nms.kcl.ac.uk filenms.kcl.ac.uk

Couplings for SDEs driven by Levy processes and theirapplications

Mateusz Majka

University of Bonn

[email protected]

5th of May 2017, London

Mateusz Majka (Bonn) Couplings for Levy-driven SDEs 5th of May 2017, London 1 / 46

Levy-driven SDE

dXt = b(Xt)dt + dLt

(Lt)t≥0 pure jump Levy process in Rd

rotationally invariant and absolutely continuous Levy measure

e.g. E exp(i〈z , Lt〉) = exp(−t|z |α), α ∈ (0, 2)

b : Rd → Rd

continuous

∃C ∈ R ∀x , y ∈ Rd : 〈b(x)− b(y), x − y〉 ≤ C |x − y |2(one-sided Lipschitz condition)


Levy-driven SDE

dXt = b(Xt)dt + dLt




b : Rd → Rd

continuous



Levy-driven SDE

dXt = b(Xt)dt + dLt




b : Rd → Rd

continuous



Levy-driven SDE

dXt = b(Xt)dt + dLt




b : Rd → Rd

continuous



Levy-driven SDE

dXt = b(Xt)dt + dLt




b : Rd → Rd

continuous



Levy-driven SDE

dXt = b(Xt)dt + dLt




b : Rd → Rd

continuous



Levy-driven SDE

dXt = b(Xt)dt + dLt




b : Rd → Rd

continuous



Dissipativity

Global dissipativity condition

There exists K > 0 such that for all x , y ∈ Rd we have〈b(x)− b(y), x − y〉 ≤ −K |x − y |2

Dissipativity at infinity condition

There exist K > 0 and R > 0 such that for all x , y ∈ Rd with |x − y | > Rwe have〈b(x)− b(y), x − y〉 ≤ −K |x − y |2


Dissipativity

Global dissipativity condition

There exists K > 0 such that for all x , y ∈ Rd we have〈b(x)− b(y), x − y〉 ≤ −K |x − y |2

Dissipativity at infinity condition

There exist K > 0 and R > 0 such that for all x , y ∈ Rd with |x − y | > Rwe have〈b(x)− b(y), x − y〉 ≤ −K |x − y |2


Examples

b(x) = −∇U(x)

x

U(x) = M|x |2

global dissipativity

x

U(x) = M(|x |2 − 1)2

dissipativity at infinity


Examples

b(x) = −∇U(x)

x

U(x) = M|x |2

global dissipativity

x

U(x) = M(|x |2 − 1)2

dissipativity at infinity


Wasserstein distances

Definition

For p ≥ 1, we define the Lp-Wasserstein distance between two probabilitymeasures µ and ν on Rd by the formula

Wp(µ, ν) :=

(inf

π∈Π(µ,ν)

∫Rd

ρ(x , y)pdπ(x , y)

) 1p

,

where ρ is a metric on Rd and Π(µ, ν) is the family of all couplings of µand ν, i.e., π ∈ Π(µ, ν) if and only if π is a measure on R2d having µ andν as its marginals.

p = 1, ρ(x , y) = f (|x − y |) where f : [0,∞)→ [0,∞) is concave,f (0) = 0 and f (x) > 0 for x > 0. Then

Wf (µ, ν) := infπ∈Π(µ,ν)

∫Rd

f (|x − y |)dπ(x , y) .



Definition


Wp(µ, ν) :=

(inf

π∈Π(µ,ν)

∫Rd


) 1p

,


p = 1, ρ(x , y) = f (|x − y |) where f : [0,∞)→ [0,∞) is concave,f (0) = 0 and f (x) > 0 for x > 0.

Then


∫Rd

f (|x − y |)dπ(x , y) .



Definition


Wp(µ, ν) :=

(inf

π∈Π(µ,ν)

∫Rd


) 1p

,


p = 1, ρ(x , y) = f (|x − y |) where f : [0,∞)→ [0,∞) is concave,f (0) = 0 and f (x) > 0 for x > 0. Then


∫Rd

f (|x − y |)dπ(x , y) .


Special cases


∫Rd

f (|x − y |)dπ(x , y) .

By taking f (x) = 1(0,∞)(x) we get Wf (µ, ν) = 12‖µ− ν‖TV .

By taking f (x) = x we get Wf (µ, ν) = W1(µ, ν).


Special cases


∫Rd

f (|x − y |)dπ(x , y) .

By taking f (x) = 1(0,∞)(x) we get Wf (µ, ν) = 12‖µ− ν‖TV .

By taking f (x) = x we get Wf (µ, ν) = W1(µ, ν).


Known results

(Xt)t≥0 Markov process with transition semigroup (pt)t≥0

If X0 ∼ µ, then for any t > 0 we have Xt ∼ µpt .Global dissipativity + any noise=⇒ Wp(µ1pt , µ2pt) ≤ e−c2tWp(µ1, µ2) for any p ∈ [1, 2]

Dissipativity at infinity + some non-degeneracy conditions on thenoise [Kulik, SPA 2009] =⇒ ‖µpt − µ0‖TV ≤ Ce−c2t

Drift of the form b(x) = Ax + F (x), where A is a matrix witheigenvalues γ1, . . . , γn such that maxk Re(γk) < 0 and F is bounded,Holder continuous + symmetric α-stable noise, α ∈ (1, 2), [Priola,Shirikyan, Xu, Zabczyk, SPA 2012] =⇒ ‖µpt − µ0‖TV ≤ Ce−c2t

Dissipativity at infinity + symmetric α-stable component of the noise[J. Wang, Bernoulli 2016] =⇒ Wp(µpt , µ0) ≤ Ce−c2t for anyp ∈ [1, 2]


Known results


If X0 ∼ µ, then for any t > 0 we have Xt ∼ µpt .

