<figure><img src="image_1.png" /></figure>

HAL
open science

# Maximal Martingale Wasserstein Inequality

Benjamin Jourdain, Kexin Shao

► To cite this version:

Benjamin Jourdain, Kexin Shao. Maximal Martingale Wasserstein Inequality. Electronic Communications in Probability, 2024, 29 (26), 10.1214/24-ECP593. hal-04241070

HAL Id: hal-04241070

https://hal.science/hal-04241070v1

Submitted on 13 Oct 2023

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

<figure><img src="image_2.png" /></figure>

Distributed under a Creative Commons CC BY 4.0 - Attribution - International License
# Maximal Martingale Wasserstein Inequality

Benjamin Jourdain*

Kexin Shao†

October 12, 2023

## Abstract

In this note, we complete the analysis of the Martingale Wasserstein Inequality started in [5] 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 $\rho \ge 2$.

## 1 Introduction

The present paper elaborates on the convergence to 0 as $n \to \infty$ of $\inf_{M \in \Pi^M(\mu_n, \nu_n)} \int_{\mathbb{R}^d \times \mathbb{R}^d} |y-x|^\rho M(dx, dy)$ with the Wasserstein distance $W_\rho(\mu_n, \nu_n)$ when for each $n \in \mathbb{N}$, $\mu_n$ and $\nu_n$ belong to the set $\mathcal{P}_\rho(\mathbb{R}^d)$ of probability measures on $\mathbb{R}^d$ with a finite moment of order $\rho \in [1, +\infty)$ and the former is smaller than the latter in the convex order. The convex order between $\mu, \nu \in \mathcal{P}_1(\mathbb{R}^d)$ which is denoted $\mu \le_{cx} \nu$ amounts to

$$ \int_{\mathbb{R}^d} f(x) \mu(dx) \le \int_{\mathbb{R}^d} f(y) \nu(dy) \quad \text{for each convex function } f: \mathbb{R}^d \to \mathbb{R}, \quad (1) $$

and, by Strassen's theorem [7], is equivalent to the non emptiness of the set of martingale couplings between $\mu$ and $\nu$ defined by

$$ \Pi^M(\mu, \nu) = \left\{ M(dx, dy) = \mu(dx)m(x, dy) \in \Pi(\mu, \nu) \mid \mu(dx)\text{-a.e., } \int_{\mathbb{R}^d} y m(x, dy) = x \right\} \text{ where} \\ \Pi(\mu, \nu) = \{ \pi \in \mathcal{P}_1(\mathbb{R}^d \times \mathbb{R}^d) \mid \forall A \in \mathcal{B}(\mathbb{R}^d), \pi(A \times \mathbb{R}^d) = \mu(A) \text{ and } \pi(\mathbb{R}^d \times A) = \nu(A) \}. $$

The Wasserstein distance with index $\rho$ is defined by

$$ W_\rho(\mu, \nu) = \left( \inf_{\pi \in \Pi(\mu, \nu)} \int_{\mathbb{R}^d \times \mathbb{R}^d} |x-y|^\rho \pi(dx, dy) \right)^{1/\rho} $$

and we also introduce $\overline{M}_\rho(\mu, \nu)$ and $\overline{\mathcal{M}}_\rho(\mu, \nu)$ respectively defined by

$$ \overline{M}_\rho^\rho(\mu, \nu) = \inf_{M \in \Pi^M(\mu, \nu)} \int_{\mathbb{R}^{2d}} |x-y|^\rho M(dx, dy), \quad \overline{\mathcal{M}}_\rho^\rho(\mu, \nu) = \sup_{M \in \Pi^M(\mu, \nu)} \int_{\mathbb{R}^{2d}} |x-y|^\rho M(dx, dy). \quad (2) $$

In dimension $d = 1$, the optimization problems defining $\overline{M}_\rho$ and $\overline{\mathcal{M}}_\rho$ are the respective subjects of [3] and [4] when $\rho = 1$, while the general case $\rho \in (0, +\infty)$ is studied in [6].

The question of interest is related to the stability of Martingale Optimal Transport problems with respect to the marginal distributions $\mu$ and $\nu$ established in dimension $d = 1$ in $[1, 8]$ while

<footnote>*CERMICS, Ecole des Ponts, INRIA, Marne-la-Vallée, France. E-mail: benjamin.jourdain@enpc.fr - This research benefited from the support of the “Chaire Risques Financiers”, Fondation du Risque.</footnote>

