arXiv:0705.1822v2 [math.PR] 26 Jun 2009
The Annals of Probability 2009, Vol. 37, No. 3, 877–902 DOI: 10.1214/08-AOP423 c Institute of Mathematical Statistics, 2009
ANTICIPATED BACKWARD STOCHASTIC DIFFERENTIAL EQUATIONS1 By Shige Peng and Zhe Yang Shandong University, and Shandong University and Cambridge University In this paper we discuss new types of differential equations which we call anticipated backward stochastic differential equations (anticipated BSDEs). In these equations the generator includes not only the values of solutions of the present but also the future. We show that these anticipated BSDEs have unique solutions, a comparison theorem for their solutions, and a duality between them and stochastic differential delay equations.
1. Introduction. Consider these types of stochastic differential delay equations (SDDEs): (1)
T ) dWt , ¯t−θ ¯t−θ Xt−θ ) dt + (Xt σtT + Xt−θ σ dXt = (µt Xt + µ
Xt = xt ,
t ∈ [t0 , T + θ]; t ∈ [t0 − θ, t0 ],
where W is a d-dimensional Brownian motion, θ > 0, xt is a deterministic function, and Q is a given FTW -measurable random variable. In the case where µ ¯=σ ¯ ≡ 0, this model is very typical in finance as the price of a stock. Then Yt0 = E[XT Q|Ft0 ] can be the price of an option valued Q at maturity time T if xt ≡ 1. It is easy to prove that (see, e.g., El Karoui, Peng and Quenez [7]) Y. is a solution to the following backward stochastic differential equation (BSDE): −dYt = (µt Yt + Zt σt ) dt − Zt dWt ,
YT = Q.
This SDE with delay, in which µ ¯ and σ ¯ are nonzero, has a solution. An interesting question is whether it can be expressed in the form of equation (1). Received April 2007; revised March 2008. Supported in part by the National Basic Research Program of China (973 Program) Grant 2007CB814900 (Financial Risk). AMS 2000 subject classifications. 60H10, 60H20, 93E03. Key words and phrases. Anticipated backward stochastic differential equation, backward stochastic differential equation, adapted process. 1
This is an electronic reprint of the original article published by the Institute of Mathematical Statistics in The Annals of Probability, 2009, Vol. 37, No. 3, 877–902. This reprint differs from the original in pagination and typographic detail. 1
2
S. PENG AND Z. YANG
The answer is positive if we can solve the following new type of “anticipated” BSDE: (2)
−dYt = (µt Yt + µ ¯t E Ft [Yt+θ ] + Zt σt Ft
[Zt+θ ]¯ σt + lt ) dt − Zt dWt ,
+E
Y t = Qt ,
Zt = Pt ,
t ∈ [t0 , T ]; t ∈ [T, T + θ]; t ∈ [T, T + θ].
We observe that the generator, that is, the dt part of the BSDE, contains the values (Y. , Z. ) for present time t as well as for future time t + θ. This is a new duality phenomenon for SDEs and BSDEs. In this paper we consider a more general form of this new type of BSDE: −dYt = f (t, Yt , Zt , Yt+δ(t) , Zt+ζ(t) ) dt − Zt dWt ,
Y = ξt , t Zt = ηt ,
t ∈ [0, T ]; t ∈ [T, T + K]; t ∈ [T, T + K].
The paper is organized as follows. In Section 2 we consider the duality between SDDEs and anticipated BSDEs. After a brief presentation of some known results that we will use in Section 3, we prove an existence and uniqueness result for anticipated BSDEs in Section 4. In Section 5 we give an important result for anticipated BSDEs: a comparison theorem. In Section 6 we use the duality between SDDEs and anticipated BSDEs mentioned in Section 2 to solve a stochastic control problem. 2. Duality between SDDEs and anticipated BSDEs. It is well known that there is perfect duality between SDEs and BSDEs (see El Karoui, Peng and Quenez [7]). In this section we consider duality between the SDDEs and the anticipated BSDEs mentioned above. We will use this duality to solve a stochastic control problem in Section 6. Theorem 2.1. Suppose θ > 0 is a given constant and µ. , µ ¯. ∈ L2F (t0 − θ, T + θ), l. ∈ L2F (t0 , T ), σ. , σ ¯. ∈ L2F (t0 − θ, T + θ; Rd×1 ), µ. , µ ¯ . , σ. , σ ¯. are uni2 (T, T + θ), P ∈ L2 (T, T + θ; Rd ), the formly bounded. Then for all Q. ∈ SF . F solution Y. of the anticipated BSDE (2) can be given by the closed formula Yt = E
Ft
XT QT +
Z
t
T
Xs ls ds +
Z
T +θ
T
(Qs µ ¯s−θ + Ps σ ¯s−θ )Xs−θ ds , a.e., a.s.,
where Xs is the solution to the SDDE (3)
T dXs = (µs Xs + µ ¯s−θ Xs−θ ) ds + (Xs σsT + Xs−θ σ ¯s−θ ) dWs ,
s ∈ [t, T + θ];
Xt = 1,
Xs = 0,
s ∈ [t − θ, t].
3
ANTICIPATED BSDES
Proof. First, we show that (3) has a unique solution. When s ∈ [t, t + θ], (3) becomes dXs = µs Xs ds + Xs σsT dWs , s ∈ [t, t + θ]; Xt = 1. We can then easily obtain a unique continuous solution ς. for (4). When s ∈ [t + θ, T + θ], (3) becomes
(4)
T ¯s−θ Xs−θ ) ds + (Xs σsT + Xs−θ σ ¯s−θ ) dWs , dXs = (µs Xs + µ
(5)
s ∈ [t + θ, T + θ], s ∈ [t, t + θ].
Xs = ςs ,
Equation (5) is a classical SDDE, thus, it has a unique solution. Applying Itˆo’s formula to Xs Ys for s ∈ [t, T ] and taking conditional expectations under Ft , we get E Ft [XT YT ] − Xt Yt =E
Ft
T
Z
t
(Ys µ ¯s−θ Xs−θ − E Fs [Ys+θ ]¯ µs Xs + Zs σ ¯s−θ Xs−θ − E
Fs
[Zs+θ ]¯ σs Xs − Xs ls ) ds .
Because Xt = 1 and Xs = 0, s ∈ [t − θ, t), we have Yt = E
Ft
−E
XT YT +
Ft
Z
t
=E
Ft
Xs ls ds − E
t
Ft
Z
t
T
(Ys µ ¯s−θ Xs−θ − Ys+θ µ ¯s Xs ) ds
(Zs σ ¯s−θ Xs−θ − Zs+θ σ ¯s Xs ) ds
= E Ft XT YT + − E Ft
T
t
Z
T
Z
T
Z
Xs ls ds −
t
Zs σ ¯s−θ Xs−θ ds −
XT QT +
Z
t
T
Xs ls ds +
Ys µ ¯s−θ Xs−θ ds +
t
T
T
Z
t+θ
Z
Z
T +θ
Z
T +θ
T
Zs σ ¯s−θ Xs−θ ds
T +θ
t+θ
Ys µ ¯s−θ Xs−θ ds
(Qs µ ¯s−θ Xs−θ + Ps σ ¯s−θ Xs−θ ) ds .
3. Preliminaries. Let (Ω, F , P, Ft , t ≥ 0) be a complete stochastic basis such that F0 contains all P -null elements of F and suppose that the filtration is generated by a d-dimensional standard Brownian motion W = (Wt )t≥0 . Given T > 0, denote the norm in Rm by | · |. We will use the following notation: • L2 (FT ; Rm ) = {Rm -valued FT -measurable random variables such that E[|ξ|2 ] < ∞};
4
S. PENG AND Z. YANG
• L2F (0, T ; Rm ) = {Rm -valued and Ft -adapted stochastic processes such that R E[ 0T |ϕt |2 dt] < ∞}; 2 (0, T ; Rm ) = {continuous processes in L2 (0, T ; Rm ) such that • SF F E[sup0≤t≤T |ϕt |2 ] < ∞}. 2 (0, T ). The above If m = 1, we denote them by L2 (FT ), L2F (0, T ) and SF are all separable Hilbert spaces. The following lemmas can be found in Peng [13], Section 3. For their originalities we refer to the notes of [13] or [7]. Our Lemma 3.1 is Lemma 3.1 of Peng [13]. Lemma 3.2, which is Theorem 3.2 of Peng [13], is a basic result of BSDEs: an existence and uniqueness theorem. Both Lemmas 3.3 and 3.4 are comparison theorems for solutions of BSDEs. Lemma 3.3 is Theorem 3.3 of Peng [13] and can also be found in El Karoui, Peng and Quenez [7]. Lemma 3.4 can be easily obtained from Lemma 3.3.