Global dissipativity + any noise=⇒ Wp(µ1pt , µ2pt) ≤ e−c2tWp(µ1, µ2) for any p ∈ [1, 2]





Known results







Known results







Known results







Known results







Couplings

Definition

If (Xt)t≥0 is a Markov process in Rd , we call the R2d -valued process(X ′t ,X

′′t )t≥0 a coupling of (Xt)t≥0 if both the marginal processes (X ′t)t≥0

and (X ′′t )t≥0 have the same transition probabilities as (Xt)t≥0, butpossibly different initial distributions.


x

x

yyy


xx

yyy


xx

y

yy


xx

y

yy


xx

yy

y


xx

yy

y


xx

y

y

y


Let (Xt ,Yt)t≥0 be a coupling with X0 ∼ µ and Y0 ∼ ν.

Then for any t ≥ 0 the joint law πt of Xt and Yt is a coupling of the lawsof Xt and Yt (i.e., a coupling of µpt and νpt)and therefore

infπ∈Π(µpt ,νpt)

∫Rd

f (|x−y |)dπ(x , y) ≤∫Rd

f (|x−y |)dπt(x , y) = Ef (|Xt−Yt |)

i.e.,Wf (µpt , νpt) ≤ Ef (|Xt − Yt |) .


Let (Xt ,Yt)t≥0 be a coupling with X0 ∼ µ and Y0 ∼ ν.Then for any t ≥ 0 the joint law πt of Xt and Yt is a coupling of the lawsof Xt and Yt (i.e., a coupling of µpt and νpt)

and therefore


∫Rd

f (|x−y |)dπ(x , y) ≤∫Rd

f (|x−y |)dπt(x , y) = Ef (|Xt−Yt |)



Let (Xt ,Yt)t≥0 be a coupling with X0 ∼ µ and Y0 ∼ ν.Then for any t ≥ 0 the joint law πt of Xt and Yt is a coupling of the lawsof Xt and Yt (i.e., a coupling of µpt and νpt)and therefore


∫Rd

f (|x−y |)dπ(x , y) ≤∫Rd

f (|x−y |)dπt(x , y) = Ef (|Xt−Yt |)



Let (Xt ,Yt)t≥0 be a coupling with X0 ∼ µ and Y0 ∼ ν.Then for any t ≥ 0 the joint law πt of Xt and Yt is a coupling of the lawsof Xt and Yt (i.e., a coupling of µpt and νpt)and therefore


∫Rd

f (|x−y |)dπ(x , y) ≤∫Rd

f (|x−y |)dπt(x , y) = Ef (|Xt−Yt |)



Example (coupling by reflection for diffusions)

dXt = b(Xt)dt + dBt

where b : Rd → Rd is locally Lipschitz and (Bt)t≥0 is a Brownian motionin Rd ;

dYt = b(Yt) + R(Xt ,Yt)dBt ,where

R(Xt ,Yt) = I − 2eteTt with et = (Xt − Yt)/|Xt − Yt | .

then (Xt ,Yt)t≥0 form a coupling [Lindvall, Rogers 1986].


Results in the diffusion case

Using the coupling by reflection for diffusions, we can construct acontinuous, concave function f and a constant c > 0 such that for thesemigroup (pt)t≥0 associated with the solution to the equation

dXt = b(Xt)dt + dBt ,

satisfying the dissipativity at infinity condition, we have

Wf (µpt , νpt) ≤ e−ctWf (µ, ν) .

Since C−1x ≤ f (x) ≤ x , we get

W1(µpt , νpt) ≤ Ce−ctW1(µ, ν) .


Results in the diffusion case

Using the coupling by reflection for diffusions, we can construct acontinuous, concave function f and a constant c > 0 such that for thesemigroup (pt)t≥0 associated with the solution to the equation

dXt = b(Xt)dt + dBt ,

satisfying the dissipativity at infinity condition, we have


Since C−1x ≤ f (x) ≤ x , we get

W1(µpt , νpt) ≤ Ce−ctW1(µ, ν) .


The function f is extended in an affine way from some point R1 > 0

f (R1)

0

f

R1

C−1xC−1x

x

C−1x

x

C−1x

x

C−1x

x

Thus from

C−1W1(µpt , νpt) ≤Wf (µpt , νpt) ≤ e−ctWf (µ, ν) ≤ e−ctW1(µ, ν)

we getW1(µpt , νpt) ≤ Ce−ctW1(µ, ν) .

See [Eberle 2016, PTRF] for details.



f (R1)

0

f

R1

C−1x

C−1x

x

C−1x

x

C−1x

x

C−1x

x

Thus from






f (R1)

0

f

R1

C−1x

C−1x

x

C−1x

x

C−1x

x

C−1x

x

Thus from






f (R1)

0

f

R1

C−1xC−1x

x

C−1x

x

C−1x

x

C−1x

x

Thus from






f (R1)

0

f

R1

C−1xC−1x

x

C−1x

x

C−1x

x

C−1x

x

Thus from






f (R1)

0

f

R1

C−1xC−1x

x

C−1x

x

C−1x

x

C−1x

x

Thus from



See [Eberle 2016, PTRF] for details.Mateusz Majka (Bonn) Couplings for Levy-driven SDEs 5th of May 2017, London 13 / 46

x

y


x

y


x

y


x

y


x

y


x

y


x

y


x

y


x

y


Optimal transport

Monge problem: Given a measure µ on S and a cost functionc : S × S → [0,∞], find a map T : S → S minimizing∫

c(x ,T (x))µ(dx) .

Kantorovich problem: Given two measures µ and ν on S and a costfunction c : S × S → [0,∞], find a coupling γ ∈ Π(µ, ν) minimizing∫

c(x , y)γ(dx dy) .


Optimal transport


c(x ,T (x))µ(dx) .


c(x , y)γ(dx dy) .


Optimal transport


c(x ,T (x))µ(dx) .


c(x , y)γ(dx dy) .


