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BSDE driven by Poisson point processes with discontinuous coefficient
J. Math. Anal. Appl. 406 (2013) 365–372
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BSDE driven by Poisson point processes with discontinuous coefficient✩ Yan Qin ∗ , Ning-Mao Xia Department of Mathematics, East China University of Science and Technology, Shanghai, China
article
abstract
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Article history: Received 9 July 2008 Available online 2 May 2013 Submitted by U. Stadtmueller
In this paper, we deal with the one-dimensional backward stochastic differential equation (BSDE) driven by Poisson processes. By means of the comparison theorem, we first prove the existence of a (minimal) solution for BSDE where the coefficient is continuous and satisfies an improved linear growth assumption. Then we extend the result to the case where the coefficient is left or right continuous. Crown Copyright © 2013 Published by Elsevier Inc. All rights reserved.
Keywords: Backward stochastic differential equation Adapted solution Comparison theorem
1. Introduction The following nonlinear backward stochastic differential equation (BSDE) was first introduced by Pardoux and Peng (1990) [16] Yt = ξ +
T
f (s, Ys , Zs )ds − t
T
Zs dWs ,
t ∈ [0, T ].
(1)
t
Because of the important applications in mathematical finance (El Karoui et al., 1997 [7]), stochastic control (El Karoui and Hamadène, 2003 [6]) and stochastic games (Hamadène and Lepeltier, 1995 [9]), this equation has been studied intensively, and many efforts have been made to relax the assumption on the generator f . For instance, Lepeltier and San Martin (1997) [13] proved the existence of a solution when f is only continuous with linear growth, Kobylanski (2000) [12] obtained the existence and uniqueness of a solution when f is continuous with a quadratic growth in z, Briand and Carmona (2000) [5] derived a solution when f is monotone Lipschitz with polynomial growth, Bahlali [1–3] studied those with locally Lipschitz coefficients, and Jia (2008) [11] dealt with the solution when f is discontinuous in y and continuous in z. The driving stochastic process Wt in Eq. (1) is a Brownian motion, which can reflect only the effect of continuous random information. So it is of significance to introduce some discontinuous process, such as the Poisson process, into the equation. Tang and Li (1994) [18] developed the BSDE as follows: Yt = ξ +
T
f (s, Ys , Zs , Us )ds − t
T
Zs dWs − t
m i =1
T
Us(i) dN˜ s(i) ,
t ∈ [0, T ],
(2)
t
and proved the existence and uniqueness of the adapted solution when f is Lipschitz continuous. Barles et al. (1997) [4] studied the BSDE, which is driven by a Brownian motion and an independent Poisson point process. Situ (1997) [17]
✩ Supported by The National Natural Science Foundation (11002055) (China) and Fundamental Research Funds for the East China University of Science and Technology (WM1114040). ∗ Corresponding author. E-mail addresses: [email protected], [email protected] (Y. Qin), [email protected] (N.-M. Xia).
0022-247X/$ – see front matter Crown Copyright © 2013 Published by Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jmaa.2013.02.071
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considered the similar problem when f is non-Lipschitz with linear growth. Ouknine [14,15] derived the existence and uniqueness of the adapted solution for BSDE driven by a Poisson random measure. El Otmani (2008) [8] discussed the onedimensional BSDE, driving by a standard Brownian motion and a family of independent Poisson processes (called the simple Lévy process in [8]). He proved the existence of a solution for the one-dimensional BSDE where the coefficient is continuous with linear growth, or left continuous, increasing and bounded. The aim of this paper is to study the solution of BSDE driven by a standard Brownian motion and a family of independent Poisson point processes. Our setting is that the generator f satisfies an improved linear growth assumption |f (t , y, z , u)| ≤ A(t )(|y| + 1) + B(|z | + ∥u∥), instead of the previous one |f (t , y, z , u)| ≤ A(|y| + |z | + 1) (in Lepeltier and San Martin (1997) [13], Jia (2008) [11], El Otmani (2008) [8]). We will first prove the existence of a (minimal) solution for BSDE where the coefficient is continuous with an improved linear growth. Then we extend the result to the case that the coefficient is discontinuous. The paper is organized as follows. The next section gives some notations and definition of the solution of BSDE (2). Section 3 contains the existence result when the coefficient is continuous. The case where coefficient is discontinuous in y is discussed in Section 4. 2. Preliminaries In this paper, we consider the BSDE (2), where W is a standard Brownian motion and N˜ = (N˜ (1) , N˜ (2) , . . . , N˜ (m) )T is a family of independent compensating Poisson point processes with definition (i)
N˜ t
= Nt(i) − t ,
t ∈ [0, T ], i = 1, 2, . . . , m. (1)
(2)
(m)
In the above expression Nt = (Nt , Nt , . . . , Nt )T is an m dimensional Poisson process with non-random intensity (t , t , . . . , t )T . (i) Let (Ω , F , {Ft }0≤t ≤T , P ) be a complete probability space with the natural filtration {Ft }0≤t ≤T (i.e. Ft = σ {Ws , Ns , 0 ≤ s ≤ t , i = 1, 2, . . . , m, N }, where N is the totality of P-null set). The filtration {Ft }0≤t ≤T is right continuous (Ft = ∩ε>0 Ft +ε = Ft + ). 1
In our paper, the Euclid-norm is defined by ∥x∥ = (xT x) 2 , for x ∈ Rm . Now we define some spaces. Let
T
M2 = {f : [0, T ] × Ω → R is progressive measurable such that E 0 |f (t )|2 dt < +∞}. S 2 = {f : [0, T ] × Ω → R is adapted and rcll, such that f ∈ M2 and E sup0≤t ≤T |f (t )|2 < +∞}. Hm2 = {f : [0, T ] × Ω → Rm is the predictable process such that E 2 H2,m = S 2 × H12 × Hm is a Banach space with the norm
∥(Y , Z , U )∥
2 H2,m
2
T
2
= E sup |Yt | + E 0≤t ≤T
0
∥f (t )∥2 dt < +∞}.
T
∥Ut ∥2 dt .
|Zt | dt + E 0
T
0
And give the data (f , ξ ) satisfying: The terminal value ξ is a given FT -measurable R-valued random variable such that E |ξ |2 < ∞. f : Ω × [0, T ] × R × R × Rm → R is a progressively measurable function such that f (t , 0, 0, 0) ∈ M 2 . To state the main theorem we need the following definition: Definition 1. (Yt , Zt , Ut ) is said to be a solution of (2) iff (Yt , Zt , Ut ) ∈ H2,m and satisfies BSDE (2), almost surely. For u = (u(1) , u(2) , . . . , u(m) )T and v = (v (1) , v (2) , . . . , v (m) )T ∈ Rm , we say that u ≥ v if u(i) ≥ v (i) for all i = 1, 2, . . . , m. 3. BSDE with continuous coefficient In this section, we study the BSDE when the generator f is continuous in (y, z ), Lipschitz in u, and satisfies an improved linear growth condition in (y, z , u). We will prove the existence of a minimal solution by means of the approximation of the function f and the comparison theorem. First of all, we can have the existence and uniqueness theorem of BSDE (2) in the simple case. Theorem 1. Assume that we can find a non-negative square-integrable function A(t ) and a positive constant B, such that
|f (t , y, z , u)| ≤ A(t )(|y| + 1) + B(|z | + ∥u∥), and
|f (t , y, z , u) − f (t , y′ , z ′ , u′ )| ≤ A(t )|y − y′ | + B(|z − z ′ | + ∥u − u′ ∥), for t ∈ [0, T ], (y, z , u), (y′ , z ′ , u′ ) ∈ R2+m . Then BSDE (2) has a unique solution. Proof. The result is a direct consequence of Situ (1997) [17].
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367
The comparison theorem is one of the principal tools in the theory of the BSDEs, but it does not hold for solutions of BSDE with jumps in general case (see the counter-example in Barles et al. (1997) [4]). With additional conditions of f , the following comparison theorem has been proved by El Otmani (2008) [8]. Theorem 2 (Comparison Theorem). Let (Y i , Z i , U i ) be a solution of the BSDE with data (f i , ξ i ), for i = 1, 2. We suppose that ξ 1 ≤ ξ 2 , and f 1 (t , y, z , u) ≤ f 2 (t , y, z , u) for all (y, z , u) ∈ R2+m ; If u ≥ v , then f 1 (t , y, z , u) ≥ f 1 (t , y, z , v) for all (t , y, z ) ∈ [0, T ] × R2 . Then Yt1 ≤ Yt2 , P-a.s. for all 0 ≤ t ≤ T . Now we give the assumptions. For all (t , y, z ) ∈ [0, T ] × R × R and u, v ∈ Rm (H1) f (t , y, z , u) is continuous in (y, z ); (H2) We can find a non-negative square-integrable function A(t ) and a positive constant B such that:
|f (t , y, z , u)| ≤ A(t )(|y| + 1) + B(|z | + ∥u∥); (H3) |f (t , y, z , u) − f (t , y, z , v)| ≤ B∥u − v∥; (H4) If u ≥ v , then f (t , y, z , u) ≥ f (t , y, z , v). Theorem 3. Assume that (H1)–(H4) hold. Then there exists a minimal solution of BSDE (2). In order to prove the theorem, we follow the method used by Lepeltier–San Martin (1997) [13], and introduce the following sequence of functions: f n ( t , y , z , u) =
inf {f (t , a, b, u) + (n + A(t ))|y − a| + (n + B)|z − b|},
(a,b)∈Q2
n ∈ N.
