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Gluing and coupling
Statistics and Probability Letters 80 (2010) 1196–1199
Contents lists available at ScienceDirect
Statistics and Probability Letters journal homepage: www.elsevier.com/locate/stapro
Gluing and coupling Shun-Xiang Ouyang ∗ Department of Mathematics, Bielefeld University, D-33501 Bielefeld, Germany
article
abstract
info
Article history: Received 14 June 2009 Received in revised form 19 March 2010 Accepted 20 March 2010 Available online 1 April 2010
We show a martingale solution for operators of the form L1 1{t <τ } + L2 1{t ≥τ } , where L1 , L2 are second order differential operators and τ is a stopping time. As applications, we study the existence of couplings. © 2010 Elsevier B.V. All rights reserved.
MSC: 60G40 60G46 Keywords: Martingale Coupling Stopping time
1. Introduction Let Ω = C ([0, ∞), Rn ) be the space of all continuous trajectories from [0, ∞) into Rn . For each ω ∈ Ω and t ∈ [0, ∞), denote the position of ω at time t by Xt (ω) = X (t , ω) = ωt ∈ Rn . For any 0 ≤ t1 < t2 ≤ ∞, set t
Mt21 = σ (Xs : t1 ≤ s ≤ t2 ).
Here we understand s ≤ t2 as s < t2 if t2 = ∞. We will also use the following simplified notation: Mt := Mt0 ,
t M t := M∞ ,
0 M := M∞ .
Let Sn represent the space of all n × n nonnegative definite real matrices. For any measurable functions a(t , x) ∈ Sn and b(t , x) ∈ Rn defined on [0, ∞) × Rn , let L(a, b) := L(a(t , x), b(t , x)) :=
n 1X
2 i ,j = 1
aij (t , x)
n X ∂2 ∂ + bi (t , x) . ∂ xi ∂ xj ∂ xi i=1
(1)
Recall that a solution to the martingale problem for L := L(a, b) starting from (s, x) ∈ [0, ∞) × Rn is a probability measure Ps,x on (Ω , M ) such that
Ps,x (Xt = x, 0 ≤ t ≤ s) = 1
(2)
and for every f ∈ C 0 (R ), a smooth function on R with compact support, ∞
f Mt
:= f (Xt ) −
d
Z t" X n 1 0
∗
d
n ∂ 2 f (Xu ) X ∂ f ( Xu ) aij (u, X (u)) + bi (u, X (u)) 2 i,j=1 ∂ xi ∂ xj ∂ xi i=1
Tel.: +49 521 1062971; fax: +49 521 1066455. E-mail addresses: [email protected], [email protected].
0167-7152/$ – see front matter © 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.spl.2010.03.016
# du
S.-X. Ouyang / Statistics and Probability Letters 80 (2010) 1196–1199
1197
is a Ps,x -martingale after time s. (In the following, we denote Px for P0,x for convenience.) For each ω0 ∈ Ω , t0 ≥ 0, and x0 ∈ Rd , δω0 ⊗t0 Pt0 ,x0 denotes the unique probability measure on (Ω , M ) such that (see Stroock and Srinivasa Varadhan, 1979, Lemma 6.1.1)
δω0 ⊗t0 Pt0 ,x0 (Xt = ω0 (t0 ), 0 ≤ t ≤ t0 ) = 1, and
δω0 ⊗t0 Pt0 ,x0 (A) = Pt0 ,x0 (A),
A ∈ M t0 .
A solution to the martingale problem for L up to a stopping time τ starting from (s, x) is a probability measure Ps,x on (Ω , M ) such that (2) holds and there exist some stopping time sequence τn ↑ τ such that for each n ≥ 1, the stopped process Mt ∧τn is a Ps,x -martingale. The main result of this paper is the following gluing theorem. f
Theorem 1.1. Let L1 and L2 be two second order differential operators as (1) on Rn . Let τ be a stopping time on Ω and define L = L1 1{t <τ } + L2 1{t ≥τ } . Assume 1. There exists a solution Px1 to the martingale problem for L1 up to τ ; τ (ω),Xτ (ω) (ω)
2. For each ω ∈ Ω , there exists a solution P2 to the martingale problem for L2 starting from (τ (ω), Xτ (ω) (ω)); 3. There exists a sequence of stopping time τn such that τn ↑ τ as n → ∞, and the following two conditions hold for each ω ∈ Ω (a) τ
Z
Lf (Xs ) ds = 0.
lim
n→∞
τn
n (b) For every f ∈ C ∞ 0 (R ), t ∧τn
Z lim
n→∞
R t ∧τn 0
L1 f (Xs ) ds is bounded and
L1 f (Xs ) ds =
0
(3)
t ∧τ
Z
L1 f (Xs ) ds.
