# coherent distributions on the square
– Extreme points and asymptotics

Stanisław Cichomski  Department of Mathematics, Informatics and Mechanics
University of Warsaw
Banacha 2, 02-097 Warsaw
Poland  and  Adam Osękowski  Department of Mathematics, Informatics and
Mechanics
University of Warsaw
Banacha 2, 02-097 Warsaw
Poland

###### Abstract.

Let $\mathcal{C}$ denote the family of all coherent distributions on the unit
square $[0,1]^{2}$, i.e. all those probability measures $\mu$ for which there
exists a random vector $(X,Y)\sim\mu$, a pair $(\mathcal{G},\mathcal{H})$ of
$\sigma$-fields and an event $E$ such that $X=\mathbb{P}(E|\mathcal{G})$,
$Y=\mathbb{P}(E|\mathcal{H})$ almost surely. In this paper we examine the set
$\mathrm{ext}(\mathcal{C})$ of extreme points of $\mathcal{C}$ and provide its
general characterisation. Moreover, we establish several structural properties
of finitely-supported elements of $\mathrm{ext}(\mathcal{C})$. We apply these
results to obtain the asymptotic sharp bound

$\lim_{\alpha\to\infty}\alpha\cdot\Big{(}\sup_{(X,Y)\in\mathcal{C}}\mathbb{E}|X-Y|^{\alpha}\Big{)}=\frac{2}{e}.$

##  1\. Introduction

Let $\mu$ be a probability measure on the unit square $[0,1]^{2}$. Following
[12], this measure is called _coherent_ , if it is the joint distribution of a
two-variate random vector $(X,Y)$ defined on some arbitrary probability space
$(\Omega,\mathcal{F},\mathbb{P})$, such that

$X=\mathbb{P}(E|\mathcal{G})\ \ \ \ \text{and}\ \ \ \
Y=\mathbb{P}(E|\mathcal{H}),\ \ \ \ \text{almost surely,}$

for some measurable event $E\in\mathcal{F}$ and some two sub-$\sigma$-fields
$\mathcal{G},\mathcal{H}\subset\mathcal{F}$. Throughout the text, the class of
all coherent probability measures will be denoted by $\mathcal{C}$; for the
sake of convenience (and with a slight abuse of notation), we will also write
$(X,Y)\in\mathcal{C}$ to indicate that the distribution of a random vector
$(X,Y)$ is coherent.

Coherent measures enjoy the following nice interpretation. Suppose that two
experts provide their personal estimates on the likelihood of some random
event $E$, and assume that the knowledge of the first and the second expert is
represented by the $\sigma$-algebras $\mathcal{G}$ and $\mathcal{H}$,
respectively. Then a natural idea to model the predictions of the experts is
to use conditional expectations: this leads to the random variables $X$ and
$Y$ as above.

The importance of coherent distributions stem from their numerous applications
in statistics (cf. [12, 13, 17, 19]) and economics (consult [1, 2, 3, 15]).
Coherent distributions are also closely related to graph theory and
combinatorial matrix theory, see for instance [4, 7, 11, 20]. Moreover, there
has been a substantial purely probabilistic advancement on this subject during
the last decade, see [5, 6, 8, 9, 10, 21]. The main interest, both in applied
and theoretical considerations, involves bounding the maximal discrepancy of
coherent vectors measured by different functionals. A canonical result of this
type is the following threshold bound of Burdzy and Pal [5].

######  Theorem 1.1.

For any parameter $\delta\in(\frac{1}{2},1]$, we have

(1.1)
$\sup_{(X,Y)\in\mathcal{C}}\mathbb{P}(|X-Y|\geq\delta)=\frac{2(1-\delta)}{2-\delta}.$

For a generalisation of (1.1) to $n$-variate coherent vectors, consult [9].
Another important example is the expectation bound established independently
in [3, 7].

######  Theorem 1.2.

For any exponent $\alpha\in(0,2]$, we have

(1.2) $\sup_{(X,Y)\in\mathcal{C}}\mathbb{E}|X-Y|^{\alpha}=2^{-\alpha}.$

The analysis of the left-hand side of (1.2) for $\alpha>2$ remains a major
open problem and constitutes one of the main motivations for this paper.
Accordingly, we investigate the asymptotic behavior of this expression and
derive an appropriate sharp estimate.