The functions fn have the following properties: Lemma 1. Assume that (H1)–(H4) hold. Then for all (y, z , u), (y′ , z ′ , u′ ) ∈ R2+m , the above functions fn have the properties (P1) (P2) (P3) (P4) (P5)
|fn (t , y, z , u) − fn (t , y′ , z ′ , u′ )| ≤ (n + A(t ))|y − y′ | + (n + B)|z − z ′ | + B∥u − u′ ∥, fn+1 (t , y, z , u) ≥ fn (t , y, z , u), If u ≥ u′ , then fn (t , y, z , u) ≥ fn (t , y, z , u′ ), |fn (t , y, z , u)| ≤ A(t )(|y| + 1) + B(|z | + ∥u∥), If (yn , zn , un ) → (y, z , u) as n → +∞, then fn (t , yn , zn , un ) → f (t , y, z , u).
Proof. (1) We first show the inequality (P1). For any real number ϵ > 0, by the definition of fn , we can find a pair
(aϵ , bϵ ) ∈ Q2 such that
fn (t , y, z , u) ≥ f (t , aϵ , bϵ , u) + (n + A(t ))|y − aϵ | + (n + B)|z − bϵ | − ϵ
= f (t , aϵ , bϵ , u) + (n + A(t ))(|y′ − aϵ | + |y − aϵ | − |y′ − aϵ |) + (n + B)(|z ′ − bϵ | + |z − bϵ | − |z ′ − bϵ |) − ϵ ≥ f (t , aϵ , bϵ , u) + (n + A(t ))(|y′ − aϵ | − |y − y′ |) + (n + B)(|z ′ − bϵ | − |z − z ′ |) − ϵ ≥ fn (t , y′ , z ′ , u) − (n + A(t ))|y − y′ | − (n + B)|z − z ′ | − ϵ. Since ϵ is arbitrary, then by interchanging the roles of (y, z ), (y′ , z ′ ), we can obtain
|fn (t , y, z , u) − fn (t , y′ , z ′ , u)| ≤ (n + A(t ))|y − y′ | + (n + B)|z − z ′ |, which implies (P1) under the assumption (H3). (2) From the definition of fn , we have, for any (y, z , u) ∈ R × R × Rm , all a, b ∈ Q2 , fn (t , y, z , u) ≤ f (t , a, b, u) + (n + A(t ))|y − a| + (n + B)|z − b|
≤ f (t , a, b, u) + ((n + 1) + A(t ))|y − a| + ((n + 1) + B)|z − b|. Notice that the left side of above inequality is independent of (a, b), then we can obtain fn (t , y, z , u) =
≤
inf
(a,b)∈Q2
fn (t , y, z , u)
inf {f (t , a, b, u) + (n + A(t ))|y − a| + (n + B)|z − b|}
(a,b)∈Q2
≤
inf {f (t , a, b, u) + ((n + 1) + A(t ))|y − a| + ((n + 1) + B)|z − b|}
(a,b)∈Q2
= fn+1 (t , y, z , u). That is (P2).
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(3) By (H4), we find that for all a, b ∈ Q2 , u ≥ u′ , fn (t , y, z , u′ ) ≤ f (t , a, b, u′ ) + (n + A(t ))|y − a| + (n + B)|z − b|
≤ f (t , a, b, u) + (n + A(t ))|y − a| + (n + B)|z − b|. Then the definition of fn implies (P3) immediately. (4) From (H1) and (H2), we know that fn (t , y, z , u) ≤ f (t , y, z , u) ≤ A(t )(|y| + 1) + B(|z | + ∥u∥). On the other hand, from (H2), we have fn (t , y, z , u) =
inf {f (t , a, b, u) + (n + A(t ))|y − a| + (n + B)|z − b|}
(a,b)∈Q2
inf {−A(t )(|a| + 1) − B(|b| + ∥u∥) + (n + A(t ))|y − a| + (n + B)|z − b|}
≥
(a,b)∈Q2
inf {−A(t )(|a| + 1) − B(|b| + ∥u∥) + A(t )|y − a| + B|z − b|}
≥
(a,b)∈Q2
≥ −A(t )(|y| + 1) − B(|z | + ∥u∥). So we can obtain the linear growth property (P4). (5) Finally, we verify the property (P5). Suppose (yn , zn , un ) → (y, z , u), as n → +∞, then for every n > 0, and ϵ = there exists a pair (an , bn ) ∈ Q2 , such that
1 , n
f (t , yn , zn , un ) ≥ fn (t , yn , zn , un )
≥ f (t , an , bn , un ) + (n + A(t ))|yn − an | + (n + B)|zn − bn | − ϵ = f (t , an , bn , un ) + (n + A(t ))|yn − an | + (n + B)|zn − bn | −
1 n
≥ −A(t )(|an | + 1) − B|bn | − B∥un ∥ + (n + A(t ))|yn − an | + (n + B)|zn − bn | −
1 n
1
≥ −A(t )(|yn | + 1) − B|zn | − B∥un ∥ + n|yn − an | + n|zn − bn | − . n
(3)
Since {(yn , zn , un )} is bounded, for any fixed t, we have that lim supn→+∞ {n|yn − an | + n|zn − bn |} < ∞. This implies that (an , bn ) → (y, z ), as n → +∞. Notice that (n + A(t ))|yn − an | + (n + B)|zn − bn | ≥ 0, from (3), we get f (t , yn , zn , un ) ≥ fn (t , yn , zn , un ) ≥ f (t , an , bn , un ) −
1 n
.
Taking n → +∞, by the hypothesis (H3), we deduce that (P5) holds.
For each fixed n ≥ 1, applying Theorem 1, there exists a unique triple solution (Y n , Z n , U n ) of BSDE (2) with data (fn , ξ ). Now we establish a priori estimate on the solution (Y n , Z n , U n ). Lemma 2. Assume that (H1)–(H4) hold. And for each fixed n ≥ 1, (Y n , Z n , U n ) is a solution of BSDE with data (fn , ξ ). Then there exists a constant C > 0, which is independent of n, such that
sup E sup |Ytn |2 + E n
0≤t ≤T
T
|Ztn |2 dt + E 0
T
∥Utn ∥2 dt
< C.
0
Proof. Ito’s formula gives, for all t ∈ [0, T ] E |Ytn |2 + E
T
|Zsn |2 ds + E
T
t
∥Usn ∥2 ds ≤ E |ξ |2 + 2E
T
t
(A(s)|Ysn |(|Ysn | + 1) + B|Ysn |(|Zsn | + ∥Usn ∥))ds. t
Then we have E |Ytn |2 +
1 2
T
1
|Zsn |2 ds + E
E
2
t
T
∥Usn ∥2 ds ≤ E |ξ |2 + t
T
A2 (s)ds + E
T
t
(2A(s) + 4B2 + 1)|Ysn |2 ds. t
Applying Gronwall’s Lemma, we get, for all n ≥ 1 sup E |Ytn |2 ≤
0≤t ≤T
E |ξ |2 +
T
A2 (s)ds exp 0
T
(2A(s) + 4B2 + 1)ds .
0
Notice that the right part of the above inequality is independent of n. The result follows directly from this and Burkholder– Davis–Gundy inequality [10].
Y. Qin, N.-M. Xia / J. Math. Anal. Appl. 406 (2013) 365–372
369
Proof of Theorem 3. Using the comparison theorem (Theorem 2), and property (P2) and (P3) in Lemma 1, we first know that {Y·n }n≥1 does not decrease. Now consider the BSDE (2) with data (f , ξ ), where f (t , y, z , u) = A(t )(|y|+ 1)+ B(|z |+∥u∥). By Theorem 1, there exists a unique triple solution (Y , Z , U ) in this case. Since |fn (t , y, z , u)| ≤ A(t )(|y| + 1) + B(|z | + ∥u∥), applying the comparison theorem (Theorem 2), we have Ytn ≤ Y t . It implies that {Y·n }n≥1 is a bounded nondecreasing sequence. Then there exists a process Yt as the limit of the monotone sequence Ytn . Since supn E sup0≤t ≤T |Ytn |2 ≤ C , we have
T
Yt ∈ S 2 and E 0 |Ytn − Yt |2 dt → 0, as n → +∞. Now we try to show that {Z·n , U·n }n≥1 is a Cauchy sequence on the Hilbert space H12 × Hm2 . By (H2), we have, for each n≥1 T
|f (s,
E
Ysn
,
Zsn
,
Usn
)| ds ≤ 4E 2
T
A (s)ds E sup | 2
0≤t ≤T
0
0
Ytn 2
2
| + 1 + 4B
p
q
T
t
| |
E
T 0
T
∥
+E
Usn 2 ds
∥
.
