# One dimensional martingale rearrangement couplings

B. Jourdain Université Paris-Est, Cermics (ENPC), INRIA F-77455 Marne-la-
Vallée, France. E-mails<EMAIL_ADDRESS><EMAIL_ADDRESS>- This research benefited from the support of the “Chaire Risques Financiers”,
Fondation du Risque. W. Margheriti11footnotemark: 1

###### Abstract

We are interested in martingale rearrangement couplings. As introduced by
Wiesel [37] in order to prove the stability of Martingale Optimal Transport
problems, these are projections in adapted Wasserstein distance of couplings
between two probability measures on the real line in the convex order onto the
set of martingale couplings between these two marginals. In reason of the lack
of relative compactness of the set of couplings with given marginals for the
adapted Wasserstein topology, the existence of such a projection is not clear
at all. Under a barycentre dispersion assumption on the original coupling
which is in particular satisfied by the Hoeffding-Fréchet or comonotone
coupling, Wiesel gives a clear algorithmic construction of a martingale
rearrangement when the marginals are finitely supported and then gets rid of
the finite support assumption by relying on a rather messy limiting procedure
to overcome the lack of relative compactness. Here, we give a direct general
construction of a martingale rearrangement coupling under the barycentre
dispersion assumption. This martingale rearrangement is obtained from the
original coupling by an approach similar to the construction we gave in [24]
of the inverse transform martingale coupling, a member of a family of
martingale couplings close to the Hoeffding-Fréchet coupling, but for a
slightly different injection in the set of extended couplings introduced by
Beiglböck and Juillet [9] and which involve the uniform distribution on
$[0,1]$ in addition to the two marginals. We last discuss the stability in
adapted Wassertein distance of the inverse transform martingale coupling with
respect to the marginal distributions.

Keywords: Martingale couplings, Martingale Optimal Transport, Adapted
Wasserstein distance, Robust finance, Convex order.

##  1 Introduction

Let $\rho\geq 1$ and $\mu,\nu$ be in the set $\mathcal{P}_{\rho}(\mathbb{R})$
of probability measures on the real line with finite order $\rho$ moment. We
denote by $\Pi(\mu,\nu)$ the set of couplings between $\mu$ and $\nu$, that is
$\pi\in\Pi(\mu,\nu)$ iff $\pi$ is a measure on $\mathbb{R}\times\mathbb{R}$
with first marginal $\mu$ and second marginal $\nu$. We denote by
$\Pi^{\mathrm{M}}(\mu,\nu)$ the set of martingale couplings between $\mu$ and
$\nu$:

$\Pi^{\mathrm{M}}(\mu,\nu)=\left\\{M\in\Pi(\mu,\nu)\mid\mu(dx)\textrm{-a.e.},\
\int_{\mathbb{R}}y\,M_{x}(dy)=x\right\\},$ (1.1)

where for any coupling $\pi\in\Pi(\mu,\nu)$ we denote by
$(\pi_{x})_{x\in\mathbb{R}}$ its disintegration with respect to its first
marginal, that is $\pi(dx,dy)=\mu(dx)\,\pi_{x}(dy)$, or with a slight abuse of
notation $\pi=\mu\times\pi_{x}$. The celebrated Strassen theorem [35] ensures
that $\Pi^{\mathrm{M}}(\mu,\nu)\neq\emptyset$ iff $\mu$ and $\nu$ are in the
convex order, which we denote $\mu\leq_{cx}\nu$, that is iff
$\int_{\mathbb{R}}f(x)\,\mu(dx)\leq\int_{\mathbb{R}}f(y)\,\nu(dy)$ for any
convex function $f:\mathbb{R}\to\mathbb{R}$.

Fix $\pi\in\Pi(\mu,\nu)$. We are interested in finding a projection of $\pi$
on the set $\Pi^{\mathrm{M}}(\mu,\nu)$ for the adapted Wasserstein distance
$\mathcal{AW}_{\rho}$ (defined in (1.5) below), that is finding a martingale
coupling $M$ between $\mu$ and $\nu$ such that

$\mathcal{AW}_{\rho}(\pi,M)=\inf_{M^{\prime}\in\Pi^{\mathrm{M}}(\mu,\nu)}\mathcal{AW}_{\rho}(\pi,M^{\prime}).$
(1.2)

This problem arose our interest when Wiesel [37] highlighted its connection
for $\rho=1$ with the stability of the Martingale Optimal Transport (MOT)
problem. The MOT problem was introduced in discrete time by Beiglböck, Henry-
Labordère and Penkner [6] and in continuous time by Galichon, Henry- Labordère
and Touzi [17] in order to get model-free bounds of an option price. It
consists in the classical Optimal Transport problem, which was formulated by
Gaspard Monge [27] in 1781 and modernised by Kantorovich [25] in 1942, to
which an additional martingale constraint is added in order to reflect the
arbitrage-free condition of the market. In our setting the MOT problem
consists in the minimisation

$\operatorname{MOT}(\mu,\nu):=\inf_{M\in\Pi^{\mathrm{M}}(\mu,\nu)}\int_{\mathbb{R}\times\mathbb{R}}C(x,y)\,M(dx,dy),$
(MOT)

where $C:\mathbb{R}\times\mathbb{R}\to\mathbb{R}_{+}$ is a nonnegative
measurable payoff function. The study of its stability, that is the continuity
of the map $(\mu,\nu)\mapsto\operatorname{MOT}(\mu,\nu)$, represents a major
stake, since it confirms the robustness of model-free bounds of an option
price. Backhoff-Veraguas and Pammer [5] gave a positive answer under mild
regularity assumptions by showing the stability of the so called martingale
$C$-monotonicity property, which is proved sufficient for optimality.
Independently, Wiesel [37] also gave a positive answer. More recently,
Beiglböck, Pammer and the two authors generalised those stability results to
the weak MOT problem [8]. For adaptations of celebrated results on classical
optimal transport theory to the MOT problem, we refer to Beiglböck and Juillet
[10], Henry-Labordère, Tan and Touzi [21] and Henry- Labordère and Touzi [22].
On duality, we refer to Beiglböck, Nutz and Touzi [12], Beiglböck, Lim and
Obłój [11] and De March [15]. We also refer to De March [14] and De March and
Touzi [16] for the multi-dimensional case.

We recall that the Wasserstein distance with index $\rho$ between $\mu$ and
$\nu$ is defined by

$\mathcal{W}_{\rho}(\mu,\nu)=\inf_{\pi\in\Pi(\mu,\nu)}\left(\int_{\mathbb{R}\times\mathbb{R}}|x-y|^{\rho}\,\pi(dx,dy)\right)^{1/\rho}.$
(1.3)

This family was meant to be as close as possible to the Hoeffding-Fréchet
coupling $\pi^{HF}$ between $\mu$ and $\nu$, that is the image of the Lebesgue
measure on $(0,1)$ by $u\mapsto(F_{\mu}^{-1}(u),F_{\nu}^{-1}(u))$, where
$F_{\eta}^{-1}$ denotes the quantile function of a probability measure $\eta$
on $\mathbb{R}$. The infimum is attained by the comonotonic or Hoeffding-
Fréchet coupling $\pi^{HF}$ between $\mu$ and $\nu$, that is the image of the
Lebesgue measure on $(0,1)$ by $u\mapsto(F_{\mu}^{-1}(u),F_{\nu}^{-1}(u))$,
where $F_{\eta}^{-1}(u)=\inf\\{x\in\mathbb{R}:\eta((-\infty,x])\geq u\\}$
denotes the quantile function of a probability measure $\eta$ on $\mathbb{R}$.
As a consequence,

$\mathcal{W}_{\rho}(\mu,\nu)=\left(\int_{(0,1)}|F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u)|^{\rho}du\right)^{1/\rho}.$
(1.4)

The topology induced by the Wasserstein distance is not always well suited for
any setting, especially in mathematical finance. Indeed, the symmetry of this
distance does not take into account the temporal structure of martingales. One
can easily get convinced that two stochastic processes very close in
Wasserstein distance can yield radically unalike information, as [3, Figure 1]
illustrates very well. Therefore, one needs to strengthen, or adapt this usual
topology. This can be done in many different ways, such as the adapted weak
topology (see below), Hellwig’s information topology [20], Aldous’s extended
weak topology [1] or the optimal stopping topology [4]. Strikingly, all those
apparently independent topologies are actually equal, at least in discrete
time [4, Theorem 1.1].

Hence it induces no loss of generality to focus on the so called adapted
Wasserstein distance. For an extensive background, we refer to [28, 29, 30,
31, 26, 13]. For all
$\mu^{\prime},\nu^{\prime}\in\mathcal{P}_{\rho}(\mathbb{R})$ and
$\pi^{\prime}\in\Pi(\mu^{\prime},\nu^{\prime})$, the adapted Wasserstein
distance with index $\rho$ between $\pi$ and $\pi^{\prime}$ is defined by

$\mathcal{AW}_{\rho}(\pi,\pi^{\prime})=\inf_{\chi\in\Pi(\mu,\mu^{\prime})}\left(\int_{\mathbb{R}\times\mathbb{R}}\left(|x-x^{\prime}|^{\rho}+\mathcal{W}_{\rho}^{\rho}(\pi_{x},\pi^{\prime}_{x^{\prime}})\right)\,\chi(dx,dx^{\prime})\right)^{1/\rho}.$
(1.5)

Note that by Lemma 5.1 below there always exists a coupling
$\chi\in\Pi(\mu,\mu)$ optimal for $\mathcal{AW}_{\rho}(\pi,\pi^{\prime})$.
Moreover it is easy to check that $\mathcal{W}_{\rho}\leq\mathcal{AW}_{\rho}$,
so that $\mathcal{AW}_{\rho}$ induces a finer topology than
$\mathcal{W}_{\rho}$. Wiesel [37] studies Problem (1.2) for $\rho=1$ and
introduces the notion of martingale rearrangement: a martingale coupling
$M\in\Pi^{\mathrm{M}}(\mu,\nu)$ is called a martingale rearrangement coupling
of $\pi$ if

$\mathcal{AW}_{1}(\pi,M)=\inf_{M^{\prime}\in\Pi^{\mathrm{M}}(\mu,\nu)}\mathcal{AW}_{1}(\pi,M^{\prime}).$
(1.6)

Actually, he works with the nested Wasserstein distance, which according to
the appendix is equal to the adapted Wasserstein distance. In the present
paper, even if we mainly concentrate on martingale rearrangements, we will
also consider a slight extension of the latter definition: a martingale
coupling $M\in\Pi^{\mathrm{M}}(\mu,\nu)$ is called an
$\mathcal{AW}_{\rho}$-minimal martingale rearrangement coupling of $\pi$ if

$\mathcal{AW}_{\rho}(\pi,M)=\inf_{M^{\prime}\in\Pi^{\mathrm{M}}(\mu,\nu)}\mathcal{AW}_{\rho}(\pi,M^{\prime}).$
(1.7)

Note that the existence of an $\mathcal{AW}_{\rho}$-minimal martingale
rearrangement coupling is not clear in the general case. Indeed, let
$(M_{n})_{n\in\mathbb{N}}$ be a sequence of martingale couplings between $\mu$
and $\nu$ such that $(\mathcal{AW}_{\rho}(\pi,M_{n}))_{n\in\mathbb{N}}$
converges to $\mathcal{AW}_{\rho}(\pi,M)$. The tightness of the marginals
$\mu$ and $\nu$ guarantees tightness and therefore relative compactness of
$(M_{n})_{n\in\mathbb{N}}$ for the $\mathcal{W}_{\rho}$-distance, but not
necessarily for the $\mathcal{AW}_{\rho}$-distance. In order to compensate
this lack of relative compactness, Wiesel [37] introduces a new assumption:
the coupling $\pi\in\Pi(\mu,\nu)$ is said to satisfy the barycentre dispersion
assumption iff

$\forall
a\in\mathbb{R},\quad\int_{\mathbb{R}}\mathds{1}_{[a,+\infty)}(x)\left(x-\int_{\mathbb{R}}y\,\pi_{x}(dy)\right)\,\mu(dx)\leq
0.$ (1.8)

The latter assumption is important in this context since it provides a
sufficient condition for a coupling $\pi$ between $\mu$ and $\nu$ to admit a
martingale rearrangement coupling. More precisely, Wiesel shows [37, Lemma
2.1] that in the general case,

$\inf_{M^{\prime}\in\Pi^{\mathrm{M}}(\mu,\nu)}\mathcal{AW}_{1}(\pi,M^{\prime})\geq\int_{\mathbb{R}}\left|\int_{\mathbb{R}}y\,\pi_{x}(dy)-x\right|\,\mu(dx),$
(1.9)

and there exists $M\in\Pi^{\mathrm{M}}(\mu,\nu)$ such that
$\mathcal{AW}_{1}(\pi,M)=\int_{\mathbb{R}}\left|\int_{\mathbb{R}}y\,\pi_{x}(dy)-x\right|\,\mu(dx)$
when $\pi$ satisfies the barycentre dispersion assumption (1.8) [37,
Proposition 2.4].

The problem (1.2) was in a certain way already considered by Rüschendorf [34],
who looked for a projection of a probability measure on a set of probability
measures with given linear constraints. Since the martingale constraint is
linear, his study encompasses our problem. Yet he considered the projection
with respect to the Kullback-Leibler distance, also known as relative entropy,
in place of $\mathcal{AW}_{\rho}$ and this does not suit our purpose. More
recently, Gerhold and Gülüm looked at a very similar problem [18, Problem 2.4]
for the infinity Wasserstein distance, the Prokhorov distance, the stop-loss
distance, the Lévy distance or modified versions of them. Once again, despite
being of great interest in their setting, in particular for their application
to the existence of a market model which is consistent with a finite set of
European call options prices on a single underlying asset [19], their choice
of distance is still inadequate for a connection with the stability of (MOT).

In Section 2 we briefly recall Wiesel’s construction [37] of a martingale
rearrangement coupling of any coupling $\pi$ which satisfies the barycentre
dispersion assumption (1.8). Then we design our own construction of a
martingale rearrangement coupling of $\pi$. This construction is actually done
through lifted couplings, in the sense of Beiglböck and Juillet [9], that is
probability measures on the enlarged space
$(0,1)\times\mathbb{R}\times\mathbb{R}$ in which the spatial domain
$\mathbb{R}\times\mathbb{R}$ of regular couplings is embedded.

Our construction in Section 2 is highly inspired of the one we did in [24],
where we designed a family $(M^{Q})_{Q\in\mathcal{Q}}$ of martingale couplings
between $\mu$ and $\nu$ parametrised by a set $\mathcal{Q}$ of probability
measures on $(0,1)^{2}$. This family was meant to be as close as possible to
the Hoeffding-Fréchet coupling $\pi^{HF}$ between $\mu$ and $\nu$. As proved
by Wiesel [37, Lemma 2.3], $\pi^{HF}$ satisfies the barycentre dispersion
assumption (1.8). We show in section 3 that the lifted coupling associated
with any element of $(M^{Q})_{Q\in\mathcal{Q}}$ is, in a very natural sense, a
lifted martingale rearrangement coupling of a lift of $\pi^{HF}$. At the level
of regular couplings on $\mathbb{R}\times\mathbb{R}$, we can conclude the same
as soon as the sign of $F_{\nu}^{-1}-F_{\mu}^{-1}$ is constant on the jumps of
$F_{\mu}$, which holds when $F_{\nu}^{-1}-F_{\mu}^{-1}$ is constant on these
jumps and $\pi^{HF}$ is concentrated on the graph of the Monge transport map
$T=F_{\nu}^{-1}\circ F_{\mu}$.

We showed in [24] that a particular element stands out from the family
$(M^{Q})_{Q\in\mathcal{Q}}$, the so called inverse transform martingale
coupling. We finally show in Section 4 the stability of the inverse transform
martingale coupling for the $\mathcal{AW}_{\rho}$-distance with respect to its
marginals. The latter stability holds in full generality at the lifted level
but a condition on the first marginals is needed at the level of regular
couplings.

Let us now recall some standard results about cumulative distribution
functions and quantile functions since they will prove very handy one-
dimensional tools. Proofs can be found for instance in [24, Appendix]. For any
probability measure $\eta$ on $\mathbb{R}$, denoting by
$F_{\eta}(x)=\eta((-\infty,x])$ and
$F_{\eta}^{-1}(u)=\inf\\{x\in\mathbb{R}:F_{\eta}(x)\geq u\\}$ the cumulative
distribution function and the quantile function of $\eta$, we have

  1. (1)

$F_{\eta}$, resp. $F_{\eta}^{-1}$, is right continuous, resp. left continuous,
and nondecreasing;

  2. (2)

For all $(x,u)\in\mathbb{R}\times(0,1)$,

$F_{\eta}^{-1}(u)\leq x\iff u\leq F_{\eta}(x),$ (1.10)

which implies

$\displaystyle F_{\eta}(x-)<u\leq F_{\eta}(x)\implies x=F_{\eta}^{-1}(u),$
(1.11) and $\displaystyle F_{\eta}(F_{\eta}^{-1}(u)-)\leq u\leq
F_{\eta}(F_{\eta}^{-1}(u));$ (1.12)

  3. (3)

For $\eta(dx)$-almost every $x\in\mathbb{R}$,

$0<F_{\eta}(x),\quad F_{\eta}(x-)<1\quad\text{and}\quad
F_{\eta}^{-1}(F_{\eta}(x))=x;$ (1.13)

  4. (4)

The image of the Lebesgue measure on $(0,1)$ by $F_{\eta}^{-1}$ is $\eta$.
This property is referred to as inverse transform sampling.

  5. (5)

Denoting by $\lambda_{(0,1)}$, resp. $\lambda_{(0,1)^{2}}$, the Lebesgue
measure on $(0,1)$, resp. $(0,1)^{2}$ and setting

$\theta(x,v)=F_{\mu}(x-)+v\mu(\\{x\\})\quad\textrm{for}\quad(x,v)\in\mathbb{R}\times[0,1],$
(1.14)

we have

$\left((u,v)\mapsto\theta(F_{\mu}^{-1}(u),v)\right)_{\sharp}\lambda_{(0,1)^{2}}=\lambda_{(0,1)},$
(1.15)

where $\sharp$ denotes the pushforward operation. Coupled with the inverse
transform sampling we also have the equivalent formulation

$\theta_{\sharp}(\mu\times\lambda_{(0,1)})=\lambda_{(0,1)}.$ (1.16)

##  2 Martingale rearrangements of couplings which satisfy the barycentre
dispersion assumption

###  2.1 Regular and lifted martingale rearrangement couplings

By (1.4) for the first equality and the inverse transform sampling for the
second one, we have for $\eta,\eta^{\prime}\in\mathcal{P}_{1}(\mathbb{R})$,

$\displaystyle\mathcal{W}_{1}(\eta,\eta^{\prime})=\int_{(0,1)}\left|F_{\eta}^{-1}(u)-F_{\eta^{\prime}}^{-1}(u)\right|du\geq\left|\int_{(0,1)}F_{\eta}^{-1}(u)du-\int_{(0,1)}F_{\eta^{\prime}}^{-1}(u)du\right|=\left|\int_{\mathbb{R}}x\,\eta(dx)-\int_{\mathbb{R}}x\,\eta^{\prime}(dx)\right|.$
(2.1)

The inequality is an equality iff either $\forall u\in(0,1)$,
$F_{\eta}^{-1}(u)\leq F_{\eta^{\prime}}^{-1}(u)$ i.e. $\eta$ is smaller then
$\eta^{\prime}$ for the stochastic order which we denote
$\eta\leq_{st}\eta^{\prime}$ or $\forall u\in(0,1)$, $F_{\eta}^{-1}(u)\geq
F_{\eta^{\prime}}^{-1}(u)$ i.e. $\eta\geq_{st}\eta^{\prime}$.

Let $\mu,\nu\in\mathcal{P}_{1}(\mathbb{R})$ such that $\mu\leq_{cx}\nu$. We
are now ready to reproduce the proof of [37, Lemma 2.1] to check (1.9). For
$M\in\Pi^{\mathrm{M}}(\mu,\nu)$ and $\chi\in\Pi(\mu,\mu)$ we have, using (2.1)
then the triangle inequality,

$\displaystyle\begin{split}\int_{\mathbb{R}\times\mathbb{R}}\left(|x-x^{\prime}|+\mathcal{W}_{1}(\pi_{x},M_{x^{\prime}})\right)\,\chi(dx,dx^{\prime})&\geq\int_{\mathbb{R}\times\mathbb{R}}\left(|x-x^{\prime}|+\left|\int_{\mathbb{R}}y\,\pi_{x}(dy)-x^{\prime}\right|\right)\,\chi(dx,dx^{\prime})\\\
&\geq\int_{\mathbb{R}}\left|\int_{\mathbb{R}}y\,\pi_{x}(dy)-x\right|\,\mu(dx).\end{split}$
(2.2)

When $\pi$ satisfies the barycentre dispersion assumption (1.8), finding a
martingale rearrangement coupling of $\pi$ amounts to find a martingale
coupling such that the inequalities in (2.2) are equalities. This observation
leads to the following lemma.

######  Lemma 2.1.

Let $\mu,\nu\in\mathcal{P}_{1}(\mathbb{R})$ and $\pi\in\Pi(\mu,\nu)$ satisfy
the barycentre dispersion assumption (1.8). Then
$M\in\Pi^{\mathrm{M}}(\mu,\nu)$ is a martingale rearrangement coupling of
$\pi$ iff there exists $\chi\in\Pi(\mu,\mu)$ such that
$\chi(dx,dx^{\prime})$-almost everywhere,

$x<x^{\prime}\implies\pi_{x}\geq_{st}M_{x^{\prime}},\quad
x>x^{\prime}\implies\pi_{x}\leq_{st}M_{x^{\prime}}\quad\textrm{and}\quad
x=x^{\prime}\implies\pi_{x}\leq_{st}M_{x}\textrm{ or }\pi_{x}\geq_{st}M_{x},$
(2.3)

in which case $\chi$ is optimal for $\mathcal{AW}_{1}(\pi,M)$.

###### Proof.

Suppose that $M$ is a martingale rearrangement coupling of $\pi$ and $\chi$ is
optimal for $\mathcal{AW}_{1}(\pi,M)$. Since $\pi$ satisfies the barycentre
dispersion assumption, we know by [37, Proposition 2.4] that
$\mathcal{AW}_{1}(M,\pi)=\int_{\mathbb{R}}\left|\int_{\mathbb{R}}y\,\pi_{x}(dy)-x\right|\,\mu(dx)$.
Then the first inequality in (2.2) is an equality, hence
$\chi(dx,dx^{\prime})$-almost everywhere,
$\mathcal{W}_{1}(\pi_{x},M_{x^{\prime}})=\left|\int_{\mathbb{R}}y\,\pi_{x}(dy)-x^{\prime}\right|$,
or equivalently $\pi_{x}$ and $M_{x^{\prime}}$ are comparable in the
stochastic order. Morever the second inequality in (2.2) is an equality as
well, hence $\chi(dx,dx^{\prime})$-almost everywhere, $x^{\prime}$ lies
between $x$ and $\int_{\mathbb{R}}y\,\pi_{x}(dy)$. We deduce that
$\chi(dx,dx^{\prime})$-almost everywhere,

$(x-x^{\prime})\left(x^{\prime}-\int_{\mathbb{R}}y\,\pi_{x}(dy)\right)\geq 0,$
(2.4)

and $\pi_{x}\leq_{st}M_{x^{\prime}}$ or $\pi_{x}\geq_{st}M_{x^{\prime}}$. Then
(2.3) is easily deduced from the fact that the map
$\eta\mapsto\int_{\mathbb{R}}z\,\eta(dz)$ is increasing for the stochastic
order.