<footnote>†INRIA Paris, 2 rue Simone Iff, CS 42112, 75589 Paris Cedex 12, France, Université Paris-Dauphine, Ecole des Ponts ParisTech. E-mail: kexin.shao@inria.fr. This project has received funding from the European Union's Horizon 2020 research and innovation programme under the Marie Skłodowska-Curie grant agreement No 945322.</footnote>
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] for $\rho \in [1, +\infty)$,

$$ \exists \underline{C}_{(\rho, \rho), 1} < \infty, \forall \mu, \nu \in \mathcal{P}_{\rho}(\mathbb{R}) \text{ with } \mu \leq_{cx} \nu, \underline{M}_{\rho}^{\rho}(\mu, \nu) \leq \underline{C}_{(\rho, \rho), 1} W_{\rho}(\mu, \nu) \sigma_{\rho}^{\rho-1}(\nu), \quad (3) $$

where the central moment $\sigma_{\rho}(\nu)$ of $\nu$ is defined by

$$ \sigma_{\rho}(\nu) = \inf_{c \in \mathbb{R}^d} \left( \int_{\mathbb{R}^d} |y - c|^{\rho} \nu(dy) \right)^{1/\rho} \quad \text{when } \rho \in [1, +\infty) \text{ and } \sigma_{\infty}(\nu) = \inf_{c \in \mathbb{R}^d} \nu - \inf_{y \in \mathbb{R}^d} |y - c|. $$

The proposition also states that $W_{\rho}(\mu, \nu)$ and $\sigma_{\rho}(\nu)$ have the right exponent in this inequality in the sense that for $1 < s < \rho$, $\sup_{\mu, \nu \in \mathcal{P}_{\rho}(\mathbb{R})}$ $\frac{\underline{M}_{\rho}^{\rho}(\mu, \nu)}{W_{\rho}^{s}(\mu, \nu)\sigma_{\rho}^{\rho-s}(\nu)} = +\infty$. The generalization of (3) to higher dimensions $d$ is also investigated in [5] where it is proved that for any $d \geq 2$,

$$ \underline{C}_{(\rho, \rho), d} := \sup_{\substack{\mu, \nu \in \mathcal{P}_{\rho}(\mathbb{R}^d) \\ \mu \leq_{cx} \nu, \mu \neq \nu}} \frac{\underline{M}_{\rho}^{\rho}(\mu, \nu)}{W_{\rho}(\mu, \nu) \sigma_{\rho}^{\rho-1}(\nu)} $$

is infinite when $\rho \in [1, \frac{1+\sqrt{5}}{2})$, while the one-dimensional constant $\underline{C}_{(\rho, \rho), 1}$ is preserved when $\mu$ and $\nu$ are products of one-dimensional probability measures or when, for $X$ distributed according to $\mu$, the conditional expectation of $X$ given the direction of $X - \mathbb{E}[X]$ is a.s. equal to $\mathbb{E}[X]$ and $\nu$ is the distribution of $X + \lambda(X - \mathbb{E}[X])$ for some $\lambda \geq 0$. The present paper answers the question of the finiteness of $\underline{C}_{(\rho, \rho), d}$ when $\rho \in [\frac{1+\sqrt{5}}{2}, +\infty)$ and $d \geq 2$, which remained open. It turns out that $\underline{C}_{(\rho, \rho), d} = +\infty$ for $d \geq 2$ when $\rho \in [1, 2)$ while for $\rho \in [2, +\infty)$ the inequality (3) generalizes in any dimension $d$ into a Maximal Martingale Wasserstein inequality with the left-hand side $\underline{M}_{\rho}^{\rho}(\mu, \nu)$ replaced by the larger $\overline{M}_{\rho}^{\rho}(\mu, \nu)$. We even replace conjugate exponents $\rho$ and $\frac{\rho}{\rho-1}$ leading to the respective indices $\rho = \rho \times 1$ and $\rho = \frac{\rho}{\rho-1} \times (\rho-1)$ in the factors $W$ and $\sigma$ in (3) by general conjugate exponents $q \in [1, +\infty]$ and $\frac{q}{q-1} \in [1, +\infty]$ leading to indices $q$ and $\frac{q(\rho-1)}{q-1}$ (equal to $+\infty$ and $\rho-1$ when $q$ is respectively equal to 1 and $+\infty$) and define

$$ \underline{C}_{(\rho, q), d} := \sup_{\substack{\mu, \nu \in \mathcal{P}_{q \vee (\rho-1)q}(\mathbb{R}^d) \\ \mu \leq_{cx} \nu, \mu \neq \nu}} \frac{\underline{M}_{\rho}^{\rho}(\mu, \nu)}{W_q(\mu, \nu) \sigma_{q(\rho-1)}^{\rho-1}(\nu)} \quad \text{and} \quad \bar{C}_{(\rho, q), d} := \sup_{\substack{\mu, \nu \in \mathcal{P}_{q \vee (\rho-1)q}(\mathbb{R}^d) \\ \mu \leq_{cx} \nu, \mu \neq \nu}} \frac{\bar{M}_{\rho}^{\rho}(\mu, \nu)}{W_q(\mu, \nu) \sigma_{q(\rho-1)}^{\rho-1}(\nu)}, $$

