi (s).
Ki (t) =
0
We make the following assumptions on the coefficients b, 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 × Rd .
(A2) For 1 ≤ i, j ≤ d, y 7→ rij (t, ω, y) is Lipschitz continuous, uniformly over (t, ω) .
(A3) For every i 6= j there exists constant vij such that |rij (t, ω, y)| ≤ vij . Set V =
(vij ) with vii = 0.We assume that σ (V ) < 1, where σ (V ) denotes the spectral
radius of V . Therefore,
(I − V )−1 = I + V + V 2 + V 3 + · · ·
In the sequel, we put β = (β1 , ..., βd ).
Remark 2.1: In view of (A3), there exists constants aj , 1 ≤ j ≤ d and 0 < α < 1
such that
X
X
ai |rij (t, ω, y)| ≤
ai vij ≤ αaj
i6=j
i6=j
for all j = 1, . . . , d and (t, ω, y) ∈ [0; T ] × Ω × Rd .
Let H stands for the space of all {Ft } −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;
d
P R T θt
(iii) E
0 e ai |Yi (t)| dt < +∞;
i=1
(iv) E
d R
P
T
i=1
0
eθt ai ϕt (Ki )dt
< +∞; where ϕt (g) denotes the total variation of g over
[t, T ] and θ > 0 is a fixed constant which will be chosen suitably later.
298
A. AMAN and M. N’ZI
b ∈ H, we define the metric
For (Y, K) , (Yb , K)
b
d((Y, K) , (Yb , K))
d Z
X
= E
i=1
T
0
d Z
X
+E
i=1
e ai |Yi (t) − Ybi |dt
θt
T
!
b i )dt .
e ai ϕt (Ki − K
θt
0
!
(2.1)
It is not difficult to see that (H, d) is a complete metric space.
e denote the collection of all (Y, K) ∈ H such that there exists an {Ft } −proLet H
gressively measurable process {D(t) = (D1 (t), ....., Dd (t)) : t ≥ 0}, with
Z t
−1
0 ≤ Di (t) ≤ ((I − V ) β)i a.s. and Ki (t) =
Di (s)ds.
0
e is a closed subset of H, (H,
e d) is a complete metric space.
Since H
P
We consider the norm kyk =
ai |yi | which is equivalent to the Euclidean norm
in Rd . So, we may assume that the local Lipschitz continuity in (A1) and Lipschitz
continuity in (A2) are with respect to this norm.
b ∈H
e with Di , D
bi
Before stating our main result, let us remark that if (Y, K), (Yb , K)
b i then
being respectively the derivatives of Ki , K
Z T
bi) =
b i (s)|ds.
ϕt (Ki − K
|Di (s) − D
t
Therefore, using integration by parts in (2.1), we have
!
d Z T
X
θt
b
d((Y, K) , (Yb , K))
= E
e ai |Yi (t) − Ybi |dt
i=1
0
!
eθt − 1
b i (t)|dt
ai |Di (t) − D
+E
θ
i=1 0
!
!
Z T
Z T θt
e
−
1
b
= E
||D(t) − D(t)||dt
.
eθt ||Y (t) − Yb (t)||dt + E
θ
0
0
d Z T
X
For every z ∈ Md (R), we put
1/2