0
|f (s, Ysn , Zsn , Usn )|2 ds ≤ C˜ . Then applying Ito’s
T
|Zsp − Zsq |2 ds + E
Zsn 2 ds
0
Lemma 2 implies that there is a constant C˜ independent of n, such that E p q formula to |Yt − Yt |2 , we get, for any positive integers p, q ≥ 1 E |Yt − Yt |2 + E
T
∥Usp − Usq ∥2 ds t
T
|Ysp − Ysq |(|fp (s, Ysp , Zsp , Usp )| + |fq (s, Ysq , Zsq , Usq )|)ds
≤ 2E t
21
T
≤ 16 C˜ E
|
Ysp
−
Ysq 2 ds
|
.
t
So {Z·n , U·n }n≥1 is a Cauchy sequence on the Hilbert space H12 × Hm2 , and then there exists a process (Yt , Zt , Ut ) as the limit of the sequence (Ytn , Ztn , Utn ). By a way similar to that in the proof of Theorem 4.1 of El Otmani (2008) [8], it is not hard to pass the limit on both sides of the BSDE with data (fn , ξ ). Then (Yt , Zt , Ut ) is a solution of BSDE (2). Using Lepeltier–San Martin’s method (see Lepeltier and San Martin (1997) [13]), we know that (Yt , Zt , Ut ) is a minimal solution. Remark 1. For example, consider the following BSDE: T
Yt = t
4 −1 s 3 Ys ds − 3
T
Zs dWs − t
m i=1
T
Us(i) dN˜ s(i) . t ∈ [0, T ].
t 2
We note that the above equation satisfies assumptions (H1)–(H4), moreover, for any c ∈ [0, T ], (Yt , Zt , Ut ) = ((max{(c 3 − 2 3
t ), 0})2 , 0, 0) satisfies the equation. So the solution is not unique. 4. BSDE with left continuous coefficient We will work under the following assumptions. For all (t , y, z ), (t , y′ , z ′ ) ∈ [0, T ] × R × R and u, v ∈ Rm (HH1) f (t , y, z , u) is left continuous in y, and continuous in z; (HH2) There are one non-negative square-integrable function A(t ) and one positive constant B such that
|f (t , y, z , u)| ≤ A(t )(|y| + 1) + B|z |; (HH3) |f (t , y, z , u) − f (t , y, z , v)| ≤ B∥u − v∥; (HH4) If u ≥ v , then f (t , y, z , u) ≥ f (t , y, z , v). (HH5) There exists a continuous function k1 (x) defined on R2 , satisfying |k1 (x)| ≤ B|x|, for any x ∈ R2 with the constant B defined in (HH2), such that
|f (t , y, z , u) − f (t , y′ , z ′ , u)| ≥ k1 (y − y′ , z − z ′ ). Theorem 4. Assume that (HH1)–(HH5) hold. Then there exists a solution of BSDE (2). To prove the theorem, we need the following lemma. Lemma 3. Let B be a given constant and k1 (x) be a continuous function defined on R2 with |k1 (x)| ≤ B|x|, for any x ∈ R2 . Let k2 (t , u) be a continuous function defined on [0, T ] × Rm with k2 (t , 0) = 0, such that for all t ∈ [0, T ], u, v ∈ Rm , we have |k2 (t , u) − k2 (t , v)| ≤ B∥u − v∥, and that if u ≥ v , then k2 (t , u) ≥ k2 (t , v). Now consider the BSDE: Yt = ξ +
T
(k1 (Ys , Zs ) + k2 (s, Us ) + φs )ds − t
T
Zs dWs − t
m i=1
t
T
Us(i) dN˜ s(i) .
(4)
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If ξ ∈ FT , E |ξ |2 < ∞ and φs ∈ M 2 , then (PP1) BSDE (4) has a minimal solution (Yt , Zt , Ut ) in H2,m . (PP2) For any solution (Yt , Zt , Ut ) of BSDE (4), ξ ≥ 0 and φt ≥ 0 implies Yt ≥ 0, P-a.s. Proof. If we notice that k1 , k2 are two continuous functions, |k1 (x)| ≤ B|x|, |k2 (t , u) − k2 (t , v)| ≤ B∥u − v∥, and that if u ≥ v , then k2 (t , u) ≥ k2 (t , v), using Theorem 3, we know that there exists a minimal solution (Yt , Zt , Ut ) of BSDE (4) in H2,m . The proof of (PP1) is completed. Now we try to prove (PP2). To use Lepeltier–San Martin’s method, we introduce the sequence {k1,n }n≥1 defined by k1,n (x) = inf {k1 (a) + (n + B)|x − a|}, a∈Q2
n ∈ N.
For all t ∈ [0, T ] and x, x′ ∈ R2 , the sequence satisfies (k1) (k2) (k3) (k4)
|k1,n (x)| ≤ B|x|, |k1,n (x) − k1,n (x′ )| ≤ (B + n)|x − x′ |, k1,n ≤ k1,n+1 If xn → x as n → +∞, then k1,n (xn ) → k1 (x).
From (k2) and the inequality |k2 (t , u) − k2 (t , v)| ≤ B∥u − v∥, for each fixed n ≥ 1, we can use Theorem 1 and find the unique solution (yn· , z·n , un· ) of the BSDE T
Yt = ξ +
(k1,n (Ys , Zs ) + k2 (s, Us ) + φs )ds −
T
Zs dWs −
t
m
t
T
Us(i) dN˜ s(i) .
(5)
t
i =1
Then by Theorem 2, we know that {yn· } is a nondecreasing sequence. To estimate the solutions of BSDE (5), we need the following equations: Yt1
=ξ+
T
(−B|
Ys1
| − B| | + k2 (s, Zs1
Us1
) + φs )ds −
t
Yt2 = ξ +
(B|Ys2 | + B|Zs2 | + k2 (s, Us2 ) + φs )ds − t
Yt3 =
−
(−B|Ys3 | − B|Zs3 | + k2 (s, Us3 ))ds −
T
Zs2 dWs −
t
T
m
m
Zs3 dWs − t
i =1 m i=1
T
T
Us1(i) dN˜ s(i) ,
(6)
t
i =1
t
T
Zs1 dWs
t T
T
T
Us2(i) dN˜ s(i) ,
(7)
t
Us3(i) dN˜ s(i) .
(8)
t
By Theorem 1, we can obtain the unique solution (Y·k , Z·k , U·k ) of the above BSDE, for every k = 1, 2, 3. If we notice that ξ ≥ 0, φt ≥ 0 and zero is the unique solution of (8), then by comparing (8), (7), (6) with (5), we have 0 = Yt3 ≤ Yt1 ≤ y1t ≤ y2t ≤ · · · ≤ ynt ≤ · · · ≤ Yt2 ,
P-a.s.
Using Lepeltier–San Martin’s method (see Lepeltier and San Martin (1997) [13]), the solution sequence {(yn· , z·n , un· )}n≥1 of the BSDE (5) converges to the minimal solution (Y· , Z· , U· ) of the BSDE (4). Then we know that Yt ≥ Yt3 = 0, for all t ∈ [0, T ]. Proof of Theorem 4. We first construct the following sequence of BSDEs Y 0t = ξ +
T
(−A(s)|Y 0s | − B|Z 0s | − A(s))ds − t
Y it = ξ +
Z 0s dWs − t
m
T
U 0s (k) dN˜ s(k) ;
(9)
t
k=1
T
t T
T
−
(f (s, Y is−1 , Z is−1 , U is ) + k1 (Y is − Y is−1 , Z is − Z is−1 ))ds m T Z is dWs − U si(k) dN˜ s(k) , i = 1, 2, . . . ;
t
k=1
(10)
t
and 0
Yt = ξ +
T
0
0
(A(s)|Y s | + B|Z s | + A(s))ds − t
T
0
Z s dWs − t
m k=1
T
0(k)
Us
dN˜ s(k) .
(11)
t 0
0
0
Using Theorem 1, BSDE (9) (or (11)) has a unique adapted solution, which is denoted by (Y 0· , Z 0· , U 0· ) (or (Y · , Z · , U · )). And for every i = 1, 2, . . . , Theorem 3 implies the existence of the minimal solution (Y i· , Z i· , U i· ) of BSDE (10).
