Maximal Martingale Wasserstein Inequality

In this note, we complete the analysis of the Martingale Wasserstein Inequality started in arXiv:2011.11599 by checking that this inequality fails in dimension $d\ge 2$ when the integrability parameter $\rho$ belongs to $[1,2)$ while a stronger Maximal Martingale Wasserstein Inequality holds whatever the dimension $d$ when $\rho\ge 2$.


Introduction
The present paper elaborates on the convergence to 0 as n → ∞ of inf M∈Π M (µn,νn) R d ×R d |y − x| ρ M (dx, dy) with the Wasserstein distance W ρ (µ n , ν n ) when for each n ∈ N, µ n and ν n belong to the set P ρ (R d ) of probability measures on R d with a finite moment of order ρ ∈ [1, +∞) and the former is smaller than the latter in the convex order.The convex order between µ, ν ∈ P 1 (R d ) which is denoted µ ≤ cx ν amounts to and, by Strassen's theorem [7], is equivalent to the non emptyness of the set of martingale couplings between µ and ν defined by The Wasserstein distance with index ρ is defined by |x − y| ρ π(dx, dy) 1/ρ and we also introduce M ρ (µ, ν) and M ρ (µ, ν) respectively defined by |x − y| ρ M (dx, dy). ( In dimension d = 1, the optimization problems defining M ρ and M ρ are the respective subjects of [3] and [4] when ρ = 1, while the general case ρ ∈ (0, +∞) is studied in [6].
The question of interest is related to the stability of Martingale Optimal Transport problems with respect to the marginal distributions µ and ν established in dimension d = 1 in [1,8] while it fails in higher dimension according to [2].A quantitative answer is given in dimension d = 1 by the Martingale Wasserstein inequality established in [5,Proposition 1] where the central moment σ ρ (ν) of ν is defined by The proposition also states that W ρ (µ, ν) and σ ρ (ν) have the right exponent in this inequality in the sense that for The generalization of (3) to higher dimensions d is also investigated in [5] where it is proved that for any d ≥ 2, 2 ) , while the one-dimensional constant C (ρ,ρ),1 is preserved when µ and ν are products of one-dimensional probability measures or when, for X distributed according to µ, the conditional expectation of X given the direction of X − E[X] is a.s.equal to E[X] and ν is the distribution of X + λ(X − E[X]) for some λ ≥ 0. The present paper answers the question of the finiteness of C (ρ,ρ),d when ρ ∈ [ 1+  3) by general conjugate exponents q ∈ [1, +∞] and q q−1 ∈ [1, +∞] leading to indices q and q(ρ−1) q−1 (equal to +∞ and ρ − 1 when q is respectively equal to 1 and +∞) and define q−1 (ν) and C (ρ,q),d := sup µ,ν∈P q∨ (ρ−1)q q−1 These constants of course depend on the norm | • | on R d (even if we do not make this dependence explicit) but, by equivalence of the norms, their finiteness does not.Since the Euclidean norm plays a particular role, we will denote it by Remark 2.
• The fact that ρ = 2 appears as a threshold is related to the equality x 2 µ(dx).

Proof
The proof of Theorem 1 (ii) relies on the next lemma, the proof of the lemma is postponed after the proof of the theorem.In what follows, to avoid making distinctions in case q ∈ {1, +∞}, we use the convention that for any probability measure γ and any measurable function f on the same probability space |f (z)| q γ(dz) q−1 γ(dz) )) when q = +∞ (resp.q = 1).
Remark 4. When ρ = 2, then (4) holds as an equality with κ ρ = 1 while, by the Cauchy-Schwarz and the triangle inequalities, Proposition 1] and we deduce that ν , one has by optimality of the comonotonic coupling and Hölder's inequality Since, by the triangle inequality and µ ≤ cx ν, one has for c ∈ R q−1 du q−1 q(ρ−1) , we deduce by minimizing over the constant c that With this inequality replacing (30) in the proof of Proposition 1 [5] and the general inequality q−1 du q−1 q , replacing the special case q = ρ in the second equation p840 in this proof, we deduce that To check that C (ρ,q),1 = +∞ for ρ ∈ [1, +∞) and q ∈ [1, +∞], let us introduce for n ≥ 2 and z > 0, This example generalizes the one introduced by Brückerhoff and Juillet in [2] which corresponds to the choice z = 1.Since belongs to Π M (µ n,z , ν n,z ), we have On the other hand, by optimality of the comonotonic coupling Let α ∈ [0, 1).The sequence n 1−α goes to ∞ with n and for ρ ∈ [1, +∞) and q ∈ [1, +∞], we have q−1 q where 1 + q(ρ−1) q−1 q−1 q = 1 by convention when q = 1 so that Let ρ ∈ [1, 2).For q = 1, the exponent of n in the equivalent of the ratio is equal to 2 − ρ > 0 so that the right-hand side goes to +∞ with n.For q ∈ (1, +∞], we may choose α ∈ q(ρ−1)−1 q−1 , 1 (with left boundary equal to ρ − 1 when q = +∞) so that q−1 q α + 1 q + 1 − ρ > 0 and the right-hand side still goes to +∞ with n.Therefore C (ρ,q),1 = +∞.To prove that C (ρ,q),d = +∞ for d ≥ 2 it is enough by [5,Lemma 1] to deal with the case d = 2, in which we use the rotation argument in [2].For n ≥ 2 and θ ∈ (0, π), M θ n defined as which is a martingale coupling between the image µ n of µ n,n −α by R ∋ x → (x, 0) ∈ R 2 and its second marginal ν θ n which, as θ → 0, converges in any W q with q ∈ [1, +∞] to the image of ν n,n −α by the same mapping.According to the proof of [2, Lemma 1 With the above analysis of the asymptotic behaviour of the right-hand side as n → ∞, we conclude that C (ρ,q),d = +∞.
(ii) Now, let ρ ∈ [2, +∞) and M ∈ Π M (µ, ν).Applying Equation (4) in Lemma 3 for the inequality and then using the martingale property of M , we obtain that for c ∈ R d , we have Denoting by π ∈ Π(µ, ν) an optimal coupling for W q (µ, ν), we have using Equation ( 5) in Lemma 3 for the inequality By the fact that all norms are equivalent in finite dimensional vector spaces, there exists λ ∈ [1, ∞) such that for all z ∈ R d , we have z λ ≤ |z| ≤ λ z .
By taking the infimum with respect to c ∈ R d , we conclude that the statement holds with C (ρ,q),d ≤ 2κ ρ κρ λ 2ρ .Finally, let us suppose that R d is endowed with the Euclidean norm.Then we can choose λ = 1, so that C (ρ,q),d ≤ 2κ ρ κρ with the right-hand side not depending on d according to Lemma 3.Moreover, by Remark 4, C (2,q),d ≤ 2 and since for α ∈ [0, 1), we have C (2,q),d = 2.
Proof of Lemma 3. We suppose that ρ > 2 since the case ρ = 2 has been addressed in Remark 4. Suppose x = 0 and y = x and set e = x x and z = y,x x 2 .The vector y x − ze is orthogonal to e and can be rewritten as ωe ⊥ with ω ≥ 0 and e ⊥ ∈ R d such that e ⊥ = 1 and e, e ⊥ = 0.One then has y x = ze + ωe ⊥ and since y = x, (z, w) = (1, 0).The first inequality (4) divided by x ρ writes: as the second factor in the right-hand side.Applying a Taylor's expansion at t = 0 to t → Since ρ > 2, we conclude that being continuous on R 2 \ {(1, 0)}, we deduce that ϕ(z, ω) < +∞.