with $W_{\infty}(\mu, \nu) = \inf_{\pi \in \Pi(\mu, \nu)} \pi - \inf_{(x,y) \in \mathbb{R}^d \times \mathbb{R}^d} |x - y|$. Since $\underline{M}_{\rho} \leq \overline{M}_{\rho}$, one has $\underline{C}_{(\rho, q), d} \leq \bar{C}_{(\rho, q), d}$. These constants of course depend on the norm $|\cdot|$ on $\mathbb{R}^d$ (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 $\|\cdot\|$ rather than $|\cdot|$.

**Theorem 1.** (i) Let $\rho \in [1, 2)$. For $q \in [1, \frac{1}{2-\rho}]$ (and even $q \in [1, +\infty]$ when $\rho = 1$), one has $\underline{C}_{(\rho, q), 1} \leq K_{\rho} < +\infty$ where the constant $K_{\rho}$ is studied in [5, Proposition 1] while, for $q \in [1, +\infty]$, $\bar{C}_{(\rho, q), 1} = +\infty$ and $\underline{C}_{(\rho, q), d} = +\infty$ for $d \geq 2$.

(ii) Let $\rho \in [2, +\infty)$ and $q \in [1, +\infty]$. One has $\bar{C}_{(\rho, q), d} < +\infty$ whatever $d$. Moreover, when $\mathbb{R}^d$ (resp. each $\mathbb{R}^d$) is endowed with the Euclidean norm, $\bar{C}_{(2,q),d} = 2$ and $\sup_{d \geq 1} \bar{C}_{(\rho,q),d} < +\infty$.

**Remark 2.** • The fact that $\rho = 2$ appears as a threshold is related to the equality $\int_{\mathbb{R}^d \times \mathbb{R}^d} \|y-x\|^2 M(dx, dy) = \int_{\mathbb{R}^d} \|y\|^2 \nu(dy) - \int_{\mathbb{R}^d} \|x\|^2 \mu(dx)$ for $M \in \Pi^M(\mu, \nu)$ when $\mu, \nu \in \mathcal{P}_2(\mathbb{R}^d)$ are such that $\mu \leq_{cx} \nu$, which implies that when $\mathbb{R}^d$ is endowed with the Euclidean norm

$$ \underline{M}_2^2(\mu, \nu) = \bar{M}_2^2(\mu, \nu) = \int_{\mathbb{R}^d} \|y\|^2 \nu(dy) - \int_{\mathbb{R}^d} \|x\|^2 \mu(dx). $$

• For $\rho \in [1, 2)$, one has $\bar{C}_{(\rho, q), d} = +\infty$ while $\sup_{\mu, \nu \in \mathcal{P}_{q \vee \frac{q}{q-1}}(\mathbb{R}^d)} \frac{\bar{M}_{\rho}^2(\mu, \nu)}{W_q(\mu, \nu) \sigma_{\frac{q}{q-1}}(\nu)} \leq \bar{C}_{(2,q), d} < +\infty$ since $\bar{M}_{\rho} \leq \bar{M}_2$.
## 2 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 $q \in \{1, +\infty\}$, we use the convention that for any probability measure $\gamma$ and any measurable function $f$ on the same probability space $(\int |f(z)|^q \gamma(dz))^{1/q}$ (resp. $((\int |f(z)|^{\frac{q}{q-1}} \gamma(dz))^{(q-1)/q}, (\int |f(z)|^{\frac{q(\rho-1)}{q-1}} \gamma(dz))^{(q-1)/q})$ is equal to $\gamma$-ess sup$_z |f(z)|$ (resp. $(\gamma$-ess sup$_z |f(z)|, \gamma$-ess sup$_z |f(z)|^{\rho-1})$) when $q = +\infty$ (resp. $q = 1$).

**Lemma 3.** Given $\rho \in [2, +\infty)$, there exist constants $\kappa_\rho, \tilde{\kappa}_\rho \in [0, +\infty)$ such that for all $d \ge 1$ and $x, y \in \mathbb{R}^d$,

$$ \|x - y\|^\rho \le \kappa_\rho ((\rho - 1)\|x\|^\rho + \|y\|^\rho - \rho\|x\|^{\rho-2}\langle x, y \rangle), \quad (4) $$

$$ \|y\|^\rho - \|x\|^\rho \le \tilde{\kappa}_\rho \|y - x\| \left( \|x\|^{\rho-1} + \|y\|^{\rho-1} \right). \quad (5) $$

**Remark 4.** When $\rho = 2$, then (4) holds as an equality with $\kappa_\rho = 1$ while, by the Cauchy-Schwarz and the triangle inequalities,

$$ \|y\|^2 - \|x\|^2 \le \langle y - x, y + x \rangle \le \|y - x\| \times \|y + x\| \le \|y - x\| (\|x\| + \|y\|) $$

so that (5) holds with $\tilde{\kappa}_\rho = 1$.