(4)
0
Define for every ω ∈ Ω τ (ω),Xτ (ω) (ω)
Qω = δω ⊗τ (ω) P2 Then
Px1
1{τ <∞} + δω 1{τ =∞} .
⊗τ Q is a solution to the martingale problem for L.
We prove Theorem 1.1 in Section 2. In Section 3, we apply this lemma to study the existence of couplings. We remark that by the relationship between martingale solution and weak solution of stochastic differential equation, we can also study the existence of weak solution of some coupled stochastic differential equations. 2. Proof of Theorem 1.1 We first prepare three lemmas. n For each f ∈ C ∞ 0 (R ) and t ≥ 0, define
θt = f (Xt ) −
t
Z
Lf (Xs ) ds, 0 t
φt = f (Xt ) −
Z
ψt = f (Xt ) −
Z
L1 f (Xs ) ds, 0 t
L2 f (Xs ) ds. 0
Lemma 2.1. For every t ≥ 0 and ω ∈ Ω , we have
θt ∧τ = φt ∧τ
(5)
θt − θt ∧τ = ψt − ψt ∧τ .
(6)
and
n Proof. By the continuity of the path Xt and the fact f ∈ C ∞ 0 (R ) and the assumption (4), we have
lim φ(t ∧ τn ) = lim f (Xt ∧τn ) − lim
n→∞
n→∞
n→∞
t ∧τn
Z 0
L1 f (Xs ) ds = φ(t ∧ τ ).
(7)
1198
S.-X. Ouyang / Statistics and Probability Letters 80 (2010) 1196–1199
On the other hand, by (3), we know lim θt ∧τ − θt ∧τn = lim
n→∞
n→∞
f (Xt ∧τ ) − f (Xt ∧τn ) +
Z
t ∧τ
Lf (Xs ) ds = 0.
(8)
t ∧τn
Note that for every t ≥ 0 and each n ∈ N, we know θt ∧τn = φt ∧τn . Hence (5) follows from (7) and (8). Now we come to the proof of (6). First, it is easy to see
(θt − θt ∧τ )1{t <τ } = 0 = (ψt − ψt ∧τ )1{t <τ } and
(θt − θt ∧τ ) 1{t ≥τ } =
t
Z
Lf (Xs ) ds 1{t ≥τ }
τ t
Z = τ
L2 f (Xs ) ds1{t ≥τ } = (ψt − ψt ∧τ ) 1{t ≥τ } .
Hence
θt − θt ∧τ = (θt − θt ∧τ )1{t <τ } + (θt − θt ∧τ )1{t ≥τ } = (ψt − ψt ∧τ )1{t <τ } + (ψt − ψt ∧τ )1{t ≥τ } = ψt − ψt ∧τ . Lemma 2.2. (θt ∧τ , Mt , Px1 ) is a martingale. Proof. By (5), we only need to show that φt ∧τ is a martingale. Since Px1 is a solution to the martingale problem for L1 up to τ , we know (φt ∧τn , Mt , Px1 ) is a martingale. Therefore, for any 0 ≤ s < t,
E(φt ∧τn |Ms ) = φs∧τn . By assumption (3), let n → ∞ we obtain E(φt ∧τ |Ms ) = φ(s ∧ τ ). This proves that φt ∧τ is a martingale.
Lemma 2.3. (θt − θt ∧τ (ω) , Mt , Qω ) is a martingale for each ω ∈ Ω . Proof. Fix an arbitrary path ω0 ∈ Ω . We set t0 := τ (ω0 ) and Xt0 (ω0 ) = x0 . For any fixed instants 0 ≤ t1 ≤ t2 , we need to prove
EQω0 (θt2 − θt2 ∧t0 |Mt1 ) = θt1 − θt1 ∧t0 .