Conversely, suppose that (2.3) and therefore (2.4) holds for some
$\chi\in\Pi(\mu,\mu)$. Then the inequalities in (2.2) are equalities, hence
$\chi$ is optimal for $\mathcal{AW}_{1}(\pi,M)$ and $M$ is a martingale
rearrangement coupling of $\pi$. ∎

To construct a martingale rearrangement coupling of $\pi$ satisfying the
barycenter dispersion assumption (1.8), we will define a probability kernel
$(m_{u})_{u\in(0,1)}$ such that
$\int_{\mathbb{R}}y\,m_{u}(dy)=F_{\mu}^{-1}(u)$ $du$-a.e. and deduce that the
probability measure

$M(dx,dy)=\int_{0}^{1}\delta_{F_{\mu}^{-1}(u)}(dx)\,m_{u}(dy)\,du$ (2.5)

is a martingale coupling between $\mu$ and $\nu$. Yet the probability kernel
$(m_{u})_{u\in(0,1)}$ is not uniquely determined from the knowledge of $M$.
Hence the definition (2.5) induces a loss of information. In order to keep
this information, one can consider like Beiglböck and Juillet [9] instead of
$M$ its lifted martingale coupling

$\widehat{M}(du,dx,dy)=\lambda_{(0,1)}(du)\,\delta_{F_{\mu}^{-1}(u)}(dx)\,m_{u}(dy)\in\Pi(\lambda_{(0,1)},\mu,\nu),$
(2.6)

where $\lambda_{(0,1)}$ denotes the Lebesgue measure on $(0,1)$. More
generally, for any $\pi\in\Pi(\mu,\nu)$, we call lifted coupling of $\pi$ any
coupling $\widehat{\pi}\in\Pi(\lambda_{(0,1)},\mu,\nu)$ such that there exists
a probability kernel $(p_{u})_{u\in(0,1)}$ which satisfies

$\widehat{\pi}(du,dx,dy)=\lambda_{(0,1)}(du)\,\delta_{F_{\mu}^{-1}(u)}(dx)\,p_{u}(dy)\quad\textrm{and}\quad\int_{u\in(0,1)}\widehat{\pi}(du,dx,dy)=\pi(dx,dy).$

We denote by $\widehat{\Pi}(\mu,\nu)$ the set of all lifted couplings between
$\mu$ and $\nu$. Notice that there exists an easy embedding

$\iota:\Pi(\mu,\nu)\to\widehat{\Pi}(\mu,\nu),\quad\pi\mapsto\lambda_{(0,1)}(du)\,\delta_{F_{\mu}^{-1}(u)}(dx)\,\pi_{F_{\mu}^{-1}(u)}(dy).$
(2.7)

For
$\widehat{\pi}=\lambda_{(0,1)}\times\widehat{\pi}_{u}=\lambda_{(0,1)}\times\delta_{F_{\mu}^{-1}(u)}\times
p_{u}$ and
$\widehat{\pi}^{\prime}=\lambda_{(0,1)}\times\widehat{\pi}^{\prime}_{u}=\lambda_{(0,1)}\times\delta_{F_{\mu}^{-1}(u)}\times
p^{\prime}_{u}$ two lifted couplings of $\pi\in\Pi(\mu,\nu)$ and
$\pi^{\prime}\in\Pi(\mu^{\prime},\nu^{\prime})$, we define their lifted
adapted Wasserstein distance of order $\rho$ by

$\displaystyle\widehat{\mathcal{AW}}_{\rho}(\pi,\pi^{\prime})$
$\displaystyle=\inf_{\chi\in\Pi(\lambda_{(0,1)},\lambda_{(0,1)})}\left(\int_{(0,1)\times(0,1)}\left(|u-u^{\prime}|^{\rho}+\mathcal{AW}_{\rho}^{\rho}(\widehat{\pi}_{u},\widehat{\pi}^{\prime}_{u^{\prime}})\right)\,\chi(du,du^{\prime})\right)^{1/\rho}$
$\displaystyle=\inf_{\chi\in\Pi(\lambda_{(0,1)},\lambda_{(0,1)})}\left(\int_{(0,1)\times(0,1)}\left(|u-u^{\prime}|^{\rho}+|F_{\mu}^{-1}(u)-F_{\mu^{\prime}}^{-1}(u^{\prime})|^{\rho}+\mathcal{W}_{\rho}^{\rho}(p_{u},p^{\prime}_{u^{\prime}})\right)\,\chi(du,du^{\prime})\right)^{1/\rho}.$

Note that by Remark 5.2 below there always exists a coupling
$\chi\in\Pi(\lambda_{(0,1)},\lambda_{(0,1)})$ optimal for
$\widehat{\mathcal{AW}}_{\rho}(\widehat{\pi},\widehat{\pi}^{\prime})$. We
denote by $\widehat{\Pi}^{\mathrm{M}}(\mu,\nu)$ the set of all lifted
martingale couplings between $\mu$ and $\nu$, that is the set of all lifted
couplings $\lambda_{(0,1)}\times\delta_{F_{\mu}^{-1}(u)}\times
m_{u}\in\widehat{\Pi}(\mu,\nu)$ such that
$\int_{\mathbb{R}}y\,m_{u}(dy)=F_{\mu}^{-1}(u)$ for $du$-almost all
$u\in(0,1)$. For $\rho\geq 1$, we then call lifted
$\widehat{\mathcal{AW}}_{\rho}$-minimal martingale rearrangement coupling (or
simply lifted martingale rearrangement coupling when $\rho=1$) of
$\widehat{\pi}\in\widehat{\Pi}(\mu,\nu)$ any lifted martingale coupling
$\widehat{M}\in\widehat{\Pi}^{\mathrm{M}}(\mu,\nu)$ such that

$\widehat{\mathcal{AW}}_{\rho}(\widehat{\pi},\widehat{M})=\inf_{\widehat{M}^{\prime}\in\widehat{\Pi}^{\mathrm{M}}(\mu,\nu)}\widehat{\mathcal{AW}}_{\rho}(\widehat{\pi},\widehat{M}^{\prime}).$

Ignoring the non-negative contribution of $|u-u^{\prime}|$ in the definition
of $\widehat{\mathcal{AW}}_{1}$ and reasoning like in (2.2), we easily check
the following lower bound analogous, at the lifted level, to (1.9).

######  Lemma 2.2.

Let $\mu,\nu\in\mathcal{P}_{1}(\mathbb{R})$ be such that $\mu\leq_{cx}\nu$.
Then for all
$\widehat{\pi}=\lambda_{(0,1)}\times\delta_{F_{\mu}^{-1}(u)}\times
p_{u}\in\widehat{\Pi}(\mu,\nu)$,

$\inf_{\widehat{M}\in\widehat{\Pi}^{\mathrm{M}}(\mu,\nu)}\widehat{\mathcal{AW}}_{1}(\widehat{\pi},\widehat{M})\geq\int_{(0,1)}\left|\int_{\mathbb{R}}y\,p_{u}(dy)-F_{\mu}^{-1}(u)\right|\,du.$

The next proposition gives a sufficient condition for the collapse through
(2.5) of a lifted martingale coupling to be a martingale rearrangement.

######  Proposition 2.3.

Let $\mu,\nu\in\mathcal{P}_{1}(\mathbb{R})$ be such that $\mu\leq_{cx}\nu$.
Let $\widehat{\pi}=\lambda_{(0,1)}\times\delta_{F_{\mu}^{-1}(u)}\times
p_{u}\in\widehat{\Pi}(\mu,\nu)$ be such that $u\mapsto p_{u}$ is constant on
the jumps of $F_{\mu}$, that is constant on the intervals
$(F_{\mu}(x-),F_{\mu}(x)]$, $x\in\mathbb{R}$, which is trivially satisfied
when $\mu$ is atomless. Suppose that
$\widehat{M}=\lambda_{(0,1)}\times\delta_{F_{\mu}^{-1}(u)}\times
m_{u}\in\widehat{\Pi}^{\mathrm{M}}(\mu,\nu)$ is such that

$\int_{(0,1)}\mathcal{W}_{1}(p_{u},m_{u})\,du\leq\int_{(0,1)}\left|\int_{\mathbb{R}}y\,p_{u}(dy)-F_{\mu}^{-1}(u)\right|\,du.$

Then the martingale coupling
$M(dx,dy)=\int_{u\in(0,1)}\delta_{F_{\mu}^{-1}(u)}(dx)\,m_{u}(dy)\,du$ is a
martingale rearrangement coupling of
$\pi=\int_{u\in(0,1)}\delta_{F_{\mu}^{-1}(u)}(dx)\,p_{u}(dy)\,du$ which
satisfies

$\mathcal{AW}_{1}(\pi,M)=\int_{\mathbb{R}}\mathcal{W}_{1}(\pi_{x},M_{x})\,\mu(dx)=\int_{\mathbb{R}}\left|\int_{\mathbb{R}}y\,\pi_{x}(dy)-x\right|\,\mu(dx).$

Of course, under the hypotheses,
$\widehat{\mathcal{AW}}_{1}(\widehat{\pi},\widehat{M})\leq\int_{(0,1)}\mathcal{W}_{1}(p_{u},m_{u})\,du\leq\int_{(0,1)}\left|\int_{\mathbb{R}}y\,p_{u}(dy)-F_{\mu}^{-1}(u)\right|\,du$
so that, by Lemma 2.2, these inequalities are equalities and $\widehat{M}$ is
a lifted martingale rearrangement of $\widehat{\pi}$.

###### Proof.

By (1.9) it suffices to show that

$\int_{\mathbb{R}}\mathcal{W}_{1}(\pi_{x},M_{x})\,\mu(dx)\leq\int_{\mathbb{R}}\left|\int_{\mathbb{R}}y\,\pi_{x}(dy)-x\right|\,\mu(dx).$
(2.8)

For $(x,v)\in\mathbb{R}\times(0,1)$, let
$\theta(x,v)=F_{\mu}(x-)+v\mu(\\{x\\})$. Using (1.16) and the fact that
$F_{\mu}^{-1}(\theta(x^{\prime},v))=x^{\prime}$ for all
$(x^{\prime},v)\in\mathbb{R}\times(0,1)$ , we get

$\displaystyle\begin{split}\pi(dx,dy)&=\int_{u\in(0,1)}\delta_{F_{\mu}^{-1}(u)}(dx)\,p_{u}(dy)\,du=\int_{(x^{\prime},v)\in\mathbb{R}\times(0,1)}\delta_{x^{\prime}}(dx)\,p_{\theta(x^{\prime},v)}(dy)\,\mu(dx^{\prime})\,dv\\\
&=\int_{v\in(0,1)}\mu(dx)\,p_{\theta(x,v)}(dy)\,dv.\end{split}$ (2.9)

Hence we have $\mu(dx)$-almost everywhere
$\pi_{x}(dy)=\int_{0}^{1}p_{\theta(x,v)}(dy)\,dv$, and similarly we find
$M_{x}(dy)=\int_{0}^{1}m_{\theta(x,v)}(dy)\,dv$. Using (1.16) for the first
and last equality, we deduce that

$\displaystyle\int_{\mathbb{R}}\mathcal{W}_{1}(\pi_{x},M_{x})\,\mu(dx)$
$\displaystyle\leq\int_{\mathbb{R}\times(0,1)}\mathcal{W}_{1}(m_{\theta(x,v)},p_{\theta(x,v)})\,\mu(dx)\,dv$
$\displaystyle=\int_{(0,1)}\mathcal{W}_{1}(m_{u},p_{u})\,du$
$\displaystyle\leq\int_{(0,1)}\left|\int_{\mathbb{R}}y\,p_{u}(dy)-F_{\mu}^{-1}(u)\right|\,du$
$\displaystyle=\int_{\mathbb{R}\times(0,1)}\left|\int_{\mathbb{R}}y\,p_{\theta(x,v)}(dy)-F_{\mu}^{-1}(\theta(x,v))\right|\,\mu(dx)\,dv.$

For $(x,v)\in\mathbb{R}\times(0,1)$, $F_{\mu}^{-1}(\theta(x,v))=x$, and since
$u\mapsto p_{u}$ is constant on the jumps of $F_{\mu}$, the map $v\mapsto
p_{\theta(x,v)}$ is constant on $(0,1)$, hence

$\displaystyle\int_{(0,1)}\left|\int_{\mathbb{R}}y\,p_{\theta(x,v)}(dy)-F_{\mu}^{-1}(\theta(x,v))\right|\,dv$
$\displaystyle=\left|\int_{\mathbb{R}\times(0,1)}y\,p_{\theta(x,v)}(dy)\,dv-x\right|.$

We deduce that

$\int_{\mathbb{R}}\mathcal{W}_{1}(\pi_{x},M_{x})\,\mu(dx)\leq\int_{\mathbb{R}}\left|\int_{\mathbb{R}\times(0,1)}y\,p_{\theta(x,v)}(dy)\,dv-x\right|\,\mu(dx)=\int_{\mathbb{R}}\left|\int_{\mathbb{R}}y\,\pi_{x}(dy)-x\right|\,\mu(dx),$

which proves (2.8) and concludes the proof. ∎

###  2.2 Construction of an explicit martingale rearrangement coupling

We recall that a coupling $\pi\in\Pi(\mu,\nu)$ between two probability
measures $\mu,\nu\in\mathcal{P}_{1}(\mathbb{R})$ in the convex order satisfies
the barycentre dispersion assumption formulated by Wiesel [37] iff

$\forall
a\in\mathbb{R},\quad\int_{\mathbb{R}}\mathds{1}_{[a,+\infty)}(x)\left(x-\int_{\mathbb{R}}y\,\pi_{x}(dy)\right)\,\mu(dx)\leq
0.$ (2.10)

First we briefly recall Wiesel’s construction [37] of a martingale
rearrangement coupling of a coupling $\pi$ which satisfies (2.10), which is
well perceivable as soon as $\pi$ has finite support but becomes rather
implicit in the general case. Then we design our own construction of such a
martingale rearrangement coupling, whose intelligibility does not depend on
the finiteness of the support of $\pi$. Since the Hoeffding-Fréchet satisfies
(2.10) [37, Lemma 2.3], this construction extends the study made in Section 3.

Let $\mu,\nu\in\mathcal{P}_{1}(\mathbb{R})$ be such that $\mu\leq_{cx}\nu$ and
$\mu\neq\nu$ and $\pi\in\Pi(\mu,\nu)$ be a coupling between $\mu$ and $\nu$
which satisfies the barycentre assumption (2.10). Suppose first that $\pi$ has
finite support and is not a martingale coupling between $\mu$ and $\nu$.
Denoting $S$ the finite support of $\mu$ and for all $x\in S$, $S_{x}$ the
finite support of $\pi_{x}$, the latter is equivalent to say that there exists
$x\in S$ such that $\int_{\mathbb{R}}y\,\pi_{x}(dy)\neq x$. As Wiesel [37]
points out, the barycentre dispersion assumption (2.10) and the convex order
between $\mu$ and $\nu$ imply the existence of $x^{-},x^{+}\in S$, $y^{-}\in
S_{x^{-}}$ and $y^{+}\in S_{x^{+}}$ such that

$\int_{\mathbb{R}}y\,\pi_{x^{-}}(dy)<x^{-},\quad
x^{+}<\int_{\mathbb{R}}y\,\pi_{x^{+}}(dy)\quad\text{and}\quad y^{-}<y^{+}.$

He then switches as much as possible the mass at $y^{-}$ and $y^{+}$ of
$\pi_{x^{-}}$ and $\pi_{x^{+}}$ in order to rectify the barycentres. More
precisely, he defines for all $x\in S$

$\pi^{(1)}_{x}=\mathds{1}_{\\{x\notin\\{x^{-},x^{+}\\}\\}}\,\pi_{x}+\mathds{1}_{\\{x=x^{-}\\}}\left(\pi_{x^{-}}+\frac{\lambda}{\mu(\\{x^{-}\\})}(\delta_{y^{+}}-\delta_{y^{-}})\right)+\mathds{1}_{\\{x=x^{+}\\}}\left(\pi_{x^{+}}+\frac{\lambda}{\mu(\\{x^{+}\\})}(\delta_{y^{-}}-\delta_{y^{+}})\right),$

where $\lambda\geq 0$ is taken as large as possible, so that

$\text{either}\quad\pi^{(1)}_{x^{-}}(\\{y^{-}\\})=0,\quad\pi^{(1)}_{x^{+}}(\\{y^{+}\\})=0,\quad\int_{\mathbb{R}}y\,\pi^{(1)}_{x^{+}}(dy)=x^{+}\quad\text{or}\quad\int_{\mathbb{R}}y\,\pi_{x^{-}}(dy)=x^{-}.$

Then the measure $\pi^{(1)}(dx,dy)=\mu(dx)\,\pi^{(1)}_{x}(dy)$ is a coupling
between $\mu$ and $\nu$ which satisfies the barycentre dispersion assumption
(2.10). Repeating inductively this process yields a sequence
$(\pi^{(n)}_{x})_{x\in\mathbb{R}}$ of couplings between $\mu$ and $\nu$ which
satisfy (2.10). By finiteness of $S$, the latter sequence is constant for $n$
large enough and the limit is precisely a martingale rearrangement coupling of
$\pi$.

In the general case, there exists by [37, Lemma 4.1] a sequence
$(\pi^{n})_{n\in\mathbb{N}^{*}}$ of finitely supported measures such that
$\mathcal{W}_{1}^{nd}(\pi^{n},\pi)\leq 1/n$ for all $n\in\mathbb{N}^{*}$. The
marginals $\mu_{n}$ and $\nu_{n}$ of $\pi^{n}$ are not in the convex order,
but a mere adaptation of the previous reasoning yields the existence of a
coupling $\pi^{n}_{mr}$ between $\mu_{n}$ and $\nu_{n}$ which is almost a
martingale rearrangement coupling of $\pi^{n}$, in the sense that

$\int_{\mathbb{R}}\left|x-\int_{\mathbb{R}}y\,(\pi^{n}_{mr})_{x}(dy)\right|\,\mu^{n}(dx)\leq\frac{1}{n}\quad\text{and}\quad\mathcal{W}_{1}^{nd}(\pi^{n}_{mr},\pi^{n})\leq\int_{\mathbb{R}}\left|x-\int_{\mathbb{R}}y\,\pi^{n}_{x}(dy)\right|\,\mu^{n}(dx).$
(2.11)

Then Wiesel shows the existence of a coupling $\pi_{mr}$ between $\mu$ and
$\nu$ such that
$\mathcal{W}_{1}^{nd}\left(\frac{1}{n}\sum_{k=1}^{n}\pi^{k}_{mr},\pi_{mr}\right)$
vanishes as $n$ goes to $+\infty$. By (2.11) taken to the limit $n\to+\infty$
and (1.9) he deduces that $\pi_{mr}$ is a martingale rearrangement coupling of
$\pi$.

We now propose an alternate construction of a martingale rearrangement
coupling of $\pi$, regardless of the finiteness of its support, deduced from a
lifted martingale coupling. For all $u\in[0,1]$, let

$G(u)=\int_{\mathbb{R}}y\,\pi_{F_{\mu}^{-1}(u)}(dy),\quad\Delta_{+}(u)=\int_{0}^{u}(F_{\mu}^{-1}-G)^{+}(v)\,dv\quad\text{and}\quad\Delta_{-}(u)=\int_{0}^{u}(F_{\mu}^{-1}-G)^{-}(v)\,dv.$
(2.12)

Let us show that (2.10) is equivalent to

$\forall u\in[0,1],\quad\Delta_{+}(u)\geq\Delta_{-}(u).$ (2.13)

Using (1.10) we see that for all $u\in(0,1)$ and $a\in\mathbb{R}$,
$u>F_{\mu}(a-)\implies F_{\mu}^{-1}(u)\geq a\implies u\geq F_{\mu}(a-)$. By
the latter implications and the inverse transform sampling we deduce that
(2.10) is equivalent to

$\forall
a\in\mathbb{R},\quad\int_{F_{\mu}(a-)}^{1}(F_{\mu}^{-1}(u)-G(u))\,du\leq 0.$

Since $\Delta_{+}(1)=\Delta_{-}(1)$, consequence of the equality of the
respective means of $\mu$ and $\nu$, we deduce that it is equivalent to

$\forall
a\in\mathbb{R},\quad\Delta_{+}(F_{\mu}(a-))\geq\Delta_{-}(F_{\mu}(a-)).$

By right continuity of $F_{\mu}$, for all $a\in\mathbb{R}$ we have
$F_{\mu}(a)=\lim_{h\to 0,h>0}F_{\mu}((a+h)-)$, so by continuity of
$\Delta_{+}$ and $\Delta_{-}$ we also have
$\Delta_{+}(F_{\mu}(a))\geq\Delta_{-}(F_{\mu}(a))$ for all $a\in\mathbb{R}$.
Moreover, for all $a\in\mathbb{R}$ such that $\mu(\\{a\\})>0$ and
$u\in(F_{\mu}(a-),F_{\mu}(a)]$, we have by (1.11) that $F_{\mu}^{-1}(u)=a$, so
$\Delta_{+}$ and $\Delta_{-}$ are affine on $(F_{\mu}(a-),F_{\mu}(a)]$. We
deduce that we also have $\Delta_{+}\geq\Delta_{-}$ on
$(F_{\mu}(a-),F_{\mu}(a)]$, hence the equivalence with (2.13).

We define

$\displaystyle\mathcal{U}^{+}=\\{u\in(0,1)\mid
F_{\mu}^{-1}(u)>G(u)\\},\quad\mathcal{U}^{-}=\\{u\in(0,1)\mid
F_{\mu}^{-1}(u)<G(u)\\},$
$\displaystyle\text{and}\quad\mathcal{U}^{0}=\\{u\in(0,1)\mid
F_{\mu}^{-1}(u)=G(u)\\},$

and thanks to the equality $\Delta_{+}(1)=\Delta_{-}(1)$ we can set for all
$u\in[0,1]$

$\phi(u)=\left\\{\begin{array}[]{rcl}\Delta_{-}^{-1}(\Delta_{+}(u))&\text{if}&u\in\mathcal{U}^{+};\\\
\Delta_{+}^{-1}(\Delta_{-}(u))&\text{if}&u\in\mathcal{U}^{-};\\\
u&\text{if}&u\in\mathcal{U}^{0}.\end{array}\right.$

Applying [24, Lemma 6.1] again with $f_{1}=(F_{\mu}^{-1}-G)^{+}$,
$f_{2}=(F_{\mu}^{-1}-G)^{-}$, $u_{0}=1$ and
$h:u\mapsto\mathds{1}_{\\{G(\phi(u))\leq F_{\mu}^{-1}(\phi(u))\\}}$ yields

$\int_{0}^{1}\mathds{1}_{\\{G(\phi(u))\leq
F_{\mu}^{-1}(\phi(u))\\}}\,d\Delta_{+}(u)=\int_{0}^{1}\mathds{1}_{\\{G(v)\leq
F_{\mu}^{-1}(v)\\}}\,d\Delta_{-}(u)=0.$

Similarly, we get $\int_{0}^{1}\mathds{1}_{\\{G(\phi(u))\geq
F_{\mu}^{-1}(\phi(u))\\}}\,d\Delta_{-}(u)=0$. We deduce that

$\phi(u)\in\mathcal{U}^{-},\quad\textrm{resp.
}\phi(u)\in\mathcal{U}^{+},\quad\text{for $du$-almost all
}u\in\mathcal{U}^{+},\quad\text{resp. }\mathcal{U}^{-}.$ (2.14)

This allows us to define for $du$-almost all
$u\in\mathcal{U}^{+}\cup\mathcal{U}^{-}$

$p(u)=\frac{G(\phi(u))-F_{\mu}^{-1}(\phi(u))}{F_{\mu}^{-1}(u)-G(u)+G(\phi(u))-F_{\mu}^{-1}(\phi(u))}\cdot$
(2.15)

Notice that (2.14) implies that for $du$-almost all $u\in\mathcal{U}^{+}$,
$\phi(\phi(u))=\Delta_{+}^{-1}(\Delta_{-}(\Delta_{-}^{-1}(\Delta_{+}(u))))$.
Since $\Delta_{-}$ is continuous we have $\Delta_{-}(\Delta_{-}^{-1}(v))=v$
for all $v\in[0,\Delta_{-}(1)]$, and using (1.13) after an appropriate
normalisation we get $\Delta_{+}^{-1}(\Delta_{+}(v))=v$ for $dv$-almost all
$v\in\mathcal{U}^{+}$. We deduce that

$u=\phi(\phi(u)),\quad\text{$du$-almost everywhere on $\mathcal{U}^{+}$}.$
(2.16)

Similarly, $\phi(\phi(u))=u$ for all $du$-almost all $u\in\mathcal{U}^{-}$. We
deduce that

$\text{for $du$-almost all
$u\in\mathcal{U}^{+}\cup\mathcal{U}^{-}$},\quad\phi(\phi(u))=u,$ (2.17)

and

$\text{for $du$-almost all $u\in\mathcal{U}^{+}\cup\mathcal{U}^{-}$},\quad
p(\phi(u))=\frac{G(u)-F_{\mu}^{-1}(u)}{F_{\mu}^{-1}(\phi(u))-G(\phi(u))+G(u)-F_{\mu}^{-1}(u)}=1-p(u).$
(2.18)

In order to define the appropriate martingale kernel, we rely on the following
lemma which allows us to inject some stochastic order in the construction, a
convenient tool for the computation of Wasserstein distances. We recall that
two probability measures $\mu$ and $\nu$ on the real line are said to be in
the stochastic order, denoted $\mu\leq_{st}\nu$, iff $F_{\mu}^{-1}(u)\leq
F_{\nu}^{-1}(u)$ for all $u\in[0,1]$. Since the Hoeffding-Fréchet coupling
between $\mu$ and $\nu$ is optimal for $\mathcal{W}_{1}(\mu,\nu)$, this
implies by the inverse transform sampling that
$\mathcal{W}_{1}(\mu,\nu)=\int_{\mathbb{R}}y\,\nu(dy)-\int_{\mathbb{R}}x\,\mu(dx)$.