######  Theorem 1.3.

We have

(1.3)
$\lim_{\alpha\to\infty}\alpha\cdot\Big{(}\sup_{(X,Y)\in\mathcal{C}}\mathbb{E}|X-Y|^{\alpha}\Big{)}=\frac{2}{e}.$

The proof of (1.3) that we present below rests on a novel, geometric-type
approach. As verified in [6], the family of coherent distributions is a
convex, compact subset of the space of probability distributions on
$[0,1]^{2}$ equipped with the usual weak topology. One of the main results of
this paper is to provide a characterisation of the extremal points of
$\mathcal{C}$, which is considered to be one of the major challenges of the
topic [6, 21].

It is instructive to take a look at the corresponding problem arising in the
theory of martingales, the solution to which is well-known. Namely (see [14]),
fix $N\in\mathbb{N}$ and consider the class of all finite martingales
$(M_{1},M_{2},\dots,M_{N})$ and the induced distributions on $\mathbb{R}^{N}$.
The extremal distributions can be characterised as follows:

  1. (i)

$M_{1}$ is concentrated in one point,

  2. (ii)

for any $n=2,\,3,\,\ldots,\,N$, the conditional distribution of $M_{n}$ given
$(M_{i})_{i=1}^{n-1}$ is concentrated on the set of cardinality at most two.

In particular, the support of a two-variate martingale with an extremal
distribution cannot exceed two points. Surprisingly, the structure of
$\mathrm{ext}(\mathcal{C})$ (the set of extreme points of $\mathcal{C}$) is
much more complex, as there exist extremal coherent measures with arbitrary
large or even countable infinite number of atoms (see [3, 21]). Conversely, as
proved in [3], elements of $\mathrm{ext}(\mathcal{C})$ are always supported on
sets of Lebesgue measure zero. The existence of non-atomic extreme points
remains a yet another open problem.

For the further discussion, we need to introduce some additional background
and notation. For a measure $\mu$ supported on $[0,1]^{2}$, we will write
$\mu^{x}$ and $\mu^{y}$ for the marginal measures of $\mu$ on $[0,1]$, i.e.
for the measures obtained by projecting $\mu$ on the first and the second
coordinate, correspondingly.

######  Definition 1.4.

Introduce the family $\mathcal{R}$, which consists of all ordered pairs
$(\mu,\nu)$ of nonnegative Borel measures on $[0,1]^{2}$ for which

$\int_{A}(1-x)\ \mathrm{d}\mu^{x}\ \ =\ \ \int_{A}x\ \mathrm{d}\nu^{x},$

and

$\int_{B}(1-y)\ \mathrm{d}\mu^{y}\ \ =\ \ \int_{B}y\ \mathrm{d}\nu^{y},$

for any Borel subsets $A,B\in\mathcal{B}([0,1])$.

It turns out that the family $\mathcal{R}$ is very closely related to the
class of coherent distributions. We will prove the following statement (a
slightly different formulation can be found in [3]).

######  Proposition 1.5.

Let $m$ be a probability measure on $[0,1]^{2}$. Then $m$ is coherent if and
only if there exists $(\mu,\nu)\in\mathcal{R}$ such that $m=\mu+\nu$.

The above result motivates the following.

######  Definition 1.6.

For a fixed $m\in\mathcal{C}$, consider the class

$\mathcal{R}(m)\ =\ \\{(\mu,\nu)\in\mathcal{R}:\ m=\mu+\nu\\}.$

Any element $(\mu,\nu)\in\mathcal{R}(m)$ will be called a _representation_ of
a coherent distribution $m$.

By the very definition, both $\mathcal{C}$ and $\mathcal{R}$, and hence also
$\mathcal{R}(m)$, are convex sets. To proceed, let us distinguish the ordering
in the class of measures, which will often be used in our considerations
below. Namely, for two Borel measures $\mu_{1},\mu_{2}$ supported on the unit
square, we will write $\mu_{1}\leq\mu_{2}$ if we have
$\mu_{1}(A)\leq\mu_{2}(A)$ for all $A\in\mathcal{B}([0,1]^{2})$.

######  Definition 1.7.

Let $m\in\mathcal{C}$. We say that the representation $(\mu,\nu)$ of $m$ is

$\cdot$ _unique_ , if for every $(\tilde{\mu},\tilde{\nu})\in\mathcal{R}$ with
$m=\tilde{\mu}+\tilde{\nu}$, we have $\tilde{\mu}=\mu$ and $\tilde{\nu}=\nu$;

$\cdot$ _minimal_ , if for all $(\tilde{\mu},\tilde{\nu})\in\mathcal{R}$ with
$\tilde{\mu}\leq\mu$ and $\tilde{\nu}\leq\nu$, there exists $\alpha\in[0,1]$
such that $(\tilde{\mu},\tilde{\nu})=\alpha\cdot(\mu,\nu)$.