Y. Qin, N.-M. Xia / J. Math. Anal. Appl. 406 (2013) 365–372
371
We can prove that the above solutions have the following properties: (kk1) For all t ∈ [0, T ] and any positive integer i, Y 0t ≤ Y it ≤ Y ti+1 , P-a.s.; 0
(kk2) For all t ∈ [0, T ] and any positive integer i, Y it ≤ Y t , P-a.s. For (kk1), we first prove Y 0t ≤ Y 1t . In fact, from (9), (10), we know Y 1t − Y 0t =
T
(k1 (Y 1s − Y 0s , Z 1s − Z 0s ) + k(20) (s, U 1s − U 0s ) + ∆1s )ds
t T
(Z 1s − Z 0s )dWs −
− t
m k=1
T
(U 1s (k) − U 0s (k) )dN˜ s(k) ,
(12)
t
(0)
where k2 (s, u) = f (s, Y 0s , Z 0s , u + U 0s ) − f (s, Y 0s , Z 0s , U 0s ) and ∆1s = f (s, Y 0s , Z 0s , U 0s ) + A(s)|Y 0s | + B|Z 0s | + A(s). From (0)
(0)
(0)
(HH4), k2 (s, u) has the property that if u ≥ v , then k2 (s, u) ≥ k2 (s, v). From (HH2), we have ∆1s ≥ 0 and ∆1s ≤ 2(A(s)|Y 0s | + B|Z 0s | + A(s)). Notice that T
E
T
(∆1s )2 ds ≤ 12 E
A2 (s)|Y 0s |2 ds + B2 E 0
0
T
|Z 0s |2 ds + E 0
≤ 12 E sup |Y 0t |2 0≤t ≤T
T
A2 (s)ds + B2 E
T
A2 (s)ds 0
T
0
|Z 0s |2 ds + E 0
T
A2 (s)ds
< +∞
0
so ∆1s ∈ M 2 . By Lemma 3, we have Y 0t ≤ Y 1t . Now we use the method of induction to show that Y it ≤ Y ti+1 under the assumption Y it−1 ≤ Y it . By (10), we have Y it+1 − Y it =
T
(k1 (Y is+1 − Y is , Z is+1 − Z is ) + k(2i) (s, U is+1 − U is ) + ∆si+1 )ds
t T
(Z is+1 − Z is )dWs −
− t
m k=1
T
(U is+1(k) − U is(k) )dN˜ s(k) ,
(13)
t
(i+1)
(i)
= f (s, Y is , Z is , U is )− f (s, Y is−1 , Z si−1 , U is )− k1 (Y is − Y is−1 , Z is − (i) (i) Z is−1 ). From (HH4), k2 (s, u) has the property that if u ≥ v , then k2 (s, u) ≥ k2 (s, v). From (HH5) and (HH2), we get that ∆is+1 ≥ 0 and ∆si+1 ∈ M2 . Then by Lemma 3, we have Y it ≤ Y ti+1 . The proof of (kk1) is completed. Now consider (kk2). For i = 1, 2, . . . , from (10) and (11), we have equations: T T m T 0 0 0 0 0(k) Y t − Y it = (−A(s)|Y s − Y is | − B|Z s − Z is | + Φsi )ds − (U s − U is(k) )dN˜ s(k) , (Z s − Z is )dWs − (14) where k2 (s, u) = f (s, Y is , Z is , u + U is )− f (s, Y is , Z is , U is ) and ∆s (i)
t
t 0
0
0
k=1
t
0
where Φsi = A(s)|Y s − Y is | + B|Z s − Z is | + A(s)|Y s | + B|Z s | + A(s) − f (s, Y is−1 , Z is−1 , U is ) − k1 (Y is − Y is−1 , Z is − Z si−1 ) ≥ f (s, Y is , Z is , U is ) − f (s, Y is−1 , Z is−1 , U is ) − k1 (Y is − Y is−1 , Z is − Z is−1 ). By (HH5) and (HH2), we get that Φsi ≥ 0 and Φsi ∈ M 2 . If we notice zero to be the unique solution of the following BSDE: Yt2 =
T
(−A(s)|Ys2 | − B|Zs2 |)ds − t
T
Zs2 dWs − t
m k=1
T
Us2(k) dN˜ s(k) .
(15)
t
0
Then by comparing (14) with (15), we have Y t ≥ Y it . This follows that the minimal solutions of Eqs. (9) and (10) form a bounded nondecreasing sequence, that is 0
Y 0t ≤ Y 1t ≤ Y 2t ≤ · · · ≤ Y it ≤ · · · ≤ Y t . P-a.s. Then we know that 0
sup E sup |Y nt |2 ≤ E sup |Y 0t |2 + E sup |Y t |2 < +∞ n
0≤t ≤T
0 ≤t ≤T
0≤t ≤T
Y nt n≥1
and that { } converges to a limit Y in S 2 . Now we consider {(Z nt , U nt )}n≥1 in H12 × Hm2 . From (10) and Ito’s formula, we have E |ξ |2 = E |Y 0i+1 |2 − 2E
T
Y is+1 (f (s, Y is , Z is , U is+1 ) + k1 (Y is+1 − Y is , Z si+1 − Z is ))ds + E 0
t
|Z is+1 |2 ds + E 0
t
∥U si+1 ∥2 ds. 0
372
Y. Qin, N.-M. Xia / J. Math. Anal. Appl. 406 (2013) 365–372
So, we can find that
t
E
t
|Z is+1 |2 ds + E
0
∥U is+1 ∥2 ds ≤ E |ξ |2 + 2E
Y is+1 (f (s, Y is , Z is , U si+1 ) + k1 (Y is+1 − Y is , Z si+1 − Z is ))ds 0
0 T
2
T
|Y is+1 |(A(s)|Y is | + B|Z is | + A(s) + B|Y is+1 − Y is | + B|Z is+1 − Z is |)ds
≤ E |ξ | + 2E 0 T
1
(|Z is+1 |2 + |Z is |2 )ds,
≤C+ E 8
0
where C = E |ξ |2 + (3
T 0
A(s)ds + 40B2 T + 4BT ) supi E sup0≤t ≤T |Y it |2 .
T |Z it |2 dt + E 0 ∥U it ∥2 dt ) < ∞. Then we know that for i = 1, 2, . . . , the functions Ψti+1 = T f (t , Y it , Z it , U it+1 ) + k1 (Y it+1 − Y it , Z it+1 − Z it ) are uniformly bounded in M 2 . Set Cˆ = supi 0 |Ψti+1 |2 dt. Applying Ito’s formula p q to |Y t − Y t |2 , we get, for any positive integers p, q ≥ 1 T T p q E |Y t − Y t |2 + E |Z ps − Z qs |2 ds + E ∥U ps − U qs ∥2 ds This implies that supi (E
T 0
t
t
T
≤ 2E
|
Y ps
−
Y qs
Ψsp
|(|
|+|
Ψsq
|)ds ≤ 4 Cˆ E
12
T
|
Y ps
−
Y qs 2 ds
|
.
t
t
It follows that {(Z nt , U nt )}n≥1 is a Cauchy sequence in H12 × Hm2 . Then there exists a process (Yt , Zt , Ut ) ∈ H2,m to be the limit of the sequence {(Y nt , Z nt , U nt )}n≥1 . By the way similar to that in Theorem 3, passing to the limit (as i → +∞) on the both sides of (10), we have the conclusion that (Yt , Zt , Ut ) solves (2). Remark 2. For example, consider the following BSDE: T
Yt =
s t
− 31
T
IYs ds −
Zs dWs − t
m i=1
T
Us(i) dN˜ s(i) . t ∈ [0, T ],
t
where Iy is an indicator function of the interval (0, +∞), that is Iy = 1 for y > 0, otherwise 0. Although all the assumptions 2
2
(HH1)–(HH4) are satisfied, but for any c ∈ [0, T ], (Yt , Zt , Ut ) = (max{ 23 (c 3 − t 3 ), 0}, 0, 0) is the solution. This means that the solution is not unique. Remark 3. In the same way as that in the Remark 6 of Jia (2008) [11], we can prove the existence of the adapted solution of BSDE (2), when the coefficient f is right continuous, decreasing and bounded. References [1] K. Bahlali, Existence and uniqueness of solutions for BSDEs with locally Lipschitz coefficient, Electron. Commun. Probab. 7 (2002) 169–179. [2] K. Bahlali, M. Eddahbi, E. Essaky, BSDE associated with Lévy processes and application to PDIE, J. Appl. Math. Stoch. Anal. 16 (1) (2003) 1–17. [3] K. Bahlali, E.H. Essaky, M. Hassani, E. Pardoux, Existence, uniqueness and stability of backward stochastic differential equations with locally monotone coefficient, C. R. Math. Acad. Sci. Paris 335 (9) (2002) 757–762. [4] G. Barles, R. Buckdahn, E. Pardoux, Backward stochastic differential equations and integral–partial differential equations, Stoch. Stoch. Rep. 60 (1997) 57–83. [5] Ph. Briand, R. Carmona, BSDEs with polynomial growth generator, J. Appl. Math. Stoch. Anal. 13 (3) (2000) 207–238. [6] N. El Karoui, S. Hamadène, BSDEs and risk control, zero-sum and nonzero-sum game problems of stochastic functional differential equations, Stoch. Process. Appl. 107 (2003) 145–169. [7] N. El Karoui, S. Peng, M.C. Quenez, Backward stochastic differential equations in finance, Math. Finance 7 (1997) 1–71. [8] M. El Otmani, BSDE driven by simple Lévy process with continuous coefficient, Statist. Probab. Lett. 78 (2) (2008) 1259–1265. [9] S. Hamadène, J.P. Lepeltier, Zero-sum stochastic differential games and BSDEs, Systems Control Lett. 24 (1995) 259–263. [10] S. He, J. Wang, J. Yan, Semimartingale Theory and Stochastic Calculus, Science Press and CRC Press, Inc., 1992. [11] G. Jia, A class of backward stochastic differential equations with discontinuous coefficients, Statist. Probab. Lett. 78 (2008) 231–237. [12] M. Kobylanski, Backward stochastic differential equations and partial differential equations with quadratic growth, Ann. Probab. 28 (2000) 259–276. [13] J.P. Lepeltier, J. San Martin, Backward stochastic differential equations with continuous coefficient, Statist. Probab. Lett. 32 (4) (1997) 425–430. [14] Y. Ouknine, Reflected backward stochastic differential equations with jumps, Stoch. Stoch. Rep. 65 (1998) 111–125. [15] Y. Ouknine, Ndiaye, Djibril, Sur l’existence de solutions d’quations diffrentielles stochastiques progrssives rtrogrades couples (Existence of solutions of coupled forward–backward stochastic differential equations), Stochastics 80 (4) (2008) 299–315 (in French). [16] E. Pardoux, S. Peng, Adapted solution of a backward stochastic differential equation, Systems Control Lett. 14 (1990) 55–61. [17] R. Situ, On solutions of backward stochastic differential equations with jumps and applications, Stoch. Process. Appl. 66 (1997) 209–236. [18] S. Tang, X. Li, Necessary conditions for optimal control of stochastic systems with random jumps, SIAM J. Control Optim. 32 (5) (1994) 1447–1475.