######  Lemma 2.4.

Let $\mathfrak{B}$ be the set of all quadruples
$(y,\widetilde{y},\mu,\widetilde{\mu})\in\mathbb{R}\times\mathbb{R}\times\mathcal{P}_{1}(\mathbb{R})\times\mathcal{P}_{1}(\mathbb{R})$
such that $\mu$ and $\widetilde{\mu}$ have respective means $x$ and
$\widetilde{x}$ and $x<y\leq\widetilde{y}<\widetilde{x}$. Endow
$\mathcal{P}_{1}(\mathbb{R})$ with the Borel $\sigma$-algebra of the weak
convergence topology and $\mathfrak{B}$ with the trace of the product
$\sigma$-algebra on
$\mathbb{R}\times\mathbb{R}\times\mathcal{P}_{1}(\mathbb{R})\times\mathcal{P}_{1}(\mathbb{R})$.

Then there exist two measurable maps
$\beta,\widetilde{\beta}:\mathfrak{B}\to\mathcal{P}_{1}(\mathbb{R})$ such that
for all $(y,\widetilde{y},\mu,\widetilde{\mu})$, denoting
$\nu=\beta(y,\widetilde{y},\mu,\widetilde{\mu})$,
$\widetilde{\nu}=\widetilde{\beta}(y,\widetilde{y},\mu,\widetilde{\mu})$ and
$p=\frac{\stackrel{{\scriptstyle\sim}}{{\smash{x}\rule{0.0pt}{1.5pt}}}-\stackrel{{\scriptstyle\sim}}{{\smash{y}\rule{0.0pt}{1.5pt}}}}{y-x+\stackrel{{\scriptstyle\sim}}{{\smash{x}\rule{0.0pt}{1.5pt}}}-\stackrel{{\scriptstyle\sim}}{{\smash{y}\rule{0.0pt}{1.5pt}}}}$
where $x$ and $\widetilde{x}$ are the respective means of $\mu$ and
$\widetilde{\mu}$, we have

$\int_{\mathbb{R}}w\,\nu(dw)=y,\quad\int_{\mathbb{R}}w\,\widetilde{\nu}(dw)=\widetilde{y},\quad\mu\leq_{st}\nu,\quad\widetilde{\nu}\leq_{st}\widetilde{\mu}\quad\text{and}\quad
p\nu+(1-p)\widetilde{\nu}=p\mu+(1-p)\widetilde{\mu}.$ (2.19)

In particular,
$p\,\delta_{y}(dz)\,\nu(dw)+(1-p)\,\delta_{\stackrel{{\scriptstyle\sim}}{{\smash{y}\rule{0.0pt}{1.5pt}}}}(dz)\,\widetilde{\nu}(dw)$
is a martingale coupling between
$p\delta_{y}(dz)+(1-p)\delta_{\stackrel{{\scriptstyle\sim}}{{\smash{y}\rule{0.0pt}{1.5pt}}}}(dz)$
and $p\mu(dw)+(1-p)\widetilde{\mu}(dw)$, and $\mathcal{W}_{1}(\mu,\nu)=y-x$,
$\mathcal{W}_{1}(\widetilde{\mu},\widetilde{\nu})=\widetilde{x}-\widetilde{y}$.
The proof, which consists in exhibiting particular maps $\beta$ and
$\tilde{\beta}$, is moved to the end of the present section.

In order to use Lemma 2.4 we need to compare $\phi$ to the identity function.
The inequality (2.13) is equivalent by appropriate normalisation of (1.10) to
$u\geq\Delta_{+}^{-1}(\Delta_{-}(u))$ for all $u\in[0,1]$, hence

$\forall u\in\mathcal{U}^{-},\quad\phi(u)\leq u.$ (2.20)

Moreover, by (2.17), [24, Lemma 6.1] applied with
$f_{1}=(F_{\mu}^{-1}-G)^{+}$, $f_{2}=(F_{\mu}^{-1}-G)^{-}$, $u_{0}=1$ and
$h:u\mapsto\mathds{1}_{\\{u<\phi(u)\\}}$ we have

$\int_{0}^{1}\mathds{1}_{\\{\phi(u)<u\\}}\,d\Delta_{+}(u)=\int_{0}^{1}\mathds{1}_{\\{\phi(u)<\phi(\phi(u))\\}}\,d\Delta_{+}(u)=\int_{0}^{1}\mathds{1}_{\\{u<\phi(u)\\}}\,d\Delta_{-}(u).$

By (2.20) the right-hand side is $0$, hence

$\text{for $du$-almost all $u\in\mathcal{U}^{+}$},\quad\phi(u)\geq u.$ (2.21)

Let

$\displaystyle A^{+}=\\{u\in\mathcal{U}^{+}\mid
F_{\mu}^{-1}(\phi(u))<G(\phi(u)),\ \phi(\phi(u))=u\text{ and
}p(\phi(u))=1-p(u)\\}$ and $\displaystyle A^{-}=\\{u\in\mathcal{U}^{-}\mid
F_{\mu}^{-1}(\phi(u))>G(\phi(u)),\ \phi(\phi(u))=u\text{ and
}p(\phi(u))=1-p(u)\\}.$

For all $u\in A^{+}$, we have by definition

$\displaystyle\phi(u)\in\mathcal{U}^{-},\quad
F_{\mu}^{-1}(\phi(\phi(u)))=F_{\mu}^{-1}(u)>G(u)=G(\phi(\phi(u))),$
$\displaystyle\phi(\phi(\phi(u)))=\phi(u)\quad\text{and}\quad
p(\phi(\phi(u)))=p(u)=1-p(\phi(u)),$

hence $\phi(u)\in A^{-}$. Similarly, for all $u\in A^{-}$, $\phi(u)\in A^{+}$.
By (2.14), (2.17), (2.18), (2.21) and the monotonicity of $F_{\mu}^{-1}$, we
deduce that $A^{+}$ and $A^{-}$ are two disjoint Borel sets such that the
Lebesgue measure of $(\mathcal{U}^{+}\backslash
A^{+})\cup(\mathcal{U}^{-}\backslash A^{-})$ is $0$ and

$\displaystyle\begin{split}&\forall u\in A^{+},\quad G(u)<F_{\mu}^{-1}(u)\leq
F_{\mu}^{-1}(\phi(u))<G(\phi(u)).\end{split}$ (2.22)

For all $u\in A^{+}$, $\pi_{F_{\mu}^{-1}(u)}$ and
$\pi_{F_{\mu}^{-1}(\phi(u))}$ have by definition respective means $G(u)$ and
$G(\phi(u))$, so by (2.22) we can apply Lemma 2.4 with

$(y,\widetilde{y},\mu,\widetilde{\mu})=(F_{\mu}^{-1}(u),F_{\mu}^{-1}(\phi(u)),\pi_{F_{\mu}^{-1}(u)},\pi_{F_{\mu}^{-1}(\phi(u))}).$

Hence there exist two probability measures
$m_{u},\widetilde{m}_{u}\in\mathcal{P}_{1}(\mathbb{R})$ with respective means
$F_{\mu}^{-1}(u)$, $F_{\mu}^{-1}(\phi(u))$ and such that

$\displaystyle\begin{split}&\pi_{F_{\mu}^{-1}(u)}\leq_{st}m_{u},\quad\widetilde{m}_{u}\leq_{st}\pi_{F_{\mu}^{-1}(\phi(u))},\\\
\text{and}\quad&p(u)m_{u}+(1-p(u))\widetilde{m}_{u}=p(u)\pi_{F_{\mu}^{-1}(u)}+(1-p(u))\pi_{F_{\mu}^{-1}(\phi(u))}.\end{split}$
(2.23)

Since $A^{+}=\phi(A^{-})$ and $A^{-}=\phi(A^{+})$, for all $u\in A^{-}$ we can
set $m_{u}=\widetilde{m}_{\phi(u)}$, so that

$\forall u\in
A^{+},\quad\pi_{F_{\mu}^{-1}(u)}\leq_{st}m_{u}\quad\text{and}\quad\forall u\in
A^{-},\quad m_{u}\leq_{st}\pi_{F_{\mu}^{-1}(u)},$ (2.24)

and ,

$\forall u\in A^{+}\cup
A^{-},\;p(u)m_{u}+p(\phi(u))m_{\phi(u)}=p(u)\pi_{F_{\mu}^{-1}(u)}+p(\phi(u))\pi_{F_{\mu}^{-1}(\phi(u))}.$
(2.25)

Finally, for all $u\in\mathcal{U}^{0}\cup(\mathcal{U}^{+}\backslash
A^{+})\cup(\mathcal{U}^{-}\backslash A^{-})$ set
$m_{u}=\pi_{F_{\mu}^{-1}(u)}$. By composition of the measurable map
$u\mapsto(F_{\mu}^{-1}(u),F_{\mu}^{-1}(\phi(u)),\pi_{F_{\mu}^{-1}(u)},\pi_{F_{\mu}^{-1}(\phi(u))})$
and the measurable map $\beta$ defined in Lemma 2.4, the map $u\mapsto m_{u}$
is measurable. By [2, Theorem 19.12] it is equivalent to say that
$(m_{u})_{u\in(0,1)}$ is a probability kernel, hence we can define

$\widehat{M}(du,dx,dy)=\lambda_{(0,1)}(du)\,\delta_{F_{\mu}^{-1}(u)}(dx)\,m_{u}(dy),$
(2.26)

and

$M(dx,dy)=\int_{0}^{1}\delta_{F_{\mu}^{-1}(u)}(dx)\,m_{u}(dy)\,du.$ (2.27)

We now state that $\widehat{M}$ is a lifted martingale rearrangement coupling
of $\widehat{\pi}=\iota(\pi)$.

######  Proposition 2.5.

Let $\mu,\nu\in\mathcal{P}_{1}(\mathbb{R})$ be such that $\mu\leq_{cx}\nu$ and
$\mu\neq\nu$ and $\pi\in\Pi(\mu,\nu)$ be a coupling between $\mu$ and $\nu$
which satisfies the barycentre dispersion assumption (2.10). Then the measure
$\widehat{M}$ defined by (2.26) is a lifted martingale rearrangement coupling
of the lifted coupling $\widehat{\pi}=\iota(\pi)$:

$\inf_{\widehat{M}^{\prime}\in\widehat{\Pi}^{\mathrm{M}}(\mu,\nu)}\widehat{\mathcal{AW}}_{1}(\widehat{\pi},\widehat{M}^{\prime})=\widehat{\mathcal{AW}}_{1}(\widehat{\pi},\widehat{M})=\int_{(0,1)}\mathcal{W}_{1}(\pi_{F_{\mu}^{-1}(u)},m_{u})\,du=\int_{(0,1)}\left|G(u)-F_{\mu}^{-1}(u)\right|\,du.$

Since $u\mapsto\pi_{F_{\mu}^{-1}(u)}$ is constant on the jumps on $F_{\mu}$ by
(1.11), we immediately deduce by Proposition 2.3 that $M$ is a martingale
rearrangement coupling of $\pi$.

######  Corollary 2.6.

Let $\mu,\nu\in\mathcal{P}_{1}(\mathbb{R})$ be such that $\mu\leq_{cx}\nu$ and
$\mu\neq\nu$ and $\pi\in\Pi(\mu,\nu)$ be a coupling between $\mu$ and $\nu$
which satisfies the barycentre dispersion assumption (2.10). Then the measure
$M$ defined by (2.27) is a martingale rearrangement coupling of $\pi$:

$\inf_{M^{\prime}\in\Pi^{\mathrm{M}}(\mu,\nu)}\mathcal{AW}_{1}(\pi,M^{\prime})=\mathcal{AW}_{1}(\pi,M)=\int_{\mathbb{R}}\mathcal{W}_{1}(\pi_{x},M_{x})\,\mu(dx)=\int_{\mathbb{R}}\left|\int_{\mathbb{R}}y\,\pi_{x}(dy)-x\right|\,\mu(dx).$

######  Remark 2.7.

As seen from the proof of Proposition 2.5 just below, for $\widehat{M}$
defined by (2.26) to be a lifted martingale rearrangement coupling of the
lifted coupling $\widehat{\pi}=\iota(\pi)$ and therefore $M$ defined by (2.27)
to be a martingale rearrangement coupling of $\pi$, it is enough that
$u\mapsto m_{u}$ is measurable, satisfies (2.24), (2.25) and
$m_{u}=\pi_{F_{\mu}^{-1}(u)}$ for all
$u\in\mathcal{U}^{0}\cup(\mathcal{U}^{+}\backslash
A^{+})\cup(\mathcal{U}^{-}\backslash A^{-})$.

###### Proof of Proposition 2.5.

Assume for a moment that $\widehat{M}\in\widehat{\Pi}^{\mathrm{M}}(\mu,\nu)$.
Then we have by (2.24) that for all $u\in(0,1)$,
$\pi_{F_{\mu}^{-1}(u)}\leq_{st}m_{u}$ or
$m_{u}\leq_{st}\pi_{F_{\mu}^{-1}(u)}$, hence
$\mathcal{W}_{1}(\pi_{F_{\mu}^{-1}(u)},m_{u})=|G(u)-F_{\mu}^{-1}(u)|$ and

$\widehat{\mathcal{AW}}_{1}(\widehat{\pi},\widehat{M})\leq\int_{(0,1)}\mathcal{W}_{1}(\pi_{F_{\mu}^{-1}(u)},m_{u})\,du=\int_{(0,1)}|G(u)-F_{\mu}^{-1}(u)|\,du,$

which proves the claim by Lemma 2.2.

It remains to show that $\widehat{M}\in\widehat{\Pi}^{\mathrm{M}}(\mu,\nu)$.
By the inverse transform sampling and the fact that $m_{u}$ has mean
$F_{\mu}^{-1}(u)$ for all $u\in(0,1)$, it is clear that $\widehat{M}$ is a
lifted martingale coupling between $\mu$ and $\int_{u\in(0,1)}m_{u}(dy)\,du$.
To conclude it is therefore sufficient to check that

$\int_{u\in(0,1)}m_{u}(dy)\,du=\nu.$ (2.28)

To this end, let $H:[0,1]\to\mathbb{R}$ be measurable and bounded. Using
(2.15), (2.17) and [24, Lemma 6.1] applied with $f_{1}=(F_{\mu}^{-1}-G)^{+}$,
$f_{2}=(F_{\mu}^{-1}-G)^{-}$, $u_{0}=1$ and
$h:u\mapsto\frac{H(\phi(u))}{F_{\mu}^{-1}(\phi(u))-G(\phi(u))+G(u)-F_{\mu}^{-1}(u)}$
for the third equality, we get

$\displaystyle\int_{\mathcal{U}^{+}}(1-p(u))H(u)\,du$
$\displaystyle=\int_{0}^{1}\frac{(F_{\mu}^{-1}-G)^{+}(u)}{F_{\mu}^{-1}(u)-G(u)+G(\phi(u))-F_{\mu}^{-1}(\phi(u))}H(u)\,du$
$\displaystyle=\int_{0}^{1}h(\phi(u))\,d\Delta_{+}(u)$
$\displaystyle=\int_{0}^{1}h(v)\,d\Delta_{-}(v)$
$\displaystyle=\int_{0}^{1}\frac{(F_{\mu}^{-1}-G)^{-}(v)}{F_{\mu}^{-1}(\phi(v))-G(\phi(v))+G(v)-F_{\mu}^{-1}(v)}H(\phi(v))\,dv$
$\displaystyle=\int_{\mathcal{U}^{-}}p(\phi(v))H(\phi(v))\,dv.$

Similarly, we have
$\int_{\mathcal{U}^{-}}(1-p(u))H(u)\,du=\int_{\mathcal{U}^{+}}p(\phi(u))H(\phi(u))\,du$.
We deduce that

$\displaystyle\begin{split}\int_{0}^{1}H(u)\,du&=\int_{\mathcal{U}^{0}}H(u)\,du+\int_{\mathcal{U}^{+}}p(u)H(u)\,du+\int_{\mathcal{U}^{+}}(1-p(u))H(u)\,du\\\
&\phantom{=}+\int_{\mathcal{U}^{-}}p(u)H(u)\,du+\int_{\mathcal{U}^{-}}(1-p(u))H(u)\,du\\\
&=\int_{\mathcal{U}^{0}}H(u)\,du+\int_{\mathcal{U}^{+}\cup\mathcal{U}^{-}}(p(u)H(u)+p(\phi(u))H(\phi(u)))\,du.\end{split}$
(2.29)

Let $f:\mathbb{R}\to\mathbb{R}$ be measurable and bounded. Using (2.29)
applied with $H:u\mapsto\int_{\mathbb{R}}f(y)\,m_{u}(dy)$ for the first
equality, the fact that $m_{u}=\pi_{F_{\mu}^{-1}(u)}$ for all
$u\in\mathcal{U}^{0}$ and (2.25) for the second equality, (2.29) again applied
with $H:u\mapsto\int_{\mathbb{R}}f(y)\,\pi_{F_{\mu}^{-1}(u)}(dy)$ for the
third equality and the inverse transform sampling for the last equality, we
get

$\displaystyle\int_{0}^{1}\int_{\mathbb{R}}f(y)\,m_{u}(dy)\,du$
$\displaystyle=\int_{\mathcal{U}^{0}}\int_{\mathbb{R}}f(y)\,m_{u}(dy)\,du+\int_{\mathcal{U}^{+}\cup\mathcal{U}^{-}}\int_{\mathbb{R}}f(y)\,(p(u)\,m_{u}(dy)+p(\phi(u))\,m_{\phi(u)}(dy))\,du$
$\displaystyle=\int_{\mathcal{U}^{0}}\int_{\mathbb{R}}f(y)\,\pi_{F_{\mu}^{-1}(u)}(dy)\,du$
$\displaystyle\phantom{=}+\int_{\mathcal{U}^{+}\cup\mathcal{U}-}\int_{\mathbb{R}}f(y)\,(p(u)\,\pi_{F_{\mu}^{-1}(u)}(dy)+p(\phi(u))\,\pi_{F_{\mu}^{-1}(\phi(u))}(dy))\,du$
$\displaystyle=\int_{0}^{1}\int_{\mathbb{R}}f(y)\,\pi_{F_{\mu}^{-1}(u)}(dy)\,du$
$\displaystyle=\int_{\mathbb{R}}f(y)\,\nu(dy),$

which shows (2.28) and concludes the proof. ∎

###### Proof of Lemma 2.4.

Let $(y,\widetilde{y},\mu,\widetilde{\mu})\in\mathfrak{B}$, $x$ and
$\widetilde{x}$ be the respective means of $\mu$ and $\widetilde{\mu}$ and
$p=\frac{\stackrel{{\scriptstyle\sim}}{{\smash{x}\rule{0.0pt}{1.5pt}}}-\stackrel{{\scriptstyle\sim}}{{\smash{y}\rule{0.0pt}{1.5pt}}}}{y-x+\stackrel{{\scriptstyle\sim}}{{\smash{x}\rule{0.0pt}{1.5pt}}}-\stackrel{{\scriptstyle\sim}}{{\smash{y}\rule{0.0pt}{1.5pt}}}}$.
First we construct two measures
$\nu,\widetilde{\nu}\in\mathcal{P}_{1}(\mathbb{R})$ which satisfy (2.19). Then
we show that $\nu$ and $\widetilde{\nu}$ are measurable in
$(y,\widetilde{y},\mu,\widetilde{\mu})$.

We set for $q\in[0,p\wedge(1-p)]$

$\displaystyle\begin{split}&J(q):=\int^{\frac{q}{p}}_{0}F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(\frac{1-pu-p}{1-p}\right)\,du+\int_{\frac{q}{p}}^{1}F_{\mu}^{-1}\left(u\right)\,du,\\\
\text{and}\quad&\widetilde{J}(q):=\int_{0}^{\frac{q}{1-p}}F_{\mu}^{-1}\left(\frac{(1-p)u}{p}\right)\,du+\int_{0}^{\frac{1-q-p}{1-p}}F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}(u)\,du.\end{split}$
(2.30)

Since the quantile functions are non-decreasing, the function $J$ is
continuous and concave as the sum of two concave functions and the function
$\widetilde{J}$ is continuous and convex as the sum of two convex functions.
We have $J(0)=\int_{0}^{1}F_{\mu}^{-1}(u)\,du=x$ and
$\widetilde{J}(0)=\int_{0}^{1}F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}(u)\,du=\widetilde{x}$
by the inverse transform sampling. If $p\leq\frac{1}{2}$, then by the change
of variables $v=\frac{1-pu-p}{1-p}$ and since
$F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}$ is
non-decreasing, we have

$\displaystyle J(p)$
$\displaystyle=\frac{1-p}{p}\int_{\frac{1-2p}{1-p}}^{1}F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(v\right)\,dv$
$\displaystyle=\int_{0}^{1}F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(v\right)\,dv+\frac{1-2p}{1-p}\left(\frac{1}{1-\frac{1-2p}{1-p}}\int_{\frac{1-2p}{1-p}}^{1}F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(v\right)\,dv-\frac{1-p}{1-2p}\int_{0}^{\frac{1-2p}{1-p}}F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(v\right)\,dv\right)$
$\displaystyle\geq\int_{0}^{1}F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(v\right)\,dv=\widetilde{x}.$

Since $x<y<\widetilde{x}$ and the function $J$ is concave, there is a unique
$q_{\star}\in(0,p)$ such that $J(q_{\star})=y$. Moreover, the left-hand
derivative of $J$ at $q_{\star}$ is positive which writes

$F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(\frac{1-q_{\star}-p}{1-p}+\right)>F_{\mu}^{-1}\left(\frac{q_{\star}}{p}\right).$
(2.31)

If $p>\frac{1}{2}$, we have
$\widetilde{J}(1-p)=\int_{0}^{1}F_{\mu}^{-1}\left(\frac{(1-p)u}{p}\right)\,du\leq\int_{0}^{1}F_{\mu}^{-1}(v)\,dv=x$
and by convexity of $\widetilde{J}$ there is a unique $q_{\star}\in(0,1-p)$
such that $\widetilde{J}(q_{\star})=\widetilde{y}$. Moreover, the left-hand
derivative of $\widetilde{J}$ at $q_{\star}$ is negative so that (2.31) still
holds.

Let $\nu$ and $\widetilde{\nu}$ be the respective images of the Lebesgue
measure on $[0,1]$ by

$\displaystyle\begin{split}&u\mapsto\mathds{1}_{\\{u<\frac{q_{\star}}{p}\\}}F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(\frac{1-pu-p}{1-p}\right)+\mathds{1}_{\\{u\geq\frac{q_{\star}}{p}\\}}F_{\mu}^{-1}(u)\\\
\text{and}\quad&u\mapsto\mathds{1}_{\\{u<\frac{q_{\star}}{1-p}\\}}F_{\mu}^{-1}\left(\frac{(1-p)u}{p}\right)+\mathds{1}_{\\{u\geq\frac{q_{\star}}{1-p}\\}}F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(u-\frac{q_{\star}}{1-p}\right).\end{split}$
(2.32)

With the definition of $J$ and $\widetilde{J}$, we easily check that
$\int_{\mathbb{R}}z\,\nu(dz)=J(q_{\star})$ and
$\int_{\mathbb{R}}z\,\widetilde{\nu}(dz)=\widetilde{J}(q_{\star})$. Moreover,
the inequality (2.31) implies that for each $u\in(0,\frac{q_{\star}}{p})$,

$F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(\frac{1-pu-p}{1-p}\right)\geq
F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(\frac{1-q_{\star}-p}{1-p}+\right)>F_{\mu}^{-1}\left(\frac{q_{\star}}{p}\right)\geq
F_{\mu}^{-1}(u),$

so that $\nu$ dominates for the stochastic order the image $\mu$ of the
Lebesgue measure on $(0,1)$ by $F_{\mu}^{-1}$. In a symmetric way,
$\widetilde{\nu}\leq_{st}\widetilde{\mu}$.

