Bernoulli 9(3), 2003, 517–534
Multidimensional backward stochastic differential equations with uniformly continuous coefficients S A ¨I D H A M A D E` N E Laboratoire de Statistique et Processus, Universite´ du Maine, 72085 Le Mans Cedex 9, France. E-mail:
[email protected] In this paper we consider the problem of the existence of a solution for backward stochastic differential equations with uniformly continuous coefficients. Keywords: backward stochastic differential equations; deterministic backward differential equations
1. Introduction A solution of a backward stochastic differential equation (BSDE) associated with a coefficient f and a terminal value on [0, 1] is an adapted process (Yt , Z t ) t0 . Furthermore, we introduce a sequence of processes (Y n , Z n ) which solve the BSDE associated with ( f n , ). Working on the components of Y n , we show that the sequence (Y n ) n is of Cauchy type. For this purpose we make use of condition (iii) and Girsanov’s theorem in order to cancel some troublesome terms by putting them in the martingale part. Assumption 2 on enables us to conclude. The solution of the BSDE associated with ( f , ) is constructed from the limit of (Y n ) n and that of ( Z n ) n . This paper is organized as follows. In Section 2 we formulate the problem accurately and give some preliminary results. Section 3 is devoted to the proof of the main theorem and investigates the conditions under which Assumption 2 is satisfied. Finally, in Section 4 we consider the problem of uniqueness.
2. Formulation of the problem and preliminary results Let B ¼ (B t ) t 0, gn is a mapping from [0, 1] 3 3 R dþd3m into R d satisfying Assumption 1 and which is Lipschitz with respect to ( y, z) uniformly in (t, ø). (ii) For all E . 0, there is an N E > 0 such that, for all n > NE , j g n (t, ø, y, z) g(t, ø, y, z)j < E for all t, y, z, a.s. 1 dþd3m , Rþ ) with compact support and which satisfies ÐProof. Let ł be a function of C (R dþd3m 7! ł n ( y, z) ¼ n2 ł(ny, nz) and R dþd3m ł( y, z)d y dz ¼ 1. For n > 0, let ł n : ( y, z) 2 R g n :¼ gł n , the convolution product of g and ł n . Therefore,
Multidimensional backward stochastic differential equations ð 8t, y, z, gn (t, ø, y, z) ¼ g(t, ø, u, v)n2 ł(n( y u), n(z v))du dv,
521
R dþd3m
u v g t, ø, y , z ¼ ł(u, v)du dv: n n R dþd3m ð
It is easily seen that the sequence ( gn ) n>0 converges pointwise to g, and, for any n > 0, g n satisfies Assumption 1 and is Lipschitz with respect to ( y, z) uniformly in (t, ø). On the other hand, for any n, m > 0, we have j g n (t, ø, y, z) g m (t, ø, y, z)j
ð
u v u v ¼
g t, ø, y , z g t, ø, y , z ł(u, v)du dv
n n m m R dþd3m
ð
u v u v
0, f n satisfies Assumption 1 and is Lipschitz with respect to ( y, z) uniformly in (t, ø). In addition, we have
Multidimensional backward stochastic differential equations j f n (t, y, z) f n (t, y9, z)j < (j y y9j),
523 8t, y, y9, z, a:s:
For any n > 0, let (Y n , Z n ) be the solution of the BSDE associated with ( f n , ) which, according to Theorem 2.1, exists. So, for any n > 0, we have (Y n , Z n ) 2 S 2,d 3 H 2,d3m , ð1 ð1 Y nt ¼ þ f n (s, Y sn , Z sn )ds Z sn dBs , t