Coupling by antimonotone rearrangement [McCann ’1999]

µ

ν

ν = µ T−1



µ ν

ν = µ T−1



µ ν

ν = µ T−1



µ ν

ν = µ T−1



µ ν

ν = µ T−1



µ ν

ν = µ T−1


Coupling by reflection

µ µ

µ = µ T−1


Mirror coupling [Hsu, Sturm ’2013]

Assume we have symmetric transition densities (p(x , ·))x∈R.How to couple p(x1, ·) and p(x2, ·) for x1 6= x2?For two symmetric transition densities p(x1, ·), p(x2, ·) on R, the mirrorcoupling m(x1, x2, ·) is the optimal coupling for all concave costs φ, i.e.,for any other coupling γ(x1, x2, ·) of p(x1, ·) and p(x2, ·) we have∫

R2

φ(|x − y |)γ(dx dy) ≥∫R2

φ(|x − y |)m(dx dy) .

Consider an R-valued random variable ζ1 with the density p(x1, ·).We want to construct a coupling (ζ1, ζ2) by defining the random variableζ2 in an appropriate way (so that it has the density p(x2, ·)).



Assume we have symmetric transition densities (p(x , ·))x∈R.

How to couple p(x1, ·) and p(x2, ·) for x1 6= x2?For two symmetric transition densities p(x1, ·), p(x2, ·) on R, the mirrorcoupling m(x1, x2, ·) is the optimal coupling for all concave costs φ, i.e.,for any other coupling γ(x1, x2, ·) of p(x1, ·) and p(x2, ·) we have∫

R2

φ(|x − y |)γ(dx dy) ≥∫R2

φ(|x − y |)m(dx dy) .




Assume we have symmetric transition densities (p(x , ·))x∈R.How to couple p(x1, ·) and p(x2, ·) for x1 6= x2?

For two symmetric transition densities p(x1, ·), p(x2, ·) on R, the mirrorcoupling m(x1, x2, ·) is the optimal coupling for all concave costs φ, i.e.,for any other coupling γ(x1, x2, ·) of p(x1, ·) and p(x2, ·) we have∫

R2

φ(|x − y |)γ(dx dy) ≥∫R2

φ(|x − y |)m(dx dy) .





R2

φ(|x − y |)γ(dx dy) ≥∫R2

φ(|x − y |)m(dx dy) .





R2

φ(|x − y |)γ(dx dy) ≥∫R2

φ(|x − y |)m(dx dy) .

Consider an R-valued random variable ζ1 with the density p(x1, ·).

We want to construct a coupling (ζ1, ζ2) by defining the random variableζ2 in an appropriate way (so that it has the density p(x2, ·)).




R2

φ(|x − y |)γ(dx dy) ≥∫R2

φ(|x − y |)m(dx dy) .



x1 x2

p(x1, ·) p(x2, ·)

z1z1

p(x2, z1) = 0

z1

p(x2, z1) = 0

z1z1

p(x2, z1)

p(x1, z1)

z1

p(x2, z1)

p(x1, z1)

p(x2, z1)/p(x1, z1) ∈ (0, 1)

z1z1

p(x2, z1) > p(x1, z1)

z1

p(x2, z1) > p(x1, z1)

P(ζ2 = ζ1|ζ1 = z1) =p(x1, z1) ∧ p(x2, z1)

p(x1, z1)

P(ζ2 = x1 + x2 − ζ1|ζ1 = z1) = 1− p(x1, z1) ∧ p(x2, z1)

p(x1, z1)


x1 x2

p(x1, ·) p(x2, ·)

z1

z1

p(x2, z1) = 0

z1

p(x2, z1) = 0

z1z1

p(x2, z1)

p(x1, z1)

z1

p(x2, z1)

p(x1, z1)

p(x2, z1)/p(x1, z1) ∈ (0, 1)

z1z1

p(x2, z1) > p(x1, z1)

z1

p(x2, z1) > p(x1, z1)

P(ζ2 = ζ1|ζ1 = z1) =p(x1, z1) ∧ p(x2, z1)

p(x1, z1)

P(ζ2 = x1 + x2 − ζ1|ζ1 = z1) = 1− p(x1, z1) ∧ p(x2, z1)

p(x1, z1)


x1 x2

p(x1, ·) p(x2, ·)

z1

z1

p(x2, z1) = 0

z1

p(x2, z1) = 0

z1z1

p(x2, z1)

p(x1, z1)

z1

p(x2, z1)

p(x1, z1)

p(x2, z1)/p(x1, z1) ∈ (0, 1)

z1z1

p(x2, z1) > p(x1, z1)

z1

p(x2, z1) > p(x1, z1)

P(ζ2 = ζ1|ζ1 = z1) =p(x1, z1) ∧ p(x2, z1)

p(x1, z1)

P(ζ2 = x1 + x2 − ζ1|ζ1 = z1) = 1− p(x1, z1) ∧ p(x2, z1)

p(x1, z1)


x1 x2

p(x1, ·) p(x2, ·)

z1z1

p(x2, z1) = 0

z1

p(x2, z1) = 0

z1z1

p(x2, z1)

p(x1, z1)

z1

p(x2, z1)

p(x1, z1)

p(x2, z1)/p(x1, z1) ∈ (0, 1)

z1z1

p(x2, z1) > p(x1, z1)

z1

p(x2, z1) > p(x1, z1)

P(ζ2 = ζ1|ζ1 = z1) =p(x1, z1) ∧ p(x2, z1)

p(x1, z1)

P(ζ2 = x1 + x2 − ζ1|ζ1 = z1) = 1− p(x1, z1) ∧ p(x2, z1)

p(x1, z1)


x1 x2

p(x1, ·) p(x2, ·)

z1z1

p(x2, z1) = 0

z1

p(x2, z1) = 0

z1

z1

p(x2, z1)

p(x1, z1)

z1

p(x2, z1)

p(x1, z1)

p(x2, z1)/p(x1, z1) ∈ (0, 1)

z1z1

p(x2, z1) > p(x1, z1)

z1

p(x2, z1) > p(x1, z1)