For $h:\mathbb{R}\to\mathbb{R}$ measurable and bounded, we have using the
changes of variables $v=\frac{1-pu-p}{1-p}$, $v=\frac{(1-p)u}{p}$ and
$v=u-\frac{q_{\star}}{1-p}$ then the inverse transform sampling for the last
equality

$\displaystyle
p\int_{\mathbb{R}}h(z)\,\nu(dz)+(1-p)\int_{\mathbb{R}}h(z)\,\widetilde{\nu}(dz)$
$\displaystyle=p\left(\int_{0}^{\frac{q_{\star}}{p}}h\left(F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(\frac{1-pu-p}{1-p}\right)\right)\,du+\int_{\frac{q_{\star}}{p}}^{1}h(F_{\mu}^{-1}(u))\,du\right)$
$\displaystyle\phantom{=}+(1-p)\left(\int_{0}^{\frac{q_{\star}}{1-p}}h\left(F_{\mu}^{-1}\left(\frac{(1-p)u}{p}\right)\right)\,du+\int^{1}_{\frac{q_{\star}}{1-p}}h\left(F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(u-\frac{q_{\star}}{1-p}\right)\right)\,du\right)$
$\displaystyle=(1-p)\int^{1}_{\frac{1-q_{\star}-p}{1-p}}h\left(F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}(v)\right)\,dv+p\int_{\frac{q_{\star}}{p}}^{1}h(F_{\mu}^{-1}(u))\,du+p\int_{0}^{\frac{q_{\star}}{p}}h\left(F_{\mu}^{-1}\left(v\right)\right)dv$
$\displaystyle\phantom{=}+(1-p)\int_{0}^{\frac{1-q_{\star}-p}{1-p}}h\left(F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(v\right)\right)\,dv$
$\displaystyle=p\int_{0}^{1}h(F_{\mu}^{-1}(u))\,du+(1-p)\int_{0}^{1}h\left(F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}\left(v\right)\right)\,dv$
$\displaystyle=p\int_{\mathbb{R}}h(z)\,\mu(dz)+(1-p)\int_{\mathbb{R}}h(z)\,\widetilde{\mu}(dz).$

Hence $p\nu+(1-p)\tilde{\nu}=p\mu+(1-p)\widetilde{\mu}$. Taking expectations
then using the definition of $p$, we get

$pJ(q_{\star})+(1-p)\widetilde{J}(q_{\star})=px+(1-p)\widetilde{x}=py+(1-p)\widetilde{y}.$

When $p\leq\frac{1}{2}$ (resp. $p>\frac{1}{2}$), $J(q_{\star})=y$ (resp.
$\widetilde{J}(q_{\star})=\widetilde{y}$) and we deduce that
$\widetilde{J}(q_{\star})=\widetilde{y}$ (resp. $J(q_{\star})=y$).

It remains to show that $\nu$ and $\widetilde{\nu}$ are measurable in
$(y,\widetilde{y},\mu,\widetilde{\mu})$. It is clear that $p$ is a measurable
function of $(y,\widetilde{y},\mu,\widetilde{\mu})$. Moreover we always have
by definition $p\in(0,1)$, so the relation
$p\nu+(1-p)\widetilde{\nu}=p\mu+(1-p)\widetilde{\mu}$ implies that it suffices
to show that $\nu$ is measurable in $(y,\widetilde{y},\mu,\widetilde{\mu})$.
Any quantile function is an element of the set $\mathcal{D}$ of the real-
valued càglàd functions on $(0,1)$. Analogously to the Skorokhod space of the
real-valued càdlàg functions on $(0,1)$, we endow $\mathcal{D}$ with the
$\sigma$-field generated by the projection maps $\alpha_{u}:\mathcal{D}\ni
f\mapsto f(u)$, $u\in(0,1)$, which coincides with the $\sigma$-field
$\sigma(\alpha_{u},u\in T)$ for any dense subset $T\subset(0,1)$. Let
$(\mu_{n})_{n\in\mathbb{N}}\in\mathcal{P}_{1}(\mathbb{R})^{\mathbb{N}}$
converge to $\mu$ for the weak convergence topology and $T$ be the complement
of the at most countable set of discontinuities of $F_{\mu}^{-1}$. Then for
all $u\in T$, $F_{\mu_{n}}^{-1}(u)=\alpha_{u}(F_{\mu_{n}}^{-1})$ converges to
$F_{\mu}^{-1}(u)=\alpha_{u}(F_{\mu}^{-1})$. We deduce that $F_{\mu}^{-1}$ and
$F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}$ are
respectively measurable in $\mu$ and $\widetilde{\mu}$. By (2.32), $\nu$ is
the image of the Lebesgue measure on $[0,1]$ by a measurable function of
$p,F_{\mu}^{-1},F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}$
and $q_{\star}$, hence it remains to prove the measurability of $q_{\star}$ in
$(y,\widetilde{y},\mu,\widetilde{\mu})$.

If $p\leq\frac{1}{2}$ we saw that $J(0)=x<y<\widetilde{x}\leq J(p)$ and
$q_{\star}$ is the only real number in $[0,p]$ such that $J(q_{\star})=y$. By
concavity of $J$ we necessarily have for all $q\in[0,p]$ that $q_{\star}\leq
q$ iff $J(q)\geq y$. Similarly, if $p>\frac{1}{2}$ then for all $q\in[0,1-p]$,
$q_{\star}\leq q$ iff $\widetilde{J}(q)\leq\widetilde{y}$. Fix then
$q\in[0,p\wedge(1-p)]$. To conclude, it suffices to show that $J(q)$ and
$\widetilde{J}(q)$ are measurable in $(y,\widetilde{y},\mu,\widetilde{\mu})$.
By [23, Lemma 4.5], the Borel $\sigma$-algebras of the weak convergence
topology and the $\mathcal{W}_{1}$-distance topology coincide on
$\mathcal{P}_{1}(\mathbb{R})$. Moreover, since
$(\mathcal{P}_{1}(\mathbb{R}),\mathcal{W}_{1})$ is separable [36, Theorem
6.18], we deduce from [2, Theorem 4.44] that the $\sigma$-field on
$\mathfrak{B}$ coincides with the trace of the Borel $\sigma$-algebra of
$\mathbb{R}\times\mathbb{R}\times\mathcal{P}_{1}(\mathbb{R})\times\mathcal{P}_{1}(\mathbb{R})$
endowed with the metric
$((y,\widetilde{y},\mu,\widetilde{\mu}),(y^{\prime},\widetilde{y}^{\prime},\mu^{\prime},\widetilde{\mu}^{\prime}))\mapsto|y-y^{\prime}|+|\widetilde{y}-\widetilde{y}^{\prime}|+\mathcal{W}_{1}(\mu,\mu^{\prime})+\mathcal{W}_{1}(\widetilde{\mu},\widetilde{\mu}^{\prime})$.
Then the measurability of $J(q)$ and $\widetilde{J}(q)$ follows from their
continuity in $(y,\widetilde{y},\mu,\widetilde{\mu})$ with respect to the
latter metric, which is clear in view of their definition (2.30) which implies
by the changes of variables $v=\frac{1-pu-p}{1-p}$ and $v=\frac{(1-p)u}{p}$
that

$\displaystyle J(q)$
$\displaystyle=\frac{1-p}{p}\int_{\frac{1-q-p}{1-p}}^{1}F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}(v)\,dv+\int_{\frac{q}{p}}^{1}F_{\mu}^{-1}(u)\,du,\quad\widetilde{J}(q)=\frac{p}{1-p}\int_{0}^{\frac{q}{p}}F_{\mu}^{-1}(v)\,dv+\int_{0}^{\frac{1-q-p}{1-p}}F_{\stackrel{{\scriptstyle\sim}}{{\smash{\mu}\rule{0.0pt}{1.5pt}}}}^{-1}(u)\,du,$

and the easy fact that if $(a_{n})_{n\in\mathbb{N}}\in[0,1]^{\mathbb{N}}$
converges to $a\in[0,1]$ and $(f_{n})_{n\in\mathbb{N}}\in
L^{1}([0,1])^{\mathbb{N}}$ converges to $f\in L^{1}([0,1])$ in $L^{1}$, then
$\int_{0}^{a_{n}}f_{n}(u)\,du$ converges to $\int_{0}^{a}f(u)\,du$ as
$n\to+\infty$. ∎

##  3 Martingale rearrangement couplings of the Hoeffding-Fréchet coupling

###  3.1 The inverse transform martingale coupling

We come back on the inverse transform martingale coupling and the family
parametrised by $\mathcal{Q}$ introduced in [24] since they will have
particular significance in the remaining of the present paper. We briefly
recall the construction and main properties and refer to [24] for an extensive
study. Let $\mu,\nu\in\mathcal{P}_{1}(\mathbb{R})$ be such that
$\mu\leq_{cx}\nu$ and $\mu\neq\nu$. For $u\in[0,1]$ we define

$\Psi_{+}(u)=\int_{0}^{u}(F_{\mu}^{-1}-F_{\nu}^{-1})^{+}(v)\,dv\quad\text{and}\quad\Psi_{-}(u)=\int_{0}^{u}(F_{\mu}^{-1}-F_{\nu}^{-1})^{-}(v)\,dv,$
(3.1)

with respective left continuous generalised inverses $\Psi_{+}^{-1}$ and
$\Psi_{-}^{-1}$. We then define $\mathcal{Q}$ as the set of probability
measures on $(0,1)^{2}$ with first marginal $\frac{1}{\Psi_{+}(1)}d\Psi_{+}$,
second marginal $\frac{1}{\Psi_{+}(1)}d\Psi_{-}$ and such that $u<v$ for
$Q(du,dv)$-almost every $(u,v)\in(0,1)^{2}$. Since $d\,\Psi_{+}$ and
$d\,\Psi_{-}$ are concentrated on two disjoint Borel sets, there exists for
each $Q\in\mathcal{Q}$ a probability kernel $(\pi^{Q}_{u})_{u\in(0,1)}$ such
that

$Q(du,dv)=\frac{1}{\Psi_{+}(1)}d\Psi_{+}(u)\,\pi^{Q}_{u}(dv)=\frac{1}{\Psi_{+}(1)}d\Psi_{-}(v)\,\pi^{Q}_{v}(du),$
(3.2)

and we exhibit a probability kernel $(\widetilde{m}^{Q}_{u})_{u\in(0,1)}$
which satisfies for $du$-almost all $u\in(0,1)$ such that $F_{\mu}^{-1}(u)\neq
F_{\nu}^{-1}(u)$

$\widetilde{m}^{Q}_{u}(dy)=\int_{v\in(0,1)}\left(\frac{F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u)}{F_{\nu}^{-1}(v)-F_{\nu}^{-1}(u)}\delta_{F_{\nu}^{-1}(v)}(dy)+\frac{F_{\nu}^{-1}(v)-F_{\mu}^{-1}(u)}{F_{\nu}^{-1}(v)-F_{\nu}^{-1}(u)}\delta_{F_{\nu}^{-1}(u)}(dy)\right)\,\pi^{Q}_{u}(dv),$
(3.3)

and $\widetilde{m}^{Q}_{u}(dy)=\delta_{F_{\nu}^{-1}(u)}(dy)$ for all
$u\in(0,1)$ such that $F_{\mu}^{-1}(u)=F_{\nu}^{-1}(u)$. Then the measure

$\widehat{M}^{Q}(du,dx,dy)=\lambda_{(0,1)}(du)\,\delta_{F_{\mu}^{-1}(u)}(dx)\,\widetilde{m}^{Q}_{u}(dy)$
(3.4)

is a lifted martingale coupling between $\mu$ and $\nu$. Moreover it was shown
by [24, Proposition 2.18] and its proof that for $du$-almost all $u\in(0,1)$,

$\int_{\mathbb{R}}|y-F_{\nu}^{-1}(u)|\,\widetilde{m}^{Q}_{u}(dy)=|F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u)|,$
(3.5)

from which we deduce that the measure

$M^{Q}(dx,dy)=\int_{0}^{1}\delta_{F_{\mu}^{-1}(u)}(dx)\,\widetilde{m}^{Q}_{u}(dy)\,du$
(3.6)

is a martingale coupling between $\mu$ and $\nu$ which satisfies
$\int_{\mathbb{R}\times\mathbb{R}}|y-x|\,M^{Q}(dx,dy)\leq
2\mathcal{W}_{1}(\mu,\nu)$. Let also

$\displaystyle\mathcal{U}^{+}=\\{u\in(0,1)\mid
F_{\mu}^{-1}(u)>F_{\nu}^{-1}(u)\\},\quad\mathcal{U}^{-}=\\{u\in(0,1)\mid
F_{\mu}^{-1}(u)<F_{\nu}^{-1}(u)\\},$ (3.7) and
$\displaystyle\mathcal{U}^{0}=\\{u\in(0,1)\mid
F_{\mu}^{-1}(u)=F_{\nu}^{-1}(u)\\}.$ (3.8)

Thanks to the equality $\Psi_{+}(1)=\Psi_{-}(1)$, consequence of the equality
of the respective means of $\mu$ and $\nu$, we can set for all $u\in[0,1]$

$\displaystyle\begin{split}\varphi(u)=\left\\{\begin{array}[]{rcl}\Psi_{-}^{-1}(\Psi_{+}(u))&\text{if}&u\in\mathcal{U}^{+};\\\
\Psi_{+}^{-1}(\Psi_{-}(u))&\text{if}&u\in\mathcal{U}^{-};\\\
u&\text{if}&u\in\mathcal{U}^{0}.\end{array}\right.\end{split}$ (3.9)

Then the measure
$Q^{IT}(du,dv)=\frac{1}{\Psi_{+}(1)}d\Psi_{+}(u)\mathds{1}_{\\{0<\varphi(u)<1\\}}\,\delta_{\varphi(u)}(dv)$
belongs to $\mathcal{Q}$. The martingale coupling $M^{IT}=M^{Q^{IT}}$ is the
so called inverse transform martingale coupling, associated to the probability
kernel $\widetilde{m}^{IT}=\widetilde{m}^{Q^{IT}}$ which satisfies for
$du$-almost all $u\in(0,1)$

$\widetilde{m}^{IT}(u,dy)=p(u)\,\delta_{F_{\nu}^{-1}(\varphi(u))}(dy)+\left(1-p(u)\right)\,\delta_{F_{\nu}^{-1}(u)}(dy),$
(3.10)

where $p(u)=\mathds{1}_{\\{F_{\mu}^{-1}(u)\neq
F_{\nu}^{-1}(u)\\}}\frac{F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u)}{F_{\nu}^{-1}(\varphi(u))-F_{\nu}^{-1}(u)}$.

###  3.2 The Hoeffding-Fréchet coupling

Let $\mu$ and $\nu$ be two probability measures on the real line with finite
first moment. We recall that the Hoeffding-Fréchet coupling between $\mu$ and
$\nu$, denoted $\pi^{HF}$, is by definition the comonotonic coupling between
$\mu$ and $\nu$, that is the image of the Lebesgue measure on $(0,1)$ by the
map $u\mapsto(F_{\mu}^{-1}(u),F_{\nu}^{-1}(u))$. Equivalently, we can write

$\pi^{HF}(dx,dy)=\int_{(0,1)}\delta_{(F_{\mu}^{-1}(u),F_{\nu}^{-1}(u))}(dx,dy)\,du.$

This coupling is of paramount importance in the classical optimal transport
theory in dimension $1$ since it attains the infimum in the minimisation
problem

$\inf_{P\in\Pi(\mu,\nu)}\int_{\mathbb{R}\times\mathbb{R}}c(x,y)\,P(dx,dy)$

as soon as $c$ satisfies the so called Monge condition, see [32, Theorem
3.1.2]. The latter condition being satisfied for any function $(x,y)\mapsto
h(|y-x|)$ where $h:\mathbb{R}\to\mathbb{R}$ is convex, we deduce that
$\pi^{HF}$ is optimal for $\mathcal{W}_{\rho}(\mu,\nu)$ for all $\rho\geq 1$.
By strict convexity, it is even the only coupling optimal for
$\mathcal{W}_{\rho}(\mu,\nu)$ for $\rho>1$. Reasoning like in (2.9), we get
that for $\mu(dx)$-almost all $x\in\mathbb{R}$,

$\pi^{HF}_{x}(dy)=\int_{(0,1)}\delta_{F_{\nu}^{-1}(\theta(x,v))}(dy)\,dv.$
(3.11)

By (3.11) and monotonicity and left continuity of $F_{\nu}^{-1}$ we recover
the well known fact that $\pi^{HF}$ is given by a measurable map, i.e. is the
image of $\mu$ by $x\mapsto(x,T(x))$ where $T:\mathbb{R}\to\mathbb{R}$ is
measurable, iff for all $x\in\mathbb{R}$ such that $\mu(\\{x\\})>0$,
$F_{\nu}^{-1}$ is constant on $(F_{\mu}(x-),F_{\mu}(x)]$. In that case, we
have $T=F_{\nu}^{-1}\circ F_{\mu}$, referred to as the Monge transport map.

###  3.3 Martingale rearrangement couplings

Our family $(M^{Q})_{Q\in\mathcal{Q}}$ of martingale couplings mentioned above
was meant to contain the closest martingale couplings from the Hoeffding-
Fréchet coupling, the latter being well known for minimising the Wasserstein
distance. Thanks to Wiesel’s definition of martingale rearrangement couplings
we can now rephrase the latter sentence in a more formal way. Let $\pi^{HF}$
be the Hoeffding-Fréchet coupling between $\mu$ and $\nu$. We will consider
the following lifted coupling of $\pi^{HF}$:

$\widehat{\pi}^{HF}(du,dx,dy)=\lambda_{(0,1)}(du)\,\delta_{F_{\mu}^{-1}(u)}(dx)\,\delta_{F_{\nu}^{-1}(u)}(dy).$
(3.12)

Recall the embedding $\iota$ defined by (2.7) and the definition of the map
$\theta$ given by (1.14). Then

$\iota(\pi^{HF})(du,dx,dy)=\lambda_{(0,1)}(du)\,\delta_{F_{\mu}^{-1}(u)}(dx)\,\int_{0}^{1}\delta_{F_{\nu}^{-1}(\theta(F_{\mu}^{-1}(u),v))}(dy)\,dv,$

which is different from $\widehat{\pi}^{HF}$ when $F_{\nu}^{-1}$ is not
constant on the jumps of $F_{\mu}$. We can actually see that
$\widehat{\pi}^{HF}=\iota^{\prime}(\pi^{HF})$, where $\iota^{\prime}$ is
another embedding $\Pi(\mu,\nu)$ to $\widehat{\Pi}(\mu,\nu)$, such that for
all $\pi\in\Pi(\mu,\nu)$, $\iota^{\prime}(\pi)$ is defined by

$\lambda_{(0,1)}(du)\,\delta_{F_{\mu}^{-1}(u)}(dx)\,\left(\mathds{1}_{\\{\mu(\\{F_{\mu}^{-1}(u)\\})>0\\}}\delta_{\left(F_{\pi_{F_{\mu}^{-1}(u)}}\right)^{-1}\left(\frac{u-F_{\mu}(F_{\mu}^{-1}(u)-)}{\mu(\\{F_{\mu}^{-1}(u)\\})}\right)}(dy)+\mathds{1}_{\\{\mu(\\{F_{\mu}^{-1}(u)\\})=0\\}}\pi_{F_{\mu}^{-1}(u)}(dy)\right).$

Although $\widehat{\pi}^{HF}$ is a very natural lifted coupling of $\pi^{HF}$,
the embedding $\iota$ used in Section 2.1 appears to be in general simpler
than $\iota^{\prime}$.

######  Proposition 3.1.

Let $\mu,\nu\in\mathcal{P}_{1}(\mathbb{R})$ be such that $\mu\leq_{cx}\nu$.
Then for all $Q\in\mathcal{Q}$, the lifted martingale coupling
$\widehat{M}^{Q}$ defined by (3.4) is a lifted martingale rearrangement
coupling of the lifted Hoeffding-Fréchet coupling $\widehat{\pi}^{HF}$ defined
by (3.12):

$\forall
Q\in\mathcal{Q},\quad\widehat{\mathcal{AW}}_{1}(\widehat{\pi}^{HF},\widehat{M}^{Q})=\inf_{\widehat{M}\in\widehat{\Pi}^{\mathrm{M}}(\mu,\nu)}\widehat{\mathcal{AW}}_{1}(\widehat{\pi}^{HF},\widehat{M}).$

###### Proof.

Let $Q\in\mathcal{Q}$. The fact that
$\widehat{M}^{Q}\in\widehat{\Pi}^{\mathrm{M}}(\mu,\nu)$ is clear. By (3.5) we
have

$\displaystyle\widehat{\mathcal{AW}_{1}}(\widehat{\pi}^{HF},\widehat{M}^{Q})$
$\displaystyle\leq\int_{(0,1)}\mathcal{W}_{1}(\delta_{F_{\nu}^{-1}(u)},\widetilde{m}^{Q}_{u})\,du=\int_{(0,1)}\int_{\mathbb{R}}|y-F_{\nu}^{-1}(u)|\,\widetilde{m}^{Q}_{u}(dy)du=\int_{(0,1)}|F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u)|\,du,$

which proves the claim by Lemma 2.2. ∎

We can also easily show that any lifted martingale coupling is a lifted
quadratic martingale rearrangement coupling of the lifted Hoeffding-Fréchet
coupling.