(5) 8t < 1:
t
The proof is based on the fact that (Y n ) n>0 and ( Z n ) n>0 are Cauchy sequences in S 2,d and H 2,d3m , respectively. Step 1. In this step we show that (Y n ) n>0 is a Cauchy sequence in S 2,d . For any n > 0, let (Y n, k , Z n,k ) k>0 be the sequence of processes of S 2,d 3 H 2,d3m defined recursively as follows: (Y n,0 , Z n,0 ) ¼ (0, 0), k ¼þ Y n, t
ð1 t
f n (s, Y sn, k1 , Z sn, k )ds
ð1 t
Z sn,k dBs ,
8t < 1, k > 1:
It is easily seen that (Y n,k , Z n, k ) k>0 converges, as k ! þ1, to (Y n , Z n ) in S 2,d 3 H 2,d3m . Now, for i ¼ 1, . . . , d, let iY n, k , i , f in and i Z n, k , be the ith components and row of respectively Y n, k, , f n and Z n, k . On the other hand, for any E . 0, let NE > 0 such that if n, m > NE then j f n (t, y, z) f m (t, y, z)j , E. n, m n, m i m, k (a) For all n, m > NE , k > 0, i ¼ 1, . . . , d, t < 1, j iY n,k t Y t j < C k,E , where C k,E is a constant which may depend on n, m, k, E. Indeed, we have i
Y n,kþ1 iY m,kþ1 t t ¼
ð1 t
¼
ð1 t
f f in (s, Y sn, k , i Z sn, kþ1 ) f im (s, Y sm, k , i Z sm, kþ1 )gds
ð1 t
( i Z sn,kþ1 i Z sm,kþ1 )dBs
f f in (s, Y sn, k , i Z sn, kþ1 ) f in (s, Y sm, k , i Z sn, kþ1 ) þ f in (s, Y sm,k , i Z sn, kþ1 )
f in (s, Y sm, k , i Z sm, kþ1 ) þ f in (s, Y sm, k , i Z sm, kþ1 ) f im (s, Y sm,k , i Z sm, kþ1 )gds
ð1 t
( i Z sn,kþ1 i Z sm,kþ1 )dBs :
But since f in is a Lipschitz mapping,
S. Hamade`ne
524 i
kþ1 Y n, t
i
kþ1 Y m, t
¼
ð1 t
f i b kn, m (s)(Y sn, k Y sm, k ) þ i a kn, m (s)( i Z sn, kþ1 i Z sm, kþ1 )
þ ( f in (s, Y sm, k , i Z sm, kþ1 ) f im (s, Y sm, k , i Z sm, kþ1 )gds
ð1 t
( i Z sn,kþ1 i Z sm,kþ1 )dBs :
Here ( i a kn, m (t)) t : 0 otherwise, and
8 i n, k i n, kþ1 k i n,kþ1 ) f in (t, Y m, ) > t , Zt < f n (t, Y t , Z t n, k m, k i k Yt Yt b n, m (t) ¼ > : 0
k m, k if Y n, 6¼ 0, t Yt
otherwise:
k Let P i,n,m be the probability on (, F ) which is equivalent to P and defined by: ! "ð # ð1 ð1 1 dP i,n,km 1 i k i k ¼E a n, m (s)dBs :¼ exp a n, m (s)dBs j i a k (s)j2 ds : 2 0 n,m dP 0 0 k the From Girsanov’s theorem (Karatzas and Shreve 1991; Revuz and Yor 1991), under P i,n,m process ðt i k B n, m (t) ¼ B t i a kn, m (s)ds, t < 1, 0
is an (F t , P i,n,km )-Brownian motion. Moreover, ð t ( i Z sn, kþ1 i Z sm, kþ1 )d i B kn, m (s) 0
is an (F t , " E i,n,km
P i,n,km )-martingale.
t 0, j iY n, t Y t j < u (t) for any t < 1 and i ¼ 1, . . . , d. Indeed, for k ¼ 0 the property holds. Suppose it also holds for some k 1, i.e., k1 k1 k1 k1 j iY n, iY m, j < u E (t), 8t < 1 and i ¼ 1, . . . , d, then jY n, Y m, j < d:u E (t), t t t t 8t < 1. Now combining that with the fact that is non-decreasing and taking into account k i m, k E (6) yields j iY n, t Y t j < u (t) for all t < 1, i ¼ 1, . . . , d: E Taking the limit as k ! 1 implies that j iY nt iY m t j < u (t), for all t < 1, i ¼ 1, . . . , d. Hence for all E . 0 there exists NE such that, for all n, m > N E , we have E E sup t0 is a Cauchy sequence in S 2,d and converges to a process which we denote by Y.