(9)
We assume t0 < ∞, otherwise Qω0 = δω0 and equality (9) is trivial. If t1 < t2 ≤ t0 , then it is cleat that (9) holds since both sides of (9) are zero. Assume t1 < t0 < t2 . Suppose that (9) holds for the case when t0 ≤ t1 < t2 . Then
EQω0 θt2 − θt2 ∧t0 |Mt1 = EQω0 EQω0 (θt2 − θt2 ∧t0 |Mt0 )|Mt1
= EQω0 (θt0 − θt0 ∧t0 |Mt1 ) = 0 = θt1 − θt1 ∧t0 . Hence we know (9) holds. In the remaining we assume t0 ≤ t1 < t2 and show (9). t ,x t ,x As P20 0 is a solution to the martingale problem for L2 , we know ψt is a P20 0 martingale starting from (t0 , x0 ). This implies t
that for any A ∈ Mt10 t ,x 0
P20
t ,x
(ψt2 − ψt0 , A) = P20 0 (ψt1 − ψt0 , A).
(10)
By (6), we only need to show that for any A ∈ Mt1 ,
Qω0 (ψt2 − ψt0 , A) = Qω0 (ψt1 − ψt0 , A).
(11)
It is enough to show (11) for A of the form: A = A1 ∩ A2 with A1 = {Xs1 ∈ Γs1 , . . . , Xsi ∈ Γsi },
A2 = {Xsi+1 ∈ Γsi+1 , . . . , Xsn ∈ Γsn },
where 0 ≤ s1 ≤ · · · ≤ si < t0 ≤ si+1 ≤ · · · ≤ sn ≤ t1 , and Γsj ∈ B (Rd ) for j = 1, 2, . . . , n. We just need to note that cylinder sets of the form A generates Mt1 and then use Dynkin’s π -λ theorem. If w0 6∈ A1 , then Qω0 (A1 ) = 0 and hence
Qω0 (ψt2 − ψt0 , A) = 0 = Qω0 (ψt1 − ψt0 , A).
S.-X. Ouyang / Statistics and Probability Letters 80 (2010) 1196–1199
1199 t
If w0 ∈ A1 , then Qω0 (A1 ) = 1 and hence Qω0 (A) = Qω0 (A2 ). Therefore, by the fact A2 ∈ Mt10 and (10), we have t ,x 0
Qω0 (ψt2 − ψt0 , A) = δω0 ⊗t0 P20
t ,x
(ψt2 − ψt0 , A2 ) = P20 0 (ψt2 − ψt0 , A2 )
t ,x
= P20 0 (ψt1 − ψt0 , A2 ) = Qω0 (ψt1 − ψt0 , A). This proves (11).
Proof of Theorem 1.1. We need to show that θt is a martingale on (Ω , M , Px1 ⊗τ Q). According to Stroock and Srinivasa Varadhan (1979, Theorem 6.1.2), it suffices to prove the following two statements. (a) (θt ∧τ , Mt , Px1 ) is a martingale. (b) (θt − θt ∧τ (ω) , Mt , Qω ) is a martingale for each ω ∈ Ω . But they are the conclusions of Lemmas 2.2 and 2.3 respectively.
3. Coupling We suppose Ω = C ([0, ∞), R2d ). Denote Zt (ω) = ωt = (Xt (ω), Yt (ω)) ∈ Rd × Rd for each ω ∈ Ω . Let σ (t , x) ∈ Sd be measurable real matrix defined on [0, ∞) × Rd . Let b(t , x), ξ (t , x, y) be Rd valued measurable functions defined on [0, ∞) × Rd and [0, ∞) × Rd × Rd respectively. Let c (t , x, y) be a d × d matrix valued measurable function defined on [0, ∞) × Rd × Rd . We assume that σ , b, c, and ξ all are locally bounded. Denote for every t ≥ 0, x, y ∈ Rd , a(t , x, y) =
σ (t , x)σ (t , x)∗ σ (t , y)σ (t , x)∗
σ (t , x)σ (t , y)∗ ∗ , σ (t , y)σ (t , y)
b(t , x, y) =
b(t , x) , b(t , y)
τ (ω),Z (τ (ω))
Let τ be a stopping time. For every ω ∈ Ω , let P2 be a solution to the martingale problem for L(a, b) starting from (τ (ω), Z (τ (ω))). Define Qω for each ω ∈ Ω as in Theorem 1.1. Applying Theorem 1.1, we see immediately the following corollary (see Chen and Li, 1989, Lemma 3.4). x ,y
Corollary 3.1. Let P1 be a solution to the martingale problem for L(a1 , b) with a1 (t , x, y) =
σ (t , x)σ (t , x)∗ c (t , x, y)∗
c (t , x, y) σ (t , y)σ (t , y)∗
x ,y
up to some stopping time τ . Then R = P1 ⊗τ Q is a solution to the martingale problem for L(a2 , b) with a2 (t , x, y) =
σ (t , x)σ (t , x)∗ ∗ c (t , x, y) 1[0,τ ) + σ (t , y)σ (t , x)∗ 1[τ ,∞)
c (t , x, y)1[0,τ ) + σ (t , x)σ (t , y)∗ 1[τ ,∞) σ (t , y)σ (t , y)∗
.