d X
d
X
ai |zij |2  .
|||z||| = 
j=1 i=1
Let H denote the space of all Ft -progressively measurable processes Z = (Zij )1≤i,j≤d
such that
!
Z T
|||Z(t)|||2 dt < +∞,
E
0
endowed with the norm
"
|Z| = E
Z
T
2
|||Z(t)||| dt
0
!#1/2
.
Backward Stochastic Differential Equations
299
It is clear that H is a Banach space.
Now, we state our main result:
Theorem 2.1: Assume (A1)-(A3). Let ξ be a bounded, FT −measurable G−valued
e
random variable. Then there is a unique couple ((Y, K), Z) ∈ H×H
solving RBSDE
(ξ, b, R).
3
Proof of the Main Result
The proof of Theorem 2.1 needs some preliminary lemmas.
Lemma 3.1: Let b 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 |bni (t, ω, y)| ≤ βi , for all 1 ≤ i ≤ d and
(t, ω, y) ∈ [0, T ] × Ω × Rd ;
(ii) for every p, ρp (bn − b) → 0 as n → +∞, where
Z
ρp (f ) = E
!
T
θs
e
sup ||f (s, x)||ds .
|x|<p
0
Proof: Let ψn be a sequence of smooth functions with support in the ball B (0, n + 1)
such that sup ψn = 1. It not difficult to see that the sequence (bn )n≥1 of truncated functions defined by bn = bψn , satisfies all the properties quoted above.
In view of Ramasubramanian [16], there exists a unique couple of processes {((Y n (t),
n
e
K (t)), Z n (t)) : t ≥ 0} ∈ H×H
solution to the RBSDE (ξ, bn , R).
We formulate some uniform estimates for the processes {((Y n (t), K n (t)), Z n (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
θt
n
e ||Y (t)||dt
!
Z
+E
0
T
0
eθt − 1
||Dn (t)||dt
θ
!
< C.
(3.1)
Proof:Let the triple (Y n , K n , Z n ) be the unique solution of RBSDE (ξ, bn , R). We
have for every i = 1, ..., d, and 0 ≤ t ≤ T
Y n (t)
= ξi +
Z
T
bni (s, Y n (s))ds −
Z
t
+
t
XZ
j6=i
T
d
X
n
Zij
(s)dBj + Kin (T ) − Kin (t)
j=1
T
rij (s, Y n (s))dKj (s).
t
Since
ϕt (Kin )
=
Z
T
|Din (s)| ds,
t
300
A. AMAN and M. N’ZI
applying Theorem 3.2 [16] and using integration by parts, we obtain
Z
E
T
θt
θe
|Yin (t)|dt
0
Z
= E
T
θeθt |Yin (t)|dt
!
!
Z
+E
− 1 |ξi | + E
e
+E 
θe
ϕt (Kin )dt
!
Z
+E
T
eθt − 1 |Din (t)|dt
!
0
θT

θt
0
0
≤ E
T
Z
T
e − 1 |bni (t, Y n (t))|dt
θt
!
0
Z

X
eθt − 1
|rij (t, Y n (t))||Djn (t)|dt .
T
0
j6=i
We know that for every (t, ω, y) and every i 6= j, |rij (t, ω, y)| ≤ vij . Moreover for every
−1
j = 1, . . . , d and n ≥ 1, |bnj (t, ω, y)| ≤ βj , |Djn (t, ω, y)| ≤ ((I − V ) β)j .
Therefore
E
Z
T
θt
|Yin (t)| dt
θe
!
+E
Z
0
θt
θe
ϕt (Kin )dt
0
θT
≤ E( e

Z
+E 
T
− 1 |ξi |) + E
Z
T
e − 1 βi dt
θt
!
!
0
T
0

X
−1
υij ((I − V ) β)j dt .
eθt − 1
(3.2)
j6=i
Let us note that
d((Y n , K n ), (0, 0)) = E
Z
T
eθt kY n (t)k dt
!
+E
0
Z
T
0
!
eθt − 1
kDn (t)k dt .
θ
Multiplying (3.2) by ai and adding leads to
n
n
θd((Y , K ), (0, 0)) ≤ E
d
X
θT
e
− 1 ai |ξi |
!
+E
i=1

Z
+E 
d Z
X
i=1
T
(eθt − 1)
0
d X
X
i=1 j6=i
In view of the inequality
X
i6=j
υij ai ≤ αaj ,
T
θt
(e − 1)ai βi dt
0
ai vij ((I − V )
−1
!