L2
Lemma 3.1. For a fixed ξ ∈ L2 (FT ) and g0 (·) which is an Ft -adapted R process satisfying E[( 0T |g0 (t)| dt)2 ] < +∞, there exists a unique pair of processes (y. , z. ) ∈ L2F (0, T ; R1+d ) satisfying the following BSDE: yt = ξ +
T
Z
g0 (s) ds −
t
Z
T
t
zs dWs ,
t ∈ [0, T ].
2 (0, T ) × L2 (0, T ; Rd ). We have the folIf g0 (·) ∈ L2F (0, T ), then (y. , z. ) ∈ SF F lowing basic estimate:
|yt |2 + E Ft (6) ≤E
Ft
Z
t
T β
2
|ys |2 + |zs |2 e β(s−t) ds
2 β(T −t)
[|ξ| e
2 ] + E Ft β
Z
T
2 β(s−t)
|g0 (s)| e
t
ds .
In particular, |y0 |2 + E (7)
Z
0
T β
2
2 βT
≤ E[|ξ| e
|ys |2 + |zs |2 e βs ds
2 ]+ E β
Z
T
2 βs
|g0 (s)| e
0
ds ,
where β > 0 is an arbitrary constant. We also have (8)
E
sup |yt |2 ≤ kE |ξ|2 +
0≤t≤T
Z
0
T
|g0 (s)|2 ds ,
where the constant k depends only on T . We assume that g = g(ω, t, y, z) : Ω × [0, T ] × Rm × Rm×d −→ Rm satisfies the following conditions:
5
ANTICIPATED BSDES
(a) g(·, y, z) is an Rm -valued and Ft -adapted process satisfying the Lipschitz condition in (y, z), that is, there exists ρ > 0 such that, for each y, y ′ ∈ Rm and z, z ′ ∈ Rm×d , |g(t, y, z) − g(t, y ′ , z ′ )| ≤ ρ(|y − y ′ | + |z − z ′ |). (b) g(·, 0, 0) ∈ L2F (0, T ; Rm ). Lemma 3.2. Assume that g satisfies (a) and (b), then for any given terminal condition ξ ∈ L2 (FT ; Rm ), BSDE Yt = ξ +
(9)
Z
T
g(s, Ys , Zs ) ds −
t
Z
T
Zs dWs ,
t
0 ≤ t ≤ T,
has a unique solution, that is, there exists a unique pair of Ft -adapted pro2 (0, T ; Rm ) × L2 (0, T ; Rm×d ) satisfying equation (9). cesses (Y. , Z. ) ∈ SF F Lemma 3.3. Assume gj (ω, t, y, z) : Ω × [0, T ] × R × Rd −→ R satisfies (a) and (b), j = 1, 2. Let (Y.(1) , Z.(1) ) and (Y.(2) , Z.(2) ) be respectively the solutions of BSDEs as follows: (j) Yt
=ξ
(j)
+
Z
T t
gj (s, Ys(j) , Zs(j) ) ds − (1)
Z
T
t
Zs(j) dWs ,
0 ≤ t ≤ T, (1)
(1)
(1)
where j = 1, 2. If ξ (1) ≥ ξ (2) and g1 (t, Yt , Zt ) ≥ g2 (t, Yt , Zt ), a.e., a.s., then (1)
Yt
(2)
≥ Yt ,
a.e., a.s.
We also have strict comparison: under the above conditions, (1)
Y0
(2)
⇐⇒
= Y0 (1)
ξ (1) = ξ (2) , (1)
(1)
a.s.,
(1)
g1 (t, Yt , Zt ) = g2 (t, Yt , Zt ),
a.e., a.s.
Lemma 3.4. We make the same assumption as in Lemma 3.3. If ξ (1) ≥ g1 (t, y, z) ≥ g2 (t, y, z), t ∈ [0, T ], y ∈ R, z ∈ Rd , then
ξ (2) ,
(1)
Yt
(2)
≥ Yt ,
a.e., a.s.
4. Existence and uniqueness theorem. We consider a new form of BSDEs as follows: (10)
−dYt = f (t, Yt , Zt , Yt+δ(t) , Zt+ζ(t) ) dt − Zt dWt ,
Yt = ξt , Zt = ηt ,
t ∈ [0, T ]; t ∈ [T, T + K]; t ∈ [T, T + K],
where δ(·) and ζ(·) are two R+ -valued continuous functions defined on [0, T ] such that:
6
S. PENG AND Z. YANG
(i) There exists a constant K ≥ 0 such that, for all s ∈ [0, T ], s + δ(s) ≤ T + K;
s + ζ(s) ≤ T + K.
(ii) There exists a constant L ≥ 0 such that, for all t ∈ [0, T ] and for all nonnegative and integrable g(·), Z
T
g(s + δ(s)) ds ≤ L
T +K
g(s) ds;
t
t
Z
Z
T
g(s + ζ(s)) ds ≤ L
t
Z
T +K
g(s) ds.
t
We call equation (10) the anticipated BSDE. The setting of our problem is as follows: to find a pair of Ft -adapted 2 (0, T + K; Rm ) × L2 (0, T + K; Rm×d ) satisfying anprocesses (Y., Z.) ∈ SF F ticipated BSDE (10). Assume that for all s ∈ [0, T ], f (s, ω, y, z, ξ, η) : Ω × Rm × Rm×d × L2 (Fr ; Rm )× L2 (Fr′ ; Rm×d ) −→ L2 (Fs , Rm ), where r, r ′ ∈ [s, T + K], and f satisfies the following conditions: (H1) There exists a constant C > 0, such that for all s ∈ [0, T ], y, y ′ ∈ z, z ′ ∈ Rm×d , ξ. , ξ.′ ∈ L2F (s, T + K; Rm ), η. , η.′ ∈ L2F (s, T + K; Rm×d ), r, r¯ ∈ [s, T + K], we have
Rm ,
|f (s, y, z, ξr , ηr¯) − f (s, y ′ , z ′ , ξr′ , ηr′¯)| ≤ C(|y − y ′ | + |z − z ′ | + E Fs [|ξr − ξr′ | + |ηr¯ − ηr′¯|]). RT
(H2) E[
0
|f (s, 0, 0, 0, 0)|2 ds] < ∞.
Remark 4.1. 1. Note that f (s, ·, ·, ·, ·) is Fs -measurable ensures the solution to the anticipated BSDE is Fs -adapted. 2. We give examples of δ(s) and f . Both examples of δ(s) satisfy (i) and (ii). Example 1: Let δ(s) ≡ c, where c > 0 is a constant. Example 2: Let s + δ(s) be a monotone nonnegative function whose converse function has a continuous differential function. We give examples of functions that satisfy (H1) and (H2): Let g satisfy (a) and (b) and let δ, ζ be two positive constants. For each t ∈ [0, T ] and (ξ. , η. ) ∈ L2F (t, T + (δ ∨ ζ); Rm × Rm×d ), define f1 , f2 such that f1 (t, ξt+δ , ηt+ζ ) = g(t, E Ft [ξt+δ ], E Ft [ηt+ζ ]), f2 (t, ξt+δ , ηt+ζ ) = g(t, Et,t+δ [ξt+δ ], Et,t+ζ [ηt+ζ ]), where Es,t [·] : L2 (Ft ) −→ L2 (Fs ), 0 ≤ s ≤ t ≤ T + K, is a Ft -consistent nonlinear evaluation (see Peng [13]). Then f1 , f2 satisfy (i) and (ii).