With these notions at hand, we will give the following general
characterisation of $\textrm{ext}(\mathcal{C})$.

######  Theorem 1.8.

Let $m$ be a coherent distribution on $[0,1]^{2}$. Then $m$ is extremal if and
only if the representation of $m$ is unique and minimal.

This statement will be established in the next section. Then, in Section 3, we
concentrate on extremal coherent measures with finite support. Let
$\textrm{ext}_{f}(\mathcal{C})=\\{\eta\in\textrm{ext}(\mathcal{C}):|\mathrm{supp}(\eta)|<\infty\\}.$
Theorem 1.8 will enable us to deduce several structural properties of
$\textrm{ext}_{f}(\mathcal{C})$; most importantly, as conjectured in [21], we
show that support of $\eta\in\textrm{ext}_{f}(\mathcal{C})$ cannot contain any
axial cycles. Here is the definition.

######  Definition 1.9.

The sequence $\big{(}(x_{i},y_{i})\big{)}_{i=1}^{2n}$ with values in
$[0,1]^{2}$ is called an _axial cycle_ , if all points $(x_{i},y_{i})$ are
distinct, the endpoint coordinates $x_{1}$ and $x_{2n}$ coincide, and we have

$x_{2i}=x_{2i+1}$ and $y_{2i-1}=y_{2i}$ for all $i$.

Remarkably, the same ‘no axial cycle’ property holds true for extremal doubly
stochastic measures (permutons) – for the relevant discussion, see [16]. Next,
in Section 4, we apply our previous results and obtain the following reduction
towards Theorem 1.3. Namely, for all $\alpha\geq 1$, we have

(1.4) $\sup_{(X,Y)\in\mathcal{C}}\mathbb{E}|X-Y|^{\alpha}\ \ =\ \
\sup_{\tilde{\textbf{z}}}\
\sum_{i=1}^{n}z_{i}\Big{|}\frac{z_{i}}{z_{i-1}+z_{i}}-\frac{z_{i}}{z_{i}+z_{i+1}}\Big{|}^{\alpha}.$

Here the supremum is taken over all $n$ and all sequences
$\tilde{\textbf{z}}=(z_{0},z_{1},\dots,z_{n+1})$ such that $z_{0}=z_{n+1}=0$,
$z_{i}>0$ for all $i=1,\,2,\,\ldots,\,n$, and $\sum_{i=1}^{n}z_{i}=1.$
Finally, using several combinatorial arguments and reductions, we prove
Theorem 1.3 by a direct analysis of the right-hand side of (1.4).

##  2\. Coherent measures, Representations

Let $\mathcal{M}([0,1]^{2})$ and $\mathcal{M}([0,1])$ denote the space of
nonnegative Borel measures on $[0,1]^{2}$ and $[0,1]$, respectively. For
$\mu\in\mathcal{M}([0,1]^{2})$, let $\mu^{x},\mu^{y}\in\mathcal{M}([0,1])$ be
defined by

$\mu^{x}(A)=\mu(A\times[0,1])$ and $\mu^{y}(B)=\mu([0,1]\times B)$,

for all Borel subsets $A,B\in\mathcal{B}([0,1])$. We begin with the following
characterisation of $\mathcal{C}$.

######  Proposition 2.1.

Let $m\in\mathcal{M}([0,1]^{2})$. The measure $m$ is a coherent distribution
if and only if it is the joint distribution of a two-variate random vector
$(X,Y)$ such that

$X=\mathbb{E}(Z|X)\ \ \ \ \text{and}\ \ \ \ Y=\mathbb{E}(Z|Y)\ \ \ \
\text{almost surely}$

for some random variable $Z$ with $0\leq Z\leq 1$.

###### Proof.

This is straightforward. See [6, 7]. ∎

Recall the definition of the class $\mathcal{R}$ formulated in the previous
section. Let us study the connection between this class and the family of all
coherent distributions.