P(ζ2 = ζ1|ζ1 = z1) =p(x1, z1) ∧ p(x2, z1)

p(x1, z1)

P(ζ2 = x1 + x2 − ζ1|ζ1 = z1) = 1− p(x1, z1) ∧ p(x2, z1)

p(x1, z1)


x1 x2

p(x1, ·) p(x2, ·)

z1z1

p(x2, z1) = 0

z1

p(x2, z1) = 0

z1

z1

p(x2, z1)

p(x1, z1)

z1

p(x2, z1)

p(x1, z1)

p(x2, z1)/p(x1, z1) ∈ (0, 1)

z1z1

p(x2, z1) > p(x1, z1)

z1

p(x2, z1) > p(x1, z1)

P(ζ2 = ζ1|ζ1 = z1) =p(x1, z1) ∧ p(x2, z1)

p(x1, z1)

P(ζ2 = x1 + x2 − ζ1|ζ1 = z1) = 1− p(x1, z1) ∧ p(x2, z1)

p(x1, z1)


x1 x2

p(x1, ·) p(x2, ·)

z1z1

p(x2, z1) = 0

z1

p(x2, z1) = 0

z1z1

p(x2, z1)

p(x1, z1)

z1

p(x2, z1)

p(x1, z1)

p(x2, z1)/p(x1, z1) ∈ (0, 1)

z1z1

p(x2, z1) > p(x1, z1)

z1

p(x2, z1) > p(x1, z1)

P(ζ2 = ζ1|ζ1 = z1) =p(x1, z1) ∧ p(x2, z1)

p(x1, z1)

P(ζ2 = x1 + x2 − ζ1|ζ1 = z1) = 1− p(x1, z1) ∧ p(x2, z1)

p(x1, z1)


x1 x2

p(x1, ·) p(x2, ·)

z1z1

p(x2, z1) = 0

z1

p(x2, z1) = 0

z1z1

p(x2, z1)

p(x1, z1)

z1

p(x2, z1)

p(x1, z1)

p(x2, z1)/p(x1, z1) ∈ (0, 1)

z1

z1

p(x2, z1) > p(x1, z1)

z1

p(x2, z1) > p(x1, z1)

P(ζ2 = ζ1|ζ1 = z1) =p(x1, z1) ∧ p(x2, z1)

p(x1, z1)

P(ζ2 = x1 + x2 − ζ1|ζ1 = z1) = 1− p(x1, z1) ∧ p(x2, z1)

p(x1, z1)


x1 x2

p(x1, ·) p(x2, ·)

z1z1

p(x2, z1) = 0

z1

p(x2, z1) = 0

z1z1

p(x2, z1)

p(x1, z1)

z1

p(x2, z1)

p(x1, z1)

p(x2, z1)/p(x1, z1) ∈ (0, 1)

z1

z1

p(x2, z1) > p(x1, z1)

z1

p(x2, z1) > p(x1, z1)

P(ζ2 = ζ1|ζ1 = z1) =p(x1, z1) ∧ p(x2, z1)

p(x1, z1)

P(ζ2 = x1 + x2 − ζ1|ζ1 = z1) = 1− p(x1, z1) ∧ p(x2, z1)

p(x1, z1)


x1 x2

p(x1, ·) p(x2, ·)

z1z1

p(x2, z1) = 0

z1

p(x2, z1) = 0

z1z1

p(x2, z1)

p(x1, z1)

z1

p(x2, z1)

p(x1, z1)

p(x2, z1)/p(x1, z1) ∈ (0, 1)

z1z1

p(x2, z1) > p(x1, z1)

z1

p(x2, z1) > p(x1, z1)

P(ζ2 = ζ1|ζ1 = z1) =p(x1, z1) ∧ p(x2, z1)

p(x1, z1)

P(ζ2 = x1 + x2 − ζ1|ζ1 = z1) = 1− p(x1, z1) ∧ p(x2, z1)

p(x1, z1)


The coupling m(x1, x2, ·) (as a measure on R2 with marginals p(x1, ·) andp(x2, ·)) is given by

m(x1, x2, dy1, dy2) = δy1(dy2) (p(x1, y1) ∧ p(x2, y1)) dy1

+ δRy1(dy2) (p(x1, y1)− p(x1, y1) ∧ p(x2, y1)) dy1

whereRy1 = x1 + x2 − y1 .

It is crucial that the densities (p(x , ·))x∈R are symmetric, in particular wehave

p(x1, y1) = p(x2,Ry1) .


The coupling m(x1, x2, ·) (as a measure on R2 with marginals p(x1, ·) andp(x2, ·)) is given by

m(x1, x2, dy1, dy2) = δy1(dy2) (p(x1, y1) ∧ p(x2, y1)) dy1

+ δRy1(dy2) (p(x1, y1)− p(x1, y1) ∧ p(x2, y1)) dy1

whereRy1 = x1 + x2 − y1 .

It is crucial that the densities (p(x , ·))x∈R are symmetric, in particular wehave

p(x1, y1) = p(x2,Ry1) .


Coupling construction in the Levy case

By the Levy-Ito decomposition,

Lt =

∫ t

0

∫|v |>1

vN(dv , ds) +

∫ t

0

∫|v |≤1

vN(dv , ds)

where N(dv , dt) = N(dv , dt)− ν(dv)dt is the compensated Poissonrandom measure and ν is the Levy (jump) measure of (Lt)t≥0.

We have

N([0, t],A)(ω) =∞∑j=1

δ(τj (ω),ξj (ω))([0, t]× A) for all ω ∈ Ω and A ∈ B(Rd)

We can embed N in R+ × Rd × [0, 1] by replacing each ξj with (ξj , ηj)where ηj is a uniformly distributed random variable on [0, 1]. We will write

Lt =

∫ t

0

∫|v |>1×[0,1]

vN(dv , du, ds) +

∫ t

0

∫|v |≤1×[0,1]

vN(dv , du, ds) .