Step 2. We now show that ( Z n ) n>0 is a Cauchy sequence in H 2,d3m . Ð1 (a) There exists a constant C such that E[ 0 k Z sn k2 ds] < C. Indeed, since the rate of growth of the functions and is at most linear, there exists a constant Æ such that, for any n > 0, we have j f n (t, ø, y, z)j < j f (t, ø, 0, 0)j þ Æ(1 þ j yj þ kzk), for all t, y, z, a.s. Now using Itoˆ’s formula with the process Y n defined in (5), we obtain ð1 ð1 ð1 jY nt j2 þ k Z sn k2 ds ¼ jj2 þ 2 Y sn f n (s, Y sn , Z sn )ds 2 Y sn Z sn dBs : t
t
t
Then, for all t < 1 and . 0, ð1 ð1 ð1 jY nt j2 þ k Z sn k2 ds < jj2 þ jY sn jfj f (s, 0, 0)j þ Æ(1 þ jY sn j þ k Z sn k)gds 2 Y sn Z sn dBs t
t
< jj2 þ
1
t
ð1 t
ð1
jY sn j2 ds þ fj f (s, 0, 0)j þ Æ(1 þ jY sn j þ k Z sn k)g2 ds t
ð1
2 Y sn Z sn dBs t
ð ð1 1 1 n2 < jj þ jY j ds þ C f1 þ j f (s, 0, 0)j2 þ jY sn j2 þ k Z sn k2 gds t s t ð1 2 Y sn Z sn dBs 2
t
ð 1 ð1 ð1 1 þ C < jj þ jY sn j2 ds þ C (1 þ j f (s, 0, 0)j2 )ds þ C k Z sn k2 ds t t t ð1 2 Y sn Z sn dBs : 2
t
Multidimensional backward stochastic differential equations
527
Through the convergence of (Y n ) n>0 in S 2,d we have sup n>0 E[sup t0 is a Cauchy sequence in H 2,d3m whose limit will be denoted by Z. Step 3. We now show that the process (Y , Z) is a solution of the BSDE associated with ( f , ). For any n > 0 and t < 1, we know from (5) that ð1 ð1 n n n Y t ¼ þ f n (s, Y s , Z s )ds Z sn dBs : t
t
Ð1
Now, for a fixedÐ t 2 [0, 1], the sequences (Y nt ) n>0 and ( t Z sn dBs ) n>0 converge in L2 (, dP) 1 towards Yt and t Z s dBs, respectively. On the other hand, "ð # " ð
# ð1 1
1 n n n n j f n (s, Y s , Z s ) f (s, Y s , Z s )jds E
f n (s, Y s , Z s )ds f (s, Y s , Z s )ds
< E t
t
"ð
0
#
1
j f n (s,
] ds "ð
0 (( Z n ) n>0 ) in S 2,d ( H 2,d3m ) and the linear growth of f , i.e., j f (t, ø, y, z)j < j f (t, ø, 0, 0)j þ Æ(1 þ j yj þ kzk) a.s. Therefore, the second term converges to 0 as n ! þ1. In the same way, it is easily seen that the third term also converges to 0 as n ! þ1. Consequently, since Y is a continuous process, we have Yt ¼ þ
ð1 t
f (s, Y s , Z s )ds
ð1 Z s dBs ,
8t < 1, a:s:,
t
i.e. (Y , Z) is a solution for the BSDE associated with ( f , ). The proof is now complete. h
Multidimensional backward stochastic differential equations
529
We now focus on the conditions under which condition the function of Assumption 3(i) satisfies Assumption 2. Here is a result in this direction. Ð1 Proposition 3.2. (i) Assume the DBE u(t) ¼ t (d:u(s))ds, t < 1, has a unique solution u 0. Then satisfies Assumption 2. Ð1 (ii) The deterministic backward equation u(t) ¼ t (d:u(s))ds, t < 1, has a Ð differential unique solution if and only if 0þ [(x)]1 dx ¼ 1. Proof. First let us emphasize that is supposed to be continuous, non-decreasing, with at most linear growth, and satisfies (0) ¼ 0 and (x) . 0 for all x . 0. (i) For any E . 0, let u E be the solution of the DBE E
u (t) ¼ E þ
ð1
(d:u E (s))ds,
t < 1:
t