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Journal of Mathematical Analysis and Applications journal homepage: www.elsevier.com/locate/jmaa
BSDE driven by Poisson point processes with discontinuous coefficient✩ Yan Qin ∗ , Ning-Mao Xia Department of Mathematics, East China University of Science and Technology, Shanghai, China
article
abstract
info
Article history: Received 9 July 2008 Available online 2 May 2013 Submitted by U. Stadtmueller
In this paper, we deal with the one-dimensional backward stochastic differential equation (BSDE) driven by Poisson processes. By means of the comparison theorem, we first prove the existence of a (minimal) solution for BSDE where the coefficient is continuous and satisfies an improved linear growth assumption. Then we extend the result to the case where the coefficient is left or right continuous. Crown Copyright © 2013 Published by Elsevier Inc. All rights reserved.
Keywords: Backward stochastic differential equation Adapted solution Comparison theorem
1. Introduction The following nonlinear backward stochastic differential equation (BSDE) was first introduced by Pardoux and Peng (1990) [16] Yt = ξ +
T
f (s, Ys , Zs )ds − t
T
Zs dWs ,
t ∈ [0, T ].
(1)
t
Because of the important applications in mathematical finance (El Karoui et al., 1997 [7]), stochastic control (El Karoui and Hamadène, 2003 [6]) and stochastic games (Hamadène and Lepeltier, 1995 [9]), this equation has been studied intensively, and many efforts have been made to relax the assumption on the generator f . For instance, Lepeltier and San Martin (1997) [13] proved the existence of a solution when f is only continuous with linear growth, Kobylanski (2000) [12] obtained the existence and uniqueness of a solution when f is continuous with a quadratic growth in z, Briand and Carmona (2000) [5] derived a solution when f is monotone Lipschitz with polynomial growth, Bahlali [1–3] studied those with locally Lipschitz coefficients, and Jia (2008) [11] dealt with the solution when f is discontinuous in y and continuous in z. The driving stochastic process Wt in Eq. (1) is a Brownian motion, which can reflect only the effect of continuous random information. So it is of significance to introduce some discontinuous process, such as the Poisson process, into the equation. Tang and Li (1994) [18] developed the BSDE as follows: Yt = ξ +
T
f (s, Ys , Zs , Us )ds − t
T
Zs dWs − t
m i =1
T
Us(i) dN˜ s(i) ,
t ∈ [0, T ],
(2)
t
and proved the existence and uniqueness of the adapted solution when f is Lipschitz continuous. Barles et al. (1997) [4] studied the BSDE, which is driven by a Brownian motion and an independent Poisson point process. Situ (1997) [17]
✩ Supported by The National Natural Science Foundation (11002055) (China) and Fundamental Research Funds for the East China University of Science and Technology (WM1114040). ∗ Corresponding author. E-mail addresses: [email protected], [email protected] (Y. Qin), [email protected] (N.-M. Xia).
0022-247X/$ – see front matter Crown Copyright © 2013 Published by Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jmaa.2013.02.071
366
Y. Qin, N.-M. Xia / J. Math. Anal. Appl. 406 (2013) 365–372
considered the similar problem when f is non-Lipschitz with linear growth. Ouknine [14,15] derived the existence and uniqueness of the adapted solution for BSDE driven by a Poisson random measure. El Otmani (2008) [8] discussed the onedimensional BSDE, driving by a standard Brownian motion and a family of independent Poisson processes (called the simple Lévy process in [8]). He proved the existence of a solution for the one-dimensional BSDE where the coefficient is continuous with linear growth, or left continuous, increasing and bounded. The aim of this paper is to study the solution of BSDE driven by a standard Brownian motion and a family of independent Poisson point processes. Our setting is that the generator f satisfies an improved linear growth assumption |f (t , y, z , u)| ≤ A(t )(|y| + 1) + B(|z | + ∥u∥), instead of the previous one |f (t , y, z , u)| ≤ A(|y| + |z | + 1) (in Lepeltier and San Martin (1997) [13], Jia (2008) [11], El Otmani (2008) [8]). We will first prove the existence of a (minimal) solution for BSDE where the coefficient is continuous with an improved linear growth. Then we extend the result to the case that the coefficient is discontinuous. The paper is organized as follows. The next section gives some notations and definition of the solution of BSDE (2). Section 3 contains the existence result when the coefficient is continuous. The case where coefficient is discontinuous in y is discussed in Section 4. 2. Preliminaries In this paper, we consider the BSDE (2), where W is a standard Brownian motion and N˜ = (N˜ (1) , N˜ (2) , . . . , N˜ (m) )T is a family of independent compensating Poisson point processes with definition (i)
N˜ t
= Nt(i) − t ,
t ∈ [0, T ], i = 1, 2, . . . , m. (1)
(2)
(m)
In the above expression Nt = (Nt , Nt , . . . , Nt )T is an m dimensional Poisson process with non-random intensity (t , t , . . . , t )T . (i) Let (Ω , F , {Ft }0≤t ≤T , P ) be a complete probability space with the natural filtration {Ft }0≤t ≤T (i.e. Ft = σ {Ws , Ns , 0 ≤ s ≤ t , i = 1, 2, . . . , m, N }, where N is the totality of P-null set). The filtration {Ft }0≤t ≤T is right continuous (Ft = ∩ε>0 Ft +ε = Ft + ). 1
In our paper, the Euclid-norm is defined by ∥x∥ = (xT x) 2 , for x ∈ Rm . Now we define some spaces. Let
T
M2 = {f : [0, T ] × Ω → R is progressive measurable such that E 0 |f (t )|2 dt < +∞}. S 2 = {f : [0, T ] × Ω → R is adapted and rcll, such that f ∈ M2 and E sup0≤t ≤T |f (t )|2 < +∞}. Hm2 = {f : [0, T ] × Ω → Rm is the predictable process such that E 2 H2,m = S 2 × H12 × Hm is a Banach space with the norm
∥(Y , Z , U )∥
2 H2,m
2
T
2
= E sup |Yt | + E 0≤t ≤T
0
∥f (t )∥2 dt < +∞}.
T
∥Ut ∥2 dt .
|Zt | dt + E 0
T
0
And give the data (f , ξ ) satisfying: The terminal value ξ is a given FT -measurable R-valued random variable such that E |ξ |2 < ∞. f : Ω × [0, T ] × R × R × Rm → R is a progressively measurable function such that f (t , 0, 0, 0) ∈ M 2 . To state the main theorem we need the following definition: Definition 1. (Yt , Zt , Ut ) is said to be a solution of (2) iff (Yt , Zt , Ut ) ∈ H2,m and satisfies BSDE (2), almost surely. For u = (u(1) , u(2) , . . . , u(m) )T and v = (v (1) , v (2) , . . . , v (m) )T ∈ Rm , we say that u ≥ v if u(i) ≥ v (i) for all i = 1, 2, . . . , m. 3. BSDE with continuous coefficient In this section, we study the BSDE when the generator f is continuous in (y, z ), Lipschitz in u, and satisfies an improved linear growth condition in (y, z , u). We will prove the existence of a minimal solution by means of the approximation of the function f and the comparison theorem. First of all, we can have the existence and uniqueness theorem of BSDE (2) in the simple case. Theorem 1. Assume that we can find a non-negative square-integrable function A(t ) and a positive constant B, such that
|f (t , y, z , u)| ≤ A(t )(|y| + 1) + B(|z | + ∥u∥), and
|f (t , y, z , u) − f (t , y′ , z ′ , u′ )| ≤ A(t )|y − y′ | + B(|z − z ′ | + ∥u − u′ ∥), for t ∈ [0, T ], (y, z , u), (y′ , z ′ , u′ ) ∈ R2+m . Then BSDE (2) has a unique solution. Proof. The result is a direct consequence of Situ (1997) [17].
Y. Qin, N.-M. Xia / J. Math. Anal. Appl. 406 (2013) 365–372
367
The comparison theorem is one of the principal tools in the theory of the BSDEs, but it does not hold for solutions of BSDE with jumps in general case (see the counter-example in Barles et al. (1997) [4]). With additional conditions of f , the following comparison theorem has been proved by El Otmani (2008) [8]. Theorem 2 (Comparison Theorem). Let (Y i , Z i , U i ) be a solution of the BSDE with data (f i , ξ i ), for i = 1, 2. We suppose that ξ 1 ≤ ξ 2 , and f 1 (t , y, z , u) ≤ f 2 (t , y, z , u) for all (y, z , u) ∈ R2+m ; If u ≥ v , then f 1 (t , y, z , u) ≥ f 1 (t , y, z , v) for all (t , y, z ) ∈ [0, T ] × R2 . Then Yt1 ≤ Yt2 , P-a.s. for all 0 ≤ t ≤ T . Now we give the assumptions. For all (t , y, z ) ∈ [0, T ] × R × R and u, v ∈ Rm (H1) f (t , y, z , u) is continuous in (y, z ); (H2) We can find a non-negative square-integrable function A(t ) and a positive constant B such that:
|f (t , y, z , u)| ≤ A(t )(|y| + 1) + B(|z | + ∥u∥); (H3) |f (t , y, z , u) − f (t , y, z , v)| ≤ B∥u − v∥; (H4) If u ≥ v , then f (t , y, z , u) ≥ f (t , y, z , v). Theorem 3. Assume that (H1)–(H4) hold. Then there exists a minimal solution of BSDE (2). In order to prove the theorem, we follow the method used by Lepeltier–San Martin (1997) [13], and introduce the following sequence of functions: f n ( t , y , z , u) =
inf {f (t , a, b, u) + (n + A(t ))|y − a| + (n + B)|z − b|},
(a,b)∈Q2
n ∈ N.