Lt =

∫ t

0

∫|v |>1

vN(dv , ds) +

∫ t

0

∫|v |≤1

vN(dv , ds)

where N(dv , dt) = N(dv , dt)− ν(dv)dt is the compensated Poissonrandom measure and ν is the Levy (jump) measure of (Lt)t≥0. We have

N([0, t],A)(ω) =∞∑j=1



Lt =

∫ t

0

∫|v |>1×[0,1]

vN(dv , du, ds) +

∫ t

0

∫|v |≤1×[0,1]

vN(dv , du, ds) .




Lt =

∫ t

0

∫|v |>1

vN(dv , ds) +

∫ t

0

∫|v |≤1

vN(dv , ds)


N([0, t],A)(ω) =∞∑j=1


We can embed N in R+ × Rd × [0, 1] by replacing each ξj with (ξj , ηj)where ηj is a uniformly distributed random variable on [0, 1].

We will write

Lt =

∫ t

0

∫|v |>1×[0,1]

vN(dv , du, ds) +

∫ t

0

∫|v |≤1×[0,1]

vN(dv , du, ds) .




Lt =

∫ t

0

∫|v |>1

vN(dv , ds) +

∫ t

0

∫|v |≤1

vN(dv , ds)


N([0, t],A)(ω) =∞∑j=1



Lt =

∫ t

0

∫|v |>1×[0,1]

vN(dv , du, ds) +

∫ t

0

∫|v |≤1×[0,1]

vN(dv , du, ds) .


Xt− Yt−

Xt

v

Xt

v

YtXt

v

Yt−

Xt

v

Yt

u ≥ ρ(v ,Xt− − Yt−)

Xt

v

Yt−

u < ρ(v ,Xt− − Yt−)


Xt− Yt−

Xt

v

Xt

v

YtXt

v

Yt−

Xt

v

Yt

u ≥ ρ(v ,Xt− − Yt−)

Xt

v

Yt−

u < ρ(v ,Xt− − Yt−)


Xt− Yt−

Xt

v

Xt

v

Yt

Xt

v

Yt−

Xt

v

Yt

u ≥ ρ(v ,Xt− − Yt−)

Xt

v

Yt−

u < ρ(v ,Xt− − Yt−)


Xt− Yt−

Xt

v

Xt

v

Yt

Xt

v

Yt−

Xt

v

Yt

u ≥ ρ(v ,Xt− − Yt−)

Xt

v

Yt−

u < ρ(v ,Xt− − Yt−)


Xt− Yt−

Xt

v

Xt

v

YtXt

v

Yt−

Xt

v

Yt

u ≥ ρ(v ,Xt− − Yt−)

Xt

v

Yt−

u < ρ(v ,Xt− − Yt−)


Xt− Yt−

Xt

v

Xt

v

YtXt

v

Yt−

Xt

v

Yt

u ≥ ρ(v ,Xt− − Yt−)

Xt

v

Yt−

u < ρ(v ,Xt− − Yt−)


ν(dv) = q(v)dv

Xt− Yt−

q(· − Xt−) q(· − Yt−)

v

q(v + Xt− − Yt−)

q(v) = q(v + Xt− − Xt−)

ρ(v ,Xt− − Yt−) =q(v + Xt− − Yt−) ∧ q(v)

q(v)


ν(dv) = q(v)dv

Xt− Yt−

q(· − Xt−) q(· − Yt−)

v

q(v + Xt− − Yt−)

q(v) = q(v + Xt− − Xt−)

ρ(v ,Xt− − Yt−) =q(v + Xt− − Yt−) ∧ q(v)

q(v)


ν(dv) = q(v)dv

Xt− Yt−

q(· − Xt−) q(· − Yt−)

v

q(v + Xt− − Yt−)

q(v) = q(v + Xt− − Xt−)

ρ(v ,Xt− − Yt−) =q(v + Xt− − Yt−) ∧ q(v)

q(v)


We can write the equation

dXt = b(Xt)dt + dLt

as

dXt = b(Xt)dt +

∫Rd×[0,1]

vN(dv , du, dt)

and we can construct the process (Yt)t≥0 by setting

dYt = b(Yt)dt +

∫Rd×[0,1]

(Xt− − Yt− + v)1u<ρ(v ,Xt−−Yt−)N(dt, dv , du)

+

∫Rd×[0,1]

R(Xt−,Yt−)v1u≥ρ(v ,Xt−−Yt−)N(dt, dv , du)

where R(Xt ,Yt) = I − 2eteTt with et = (Xt − Yt)/|Xt − Yt | and ρ is a

function with values in [0, 1].


We can write the equation

dXt = b(Xt)dt + dLt

as

dXt = b(Xt)dt +

∫Rd×[0,1]

vN(dv , du, dt)

and we can construct the process (Yt)t≥0 by setting

dYt = b(Yt)dt +

∫Rd×[0,1]


+

∫Rd×[0,1]


where R(Xt ,Yt) = I − 2eteTt with et = (Xt − Yt)/|Xt − Yt | and ρ is a

function with values in [0, 1].


The system of equations:

dXt = b(Xt)dt +

∫Rd×[0,1]

vN(dv , du, dt)

dYt = b(Yt)dt +

∫Rd×[0,1]


+

∫Rd×[0,1]


has a solution

the solution is a coupling

Then we look for a concave function f such that

Ef (|Xt − Yt |) ≤ Ef (|X0 − Y0|)− c

∫ t

0f (|Xs − Ys |)ds

and thus

Wf (µ1pt , µ2pt) ≤ Ef (|Xt − Yt |) ≤ Ef (|X0 − Y0|)e−ct



dXt = b(Xt)dt +

∫Rd×[0,1]

vN(dv , du, dt)

dYt = b(Yt)dt +

∫Rd×[0,1]


+

∫Rd×[0,1]


has a solution



Ef (|Xt − Yt |) ≤ Ef (|X0 − Y0|)− c

∫ t

0f (|Xs − Ys |)ds

and thus




dXt = b(Xt)dt +

∫Rd×[0,1]

vN(dv , du, dt)

dYt = b(Yt)dt +

∫Rd×[0,1]