###### Proof of Proposition 1.5.

First, we show that the decomposition $m=\mu+\nu$ exists for all
$m\in\mathcal{C}$. Indeed, by virtue of Proposition 2.1, we can find a random
vector $(X,Y)\sim m$ defined on some probability space
$(\Omega,\mathcal{F},\mathbb{P})$, such that $X=\mathbb{E}(Z|X)$ and
$Y=\mathbb{E}(Z|Y)$ for some random variable $Z\in[0,1]$. For a set
$C\in\mathcal{B}([0,1]^{2})$, we put

(2.1) $\mu(C)=\int_{\\{(X,Y)\in C\\}}Z\ \mathrm{d}\mathbb{P}\ \ \ \ \
\text{and}\ \ \ \ \ \nu(C)=\int_{\\{(X,Y)\in C\\}}(1-Z)\
\mathrm{d}\mathbb{P}.$

Then the equality $m=\mu+\nu$ is evident. Furthermore, for a fixed
$A\in\mathcal{B}([0,1])$, we have

(2.2) $\int_{\\{X\in A\\}}X\ \mathrm{d}\mathbb{P}\ =\ \int_{\\{X\in A\\}}Z\
\mathrm{d}\mathbb{P}\ =\ \int_{A}1\ \mathrm{d}\mu^{x},$

where the first equality is due to $X=\mathbb{E}(Z|X)$ and the second is a
consequence of (2.1). Moreover, we may also write

(2.3) $\int_{\\{X\in A\\}}X\ \mathrm{d}\mathbb{P}\ =\ \int_{A\times[0,1]}x\
\mathrm{d}m\ =\ \int_{A}x\ \mathrm{d}\mu^{x}+\int_{A}x\ \mathrm{d}\nu^{x}.$

Combining (2.2) and (2.3), we get

$\ \int_{A}(1-x)\ \mathrm{d}\mu^{x}\ =\ \int_{A}x\ \mathrm{d}\nu^{x},$

for all $A\in\mathcal{B}([0,1])$. The symmetric condition (the second
requirement in Definition 1.4) is shown analogously. This completes the first
part of the proof.

Now, pick a probability measure $m$ on $[0,1]^{2}$ such that $m=\mu+\nu$ for
some $(\mu,\nu)\in\mathcal{R}$. We need to show that $m$ is coherent. To this
end, consider the probability space $([0,1]^{2},\mathcal{B}([0,1]^{2}),m)$ and
the random variables $X,Y:[0,1]^{2}\rightarrow[0,1]$ defined by

$X(x,y)=x\ \ \ \ \text{and}\ \ \ \ Y(x,y)=y,\ \ \ \ x,\,y\in[0,1].$

Additionally, let $Z$ denote the Radon–Nikodym derivative of $\mu$ with
respect to $m$: we have $0\leq Z\leq 1$ $m$–almost surely and
$\mu(C)=\int_{C}Z\mathrm{d}m$ for all $C\in\mathcal{B}([0,1]^{2})$. Again by
Proposition 2.1, it is sufficient to verify that $X=\mathbb{E}(Z|X)$ and
$Y=\mathbb{E}(Z|Y)$. By symmetry, it is enough to show the first equality. Fix
$A\in\mathcal{B}([0,1])$ and note that

(2.4) $\int_{\\{X\in A\\}}X\ \mathrm{d}m\ =\ \int_{A\times[0,1]}x\
\mathrm{d}m\ =\ \int_{A}x\ \mathrm{d}\mu^{x}+\int_{A}x\ \mathrm{d}\nu^{x}.$

Similarly, we also have

(2.5) $\int_{\\{X\in A\\}}Z\ \mathrm{d}m\ =\ \int_{A\times[0,1]}Z\
\mathrm{d}m\ =\ \mu(A\times[0,1])\ =\ \int_{A}1\ \mathrm{d}\mu^{x}.$

Finally, note that by $(\mu,\nu)\in\mathcal{R}$, the right-hand sides of (2.4)
and (2.3) are equal. Therefore we obtain the identity

$\int_{\\{X\in A\\}}X\ \mathrm{d}m\ =\ \int_{\\{X\in A\\}}Z\ \mathrm{d}m$

for arbitrary $A\in\mathcal{B}([0,1])$. This yields the claim. ∎

We turn our attention to the characterisation of $\mathrm{ext}(\mathcal{C})$
stated in the previous section.

