Bernoulli 13(1), 2007, 80–91 DOI: 10.3150/07-BEJ5004
One-dimensional backward stochastic differential equations whose coefficient is monotonic in y and non-Lipschitz in z P H I L I P P E B R I A N D 1, J E A N - P I E R R E L E P E LT I E R 2 and JA I ME S A N MA RT I´ N 3 1 IRMAR, Universite´ de Rennes 1, 35042 Rennes Cedex, France. E-mail:
[email protected] 2 Universite´ du Maine, Laboratoire de Statistique et Processus, BP 535, 72017 Le Mans Cedex, Fance. E-mail:
[email protected] 3 Universidad de Chile, Facultad de Ciencias Fı´sicas y Matema´ ticas, Departamento de Ingenierı´a Matema´tica y Centro de Modelamiento Matema´ tico UMR2071-CNRS, Casilla 170-3, Correo 3, Santiago, Chile. E-mail:
[email protected]
In this paper we study one-dimensional BSDE’s whose coefficient f is monotonic in y and nonLipschitz in z. We obtain a general existence result when f has at most quadratic growth in z and is bounded. We study the special case f (t, y, z) ¼ jzj p where p 2 (1, 2]. Finally, we study the case f has a linear growth in z, general growth in y and is not necessarily bounded. Keywords: backward stochatic differential equations; monotonic non-Lipschitz coefficient
1. Introduction Nonlinear backward stochastic differential equations (BSDE) were first introduced by Pardoux and Peng (1990). They proved that when the terminal value is square integrable and the coefficient f is uniformly Lipschitz in ( y, z) there exists a unique solution with smooth square integrability properties. In the one-dimensional case, thanks to comparison properties, it is possible to obtain better existence results. For example, when f is only continuous in ( y, z) there is a solution to the BSDE under the following assumptions: •
is square integrable, and f has a uniform linear growth in y, z (see Lepeltier and San Martı´n 1996), j f ð t, y, zÞj < C(1 þ j yj þ jzj);
•
is bounded and f has a superlinear growth in y and quadratic growth in z (see Lepeltier and San Martı´n 1998; Kobylanski 2000) j f (t, y, z)j < l( y) þ Cjzj2 ,
1350–7265 # 2007 ISI/BS
81
One-dimensional BSDEs where l . 0 satisfies
Ð1 0
dx=l(x) ¼
Ð0
1
dx=l(x) ¼ 1.
We mention also the recent results obtained by Briand and Hu (2005), who assume superlinear growth in y, quadratic growth in z and an unbounded terminal random variable which should satisfy an exponential integrability condition. We also use in section 4 a localization method taken from that paper. When the generator f has general growth in y there is no guarantee that a solution exists, even in the case where is bounded. This is most easily seen when is non-random and f (t, ø, y, z) ¼ f ( y), because in this case the BSDE is just an ordinary differential equation. We would like to mention the existence of local solutions for BSDE, proved in Lepeltier and San Martı´n (2002), where explosions may occur and a global solution does not exist. This paper is related to the work of Pardoux (1999), who he studied the multidimensional case where f is assumed to be monotonic in y, h f (t, y, z) f (t, y^, z), y y^i < j y y^j2 ,
(1:1)
for some > 0. He also assumed f is Lipschitz in z and j f (t, y, z)j < j(j yj) þ Cjzj
(1:2)