*Proof of Theorem 1.* (i) In dimension $d = 1$, one has $M_1 \le K_1 W_1$ with $K_1 = 2$ according to [5, Proposition 1] and we deduce that $C_{(1,q),1} \le K_1$ for $q \in [1, +\infty]$ since $W_1 \le W_q$. Let now $\rho \in (1, 2)$ and $q \in [1, \frac{1}{2-\rho}]$. One has $\frac{q(\rho-1)}{q-1} \ge 1$ since, when $q > 1$, $\frac{q}{q-1} = 1 + \frac{1}{q-1} \ge 1 + \frac{2-\rho}{\rho-1} = \frac{1}{\rho-1}$. For $\mu, \nu \in P_{q \vee \frac{q(\rho-1)}{q-1}}(\mathbb{R})$ with respective quantile functions $F_\mu^{-1}$ and $F_\nu^{-1}$, one has by optimality of the comonotonic coupling and Hölder's inequality

$$ \begin{aligned} W_\rho^\rho(\mu, \nu) &= \int_0^1 |F_\nu^{-1}(u) - F_\mu^{-1}(u)| \times |F_\nu^{-1}(u) - F_\mu^{-1}(u)|^{\rho-1} du \\ &\le \left( \int_0^1 |F_\nu^{-1}(u) - F_\mu^{-1}(u)|^q du \right)^{1/q} \left( \left( \int_0^1 |F_\nu^{-1}(u) - F_\mu^{-1}(u)|^{\frac{q(\rho-1)}{q-1}} du \right)^{\frac{q-1}{q(\rho-1)}} \right)^{\rho-1}. \end{aligned} $$

Since, by the triangle inequality and $\mu \le cx$, one has for $c \in \mathbb{R}$

$$ \begin{aligned} \left( \int_0^1 |F_\nu^{-1}(u) - F_\mu^{-1}(u)|^{\frac{q(\rho-1)}{q-1}} du \right)^{\frac{q-1}{q(\rho-1)}} &\le \left( \int_0^1 |F_\nu^{-1}(u) - c|^{\frac{q(\rho-1)}{q-1}} du \right)^{\frac{q-1}{q(\rho-1)}} + \left( \int_0^1 |F_\mu^{-1}(u) - c|^{\frac{q(\rho-1)}{q-1}} du \right)^{\frac{q-1}{q(\rho-1)}} \\ &\le 2 \left( \int_0^1 |F_\nu^{-1}(u) - c|^{\frac{q(\rho-1)}{q-1}} du \right)^{\frac{q-1}{q(\rho-1)}}, \end{aligned} $$

we deduce by minimizing over the constant $c$ that

$$ W_\rho^\rho(\mu, \nu) \le W_q(\mu, \nu) \times 2^{\rho-1} \sigma_{\frac{q(\rho-1)}{q-1}}^{\rho-1}(\nu). $$

With this inequality replacing (30) in the proof of Proposition 1 [5] and the general inequality

$$ \int_0^1 |F_\nu^{-1}(u) - F_\mu^{-1}(u)| |F_\nu^{-1}(u) - c|^{\rho-1} du \le W_q(\mu, \nu) \left( \int_0^1 |F_\nu^{-1}(u) - c|^{\frac{q(\rho-1)}{q-1}} du \right)^{\frac{q-1}{q}}, $$

replacing the special case $q = \rho$ in the second equation p840 in this proof, we deduce that $W_\rho^\rho(\mu, \nu) \le K_\rho W_q(\mu, \nu) \sigma_{\frac{q(\rho-1)}{q-1}}^{\rho-1}(\nu)$.
To check that $\bar{C}_{(\rho,q),1} = +\infty$ for $\rho \in [1, +\infty)$ and $q \in [1, +\infty]$, let us introduce for $n \ge 2$ and $z > 0$,

$$ \mu_{n,z} = \frac{1}{2((n-1)z+1)} \left( (1+z)(\delta_1 + \delta_n) + 2z \sum_{i=2}^{n-1} \delta_i \right) \\ \text{and } \nu_{n,z} = \frac{1}{2((n-1)z+1)} \left( \delta_{1-z} + \delta_{n+z} + z(\delta_1 + \delta_n) + 2z \sum_{i=2}^{n-1} \delta_i \right). $$