7
ANTICIPATED BSDES
The following is the main result of this section: an existence and uniqueness theorem for adapted solutions for anticipated BSDEs. Theorem 4.2. Suppose that f satisfies (H1) and (H2), and δ, ζ satisfy 2 (T, T + K; Rm ) (i) and (ii). Then for any given terminal conditions ξ. ∈ SF 2 m×d and η. ∈ LF (T, T + K; R ), the anticipated BSDE (10) has a unique solution, that is, there exists a unique pair of Ft -adapted processes (Y., Z.) ∈ 2 (0, T + K; Rm ) × L2 (0, T + K; Rm×d ) satisfying (10). SF F Proof. We fix β = 12C 2 (2L + 1) + 2, where C is the Lipschitz constant of f given in (H1), and introduce a norm in the Banach space L2F (0, T + K; Rm ): Z
kν(·)kβ = E
T +K
0
2 βs
|νs | e
ds
1/2
.
Clearly, it is equivalent to the original norm of L2F (0, T + K; Rm ). But it is more convenient to use this norm to construct a contraction mapping that allows us to apply the Fixed Point Theorem. Set Z T Z T Zs dWs , f (s, y , z , y , z ) ds − Y = ξ + s s s+δ(s) s+ζ(s) T t t t
t ∈ [0, T ]; t ∈ [T, T + K]; t ∈ [T, T + K].
Y = ξt , t
Zt = ηt ,
Define a mapping h : L2F (0, T +K; Rm ×Rm×d ) −→ L2F (0, T +K; Rm ×Rm×d) such that h[(y. , z. )] = (Y. , Z. ). Now we prove that h is a contraction mapping under the norm k · kβ . For two arbitrary elements (y. , z. ) and (y.′ , z.′ ) in L2F (0, T + K; Rm × Rm×d), set (Y. , Z. ) = h[(y. , z. )] and (Y.′ ., Z.′ ) = h[(y.′ , z.′ )]. Denote their differences by (Yˆ. , Zˆ. ) = ((Y − Y ′ ). , (Z − Z ′ ). ).
(ˆ y. , zˆ. ) = ((y − y ′ ). , (z − z ′ ). ), By basic estimate (7), we have E
Z
0
T β
2
|Yˆs |2 + |Zˆs |2 e βs ds
2 ≤ E β
Z
T 0
|f (s, ys , zs , ys+δ(s) , zs+ζ(s))
′ ′ − f (s, ys′ , zs′ , ys+δ(s) , zs+ζ(s) )|2 e βs ds .
8
S. PENG AND Z. YANG
Since δ(s) and ζ(s) satisfy (ii) and f satisfies (H1), by the Fubini Theorem, we have Z T β ˆ 2 2 βs ˆ E |Ys | + |Zs | e ds 2 0 2C 2 E β
Z
T
≤
6C 2 E β
T
≤
Z
≤
0
0
(|ˆ ys | + |ˆ zs | + E Fs [|ˆ ys+δ(s) | + |ˆ zs+ζ(s) |])2 e βs ds (|ˆ ys |2 + |ˆ zs |2 + 2|ˆ ys+δ(s) |2 + 2|ˆ zs+ζ(s) |2 )e βs ds
6C 2 (2L + 1) β
E
Z
T +K
2
2
(|ˆ ys | + |ˆ zs | )e
0
βs
ds .
Because β = 12C 2 (2L + 1) + 2, then E
Z
T +K
0
(|Yˆs |2 + |Zˆs |2 )e βs ds
1 ≤ E 2 or
Z
0
T +K
2
2
(|ˆ ys | + |ˆ zs | )e
k(Yˆ. , Zˆ. )kβ ≤
βs
ds ,
√1 k(ˆ y. , zˆ. )kβ . 2
Consequently, h is a strict contraction mapping of L2F (0, T + K; Rm × Rm×d ). It follows by the Fixed Point Theorem that (10) has a unique solution (Y. , Z. ) ∈ L2F (0, T + K; Rm × Rm×d ). Since f satisfies (H1) and (H2) and since δ, ζ satisfy (i) and (ii), we have f (·, Y. , Z. , Y.+δ(.) , Z.+ζ(.) ) ∈ L2F (0, T ; Rm ). 2 (0, T + K; Rm ). Thus, by Lemma 3.1, we obtain Y. ∈ SF The following example shows that a simple case of the anticipated BSDE (10) has a solution. Consider the following anticipated BSDE:
Example 4.3.
Yt = T WT − Yt = tWt ,
Z
t
T
1 E Fs [Ys+δ ] ds − s+δ
Z
T t
Zs dWs ,
t ∈ [0, T ]; t ∈ [T, T + δ],
where δ ≥ 0 is a given constant. Then (tWt , t)t∈[0,T +δ] is its solution. The following proposition is an estimate of the solution of the anticipated BSDE (10). Proposition 4.4. Assume that f satisfies (H1) and (H2), and also δ and ζ satisfy (i) and (ii). Then there exists a positive constant C0 that only
9
ANTICIPATED BSDES
2 (T, T + depends on C in (H1), L in (ii), and T such that for each ξ. ∈ SF m 2 m×d K; R ) and each η. ∈ LF (T, T + K; R ), the solution (Y ., Z.) of the anticipated BSDE (10) satisfies
E Ft (11)
sup |Ys |2 +
t≤s≤T
T
Z
|Zs |2 ds
t
Z
≤ C0 E Ft |ξT |2 +
T +K
T
(|ξs |2 + |ηs |2 ) ds +
Z
T
|f (s, 0, 0, 0, 0)| ds
t
2
for each t ∈ [0, T ]. Proof. For s ∈ [0, T ], applying Itˆo’s formula to eβs |Ys |2 , we obtain eβs |Ys |2 +
T
Z
s
eβr (β|Yr |2 + |Zr |2 ) dr
= eβT |ξT |2 − 2 +2
Z
T
s
Z
s
T
eβr (Yr , Zr dWr )
eβr (f (r, Yr , Zr , Yr+δ(r) , Zr+ζ(r) ), Yr ) dr.
Since 2(f (r, Yr , Zr , Yr+δ(r) , Zr+ζ(r) ), Yr ) = 2(f (r, Yr , Zr , Yr+δ(r) , Zr+ζ(r) ) − f (r, Yr , Zr , Yr+δ(r) , 0), Yr ) + 2(f (r, Yr , Zr , Yr+δ(r) , 0) − f (r, Yr , Zr , 0, 0), Yr ) + 2(f (r, Yr , Zr , 0, 0) − f (r, Yr , 0, 0, 0), Yr ) + 2(f (r, Yr , 0, 0, 0) − f (r, 0, 0, 0, 0), Yr ) + 2(f (r, 0, 0, 0, 0), Yr ) ≤ 2CE Fr [|Zr+ζ(r) |]|Yr | + 2CE Fr [|Yr+δ(r) |]|Yr | + 2C|Yr ||Zr | + 2C|Yr |2 + 2(f (r, 0, 0, 0, 0), Yr ) ≤ (3LC 2 + 4LT C 2 + 3C 2 + 2C)|Yr |2 +
1 Fr E [|Zr+ζ(r) |2 ] 3L
1 1 E Fr [|Yr+δ(r) |2 ] + |Zr |2 + 2(f (r, 0, 0, 0, 0), Yr ), 4LT 3
+
we get, for s ∈ [0, T ], βs
2
e |Ys | + (12)
Z
s
T
βr
e
"
#
2 (β − 3LC − 4LT C − 3C − 2C)|Yr | + |Zr |2 dr 3
≤ eβT |ξT |2 + 2
2
Z
s
T
2
2
eβr (f (r, 0, 0, 0, 0), Yr ) dr − 2
2
Z
s
T
eβr (Yr , Zr dWr )
,
10
S. PENG AND Z. YANG
T T 1 1 eβr E Fr [|Zr+ζ(r) |2 ] dr + eβr E Fr [|Yr+δ(r) |2 ] dr. 3L s 4LT s Taking conditional expectations under Fs on both sides of (12), we have
Z
+
βs
2
e |Ys | + E
Z
Fs
Z
T
βr
e s
(β − 3LC 2 − 4LT C 2 − 3C 2 − 2C)|Yr |2 2 + |Zr |2 dr 3
≤ E Fs eβT |ξT |2 + 2 +
1 E Fs 4LT
+
1 Fs E 3L
≤E
Fs
βT
e
T
Z s
T s
eβr (f (r, 0, 0, 0, 0), Yr ) dr
eβr E Fr [|Yr+δ(r) |2 ] dr
s
T
Z
Z
eβr E Fr [|Zr+ζ(r) |2 ] dr
2
|ξT | + 2
Z
T s
T +K
Z
βr
E
Z
s
≤E
(13)
βr
2
e |Zr | dr Fs
+E
βT
3e
Fs
Since for t ≤ s ≤ T , s
T
T +K
Z
s
βr
2
e |Yr | dr
Z
1 , then 4T
2
|ξT | + 6 T +K
βr
e
T
Z
e |Zr | dr .