for some non-decreasing continuous function j : Rþ ! Rþ . One of our objectives is to extend this (in the one-dimensional case) when f has at most quadratic growth in z and is bounded (see Section 2). This result can also be seen as a generalization of the results in Lepeltier and San Martı´n (1998) where the monotone condition on f allows us to extend the superlinear condition in y to a quite general growth condition. The purpose of Section 3 is to study, in a very particular setting, the situation when is not bounded. For this purpose we study the special case f (t, y, z) ¼ jzj2 : Interestingly, it turns out that for 2 L2 the condition E(exp(2)) , 1 is necessary and sufficient for the existence of a solution. Section 4 is devoted to the general case where f is monotonic in y, has a linear growth in z, and satisfies a growth condition similar to (1.2), and 2 L p . Here again is the monotonic condition that makes it possible to have a solution, if we think that we have a very general growth condition in y. Let us introduce the basic notation and definitions we need for this paper. Let (B t )0< t 2, and a 2 R, we have from Itoˆ’s formula, ðT exp(at)(Y Ct ) n ¼ exp(aT ) n þ n exp(as)(Y Ct ) n1 g C (Y Cs ) f (s, Y Cs , Z Cs )ds t
n
ðT t
a
ðT t
exp(as)(Y Cs ) n1 Z Cs
n(n 1) dBs 2
ðT t
exp(as)(Y Cs ) n2 j Z Cs j2 ds
exp(as)(Y Cs ) n ds:
Since yf (s, y, z) < yf (s, 0, z) þ y 2 and y n2 > 0, we obtain y n1 f (s, y, z) < y n1 f (s, 0, z) þ y n and, moreover, g C ( y) y n1 f (s, y, z) < j yj n1 (j(0) þ Ajzj2 ) g C ( y) þ y n < (1 þ y n )j(0) þ Ajzj2 j yj n1 g C ( y) þ y n < (1 þ y n )j(0) þ 2CAjzj2 y n2 þ y n : We then obtain the following bound: exp(at)(Y Ct ) n < exp(aT ) n þ n þ 2CnA
ðT t
n(n 1) 2
ðT t
exp(as)(1 þ (Y Cs ) n )j(0)ds
exp(as)jY Cs j n2 j Z s j2 ds ðT t
ðT þ n exp(as)(Y Cs ) n ds t
exp(as)(Y Cs ) n2 j Z Cs j2 ds a
ðT t
exp(as)(Y Cs ) n ds
þ M Tn M nt , where M n is a martingale. If we choose n, a such that n 1 > 4CA and a ¼ (j(0) þ )n, we obtain exp(n(j(0) þ )t)(Y Ct ) n < exp(n(j(0) þ )T )( n þ 1) þ M Tn M nt , and taking the conditional expectation with respect to F t yields exp(n(j(0) þ )t)(Y Ct ) n < exp(n(j(0) þ )T )E( n þ 1jF t ) n þ 1) < exp(n(j(0) þ )T )(kk1
This gives the a priori bound
84
Ph. Briand, J.-P. Lepeltier and J. San Martín n jY Ct j < exp((j(0) þ )T )(kk1 þ 1)1= n < exp((j(0) þ )T)(kk1 þ 1):
If C is chosen such that C > exp((j(0) þ )T )(kk1 þ 1), then jY Ct j < C which implies g(Y Ct ) ¼ 1. Therefore (Y C , Z C ) satisfies ðT ðT C C C Yt ¼ þ f (s, Y s , Z s )ds Z Cs dBs t
t
and is a bounded solution of the desired BSDE. The rest of the result follows immediately. h Remark 2.1. We have in fact used the monotonicity only for the pair y and y^ ¼ 0.
3. The case f (t, y, z) ¼ jzj2 We consider the BSDE given by Yt ¼ þ
ðT
j Z s j2 ds
t
ðT
Z s dBs :
(3:1)
t
Theorem 3.1. Assume 2 L2 . Then, there exists a solution (Y , Z) 2 H 2 (R dþ1 ) to (3.1) if and only if E(exp(2)) , 1. Proof. Let (Y , Z) be a solution of (3.1). By Itoˆ’s formula we have ðt exp(2Y t ) ¼ exp(2Y0 ) þ 2 exp(2Y s ) Z s dBs ¼ exp(2Y0 ) þ M t , 0
where M is a local martingale. Consider n ¼ inf ft . 0 : Y t > ng. We have that n "T , when n ! 1. Then M t^ n is a bounded martingale and we have E(exp(2Y n )) ¼ E(exp(2Y0 )): Finally, from Fatou’s lemma we obtain E lim inf exp(2Y n ) ¼ E(exp(2)) < E(exp(2Y0 )) , 1: n!1
We now assume that E(exp(2)) , 1. Let M t ¼: E(exp(2)jF t ) ¼ M 0 þ
ðt
Z s dBs :
0
By Itoˆ’s formula we have 1 1 1 log M t ¼ log M 0 þ 2 2 2 and 12 log M T ¼ .