######  Proposition 3.2.

Let $\mu,\nu\in\mathcal{P}_{2}(\mathbb{R})$ be such that $\mu\leq_{cx}\nu$.
Then any lifted martingale coupling between $\mu$ and $\nu$ is a
$\widehat{\cal AW}_{2}$-minimal lifted martingale rearrangement coupling of
the lifted Hoeffding-Fréchet coupling $\widehat{\pi}^{HF}$ defined by (3.12):

$\forall
M,M^{\prime}\in\widehat{\Pi}^{\mathrm{M}}(\mu,\nu),\quad\widehat{\mathcal{AW}}_{2}(\widehat{M},\widehat{\pi}^{HF})=\widehat{\mathcal{AW}}_{2}(\widehat{M}^{\prime},\widehat{\pi}^{HF}).$

###### Proof.

Let $\widehat{M}=\lambda_{(0,1)}\times\delta_{F_{\mu}^{-1}(u)}\times
m_{u}\in\widehat{\Pi}^{\mathrm{M}}(\mu,\nu)$ and
$\chi\in\Pi(\lambda_{(0,1)},\lambda_{(0,1)})$ be optimal for
$\widehat{\mathcal{AW}}_{2}(\widehat{M},\widehat{\pi}^{HF})$, so that

$\displaystyle\widehat{\mathcal{AW}}_{2}^{2}(\widehat{M},\widehat{\pi}^{HF})$
$\displaystyle=\int_{(0,1)\times(0,1)}\left(|u-u^{\prime}|^{2}+|F_{\mu}^{-1}(u)-F_{\mu}^{-1}(u^{\prime})|^{2}+\mathcal{W}_{2}^{2}(m_{u},\delta_{F_{\nu}^{-1}(u^{\prime})})\right)\,\chi(du,du^{\prime})$
$\displaystyle\geq\int_{(0,1)\times(0,1)}\mathcal{W}_{2}^{2}(m_{u},\delta_{F_{\nu}^{-1}(u^{\prime})})\,\chi(du,du^{\prime}).$

By bias-variance decomposition for the first equality, the fact that the image
of $\lambda_{(0,1)}$ by $u\mapsto(F_{\mu}^{-1}(u),F_{\nu}^{-1}(u))$ is optimal
for $\mathcal{W}_{2}^{2}(\mu,\nu)$ for the inequality, and by bias-variance
decomposition again for the second equality, we have that

$\displaystyle\begin{split}&\int_{(0,1)\times(0,1)}\mathcal{W}_{2}^{2}(m_{u},\delta_{F_{\nu}^{-1}(u^{\prime})})\,\chi(du,du^{\prime})\\\
&=\int_{(0,1)\times(0,1)}\left(|F_{\nu}^{-1}(u^{\prime})-F_{\mu}^{-1}(u)|^{2}+\int_{\mathbb{R}}|F_{\mu}^{-1}(u)-y|^{2}\,m_{u}(dy)\right)\,\chi(du,du^{\prime})\\\
&\geq\int_{(0,1)}\left(|F_{\nu}^{-1}(u)-F_{\mu}^{-1}(u)|^{2}+\int_{\mathbb{R}}|F_{\mu}^{-1}(u)-y|^{2}\,m_{u}(dy)\right)\,du\\\
&=\int_{(0,1)}\int_{\mathbb{R}}|F_{\nu}^{-1}(u)-y|^{2}\,m_{u}(dy)\,du\\\
&=\int_{(0,1)}\mathcal{W}_{2}^{2}(m_{u},\delta_{F_{\nu}^{-1}(u)})\,du\geq\widehat{\mathcal{AW}}_{2}^{2}(\widehat{M},\widehat{\pi}^{HF}).\end{split}$
(3.13)

Using the fact that
$\int_{(0,1)}\int_{\mathbb{R}}|F_{\mu}^{-1}(u)-y|^{2}\,m_{u}(dy)\,du=\int_{\mathbb{R}}|y|^{2}\,\nu(dy)-\int_{\mathbb{R}}|x|^{2}\,\mu(dx)$,
we deduce that

$\widehat{\mathcal{AW}}_{2}^{2}(\widehat{M},\widehat{\pi}^{HF})=\int_{(0,1)}\mathcal{W}_{2}^{2}(m_{u},\delta_{F_{\nu}^{-1}(u)})\,du=\mathcal{W}_{2}^{2}(\mu,\nu)+\int_{\mathbb{R}}|y|^{2}\,\nu(dy)-\int_{\mathbb{R}}|x|^{2}\,\mu(dx),$

hence $\widehat{\mathcal{AW}}_{2}^{2}(\widehat{M},\widehat{\pi}^{HF})$ does
not depend on the choice of $M$. ∎

A similar conclusion holds for regular couplings. Just this once, we provide a
proof valid in any dimension. In the following statement,
$d\in\mathbb{N}^{*}$. The definitions (1.1), (1.3), (1.5) (1.7) given in
$\mathbb{R}$ have straightfoward extensions to $\mathbb{R}^{d}$ endowed with
the Euclidean norm $|\cdot|$.

######  Proposition 3.3.

Let $\mu,\nu\in\mathcal{P}_{2}(\mathbb{R}^{d})$ be such that $\mu\leq_{cx}\nu$
and $\pi\in\Pi(\mu,\nu)$ be optimal for $\mathcal{W}_{2}(\mu,\nu)$ and
concentrated on the graph of a measurable map
$T:\mathbb{R}^{d}\to\mathbb{R}^{d}$. Then any $M\in\Pi^{M}(\mu,\nu)$ is an
${\cal AW}_{2}$-minimal martingale rearrangement coupling of $\pi$.

###### Proof.

Let $M\in\Pi^{\mathrm{M}}(\mu,\nu)$ and $\chi\in\Pi(\mu,\mu)$ be optimal for
$\mathcal{AW}_{2}(M,\pi)$, so that

$\mathcal{AW}_{2}^{2}(M,\pi)=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\left(|x-x^{\prime}|^{2}+\mathcal{W}_{2}^{2}(M_{x},\delta_{T(x^{\prime})})\right)\,\chi(dx,dx^{\prime})\geq\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\mathcal{W}_{2}^{2}(M_{x},\delta_{T(x^{\prime})})\,\chi(dx,dx^{\prime}).$

By bias-variance decomposition for the first equality and the fact that the
image of $\chi$ by $(x,x^{\prime})\mapsto(x,T(x^{\prime}))$ is a coupling
between $\mu$ and $\nu$ for the first inequality, and by bias-variance
decomposition again for the second equality, we have that

$\displaystyle\begin{split}\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\mathcal{W}_{2}^{2}(M_{x},\delta_{T(x^{\prime})})\,\chi(dx,dx^{\prime})&=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\left(|T(x^{\prime})-x|^{2}+\int_{\mathbb{R}^{d}}|x-y|^{2}\,M_{x}(dy)\right)\,\chi(dx,dx^{\prime})\\\
&\geq\mathcal{W}_{2}^{2}(\mu,\nu)+\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}|x-y|^{2}\,M(dx,dy)\\\
&=\int_{\mathbb{R}^{d}}\left(|x-T(x)|^{2}+\int_{\mathbb{R}^{d}}|x-y|^{2}\,M_{x}(dy)\right)\,\mu(dx)\\\
&=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}|y-T(x)|^{2}\,M(dx,dy)\\\
&=\int_{\mathbb{R}^{d}}\mathcal{W}_{2}^{2}(M_{x},\delta_{T(x)})\,\mu(dx)\geq\mathcal{AW}_{2}^{2}(M,\pi).\end{split}$
(3.14)

Using the fact that
$\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}|x-y|^{2}\,M(dx,dy)=\int_{\mathbb{R}^{d}}|y|^{2}\,\nu(dy)-\int_{\mathbb{R}^{d}}|x|^{2}\,\mu(dx)$,
we deduce that

$\mathcal{AW}_{2}^{2}(M,\pi)=\int_{\mathbb{R}^{d}}\mathcal{W}_{2}^{2}(M_{x},\delta_{T(x)})\,\mu(dx)=\mathcal{W}_{2}^{2}(\mu,\nu)+\int_{\mathbb{R}^{d}}|y|^{2}\,\nu(dy)-\int_{\mathbb{R}^{d}}|x|^{2}\,\mu(dx),$

hence any martingale coupling $M\in\Pi^{\mathrm{M}}(\mu,\nu)$ is an
$\mathcal{AW}_{2}$-minimal martingale rearrrangement coupling of $\pi$. ∎

The use of Lemma 2.1 allows us to easily prove that the analogue of
Proposition 3.1 holds for regular couplings as soon as on each interval
$(F_{\mu}(x-),F_{\mu}(x)]$, where $x\in\mathbb{R}$, the sign of $u\mapsto
F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u)$ is constant. Of course this includes the case
where $F_{\nu}^{-1}$ is constant on the intervals of the form
$(F_{\mu}(x-),F_{\mu}(x)]$ for $x\in\mathbb{R}$, or equivalently the
Hoeffding-Fréchet coupling $\pi^{HF}$ between $\mu$ and $\nu$ is concentrated
on the graph of the Monge transport map $T=F_{\nu}^{-1}\circ F_{\mu}$. In the
latter case, the conclusion of Proposition 3.4 below can also be seen as an
immediate consequence of Proposition 2.3 and the proof of Proposition 3.1.

######  Proposition 3.4.

Let $\mu,\nu\in\mathcal{P}_{1}(\mathbb{R})$ be such that $\mu\leq_{cx}\nu$ and
on each interval $(F_{\mu}(x-),F_{\mu}(x)]$, where $x\in\mathbb{R}$, the sign
of $u\mapsto F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u)$ is constant. Then for all
$Q\in\mathcal{Q}$, the martingale coupling $M^{Q}$ defined by (3.6) is a
martingale rearrangement coupling of the Hoeffding-Fréchet coupling
$\pi^{HF}$:

$\forall
Q\in\mathcal{Q},\quad\mathcal{AW}_{1}(\pi^{HF},M^{Q})=\inf_{M\in\Pi^{\mathrm{M}}(\mu,\nu)}\mathcal{AW}_{1}(\pi^{HF},M).$

###### Proof.

By Lemma 2.1 applied with $\chi=(x\mapsto(x,x))_{\sharp}\mu$, it suffices to
show that $\mu(dx)$-almost everywhere,

$\pi^{HF}_{x}\leq_{st}M^{Q}_{x}\quad\textrm{or}\quad\pi^{HF}_{x}\geq_{st}M^{Q}_{x}.$
(3.15)

Reasoning like in (2.9) we get that $\mu(dx)$-almost everywhere,

$\pi^{HF}_{x}(dy)=\int_{(0,1)}\delta_{F_{\nu}^{-1}(\theta(x,v))}(dy)\,dv\quad\textrm{and}\quad
M^{Q}_{x}(dy)=\int_{(0,1)}\widetilde{m}^{Q}_{\theta(x,v)}(dy)\,dv.$

We deduce from [24, Lemma 2.5] that for $du$-almost all $u\in(0,1)$ such that
$F_{\mu}^{-1}(u)\geq F_{\nu}^{-1}(u)$ (resp $F_{\mu}^{-1}(u)\leq
F_{\nu}^{-1}(u)$), $\delta_{F_{\nu}^{-1}(u)}\leq_{st}\widetilde{m}^{Q}_{u}$
(resp. $\delta_{F_{\nu}^{-1}(u)}\geq_{st}\widetilde{m}^{Q}_{u}$). This implies
that for $du$-almost all $u\in(0,1)$ such that $\mu(\\{F_{\mu}^{-1}(u)\\})=0$,
$\pi^{HF}_{F_{\mu}^{-1}(u)}=\delta_{F_{\nu}^{-1}(u)}$ and
$M^{Q}_{F_{\mu}^{-1}(u)}=\widetilde{m}^{Q}_{u}$ are comparable under the
stochastic order. Moreover, the assumption made on the sign of the map
$F_{\mu}^{-1}-F_{\nu}^{-1}$ on the jumps of $F_{\mu}$ implies that for
$du$-almost all $u\in(0,1)$ such that $\mu(\\{F_{\mu}^{-1}(u)\\})>0$, we have
either
$(\theta(F_{\mu}^{-1}(u),0),\theta(F_{\mu}^{-1}(u),1)]\subset\\{F_{\mu}^{-1}\geq
F_{\nu}^{-1}\\}$ so that, using the characterization of the stochastic order
in terms of the cumulative disribution functions,

$\pi^{HF}_{F^{-1}_{\mu}(u)}(dy)=\int_{(0,1)}\delta_{F_{\nu}^{-1}(\theta(F^{-1}_{\mu}(u),v))}(dy)\,dv\leq_{st}\int_{(0,1)}\widetilde{m}^{Q}_{\theta(F^{-1}_{\mu}(u),v)}(dy)\,dv=M^{Q}_{F^{-1}_{\mu}(u)}(dy),$

or
$(\theta(F_{\mu}^{-1}(u),0),\theta(F_{\mu}^{-1}(u),1)]\subset\\{F_{\mu}^{-1}\leq
F_{\nu}^{-1}\\}$ so that
$\pi^{HF}_{F^{-1}_{\mu}(u)}\geq_{st}M^{Q}_{F^{-1}_{\mu}(u)}$. By the inverse
transform sampling, this shows (3.15) and completes the proof. ∎

In the next example where the above constant sign condition fails, the inverse
transform martingale coupling between $\mu$ and $\nu$ is not a martingale
rearrangement coupling of $\pi^{HF}$. Therefore, in general, we cannot say
that every element of our family $(M^{Q})_{Q\in\mathcal{Q}}$ is a martingale
rearrangement coupling of the Hoeffding-Fréchet coupling. However, we show in
the next proposition that we can always find a specific parameter
$Q\in\mathcal{Q}$ such that the martingale coupling $M^{Q}$ is a martingale
rearrangement coupling of $\pi^{HF}$.

######  Example 3.5.

Let $\mu=\frac{1}{4}(\delta_{-1}+2\delta_{0}+\delta_{1})$ and
$\nu=\frac{1}{4}(\delta_{-2}+\delta_{-1}+\delta_{1}+\delta_{2})$. The
Hoeffding-Fréchet coupling $\pi^{HF}$ between $\mu$ and $\nu$ is given by

$\pi^{HF}=\frac{1}{4}\left(\delta_{(-1,-2)}+\delta_{(0,-1)}+\delta_{(0,1)}+\delta_{(1,2)}\right).$

To see that the inverse transform martingale coupling

$M^{IT}=\frac{1}{6}\delta_{(-1,-2)}+\frac{1}{12}\delta_{(-1,1)}+\frac{1}{12}\delta_{(0,-2)}+\frac{1}{6}\delta_{(0,-1)}+\frac{1}{6}\delta_{(0,1)}+\frac{1}{12}\delta_{(0,2)}+\frac{1}{12}\delta_{(1,-1)}+\frac{1}{6}\delta_{(1,2)}$

is not a martingale rearrangement coupling of $\pi^{HF}$, we rely on the
equivalent condition provided by Lemma 2.1. One can readily compute
$M^{IT}_{0}=\frac{1}{6}\delta_{-2}+\frac{1}{3}\delta_{-1}+\frac{1}{3}\delta_{1}+\frac{1}{6}\delta_{2}$,
$\pi^{HF}_{-1}=\delta_{-2}$,
$\pi^{HF}_{0}=\frac{1}{2}(\delta_{-1}+\delta_{1})$ and
$\pi^{HF}_{1}=\delta_{2}$. Then $-1<0$, $1>0$ and $0=0$, but we have neither
$\pi^{HF}_{-1}\geq_{st}M^{IT}_{0}$, $\pi^{HF}_{1}\leq_{st}M^{IT}_{0}$,
$\pi^{HF}_{0}\leq_{st}M^{IT}_{0}$ nor $\pi^{HF}_{0}\geq_{st}M^{IT}_{0}$. We
deduce by Lemma 2.1 that $M^{IT}$ is not a martingale rearrangement coupling
of $\pi^{HF}$.

Note that the martingale rearrangement constructed in Section 2.2 is

$\frac{3}{16}\delta_{(-1,-2)}+\frac{1}{16}\delta_{(-1,2)}+\frac{1}{4}\delta_{(0,-1)}+\frac{1}{4}\delta_{(0,1)}+\frac{1}{16}\delta_{(1,-2)}+\frac{3}{16}\delta_{(1,2)}.$

######  Proposition 3.6.

Let $\mu,\nu\in\mathcal{P}_{1}(\mathbb{R})$ be such that $\mu\leq_{cx}\nu$.
Let $\pi^{HF}$ be the Hoeffding-Fréchet coupling between $\mu$ and $\nu$. Then
there exists $Q\in\mathcal{Q}$ such that the martingale coupling $M^{Q}$
defined by (3.6) is a martingale rearrangement coupling of $\pi^{HF}$:

$\mathcal{AW}_{1}(\pi^{HF},M^{Q})=\inf_{M\in\Pi^{\mathrm{M}}(\mu,\nu)}\mathcal{AW}_{1}(\pi^{HF},M).$

######  Remark 3.7.

We show in the proof that as soon as on each interval
$(F_{\mu}(x-),F_{\mu}(x)]$ for $x\in\mathbb{R}$ the sign of $u\mapsto
F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u)$ is constant, the martingale rearrangement
coupling $M^{Q}$ is the inverse transform martingale coupling, in coherence
with Proposition 3.4.

###### Proof of Proposition 3.6.

By (1.9), we have

$\inf_{M\in\Pi^{\mathrm{M}}(\mu,\nu)}\mathcal{AW}_{1}(\pi^{HF},M)\geq\int_{\mathbb{R}}\left|x-\int_{\mathbb{R}}y\,\pi^{HF}_{x}(dy)\right|\,\mu(dx),$
(3.16)

hence it is sufficient to show that there exists $Q\in\mathcal{Q}$ such that

$\mathcal{AW}_{1}(\pi^{HF},M^{Q})\leq\int_{\mathbb{R}}\left|x-\int_{\mathbb{R}}y\,\pi^{HF}_{x}(dy)\right|\,\mu(dx).$
(3.17)

If $\mu=\nu$ then the statement is straightforward, hence we suppose
$\mu\neq\nu$. The proof is achieved in four steps. First we exhibit an
appropriate subdivision of the intervals $(0,1)$ in order to define a measure
$Q$ on $(0,1)^{2}$. Second we show that $Q$ belongs to $\mathcal{Q}$ and is
therefore associated to the martingale coupling $M^{Q}$ between $\mu$ and
$\nu$. Then we find for $\mu(dx)$-almost all $x\in\mathbb{R}$ a coupling
$\eta_{x}\in\Pi(\pi^{HF}_{x},M^{Q}_{x})$, so that

$\mathcal{AW}_{1}(\pi^{HF},M^{Q})\leq\int_{\mathbb{R}}\mathcal{W}_{1}(\pi^{HF}_{x},M^{Q}_{x})\,\mu(dx)\leq\int_{\mathbb{R}}\left(\int_{\mathbb{R}}|y-y^{\prime}|\,\eta_{x}(dy,dy^{\prime})\right)\,\mu(dx).$
(3.18)

Last, we show that for $\mu(dx)$-almost all $x\in\mathbb{R}$,

$\int_{\mathbb{R}\times\mathbb{R}}|y-y^{\prime}|\,\eta_{x}(dy,dy^{\prime})=\left|x-\int_{\mathbb{R}}y\,\pi^{HF}_{x}(dy)\right|,$
(3.19)

which implies (3.17) and completes the proof.

Step 1. Recall the definitions of $\Psi_{+}$, $\Psi_{-}$, $\mathcal{U}^{+}$
and $\mathcal{U}^{-}$ from (3.1) and (3.7). Let
$A_{\mu}=\\{x\in\mathbb{R}\mid\mu(\\{x\\})>0\\}$. For all $x\in A_{\mu}$, let
$\mathcal{U}_{x}=(F_{\mu}(x-),F_{\mu}(x)]$,
$\mathcal{U}^{+}_{x}=\mathcal{U}^{+}\cap\mathcal{U}_{x}$,
$\mathcal{U}^{-}_{x}=\mathcal{U}^{-}\cap\mathcal{U}_{x}$ and

$\displaystyle a_{x}$
$\displaystyle=\inf\left\\{a\in[F_{\mu}(x-),F_{\mu}(x)]\mid\int_{F_{\mu}(x-)}^{a}(x-F_{\nu}^{-1}(u))^{+}\,du=\left(\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(u))^{+}\,du\right)\right.$
$\displaystyle\phantom{=\inf\left\\{a\in[F_{\mu}(x-),F_{\mu}(x)]\mid\int_{F_{\mu}(x-)}^{a}(x-F_{\nu}^{-1}(u))^{+}\,du=\right.\
}\left.\wedge\left(\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(u))^{-}\,du\right)\right\\},$
$\displaystyle b_{x}$
$\displaystyle=\sup\left\\{b\in[F_{\mu}(x-),F_{\mu}(x)]\mid\int_{b}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(u))^{-}\,du=\left(\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(u))^{+}\,du\right)\right.$
$\displaystyle\phantom{=\sup\left\\{b\in[F_{\mu}(x-),F_{\mu}(x)]\mid\int_{b}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(u))^{-}\,du=\right.\
}\left.\wedge\left(\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(u))^{-}\,du\right)\right\\},$
$\displaystyle\mathcal{V}^{+}_{x}$
$\displaystyle=(F_{\mu}(x-),a_{x}],\quad\mathcal{V}^{-}_{x}=(b_{x},F_{\mu}(x)],\quad\widetilde{\mathcal{U}}^{+}=\mathcal{U}^{+}\backslash(\bigcup_{x\in
A_{\mu}}\mathcal{V}^{+}_{x}),\quad\text{and}\quad\widetilde{\mathcal{U}}^{-}=\mathcal{U}^{-}\backslash(\bigcup_{x\in
A_{\mu}}\mathcal{V}^{-}_{x}).$

By monotonicity of $u\mapsto x-F_{\nu}^{-1}(u)$ and by definition of $a_{x}$
and $b_{x}$, we have $F_{\mu}(x-)\leq a_{x}\leq b_{x}\leq F_{\mu}(x)$,

$\displaystyle\begin{split}\int_{\mathcal{V}^{+}_{x}}(x-F_{\nu}^{-1}(u))\,du&=\int_{\mathcal{V}^{-}_{x}}(x-F_{\nu}^{-1}(u))\,du\\\
&=\left(\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(u))^{+}\,du\right)\wedge\left(\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(u))^{-}\,du\right),\end{split}$
(3.20)

$\mathcal{V}^{+}_{x}\subset\mathcal{U}^{+}_{x}$ and
$\mathcal{V}^{-}_{x}\subset\mathcal{U}^{-}_{x}$. Moreover, we have

$\displaystyle\begin{split}\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(u))^{+}\,du\geq\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(u))^{-}\,du&\iff(F_{\mu}(x-),b_{x}]\cap\mathcal{U}^{-}_{x}=\emptyset\iff\mathcal{V}^{-}_{x}=\mathcal{U}^{-}_{x},\\\
\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(u))^{+}\,du\leq\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(u))^{-}\,du&\iff(a_{x},F_{\mu}(x)]\cap\mathcal{U}^{+}_{x}=\emptyset\iff\mathcal{V}^{+}_{x}=\mathcal{U}^{+}_{x}.\end{split}$
(3.21)

$u$$|x-F_{\nu}^{-1}(u)|$$F_{\mu}(x-)$$F_{\mu}(x)$$a_{x}$$b_{x}$$\mathcal{U}^{+}_{x}$$\mathcal{V}^{+}_{x}$$\mathcal{U}^{-}_{x}=\mathcal{V}^{-}_{x}$
Figure 1: Points and intervals involved in the proof in the case where
$\int_{\mathcal{U}_{x}}(x-F_{\nu}^{-1}(u))^{+}\,du>\int_{\mathcal{U}_{x}}(x-F_{\nu}^{-1}(u))^{-}\,du$.

For $x\in A_{\mu}$, let $Q_{x}$ be any measure on $(0,1)^{2}$ such that its
first and second marginal are respectively

$\frac{1}{\Psi_{+}(1)}\mathds{1}_{\mathcal{V}^{+}_{x}}(u)\,(x-F_{\nu}^{-1}(u))^{+}\,du\quad\text{and}\quad\frac{1}{\Psi_{+}(1)}\mathds{1}_{\mathcal{V}^{-}_{x}}(v)\,(x-F_{\nu}^{-1}(v))^{-}\,dv.$
(3.22)

Notice that $a_{x}\leq b_{x}$ implies

$Q_{x}(\\{(u,v)\in(0,1)^{2}\mid u<v\\})=Q_{x}((0,1)^{2}).$ (3.23)

Let now $\chi_{+},\chi_{-}:[0,1]\to\mathbb{R}$ be defined for all $u\in[0,1]$
by

$\chi_{+}(u)=\int_{0}^{u}(F_{\mu}^{-1}-F_{\nu}^{-1})^{+}(v)\mathds{1}_{\widetilde{\mathcal{U}}^{+}}(v)\,dv\quad\text{and}\quad\chi_{-}(u)=\int_{0}^{u}(F_{\mu}^{-1}-F_{\nu}^{-1})^{-}(v)\mathds{1}_{\widetilde{\mathcal{U}}^{-}}(v)\,dv.$
(3.24)

For all $u\in[0,1]$, $[0,u]\cap\mathcal{U}^{+}$, resp.
$[0,u]\cap\mathcal{U}^{-}$, is the disjoint union of
$[0,u]\cap\widetilde{\mathcal{U}}^{+}$ and $[0,u]\cap\left(\bigcup_{x\in
A_{\mu}}\mathcal{V}^{+}_{x}\right)$, resp.
$[0,u]\cap\widetilde{\mathcal{U}}^{-}$ and $[0,u]\cap\left(\bigcup_{x\in
A_{\mu}}\mathcal{V}^{-}_{x}\right)$. Therefore, for all $u\in[0,1]$,

$\displaystyle\begin{split}\Psi_{+}(u)&=\chi_{+}(u)+\sum_{x\in
A_{\mu}}\int_{[0,u]\cap\mathcal{V}^{+}_{x}}(F_{\mu}^{-1}-F_{\nu}^{-1})^{+}(v)\,dv,\\\
\text{and}\quad\Psi_{-}(u)&=\chi_{-}(u)+\sum_{x\in
A_{\mu}}\int_{[0,u]\cap\mathcal{V}^{-}_{x}}(F_{\mu}^{-1}-F_{\nu}^{-1})^{-}(v)\,dv.\end{split}$
(3.25)

Applying (3.25) with $u=1$, (3.20) and the equality $\Psi_{+}(1)=\Psi_{-}(1)$,
we get $\chi_{+}(1)=\chi_{-}(1)$, hence we can define the map
$\Gamma:[0,1]\to[0,1]$ for all $u\in[0,1]$ by

$\displaystyle\begin{split}\Gamma(u)=\left\\{\begin{array}[]{rl}\chi_{-}^{-1}(\chi_{+}(u))&\text{if}\
u\in\widetilde{\mathcal{U}}^{+};\\\ \chi_{+}^{-1}(\chi_{-}(u))&\text{if}\
u\in\widetilde{\mathcal{U}}^{-};\\\
u&\text{otherwise.}\end{array}\right.\end{split}$

Let then $\widetilde{Q}$ be the measure on $(0,1)^{2}$ defined by

$\widetilde{Q}(du,dv)=\frac{1}{\Psi_{+}(1)}\,d\chi_{+}(u)\,\delta_{\Gamma(u)}(dv).$
(3.26)

Step 2. We now show that the measure $Q$ on $(0,1)^{2}$ defined by

$Q=\widetilde{Q}+\sum_{x\in A_{\mu}}Q_{x}$

is an element of $\mathcal{Q}$. Notice that if for all $x\in\mathbb{R}$,
$u\mapsto F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u)$ does not change sign on
$\mathcal{U}_{x}$, then $\mathcal{V}^{+}_{x}=\mathcal{V}^{-}_{x}=\emptyset$.
In that case, $\widetilde{\mathcal{U}}^{+}=\mathcal{U}^{+}$ and
$\widetilde{\mathcal{U}}^{-}=\mathcal{U}^{-}$, hence $Q=\widetilde{Q}=Q^{IT}$,
which proves the statement of Remark 3.7, provided that we complete the
present proof.