Remark 3.2. A typical use of this fact is the following. First we obtain successful coupling (the marginal processes meet each other) by choosing c (t , x, y) properly. Then the marginal processes will move together after the coupling time. Similarly, we can obtain coupling by choosing proper drift. x ,y
˜ 1 be a solution to the martingale problem for L(a, b1 ) with Corollary 3.3. Let P b1 (t , x, y) =
b(t , x) b(t , y) + ξ (t , x, y)
x ,y
˜ 1 ⊗τ Q is a solution to the martingale problem for L(a, b2 ) with up to some stopping time τ . Then R = P b2 (t , x, y) =
b( t , x) b(t , y) + ξ (t , x, y)1{τ
.
Acknowledgements This work was supported in part by the DFG through the International Graduate College ‘‘Stochastics and Real World Models’’ in Bielefeld University. The author thanks a referee for suggestions which make the paper more clear. References Chen, Mu-Fa, Li, Shao-Fu, 1989. Coupling methods for multidimensional diffusion processes. Ann. Probab. 17 (1), 151–177. MR MR972776 (90a:60134). Stroock, Daniel W., Srinivasa Varadhan, S.R., 1979. Multidimensional Diffusion Processes. In: Grundlehren der Mathematischen Wissenschaften (Fundamental Principles of Mathematical Sciences), vol. 233. Springer-Verlag, Berlin. MR532498 (81f:60108).
Contents lists available at ScienceDirect
Statistics and Probability Letters journal homepage: www.elsevier.com/locate/stapro
Gluing and coupling Shun-Xiang Ouyang ∗ Department of Mathematics, Bielefeld University, D-33501 Bielefeld, Germany
article
abstract
info
Article history: Received 14 June 2009 Received in revised form 19 March 2010 Accepted 20 March 2010 Available online 1 April 2010
We show a martingale solution for operators of the form L1 1{t <τ } + L2 1{t ≥τ } , where L1 , L2 are second order differential operators and τ is a stopping time. As applications, we study the existence of couplings. © 2010 Elsevier B.V. All rights reserved.
MSC: 60G40 60G46 Keywords: Martingale Coupling Stopping time
1. Introduction Let Ω = C ([0, ∞), Rn ) be the space of all continuous trajectories from [0, ∞) into Rn . For each ω ∈ Ω and t ∈ [0, ∞), denote the position of ω at time t by Xt (ω) = X (t , ω) = ωt ∈ Rn . For any 0 ≤ t1 < t2 ≤ ∞, set t
Mt21 = σ (Xs : t1 ≤ s ≤ t2 ).
Here we understand s ≤ t2 as s < t2 if t2 = ∞. We will also use the following simplified notation: Mt := Mt0 ,
t M t := M∞ ,
0 M := M∞ .
Let Sn represent the space of all n × n nonnegative definite real matrices. For any measurable functions a(t , x) ∈ Sn and b(t , x) ∈ Rn defined on [0, ∞) × Rn , let L(a, b) := L(a(t , x), b(t , x)) :=
n 1X
2 i ,j = 1
aij (t , x)
n X ∂2 ∂ + bi (t , x) . ∂ xi ∂ xj ∂ xi i=1
(1)
Recall that a solution to the martingale problem for L := L(a, b) starting from (s, x) ∈ [0, ∞) × Rn is a probability measure Ps,x on (Ω , M ) such that
Ps,x (Xt = x, 0 ≤ t ≤ s) = 1
(2)
and for every f ∈ C 0 (R ), a smooth function on R with compact support, ∞
f Mt
:= f (Xt ) −
d
Z t" X n 1 0
∗
d
n ∂ 2 f (Xu ) X ∂ f ( Xu ) aij (u, X (u)) + bi (u, X (u)) 2 i,j=1 ∂ xi ∂ xj ∂ xi i=1
Tel.: +49 521 1062971; fax: +49 521 1066455. E-mail addresses: [email protected], [email protected].