ðt 0
Zs 1 dBs Ms 4
ð t Z s 2 M s ds 0
(3:2)
85
One-dimensional BSDEs
In order to prove that (12 log M t , 12 ( Z t =M t )) is solution of the BSDE, it is enough to prove that Zt 2 H 2 (R d ), Mt
1 log M t 2 S 2 : 2
Since log is concave we have 1 1 log M t ¼ log E(exp(2)jF t ) > E(jF t ) > E( jF t ) ¼: N t : 2 2 Consider for a . 0 the stopping time ( ) ð t ð t Z s 2 T a ¼ inf t . 0 : jN t j . a, ds . a, Z s dBs . a ^ T : 0 Ms 0 Using (3.2) and the fact (a þ b þ c)2 < 3(a2 þ b2 þ c2 ), we have ð Ta ð Ta Z s 2 Zs ds < 2 log M 0 4N T þ 2 dBs a Ms M s 0 0 and !2 ð T a 2 ð Ta Z s 2 Zs ds < 12(log M 0 )2 þ 48(N T )2 þ 12 dB : s a Ms 0 0 Ms Taking the expectation and using the fact that N 2t is a submartingale, we obtain !2 ! ð Ta ð Ta Z s 2 Z s 2 2 2 E M s ds < 12(log M 0 ) þ 48E( ) þ 12E M s ds , 0 0 and finally !2 ð Ta Z s 2 2 2 E M s ds < C1 (log M 0 ) þ E( ) þ 1 < C2 , 0
for some C2 . 0. To obtain the last inequality we have used the fact 12x < x 2 =2 þ 72. Since T a % T as a ! 1 we obtain from the monotone convergence theorem ! ðT pffiffiffiffiffiffi Z s 2 (3:3) E M s ds < C2 , 0 that is, Z=M 2 H 2 (R d ). Now from (3.2) we have 0 1 2 !2 ð t 2 ð t Z 1 Z s s ds A, dBs þ (log M t )2 < 3@(log M 0 )2 þ 4 0 M s 0 Ms which, together with the Burkholder–Davis–Gundy inequality, finally gives
86
Ph. Briand, J.-P. Lepeltier and J. San Martín
2
E
sup log M t
, 1:
0< t 0 such that, for all t, ø, y, y^, z, ( f (t, ø, y, z) f (t, ø, y^, z))( y y^) < ( y y^)2 ;
•
there exist a non-decreasing and continuous function j : Rþ ! Rþ such that
87
One-dimensional BSDEs
j(0) ¼ 0, a constant A > 0 and a nonnegative continuous adapted process f g t g t2[0,T ] such that P-almost surely j f (t, ø, y, z)j < g t (ø) þ j(j yj) þ Ajzj; •
for some p . 1, ðT E jj p þ g sp ds , 1: 0
Then the BSDE (1.3) has a minimal solution in S which belongs to S p 3 H p . Moreover, we have, if c ¼ þ A2 =(1 ^ ( p 1)), " ð T p=2 # ð T p cpt p 2cs 2 , e j Z s j ds e cs g s ds < C p E e cpT jj p þ E sup e jY t j þ t2[0,T ]
0
where the constant þ A2 =2[1 ^ ( p 1)],
Cp
0
p and also, for a ¼ jj þ (1 1= p)
depends only on
8t 2 [0, T ],
p
jY t j < e
apT
ðT p p E jj þ g s dsF t : 0
Remark 4.1. We can also construct a maximal solution in S with the same properties. Proof. Let us assume for the moment that ¼ 0. Thus, for each (t, z), y 7! f (t, y, z) is nonincreasing. For n > A, let us introduce the function
f n (t, y, z) ¼ inf f (t, y, q) þ njz qj : q 2 Q d : Then it is easy to check that, for each n > A, P-almost surely, • • •
for all (t, z), y 7! f n (t, y, z) is non-increasing; for all (t, y), z 7! f n (t, y, z) is n–Lipschitz; for all (t, y, z), j f n (t, y, z)j < g t þ j(j yj) þ Ajzj.