To prove that $Q\in\mathcal{Q}$ we begin to show that

$Q(\\{(u,v)\in(0,1)^{2}\mid u<v\\})=Q((0,1)^{2}).$ (3.27)

In view of (3.23) it suffices to show that

$\widetilde{Q}(\\{(u,v)\in(0,1)^{2}\mid u<v\\})=\widetilde{Q}((0,1)^{2}),$

which by (3.26) is equivalent to

$\Gamma(u)>u,\quad d\chi_{+}(u)\text{-almost everywhere}.$ (3.28)

Let $u\in[0,1]$. Suppose first that $u\notin B_{\mu}:=\bigcup_{x\in
A_{\mu}}(F_{\mu}(x-),F_{\mu}(x))$. Then for all $x\in A_{\mu}$, we have either
$u\leq F_{\mu}(x-)$ or $F_{\mu}(x)\leq u$, and since
$\mathcal{V}^{+}_{x},\mathcal{V}^{-}_{x}\subset\mathcal{U}_{x}$, either
$\mathcal{V}^{+}_{x},\mathcal{V}^{-}_{x}\subset(u,1)$ or
$\mathcal{V}^{+}_{x},\mathcal{V}^{-}_{x}\subset[0,u]$. Equivalently,

$[0,u]\cap\mathcal{V}^{+}_{x}=\mathcal{V}^{+}_{x}\quad\text{and}\quad[0,u]\cap\mathcal{V}^{-}_{x}=\mathcal{V}^{-}_{x},\quad\text{or}\quad[0,u]\cap\mathcal{V}^{+}_{x}=[0,u]\cap\mathcal{V}^{-}_{x}=\emptyset.$

If $[0,u]\cap\mathcal{V}^{+}_{x}=\mathcal{V}^{+}_{x}$ and
$[0,u]\cap\mathcal{V}^{-}_{x}=\mathcal{V}^{-}_{x}$, (3.20) yields

$\displaystyle\int_{[0,u]\cap\mathcal{V}^{+}_{x}}(F_{\mu}^{-1}-F_{\nu}^{-1})^{+}(v)\,dv$
$\displaystyle=\int_{\mathcal{V}^{+}_{x}}(x-F_{\nu}^{-1}(v))\,dv=\int_{\mathcal{V}^{-}_{x}}(x-F_{\nu}^{-1}(v))\,dv$
$\displaystyle=\int_{[0,u]\cap\mathcal{V}^{-}_{x}}(F_{\mu}^{-1}-F_{\nu}^{-1})^{-}(v)\,dv.$

Else if $[0,u]\cap\mathcal{V}^{+}_{x}=[0,u]\cap\mathcal{V}^{-}_{x}=\emptyset$,
then we clearly have
$\int_{[0,u]\cap\mathcal{V}^{+}_{x}}(F_{\mu}^{-1}-F_{\nu}^{-1})^{+}(v)\,dv=\int_{[0,u]\cap\mathcal{V}^{-}_{x}}(F_{\mu}^{-1}-F_{\nu}^{-1})^{-}(v)\,dv$
too. We then deduce from (3.25) that
$\chi_{+}(u)-\chi_{-}(u)=\Psi_{+}(u)-\Psi_{-}(u)$. By [24, (3.8)],
$\Psi_{+}(u)>\Psi_{-}(u)$ for $du$-almost every $u\in\mathcal{U}^{+}$ and
therefore $u\in\widetilde{\mathcal{U}}^{+}$. We deduce that

$\chi_{+}(u)\geq\chi_{-}(u),\quad\text{for all $u\in[0,1]\backslash B_{\mu}$,
and the inequality is strict for $du$-almost every
}u\in\widetilde{\mathcal{U}}^{+}\backslash B_{\mu}.$ (3.29)

Suppose now that $u\in\widetilde{\mathcal{U}}^{+}\cap B_{\mu}$. Then there
exists $x\in A_{\mu}$ such that $u\in(F_{\mu}(x-),F_{\mu}(x))$, or
equivalently $u\in(a_{x},F_{\mu}(x))\cap\mathcal{U}^{+}$. Since $F_{\mu}^{-1}$
and $F_{\nu}^{-1}$ are left continuous, we have for $\varepsilon>0$ small
enough $[u-\varepsilon,u]\subset\mathcal{U}^{+}$ and of course
$[u-\varepsilon,u]\subset(a_{x},F_{\mu}(x))$, hence
$[u-\varepsilon,u]\subset\widetilde{\mathcal{U}}^{+}$ and

$\chi_{+}(u)>\chi_{+}(u-\varepsilon)\geq\chi_{+}(F_{\mu}(x-)).$

Using (3.29) with $F_{\mu}(x-)\in[0,1]\backslash B_{\mu}$, we get
$\chi_{+}(F_{\mu}(x-))\geq\chi_{-}(F_{\mu}(x-))$. Moreover, the existence of
$u$ implies that $\mathcal{V}^{+}_{x}\neq\mathcal{U}^{+}_{x}$ and therefore
$\mathcal{V}^{-}_{x}=\mathcal{U}^{-}_{x}$ by (3.21), hence
$\chi_{-}(u)=\chi_{-}(F_{\mu}(x-))$. We deduce that

$\chi_{+}(u)>\chi_{-}(u),\quad\text{for all
}u\in\widetilde{\mathcal{U}}^{+}\cap B_{\mu}.$ (3.30)

Since for all $u\in(0,1)$, $\chi_{+}(u)>\chi_{-}(u)\iff\Gamma(u)>u$, (3.28)
follows directly from (3.29) and (3.30), which proves (3.27).

We now show that $Q$ has the right marginals. On the one hand,
$\mathcal{U}^{+}=\widetilde{\mathcal{U}}^{+}\cup\left(\bigcup_{x\in
A_{\mu}}\mathcal{V}^{+}_{x}\right)$ where the unions are disjoint, hence its
first marginal is

$\displaystyle\begin{split}&\frac{1}{\Psi_{+}(1)}d\chi_{+}(u)+\sum_{x\in
A_{\mu}}\frac{1}{\Psi_{+}(1)}\mathds{1}_{\mathcal{V}^{+}_{x}}(u)(x-F_{\nu}^{-1}(u))^{+}\,du\\\
&=\frac{1}{\Psi_{+}(1)}\left((F_{\mu}^{-1}-F_{\nu}^{-1})^{+}(u)\mathds{1}_{\widetilde{\mathcal{U}}^{+}}(u)\,du+\sum_{x\in
A_{\mu}}(F_{\mu}^{-1}-F_{\nu}^{-1})^{+}(u)\mathds{1}_{\mathcal{V}^{+}_{x}}(u)\,du\right)\\\
&=\frac{1}{\Psi_{+}(1)}(F_{\mu}^{-1}-F_{\nu}^{-1})^{+}(u)\mathds{1}_{\mathcal{U}^{+}}(u)\,du=\frac{1}{\Psi_{+}(1)}d\Psi_{+}(u).\end{split}$
(3.31)

On the other hand, by [24, Lemma 6.1] applied with
$f_{1}=(F_{\mu}^{-1}-F_{\nu}^{-1})^{+}\mathds{1}_{\widetilde{\mathcal{U}}^{+}}$,
$f_{2}=(F_{\mu}^{-1}-F_{\nu}^{-1})^{-}\mathds{1}_{\widetilde{\mathcal{U}}^{-}}$,
$u_{0}=1$ and
$h:u\mapsto\mathds{1}_{\\{u\notin\widetilde{\mathcal{U}}^{-}\\}}$, we get

$\int_{0}^{1}\mathds{1}_{\\{\Gamma(u)\notin\widetilde{\mathcal{U}}^{-}\\}}\,d\chi_{+}(u)=\int_{0}^{1}\mathds{1}_{\\{v\notin\widetilde{\mathcal{U}}^{-}\\}}\,d\chi_{-}(v)=0,$

hence $\Gamma(u)\in\widetilde{\mathcal{U}}^{-}$ for $du$-almost all
$u\in\widetilde{\mathcal{U}}^{+}$. Reasoning like in the derivation of (2.16)
with
$(\Gamma,\chi_{+},\chi_{-},\widetilde{\mathcal{U}}^{+},\widetilde{\mathcal{U}}^{-})$
replacing $(\varphi,\Psi_{+},\Psi_{-},\mathcal{U}^{+},\mathcal{U}^{-})$, we
get that

$u=\Gamma(\Gamma(u)),\quad\text{$du$-almost everywhere on
$\widetilde{\mathcal{U}}^{+}$}.$ (3.32)

Let $H:(0,1)^{2}\to\mathbb{R}$ be a measurable and bounded map. Applying [24,
Lemma 6.1] with
$f_{1}=(F_{\mu}^{-1}-F_{\nu}^{-1})^{+}\mathds{1}_{\widetilde{\mathcal{U}}^{+}}$,
$f_{2}=(F_{\mu}^{-1}-F_{\nu}^{-1})^{-}\mathds{1}_{\widetilde{\mathcal{U}}^{-}}$,
$u_{0}=1$ and $h:u\mapsto H(\Gamma(u),u)$ then yields

$\displaystyle\int_{(0,1)^{2}}H(u,v)\,\widetilde{Q}(du,dv)$
$\displaystyle=\frac{1}{\Psi_{+}(1)}\int_{0}^{1}H(u,\Gamma(u))\,d\chi_{+}(u)$
$\displaystyle=\frac{1}{\Psi_{+}(1)}\int_{0}^{1}H(\Gamma(\Gamma(u)),\Gamma(u))(F_{\mu}^{-1}-F_{\nu}^{-1})^{+}(u)\mathds{1}_{\widetilde{\mathcal{U}}^{+}}(u)\,du$
$\displaystyle=\frac{1}{\Psi_{+}(1)}\int_{0}^{1}H(\Gamma(v),v)(F_{\mu}^{-1}-F_{\nu}^{-1})^{-}(v)\mathds{1}_{\widetilde{\mathcal{U}}^{-}}(v)\,dv$
$\displaystyle=\frac{1}{\Psi_{-}(1)}\int_{0}^{1}H(\Gamma(v),v)\,d\chi_{-}(v).$

We deduce that

$\widetilde{Q}(du,dv)=\frac{1}{\Psi_{+}(1)}d\chi_{-}(v)\,\delta_{\Gamma(v)}(du),$
(3.33)

and we show with a calculation similar to (3.31) that its second marginal is
$\frac{1}{\Psi_{+}(1)}d\Psi_{-}$. We conclude that $Q\in\mathcal{Q}$.

Step 3. For all $x\in\mathbb{R}$, let
$(\eta_{x}(dy,dy^{\prime}))_{x\in\mathbb{R}}$ be the probability kernel
defined by

$\left\\{\begin{array}[]{r}\displaystyle\delta_{x}(dy)\,\delta_{x}(dy^{\prime})\\\
\displaystyle\textrm{if}\ F_{\mu}(x)=0\ \textrm{or}\ F_{\mu}(x-)=1;\\\ \\\
\displaystyle\frac{1}{\mu(\\{x\\})}\left(\int_{u\in\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}}\widetilde{m}^{Q}_{u}(dy^{\prime})\,\delta_{y^{\prime}}(dy)\,du+\int_{u\in(a_{x},b_{x})}\widetilde{m}^{Q}_{u}(dy^{\prime})\,\delta_{F_{\nu}^{-1}(u)}(dy)\,du\right)\\\
\displaystyle\textrm{if}\ \mu(\\{x\\})>0;\\\ \\\
\displaystyle\delta_{F_{\nu}^{-1}(F_{\mu}(x))}(dy)\,\widetilde{m}^{Q}_{F_{\mu}(x)}(dy^{\prime})\\\
\textrm{otherwise},\end{array}\right.$

where the probability kernel $(\widetilde{m}^{Q}_{u})_{u\in(0,1)}$ is given by
(3.3). Notice that in view of the definition (3.6) of $M^{Q}$, one can check
that for $\mu(dx)$-almost all $x\in\mathbb{R}$,

$M^{Q}_{x}(dy)=\left\\{\begin{array}[]{cr}\delta_{x}(dy)&\text{if}\
F_{\mu}(x)=0\ \text{or}\ F_{\mu}(x_{-})=1;\\\ \\\
\displaystyle\frac{1}{\mu(\\{x\\})}\int_{u=F_{\mu}(x-)}^{F_{\mu}(x)}\widetilde{m}^{Q}_{u}(dy)\,du&\text{if}\
\mu(\\{x\\})>0;\\\ \\\
\displaystyle\widetilde{m}^{Q}_{F_{\mu}(x)}(dy)&\text{otherwise}.\end{array}\right.$
(3.34)

Let us show that for $\mu(dx)$-almost all $x\in\mathbb{R}$,
$\eta_{x}(dy,dy^{\prime})$ is a coupling between $\pi^{HF}_{x}$ and
$M^{Q}_{x}$. Let $x\in\mathbb{R}$. By (1.13) we may suppose without loss of
generality that $F_{\mu}(x)>0$ and $F_{\mu}(x-)<1$. Let
$h:\mathbb{R}\to\mathbb{R}$ be a measurable and bounded map. Suppose first
that $\mu(\\{x\\})=0$. Then by (3.11) for the second equality,

$\displaystyle\int_{\mathbb{R}\times\mathbb{R}}h(y)\,\eta_{x}(dy,dy^{\prime})$
$\displaystyle=h(F_{\nu}^{-1}(F_{\mu}(x)))=\int_{\mathbb{R}}h(y)\,\pi^{HF}_{x}(dy),$
$\displaystyle\text{and}\quad\int_{\mathbb{R}\times\mathbb{R}}h(y^{\prime})\,\eta_{x}(dy,dy^{\prime})$
$\displaystyle=\int_{\mathbb{R}}h(y^{\prime})\,\widetilde{m}^{Q}_{F_{\mu}(x)}(dy^{\prime})=\int_{\mathbb{R}}h(y^{\prime})\,M^{Q}_{x}(dy^{\prime}).$

Suppose now that that $\mu(\\{x\\})>0$. On the one hand, using the fact that
$\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}\cup(a_{x},b_{x}]=(F_{\mu}(x-),F_{\mu}(x)]$
for the second equality, we have

$\displaystyle\int_{\mathbb{R}\times\mathbb{R}}h(y^{\prime})\,\eta_{x}(dy,dy^{\prime})$
$\displaystyle=\frac{1}{\mu(\\{x\\})}\left(\int_{\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}}\int_{\mathbb{R}}h(y^{\prime})\,\widetilde{m}^{Q}_{u}(dy^{\prime})\,du+\int_{(a_{x},b_{x})}\int_{\mathbb{R}}h(y^{\prime})\,\widetilde{m}^{Q}_{u}(dy^{\prime})\,du\right)$
$\displaystyle=\frac{1}{\mu(\\{x\\})}\int_{(F_{\mu}(x-),F_{\mu}(x)]}\int_{\mathbb{R}}h(y^{\prime})\,\widetilde{m}^{Q}_{u}(dy^{\prime})\,du=\int_{\mathbb{R}}h(y^{\prime})\,M^{Q}_{x}(dy^{\prime}).$

On the other hand,

$\int_{\mathbb{R}\times\mathbb{R}}h(y)\,\eta_{x}(dy,dy^{\prime})=\frac{1}{\mu(\\{x\\})}\left(\int_{\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}}\int_{\mathbb{R}}h(y^{\prime})\,\widetilde{m}^{Q}_{u}(dy^{\prime})\,du+\int_{(a_{x},b_{x})}h(F_{\nu}^{-1}(u))\,du\right).$
(3.35)

Assume for a moment that

$\int_{\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}}\int_{\mathbb{R}}h(y^{\prime})\,\widetilde{m}^{Q}_{u}(dy^{\prime})\,du=\int_{\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}}h(F_{\nu}^{-1}(u))\,du.$
(3.36)

We then deduce with (3.35) and (3.11) that

$\displaystyle\int_{\mathbb{R}\times\mathbb{R}}h(y)\,\eta_{x}(dy,dy^{\prime})$
$\displaystyle=\frac{1}{\mu(\\{x\\})}\left(\int_{\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}}h(F_{\nu}^{-1}(u))\,du+\int_{(a_{x},b_{x})}h(F_{\nu}^{-1}(u))\,du\right)$
$\displaystyle=\frac{1}{\mu(\\{x\\})}\int_{(F_{\mu}(x-),F_{\mu}(x)]}h(F_{\nu}^{-1}(u))\,du=\int_{\mathbb{R}}h(y)\,\pi^{HF}_{x}(dy).$

This proves that we indeed have
$\eta_{x}(dy,dy^{\prime})\in\Pi(\pi^{HF}_{x},M^{Q}_{x})$ for $\mu(dx)$-almost
all $x\in\mathbb{R}$, hence (3.18) holds.

Let us then prove (3.36). By (3.2) there exists a probability kernel
$(\pi^{Q}_{u})_{u\in(0,1)}$ such that

$Q(du,dv)=\frac{1}{\Psi_{+}(1)}\,d\Psi_{+}(u)\,\pi^{Q}_{u}(dv)=\frac{1}{\Psi_{+}(1)}\,d\Psi_{-}(v)\,\pi^{Q}_{v}(du).$

Since the first, resp. second marginals of $Q-Q_{x}$ and $Q_{x}$ are singular,
i.e. supported by disjoint measurable subsets of $(0,1)$, we have

$Q_{x}(du,dv)=\frac{1}{\Psi_{+}(1)}\mathds{1}_{\mathcal{V}^{+}_{x}}(u)\,d\Psi_{+}(u)\,\pi^{Q}_{u}(dv)=\frac{1}{\Psi_{+}(1)}\mathds{1}_{\mathcal{V}^{-}_{x}}(v)\,d\Psi_{-}(v)\,\pi^{Q}_{v}(du).$
(3.37)

For $u,v\in(0,1)$ such that $F_{\nu}^{-1}(u)\neq F_{\nu}^{-1}(v)$, let
$\Delta(u,v)=\frac{h(F_{\nu}^{-1}(v))-h(F_{\nu}^{-1}(u))}{F_{\nu}^{-1}(v)-F_{\nu}^{-1}(u)}$.
By [24, Lemma 2.5], for $du$-almost all $u\in(0,1)$ we have

$\displaystyle\begin{split}\int_{\mathbb{R}}h(y)\,\widetilde{m}^{Q}_{u}(dy)&=h(F_{\nu}^{-1}(u))+\int_{(0,1)}\Delta(u,v)(F_{\mu}^{-1}-F_{\nu}^{-1})^{+}(u)\,\pi^{Q}_{u}(dv)\\\
&\phantom{=\
}-\int_{(0,1)}\Delta(u,v)(F_{\mu}^{-1}-F_{\nu}^{-1})^{-}(u)\,\pi^{Q}_{u}(dv),\end{split}$
(3.38)

where the integrals are well defined. Using the facts that
$\mathcal{V}^{+}_{x}\subset\mathcal{U}^{+}$,
$\mathcal{V}^{-}_{x}\subset\mathcal{U}^{-}$ and the symmetry of the function
$\Delta$, we get

$\displaystyle\int_{\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}}\int_{(0,1)}\Delta(u,v)(F_{\mu}^{-1}-F_{\nu}^{-1})^{+}(u)\,\pi^{Q}_{u}(dv)\,du$
$\displaystyle=\int_{(0,1)^{2}}\Delta(u,v)(F_{\mu}^{-1}-F_{\nu}^{-1})^{+}(u)\mathds{1}_{\mathcal{V}^{+}_{x}}(u)\,\pi^{Q}_{u}(dv)\,du$
$\displaystyle=\Psi_{+}(1)\int_{(0,1)^{2}}\Delta(u,v)\,Q_{x}(du,dv)$
$\displaystyle=\int_{(0,1)^{2}}\Delta(u,v)(F_{\mu}^{-1}-F_{\nu}^{-1})^{-}(v)\mathds{1}_{\mathcal{V}^{-}_{x}}(v)\,\pi^{Q}_{v}(du)\,dv$
$\displaystyle=\int_{\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}}\int_{(0,1)}\Delta(u,v)(F_{\mu}^{-1}-F_{\nu}^{-1})^{-}(u)\,\pi^{Q}_{u}(dv)\,du.$

Then (3.36) is a direct consequence of (3.38) integrated on
$\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}$ with respect to the Lebesgue
measure.

Step 4. As mentioned at the beginning of the proof, it remains only to show
that for $\mu(dx)$-almost all $x\in\mathbb{R}$, (3.19) is satisfied. By (3.5)
and (1.16) we have for $\mu(dx)$-almost all $x\in\mathbb{R}$ and $dv$-almost
$v\in(0,1)$

$\int_{\mathbb{R}}|y^{\prime}-F_{\nu}^{-1}(F_{\mu}(x-)+v\mu(\\{x\\}))|\,\widetilde{m}^{Q}_{F_{\mu}(x-)+v\mu(\\{x\\})}(dy^{\prime})=|x-F_{\nu}^{-1}(F_{\mu}(x-)+v\mu(\\{x\\}))|.$

The latter equality implies that for $\mu(dx)$-almost all $x\in\mathbb{R}$
such that $\mu(\\{x\\})=0$,

$\displaystyle\int_{\mathbb{R}\times\mathbb{R}}|y-y^{\prime}|\eta_{x}(dy,dy^{\prime})$
$\displaystyle=\int_{\mathbb{R}}|y^{\prime}-F_{\nu}^{-1}(F_{\mu}(x))|\,\widetilde{m}^{Q}_{F_{\mu}(x)}(dy^{\prime})=|x-F_{\nu}^{-1}(F_{\mu}(x))|$
$\displaystyle=\left|x-\int_{\mathbb{R}}y\,\pi^{HF}_{x}(dy)\right|.$

It remains to show (3.19) for $x\in\mathbb{R}$ such that $\mu(\\{x\\})>0$. For
such an element $x$, we have by (3.21) either
$\mathcal{V}^{+}_{x}=\mathcal{U}^{+}_{x}$, which implies
$(a_{x},b_{x})\cap\mathcal{U}^{+}_{x}=\emptyset$, or
$\mathcal{V}^{-}_{x}=\mathcal{U}^{-}_{x}$, which implies
$(a_{x},b_{x})\cap\mathcal{U}^{-}_{x}=\emptyset$. In both cases we have that
$u\mapsto F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u)$ does not change sign on
$(a_{x},b_{x})$. Added to (3.5) and (3.20), we deduce that

$\displaystyle\int_{\mathbb{R}\times\mathbb{R}}|y-y^{\prime}|\,\eta_{x}(dy,dy^{\prime})$
$\displaystyle=\frac{1}{\mu(\\{x\\})}\int_{(a_{x},b_{x})}\left(\int_{\mathbb{R}}|y^{\prime}-F_{\nu}^{-1}(u)|\,\widetilde{m}^{Q}_{u}(dy^{\prime})\right)\,du$
$\displaystyle=\frac{1}{\mu(\\{x\\})}\int_{(a_{x},b_{x})}\left|F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u)\right|\,du$
$\displaystyle=\frac{1}{\mu(\\{x\\})}\left|\int_{(a_{x},b_{x})}(F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u))\,du\right|$
$\displaystyle=\frac{1}{\mu(\\{x\\})}\left|\int_{\mathcal{V}^{+}_{x}}(F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u))\,du+\int_{(a_{x},b_{x})}(F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u))\,du+\int_{\mathcal{V}^{-}_{x}}(F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u))\,du\right|$
$\displaystyle=\frac{1}{\mu(\\{x\\})}\left|\int_{(F_{\mu}(x-),F_{\mu}(x)]}(x-F_{\nu}^{-1}(u))\,du\right|$
$\displaystyle=\left|x-\frac{1}{\mu(\\{x\\})}\int_{(F_{\mu}(x-),F_{\mu}(x)]}F_{\nu}^{-1}(u)\,du\right|$
$\displaystyle=\left|x-\int_{\mathbb{R}}y\,\pi^{HF}_{x}(dy)\right|,$

which shows (3.19) and completes the proof. ∎

######  Remark 3.8.

  1. (i)

Despite appearances, the martingale coupling $M^{Q}$ constructed in the proof
of Proposition 3.6 does not depend on the choice of the measures $Q_{x}$,
$x\in A_{\mu}$, whose marginals are given by (3.22). Informally, we see that
$(Q_{x})_{x\in A_{\mu}}$ does not affect $(M^{Q}_{x})_{x\in\mathbb{R}}$
outside the jumps of $F_{\mu}$. Moreover, for all $x\in A_{\mu}$, $Q_{x}$
describes the way the elements of $\mathcal{V}^{+}_{x}$ are matched with the
elements of $\mathcal{V}^{-}_{x}$, but this level of detail is not seen by
$M^{Q}_{x}$, which only retains the contribution
$\frac{1}{\mu(\\{x\\})}\int_{\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}}\delta_{F_{\nu}^{-1}(u)}(dy)\,du$,
as shown below.