###### Proof of Theorem 1.8, the implication ‘$\Rightarrow$’.

Let $m$ be an extremal coherent measure and suppose, on contrary, that
$(\mu_{1},\nu_{1})$ and $(\mu_{2},\nu_{2})$ are two different elements of
$\mathcal{R}(m)$. We will prove that $m-\mu_{1}+\mu_{2}$ and
$m-\mu_{2}+\mu_{1}$ are also coherent distributions. Because of

$m\ =\ \frac{1}{2}(m-\mu_{1}+\mu_{2})\ +\ \frac{1}{2}(m-\mu_{2}+\mu_{1}),$

we will obtain the contradiction with the assumed extremality of $m$. By
symmetry, it is enough to show that $(m-\mu_{1}+\mu_{2})\in\mathcal{C}$. To
this end, by virtue of Proposition 1.5, it suffices to check that
$m-\mu_{1}+\mu_{2}$ is a probability measure and
$(\mu_{2},m-\mu_{1})\in\mathcal{R}$. First, note that $\nu_{1}=m-\mu_{1}$ is
nonnegative and fix an arbitrary $A\in\mathcal{B}([0,1])$. As
$(\mu_{1},\nu_{1})$ and $(\mu_{2},\nu_{2})$ are representations of $m$,
Definition 1.4 gives

$\int_{A}1\ \mathrm{d}\mu_{1}^{x}\ =\ \int_{A}x\
(\mathrm{d}\nu_{1}^{x}+\mathrm{d}\mu_{1}^{x})\ =\ \int_{A}x\ \mathrm{d}m^{x},$

and

(2.6) $\int_{A}1\ \mathrm{d}\mu_{2}^{x}\ =\ \int_{A}x\
(\mathrm{d}\nu_{2}^{x}+\mathrm{d}\mu_{2}^{x})\ =\ \int_{A}x\ \mathrm{d}m^{x},$

so $\mu_{1}^{x}(A)=\mu_{2}^{x}(A)$. Similarly, we can deduce that
$\mu_{1}^{y}=\mu_{2}^{y}$, which means that marginal distributions of
$\mu_{1}$ and $\mu_{2}$ are equal. This, together with $m-\mu_{1}\geq 0$,
proves that $m-\mu_{1}+\mu_{2}$ is a probability measure. Next, using (2.6)
and $\mu_{1}^{x}=\mu_{2}^{x}$, we can also write

(2.7) $\int_{A}(1-x)\ \mathrm{d}\mu_{2}^{x}\ =\ \int_{A}x\
\mathrm{d}m^{x}-\int_{A}x\ \mathrm{d}\mu_{1}^{x}\ =\ \int_{A}x\
\mathrm{d}(m-\mu_{1})^{x}.$

In the same way we get

(2.8) $\int_{B}(1-y)\ \mathrm{d}\mu_{2}^{y}\ =\ \int_{B}y\
\mathrm{d}(m-\mu_{1})^{y},$

for all $B\in\mathcal{B}([0,1])$. By (2.7) and (2.8), we obtain that
$(\mu_{2},m-\mu_{1})\in\mathcal{R}$ and this completes the proof of the
uniqueness.

To show the minimality, let $m$ be an extremal coherent measure with the
representation $(\mu,\nu)$ (which is unique, as we have just proved). Consider
any nonzero $(\tilde{\mu},\tilde{\nu})\in\mathcal{R}$ with
$\tilde{\mu}\leq\mu$ and $\tilde{\nu}\leq\nu$. Then, by the very definition of
$\mathcal{R}$, we have $(\mu-\tilde{\mu},\nu-\tilde{\nu})\in\mathcal{R}$.
Therefore, by Proposition 1.5, we get

$\alpha^{-1}(\tilde{\mu}+\tilde{\nu}),\
(1-\alpha)^{-1}(m-\tilde{\mu}-\tilde{\nu})\ \in\mathcal{C},$

where $\alpha=(\tilde{\mu}+\tilde{\nu})([0,1]^{2})\in(0,1]$. We have the
identity

(2.9) $m\ =\
\alpha\cdot\Big{(}\alpha^{-1}(\tilde{\mu}+\tilde{\nu})\Big{)}+(1-\alpha)\cdot\Big{(}(1-\alpha)^{-1}(m-\tilde{\mu}-\tilde{\nu})\Big{)},$

which combined with the extremality of $m$ yields
$m=\alpha^{-1}(\tilde{\mu}+\tilde{\nu})=\alpha^{-1}\tilde{\mu}+\alpha^{-1}\tilde{\nu}$.
But $(\alpha^{-1}\tilde{\mu},\alpha^{-1}\tilde{\nu})$ belongs to
$\mathcal{R}$, since $(\tilde{\mu},\tilde{\nu})$ does, and hence
$(\alpha^{-1}\tilde{\mu},\alpha^{-1}\tilde{\nu})$ is a representation of $m$.
By the uniqueness, we deduce that
$(\tilde{\mu},\tilde{\nu})=\alpha\cdot(\mu,\nu)$. ∎