The functions fn have the following properties: Lemma 1. Assume that (H1)–(H4) hold. Then for all (y, z , u), (y′ , z ′ , u′ ) ∈ R2+m , the above functions fn have the properties (P1) (P2) (P3) (P4) (P5)
|fn (t , y, z , u) − fn (t , y′ , z ′ , u′ )| ≤ (n + A(t ))|y − y′ | + (n + B)|z − z ′ | + B∥u − u′ ∥, fn+1 (t , y, z , u) ≥ fn (t , y, z , u), If u ≥ u′ , then fn (t , y, z , u) ≥ fn (t , y, z , u′ ), |fn (t , y, z , u)| ≤ A(t )(|y| + 1) + B(|z | + ∥u∥), If (yn , zn , un ) → (y, z , u) as n → +∞, then fn (t , yn , zn , un ) → f (t , y, z , u).
Proof. (1) We first show the inequality (P1). For any real number ϵ > 0, by the definition of fn , we can find a pair
(aϵ , bϵ ) ∈ Q2 such that
fn (t , y, z , u) ≥ f (t , aϵ , bϵ , u) + (n + A(t ))|y − aϵ | + (n + B)|z − bϵ | − ϵ
= f (t , aϵ , bϵ , u) + (n + A(t ))(|y′ − aϵ | + |y − aϵ | − |y′ − aϵ |) + (n + B)(|z ′ − bϵ | + |z − bϵ | − |z ′ − bϵ |) − ϵ ≥ f (t , aϵ , bϵ , u) + (n + A(t ))(|y′ − aϵ | − |y − y′ |) + (n + B)(|z ′ − bϵ | − |z − z ′ |) − ϵ ≥ fn (t , y′ , z ′ , u) − (n + A(t ))|y − y′ | − (n + B)|z − z ′ | − ϵ. Since ϵ is arbitrary, then by interchanging the roles of (y, z ), (y′ , z ′ ), we can obtain
|fn (t , y, z , u) − fn (t , y′ , z ′ , u)| ≤ (n + A(t ))|y − y′ | + (n + B)|z − z ′ |, which implies (P1) under the assumption (H3). (2) From the definition of fn , we have, for any (y, z , u) ∈ R × R × Rm , all a, b ∈ Q2 , fn (t , y, z , u) ≤ f (t , a, b, u) + (n + A(t ))|y − a| + (n + B)|z − b|
≤ f (t , a, b, u) + ((n + 1) + A(t ))|y − a| + ((n + 1) + B)|z − b|. Notice that the left side of above inequality is independent of (a, b), then we can obtain fn (t , y, z , u) =
≤
inf
(a,b)∈Q2
fn (t , y, z , u)
inf {f (t , a, b, u) + (n + A(t ))|y − a| + (n + B)|z − b|}
(a,b)∈Q2
≤
inf {f (t , a, b, u) + ((n + 1) + A(t ))|y − a| + ((n + 1) + B)|z − b|}
(a,b)∈Q2
= fn+1 (t , y, z , u). That is (P2).
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Y. Qin, N.-M. Xia / J. Math. Anal. Appl. 406 (2013) 365–372
(3) By (H4), we find that for all a, b ∈ Q2 , u ≥ u′ , fn (t , y, z , u′ ) ≤ f (t , a, b, u′ ) + (n + A(t ))|y − a| + (n + B)|z − b|
≤ f (t , a, b, u) + (n + A(t ))|y − a| + (n + B)|z − b|. Then the definition of fn implies (P3) immediately. (4) From (H1) and (H2), we know that fn (t , y, z , u) ≤ f (t , y, z , u) ≤ A(t )(|y| + 1) + B(|z | + ∥u∥). On the other hand, from (H2), we have fn (t , y, z , u) =
inf {f (t , a, b, u) + (n + A(t ))|y − a| + (n + B)|z − b|}
(a,b)∈Q2
inf {−A(t )(|a| + 1) − B(|b| + ∥u∥) + (n + A(t ))|y − a| + (n + B)|z − b|}
≥
(a,b)∈Q2
inf {−A(t )(|a| + 1) − B(|b| + ∥u∥) + A(t )|y − a| + B|z − b|}
≥
(a,b)∈Q2
≥ −A(t )(|y| + 1) − B(|z | + ∥u∥). So we can obtain the linear growth property (P4). (5) Finally, we verify the property (P5). Suppose (yn , zn , un ) → (y, z , u), as n → +∞, then for every n > 0, and ϵ = there exists a pair (an , bn ) ∈ Q2 , such that
1 , n
f (t , yn , zn , un ) ≥ fn (t , yn , zn , un )
≥ f (t , an , bn , un ) + (n + A(t ))|yn − an | + (n + B)|zn − bn | − ϵ = f (t , an , bn , un ) + (n + A(t ))|yn − an | + (n + B)|zn − bn | −
1 n
≥ −A(t )(|an | + 1) − B|bn | − B∥un ∥ + (n + A(t ))|yn − an | + (n + B)|zn − bn | −
1 n
1
≥ −A(t )(|yn | + 1) − B|zn | − B∥un ∥ + n|yn − an | + n|zn − bn | − . n
(3)
Since {(yn , zn , un )} is bounded, for any fixed t, we have that lim supn→+∞ {n|yn − an | + n|zn − bn |} < ∞. This implies that (an , bn ) → (y, z ), as n → +∞. Notice that (n + A(t ))|yn − an | + (n + B)|zn − bn | ≥ 0, from (3), we get f (t , yn , zn , un ) ≥ fn (t , yn , zn , un ) ≥ f (t , an , bn , un ) −
1 n
.
Taking n → +∞, by the hypothesis (H3), we deduce that (P5) holds.
For each fixed n ≥ 1, applying Theorem 1, there exists a unique triple solution (Y n , Z n , U n ) of BSDE (2) with data (fn , ξ ). Now we establish a priori estimate on the solution (Y n , Z n , U n ). Lemma 2. Assume that (H1)–(H4) hold. And for each fixed n ≥ 1, (Y n , Z n , U n ) is a solution of BSDE with data (fn , ξ ). Then there exists a constant C > 0, which is independent of n, such that
sup E sup |Ytn |2 + E n
0≤t ≤T
T
|Ztn |2 dt + E 0
T
∥Utn ∥2 dt
< C.
0
Proof. Ito’s formula gives, for all t ∈ [0, T ] E |Ytn |2 + E
T
|Zsn |2 ds + E
T
t
∥Usn ∥2 ds ≤ E |ξ |2 + 2E
T
t
(A(s)|Ysn |(|Ysn | + 1) + B|Ysn |(|Zsn | + ∥Usn ∥))ds. t
Then we have E |Ytn |2 +
1 2
T
1
|Zsn |2 ds + E
E
2
t
T
∥Usn ∥2 ds ≤ E |ξ |2 + t
T
A2 (s)ds + E
T
t
(2A(s) + 4B2 + 1)|Ysn |2 ds. t
Applying Gronwall’s Lemma, we get, for all n ≥ 1 sup E |Ytn |2 ≤
0≤t ≤T
E |ξ |2 +
T
A2 (s)ds exp 0
T
(2A(s) + 4B2 + 1)ds .
0
Notice that the right part of the above inequality is independent of n. The result follows directly from this and Burkholder– Davis–Gundy inequality [10].
Y. Qin, N.-M. Xia / J. Math. Anal. Appl. 406 (2013) 365–372
369
Proof of Theorem 3. Using the comparison theorem (Theorem 2), and property (P2) and (P3) in Lemma 1, we first know that {Y·n }n≥1 does not decrease. Now consider the BSDE (2) with data (f , ξ ), where f (t , y, z , u) = A(t )(|y|+ 1)+ B(|z |+∥u∥). By Theorem 1, there exists a unique triple solution (Y , Z , U ) in this case. Since |fn (t , y, z , u)| ≤ A(t )(|y| + 1) + B(|z | + ∥u∥), applying the comparison theorem (Theorem 2), we have Ytn ≤ Y t . It implies that {Y·n }n≥1 is a bounded nondecreasing sequence. Then there exists a process Yt as the limit of the monotone sequence Ytn . Since supn E sup0≤t ≤T |Ytn |2 ≤ C , we have
T
Yt ∈ S 2 and E 0 |Ytn − Yt |2 dt → 0, as n → +∞. Now we try to show that {Z·n , U·n }n≥1 is a Cauchy sequence on the Hilbert space H12 × Hm2 . By (H2), we have, for each n≥1 T
|f (s,
E
Ysn
,
Zsn
,
Usn
)| ds ≤ 4E 2
T
A (s)ds E sup | 2
0≤t ≤T
0
0
Ytn 2
2
| + 1 + 4B
p
q
T
t
| |
E
T 0
T
∥
+E
Usn 2 ds
∥
.