Formally, since for all $x\in A_{\mu}$, the first, resp. second marginals of
$\widetilde{Q}$ and $Q_{x}$ are singular, i.e. supported by disjoint
measurable subsets of $(0,1)$, we have $\pi^{Q}_{u}=\delta_{\Gamma(u)}$ for
$du$-almost all
$u\in\widetilde{\mathcal{U}}^{+}\cup\widetilde{\mathcal{U}}^{-}$. In view of
the definition (3.3), we deduce that for all $du$-almost all
$u\in\widetilde{\mathcal{U}}^{+}\cup\widetilde{\mathcal{U}}^{-}$,
$\widetilde{m}^{Q}_{u}$ does not depend on $(Q_{x})_{x\in A_{\mu}}$, neither
does it for all $u\in(0,1)$ such that $F_{\mu}^{-1}(u)=F_{\nu}^{-1}(u)$.
Moreover, the image of the continuous part $\mu-\sum_{x\in
A_{\mu}}\mu(\\{x\\})\delta_{x}$ of $\mu$ by $F_{\mu}^{-1}$ is absolutely
continuous with respect to the Lebesgue measure on $(0,1)$. This implies that
for $\mu(dx)$-almost all $x\in\mathbb{R}$ such that $\mu(\\{x\\})=0$,
$M^{Q}_{x}=\widetilde{m}^{Q}_{F_{\mu}(x)}$ does not depend on
$(Q_{x^{\prime}})_{x^{\prime}\in A_{\mu}}$.

Let now $x\in A_{\mu}$. Since
$(F_{\mu}(x-),F_{\mu}(x)]=\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}\cup(a_{x},b_{x}]$,
we have

$\displaystyle M^{Q}_{x}(dy)$
$\displaystyle=\frac{1}{\mu(\\{x\\})}\int_{F_{\mu}(x-)}^{F_{\mu}(x)}\widetilde{m}^{Q}_{u}(dy)\,du$
$\displaystyle=\frac{1}{\mu(\\{x\\})}\left(\int_{\mathcal{V}^{+}_{x}}\widetilde{m}^{Q}_{u}(dy)\,du+\int_{\mathcal{V}^{-}_{x}}\widetilde{m}^{Q}_{u}(dy)\,du+\int_{(a_{x},b_{x}]}\widetilde{m}^{Q}_{u}(dy)\,du\right).$

We can check that

$\int_{\mathcal{V}^{+}_{x}}\widetilde{m}^{Q}_{u}(dy)\,du+\int_{\mathcal{V}^{-}_{x}}\widetilde{m}^{Q}_{u}(dy)\,du=\int_{\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}}\delta_{F_{\nu}^{-1}(u)}(dy)\,du$

and deduce that
$M^{Q}_{x}(dy)=\frac{1}{\mu(\\{x\\})}\left(\int_{\mathcal{V}^{+}_{x}\cup\mathcal{V}^{-}_{x}}\delta_{F_{\nu}^{-1}(u)}(dy)\,du+\int_{(a_{x},b_{x}]}\widetilde{m}^{Q}_{u}(dy)\,du\right)$.
Since
$(a_{x},b_{x}]\subset\widetilde{\mathcal{U}}^{+}\cup\widetilde{\mathcal{U}}^{-}$,
we have from the foregoing that $\pi^{Q}_{u}$ and therefore
$\widetilde{m}^{Q}_{u}$ does not depend on $(Q_{x})_{x\in A_{\mu}}$ for
$du$-almost all $u\in(a_{x},b_{x}]$, hence $M^{Q}_{x}$ is independent of
$(Q_{x^{\prime}})_{x^{\prime}\in A_{\mu}}$.

  2. (ii)

The continuous functions $\Delta_{+}$ and $\Delta_{-}$ defined in (2.12)
respectively coincide with $\chi_{+}$ and $\chi_{-}$ outside the jumps of
$F_{\mu}$. Let $u\notin\bigcup_{x\in A_{\mu}}[F_{\mu}(x-),F_{\mu}(x)]$. We
have

$\displaystyle\\{v\in(0,u)\mid\mu(\\{F_{\mu}^{-1}(v)\\})=0\\}\cup\bigcup_{x\in
A_{\mu}\cap(-\infty,F_{\mu}^{-1}(u))}(F_{\mu}(x-),F_{\mu}(x)]$
$\displaystyle\subset(0,u)$
$\displaystyle\subset\\{v\in(0,u)\mid\mu(\\{F_{\mu}^{-1}(v)\\})=0\\}\cup\bigcup_{x\in
A_{\mu}\cap(-\infty,F_{\mu}^{-1}(u))}[F_{\mu}(x-),F_{\mu}(x)],$

where the sets in the first union are disjoint. Hence

$\displaystyle\Delta_{\pm}(u)$
$\displaystyle=\int_{0}^{u}(F_{\mu}^{-1}-G)^{\pm}(v)\mathds{1}_{\\{\mu(\\{F_{\mu}^{-1}(v)\\})=0\\}}\,dv+\sum_{x\in
A_{\mu}\cap(-\infty,F_{\mu}^{-1}(u))}\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(F_{\mu}^{-1}-G)^{\pm}(v)\,dv,$
$\displaystyle\chi_{\pm}(u)$
$\displaystyle=\int_{0}^{u}(F_{\mu}^{-1}-F_{\nu}^{-1})^{\pm}(v)\mathds{1}_{\\{\mu(\\{F_{\mu}^{-1}(v)\\})=0\\}}\,dv+\sum_{x\in
A_{\mu}\cap(-\infty,F_{\mu}^{-1}(u))}\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(F_{\mu}^{-1}-F_{\nu}^{-1})^{\pm}(v)\mathds{1}_{\widetilde{\mathcal{U}}^{\pm}}(v)\,dv.$

For all $v\in(0,1)$ such that $\mu(\\{F_{\mu}^{-1}(v)\\})=0$,
$G(v)=F_{\nu}^{-1}(v)$. Moreover, for $x\in A_{\mu}$, $F_{\mu}^{-1}$ and $G$
are constant and respectively equal to $x$ and
$\frac{1}{F_{\mu}(x)-F_{\mu}(x-)}\int_{F_{\mu}(x-)}^{F_{\mu}(x)}F_{\nu}^{-1}(v)\,dv$
on $(F_{\mu}(x-),F_{\mu}(x)]$, so that, using the definition of
$\widetilde{\mathcal{U}}^{\pm}$ for the second equality, we obtain

$\displaystyle\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(F_{\mu}^{-1}-F_{\nu}^{-1})^{\pm}(v)\mathds{1}_{\widetilde{\mathcal{U}}^{\pm}}(v)\,dv$
$\displaystyle=\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(v))^{\pm}\mathds{1}_{\widetilde{\mathcal{U}}^{\pm}}(v)\,dv=\left(\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(x-F_{\nu}^{-1}(v))\,dv\right)^{\pm}$
$\displaystyle=\left(\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(x-G(v))\,dv\right)^{\pm}=\int_{F_{\mu}(x-)}^{F_{\mu}(x)}(F_{\mu}^{-1}-G)^{\pm}(v)\,dv.$

We deduce that $\Delta_{\pm}(u)=\chi_{\pm}(u)$. For all $x\in A_{\mu}$,
$u\mapsto\Delta_{\pm}$ is affine on $[F_{\mu}(x-),F_{\mu}(x)]$, whereas
$\chi_{\pm}$ realises an a priori different continuous interpolation. However
the fact that $\Delta_{\pm}$ and $\chi_{\pm}$ coincide outside the jumps of
$F_{\mu}$ indicates that the constructions of the martingale couplings $M$
defined in Section 2.2 and $M^{Q}$ defined in the proof of Proposition 3.6 are
very close.

Let us now illustrate with the following example that the martingale coupling
$M^{Q}$ constructed in the proof of Proposition 3.6 is in general different
from the inverse transform martingale coupling $M^{IT}$.

######  Example 3.9.

Let $\mu=\frac{1}{2}(\delta_{-2}+\delta_{2})$ and
$\nu=\frac{1}{3}\delta_{-4}+\frac{1}{6}\delta_{-1}+\frac{1}{6}\delta_{1}+\frac{1}{3}\delta_{4}$.
Let $Q\in\mathcal{Q}$, $(\widetilde{m}^{Q}_{u})_{u\in(0,1)}$ be defined by
(3.3) and $M^{Q}$ be defined by (3.6). With the definition (3.7) in mind, we
have

$\mathcal{U}^{+}=\left(0,\frac{1}{3}\right]\cup\left(\frac{1}{2},\frac{2}{3}\right]\quad\textrm{and}\quad\mathcal{U}^{-}=\left(\frac{1}{3},\frac{1}{2}\right]\cup\left(\frac{2}{3},1\right).$

Let $u\in(0,1)$ be such that $F_{\mu}^{-1}(u)=-2$, or equivalently
$u\leq\frac{1}{2}$. If $u\leq\frac{1}{3}$ then $u\in\mathcal{U}^{+}$. For all
$v\in(0,1)$, $F_{\nu}^{-1}(v)=1\iff\frac{1}{2}<v\leq\frac{2}{3}\implies
v\in\mathcal{U}^{+}$, so $\widetilde{m}^{Q}_{u}(\\{1\\})=0$. Else if
$u>\frac{1}{3}$, then $F_{\nu}^{-1}(u)=-1$, so
$\widetilde{m}^{Q}_{u}(\\{1,4\\})=0$. Since $\widetilde{m}^{Q}_{u}$ must have
mean $-2$, $\widetilde{m}^{Q}_{u}(\\{1,4\\})=0$. We deduce that
$M^{Q}(\\{(-2,1)\\})=0$. Similarly we find $M^{Q}(\\{(2,-1)\\})=0$. Since
$\nu(\\{-1\\})=\nu(\\{1\\})=\frac{1}{6}$, we deduce that
$M^{Q}(\\{(-2,-1)\\})=M^{Q}(\\{(2,1)\\})=\frac{1}{6}$. Then the martingale
constraint imposes

$M^{Q}=\frac{13}{48}\delta_{(-2,-4)}+\frac{1}{6}\delta_{(-2,-1)}+\frac{1}{16}\delta_{(-2,4)}+\frac{1}{16}\delta_{(2,-4)}+\frac{1}{6}\delta_{(2,1)}+\frac{13}{48}\delta_{(2,4)},$

which coincides therefore with the martingale coupling $M^{Q}$ constructed in
the proof of Proposition 3.6.

We easily find that the Hoeffding-Fréchet coupling $\pi^{HF}$ between $\mu$
and $\nu$ is given by

$\pi^{HF}=\frac{1}{3}\delta_{(-2,-4)}+\frac{1}{6}\delta_{(-2,-1)}+\frac{1}{6}\delta_{(2,1)}+\frac{1}{3}\delta_{(2,4)},$

and for all $u\in(0,\frac{1}{2})$, resp. $u\in(\frac{1}{2},1)$, we have
$\varphi(u)=u+\frac{1}{2}$, resp. $\varphi(u)=u-\frac{1}{2}$. Then one can
easily check that the probability kernel $(m_{u})_{u\in(0,1)}$ defined for all
$u\in(0,\frac{1}{2})$ by

$m_{u}=\frac{5}{9}\delta_{-4}+\frac{5}{18}\delta_{-1}+\frac{1}{18}\delta_{1}+\frac{1}{9}\delta_{4},$

and for all $u\in(\frac{1}{2},1)$ by

$m_{u}=\frac{1}{9}\delta_{-4}+\frac{1}{18}\delta_{-1}+\frac{5}{18}\delta_{1}+\frac{5}{9}\delta_{4},$

satisfies (2.23) with $\pi=\pi^{HF}$. Then the martingale coupling $M$ defined
by (2.27) is

$M=\frac{5}{18}\delta_{(-2,-4)}+\frac{5}{36}\delta_{(-2,-1)}+\frac{1}{36}\delta_{(-2,1)}+\frac{1}{18}\delta_{(-2,4)}+\frac{1}{18}\delta_{(2,-4)}+\frac{1}{36}\delta_{(2,-1)}+\frac{5}{36}\delta_{(2,1)}+\frac{5}{18}\delta_{(2,4)},$

hence $M^{Q}\neq M$.

###  3.4 An example of $\mathcal{AW}_{\rho}$-minimal martingale rearrangement
for $\rho>2$

Let $f:\mathbb{R}\to\mathbb{R}$ and $q:\mathbb{R}\to[0,1]$ be defined for all
$y\in\mathbb{R}$ by

$\displaystyle f(y)$
$\displaystyle=\frac{1+\textrm{e}}{6}\left(\textrm{e}^{-|y|}\mathds{1}_{\\{|y|\geq
1\\}}+\frac{\textrm{e}^{-|y|}+1}{1+\textrm{e}}\mathds{1}_{\\{|y|<1\\}}\right);$
$\displaystyle q(y)$
$\displaystyle=\frac{\textrm{e}}{1+\textrm{e}}\mathds{1}_{\\{y\leq-1\\}}+\frac{1}{1+\textrm{e}^{y}}\mathds{1}_{\\{-1<y<1\\}}+\frac{1}{1+\textrm{e}}\mathds{1}_{\\{y\geq
1\\}}.$

Let $T:\mathbb{R}\to\mathbb{R}$ be the inverse of the continuous increasing
map $y\mapsto y+2q(y)-1$, so that for all $y\in\mathbb{R}$,
$q(y)=\frac{1+T^{-1}(y)-y}{2}$. Let $\nu(dy)=f(y)\,dy$ and
$\mu=(T^{-1})_{\sharp}\nu$. We can easily compute

$\sup_{x\in\mathbb{R}}|x-T(x)|=\sup_{y\in\mathbb{R}}|T^{-1}(y)-y|=\sup_{y\in\mathbb{R}}|2q(y)-1|=\frac{\operatorname{\operatorname{e}}-1}{\operatorname{\operatorname{e}}+1}<1.$

By considering the cases $y\leq-2$, $-2<y\leq-1$, $-1<y\leq 0$, $0<y\leq 1$,
$1<y\leq 2$ and $2\leq y$, it is easy to check that

$\forall y\in\mathbb{R},\quad q(y-1)f(y-1)+(1-q(y+1))f(y+1)=f(y).$ (3.39)

Let

$m^{0}_{u}=q(F_{\nu}^{-1}(u))\,\delta_{F_{\nu}^{-1}(u)+1}+(1-q(F_{\nu}^{-1}(u)))\,\delta_{F_{\nu}^{-1}(u)-1}$
(3.40)

and $h:\mathbb{R}\to\mathbb{R}$ be measurable and bounded. Then

$\displaystyle\int_{(0,1)\times\mathbb{R}}h(y)\,du\,m^{0}_{u}(dy)$
$\displaystyle=\int_{(0,1)}\left(q(F_{\nu}^{-1}(u))h(F_{\nu}^{-1}(u)+1)+(1-q(F_{\nu}^{-1}(u)))h(F_{\nu}^{-1}(u)-1)\right)\,du$
$\displaystyle=\int_{\mathbb{R}}\left(q(y)h(y+1)+(1-q(y))h(y-1)\right)\,\nu(dy)$
$\displaystyle=\int_{\mathbb{R}}q(y)h(y+1)f(y)\,dy+\int_{\mathbb{R}}(1-q(y))h(y-1)f(y)\,dy$
$\displaystyle=\int_{\mathbb{R}}\left(q(y-1)f(y-1)+(1-q(y+1))f(y+1)\right)h(y)\,dy$
$\displaystyle=\int_{\mathbb{R}}f(y)h(y)\,dy,$

where we used (3.39) for the last equality. We deduce that
$\int_{u\in(0,1)}m^{0}_{u}(dy)\,du=\nu(dy)$. Hence
$\widehat{M}^{0}=\lambda_{(0,1)}\times\delta_{F_{\mu}^{-1}(u)}\times
m^{0}_{u}\in\widehat{\Pi}^{\mathrm{M}}(\mu,\nu)$.

Let us now show that $\widehat{M}^{0}$ is the only
$\widehat{\mathcal{AW}}_{\rho}$-minimal martingale rearrangement coupling of
$\widehat{\pi}^{HF}$ for $\rho>2$. Since $|y-F_{\nu}^{-1}(u)|$ is
$du\,m^{0}_{u}(dy)$-almost everywhere constant, we have

$\displaystyle\begin{split}\left(\int_{(0,1)}\mathcal{W}_{2}^{2}(m^{0}_{u},\delta_{F_{\nu}^{-1}(u)})\,du\right)^{\rho/2}&=\left(\int_{(0,1)}\int_{\mathbb{R}}|F_{\nu}^{-1}(u)-y|^{2}\,m^{0}_{u}(dy)\,du\right)^{\rho/2}\\\
&=\int_{(0,1)}\int_{\mathbb{R}}|F_{\nu}^{-1}(u)-y|^{\rho}\,m^{0}_{u}(dy)\,du\geq\widehat{\mathcal{AW}}_{\rho}^{\rho}(\widehat{M}^{0},\widehat{\pi}^{HF}).\end{split}$
(3.41)

Since by Proposition 3.2 and its proof,
$\int_{(0,1)}\mathcal{W}_{2}^{2}(m_{u},\delta_{F_{\nu}^{-1}(u)})\,du$ does not
depend on $\widehat{M}=\lambda_{(0,1)}\times\delta_{F_{\mu}^{-1}(u)}\times
m_{u}\in\widehat{\Pi}^{\mathrm{M}}(\mu,\nu)$, to conclude it is enough to show
that for $\widehat{M}\neq\widehat{M}^{0}$,

$\widehat{\mathcal{AW}}_{\rho}^{\rho}(\widehat{M},\widehat{\pi}^{HF})>\left(\int_{(0,1)}\mathcal{W}_{2}^{2}(m_{u},\delta_{F_{\nu}^{-1}(u)})\,du\right)^{\rho/2}.$
(3.42)

Let $\chi\in\Pi(\lambda_{(0,1)},\lambda_{(0,1)})$ be optimal for
$\widehat{\mathcal{AW}}_{\rho}(\widehat{M},\widehat{\pi}^{HF})$. Suppose first
that $\chi(du,du^{\prime})=\lambda_{(0,1)}(du)\,\delta_{u}(du^{\prime})$.
Since

$\int_{(0,1)}\int_{\mathbb{R}}|y-F_{\nu}^{-1}(u)|^{2}\,m_{u}(dy)\,du=\int_{(0,1)}\mathcal{W}_{2}^{2}(m_{u},\delta_{F_{\nu}^{-1}(u)})\,du=\int_{(0,1)}\mathcal{W}_{2}^{2}(m^{0}_{u},\delta_{F_{\nu}^{-1}(u)})\,du=1,$

and $\widehat{M}\neq\widehat{M}^{0}$, $|y-F_{\nu}^{-1}(u)|$ is not
$du\,m_{u}(dy)$-almost everywhere constant, so by Jensen’s strict inequality
we have

$\displaystyle\widehat{\mathcal{AW}}_{\rho}^{\rho}(\widehat{M},\widehat{\pi}^{HF})$
$\displaystyle=\int_{(0,1)}\mathcal{W}_{\rho}^{\rho}(m_{u},\delta_{F_{\nu}^{-1}(u)})\,du=\int_{\mathbb{R}\times(0,1)}|y-F_{\nu}^{-1}(u)|^{\rho}\,m_{u}(dy)\,du$
$\displaystyle>\left(\int_{\mathbb{R}\times(0,1)}|y-F_{\nu}^{-1}(u)|^{2}\,m_{u}(dy)\,du\right)^{\rho/2}=\left(\int_{(0,1)}\mathcal{W}_{2}^{2}(m_{u},\delta_{F_{\nu}^{-1}(u)})\,du\right)^{\rho/2}.$

Else if
$\chi(du,du^{\prime})\neq\lambda_{(0,1)}(du)\,\delta_{u}(du^{\prime})$, then
using Jensen’s inequality for the third inequality and (3.13) for the fourth,
we have

$\displaystyle\begin{split}\widehat{\mathcal{AW}}_{\rho}^{\rho}(\widehat{M},\widehat{\pi}^{HF})&>\int_{(0,1)\times(0,1)}\mathcal{W}_{\rho}^{\rho}(m_{u},\delta_{F_{\nu}^{-1}(u^{\prime})})\,\chi(du,du^{\prime})\geq\int_{(0,1)\times(0,1)}\mathcal{W}_{2}^{\rho}(m_{u},\delta_{F_{\nu}^{-1}(u^{\prime})})\,\chi(du,du^{\prime})\\\
&\geq\left(\int_{(0,1)\times(0,1)}\mathcal{W}_{2}^{2}(m_{u},\delta_{F_{\nu}^{-1}(u^{\prime})})\,\chi(du,du^{\prime})\right)^{\rho/2}\geq\left(\int_{(0,1)}\mathcal{W}_{2}^{2}(m_{u},\delta_{F_{\nu}^{-1}(u)})\,du\right)^{\rho/2},\end{split}$
(3.43)

which proves (3.42) and therefore that $\widehat{M}^{0}$ is the only
$\widehat{\mathcal{AW}}_{\rho}$-minimal martingale rearrangement coupling of
$\widehat{\pi}^{HF}$. Note that (3.43) is valid for
$\widehat{M}=\widehat{M}^{0}$, which in view of (3.41) shows that
$\lambda_{(0,1)}(du)\,\delta_{u}(du^{\prime})$ is the only coupling between
$\lambda_{(0,1)}$ and $\lambda_{(0,1)}$ optimal for
$\widehat{\mathcal{AW}}_{\rho}(\widehat{M}^{0},\widehat{\pi}^{HF})$. With
similar arguments we prove that

$M^{0}(dx,dy)=\mu(dx)\left(q(T(x))\,\delta_{T(x)+1}(dy)+(1-q(T(x)))\,\delta_{T(x)-1}(dy)\right)$

is the only $\mathcal{AW}_{\rho}$-minimal martingale rearrangement coupling of
$\pi^{HF}$, and $\mu(dx)\,\delta_{x}(dx^{\prime})$ is the only coupling
between $\mu$ and $\mu$ optimal for $\mathcal{AW}_{\rho}(M^{0},\pi^{HF})$.