β)j dt .
Backward Stochastic Differential Equations
301
we have
d
X
θd((Y n , K n ), (0, 0)) ≤ E
− 1 ai |ξi |
eθT
i=1

+αE 
!
d Z
X
+E
T
(eθt − 1)
0
d
X
j=1
(eθt − 1)ai βi dt
0
i=1
Z
T
!

aj ((I − V )−1 β)j dt
≤ eθT − 1 E||ξ||
Z T
−1
+(||β|| + α||((I − V ) β)||)
(eθt − 1)dt
0
≤ C.
Hence inequality (3.1) is proved.
Now, we shall prove the convergence of the sequence (Y n , K n , Z n )n≥1 .
e
Theorem 3.1: Assume (A1)-(A3). Then there exists ((Y, K), Z) ∈ H×H
such
that
(
!
)
Z T
Z T θt
e −1
θt
n
n
kD (t) − D(t)k dt = 0
lim
E
e kY (t) − Y (t)k dt + E
n→+∞
θ
0
0
and
Z
lim E
n→+∞
T
|||Z n (t) − Z(t)|||2 dt
!
= 0,
0
where
Ki (t) =
Z
t
Di (s)ds, i = 1, . . . , d.
0
Proof: It follows from the same idea used in the proof of inequality (3.1) that
!
!
Z T
Z T
E
θeθt |Yim (t) − Yin (t)|dt + E
θeθt ϕt (Kim − Kin )dt
0
= E
Z
T
θt
θe
|Yim (t)
−
Yin (t)|dt
!
0
≤ E
Z
+E
T
e − 1 |Dim (t) − Din (t)|dt
θt
0
T
θt
(e −
0

Z

+E
≤ E
0
Z
Z
T
0
T
m
1)|bm
i (t, Y (t))
−
bni (t, Y n (t))|dt
!

X
θt
m
m
n
n
(e − 1) rij (t, Y (t))Dj (t) − rij (t, Y (t)Dj (t) dt
j6=i
!
m
n
n
(eθt − 1)|bm
i (t, Y (t)) − bi (t, Y (t))|dt
0

Z
+E 
T
(eθt − 1)
0
X
j6=i

|rij (t, Y m (t) − rij (t, Y n (t))||Djn (t)|dt
!
302
A. AMAN and M. N’ZI

+E 
Z
T
(eθt − 1)
0
X
j6=i

|rij (t, Y m (t))||Djm (t) − Djn (t)|dt .
For an arbitrary number N > 1, let LN be the Lipschitz constant of b in the ball
B(0, N ). We put
N
m
n
N
AN
m,n = {ω ∈ Ω, ||Y (t, ω)|| + ||Y (t, ω)|| > N } , Am,n = Ω\Am,n .
It follows that
E
Z
T
θeθt |Yim (t) − Yin (t)|dt
!
+E
0
≤ E
Z
T
(eθt − 1)|Dim (t) − Din (t)|dt
0
T
θt
(e −
m
1)|bm
i (t, Y (t))
0
+E
Z
Z

+E 

+E 
−
bni (t, Y n (t))|1AN
dt
m,n
!
T
m
n
n
(eθt − 1)|bm
i (t, Y (t)) − bi (t, Y (t))|1AN dt
T
(eθt − 1)
0
Z
!
m,n
0
Z
!
X
j6=i
T
(eθt − 1)
0
X
j6=i

|rij (t, Y m (t) − rij (t, Y n (t))||Djn (t)|dt

|rij (t, Y m (t))| |Djm (t) − Djn (t)|dt
= I1 + I2 + I3 + I4 .
(3.3)
It not difficult to check that
!
Z T
θt
m
m
n
n
(e − 1)|bi (t, Y (t)) − bi (t, Y (t))|1AN dt
I2 = E
m,n
0
≤ E
Z
T
0
+E
m
m
(eθt − 1)|bm
i (t, Y (t)) − bi (t, Y (t))|1AN dt
m,n
Z
T
(eθt − 1)|bi (t, Y m (t)) − bi (t, Y n (t))|1AN dt
m,n
0
+E
Z
T
θt
n
(e − 1)|bi (t, Y (t)) −
0
!
bni (t, Y n (t))|1AN dt
m,n
!
.
Backward Stochastic Differential Equations
303
LN
−locally Lipschitz, we get
ai
Since bi is
Z
≤ E
I2
T
m
m
e − 1 |bm
i (t, Y (t)) − bi (t, Y (t))| 1AN dt
θt
m,n
0
Z
+E
T
e − 1 |bi (t, Y n (t)) − bni (t, Y n (t))| 1AN dt
θt
!
m,n
0
+
!
Z
LN
E
ai
T
!
eθt − 1 kY m (t) − Y n (t)k dt .
(3.4)
0
In view of the Lipschitz condition on R and the boundedness of Djn (t), we obtain that
there exists C1 > 0 such that
!
Z T
n m
θt
n
I3 ≤ LE
e − 1 kY (t) − Y (t)k Dj (t) dt
0