0167-7152/$ – see front matter © 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.spl.2010.03.016
# du
S.-X. Ouyang / Statistics and Probability Letters 80 (2010) 1196–1199
1197
is a Ps,x -martingale after time s. (In the following, we denote Px for P0,x for convenience.) For each ω0 ∈ Ω , t0 ≥ 0, and x0 ∈ Rd , δω0 ⊗t0 Pt0 ,x0 denotes the unique probability measure on (Ω , M ) such that (see Stroock and Srinivasa Varadhan, 1979, Lemma 6.1.1)
δω0 ⊗t0 Pt0 ,x0 (Xt = ω0 (t0 ), 0 ≤ t ≤ t0 ) = 1, and
δω0 ⊗t0 Pt0 ,x0 (A) = Pt0 ,x0 (A),
A ∈ M t0 .
A solution to the martingale problem for L up to a stopping time τ starting from (s, x) is a probability measure Ps,x on (Ω , M ) such that (2) holds and there exist some stopping time sequence τn ↑ τ such that for each n ≥ 1, the stopped process Mt ∧τn is a Ps,x -martingale. The main result of this paper is the following gluing theorem. f
Theorem 1.1. Let L1 and L2 be two second order differential operators as (1) on Rn . Let τ be a stopping time on Ω and define L = L1 1{t <τ } + L2 1{t ≥τ } . Assume 1. There exists a solution Px1 to the martingale problem for L1 up to τ ; τ (ω),Xτ (ω) (ω)
2. For each ω ∈ Ω , there exists a solution P2 to the martingale problem for L2 starting from (τ (ω), Xτ (ω) (ω)); 3. There exists a sequence of stopping time τn such that τn ↑ τ as n → ∞, and the following two conditions hold for each ω ∈ Ω (a) τ
Z
Lf (Xs ) ds = 0.
lim
n→∞
τn
n (b) For every f ∈ C ∞ 0 (R ), t ∧τn
Z lim
n→∞
R t ∧τn 0
L1 f (Xs ) ds is bounded and
L1 f (Xs ) ds =
0
(3)
t ∧τ
Z
L1 f (Xs ) ds.
(4)
0
Define for every ω ∈ Ω τ (ω),Xτ (ω) (ω)
Qω = δω ⊗τ (ω) P2 Then
Px1
1{τ <∞} + δω 1{τ =∞} .
⊗τ Q is a solution to the martingale problem for L.
We prove Theorem 1.1 in Section 2. In Section 3, we apply this lemma to study the existence of couplings. We remark that by the relationship between martingale solution and weak solution of stochastic differential equation, we can also study the existence of weak solution of some coupled stochastic differential equations. 2. Proof of Theorem 1.1 We first prepare three lemmas. n For each f ∈ C ∞ 0 (R ) and t ≥ 0, define
θt = f (Xt ) −
t
Z
Lf (Xs ) ds, 0 t
φt = f (Xt ) −
Z
ψt = f (Xt ) −
Z
L1 f (Xs ) ds, 0 t
L2 f (Xs ) ds. 0
Lemma 2.1. For every t ≥ 0 and ω ∈ Ω , we have
θt ∧τ = φt ∧τ
(5)
θt − θt ∧τ = ψt − ψt ∧τ .
(6)
and
n Proof. By the continuity of the path Xt and the fact f ∈ C ∞ 0 (R ) and the assumption (4), we have
lim φ(t ∧ τn ) = lim f (Xt ∧τn ) − lim
n→∞
n→∞
n→∞
t ∧τn
Z 0
L1 f (Xs ) ds = φ(t ∧ τ ).
(7)
1198
S.-X. Ouyang / Statistics and Probability Letters 80 (2010) 1196–1199
On the other hand, by (3), we know lim θt ∧τ − θt ∧τn = lim
n→∞
n→∞
f (Xt ∧τ ) − f (Xt ∧τn ) +
Z
t ∧τ
Lf (Xs ) ds = 0.