This example generalizes the one introduced by Brückerhoff and Juillet in [2] which corresponds to the choice $z = 1$. Since

$$ M_{n,z} = \frac{1}{2((n-1)z+1)} \left( \delta_{(1,1-z)} + z\delta_{(1,2)} + z\delta_{(n,n-1)} + \delta_{(n,n+z)} + z \sum_{i=2}^{n-1} (\delta_{(i,i-1)} + \delta_{(i,i+1)}) \right) $$

belongs to $\Pi^M(\mu_{n,z}, \nu_{n,z})$, we have

$$ \overline{M}_{\rho}^{\rho}(\mu_{n,z}, \nu_{n,z}) \geq \int_{\mathbb{R} \times \mathbb{R}} |y-x|^{\rho} M_{n,z}(dx, dy) = \frac{(n-1)z + z^{\rho}}{(n-1)z+1}. $$

On the other hand, by optimality of the comonotonic coupling $W_{\rho}^{\rho}(\mu_{n,z}, \nu_{n,z}) = \frac{z^{\rho}}{(n-1)z+1}$ for $\rho \in [1, +\infty)$ and $W_{\infty}(\mu_{n,z}, \nu_{n,z}) = z$. Last $\sigma_{\infty}(\nu_{n,z}) = \frac{n-1+2z}{2}$ and, when $\rho \in [1, +\infty)$,

$$ \sigma_{\rho}^{\rho}(\nu_{n,z}) = \frac{1}{2^{\rho}((n-1)z+1)} \left( (n-1+2z)^{\rho} + z(n-1)^{\rho} + 2z \sum_{i=2}^{\lfloor \frac{n+1}{2} \rfloor} (n+1-2i)^{\rho} \right). $$

Let $\alpha \in [0, 1)$. The sequence $n^{1-\alpha}$ goes to $\infty$ with $n$ and for $\rho \in [1, +\infty)$ and $q \in [1, +\infty]$, we have

$$ \int_{\mathbb{R} \times \mathbb{R}} |y-x|^{\rho} M_{n,n-\alpha}(dx, dy) \to 1, \quad W_q(\mu_{n,n-\alpha}, \nu_{n,n-\alpha}) \sim n^{\alpha \frac{(1-q)}{q} - \frac{1}{q}} $$

and $\sigma_{\frac{q(\rho-1)}{q-1}}^{\rho-1}(\nu_{n,n-\alpha}) \sim \frac{n^{\rho-1}}{2^{\rho-1}(1+\frac{q(\rho-1)}{q-1})^{\frac{q-1}{q}}}$ where $(1+\frac{q(\rho-1)}{q-1})^{\frac{q-1}{q}} = 1$ by convention when $q=1$ so that

$$ \frac{\int_{\mathbb{R} \times \mathbb{R}} |y-x|^{\rho} M_{n,n-\alpha}(dx, dy)}{W_q(\mu_{n,n-\alpha}, \nu_{n,n-\alpha}) \sigma_{\frac{q(\rho-1)}{q-1}}^{\rho-1}(\nu_{n,n-\alpha})} \sim 2^{\rho-1} \left(1 + \frac{q(\rho-1)}{q-1}\right)^{\frac{q-1}{q}} n^{\frac{q-1}{q}\alpha + \frac{1}{q} + 1 - \rho}. $$

Let $\rho \in [1, 2)$. For $q=1$, the exponent of $n$ in the equivalent of the ratio is equal to $2-\rho > 0$ so that the right-hand side goes to $+\infty$ with $n$. For $q \in (1, +\infty]$, we may choose $\alpha \in (\frac{q(\rho-1)-1}{q-1}, 1)$ (with left boundary equal to $\rho-1$ when $q = +\infty$) so that $\frac{q-1}{q}\alpha + \frac{1}{q} + 1 - \rho > 0$ and the right-hand side still goes to $+\infty$ with $n$. Therefore $\bar{C}_{(\rho,q),1} = +\infty$. To prove that $\underline{C}_{(\rho,q),d} = +\infty$ for $d \ge 2$ it is enough by [5, Lemma 1] to deal with the case $d=2$, in which we use the rotation argument in [2]. For $n \ge 2$ and $\theta \in (0, \pi)$, $M_n^{\theta}$ defined as $\frac{1}{2((n-1)n^{-\alpha}+1)}$ times

$$ \delta_{((1,0),(1-n^{-\alpha}\cos\theta, -n^{-\alpha}\sin\theta))} + n^{-\alpha}\delta_{((1,0),(1+\cos\theta,\sin\theta))} + n^{-\alpha}\delta_{((n,0),(n-\cos\theta, -\sin\theta))} \\ + \delta_{((n,0),(n+n^{-\alpha}\cos\theta, n^{-\alpha}\sin\theta))} + n^{-\alpha}\sum_{i=2}^{n-1} (\delta_{((i,0),(i-\cos\theta, -\sin\theta))} + \delta_{((i,0),(i+\cos\theta, \sin\theta))}) $$