Z
≤ LE 
≤ LC1 E
T
0
Z

X
−1
eθt − 1 kY m (t) − Y n (t)k
((I − V ) β)j dt
j6=i
T
eθt − 1 kY m (t) − Y n (t)k dt.
(3.5)
0
Now, from the boundness of R, we have


Z T
X
I4 ≤ E 
eθt − 1
vij Djm (t) − Djn (t) dt .
0
(3.6)
j6=i
By virtue of (3.3)-(3.6), we deduce that
Z
E
T
θt
θe
|Yim (t)
−
Yin (t)| dt
0
Z
≤ E
T
T
e − 1 |Dim (t) − Din (t)| dt
θt
m
n
n
dt
e − 1 |bm
i (t, Y (t)) − bi (t, Y (t))| 1AN
m,n
θt
Z
T
Z
!
!
m
m
e − 1 |bm
i (t, Y (t)) − bi (t, Y (t))| 1AN dt
θt
!
m,n
0
+E
+E
Z
0
0
+E
!
T
e − 1 |bi (t, Y n (t)) − bni (t, Y n (t))| 1AN dt
θt
0
m,n
Z
T
LN
+
+ LC1 E
eθt − 1 kY m (t) − Y n (t)k dt
ai
0


Z T
X
vij Djm (t) − Djn (t) dt .
+E 
eθt − 1
0
!
j6=i
!
(3.7)
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A. AMAN and M. N’ZI
Multiplying (3.7) by ai , adding and using
P
i6=j
ai vij ≤ αaj , we obtain
θd((Y m , K m ), (Y n , K n ))
!
d Z T
X
θt
m
m
n
n
e − 1 ai |bi (t, Y (t)) − bi (t, Y (t))| 1AN
dt
≤ E
m,n
0
i=1
d Z
X
+E
T
d Z
X
T
d
X
ai
!
!
LC1 E
Z
T
e − 1 ||Y m (t) − Y n (t)||dt
θt
!
0
i=1
Z
!
m,n
+ dLN +
+αE
e − 1 ai |bi (t, Y n (t)) − bni (t, Y n (t))| 1AN dt
θt
0
i=1
!
m,n
0
i=1
+E
m
m
eθt − 1 ai |bm
i (t, Y (t)) − bi (t, Y (t))| 1AN dt
!
m
n
e − 1 ||D (t) − D (t)||dt .
T
θt
0
Choosing θ large enough such that
1
θ
d((Y m , K m ), (Y n , K n ))
dLN +
P
d
i=1
ai LC1 ≤ α leads to
≤ αd((Y m , K m ), (Y n , K n ))
1
1
+ ρN (bn − b) + ρN (bm − b)
θ
θ
1 N
+ Cm,n
θ
(3.8)
where
N
Cm,n
= E
d Z
X
≤ 2
d Z T
X
≤
2
N
0
d
X
i=1
m
n
n
dt
eθt − 1 ai |bm
i (t, Y (t)) − bi (t, Y (t))| 1AN
m,n
0
i=1
i=1
T
dt
eθt − 1 ai βi E 1AN
m,n
ai βi
Z
T
eθt − 1 E (kY n (t)k + kY m (t)k) dt.
0
Let C2 be such that
d
X
ai βi < C2 .
i=1
We have
N
≤
Cm,n
2C2
E
N
Z
T
eθt − 1 (kY n (t)k + kY m (t)k) dt.
0
By virtue of (3.1), there exists C > 0 such that
N
Cm,n
≤
C
.
N
!
Backward Stochastic Differential Equations
305
(3.9)
Therefore
1
(1 − α) d((Y , K ), (Y , K )) ≤
θ
n
n
m
m
C
N
1
1
+ ρN (bn − b) + ρN (bm − b).
θ
θ
Passing to the limit on n, m and N in (3.9 ), we deduce that (Y n , K n )n∈N is a Cauchy
e Since H
e is a Banach space, we set
sequence in H.
lim Y n = Y , and
lim K n = K.
n→+∞
n→+∞
If we return to the equation satisfied by the triple (Y n , K n , Z n )n∈N and use Itô’s formula,
we have