+

∫Rd×[0,1]


has a solution



Ef (|Xt − Yt |) ≤ Ef (|X0 − Y0|)− c

∫ t

0f (|Xs − Ys |)ds

and thus




dXt = b(Xt)dt +

∫Rd×[0,1]

vN(dv , du, dt)

dYt = b(Yt)dt +

∫Rd×[0,1]


+

∫Rd×[0,1]


has a solution



Ef (|Xt − Yt |) ≤ Ef (|X0 − Y0|)− c

∫ t

0f (|Xs − Ys |)ds

and thus




dXt = b(Xt)dt +

∫Rd×[0,1]

vN(dv , du, dt)

dYt = b(Yt)dt +

∫Rd×[0,1]


+

∫Rd×[0,1]


has a solution



Ef (|Xt − Yt |) ≤ Ef (|X0 − Y0|)− c

∫ t

0f (|Xs − Ys |)ds

and thus



f = a1(0,∞) + f1

where a is a positive constant and f1 is defined as a concave, strictlyincreasing C 2 function up to some R1 > 0 and from f (R1) is extended inan affine way.

f (R1)

a

0

f1

R1


f = a1(0,∞) + f1

where a is a positive constant and f1 is defined as a concave, strictlyincreasing C 2 function up to some R1 > 0 and from f (R1) is extended inan affine way.

f (R1)

a

0

f1

R1


Then1(0,∞)(x) ≤ a−1f (x) and x ≤ Cf (x)

for some C > 0 and for all x ≥ 0.

f (R1)

a

0

f1

f (x)

a1(0,∞)(x)C−1x

R1

W1(µ1pt , µ2pt) ≤ C (µ1, µ2)e−ct

‖µ1pt − µ2pt‖TV ≤ C (µ1, µ2)e−ct




f (R1)

a

0

f1

f (x)

a1(0,∞)(x)C−1x

R1






f (R1)

a

0

f1

f (x)

a1(0,∞)(x)

C−1x

R1






f (R1)

a

0

f1

f (x)

a1(0,∞)(x)

C−1x

R1




Assumption 1

The Levy measure ν is rotationally invariant, i.e.

ν(AB) = ν(B)

for any Borel set B ∈ B(Rd) and any d × d orthogonal matrix A.

Assumption 2

ν is absolutely continuous with respect to the Lebesgue measure on Rd ,i.e.,

ν(dv) = q(v)dv .

Assumption 3

There exist constants m, δ > 0 such that δ < 2m and

infx∈Rd :0<|x |≤δ

∫|v |≤m∩|v+x |≤m

q(v) ∧ q(v + x)dv > 0 .


Assumption 4

There exists a constant ε > 0 such that ε ≤ δ and∫|v |≤ε/2

q(v)dv > 0 .

Then we obtain



Assumption 5

ε/(∫ ε

0|y |2ν1(dy)

)bounded as ε→ 0 ,

where ν1 is the first marginal of the rotationally invariant measure ν.

W1(µ1pt , µ2pt) ≤ Ce−ctW1(µ1, µ2) .


Assumption 4


q(v)dv > 0 .

Then we obtain



Assumption 5

ε/(∫ ε

0|y |2ν1(dy)





Assumption 4


q(v)dv > 0 .

Then we obtain



Assumption 5

ε/(∫ ε

0|y |2ν1(dy)





Assumption 4


q(v)dv > 0 .

Then we obtain



Assumption 5

ε/(∫ ε

0|y |2ν1(dy)





Some extensions

Consider the equation

dXt = b(Xt)dt + dB1t + σ(Xt)dB

2t + dLt +

∫Ug(Xt−, u)N(dt, du) ,

where all the sources of noise are independent and there exist constants K ,R > 0 such that

〈b(x)− b(y), x − y〉+ ‖σ(x)− σ(y)‖2HS +

∫U|g(x , u)− g(y , u)|2ν(du)

≤ −K |x − y |2 ,

for all x , y ∈ Rd with |x − y | > R.



2t + dLt +


We apply the coupling from [M. ’2015] to (Lt)t≥0, by using the operatorM(·, ·), reflection coupling from [Lindvall, Rogers ’1986] to (B1

t )t≥0 andsynchronous coupling to the other two noises. Hence we have

dYt = b(Yt)dt + R(Xt ,Yt)dB1t + σ(Yt)dB

2t

+ M(Xt−,Yt−)dLt +

∫Ug(Yt−, u)N(dt, du) .

This allows us to get




2t + dLt +


We apply the coupling from [M. ’2015] to (Lt)t≥0, by using the operatorM(·, ·), reflection coupling from [Lindvall, Rogers ’1986] to (B1

t )t≥0 andsynchronous coupling to the other two noises. Hence we have

dYt = b(Yt)dt + R(Xt ,Yt)dB1t + σ(Yt)dB

2t

+ M(Xt−,Yt−)dLt +

∫Ug(Yt−, u)N(dt, du) .

This allows us to get



Transportation inequalities

Definition

We define the relative entropy (Kullback-Leibler information) of µ1 withrespect to µ2 by

H(µ1|µ2) :=

∫log dµ1

dµ2dµ1 if µ1 µ2 ,

+∞ otherwise .

Definition

Let α be a non-decreasing, left continuous function on R+ with α(0) = 0.We say that a probability measure µ satisfies a W1H-inequality withdeviation function α (or simply α-W1H inequality) if for any probabilitymeasure η we have

α(W1(η, µ)) ≤ H(η|µ) .