which is a martingale coupling between the image $\mu_n$ of $\mu_{n,n^{-\alpha}}$ by $\mathbb{R} \ni x \mapsto (x, 0) \in \mathbb{R}^2$ and its second marginal $\nu_n^{\theta}$ which, as $\theta \to 0$, converges in any $W_q$ with $q \in [1, +\infty]$ to the image of $\nu_{n,n^{-\alpha}}$
by the same mapping. According to the proof of [2, Lemma 1.1], $\Pi^M(\mu_n, \nu_n^\theta) = \{M_n^\theta\}$ so that
$$ \underline{M}_\rho^\rho(\mu_n, \nu_n^\theta) = \int_{\mathbb{R}^2 \times \mathbb{R}^2} |y-x|^\rho M_n^\theta(dx, dy) \text{ and} $$

$$
\lim_{\theta \to 0} \frac{\underline{M}_{\rho}^{\rho}(\mu_n, \nu_n^{\theta})}{W_q(\mu_n, \nu_n^{\theta}) \sigma_{\frac{q(\rho-1)}{q-1}}^{\rho-1}(\nu_n^{\theta})} = \frac{\int_{\mathbb{R} \times \mathbb{R}} |y-x|^{\rho} M_{n,n-\alpha}(dx, dy)}{W_q(\mu_{n,n-\alpha}, \nu_{n,n-\alpha}) \sigma_{\frac{q(\rho-1)}{q-1}}^{\rho-1}(\nu_{n,n-\alpha})}.
$$

With the above analysis of the asymptotic behaviour of the right-hand side as $n \to \infty$, we conclude
that $\underline{C}_{(\rho,q),d} = +\infty$.

(ii) Now, let $\rho \in [2, +\infty)$ and $M \in \Pi^M(\mu, \nu)$. Applying Equation (4) in Lemma 3 for the inequality and then using the martingale property of $M$, we obtain that for $c \in \mathbb{R}^d$, we have

$$
\begin{align}
\int_{\mathbb{R}^d \times \mathbb{R}^d} \|x - y\|^\rho M(dx, dy) &= \int_{\mathbb{R}^d \times \mathbb{R}^d} \|(x - c) - (y - c)\|^\rho M(dx, dy) \nonumber \\
&\le \kappa_\rho \int_{\mathbb{R}^d \times \mathbb{R}^d} ((\rho - 1)\|x - c\|^\rho + \|y - c\|^\rho - \rho\|x - c\|^{\rho-2} \langle x - c, y - c \rangle) M(dx, dy) \nonumber \\
&= \kappa_\rho \left( \int_{\mathbb{R}^d} \|y - c\|^\rho \nu(dy) - \int_{\mathbb{R}^d} \|x - c\|^\rho \mu(dx) \right). \tag{6}
\end{align}
$$

Denoting by $\pi \in \Pi(\mu, \nu)$ an optimal coupling for $W_q(\mu, \nu)$, we have using Equation (5) in Lemma 3 for the inequality

$$
\begin{align}
\int_{\mathbb{R}^d} \|y-c\|^\rho \nu(dy) - \int_{\mathbb{R}^d} \|x-c\|^\rho \mu(dx) &= \int_{\mathbb{R}^d \times \mathbb{R}^d} (\|y-c\|^\rho - \|x-c\|^\rho) \pi(dx, dy) \nonumber \\
&\le \tilde{\kappa}_\rho \int_{\mathbb{R}^d \times \mathbb{R}^d} \|y-x\| \left( \|x-c\|^{\rho-1} + \|y-c\|^{\rho-1} \right) \pi(dx, dy). \tag{7}
\end{align}
$$

By the fact that all norms are equivalent in finite dimensional vector spaces, there exists $\lambda \in [1, \infty)$ such that for all $z \in \mathbb{R}^d$, we have

$$
\frac{\|z\|}{\lambda} \leq |z| \leq \lambda \|z\|.
$$

Therefore, using (6) and (7) for the second inequality, Hölder's inequality for the fourth, the triangle inequality for the fifth and $\mu \leq_{cx} \nu$ for the sixth, we get that for $c \in \mathbb{R}^d$,