0
|f (s, Ysn , Zsn , Usn )|2 ds ≤ C˜ . Then applying Ito’s
T
|Zsp − Zsq |2 ds + E
Zsn 2 ds
0
Lemma 2 implies that there is a constant C˜ independent of n, such that E p q formula to |Yt − Yt |2 , we get, for any positive integers p, q ≥ 1 E |Yt − Yt |2 + E
T
∥Usp − Usq ∥2 ds t
T
|Ysp − Ysq |(|fp (s, Ysp , Zsp , Usp )| + |fq (s, Ysq , Zsq , Usq )|)ds
≤ 2E t
21
T
≤ 16 C˜ E
|
Ysp
−
Ysq 2 ds
|
.
t
So {Z·n , U·n }n≥1 is a Cauchy sequence on the Hilbert space H12 × Hm2 , and then there exists a process (Yt , Zt , Ut ) as the limit of the sequence (Ytn , Ztn , Utn ). By a way similar to that in the proof of Theorem 4.1 of El Otmani (2008) [8], it is not hard to pass the limit on both sides of the BSDE with data (fn , ξ ). Then (Yt , Zt , Ut ) is a solution of BSDE (2). Using Lepeltier–San Martin’s method (see Lepeltier and San Martin (1997) [13]), we know that (Yt , Zt , Ut ) is a minimal solution. Remark 1. For example, consider the following BSDE: T
Yt = t
4 −1 s 3 Ys ds − 3
T
Zs dWs − t
m i=1
T
Us(i) dN˜ s(i) . t ∈ [0, T ].
t 2
We note that the above equation satisfies assumptions (H1)–(H4), moreover, for any c ∈ [0, T ], (Yt , Zt , Ut ) = ((max{(c 3 − 2 3
t ), 0})2 , 0, 0) satisfies the equation. So the solution is not unique. 4. BSDE with left continuous coefficient We will work under the following assumptions. For all (t , y, z ), (t , y′ , z ′ ) ∈ [0, T ] × R × R and u, v ∈ Rm (HH1) f (t , y, z , u) is left continuous in y, and continuous in z; (HH2) There are one non-negative square-integrable function A(t ) and one positive constant B such that
|f (t , y, z , u)| ≤ A(t )(|y| + 1) + B|z |; (HH3) |f (t , y, z , u) − f (t , y, z , v)| ≤ B∥u − v∥; (HH4) If u ≥ v , then f (t , y, z , u) ≥ f (t , y, z , v). (HH5) There exists a continuous function k1 (x) defined on R2 , satisfying |k1 (x)| ≤ B|x|, for any x ∈ R2 with the constant B defined in (HH2), such that
|f (t , y, z , u) − f (t , y′ , z ′ , u)| ≥ k1 (y − y′ , z − z ′ ). Theorem 4. Assume that (HH1)–(HH5) hold. Then there exists a solution of BSDE (2). To prove the theorem, we need the following lemma. Lemma 3. Let B be a given constant and k1 (x) be a continuous function defined on R2 with |k1 (x)| ≤ B|x|, for any x ∈ R2 . Let k2 (t , u) be a continuous function defined on [0, T ] × Rm with k2 (t , 0) = 0, such that for all t ∈ [0, T ], u, v ∈ Rm , we have |k2 (t , u) − k2 (t , v)| ≤ B∥u − v∥, and that if u ≥ v , then k2 (t , u) ≥ k2 (t , v). Now consider the BSDE: Yt = ξ +
T
(k1 (Ys , Zs ) + k2 (s, Us ) + φs )ds − t
T
Zs dWs − t
m i=1
t
T
Us(i) dN˜ s(i) .
(4)
370
Y. Qin, N.-M. Xia / J. Math. Anal. Appl. 406 (2013) 365–372
If ξ ∈ FT , E |ξ |2 < ∞ and φs ∈ M 2 , then (PP1) BSDE (4) has a minimal solution (Yt , Zt , Ut ) in H2,m . (PP2) For any solution (Yt , Zt , Ut ) of BSDE (4), ξ ≥ 0 and φt ≥ 0 implies Yt ≥ 0, P-a.s. Proof. If we notice that k1 , k2 are two continuous functions, |k1 (x)| ≤ B|x|, |k2 (t , u) − k2 (t , v)| ≤ B∥u − v∥, and that if u ≥ v , then k2 (t , u) ≥ k2 (t , v), using Theorem 3, we know that there exists a minimal solution (Yt , Zt , Ut ) of BSDE (4) in H2,m . The proof of (PP1) is completed. Now we try to prove (PP2). To use Lepeltier–San Martin’s method, we introduce the sequence {k1,n }n≥1 defined by k1,n (x) = inf {k1 (a) + (n + B)|x − a|}, a∈Q2
n ∈ N.
For all t ∈ [0, T ] and x, x′ ∈ R2 , the sequence satisfies (k1) (k2) (k3) (k4)
|k1,n (x)| ≤ B|x|, |k1,n (x) − k1,n (x′ )| ≤ (B + n)|x − x′ |, k1,n ≤ k1,n+1 If xn → x as n → +∞, then k1,n (xn ) → k1 (x).
From (k2) and the inequality |k2 (t , u) − k2 (t , v)| ≤ B∥u − v∥, for each fixed n ≥ 1, we can use Theorem 1 and find the unique solution (yn· , z·n , un· ) of the BSDE T
Yt = ξ +
(k1,n (Ys , Zs ) + k2 (s, Us ) + φs )ds −
T
Zs dWs −
t
m
t
T
Us(i) dN˜ s(i) .
(5)
t
i =1
Then by Theorem 2, we know that {yn· } is a nondecreasing sequence. To estimate the solutions of BSDE (5), we need the following equations: Yt1
=ξ+
T
(−B|
Ys1
| − B| | + k2 (s, Zs1
Us1
) + φs )ds −
t
Yt2 = ξ +
(B|Ys2 | + B|Zs2 | + k2 (s, Us2 ) + φs )ds − t
Yt3 =
−
(−B|Ys3 | − B|Zs3 | + k2 (s, Us3 ))ds −
T
Zs2 dWs −
t
T
m
m
Zs3 dWs − t
i =1 m i=1
T
T
Us1(i) dN˜ s(i) ,
(6)
t
i =1
t
T
Zs1 dWs
t T
T
T
Us2(i) dN˜ s(i) ,
(7)
t
Us3(i) dN˜ s(i) .
(8)
t
By Theorem 1, we can obtain the unique solution (Y·k , Z·k , U·k ) of the above BSDE, for every k = 1, 2, 3. If we notice that ξ ≥ 0, φt ≥ 0 and zero is the unique solution of (8), then by comparing (8), (7), (6) with (5), we have 0 = Yt3 ≤ Yt1 ≤ y1t ≤ y2t ≤ · · · ≤ ynt ≤ · · · ≤ Yt2 ,
P-a.s.
Using Lepeltier–San Martin’s method (see Lepeltier and San Martin (1997) [13]), the solution sequence {(yn· , z·n , un· )}n≥1 of the BSDE (5) converges to the minimal solution (Y· , Z· , U· ) of the BSDE (4). Then we know that Yt ≥ Yt3 = 0, for all t ∈ [0, T ]. Proof of Theorem 4. We first construct the following sequence of BSDEs Y 0t = ξ +
T
(−A(s)|Y 0s | − B|Z 0s | − A(s))ds − t
Y it = ξ +
Z 0s dWs − t
m
T
U 0s (k) dN˜ s(k) ;
(9)
t
k=1
T
t T
T
−
(f (s, Y is−1 , Z is−1 , U is ) + k1 (Y is − Y is−1 , Z is − Z is−1 ))ds m T Z is dWs − U si(k) dN˜ s(k) , i = 1, 2, . . . ;
t
k=1
(10)
t
and 0
Yt = ξ +
T
0
0
(A(s)|Y s | + B|Z s | + A(s))ds − t
T
0
Z s dWs − t
m k=1
T
0(k)
Us
dN˜ s(k) .
(11)
t 0
0
0
Using Theorem 1, BSDE (9) (or (11)) has a unique adapted solution, which is denoted by (Y 0· , Z 0· , U 0· ) (or (Y · , Z · , U · )). And for every i = 1, 2, . . . , Theorem 3 implies the existence of the minimal solution (Y i· , Z i· , U i· ) of BSDE (10).