Theorem [Gozlan-Leonard, PTRF 2007]

Let α : R+ → R+ be a deviation function like above. The followingproperties are equivalent:

1 the α-W1H inequality holds for the probability measure µ,

2 for every f : Rd → R bounded and Lipschitz with ‖f ‖Lip ≤ 1 we have∫eλ(f−µ(f ))dµ ≤ eα

∗(λ) for any λ > 0 ,

where α∗(λ) := supr≥0(rλ− α(r)) is the semi-Legendretransformation (convex conjugate) of α,

3 if (ξk)k≥1 is a sequence of i.i.d Rd -valued random variables withcommon law µ, then for every f : Rd → R bounded and Lipschitzwith ‖f ‖Lip ≤ 1 we have

P

(1

n

n∑k=1

f (ξk)− µ(f ) > r

)≤ e−nα(r) for any r > 0, n ≥ 1 .



dXt = b(Xt)dt + σ(Xt)dWt +

∫Ug(Xt−, u)N(dt, du) .

Assume there is a function g∞ on U s.t. |g(x , u)| ≤ g∞(u) for all x ∈ Rd ,u ∈ U and there is a constant σ∞ s.t. ‖σ(x)‖HS ≤ σ∞ for all x ∈ Rd .Assume further that there exists λ > 0 such that

β(λ) :=

∫U

(eλg∞(u) − λg∞(u)− 1)ν(du) <∞

where ν is the intensity measure associated with N.Finally, suppose that there exists K > 0 s.t. for all x , y ∈ Rd

〈b(x)− b(y), x − y〉+ ‖σ(x)− σ(y)‖2HS +

∫U|g(x , u)− g(y , u)|2ν(du)

≤ −K |x − y |2





Assume there is a function g∞ on U s.t. |g(x , u)| ≤ g∞(u) for all x ∈ Rd ,u ∈ U and there is a constant σ∞ s.t. ‖σ(x)‖HS ≤ σ∞ for all x ∈ Rd .

Assume further that there exists λ > 0 such that

β(λ) :=

∫U

(eλg∞(u) − λg∞(u)− 1)ν(du) <∞


〈b(x)− b(y), x − y〉+ ‖σ(x)− σ(y)‖2HS +

∫U|g(x , u)− g(y , u)|2ν(du)

≤ −K |x − y |2






β(λ) :=

∫U

(eλg∞(u) − λg∞(u)− 1)ν(du) <∞

where ν is the intensity measure associated with N.

Finally, suppose that there exists K > 0 s.t. for all x , y ∈ Rd

〈b(x)− b(y), x − y〉+ ‖σ(x)− σ(y)‖2HS +

∫U|g(x , u)− g(y , u)|2ν(du)

≤ −K |x − y |2






β(λ) :=

∫U

(eλg∞(u) − λg∞(u)− 1)ν(du) <∞


〈b(x)− b(y), x − y〉+ ‖σ(x)− σ(y)‖2HS +

∫U|g(x , u)− g(y , u)|2ν(du)

≤ −K |x − y |2


Theorem [Wu, AIHP 2010], [Ma, SPA 2010]

Under the above assumptions for any T > 0 and any x ∈ Rd we have theW1H transportation inequality with deviation function αT for the measureδxpT , which is the law of the random variable XT (x), where (Xt(x))t≥0 isa solution to


∫Ug(Xt−, u)N(dt, du)

starting from x , i.e., we have

αT (W1(η, δxpT )) ≤ H(η|δxpT )

for any probability measure η on Rd .


Malliavin calculus

Consider a probability space (Ω,F , (Ft)t≥0,P) with a Brownian motion(Wt)t≥0 and a Poisson random measure N. For smooth random variablesF we have∫ t

0〈∇sF , hs〉ds = lim

ε→0

1

ε

(F (W· + ε

∫ ·0hsds)− F (W·)

)where the process (∇tF )t≥0 is the Malliavin derivative of F with respectto the underlying Brownian motion.

Let N =∑∞

j=1 δ(τj ,ξj ). Then for any random variable F we define

Dt,uF = F (N + δ(t,u))− F (N)

which is the Malliavin derivative of F with respect to N.


Malliavin calculus

Consider a probability space (Ω,F , (Ft)t≥0,P) with a Brownian motion(Wt)t≥0 and a Poisson random measure N. For smooth random variablesF we have∫ t

0〈∇sF , hs〉ds = lim

ε→0

1

ε

(F (W· + ε

∫ ·0hsds)− F (W·)

)where the process (∇tF )t≥0 is the Malliavin derivative of F with respectto the underlying Brownian motion.Let N =

∑∞j=1 δ(τj ,ξj ). Then for any random variable F we define

Dt,uF = F (N + δ(t,u))− F (N)

which is the Malliavin derivative of F with respect to N.


Clark-Ocone formula

Theorem

Assume that F is such that

E∫ T

0|∇tF |2dt + E

∫ T

0

∫U|Dt,uF |2ν(du)dt <∞ .

Then

F = EF +

∫ T

0E[∇tF |Ft ]dWt +

∫ T

0

∫UE[Dt,uF |Ft ]N(dt, du) .


Lemma [Wu, AIHP 2010 and Ma, SPA 2010]

If there exists h : [0,T ]× U → R such that∫ T

0

∫U h(t, u)2ν(du)dt <∞

andE[Dt,uF |Ft ] ≤ h(t, u)

and there exists g : [0,T ]→ Rm such that∫ T

0 |g(t)|2dt <∞ and

|E[∇tF |Ft ]| ≤ |g(t)|

then for any C 2 convex function φ : R→ R such that φ′ is also convex, wehave

Eφ(F − EF ) ≤ Eφ(∫ T

0

∫Uh(t, u)N(dt, du) +

∫ T

0g(t)dWt

).


In particular, for any λ > 0 we have

Eeλ(F−EF ) ≤ exp

(∫ T

0

∫U

(eλh(t,u) − λh(t, u)− 1)ν(du)dt

+

∫ T

0

1

2|g(t)|2dt

)

But by the Gozlan-Leonard lemma we know that∫eλ(f−µ(f ))dµ ≤ eα

∗(λ) for any λ > 0 ,

implies α-W1H.