$$
\begin{align*}
\int_{\mathbb{R}^d \times \mathbb{R}^d} |x - y|^\rho M(dx, dy) &\le \lambda^\rho \int_{\mathbb{R}^d \times \mathbb{R}^d} \|x - y\|^\rho M(dx, dy) \\
&\le \kappa_\rho \tilde{\kappa}_\rho \lambda^\rho \int_{\mathbb{R}^d \times \mathbb{R}^d} \|x - y\| \left( \|x - c\|^{\rho-1} + \|y - c\|^{\rho-1} \right) \pi(dx, dy) \\
&\le \kappa_\rho \tilde{\kappa}_\rho \lambda^{2\rho} \int_{\mathbb{R}^d \times \mathbb{R}^d} |x - y| \left( |x - c|^{\rho-1} + |y - c|^{\rho-1} \right) \pi(dx, dy) \\
&\le \kappa_\rho \tilde{\kappa}_\rho \lambda^{2\rho} W_q(\mu, \nu) \left( \int_{\mathbb{R}^d \times \mathbb{R}^d} \left( |x - c|^{\rho-1} + |y - c|^{\rho-1} \right)^{\frac{q}{q-1}} \pi(dx, dy) \right)^{\frac{q-1}{q}} \\
&\le \kappa_\rho \tilde{\kappa}_\rho \lambda^{2\rho} W_q(\mu, \nu) \left( \left( \int_{\mathbb{R}^d} |x - c|^{\frac{q(\rho-1)}{q-1}} \mu(dx) \right)^{(q-1)/q} + \left( \int_{\mathbb{R}^d} |y - c|^{\frac{q(\rho-1)}{q-1}} \nu(dy) \right)^{(q-1)/q} \right) \\
&\le 2\kappa_\rho \tilde{\kappa}_\rho \lambda^{2\rho} W_q(\mu, \nu) \left( \int_{\mathbb{R}^d} |y - c|^{\frac{q(\rho-1)}{q-1}} \nu(dy) \right)^{\frac{q-1}{q}}.
\end{align*}
$$

By taking the infimum with respect to $c \in \mathbb{R}^d$, we conclude that the statement holds with $\bar{C}_{(\rho,q),d} \le 2\kappa_\rho \tilde{\kappa}_\rho \lambda^{2\rho}$. Finally, let us suppose that $\mathbb{R}^d$ is endowed with the Euclidean norm. Then we can choose $\lambda = 1$, so that $\bar{C}_{(\rho,q),d} \le 2\kappa_\rho \tilde{\kappa}_\rho$ with the right-hand side not depending on $d$ according to Lemma 3. Moreover, by Remark 4, $\bar{C}_{(2,q),d} \le 2$ and since for $\alpha \in [0, 1)$,

$$
\lim_{n \to \infty} \frac{\overline{M}_2^2(\mu_{n,n-\alpha}, \nu_{n,n-\alpha})}{\sqrt{W_1(\mu_{n,n-\alpha}, \nu_{n,n-\alpha}) \sigma_{\infty}(\nu_{n,n-\alpha})}} = 2,
$$

we have $\bar{C}_{(2,q),d} = 2$.

<figure><img src="image_3.png" /></figure>
Proof of Lemma 3. We suppose that $\rho > 2$ since the case $\rho = 2$ has been addressed in Remark 4.

Suppose $x \neq 0$ and $y \neq x$ and set $e = \frac{x}{\|x\|}$ and $z = \frac{y,x}{\|x\|^2}$. The vector $\frac{y}{\|x\|} - ze$ is orthogonal to $e$ and can be rewritten as $\omega e^{\perp}$ with $\omega \ge 0$ and $e^{\perp} \in \mathbb{R}^d$ such that $\|e^{\perp}\| = 1$ and $\langle e, e^{\perp} \rangle = 0$. One then has $\frac{y}{\|x\|} = ze + \omega e^{\perp}$ and since $y \neq x$, $(z, w) \neq (1, 0)$.

The first inequality (4) divided by $\|x\|^\rho$ writes:

$$((1-z)^2 + \omega^2)^{\frac{\rho}{2}} \le \kappa_{\rho} ((\rho-1) + (z^2 + \omega^2)^{\frac{\rho}{2}} - \rho z).$$

Let us define $\varphi(z, \omega) = \rho-1 + (z^2+\omega^2)^{\frac{\rho}{2}} - \rho z = -\rho(z-1) - 1 + (1+2(z-1)+(z-1)^2+\omega^2)^{\frac{\rho}{2}}$ as the second factor in the right-hand side. Applying a Taylor's expansion at $t=0$ to $t \mapsto (1+t)^{\frac{\rho}{2}}$, we obtain

$$\varphi(z, \omega) = \frac{\rho}{2}\omega^2 + \frac{\rho}{2}(\rho-1)(z-1)^2 + o((z-1)^2 + \omega^2).$$

Since $\rho > 2$, we conclude that

$$\lim_{(z, \omega) \to (1,0)} \frac{((1-z)^2 + \omega^2)^{\frac{\rho}{2}}}{\varphi(z, \omega)} = 0.$$