(8)
t ∧τn
Note that for every t ≥ 0 and each n ∈ N, we know θt ∧τn = φt ∧τn . Hence (5) follows from (7) and (8). Now we come to the proof of (6). First, it is easy to see
(θt − θt ∧τ )1{t <τ } = 0 = (ψt − ψt ∧τ )1{t <τ } and
(θt − θt ∧τ ) 1{t ≥τ } =
t
Z
Lf (Xs ) ds 1{t ≥τ }
τ t
Z = τ
L2 f (Xs ) ds1{t ≥τ } = (ψt − ψt ∧τ ) 1{t ≥τ } .
Hence
θt − θt ∧τ = (θt − θt ∧τ )1{t <τ } + (θt − θt ∧τ )1{t ≥τ } = (ψt − ψt ∧τ )1{t <τ } + (ψt − ψt ∧τ )1{t ≥τ } = ψt − ψt ∧τ . Lemma 2.2. (θt ∧τ , Mt , Px1 ) is a martingale. Proof. By (5), we only need to show that φt ∧τ is a martingale. Since Px1 is a solution to the martingale problem for L1 up to τ , we know (φt ∧τn , Mt , Px1 ) is a martingale. Therefore, for any 0 ≤ s < t,
E(φt ∧τn |Ms ) = φs∧τn . By assumption (3), let n → ∞ we obtain E(φt ∧τ |Ms ) = φ(s ∧ τ ). This proves that φt ∧τ is a martingale.
Lemma 2.3. (θt − θt ∧τ (ω) , Mt , Qω ) is a martingale for each ω ∈ Ω . Proof. Fix an arbitrary path ω0 ∈ Ω . We set t0 := τ (ω0 ) and Xt0 (ω0 ) = x0 . For any fixed instants 0 ≤ t1 ≤ t2 , we need to prove
EQω0 (θt2 − θt2 ∧t0 |Mt1 ) = θt1 − θt1 ∧t0 .
(9)
We assume t0 < ∞, otherwise Qω0 = δω0 and equality (9) is trivial. If t1 < t2 ≤ t0 , then it is cleat that (9) holds since both sides of (9) are zero. Assume t1 < t0 < t2 . Suppose that (9) holds for the case when t0 ≤ t1 < t2 . Then
EQω0 θt2 − θt2 ∧t0 |Mt1 = EQω0 EQω0 (θt2 − θt2 ∧t0 |Mt0 )|Mt1
= EQω0 (θt0 − θt0 ∧t0 |Mt1 ) = 0 = θt1 − θt1 ∧t0 . Hence we know (9) holds. In the remaining we assume t0 ≤ t1 < t2 and show (9). t ,x t ,x As P20 0 is a solution to the martingale problem for L2 , we know ψt is a P20 0 martingale starting from (t0 , x0 ). This implies t
that for any A ∈ Mt10 t ,x 0
P20
t ,x
(ψt2 − ψt0 , A) = P20 0 (ψt1 − ψt0 , A).
(10)
By (6), we only need to show that for any A ∈ Mt1 ,
Qω0 (ψt2 − ψt0 , A) = Qω0 (ψt1 − ψt0 , A).
(11)
It is enough to show (11) for A of the form: A = A1 ∩ A2 with A1 = {Xs1 ∈ Γs1 , . . . , Xsi ∈ Γsi },
A2 = {Xsi+1 ∈ Γsi+1 , . . . , Xsn ∈ Γsn },
where 0 ≤ s1 ≤ · · · ≤ si < t0 ≤ si+1 ≤ · · · ≤ sn ≤ t1 , and Γsj ∈ B (Rd ) for j = 1, 2, . . . , n. We just need to note that cylinder sets of the form A generates Mt1 and then use Dynkin’s π -λ theorem. If w0 6∈ A1 , then Qω0 (A1 ) = 0 and hence
Qω0 (ψt2 − ψt0 , A) = 0 = Qω0 (ψt1 − ψt0 , A).