Z TX
d
2
m
n
E( |Yim (t) − Yin (t)| ) + E 
|Zij
(s) − Zij
(s)|2 ds
0
=
Z
2E
j=1
!
T
|Yim (s)
−
m
Yin (s)| |bm
i (s, Y (s))
−
bni (s, Y n (s))| ds
0
+2E
Z
!
T
|Yim (s)
−
Yin (s)|
|Dim (s)
−
Din (s)| ds
0

X
m
n
m
m
n
n
|Yi (s) − Yi (s)| rij (s, Y (s))Dj (s) − rij (s, Y (s))Dj (s) ds
0
j6=i
!
!
Z T
Z T
−1
m
n
m
n
|Yi (s) − Yi (s)| ds + 4 (I − V ) β E
|Yi (s) − Yi (s)| ds
≤ 4βi E

Z
+2E 
T
i
0
+4
X
Z
−1
vij (I − V ) β E
j
j6=i
!
T
0
|Yim (s) − Yin (s)| ds .
(3.10)
0
Multiplying (3.10) by ai and adding leads to the existence of C > 0 such that


!
Z TX
d X
d
d
X
2
2
m
n
E
ai |Yim (t) − Yin (t)| + E 
ai Zij
(s) − Zij
(s) ds
0
i=1
≤ 4E
Z
d
T X
0
+4E
T
0

Z
+4E 
≤ CE
ai βi |Yim (s) − Yin (s)| ds
0
!
d X
−1
m
n
(I − V ) β ai |Yi (s) − Yi (s)| ds
i
i=1
T
T
0
i=1 j=1
i=1
Z
Z
!
d X
X
ai vij ((I − V )
i=1 j6=i
d
X
i=1
ai |Yim (s)
−
−1

β)j |Yim (s) − Yin (s)| ds .
!
Yin (s)| ds
.
(3.11)
306
A. AMAN and M. N’ZI
Passing to the limit on m, n, we deduce that (Z n )n≥1 is a Cauchy sequence in the
Banach H. Since H is a Banach space, we put
Z = lim Z n .
n→+∞
n
n
n
Lemma 3.3:Let (Y , K , Z )n≥1 be the unique solution of the RBSDE (ξ, bn , R) .
Then
bn (., Y n ) converges to b(., Y ) in (L1+ (Ω × [0, T ] , dP × eθt dt)).
Proof:Set
N
n
AN
n = {ω ∈ Ω, ||Y (t, ω)|| + ||Y (t, ω)|| > N } , An = Ω\A.
We have
Z
E
T
θt
e
|bni (t, Y n (t))
− bi (t, Y (t))| dt
0
≤E
Z
T
eθt |bni (t, Y n (t)) − bi (t, Y (t))| 1AN
dt
n
0
+E
Z
T
θt
e
|bni (t, Y n (t))
0
Z
+E
!
!
n
eθt |bi (t, Y n (t)) − bi (t, Y (t))| 1AN dt
!
n
2βi
E
≤
N
+E
− bi (t, Y (t))| 1AN dt
T
0
Z
n
!
Z
T
θt
n
e (kY (t)k + kY (t)k) dt
0
T
θt
e
|bni (t, Y n (t))
0
Z
LN
E
ai
n
− bi (t, Y (t))| 1AN dt
!
!
n
!
T
θt
n
e kY (t) − Y (t)k dt .
(3.12)
0
Multiplying (3.12) by ai and adding, we get
!
Z T
θt
n
n
E
e kb (t, Y (t)) − b(t, Y (t))k dt
0
!
Z T
d
2 X
θt
n
≤ ρN (b − b) +
βi ai E
e (kY (t)k + kY (t)k) dt
N i
0
!
Z T
θt
n
+dLN E
e kY (t) − Y (t)k dt .
n
0
By virtue of (3.1), we deduce that there exists C > 0 such that
!
Z T
θt
n
n
e kb (t, Y (t)) − b(t, Y (t))k dt
E
0
C
+ dLN E
≤ ρN (b − b) +
N
n
Z
T
θt
n
!
e kY (t) − Y (t)k dt .
0
Backward Stochastic Differential Equations
307
Passing to the limit on n, 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).
n 0
o
0
0
Uniqueness: Let {(Y (t), K(t), Z(t)), 0 ≤ t ≤ T } and (Y (t), K (t), Z (t)), 0 ≤ t ≤ T
be two solutions of our RBSDE. For every t ≥ 0, define
0
0
0
0
(∆Y (t), ∆K(t), ∆Z(t), ∆D(t)) = (Y (t)−Y (t), K(t)−K (t), Z(t)−Z (t), D(t)−D (t)).
We have
E
Z
T
θt
θe |∆Yi (t)| dt + E
0
≤ E
Z
Z
T
e − 1 |∆Di (t)| dt
θt
0
T
e − 1 |bi (t, Y (t)) − bi (t, Y 0 (t))| dt
θt
0