As $|(z, \omega)| \to +\infty$, $\varphi(z, \omega) \sim (z^2 + \omega^2)^{\frac{\rho}{2}} \sim ((z-1)^2 + \omega^2)^{\rho}$. Therefore,

$$\lim_{|(z, \omega)| \to +\infty} \frac{((z-1)^2 + \omega^2)^{\frac{\rho}{2}}}{\varphi(z, \omega)} = 1.$$

The function $(z, w) \mapsto \frac{((z-1)^2+\omega^2)^{\frac{\rho}{2}}}{\varphi(z, w)}$ being continuous on $\mathbb{R}^2 \setminus \{(1,0)\}$, we deduce that

$$1 \le \sup_{(z, \omega) \neq (1,0)} \frac{((z-1)^2 + \omega^2)^{\frac{\rho}{2}}}{\varphi(z, \omega)} < +\infty.$$

Since when $x=0$ or $y=x$, (4) holds with $\kappa_\rho$ replaced by 1, we conclude that the optimal constant is $\kappa_\rho = \sup_{(z,\omega) \neq (1,0)} \frac{((z-1)^2+\omega^2)^{\frac{\rho}{2}}}{\varphi(z, \omega)}$.

For the second inequality (5), we can apply the same approach: divided by $\|x\|^\rho$, it writes

$$(z^2 + \omega^2)^{\frac{\rho}{2}} - 1 \le \tilde{\kappa}_{\rho} ((z-1)^2 + \omega^2)^{\frac{1}{2}} ((z^2 + \omega^2)^{\frac{\rho-1}{2}} + 1).$$

As $(z, \omega) \to (1,0)$, $(z^2 + \omega^2)^{\frac{\rho}{2}} - 1 = (1 + 2(z-1) + (z-1)^2 + \omega^2)^{\frac{\rho}{2}} - 1 \sim \frac{\rho}{2}(2(z-1) + \omega^2)$

$$\limsup_{(z, \omega) \to (1,0)} \frac{(z^2 + \omega^2)^{\frac{\rho}{2}} - 1}{((z-1)^2 + \omega^2)^{\frac{1}{2}} \left(1 + (z^2 + \omega^2)^{\frac{\rho-1}{2}}\right)} = \limsup_{z \to 1} \frac{\rho(z-1)}{2|z-1|} = \frac{\rho}{2}.$$

On the other hand,

$$\lim_{|(z, \omega)| \to +\infty} \frac{(z^2 + \omega^2)^{\frac{\rho}{2}} - 1}{((z-1)^2 + \omega^2)^{\frac{1}{2}} \left(1 + (z^2 + \omega^2)^{\frac{\rho-1}{2}}\right)} = 1.$$

By continuity of the considered function over $\mathbb{R}^2 \setminus \{(1,0)\}$, we deduce that

$$\frac{\rho}{2} \vee 1 \le \sup_{(z, \omega) \ne (1,0)} \frac{(z^2 + \omega^2)^{\frac{\rho}{2}} - 1}{((z-1)^2 + \omega^2)^{\frac{1}{2}} \left(1 + (z^2 + \omega^2)^{\frac{\rho-1}{2}}\right)} < +\infty.$$

Since when $x=0$ or $y=x$, (5) holds with $\tilde{\kappa}_\rho$ replaced by 1, we conclude that the optimal constant is $\tilde{\kappa}_\rho = \sup_{(z,\omega) \neq (1,0)} \frac{(z^2+\omega^2)^{\frac{\rho}{2}}-1}{((z-1)^2+\omega^2)^{\frac{1}{2}} \left(1+(z^2+\omega^2)^{\frac{\rho-1}{2}}\right)}$. $\square$
References

[1] J. Backhoff-Veraguas and G. Pammer. Stability of martingale optimal transport and weak optimal transport. *Ann. Appl. Probab.*, 32(1):721-752, 2022.

[2] Martin Brückerhoff and Nicolas Juillet. Instability of martingale optimal transport in dimension $d \ge 2$. *Electron. Commun. Probab.*, 27:Paper No. 24, 10, 2022.

[3] D. Hobson and M. Klimmek. Robust price bounds for the forward starting straddle. *Finance and Stochastics*, 19(1):189-214, 2015.

[4] D. Hobson and A. Neuberger. Robust Bounds for Forward Start Options. *Mathematical Finance*, 22(1):31-56, 2012.

[5] B. Jourdain and W. Margheriti. Martingale Wasserstein inequality for probability measures in the convex order. *Bernoulli*, 28(2):830-858, 2022.

[6] B. Jourdain and K. Shao. Non-decreasing martingale couplings. arXiv:2305.00565, 2023.

[7] V. Strassen. The existence of probability measures with given marginals. *Annals of Mathematical Statistics*, 36(2):423-439, 1965.

[8] J. Wiesel. Continuity of the martingale optimal transport problem on the real line. *Ann. Appl. Probab.*, to appear.