s
T
2
Set β = 3LC 2 + 4LT C 2 + 3C 2 + 2C + Fs
eβr (f (r, 0, 0, 0, 0), Yr ) dr 1 + 4T
1 + E Fs 3
Z
T s
βr
e (f (r, 0, 0, 0, 0), Yr ) dr
3 |ξr |2 dr . |ηr | + 4T
2
Z T Z s βr βr e (Yr , Zr dWr ) e (Yr , Zr dWr ) − e (Yr , Zr dWr ) = t t Z s Z T βr βr e (Yr , Zr dWr ) + e (Yr , Zr dWr ) , ≤ βr
t
t
by the Burkholder–Davis–Gundy inequality, we have E
Ft
Z sup
t≤s≤T
s
T
e (Yr , Zr dWr ) βr
11
ANTICIPATED BSDES
≤ 2E Ft (14)
≤ 6E
Z sup
Ft
t
t≤s≤T T
Z
s βr e (Yr , Zr dWr )
2βr
e
t
≤ 6E Ft
2
2
|Yr | |Zr | dr
1 ≤ E Ft 4
T
Z
sup e1/2βr |Yr |
t≤r≤T
1/2
t
eβr |Zr |2 dr
sup eβr |Yr |2 + 36E Ft
t≤r≤T
Z
t
T
1/2
eβr |Zr |2 dr .
From estimates (12) and (14) we have E Ft
sup eβs |Ys |2
t≤s≤T
≤ E Ft eβT |ξT |2 + 2
Z
T
t
eβr |f (r, 0, 0, 0, 0)||Yr | dr
t≤s≤T
+ E Ft
1 3L
T
Z
t
1 + 4LT ≤E
Ft
βT
[e
T
t
+ E Ft
+ 2E Ft ≤E
Ft
βT
[e
+ E Ft
1 3L Z
Z
T
βr
e E
t
t
T +K
t T
t
2
[|Yr+δ(r) | ] dr βr
2
sup e |Yr |
t≤r≤T
1 4LT
eβr |f (r, 0, 0, 0, 0)||Yr | dr
1 |ξT | ] + E Ft 2
Z
s
eβr (Yr , Zr dWr )
eβr |Zr+ζ(r) |2 dr +
2
1 3
Fr
eβr |Zr |2 dr T
t
T
Z
+ 2E Ft
Z
1 |ξT | ] + E Ft 2 Z
T
eβr E Fr [|Zr+ζ(r) |2 ] dr
2
+ 72E Ft
Z
+ 2 sup
βr
sup e |Yr |
eβr |Zr |2 dr +
1 4T
Z
t
eβr |Yr+δ(r) |2 dr
+ 72E
T +K
t
eβr |f (r, 0, 0, 0, 0)||Yr | dr
T
2
t≤r≤T
Z
Ft
Z
T
2
e |Zr | dr
t
eβr |Yr |2 dr
βr
12
S. PENG AND Z. YANG
≤E
Ft
βT
[e
3 |ξT | ] + E Ft 4 2
1 + 72 + E Ft 3
+E
Ft
Z
T +K
e
+ 2E
Z
T
βr
2
sup e |Yr |
t≤r≤T
βr
2
e |Zr | dr
t
T
Ft
T
Z
βr
1 1 |ηr |2 + |ξr |2 dr 3 4T
βr
e |f (r, 0, 0, 0, 0)||Yr | dr .
t
Denote by C0 > 0 a constant that depends only on T, L and C, which we allow to change from line to line. From the estimate above and estimate (13), 1 Ft E 4
sup eβs |Ys |2
t≤s≤T
≤ C0 E Ft eβT |ξT |2 + + C0 E Ft ≤ C0 E
Ft
≤ C0 E
e
T
eβr (|ηr |2 + |ξr |2 ) dr Z
sup e1/2βr |Yr |
2
|ξT | +
T +K
2
sup e |Yr |
T +K
Z
T
t
βr
2
2
+ 2C02 E Ft 2
2
Z
βr
2
sup e |Yr |
t≤r≤T
+ 2C02 E Ft
T
1/2βr
e
|f (r, 0, 0, 0, 0)| dr
t
(|ηr | + |ξr | ) dr
T
e1/2βr |f (r, 0, 0, 0, 0)| dr
e (|ηr | + |ξr | ) dr
T
t≤r≤T 2
Z
βr
|ξT | +
1 + E Ft 8
T +K
t≤r≤T
βT
1 + E Ft 8 Ft
Z
2
Z
T
|f (r, 0, 0, 0, 0)| dr
t
2
.
Then E Ft
sup |Ys |2 + E Ft
t≤s≤T
≤ C0 E
Ft
2
|ξT | +
Z
Z
T t
T +K
T
|Zs |2 ds 2
2
(|ξs | + |ηs | ) ds +
Z
t
T
|f (s, 0, 0, 0, 0)| ds
2
.
The following proposition shows the importance of the effect of anticipated time on the solution to anticipated BSDEs.
13
ANTICIPATED BSDES
Proposition 4.5. Let (Y.(1) , Z.(1) ) and (Y.(2) , Z.(2) ) be respectively solutions of the following two anticipated BSDEs: −dY (j) = f (t, Y (j) , Z (j) , Y (j) (j) Yt
(j) t+δj (t) ) dt − Zt dWt ,
t
t
t
t ∈ [0, T ];
= ξt ,
t ∈ [T, T + K],
2 (T, T + K; Rm ), δ and δ satisfy (i) and (ii), where j = 1, 2. Assume ξ. ∈ SF 1 2 f satisfies (H2), and there exists a constant C¯ > 0, such that for all s ∈ [0, T ], y, y ′ ∈ Rm , z, z ′ ∈ Rm×d , θ, θ ′ ∈ L2F (s, T + K; Rm ) and r ∈ [s, T + K],
¯ − y ′ | + |z − z ′ | + |E Fs [θr − θ ′ ]|). |f (s, y, z, θr ) − f (s, y ′ , z ′ , θr′ )| ≤ C(|y r ˜ > 0 only If for any t ∈ [0, T ], δ1 (t) ≤ δ2 (t), then there exists a constant M ¯ L and T such that depending on C, (1) |Yt
(2) − Yt | 2
˜ ≤M
Z
T
t
(δ2 (s) − δ1 (s)) ds ×E
Ft
2
|ξT | +
Z
T +K
2
|ξs | ds +
T
Z
T
2
|f (s, 0, 0, 0)| ds .