Z
+E 

Z
+E 
!
!

X
|rij (t, Y (t) − rij (t, Y 0 (t))| |Dj (t)| dt
eθt − 1
T
0
j6=i

X
|rij (t, Y 0 (t))| |∆Dj (t)| dt .
eθt − 1
T
0
(3.13)
j6=i
For an arbitrary number N > 1, let LN be Lipschitz constant of b in the ball B(0, N ).
We put
n
o
0
N
AN = ω ∈ Ω, kY (t, ω)k + ||Y (t, ω)|| > N , A = Ω\AN .
By virtue of (3.13) and the Lipschitz continuity of R, we deduce that there exists C1 > 0
such that
E
Z
T
θt
θe |∆Yi (t)| dt
0
≤ E
Z
!
+E
Z
T
e − 1 |∆Di (t)| dt
θt
0
T
e − 1 |bi (t, Y (t)) − bi (t, Y 0 (t))| 1AN dt
θt
!
0
+E
Z
T
e − 1 |bi (t, Y (t)) − bi (t, Y 0 (t))| 1AN dt
θt
0
Z
+LC1 E

Z
+E 
T
e − 1 k∆Y (t)k dt
θt
0
T
0
!

X
υij |∆Dj (t)| dt .
eθt − 1
j6=i
!
!
308
A. AMAN and M. N’ZI
From the boundeness condition on the coefficient b, we get
!
Z
Z
T
T
θeθt |∆Yi (t)| dt
E
+E
0
eθt − 1 |∆Di (t)| dt
!
0
!
0
2βi
θt
E
≤
e − 1 kY (t)k + ||Y (t)||dt
N
0
!
Z T
LN
θt
E
+ LC1 +
(e − 1||∆Z(t)||dt
ai
0
Z T
X
+E
υij |∆Dj (t)| dt.
eθt − 1
Z
T
0
(3.14)
j6=i
Multiplying (3.14) by ai , adding and using (3.1) and the inequality
we get the existence of C > 0 such that
0
0
θd((Y, Z), (Y , Z ))
≤
C
N
+ dLN +
ai
!
! Z
LC1 E
T
T
i6=j
ai υij ≤ αaj ,
eθt − 1 k∆Y (t)k dt
0
i=1
Z
+αE
d
X
P
!
eθt − 1 k∆D(t)k dt .
0
1
θ
Choosing θ large enough such that
0
0
dLN +
d((Y, K), (Y , K )) ≤
LC
≤ α, we get
a
i
1
i=1
P
d
0
0
C
+ αd((Y, K), (Y , K )).
θN
Finally
0
0
(1 − α) d((Y, K), (Y , K )) ≤
C
,
θN
which leads to
Y = Y 0 and K = K 0 ,
by letting N going to +∞.
By the same calculations as in (3.10) and (3.11), we obtain the existence of C > 0
such that


!
Z TX
d X
d
d
X
2
2
E
ai |∆Yi (t)| + E 
ai |∆Zij (s)| ds
0
i=1
≤ CE
Z
T
0
Therefore
d
X
!
i=1 j=1
ai |∆Yi (s)| ds .
i=1
0
Z=Z .
Backward Stochastic Differential Equations
309
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