Y. Qin, N.-M. Xia / J. Math. Anal. Appl. 406 (2013) 365–372
371
We can prove that the above solutions have the following properties: (kk1) For all t ∈ [0, T ] and any positive integer i, Y 0t ≤ Y it ≤ Y ti+1 , P-a.s.; 0
(kk2) For all t ∈ [0, T ] and any positive integer i, Y it ≤ Y t , P-a.s. For (kk1), we first prove Y 0t ≤ Y 1t . In fact, from (9), (10), we know Y 1t − Y 0t =
T
(k1 (Y 1s − Y 0s , Z 1s − Z 0s ) + k(20) (s, U 1s − U 0s ) + ∆1s )ds
t T
(Z 1s − Z 0s )dWs −
− t
m k=1
T
(U 1s (k) − U 0s (k) )dN˜ s(k) ,
(12)
t
(0)
where k2 (s, u) = f (s, Y 0s , Z 0s , u + U 0s ) − f (s, Y 0s , Z 0s , U 0s ) and ∆1s = f (s, Y 0s , Z 0s , U 0s ) + A(s)|Y 0s | + B|Z 0s | + A(s). From (0)
(0)
(0)
(HH4), k2 (s, u) has the property that if u ≥ v , then k2 (s, u) ≥ k2 (s, v). From (HH2), we have ∆1s ≥ 0 and ∆1s ≤ 2(A(s)|Y 0s | + B|Z 0s | + A(s)). Notice that T
E
T
(∆1s )2 ds ≤ 12 E
A2 (s)|Y 0s |2 ds + B2 E 0
0
T
|Z 0s |2 ds + E 0
≤ 12 E sup |Y 0t |2 0≤t ≤T
T
A2 (s)ds + B2 E
T
A2 (s)ds 0
T
0
|Z 0s |2 ds + E 0
T
A2 (s)ds
< +∞
0
so ∆1s ∈ M 2 . By Lemma 3, we have Y 0t ≤ Y 1t . Now we use the method of induction to show that Y it ≤ Y ti+1 under the assumption Y it−1 ≤ Y it . By (10), we have Y it+1 − Y it =
T
(k1 (Y is+1 − Y is , Z is+1 − Z is ) + k(2i) (s, U is+1 − U is ) + ∆si+1 )ds
t T
(Z is+1 − Z is )dWs −
− t
m k=1
T
(U is+1(k) − U is(k) )dN˜ s(k) ,
(13)
t
(i+1)
(i)
= f (s, Y is , Z is , U is )− f (s, Y is−1 , Z si−1 , U is )− k1 (Y is − Y is−1 , Z is − (i) (i) Z is−1 ). From (HH4), k2 (s, u) has the property that if u ≥ v , then k2 (s, u) ≥ k2 (s, v). From (HH5) and (HH2), we get that ∆is+1 ≥ 0 and ∆si+1 ∈ M2 . Then by Lemma 3, we have Y it ≤ Y ti+1 . The proof of (kk1) is completed. Now consider (kk2). For i = 1, 2, . . . , from (10) and (11), we have equations: T T m T 0 0 0 0 0(k) Y t − Y it = (−A(s)|Y s − Y is | − B|Z s − Z is | + Φsi )ds − (U s − U is(k) )dN˜ s(k) , (Z s − Z is )dWs − (14) where k2 (s, u) = f (s, Y is , Z is , u + U is )− f (s, Y is , Z is , U is ) and ∆s (i)
t
t 0
0
0
k=1
t
0
where Φsi = A(s)|Y s − Y is | + B|Z s − Z is | + A(s)|Y s | + B|Z s | + A(s) − f (s, Y is−1 , Z is−1 , U is ) − k1 (Y is − Y is−1 , Z is − Z si−1 ) ≥ f (s, Y is , Z is , U is ) − f (s, Y is−1 , Z is−1 , U is ) − k1 (Y is − Y is−1 , Z is − Z is−1 ). By (HH5) and (HH2), we get that Φsi ≥ 0 and Φsi ∈ M 2 . If we notice zero to be the unique solution of the following BSDE: Yt2 =
T
(−A(s)|Ys2 | − B|Zs2 |)ds − t
T
Zs2 dWs − t
m k=1
T
Us2(k) dN˜ s(k) .
(15)
t
0
Then by comparing (14) with (15), we have Y t ≥ Y it . This follows that the minimal solutions of Eqs. (9) and (10) form a bounded nondecreasing sequence, that is 0
Y 0t ≤ Y 1t ≤ Y 2t ≤ · · · ≤ Y it ≤ · · · ≤ Y t . P-a.s. Then we know that 0
sup E sup |Y nt |2 ≤ E sup |Y 0t |2 + E sup |Y t |2 < +∞ n
0≤t ≤T
0 ≤t ≤T
0≤t ≤T
Y nt n≥1
and that { } converges to a limit Y in S 2 . Now we consider {(Z nt , U nt )}n≥1 in H12 × Hm2 . From (10) and Ito’s formula, we have E |ξ |2 = E |Y 0i+1 |2 − 2E
T
Y is+1 (f (s, Y is , Z is , U is+1 ) + k1 (Y is+1 − Y is , Z si+1 − Z is ))ds + E 0
t
|Z is+1 |2 ds + E 0
t
∥U si+1 ∥2 ds. 0
372
Y. Qin, N.-M. Xia / J. Math. Anal. Appl. 406 (2013) 365–372
So, we can find that
t
E
t
|Z is+1 |2 ds + E
0
∥U is+1 ∥2 ds ≤ E |ξ |2 + 2E
Y is+1 (f (s, Y is , Z is , U si+1 ) + k1 (Y is+1 − Y is , Z si+1 − Z is ))ds 0
0 T
2
T
|Y is+1 |(A(s)|Y is | + B|Z is | + A(s) + B|Y is+1 − Y is | + B|Z is+1 − Z is |)ds
≤ E |ξ | + 2E 0 T
1
(|Z is+1 |2 + |Z is |2 )ds,
≤C+ E 8
0
where C = E |ξ |2 + (3
T 0
A(s)ds + 40B2 T + 4BT ) supi E sup0≤t ≤T |Y it |2 .
T |Z it |2 dt + E 0 ∥U it ∥2 dt ) < ∞. Then we know that for i = 1, 2, . . . , the functions Ψti+1 = T f (t , Y it , Z it , U it+1 ) + k1 (Y it+1 − Y it , Z it+1 − Z it ) are uniformly bounded in M 2 . Set Cˆ = supi 0 |Ψti+1 |2 dt. Applying Ito’s formula p q to |Y t − Y t |2 , we get, for any positive integers p, q ≥ 1 T T p q E |Y t − Y t |2 + E |Z ps − Z qs |2 ds + E ∥U ps − U qs ∥2 ds This implies that supi (E
T 0
t
t
T
≤ 2E
|
Y ps
−
Y qs
Ψsp
|(|
|+|
Ψsq
|)ds ≤ 4 Cˆ E
12
T
|
Y ps
−
Y qs 2 ds
|
.
t
t
It follows that {(Z nt , U nt )}n≥1 is a Cauchy sequence in H12 × Hm2 . Then there exists a process (Yt , Zt , Ut ) ∈ H2,m to be the limit of the sequence {(Y nt , Z nt , U nt )}n≥1 . By the way similar to that in Theorem 3, passing to the limit (as i → +∞) on the both sides of (10), we have the conclusion that (Yt , Zt , Ut ) solves (2). Remark 2. For example, consider the following BSDE: T
Yt =
s t
− 31
T
IYs ds −
Zs dWs − t
m i=1
T
Us(i) dN˜ s(i) . t ∈ [0, T ],
t
where Iy is an indicator function of the interval (0, +∞), that is Iy = 1 for y > 0, otherwise 0. Although all the assumptions 2
2
(HH1)–(HH4) are satisfied, but for any c ∈ [0, T ], (Yt , Zt , Ut ) = (max{ 23 (c 3 − t 3 ), 0}, 0, 0) is the solution. This means that the solution is not unique. Remark 3. In the same way as that in the Remark 6 of Jia (2008) [11], we can prove the existence of the adapted solution of BSDE (2), when the coefficient f is right continuous, decreasing and bounded. References [1] K. Bahlali, Existence and uniqueness of solutions for BSDEs with locally Lipschitz coefficient, Electron. Commun. Probab. 7 (2002) 169–179. [2] K. Bahlali, M. Eddahbi, E. Essaky, BSDE associated with Lévy processes and application to PDIE, J. Appl. Math. Stoch. Anal. 16 (1) (2003) 1–17. [3] K. Bahlali, E.H. Essaky, M. Hassani, E. Pardoux, Existence, uniqueness and stability of backward stochastic differential equations with locally monotone coefficient, C. R. Math. Acad. Sci. Paris 335 (9) (2002) 757–762. [4] G. Barles, R. Buckdahn, E. Pardoux, Backward stochastic differential equations and integral–partial differential equations, Stoch. Stoch. Rep. 60 (1997) 57–83. [5] Ph. Briand, R. Carmona, BSDEs with polynomial growth generator, J. Appl. Math. Stoch. Anal. 13 (3) (2000) 207–238. [6] N. El Karoui, S. Hamadène, BSDEs and risk control, zero-sum and nonzero-sum game problems of stochastic functional differential equations, Stoch. Process. Appl. 107 (2003) 145–169. [7] N. El Karoui, S. Peng, M.C. Quenez, Backward stochastic differential equations in finance, Math. Finance 7 (1997) 1–71. [8] M. El Otmani, BSDE driven by simple Lévy process with continuous coefficient, Statist. Probab. Lett. 78 (2) (2008) 1259–1265. [9] S. Hamadène, J.P. Lepeltier, Zero-sum stochastic differential games and BSDEs, Systems Control Lett. 24 (1995) 259–263. [10] S. He, J. Wang, J. Yan, Semimartingale Theory and Stochastic Calculus, Science Press and CRC Press, Inc., 1992. [11] G. Jia, A class of backward stochastic differential equations with discontinuous coefficients, Statist. Probab. Lett. 78 (2008) 231–237. [12] M. Kobylanski, Backward stochastic differential equations and partial differential equations with quadratic growth, Ann. Probab. 28 (2000) 259–276. [13] J.P. Lepeltier, J. San Martin, Backward stochastic differential equations with continuous coefficient, Statist. Probab. Lett. 32 (4) (1997) 425–430. [14] Y. Ouknine, Reflected backward stochastic differential equations with jumps, Stoch. Stoch. Rep. 65 (1998) 111–125. [15] Y. Ouknine, Ndiaye, Djibril, Sur l’existence de solutions d’quations diffrentielles stochastiques progrssives rtrogrades couples (Existence of solutions of coupled forward–backward stochastic differential equations), Stochastics 80 (4) (2008) 299–315 (in French). [16] E. Pardoux, S. Peng, Adapted solution of a backward stochastic differential equation, Systems Control Lett. 14 (1990) 55–61. [17] R. Situ, On solutions of backward stochastic differential equations with jumps and applications, Stoch. Process. Appl. 66 (1997) 209–236. [18] S. Tang, X. Li, Necessary conditions for optimal control of stochastic systems with random jumps, SIAM J. Control Optim. 32 (5) (1994) 1447–1475.