t
Proof. Setting y. = Y.(1) − Y.(2) , z. = Z.(1) − Z.(2) , then by estimate (6), we obtain, for all t ∈ [0, T ], 2
|yt | + E
Ft
Z
2 ≤ E Ft β
t
T β
Z
2
T t
2
2
|ys | + |zs |
e
β(s−t)
ds
(1)
(2)
|f (s, Ys(1) , Zs(1) , Ys+δ1 (s) ) − f (s, Ys(2) , Zs(2) , Ys+δ2 (s) )|2 ×e
β(s−t)
ds
2C¯ 2 Ft T (1) (2) (|ys | + |zs | + |E Fs [Ys+δ1 (s) − Ys+δ2 (s) ]|)2 e β(s−t) ds E ≤ β t Z T 2 ¯ 6C Ft ≤ E (|ys |2 + |zs |2 β t Z
+ |E
Fs
[ys+δ1 (s) ] + E
Fs
(2) [Ys+δ1 (s)
(2) − Ys+δ2 (s) ]|2 )e β(s−t) ds
Z Z 6C¯ 2 + 12C¯ 2 L Ft T 6C¯ 2 Ft T 2 β(s−t) 2 β(s−t) ≤ E E |ys | e ds + |zs | e ds β β t t Z 12C¯ 2 Ft T Fs (2) (2) |E [Ys+δ1 (s) − Ys+δ2 (s) ]|2 e β(s−t) ds . E + β t
14
S. PENG AND Z. YANG
But (2)
(2)
E Fs [Ys+δ1 (s) − Ys+δ2 (s) ] = E Fs
s+δ2 (s)
Z
s+δ1 (s)
(2)
f (r, Yr(2) , Zr(2) , Yr+δ2 (r) ) dr ,
and set β = 6C¯ 2 , hence, 2
|yt | ≤ (1 + 2L)E + 2E Ft
Ft T
t
+ 2E
Ft
2 β(s−t)
|ys | e
t
Z
≤ (1 + 2L)e
T
Z
Z
E Fs
β(T −t)
Z
E
Ft
s+δ2 (s)
s+δ1 (s)
Z
T
(2)
|ys | ds
t
(δ2 (s) − δ1 (s))
2
f (r, Yr(2) , Zr(2) , Yr+δ2 (r) ) dr e β(s−t) ds 2
T
t
ds
Z
s+δ2 (s) s+δ1 (s)
(2)
|f (r, Yr(2) , Zr(2) , Yr+δ2 (r) )|2 dr ×e
≤ (1 + 2L)e β(T −t) E Ft + 8E Ft
Z
T
t
Z
T
t
|ys |2 ds
β(s−t)
ds
Z
t
T
(2)
(C¯ 2 |Yr(2) |2 + C¯ 2 |Zr(2) |2 + C¯ 2 |Yr+δ2 (r) |2 2
+ |f (r, 0, 0, 0)| ) dr
+8
Z
T t
× E Ft
(δ2 (s) − δ1 (s))e β(s−t) ds ×
≤ (1 + 2L)e
β(T −t)
E
Ft
Z
T
t
2
|ys | ds
(δ2 (s) − δ1 (s))e β(s−t) ds
Z
t
T
((1 + L)C¯ 2 |Yr(2) |2 + C¯ 2 |Zr(2) |2 + |f (r, 0, 0, 0)|2 ) dr +
Z
T +K
T
LC¯ 2 |ξr |2 dr .
¯ > 0 depending only on C, ¯ L From estimate (11), we can find a constant M and T such that ¯ E Ft |yt |2 ≤ M
Z
t
T
|ys |2 ds
15
ANTICIPATED BSDES
¯ +M
Z
T
(δ2 (s) − δ1 (s)) ds
t
× E Ft |ξT |2 + Thus, by Gronwall’s inequality, Z
¯ |yt |2 ≤ M
T
|ξs |2 ds +
Z
T
|f (s, 0, 0, 0)|2 ds .
t
(δ2 (s) − δ1 (s)) ds
× E Ft |ξT |2 + ¯ ˜ =M ¯ e MT Fix M , therefore,
˜ |yt | ≤ M
T +K
T
t
2
Z
Z
T +K
T
|ξs |2 ds +
Z
T
¯
|f (s, 0, 0, 0)|2 ds e Mt .
t
T
t
×E
Z
(δ2 (s) − δ1 (s)) ds
Ft
2
|ξT | +
Z
T +K
T
2
|ξs | ds +
Z
T
2
|f (s, 0, 0, 0)| ds .
t
5. Comparison theorem for 1-dimensional anticipated BSDEs. Lemma 3.3 is a typical version of a comparison theorem. It is a fundamentally important result in BSDE theory. Some further developments in this direction are Cao and Yan [3], Lin [10], Liu and Ren [11], Zhang [16] and Situ [15], without mentioning many other widely circulated papers listed in [13]. Recently Hu and Peng [8] gave a comparison theorem for multidimensional BSDEs. Comparison theorems for BSDEs have received a lot of attention because of their importance. For example, the punishment method in reflected BSDEs is based on a comparison theorem (see [4, 6, 9] and [14]). Moreover, research on properties of g-expectations (see Peng [13]) and the proof of a monotonic limit theorem for BSDEs (see Peng [12]) both depend on comparison theorems. It is well known that 1-dimensional BSDEs have comparison theorems (see Lemmas 3.3 and 3.4) when their generators satisfy the conditions of existence and uniqueness theorems for BSDEs. It is very important to notice that the conditions on f needed for the comparison theorem for anticipated BSDEs are stronger than those needed for the existence and uniqueness theorem. Using the comparison theorem for anticipated BSDEs, we will solve a stochastic control problem in Section 6. Let (Y.(1) , Z.(1) ), (Y.(2) , Z.(2) ) be respectively solutions of the following two 1-dimensional anticipated BSDEs: (
(j)
−dYt (j) Yt
(j)
(j)
(j)
(j)
= fj (t, Yt , Zt , Yt+δ(t) ) dt − Zt dWt ,
(j) = ξt ,
where j = 1, 2.
0 ≤ t ≤ T; T ≤ t ≤ T + K,
16
S. PENG AND Z. YANG
Theorem 5.1. Assume that f1 , f2 satisfies (H1) and (H2), ξ.(1) , ξ.(2) ∈ + K), δ satisfies (i), (ii), and for all t ∈ [0, T ], y ∈ R, z ∈ Rd , f2 (t, y, z, ·) is increasing, that is, f2 (t, y, z, θr ) ≥ f2 (t, y, z, θr′ ), if θr ≥ θr′ , θ, θ ′ ∈ (1) (2) L2F (t, T + K), r ∈ [t, T + K]. If ξs ≥ ξs , s ∈ [T, T + K] and f1 (t, y, z, θr ) ≥ f2 (t, y, z, θr ), t ∈ [0, T ], y ∈ R, z ∈ Rd , θ ∈ L2F (t, T + K), r ∈ [t, T + K], then 2 (T, T SF
(1)
Yt
(2)
≥ Yt ,
a.e., a.s.
Proof. Set Z (2) (3) Yt = ξT +
(3) Yt
T t
(2) = ξt ,
(1) f2 (s, Ys(3) , Zs(3) , Ys+δ(s) ) ds −
t ∈ [0, T ]; t ∈ [T, T + K].
Z
T t
Zs(3) dWs ,
By Lemma 3.2, we know there exists a unique pair of Ft -adapted processes 2 (0, T ) × L2 (0, T ; Rd ) that satisfies the above BSDE. Since (Y.(3) , Z.(3) ) ∈ SF F (1) (1) f1 (s, y, z, Ys+δ(s) ) ≥ f2 (s, y, z, Ys+δ(s) ), s ∈ [0, T ], y ∈ R, z ∈ Rd , by Lemma 3.4, we obtain (1)
Yt
(3)
≥ Yt ,
a.e., a.s.
Set Z (2) (4) Yt = ξT +
(4) Yt
T t
(2) = ξt ,
(3) f2 (s, Ys(4) , Zs(4) , Ys+δ(s) ) ds −
t ∈ [0, T ]; t ∈ [T, T + K].
Z
T t
Zs(4) dWs ,
(1)
Since for all t ∈ [0, T ], y ∈ R, z ∈ Rd , f2 (t, y, z, ·) is increasing and Yt a.e., a.s., by Lemma 3.4, we know (3)
Yt
(4)
≥ Yt ,
(3)
≥ Yt ,
a.e., a.s.