If . E then 0 < u E < u ,Ð and it follows that, for all t < 1, u E (t) & u~(t) as E & 0. In 1 addition, u~ satisfies u~(t) ¼ t (d:~ u(s))ds, t < 1. As the solution of this latter equation is unique then u~(t) ¼ 0, for all t < 1, and so u E (0) ! 0 asÐ E & 0. 1 (ii) The condition is sufficient. For z . 0, let G(z) ¼ z [(d:x)]1 dx. If u 6¼ 0 then there exists some t0 2]0, 1] such that u(t0 ) ¼ 0 and u(t) . 0 for any t , t0, and then (G(u(t))9 ¼ 1 for all t , t0 . This implies that u(t) ¼ G1 (G(u(t1 ) t1 þ t) for any t < t1 , t0 . Now taking the limit as t1 % t0 yields G(u(t1 )) ! þ1 and G1 (þ1) ¼ 0, whence u(t) ¼ 0 for all t < t0 which is a contradiction. Ð The condition is also necessary. Let us suppose 0þ [(x)]1 dx , 1. For E . 0, let Ð1 (d:x)]1 dx, z > 0, and let u E be a function such that u E (t) ¼ ÐG1E (z) ¼ z [E þ E [E þ (d:u (s))]ds, t < 1. u E is unique since GE (u E (t))9 ¼ 1 for all t < 1, and then t GE (u E (1)) GE (u E (t)) ¼ 1 t, t < 1, which implies u E (t) ¼ GE 1 (GE (0) 1 þ t), t < 1. Now ifÐ . E then u > u E , hence u E & v pointwise as E & 0. But v satisfies 1 v(t) ¼ t (d:v(s))ds, and then v ¼ 0, since the solution of this latter equation is unique. It follows that u E & 0 as E & 0 pointwiseÐ and uniformly by virtue of Dini’s theorem. 1 Henceforth, for all t < 1, GE (u E (t)) ! 0 [(x)]1 dx , 1 as E ! 0. Now since E E GE (u (1)) GE (u (t)) ¼ 1 t, for all t < 1, taking the limit as E ! 0, we obtain 0 ¼ 1 t, for all t < 1 which is a contradiction. h We are now ready to give the following result whose proof is a direct consequence of Theorem 3.1 and Proposition 3.2. Theorem 3.3. Assume the mapping ( t, ø, y, z) 7! f ( t, ø, y, z) satisfies Assumptions 1, 3(ii) and 3(iii), and j f ( t, y, z) f ( t, y9, z)j < ( j y y9j), for all t, z, y, y9, where is a continuous non-decreasing function from Rþ into itself with at most linear growth and such Ð that ( 0) ¼ 0, 0þ [ ( x)]1 dx ¼ þ1 and ( x) . 0 for all x . 0. Then the BSDE associated with ( f , ) has a solution. According to this theorem, if f is as in Example 2 above then the BSDE associated with ( f , ) has a solution.
S. Hamade`ne
530
4. Uniqueness In this section we deal with the issue of the uniqueness of the solution for the BSDE (4) associated with ( f , ). Let us assume that the mapping (t, ø, y, z) 7! f (t, ø, y, z) satisfies the following assumption. Assumption 4. (i) For all t, y, y9, z, j f (t, y, z) f (t, y9, z)j < (j y y9j), where is continuous and non-decreasing, grows at most linearly and satisfies (0) ¼ 0, (x) . 0, for all Ð x . 0, and 0þ [(x)]1 dx ¼ 1. (ii) The function z 7! f (t, y, z) is Lipschitz uniformly with respect to (t, ø, y). In addition, for any i ¼ 1, . . . , d, f i (t, y, z), the ith component of f , depends only on the ith row of z. Then we have the following result. Theorem 4.1. Under Assumptions 1 and 4, the solution (Y , Z) of the BSDE (4) associated with ( f , ) is unique. Proof. The existence follows from Theorem 3.3. Let us focus on the uniqueness. Let (Y 9, Z9) 2 S 2,d 3 H 2,d3m be another solution of the BSDE associated with ( f , ), i.e. ð1 ð1 8t < 1, a:s: Y 9t ¼ þ f (s, Y 9s , Z9s )ds Z9s dBs , t
t
(i) The process Y Y 9 is uniformly bounded, i.e. there exists a constant C~ such that jYt Y 9t j < C~, for all t < 1. Indeed, using Itoˆ’s formula we arrive, for all t < 1, at ð1 ð1 2 2 jYt Y 9t j þ k Z9s Z s k ds ¼ 2 (Y s Y 9s )( f (s, Y s , Z s ) f (s, Y 9s , Z9s ))ds t
t
ð1 2 (Y s Y 9s )( Z s Z9s )dBs t
ð1
< 2 jY s Y 9s jf(jY s Y 9s j) þ kk Z s Z9s kgds t
ð1 2 (Y s Y 9s )( Z s Z9s )dBs t
ð1
ð1