Obtaining bounds for Malliavin derivatives

Consider the solution (Xt(x))t≥0 to


∫Ug(Xt−, u)N(dt, du)

with initial condition x ∈ Rd as a functional of the underlying Poissonrandom measure N. Then define

X (t,u)(x ,N(ω)) := X (x ,N(ω) + δ(t,u)) ,

which means that we add a jump of size g(Xt−, u) at time t to every pathof X .


We haveX

(t,u)s (x) = Xs(x) for s < t

and

X(t,u)s (x) = Xt(x) + g(Xt−, u) +

∫ s

tb(X

(t,u)r (x))dr

+

∫ s

tσ(X

(t,u)r (x))dWr +

∫ s

t

∫Ug(X

(t,u)r− (x , u))N(dt, du) for s ≥ t .

This means that after time t, the process (X(t,u)s (x))s≥t is the solution of

the same SDE but with different initial condition, i.e.,X

(t,u)t (x) = Xt(x) + g(Xt−, u).


When t > T , we have D(t,u)(XT (x)) = 0.

If our SDE is globally dissipative, then for any x , y ∈ Rd

E|Xt(x)− Xt(y)| ≤ e−Kt |x − y |

Hence for any Lipschitz function f : Rd → R with ‖f ‖Lip ≤ 1 if t < T wehave

E[D(t,u)f (XT (x))|Ft ] = E[f (X(t,u)T (x))− f (XT (x))|Ft ]

≤ E[∣∣∣f (X

(t,u)T (x))− f (XT (x))

∣∣∣ |Ft

]≤ E

[∣∣∣X (t,u)T (x)− XT (x)

∣∣∣ |Ft

]≤ e−K(T−t)|g(Xt−, u)|

If |g(x , u)| ≤ g∞(u) and∫U |g∞(u)|2ν(du) <∞, then we get the required

bound on E[D(t,u)f (XT (x))|Ft ].




E|Xt(x)− Xt(y)| ≤ e−Kt |x − y |



≤ E[∣∣∣f (X

(t,u)T (x))− f (XT (x))

∣∣∣ |Ft

]≤ E

[∣∣∣X (t,u)T (x)− XT (x)

∣∣∣ |Ft

]≤ e−K(T−t)|g(Xt−, u)|






E|Xt(x)− Xt(y)| ≤ e−Kt |x − y |



≤ E[∣∣∣f (X

(t,u)T (x))− f (XT (x))

∣∣∣ |Ft

]≤ E

[∣∣∣X (t,u)T (x)− XT (x)

∣∣∣ |Ft

]≤ e−K(T−t)|g(Xt−, u)|






E|Xt(x)− Xt(y)| ≤ e−Kt |x − y |



≤ E[∣∣∣f (X

(t,u)T (x))− f (XT (x))

∣∣∣ |Ft

]≤ E

[∣∣∣X (t,u)T (x)− XT (x)

∣∣∣ |Ft

]≤ e−K(T−t)|g(Xt−, u)|




Assume we have a coupling (Xt ,Yt)t≥0 (and thus (Xt)t≥0 and (Yt)t≥0

have the same transition probabilities) for which we can prove

E|Xt(x)− Yt(y)| ≤ C e−Kt |x − y | .

Then


= E[f (X(t,u)T (x))− f (YT (x))|Ft ]

≤ E[∣∣∣f (X

(t,u)T (x))− f (YT (x))

∣∣∣ |Ft

]≤ E

[∣∣∣X (t,u)T (x)− YT (x)

∣∣∣ |Ft

]≤ C e−K(T−t)|g(Xt−, u)|


Assume we have a coupling (Xt ,Yt)t≥0 (and thus (Xt)t≥0 and (Yt)t≥0

have the same transition probabilities) for which we can prove

E|Xt(x)− Yt(y)| ≤ C e−Kt |x − y | .

Then


= E[f (X(t,u)T (x))− f (YT (x))|Ft ]

≤ E[∣∣∣f (X

(t,u)T (x))− f (YT (x))

∣∣∣ |Ft

]≤ E

[∣∣∣X (t,u)T (x)− YT (x)

∣∣∣ |Ft

]≤ C e−K(T−t)|g(Xt−, u)|


For the Gaussian case we have

E〈∇Xt , h〉H = limε→0

1

εE(Xt(W· + ε

∫ ·0hsds)− Xt(W·)

),

where

Xt(W· + ε

∫ ·0hsds) = X0 +

∫ t

0b(Xs)ds + ε

∫ t

0σ(Xs)hsds

+

∫ t

0σ(Xs)dWs +

∫Ug(Xs−, u)N(ds, du) .

Question: How to couple two jump diffusions with different drifts?




1

εE(Xt(W· + ε


),

where

Xt(W· + ε

∫ ·0hsds) = X0 +

∫ t

0b(Xs)ds + ε

∫ t

0σ(Xs)hsds

+

∫ t

0σ(Xs)dWs +






1

εE(Xt(W· + ε


),

where

Xt(W· + ε

∫ ·0hsds) = X0 +

∫ t

0b(Xs)ds + ε

∫ t

0σ(Xs)hsds

+

∫ t

0σ(Xs)dWs +




A. Eberle (2016)

Reflection couplings and contraction rates for diffusions.

Probab. Theory Related Fields 166 (2016), no. 3-4, 851 – 886.

T. Lindvall and L. C. G. Rogers (1986)

Coupling of multidimensional diffusions by reflection.

Ann. Probab. 14 (1986), no. 3, 860 – 872.

M. B. Majka (2015)

Coupling and exponential ergodicity for stochastic differential equations driven byLevy processes.

To appear in Stochastic Process. Appl..

M. B. Majka (2016)

Transportation inequalities for non-globally dissipative SDEs with jumps viaMalliavin calculus and coupling.

arXiv:1610.06916.

L. Wu (2010)

Transportation inequalities for stochastic differential equations of pure jumps.

Ann. Inst. Henri Poincare: Probab. Stat. 46 (2010), no. 2, 465 – 479.


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