For n = 5, 6, . . . , we consider the following classical BSDE: Z (2) (n) Yt = ξ + T
(n)
Yt
t
(2)
= ξt ,
(4)
T
(n−1) f2 (s, Ys(n) , Zs(n) , Ys+δ(s) ) ds −
t ∈ [0, T ]; t ∈ [T, T + K]. (5)
(n)
Z
t
T
Zs(n) dWs ,
Similarly, we have Yt ≥ Yt ≥ · · · ≥ Yt ≥ · · · , a.e., a.s. We use kν(·)kβ in the proof of Theorem 4.2 as the norm in the Banach space L2F (0, T + (n) (n) (n−1) ˆ (n) (n) (n−1) K; R) × L2F (0, T ; Rd ). Set Yˆs = Ys − Ys , Zs = Zs − Zs , n ≥ 4.
17
ANTICIPATED BSDES
Then, by (7), we have E
Z
T β
|Yˆs(n) |2 + |Zˆs(n) |2 e βs ds
2 Z T 2 (n−1) ≤ E |f2 (s, Ys(n) , Zs(n) , Ys+δ(s) ) β 0
0
(n−2)
− f2 (s, Ys(n−1) , Zs(n−1) , Ys+δ(s) )|2 e βs ds 6C 2 E β
≤
Z
T 0
2
6C L (|Yˆs(n) |2 + |Zˆs(n) |2 )e βs ds + E β
Z
0
T
|Yˆs(n−1) |2 e βs ds .
Set β = 18C 2 L + 18C 2 + 3. Then 2 E 3
Z
T
0
(|Yˆs(n) |2 + |Zˆs(n) |2 )e βs ds
1 ≤ E 3
T
Z
0
1 |Yˆs(n−1) |2 e βs ds ≤ E 3
Z
0
T
(|Yˆs(n−1) |2 + |Zˆs(n−1) |2 )e βs ds .
Hence, E
Z
T
0
n−4 Z
1 (|Yˆs(n) |2 + |Zˆs(n) |2 )e βs ds ≤ 2
T
E
0
(|Yˆs(4) |2 +|Zˆs(4) |2 )e βs ds .
It follows that (Y.(n) )n≥4 and (Z.(n) )n≥4 are respectively Cauchy sequences in L2F (0, T + K) and in L2F (0, T ; Rd ). Denote their limits by Y. and Z. , respectively. Since L2F (0, T + K) and L2F (0, T ; Rd ) are both Banach spaces, we obtain (Y. , Z. ) ∈ L2F (0, T + K) × L2F (0, T ; Rd ). Note for all t ∈ [0, T ], E
Z
T
t
(n−1)
|f2 (s, Ys(n) , Zs(n) , Ys+δ(s) ) − f2 (s, Ys , Zs , Ys+δ(s) )|2 e βs ds
≤ 3C 2 E
Z
T
t
(|Ys(n) − Ys |2 + |Zs(n) − Zs |2 + L|Ys(n−1) − Ys |2 )e βs ds → 0,
when n → ∞. Therefore, (Y. , Z. ) satisfies the following anticipated BSDE: Z T Z T Y = ξ (2) + Zs dWs , f (s, Y , Z , Y ) ds − 2 s s s+δ(s) t T t t (2)
Yt = ξt ,
0 ≤ t ≤ T; T ≤ t ≤ T + K.
By Theorem 4.2, we know
(2)
Yt = Yt , (1)
Since Yt
(3)
≥ Yt
(4)
≥ Yt
a.e., a.s.
≥ Yt , it holds immediately (1)
Yt
(2)
≥ Yt ,
a.e., a.s.
18
S. PENG AND Z. YANG
If f2 is nonincreasing in the anticipated term of Y. , Theorem 5.1 does not hold. The following example shows this. Given T > δ > 0, consider the following two anticipated
Example 5.2. BSDEs:
(15)
and (16)
Yt = c + Yt = c,
Yt′ =
Z
t
T
Z
T
t
aE Fs [Ys+δ ] ds −
Z
Zs dWs ,
t
′ ′ aE Fs [I[Ys+δ 0 when t ∈ [T − δ, T − δ/2). If f2 contains the anticipated term of Z. , Theorem 5.1 does not hold. This is shown in the following example. Example 5.3.
(17)
Given T > δ > 0, consider the two anticipated BSDEs
Z T Z Tr π Fs 2 Zs dWs , Y = WT − T − E [|Zs+δ − Zs |] ds − t 2δ t t Y = Wt2 − (T − t), t
Zt = 2Wt ,
and
(18)
t ∈ [0, T ]; t ∈ [T, T + δ]; t ∈ [T, T + δ],
Z Tr Z T π Fs ′ ′ 2 ′ Y = 4WT − Zs′ dWs , E [|Zs+δ − Zs |] ds − t 2δ t t Y ′ = 4Wt2 − 4(T − t), t′
Zt = 8Wt ,
t ∈ [0, T ]; t ∈ [T, T + δ]; t ∈ [T, T + δ].
We can check that the solution of (17) is (Yt , Zt ) = (Wt2 − T − (T − t), 2Wt ) and that the solution of (18) is (Yt′ , Zt′ ) = (4Wt2 − 4(T − t), 8Wt ). We have YT < YT′ but Y0 > Y0′ .
19
ANTICIPATED BSDES (1)
(2)
Theorem 5.4. Under the assumptions of Theorem 5.1, if ξs ≥ ξs , s ∈ (1) (1) (1) (1) (1) (1) [T, T + K] and f1 (t, Yt , Zt , Yt+δ(t) ) ≥ f2 (t, Yt , Zt , Yt+δ(t) ), t ∈ [0, T ], then (1)
Yt
(2)
≥ Yt ,
a.e., a.s.
We also have a strict comparison. Given the assumptions of Theorem 5.1, suppose [T, T + K] ⊂ {t + δ(t), t ∈ [0, T ]} and f2 is strictly increasing in θ. Then (1)
Y0
(2)
(1) (1) (1) (1) (1) (1) f1 (t, Yt , Zt , Yt+δ(t) ) = f2 (t, Yt , Zt , Yt+δ(t) ),
⇐⇒
= Y0
Proof. Set
Z (2) (3) Yt = ξT +
(3)
Yt
(1) ξs
T t
(2)
= ξt ,
t ∈ [0, T ], s ∈ [T, T + K].
(2) = ξs ,
(1) f2 (s, Ys(3) , Zs(3) , Ys+δ(s) ) ds −
t ∈ [0, T ]; t ∈ [T, T + K].
Z
T t
Zs(3) dWs ,
(1) (1) (1) (1) (1) (1) Set f˜t = f1 (t, Yt , Zt , Yt+δ(t) )−f2 (t, Yt , Zt , Yt+δ(t) ) and y. = Y.(1) −Y.(3) , z. = Z.(1) − Z.(3) , ξ˜. = ξ.(1) − ξ.(2) . Then the pair (y. , z. ) can be regarded as the solution to the linear BSDE
Z ˜ y t = ξT +
T
t
where as =
t ∈ [0, T ]; t ∈ [T, T + K],
T
Z
t
zs dWs ,
(1) (1) (1) (3) (1) (1) f2 (s, Ys , Zs , Ys+δ(s) ) − f2 (s, Ys , Zs , Ys+δ(s) )
,
(3) (1) (1) (3) (3) (1) f2 (s, Ys , Zs , Ys+δ(s) ) − f2 (s, Ys , Zs , Ys+δ(s) )
,
bs =
yt = ξ˜t ,
(as ys + bs zs + f˜s ) ds −
(1) Ys
(3) − Ys
0,
(1) Zs
(3) − Zs
0
t
bs dWs −
1 2
Z
0
t
|bs |2 ds +
Z
0
t
(3)
(1)
= Ys ,
(3)
(1)
(3)
(1)
(3)
if Zs 6= Zs ; if Zs = Zs .
Since f2 satisfies (H1), |as | ≤ C and |bs | ≤ C. Set Z
6= Ys ;
if Ys
0,
Xt := exp
(1)
if Ys
as ds ≥ 0.
20
S. PENG AND Z. YANG
We apply Itˆo’s formula to Xs ys on [t, T ] and take conditional expectations on both sides: yt = E
Ft
ξ˜T XT +
Z
T
t
f˜s Xs ds .