S.-X. Ouyang / Statistics and Probability Letters 80 (2010) 1196–1199
1199 t
If w0 ∈ A1 , then Qω0 (A1 ) = 1 and hence Qω0 (A) = Qω0 (A2 ). Therefore, by the fact A2 ∈ Mt10 and (10), we have t ,x 0
Qω0 (ψt2 − ψt0 , A) = δω0 ⊗t0 P20
t ,x
(ψt2 − ψt0 , A2 ) = P20 0 (ψt2 − ψt0 , A2 )
t ,x
= P20 0 (ψt1 − ψt0 , A2 ) = Qω0 (ψt1 − ψt0 , A). This proves (11).
Proof of Theorem 1.1. We need to show that θt is a martingale on (Ω , M , Px1 ⊗τ Q). According to Stroock and Srinivasa Varadhan (1979, Theorem 6.1.2), it suffices to prove the following two statements. (a) (θt ∧τ , Mt , Px1 ) is a martingale. (b) (θt − θt ∧τ (ω) , Mt , Qω ) is a martingale for each ω ∈ Ω . But they are the conclusions of Lemmas 2.2 and 2.3 respectively.
3. Coupling We suppose Ω = C ([0, ∞), R2d ). Denote Zt (ω) = ωt = (Xt (ω), Yt (ω)) ∈ Rd × Rd for each ω ∈ Ω . Let σ (t , x) ∈ Sd be measurable real matrix defined on [0, ∞) × Rd . Let b(t , x), ξ (t , x, y) be Rd valued measurable functions defined on [0, ∞) × Rd and [0, ∞) × Rd × Rd respectively. Let c (t , x, y) be a d × d matrix valued measurable function defined on [0, ∞) × Rd × Rd . We assume that σ , b, c, and ξ all are locally bounded. Denote for every t ≥ 0, x, y ∈ Rd , a(t , x, y) =
σ (t , x)σ (t , x)∗ σ (t , y)σ (t , x)∗
σ (t , x)σ (t , y)∗ ∗ , σ (t , y)σ (t , y)
b(t , x, y) =
b(t , x) , b(t , y)
τ (ω),Z (τ (ω))
Let τ be a stopping time. For every ω ∈ Ω , let P2 be a solution to the martingale problem for L(a, b) starting from (τ (ω), Z (τ (ω))). Define Qω for each ω ∈ Ω as in Theorem 1.1. Applying Theorem 1.1, we see immediately the following corollary (see Chen and Li, 1989, Lemma 3.4). x ,y
Corollary 3.1. Let P1 be a solution to the martingale problem for L(a1 , b) with a1 (t , x, y) =
σ (t , x)σ (t , x)∗ c (t , x, y)∗
c (t , x, y) σ (t , y)σ (t , y)∗
x ,y
up to some stopping time τ . Then R = P1 ⊗τ Q is a solution to the martingale problem for L(a2 , b) with a2 (t , x, y) =
σ (t , x)σ (t , x)∗ ∗ c (t , x, y) 1[0,τ ) + σ (t , y)σ (t , x)∗ 1[τ ,∞)
c (t , x, y)1[0,τ ) + σ (t , x)σ (t , y)∗ 1[τ ,∞) σ (t , y)σ (t , y)∗
.
Remark 3.2. A typical use of this fact is the following. First we obtain successful coupling (the marginal processes meet each other) by choosing c (t , x, y) properly. Then the marginal processes will move together after the coupling time. Similarly, we can obtain coupling by choosing proper drift. x ,y
˜ 1 be a solution to the martingale problem for L(a, b1 ) with Corollary 3.3. Let P b1 (t , x, y) =
b(t , x) b(t , y) + ξ (t , x, y)
x ,y
˜ 1 ⊗τ Q is a solution to the martingale problem for L(a, b2 ) with up to some stopping time τ . Then R = P b2 (t , x, y) =
b( t , x) b(t , y) + ξ (t , x, y)1{τ
.
Acknowledgements This work was supported in part by the DFG through the International Graduate College ‘‘Stochastics and Real World Models’’ in Bielefeld University. The author thanks a referee for suggestions which make the paper more clear. References Chen, Mu-Fa, Li, Shao-Fu, 1989. Coupling methods for multidimensional diffusion processes. Ann. Probab. 17 (1), 151–177. MR MR972776 (90a:60134). Stroock, Daniel W., Srinivasa Varadhan, S.R., 1979. Multidimensional Diffusion Processes. In: Grundlehren der Mathematischen Wissenschaften (Fundamental Principles of Mathematical Sciences), vol. 233. Springer-Verlag, Berlin. MR532498 (81f:60108).