It follows from Briand et al. (2003: Theorem 4.2) that, for each n > A, the BSDE ðT ðT n n n Yt ¼ þ f n (s, Y s , Z s ) ds Z sn dBs t
n
n
p
t
p
has a unique solution (Y , Z ) 2 S 3 H . On the other hand, we have, for each n > A, y f n (t, y, z) < g t j yj þ Aj yj jzj: Thus, from Briand et al. (2003: Proposition 3.2), we have the following inequality, with c ¼ A2 =(1 ^ ( p 1)):
88
Ph. Briand, J.-P. Lepeltier and J. San Martín 2 E4 sup e cpt jY nt j p þ
ð T e
t2[0,T ]
2cs
0
j Z sn j2
2p ds
3
ð T p cs 5 < C p E e cpT jj p þ , e g s ds
(4:1)
0
where the constant C p depends only on p. In particular, the sequence ((Y n , Z n )) n> A is bounded in S p 3 H p . We now turn to the key estimate. Let us fix n > A. Let a be a real number to be chosen later. We set U t ¼ e at Y nt and V t ¼ e at Z nt . Then (U , V ) solves the BSDE associated with ¼ e aT and F(s, y, z) ¼ e at f n (t, eat y, eat z) ay. It follows from Briand et al. (2003: Corollary 2.3) that, if we let c( p) ¼ p[( p 1) ^ 1]=2, ðT jU t j p þ c( p) jU s j p2 1 U s 6¼0 jV s j2 ds t p
< jj þ p
ðT
jU s j
p1
U^ s F(s, U s , V s ) ds p
t
ðT
jU s j p1 U^s V s dBs ,
t
where u^ ¼ juj1 u1juj6¼0 . But we have u^F(t, u, v) < G t ajuj þ Ajvj, where G t ¼ e at g t . The previous inequality shows that, with probability one, ðT jU s j p2 1 U s 6¼0 jV s j2 ds , þ1: 0
Moreover, we have pjuj p1 u^F(t, u, v) < pjuj p1 G t apjuj p þ Ajuj p1 jvj and, using Young’s inequality for the first term and the fact that ab < a2 =2 þ b2 =2 for the third, we deduce that pjuj p1 u^F(t, u, v) < ( p 1)juj p þ G tp apjuj p þ
pA2 juj p þ c( p)juj p2 1juj.0 jvj2 : 2[( p 1) ^ 1]
Choosing a ¼ (1 1= p) þ A2 =2[( p 1) ^ 1], we obtain ðT ðT G sp ds p jU s j p1 U^ s V s dBs : jU t j p < p þ 0
p
t
p
But Ð t since (U , V ) 2 S 3 H , it follows from the Burkholder–Davis–Gundy inequality that f 0 jU s j p1 U^s V s dBs g t2[0, t] is a uniformly integrable martingale. Thus, taking the conditional expectation of the previous estimate, we obtain ðT G sp dsF t jU t j p < E p þ 0
89
One-dimensional BSDEs
from which we deduce, coming back to the definition of U , and (G t ) t2[0,T], that, with the notation a ¼ (1 1= p) þ A2 =2[( p 1) ^ 1], ðT (4:2) 8t 2 [0, T ], 8n > A, jY nt j p < e apT E jj p þ g sp dsF t : 0
For the rest of the proof, we set Mt ¼ e
apT
ðT p p E jj þ g s dsF t : 0
With inequality (4.2) to hand, let us construct the minimal solution to our BSDE. For this purpose let us observe that the sequence ( f n ) n> A is non-decreasing. Thus, it follows from the comparison theorem (see Briand and Hu 2005: Proposition 5) that 8n > A, 8t 2 [0, T ],
, Y nt < Y nþ1 t
so that we set Y t ¼ sup n> A Y nt . Now, we use the same localization procedure as in Briand and Hu (2005). For k > 1, let k be the following stopping time: k ¼ inf f t 2 [0, T ] : M t þ g t > k g ^ T , and we introduce the stopped process Y nk (t) ¼ Y nt^ k together with Z nk (t) ¼ Z nt 1 t A t2[0,T ]
Thus, if r( y) ¼ yk=max(j yj, k), we have ðT ðT Z nk (s) dBs Y nk (t) ¼ nk þ 1 s