(1) (3) Since ξ˜T ≥ 0, f˜t ≥ 0, a.e., a.s., we get Yt ≥ Yt , a.e., a.s. Then similarly to the proof of Theorem 5.1, we obtain (1)
Yt
(2)
≥ Yt ,
a.e., a.s.
Now we only need to prove the strict comparison theorem. (1)
(=⇒) Suppose Y0
(2)
= Y0 , by Lemma 3.3, we get (1)
(1)
(1)
(1)
(2)
(1)
f1 (t, Yt , Zt , Yt+δ(t) ) = f2 (t, Yt , Zt , Yt+δ(t) ), (1)
Since Y0
(3)
≥ Y0
(1)
(2)
≥ Y0 , we know Y0
t ∈ [0, T ].
(3)
= Y0 . Also by Lemma 3.3, we get
(1)
(1)
(1)
(1)
(1)
(1)
t ∈ [0, T ].
(1)
(1)
(1)
(1)
(1)
(2)
t ∈ [0, T ].
f1 (t, Yt , Zt , Yt+δ(t) ) = f2 (t, Yt , Zt , Yt+δ(t) ), Therefore, f2 (t, Yt , Zt , Yt+δ(t) ) = f2 (t, Yt , Zt , Yt+δ(t) ),
Note that for all t ∈ [0, T ], y ∈ R, z ∈ Rd , f2 (t, y, z, ·) is strictly increasing, (1) (2) (1) (2) hence, Yt+δ(t) = Yt+δ(t) , t ∈ [0, T ]. In particular, ξt = ξt , t ∈ [T, T + K]. (1)
(1)
(1)
(1)
(1)
(1)
(⇐=) Suppose f1 (t, Yt , Zt , Yt+δ(t) ) = f2 (t, Yt , Zt , Yt+δ(t) ), t ∈ [0, T ] (1)
(2)
and ξs = ξs , s ∈ [T, T + K]. Then (1) y t = Yt
(3) − Yt
=E
Ft
ξ˜T XT +
Z
t
T
f˜s Xs ds ≡ 0.
Therefore, Z (1) (2) Yt = ξ + T
(1) Yt
(2) = ξt ,
(1)
By Theorem 4.2, Yt
T t
(1)
f2 (s, Ys(1) , Zs(3) , Ys+δ(s) ) ds −
t ∈ [0, T ]; t ∈ [T, T + K]. (2)
Z
T t
Zs(3) dWs ,
(1)
= Yt , a.e., a.s., in particular, Y0 (1)
(1)
(2)
(2)
= Y0 .
(2)
Corollary 5.5. Let (Y· , Z· ) and (Y· , Z· ) be respectively the solutions for the following two 1-dimensional anticipated BSDEs: −dY (j) = f (t, Y (j) , Z (j) , Y (j) t
Y (j) = ξ , t t
t
t
(j) t+δj (t) ) dt − Zt dWt ,
0 ≤ t ≤ T; T ≤ t ≤ T + K,
21
ANTICIPATED BSDES
2 (T, T + K), f satisfies (H1) and (H2), for where j = 1, 2. Suppose ξ. is in SF d all t ∈ [0, T ], y ∈ R, z ∈ R , f (t, y, z, ·) is increasing, and δ1 , δ2 satisfy (i) and (1) (1) (ii). If Yt+δ1 (t) ≥ Yt+δ2 (t) , a.e., a.s., then (1)
Yt
(2)
≥ Yt ,
a.e., a.s.
Proof. Set Z (3) Yt = ξT +
(3) Yt
= ξt ,
T t
(1)
f (s, Ys(3) , Zs(3) , Ys+δ2 (s) ) ds −
t ∈ [0, T ]; t ∈ [T, T + K].
Z
t
T
Zs(3) dWs ,
From Lemma 3.2, there exists a unique pair of Ft -adapted processes (Y.(3) , 2 (0, T ) × L2 (0, T ; R1×d ) that satisfies the above BSDE. Since Z.(3) ) ∈ SF F (1) (1) f (s, y, z, Ys+δ1(s) ) ≥ f (s, y, z, Ys+δ2(s) ), by Lemma 3.4, we know (1)
Yt
(3)
≥ Yt ,
a.e., a.s.
The remaining proof is similar to Theorem 5.1, we omit it. 6. Stochastic control problems. El Karoui, Peng and Quenez [7] applied the duality between SDEs and BSDEs to stochastic control problems. Now we consider if it is feasible to use the duality between SDDEs and anticipated BSDEs to solve these problems. Let θ > 0 be a given constant. Now we consider the following stochastic control problem: the laws of the controlled process belong to a family of equivalent measures whose densities are u dXsu = (α(s, us )Xsu + b(s − θ, us−θ )Xs−θ ) ds + Xsu σ T (s, us ) dWs ,
s ∈ [t, T + θ];
u Xtu = 1,
Xs = 0,
s ∈ [t − θ, t),
where the coefficients α(s, u) : R × Rk −→ R, b(s, u) : R × Rk −→ R+ and σ(s, u) : R×Rk −→ Rd×1 are adapted processes uniformly continuous with respect to (s, u). A feasible control (us , s ∈ [−θ, T + θ]) is a continuous adapted process valued in a compact subset U in Rk . The set of feasible controls is denoted by U. The problem is to maximize over all feasible control processes u the objective function
J(u) = E XTu Q(T ) +
Z
T +θ
T
u Q(s)b(s − θ, us−θ ) ds Xs−θ
+
Z
0
T
Xsu l(s, us ) ds ,
22
S. PENG AND Z. YANG
2 (T, T + θ) is the terminal condition, (l(ω, s, u ), s ∈ [0, T ]) where Q(·) ∈ SF s is the running cost associated with the control process u and l(s, u) is an adapted process uniformly continuous with respect to (s, u). Assume α(s, u), b(s, u), |σ(s, u)| and l(s, u) are uniformly bounded by µ. Notice that, by Theorem 2.1, J(u) = Y0u , where (Y.u , Z.u ) is the solution to the following linear anticipated BSDE:
u −dYtu = f u (t, Ytu , Ztu , Yt+θ ) dt − Ztu dWt , u Yt = Q(t),
t ∈ [0, T ]; t ∈ [T, T + θ],
where f u (t, y, z, ηr ) = α(t, ut )y +zσ(t, ut )+b(t, ut )E Ft [ηr ]+l(t, ut ), t ∈ [0, T ], y ∈ R, z ∈ Rd , η. ∈ L2F (t, T + θ), r ∈ [t, T + θ] and
Ytu = E Ft XTu Q(T ) +
Z
T +θ
T
u Xs−θ Q(s)b(s − θ, us−θ ) ds
+
Z
t
T
Xsu l(s, us ) ds
.