######  Remark 3.10.

According to [24, Proposition 2.11], any element of our family
$(M^{Q})_{Q\in\mathcal{Q}}$ of martingale couplings minimises
$\int_{\mathbb{R}\times\mathbb{R}}|y-T(x)|\,M(dx,dy)$ among all martingale
couplings $M$ between $\mu$ and $\nu$ and satisfies
$\int_{\mathbb{R}\times\mathbb{R}}|y-T(x)|\,M^{Q}(dx,dy)=\mathcal{W}_{1}(\mu,\nu)$.
According to [24, Proposition 3.5], since $\rho>2$, the inverse transform
martingale coupling $M^{IT}$ minimises
$\int_{\mathbb{R}\times\mathbb{R}}|y-T(x)|^{\rho}\,M^{Q}(dx,dy)$ (it maximises
this integral for $1<\rho<2$) among all martingale couplings $M^{Q}$
parametrised by $Q\in\mathcal{Q}$. Yet the optimum over the whole set of
martingale couplings between $\mu$ and $\nu$ is not $M^{IT}$ but $M^{0}$.

Indeed, by construction we have $M^{IT}_{x}(\\{T(x)\\})>0$, $\mu(dx)$-almost
everywhere, hence $M^{IT}\neq M^{0}$ and $|y-T(x)|$ is not
$M^{IT}(dx,dy)$-almost everywhere constant. Then by Jensen’s strict inequality
and the fact that $\int_{\mathbb{R}\times\mathbb{R}}|y-T(x)|^{2}\,M(dx,dy)$
does not depend on the choice of $M\in\Pi^{\mathrm{M}}(\mu,\nu)$, we get

$\displaystyle\begin{split}\left(\int_{\mathbb{R}\times\mathbb{R}}|y-T(x)|^{\rho}\,M^{IT}(dx,dy)\right)^{1/\rho}&>\left(\int_{\mathbb{R}\times\mathbb{R}}|y-T(x)|^{2}\,M^{0}(dx,dy)\right)^{1/2}\\\
&=\left(\int_{\mathbb{R}\times\mathbb{R}}|y-T(x)|^{\rho}\,M^{0}(dx,dy)\right)^{1/\rho}.\end{split}$
(3.44)

Note that when $1<\rho<2$, one can readily adapt the above arguments to show
that $M^{0}$ is the only $\mathcal{AW}_{\rho}$-maximal martingale
rearrangement coupling of $\pi^{HF}$, i.e. it maximises
$\mathcal{AW}_{\rho}(\pi^{HF},M)$ over $M\in\Pi^{\mathrm{M}}(\mu,\nu)$, and
reasoning similarly to (3.44) yields

$\left(\int_{\mathbb{R}\times\mathbb{R}}|y-T(x)|^{\rho}\,M^{IT}(dx,dy)\right)^{1/\rho}<\left(\int_{\mathbb{R}\times\mathbb{R}}|y-T(x)|^{\rho}\,M^{0}(dx,dy)\right)^{1/\rho}.$

##  4 Stability of the inverse transform martingale coupling

In the next proposition we prove the stability in $\mathcal{AW}_{\rho}$, for
$\rho\geq 1$, of the lifted inverse transform martingale coupling, defined for
all $\mu,\nu\in\mathcal{P}_{1}(\mathbb{R})$ in the convex order by

$\widehat{M}^{IT}(du,dx,dy)=\lambda_{(0,1)}(du)\,\delta_{F_{\mu}^{-1}(u)}(dx)\,\widetilde{m}^{IT}_{u}(dy),$

where $(\widetilde{m}^{IT}_{u})_{u\in(0,1)}$ is defined by (3.10). In another
proposition, we give a condition on the first marginals ensuring that the
inverse transform martingale coupling is stable in $\mathcal{AW}_{\rho}$.

######  Proposition 4.1.

Let $\rho\geq 1$ and $\mu_{n},\nu_{n}\in\mathcal{P}_{\rho}(\mathbb{R})$,
$n\in\mathbb{N}$, be in convex order and respectively converge to $\mu$ and
$\nu$ in $\mathcal{W}_{\rho}$ as $n\to+\infty$. Then

$\widehat{\mathcal{AW}}^{\rho}_{\rho}(\widehat{M}^{IT}_{n},\widehat{M}^{IT})\leq\int_{(0,1)}\mathcal{W}_{\rho}^{\rho}((\widetilde{m}^{IT}_{n})_{u},\widetilde{m}^{IT}_{u})\,du\underset{n\to+\infty}{\longrightarrow}0,$
(4.1)

where
$\widehat{M}^{IT}_{n}=\lambda_{(0,1)}\times\delta_{F_{\mu_{n}}^{-1}(u)}\times(\widetilde{m}^{IT}_{n})_{u}$,
resp.
$M^{IT}=\lambda_{(0,1)}\times\delta_{F_{\mu}^{-1}(u)}\times\widetilde{m}^{IT}_{u}$,
denotes the lifted inverse transform martingale coupling between $\mu_{n}$ and
$\nu_{n}$, resp. $\mu$ and $\nu$.

###### Proof.

Since

$\widehat{\mathcal{AW}}^{\rho}_{\rho}(\widehat{M}^{IT}_{n},\widehat{M}^{IT})\leq\int_{(0,1)}\mathcal{W}_{\rho}^{\rho}((\widetilde{m}^{IT}_{n})_{u},\widetilde{m}^{IT}_{u})\,du,$

it suffices to show that the right-hand side vanishes as $n$ goes to
$+\infty$. This is achieved in two steps. First, we prove that, on the
probability space $(0,1)$ endowed with the Lebesgue measure, the family of
random variables
$\left(\mathcal{W}_{\rho}^{\rho}\left((\widetilde{m}^{IT}_{n})_{u},\widetilde{m}^{IT}_{u}\right)\right)_{n\in\mathbb{N}}$
is uniformly integrable. Second we show for $du$-almost all $u\in(0,1)$ that

$\mathcal{W}_{\rho}^{\rho}\left((\widetilde{m}^{IT}_{n})_{u},\widetilde{m}^{IT}_{u}\right)\underset{n\to+\infty}{\longrightarrow}0$
(4.2)

Let us begin with the uniform integrability. For $u\in(0,1)$, we can estimate

$\mathcal{W}_{\rho}^{\rho}\left((\widetilde{m}^{IT}_{n})_{u},\widetilde{m}^{IT}_{u}\right)\leq
2^{\rho-1}\int_{\mathbb{R}}|y|^{\rho}\,\left((\widetilde{m}^{IT}_{n})_{u}(dy)+\widetilde{m}^{IT}_{u}(dy)\right).$
(4.3)

According to [24, Lemma 2.6], $M^{IT}$ is the image of
$\mathds{1}_{(0,1)}(u)\,du\,\widetilde{m}^{IT}_{u}(dy)$ by
$(u,y)\mapsto(F_{\mu}^{-1}(u),y)$ so that the second marginal of this measure
is $\nu(dy)$. Therefore

$\int_{(0,1)}\int_{\mathbb{R}}|y|^{\rho}\,\widetilde{m}^{IT}_{u}(dy)\,du=\int_{\mathbb{R}}|y|^{\rho}\,\nu(dy)<+\infty.$

Hence it is enough to check the uniform integrability of
$\left(\int_{\mathbb{R}}|y|^{\rho}\,(\widetilde{m}^{IT}_{n})_{u}(dy)\right)_{n\in\mathbb{N}}$
to ensure that of
$\left(\mathcal{W}_{\rho}^{\rho}\left((\widetilde{m}^{IT}_{n})_{u},\widetilde{m}^{IT}_{u}\right)\right)_{n\in\mathbb{N}}$.
Since the second marginal of the measure
$\mathds{1}_{(0,1)}(u)\,du\,(\widetilde{m}^{IT}_{n})_{u}(dy)$ is
$\nu_{n}(dy)$, this measure also writes $\nu_{n}(dy)k^{n}_{y}(du)$ for some
probability kernel $k^{n}$ on $\mathbb{R}\times(0,1)$. Let $\varepsilon>0$ and
$A$ be a measurable subset of $(0,1)$ such that $\lambda(A)<\varepsilon$. For
all $n\in\mathbb{N}$, we have

$J_{n}(A):=\int_{A}\int_{\mathbb{R}}|y|^{\rho}\,(\widetilde{m}^{IT}_{n})_{u}(dy)\,du=\int_{\mathbb{R}}|y|^{\rho}\,\tau_{n}(dy),$

where
$\tau_{n}(dy)=\int_{u=0}^{1}\mathds{1}_{A}(u)\,k^{n}_{y}(du)\,\nu_{n}(dy)$ is
such that $\tau_{n}\leq\nu_{n}$ and $\tau_{n}(\mathbb{R})=\lambda(A)$. Hence

$\sup_{A\in\mathcal{B}((0,1)),\ \lambda(A)\leq\varepsilon}J_{n}(A)\leq
I_{\varepsilon}^{\rho}(\nu_{n}),$

where $I^{\rho}_{\varepsilon}(\zeta)$ is defined for all
$\zeta\in\mathcal{P}_{\rho}(\mathbb{R})$ as the supremum of
$\int_{\mathbb{R}}|y|^{\rho}\,\tau(dy)$ over all finite measures $\tau$ on
$\mathbb{R}$ such that $\tau\leq\zeta$ and $\tau(\mathbb{R})\leq\varepsilon$.
Let $\eta>0$. By [7, Lemma 3.1 (b)], since
$\nu\in{\mathcal{P}}_{\rho}(\mathbb{R})$, there exists
$\varepsilon^{\prime}>0$ such that
$I^{\rho}_{\varepsilon^{\prime}}(\nu)<\eta$. Let then $N\in\mathbb{N}$ be such
that for all $n>N$, $\mathcal{W}_{\rho}^{\rho}(\nu_{n},\nu)<\eta$, so that by
[7, Lemma 3.1 (c)], $I^{\rho}_{\varepsilon^{\prime}}(\nu_{n})\leq
2^{\rho-1}(\mathcal{W}_{\rho}^{\rho}(\nu_{n},\nu)+I^{\rho}_{\varepsilon^{\prime}}(\nu))<2^{\rho}\eta$.
By [7, Lemma 3.1 (b)] again there exists $\varepsilon^{\prime\prime}>0$ such
that for all $n\leq N$,
$I^{\rho}_{\varepsilon^{\prime\prime}}(\nu_{n})<2^{\rho}\eta$. We deduce that
for all
$\varepsilon\in(0,\varepsilon^{\prime}\wedge\varepsilon^{\prime\prime})$,

$\sup_{n\in\mathbb{N}}\sup_{A\in\mathcal{B}((0,1)),\
\lambda(A)\leq\varepsilon}J_{n}(A)\leq 2^{\rho}\eta,$

which yields uniform integrability of
$\left(\int_{\mathbb{R}}|y|^{\rho}\,(\widetilde{m}^{IT}_{n})_{u}(dy)\right)_{n\in\mathbb{N}}$.

Next, we show the $du$-almost everywhere pointwise convergence of (4.2).
Since, by monotonicity, $u\mapsto(F_{\mu}^{-1}(u),F_{\nu}^{-1}(u))$ is
continuous $du$-almost everywhere on $(0,1)$ and, then, the weak convergence
implies that

$(F_{\mu_{n}}^{-1}(u),F_{\nu_{n}}^{-1}(u))\underset{n\to+\infty}{\longrightarrow}(F_{\mu}^{-1}(u),F_{\nu}^{-1}(u)),$
(4.4)

we suppose without loss of generality that this convergence holds. Let
$n\in\mathbb{N}$. Let $\Psi_{n+}$, resp. $\Psi_{n-}$, be the map defined by
the left-hand, resp. right-hand side of (3.1), with $(\mu_{n},\nu_{n})$
replacing $(\mu,\nu)$. By (3.10),

$(\widetilde{m}_{n}^{IT})_{u}=p_{n}(u)\delta_{F_{\nu_{n}}^{-1}(\varphi_{n}(u))}+(1-p_{n}(u))\delta_{F_{\nu_{n}}^{-1}(u)}\quad\mbox{
with }\quad p_{n}(u)=\mathds{1}_{\\{F_{\mu_{n}}^{-1}(u)\neq
F_{\nu_{n}}^{-1}(u)\\}}\frac{F_{\mu_{n}}^{-1}(u)-F_{\nu_{n}}^{-1}(u)}{F_{\nu_{n}}^{-1}(\varphi_{n}(u))-F_{\nu_{n}}^{-1}(u)}\in[0,1]$

and $\varphi_{n}(u)=\Psi_{n-}^{-1}(\Psi_{n+}(u))$.

Suppose first that $u\in\mathcal{U}_{0}$ i.e.
$F_{\mu}^{-1}(u)=F_{\nu}^{-1}(u)$, so that
$\widetilde{m}^{IT}_{u}=\delta_{F_{\nu}^{-1}(u)}$. We have

$\displaystyle\begin{split}\mathcal{W}_{\rho}^{\rho}((\widetilde{m}_{n}^{IT})_{u},\widetilde{m}^{IT}_{u})&=p_{n}(u)|F_{\nu_{n}}^{-1}(\varphi_{n}(u))-F_{\nu}^{-1}(u)|^{\rho}+(1-p_{n}(u))|F_{\nu_{n}}^{-1}(u)-F_{\nu}^{-1}(u)|^{\rho}\\\
&\leq
2^{\rho-1}p_{n}(u)|F_{\nu_{n}}^{-1}(\varphi_{n}(u))-F_{\nu_{n}}^{-1}(u)|^{\rho}+(2^{\rho-1}p_{n}(u)+1-p_{n}(u))|F_{\nu_{n}}^{-1}(u)-F_{\nu}^{-1}(u)|^{\rho}\\\
&\leq 2^{\rho-1}\mathds{1}_{\\{F_{\mu_{n}}^{-1}(u)\neq
F_{\nu_{n}}^{-1}(u)\\}}|F_{\mu_{n}}^{-1}(u)-F_{\nu_{n}}^{-1}(u)|^{\rho}+2^{\rho-1}|F_{\nu_{n}}^{-1}(u)-F_{\nu}^{-1}(u)|^{\rho}\\\
&\leq
2^{2\rho-2}|F_{\mu_{n}}^{-1}(u)-F_{\mu}^{-1}(u)|^{\rho}+2^{\rho-1}(2^{\rho-1}+1)|F_{\nu_{n}}^{-1}(u)-F_{\nu}^{-1}(u)|^{\rho},\end{split}$
(4.5)

where the right-hand side goes to $0$ as $n\to\infty$ by (4.4).

Suppose next that $u\in\mathcal{U}_{+}$ i.e.
$F_{\mu}^{-1}(u)>F_{\nu}^{-1}(u)$, the case $u\in\mathcal{U}_{-}$ being
treated in a similar way. Then without loss of generality

$\widetilde{m}^{IT}_{u}=p(u)\delta_{F_{\nu}^{-1}(\varphi(u))}+(1-p(u))\delta_{F_{\nu}^{-1}(u)}\mbox{
with
}p(u)=\frac{F_{\mu}^{-1}(u)-F_{\nu}^{-1}(u)}{F_{\nu}^{-1}(\varphi(u))-F_{\nu}^{-1}(u)}$

and $\varphi(u)=\Psi_{-}^{-1}(\Psi_{+}(u))$. By (4.4), for $n$ large enough,
$u\in\mathcal{U}_{n+}$ so that without loss of generality,
$p_{n}(u)=\frac{F_{\mu_{n}}^{-1}(u)-F_{\nu_{n}}^{-1}(u)}{F_{\nu_{n}}^{-1}(\varphi_{n}(u))-F_{\nu_{n}}^{-1}(u)}$
and checking (4.2) amounts to show that

$F_{\nu_{n}}^{-1}(\varphi_{n}(u))\underset{n\to+\infty}{\longrightarrow}F_{\nu}^{-1}(\varphi(u)).$
(4.6)

It was shown in the proof of [24, Proposition 5.10] that $\Psi_{n+}$ converges
uniformly to $\Psi_{+}$ on $[0,1]$ and for $dv$-almost every $v\in(0,1)$,

$F_{\nu_{n}}^{-1}(\Psi_{n-}^{-1}(\Psi_{n+}(1)v))\underset{n\to+\infty}{\longrightarrow}F_{\nu}^{-1}(\Psi_{-}^{-1}(\Psi_{+}(1)v)).$
(4.7)

Let $\mathcal{D}$ be the set of discontinuities of
$F_{\nu}^{-1}\circ\Psi_{-}^{-1}$, which is at most countable by monotonicity.
Then [33, Proposition 4.10, Chapter 0] yields

$0=\int_{\Psi_{+}(0)}^{\Psi_{+}(1)}\mathds{1}_{\mathcal{D}}(v)\,dv=\int_{0}^{1}\mathds{1}_{\\{\Psi_{+}(u)\in\mathcal{D}\\}}\,d\Psi_{+}(u).$

We deduce that for $du$-almost all $u\in\mathcal{U}_{+}$,
$F_{\nu}^{-1}\circ\Psi_{-}^{-1}$ is continuous at $\Psi_{+}(u)$, which we
suppose from now. According to (4.7), there exists $\varepsilon>0$ arbitrarily
small such that

$\displaystyle
F_{\nu_{n}}^{-1}\left(\Psi_{n-}^{-1}\left(\Psi_{n+}(1)\frac{\Psi_{+}(u)-\varepsilon}{\Psi_{+}(1)}\right)\right)$
$\displaystyle\underset{n\to+\infty}{\longrightarrow}F_{\nu}^{-1}\left(\Psi_{-}^{-1}\left(\Psi_{+}(u)-\varepsilon\right)\right)$
$\displaystyle\text{and}\quad
F_{\nu_{n}}^{-1}\left(\Psi_{n-}^{-1}\left(\Psi_{n+}(1)\frac{\Psi_{+}(u)+\varepsilon}{\Psi_{+}(1)}\right)\right)$
$\displaystyle\underset{n\to+\infty}{\longrightarrow}F_{\nu}^{-1}\left(\Psi_{-}^{-1}\left(\Psi_{+}(u)+\varepsilon\right)\right).$

For $n$ large enough, we have
$\Psi_{+}(u)\in\left[\Psi_{n+}(1)\frac{\Psi_{+}(u)-\varepsilon}{\Psi_{+}(1)},\Psi_{n+}(1)\frac{\Psi_{+}(u)+\varepsilon}{\Psi_{+}(1)}\right]$.
Therefore, by monotonicity, we have

$\displaystyle
F_{\nu}^{-1}\left(\Psi_{-}^{-1}\left(\Psi_{+}(u)-\varepsilon\right)\right)$
$\displaystyle=\liminf_{n\to+\infty}F_{\nu_{n}}^{-1}\left(\Psi_{n-}^{-1}\left(\Psi_{n+}(1)\frac{\Psi_{+}(u)-\varepsilon}{\Psi_{+}(1)}\right)\right)$
$\displaystyle\leq\liminf_{n\to+\infty}F_{\nu_{n}}^{-1}(\Psi_{n-}^{-1}(\Psi_{n+}(u)))$
$\displaystyle\leq\limsup_{n\to+\infty}F_{\nu_{n}}^{-1}(\Psi_{n-}^{-1}(\Psi_{n+}(u)))$
$\displaystyle\leq\limsup_{n\to+\infty}F_{\nu_{n}}^{-1}\left(\Psi_{n-}^{-1}\left(\Psi_{n+}(1)\frac{\Psi_{+}(u)+\varepsilon}{\Psi_{+}(1)}\right)\right)$
$\displaystyle=F_{\nu}^{-1}\left(\Psi_{-}^{-1}\left(\Psi_{+}(u)+\varepsilon\right)\right).$

Since $F_{\nu}^{-1}\circ\Psi_{-}^{-1}$ is continuous at $\Psi_{+}(u)$, we get
when $\varepsilon$ vanishes the convergence (4.6), which concludes the proof
of (4.2) and therefore (4.1) ∎

######  Proposition 4.2.

Let $\rho\geq 1$ and $\mu_{n},\nu_{n}\in\mathcal{P}_{\rho}(\mathbb{R})$,
$n\in\mathbb{N}$, be in convex order and respectively converge to $\mu$ and
$\nu$ in $\mathcal{W}_{\rho}$ as $n\to+\infty$. Suppose that asymptotically,
any jump of $F_{\mu}$ is included in a jump of $F_{\mu_{n}}$, that is

$\forall
x\in\mathbb{R},\quad\mu(\\{x\\})>0\implies\exists(x_{n})_{n\in\mathbb{N}}\in\mathbb{R}^{\mathbb{N}},\quad
F_{\mu_{n}}(x_{n})\wedge F_{\mu}(x)-F_{\mu_{n}}(x_{n}-)\vee
F_{\mu}(x-)\underset{n\to+\infty}{\longrightarrow}\mu(\\{x\\}),$ (4.8)

which is for instance satisfied if $\mu$ is non-atomic. Then

$\mathcal{AW}_{\rho}(M_{n}^{IT},M^{IT})\underset{n\to+\infty}{\longrightarrow}0,$
(4.9)

where $M_{n}^{IT}$, resp. $M^{IT}$, denotes the inverse transform martingale
coupling between $\mu_{n}$ and $\nu_{n}$, resp. $\mu$ and $\nu$.

######  Remark 4.3.

If (4.8) is not satisfied, then $\eqref{eq:stabilityAW1ITMC}$ may not hold.
Indeed, for $n\in\mathbb{N}^{*}$, let $\mu_{n}=\mathcal{U}((-1/n,1/n))$,
$\mu=\delta_{0}$ and $\nu_{n}=\nu=\mathcal{U}((-1,1))$. We trivially have
$M^{IT}(dx,dy)=\mu(dx)\,\nu(dy)$, so
$\mathcal{AW}_{1}(M_{n}^{IT},M^{IT})\geq\int_{x\in\mathbb{R}}\mathcal{W}_{1}((M_{n}^{IT})_{x},\nu)\,\mu_{n}(dx)$.
However, for $n\in\mathbb{N}^{*}$, since $F_{\mu_{n}}$ is continuous, we have
that for all $x\in\mathbb{R}$,
$(M_{n}^{IT})_{x}=(\widetilde{m}^{IT}_{n})_{F_{\mu}(x)}$, where according to
(3.10), $((\widetilde{m}^{IT}_{n})_{u}(dy))_{u\in(0,1)}$ is a probability
kernel such that for all $u\in(0,1)$, there exist $a,b\in[-1,1]$ and
$p\in[0,1]$ which satisfy
$\widetilde{m}^{IT}_{n}(u,dy)=p\delta_{a}+(1-p)\delta_{b}$. Using the fact
that the comonotonic coupling is optimal for the $\mathcal{W}_{1}$-distance,
we get

$\mathcal{W}_{1}(p\delta_{a}+(1-p)\delta_{b},\nu)=\int_{0}^{p}|a+1-2u|\,du+\int_{p}^{1}|b+1-2u|\,du.$

It is easy to show that $\int_{0}^{p}|a+1-2u|\,du$ is equal to $p(a+1-p)\geq
p^{2}$ if $(a+1)/2>p$, and equal to $(a+1)^{2}/2-p(a+1)+p^{2}\leq p^{2}$ if
$(a+1)/2\leq p$. Therefore, one can readily show that
$\int_{0}^{p}|a+1-2u|\,du\geq p^{2}/2$, attained for $a=p-1$. Similarly, we
have $\int_{p}^{1}|b+1-2u|\,du\geq(1-p)^{2}/2$, attained for $b=p$. We deduce
that for all $(a,b,p)\in\mathbb{R}^{2}\times[0,1]$,
$\mathcal{W}_{1}(p\delta_{a}+(1-p)\delta_{b},\nu)\geq(p^{2}+(1-p)^{2})/2\geq
1/4$, attained for $p=1/2$, hence
$\int_{x\in\mathbb{R}}\mathcal{W}_{1}((M^{IT}_{n})_{x},\nu)\,\mu_{n}(dx)\geq
1/4$, which proves that (4.9) is not satisfied.

###### Proof of Proposition 4.2.

By Lemma 5.3 below we may suppose without loss of generality that $\rho=1$. We
have

$\displaystyle\mathcal{AW}_{1}(M^{IT}_{n},M^{IT})$
$\displaystyle\leq\int_{0}^{1}\left(|F_{\mu_{n}}^{-1}(u)-F_{\mu}^{-1}(u)|+\mathcal{W}_{1}\left((M^{IT}_{n})_{F_{\mu_{n}}^{-1}(u)},M^{IT}_{F_{\mu}^{-1}(u)}\right)\right)\,du$