t
t
< C jY s Y 9s j2 ds þ
(jY s Y 9s j)2 ds þ
1 2
ð1
k Z s Z9s k2 ds
t
ð1
2 (Y s Y 9s )( Z s Z9s )dBs t
(7)
Multidimensional backward stochastic differential equations
531
since, for all a, b 2 R and E . 0, jabj < Ea2 þ E1 b2 . The growth of is at most linear, then (j yj)2 < C(1 þ j yj2 ) for all y 2 R. On the other hand, since (Y , Z) and (Y 9, Z9) belong to SÐ2,d 3 H 2,d3m then, using the Burkholder–Davis–Gundy inequality, we deduce that t ( 0 (Y s Y 9s )( Z s Z9s )dBs ) t t > 0, we have ( E[jY s Y 9s j2 jF t ] < C 1 þ
ð1
) E[jY u Y 9u j2 jF t ]du :
s
Now by Gronwall’s inequality we obtain E[jY s Y 9s j2 jF t ] < C~, which yields the desired result after taking s ¼ t. (ii) We show that the solution of the BSDE associated with ( f , ) is unique. For any i ¼ 1, . . . , d and t < 1, we have ð1 ð1 i Yt iY 9t ¼ ( f i (s, Y s , i Z s ) f i (s, Y 9s , i Z9s ))ds ( i Z s i Z9s )dBs t
t
where, once again, iY , iY 9, f i , i Z and i Z9 are the ith components and rows of respectively Y , Y 9, f , Z and Z9. Then, using Tanaka’s formula, we obtain ð1 j i Yt iY 9t j þ 2(¸1i (0) ¸ it (0)) ¼ sgn( iY s iY 9s )( f i (s, Y s , i Z s ) f i (s, Y 9s , i Z9s ))ds t
ð1
sgn( iY s iY 9s )( i Z s i Z9s )dBs
t < 1,
t
where (¸ it (0)) t : 0 otherwise: Then i
i
j Yt Y 9t j
0. This implies that the sequence (vEn ) n>0 converges pointwise to a function vE : Rþ ! Rþ which satisfies ð1 vE (t) ¼ E þ (d:vE (s))ds, t < 1: t
Now if E < then &0 where, for any t < 3.2). Therefore, we have vE (0) & 0 as E & 0. Now for E, n and k > 0, let vE n, k be the function defined recursively as follows: vEn
< vn for Ð 1 any n > 0, and then vE < v . It follows that vE & v as E 1, v t ¼ t (d:v(s))ds, so that v 0 (according to Proposition
vEn,0 ¼ C~, vEn, k (t)
¼Eþ
ð1 t
n (d:vEn, k1 (s))ds,
k > 1, t < 1:
(9)
Since n is Lipschitz, vEn, k ! vEn as k ! þ1. On the other hand, it is easily seen by induction that for all k > 0, j i Yt iY 9t j < vEn, k (t), t < 1, i ¼ 1, . . . , d. Indeed, for k ¼ 0 the formula holds. Suppose it also holds for some k 1, then (jYt Y 9t j) < (d:vEn, k1 (t)) < n (d:vEn, k1 (t)) for all t < 1. Now, using (8) and (9), we have j i Yt iY 9t j < vEn, k (t) for all t < 1, i ¼ 1, . . . , d: Taking the limit as first k ! 1, then n ! 1, and finally E ! 0, we obtain j i Yt iY 9t j ¼ 0 for all t < 1. Therefore the solution is unique. h Finally a word about the work of Mao (1995) on the same subject. He shows that if the coefficient f satisfies j f (t, ø, y, z) f (t, ø, y9, z9)j2 < (j y y9j2 ) þ kjz z9j2 , where k is constant and satisfies Assumption 4(i) and (importantly) is concave, then the BSDE associated with ( f , ) has a unique solution. He does not require the second part of Assumption 4(ii). In his proof he uses Bihari’s inequality.
Multidimensional backward stochastic differential equations
533
Using our approach and under the same hypotheses as in Mao (1995), but without requiring the concavity of , it could be possible to obtain the same result as he does. The issue of the existence and uniqueness of the solution for the BSDE associated with ( f , ) when f satisfies Assumption 4(i) and the mapping z 7! f (t, ø, y, z) is uniformly Lipschitz is still open.
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