Theorem 6.1. Set f (t, y, z, ηr ) = esssup{f u (t, y, z, ηr ), u ∈ U }, t ∈ [0, T ], y ∈ R, z ∈ Rd , η. ∈ L2F (t, T + θ), r ∈ [t, T + θ]. Then anticipated BSDE (19)
−dYt = f (t, Yt , Zt , Yt+θ ) dt − Zt dWt , Yt = Q(t),
t ∈ [0, T ]; t ∈ [T, T + θ],
has a unique solution (Y. , Z. ). Moreover, Y. is the value function Y.∗ of the control problem, that is, for each t ∈ [0, T ], Yt = Yt∗ = esssup{Ytu , u ∈ U }. Proof. On one hand, since α, b, |σ| and l are uniformly bounded by µ, for all t ∈ [0, T ], s ∈ [T, T + θ], y, y ′ ∈ R, z, z ′ ∈ Rd , η. , η.′ ∈ L2F (t, T + θ), and r ∈ [t, T + θ], f (t, y, z, ηr ) − f (t, y ′ , z ′ , ηr′ ) ≤ esssup{α(t, ut )(y − y ′ ) + (z − z ′ )σ(t, ut ) + b(t, ut )E Ft [ηr − ηr′ ], u ∈ U } ≤ µ(|y − y ′ | + |z − z ′ | + E Ft [|ηr − ηr′ |]). Notice E[ 0T |f (t, 0, 0, 0)|2 dt] ≤ µ2 T, then by Theorem 4.2, the anticipated 2 (0, T + K) × L2 (0, T ; Rd ). BSDE (19) has a unique solution (Y., Z.) ∈ SF F u Since for all u ∈ U, f (t, y, z, η) ≤ f (t, y, z, η) and f u (t, y, z, η) is increasing in η, by Theorem 5.1, we get Yt ≥ Ytu , a.e., a.s. Thus, Yt ≥ Yt∗ , a.e., a.s. R
23
ANTICIPATED BSDES
On the other hand, by the definition of f , we know for all ε > 0, for each (ω, t) ∈ Ω × [0, T ), {u ∈ U; f (ω, t, Yt(ω), Zt (ω), Yt+θ (ω)) ≤ α(t, u)Yt (ω) + Zt (ω)σ(t, u) 6 ∅. + b(t, u)E Ft [Yt+θ (ω)] + l(ω, t, u) + ε} = Then by a Measurable Selection Theorem, for example, that can be found in Dellacherie [5] or in Beneˇs [1, 2], there exists a uε ∈ U such that ε
f (t, Yt, Zt , Yt+θ ) ≤ f u (t, Yt , Zt , Yt+θ ) + ε,
a.e., a.s. ε
Denote the solution to the anticipated BSDE corresponding to (f u , Q. ) by ε ε (Y.u , Z.u ). First, consider the case when t ∈ [T − θ, T ]. Thus, t + θ ∈ [T, T + θ], Yt+θ = uε and Yt+θ ε
ε
ε
ε
u ) − f (t, Yt , Zt , Yt+θ ) f u (t, Ytu , Ztu , Yt+θ ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
u ) − f u (t, Yt , Zt , Yt+θ ) − ε ≥ f u (t, Ytu , Ztu , Yt+θ ε
u u )−ε ) − f u (t, Yt , Zt , Yt+θ = f u (t, Ytu , Ztu , Yt+θ (1)
(2)
ε
ε
= gt (Ytu − Yt ) + gt (Ztu − Zt ) − ε, where, for t ∈ [T − θ, T ], (1)
gt
(2)
gt
=
ε uε uε uε uε uε uε u f (t, Yt , Zt , Yt+θ ) − f (t, Yt , Zt , Yt+θ ) ε
=
Ytu − Yt
0,
ε uε uε uε uε u f (t, Yt , Zt , Yt+θ ) − f (t, Yt , Zt , Yt+θ )
ε Ztu
− Zt
,
0,
,
ε
if Ytu 6= Yt ; ε
if Ytu = Yt , ε
if Ztu 6= Zt ; ε
if Ztu = Zt .
That is, for t ∈ [T − θ, T ], ε
Ytu − Yt ≥
Z
T
t
−
Z
t
ε
ε
(gs(1) (Ysu − Ys ) + gs(2) (Zsu − Zs ) − ε) ds T
ε
(Zsu − Zs ) dWs .
ε (1) (1) Hence, Ytu − Yt ≥ Y˜t , where Y˜t is the solution of BSDE:
(1) Y˜t =
Z
t
T
(gs(1) Y˜s(1) + gs(2) Z˜s(1) − ε) ds −
Z
t
T
Z˜s(1) dWs ,
t ∈ [T − θ, T ].
24
S. PENG AND Z. YANG (1)
(2)
Since |gt | ≤ µ, |gt | ≤ µ, we get (1) Y˜t = −εE Ft
where
Z
T
t
˜ s(1) ds , X
t ∈ [T − θ, T ],
t 1 t (2) 2 gs(1) ds ≥ 0. |gs | ds + 2 0 0 0 Therefore, there exists a constant ρ1 > 0 depending only on µ, θ and T such that ε (1) Ytu − Yt ≥ Y˜t ≥ −ρ1 ε, t ∈ [T − θ, T ].
˜ t(1) = exp X
t
Z
gs(2) dWs −
Z
Z
Second, consider the case when t ∈ [T − 2θ, T − θ]. Then t + θ ∈ [T − θ, T ], uε + ρ ε. Since for all t ∈ [0, T ], y ∈ R, z ∈ Rd , f u (t, y, z, ·) is a inYt+θ ≤ Yt+θ 1 creasing and linear function, we have ε
ε
ε
ε
u f u (t, Ytu , Ztu , Yt+θ ) − f (t, Yt , Zt , Yt+θ ) ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
u ≥ f u (t, Ytu , Ztu , Yt+θ ) − f u (t, Yt , Zt , Yt+θ ) − ε u u + ρ1 ε) − ε ) − f u (t, Yt , Zt , Yt+θ ≥ f u (t, Ytu , Ztu , Yt+θ u u ) − µρ1 ε − ε ) − f u (t, Yt , Zt , Yt+θ ≥ f u (t, Ytu , Ztu , Yt+θ (1)
(2)
ε
ε
= gt (Ytu − Yt ) + gt (Ztu − Zt ) − (µρ1 + 1)ε, where, for t ∈ [T − 2θ, T − θ], (1)
gt
(2)
gt
=
ε uε uε uε uε uε uε u f (t, Yt , Zt , Yt+θ ) − f (t, Yt , Zt , Yt+θ ) ε
=
Ytu − Yt
ε
if Ytu = Yt ,
0,
ε uε uε uε uε u f (t, Yt , Zt , Yt+θ ) − f (t, Yt , Zt , Yt+θ )
ε
if Ytu 6= Yt ;
,
ε Ztu
− Zt
,
0,
ε
if Ztu 6= Zt ; ε
if Ztu = Zt .
Therefore, for t ∈ [T − 2θ, T − θ], ε
ε
Ytu − Yt ≥ YTu−θ − YT −θ − +
Z
T −θ
t
Z
t
ε
T −θ
ε
(Zsu − Zs ) dWs ε
(gs(1) (Ysu − Ys ) + gs(2) (Zsu − Zs ) − (µρ1 + 1)ε) ds.
(2) (2) Hence, Yt − Yt ≥ Y˜t , where Y˜t is the solution to the following BSDE: For t ∈ [T − 2θ, T − θ], uε
ε (2) Y˜t = YTu−θ − YT −θ
+
Z
t
T −θ
(gs(1) Y˜s(2) + gs(2) Z˜s(2) − (µρ1 + 1)ε) ds −
Z
t
T −θ
Z˜s(2) dWs .
25
ANTICIPATED BSDES (1)
(2)
Note |gt | ≤ µ and |gt | ≤ µ. We have, for all t ∈ [T − 2θ, T − θ],
ε (2) ˜ (2) − Y˜t = E Ft (YTu−θ − YT −θ )X T −θ
Z
T −θ
t
˜ (2) ds , (µρ1 + 1)εX s
where ˜ (2) = exp X t
Z
t
0
gs(2) dWs −
uε
1 2
Z
0
t
|gs(2) |2 ds +
Z
t 0
gs(1) ds ≥ 0.
Since YT −θ − YT −θ ≥ −ρ1 ε, there exists a constant ρ2 > 0 depending only on µ, θ and T such that ε (2) Ytu − Yt ≥ Y˜t ≥ −ρ2 ε,
t ∈ [T − 2θ, T − θ].
Similarly, we get constants ρ3 , ρ4 , . . . , ρ[ T ]+1 > 0 such that θ
T ; − Yt ≥ −ρn ε, t ∈ [T − nθ, T − (n − 1)θ], n = 3, 4, . . . , θ T uε Yt − Yt ≥ −ρ[T /θ]+1 ε, t ∈ 0, T − θ . θ
ε Ytu
Setting ρ = max{ρ2 , ρ3 , . . . , ρ[T /θ]+1 }, we obtain ε
Ytu − Yt ≥ −ρε, uε
Since Yt
t ∈ [0, T ]. ε
≤ Yt , a.e., a.s., setting ε → 0, we get Ytu → Yt , a.e., a.s. Thus, Yt = Yt∗ ,
a.e., a.s.
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[email protected]
School of Mathematics Shandong University Jinan, 250100 China And Present Address: Department of Engineering Cambridge University Trumpington Street Cambridge CB2 1PZ United Kingdom E-mail:
[email protected]