###### Proof of Theorem 1.8, the implication ‘$\Leftarrow$’.

Let $m$ be a coherent distribution with the unique and minimal representation
$(\mu,\nu)$. To show that $m$ is extremal, consider the decomposition
$m=\beta\cdot m_{1}+(1-\beta)\cdot m_{2}$ for some $m_{1},m_{2}\in\mathcal{C}$
and $\beta\in(0,1)$. Moreover, let $(\mu_{1},\nu_{1})\in\mathcal{R}(m_{1})$
and $(\mu_{2},\nu_{2})\in\mathcal{R}(m_{2})$. By the convexity of
$\mathcal{R}$, we have

(2.10) $(\mu^{\prime},\nu^{\prime}):=\ (\beta\mu_{1}+(1-\beta)\mu_{2},\
\beta\nu_{1}+(1-\beta)\nu_{2})\ \in\mathcal{R}(m)$

and hence, by the uniqueness, we get $(\mu^{\prime},\nu^{\prime})=(\mu,\nu)$.
Then, directly by (2.10), we have

(2.11) $\beta\mu_{1}\leq\mu\ \ \ \textrm{and}\ \ \ \beta\nu_{1}\leq\nu.$

Combining this with the minimality of $(\mu,\nu)$, we get
$(\beta\mu_{1},\beta\nu_{1})=\alpha(\mu,\nu)$ for some $\alpha\in[0,1]$. Since
$m=\mu+\nu$ and $m_{1}=\mu_{1}+\nu_{1}$ are probability measures, this gives
$\alpha=\beta$ and hence $(\mu_{1},\nu_{1})=(\mu,\nu)$. This implies $m=m_{1}$
and completes the proof. ∎

##  3\. Extreme points with finite support

In this section we study the geometric structure of the supports of measures
belonging to
$\textrm{ext}_{f}(\mathcal{C})=\\{\eta\in\textrm{ext}(\mathcal{C}):|\mathrm{supp}(\eta)|<\infty\\}$.
Our key result is presented in Theorem 3.7 – we prove that the support of an
extremal coherent distribution cannot contain any axial cycles (see Definition
1.9). Let us emphasize that this property has been originally conjectured in
[21]. We start with a simple combinatorial observation: it is straightforward
to check that certain special ‘alternating’ cycles are forbidden.

######  Definition 3.1.

Let $\eta$ be a coherent distribution with a unique representation $(\mu,\nu)$
and let $\big{(}(x_{i},y_{i})\big{)}_{i=1}^{2n}$ be an axial cycle contained
in $\mathrm{supp}(\eta)$. Then $\big{(}(x_{i},y_{i})\big{)}_{i=1}^{2n}$ is an
alternating cycle if

$(x_{2i+1},y_{2i+1})\in\mathrm{supp}(\mu)\ \ \ \ \text{and}\ \ \ \
(x_{2i},y_{2i})\in\mathrm{supp}(\nu),$

for all $i=1,2,\dots,n$ (with the convention
$x_{2n+1}=x_{1},\,y_{2n+1}=y_{1}$).

Figure 1. An example of an alternating cycle. Red points represent probability
masses in $\mathrm{supp}(\mu)$, while blue points indicate probability masses
in $\mathrm{supp}(\nu)$. Arrows outline a possible transformation of the
representation $(\mu,\nu)$.

######  Proposition 3.2.

If $\eta\in\mathrm{ext}_{f}(\mathcal{C})$, then $\mathrm{supp}(\eta)$ does not
contain any alternating cycles.

###### Proof.

Let $\eta$ be a coherent distribution with a unique representation $(\mu,\nu)$
and a finite support. Additionally, assume that
$\big{(}(x_{i},y_{i})\big{)}_{i=1}^{2n}$ is an alternating cycle contained in
$\mathrm{supp}(\eta)$. Let $\delta$ be the smaller of the two numbers

$\min_{0\leq i\leq n-1}\mu(x_{2i+1},y_{2i+1})\ \ \ \ \ \text{and}\ \ \ \ \
\min_{1\leq i\leq n}\nu(x_{2i},y_{2i})$

(for brevity, in what follows we will skip the parentheses and write
$\mu(a,b)$, $\nu(a,b)$ instead of $\mu(\\{a,b\\})$, $\nu(\\{a,b\\})$,
respectively). By Definition 3.1, we have $\delta>0$. Now, consider the
transformation $(\mu,\nu)\mapsto(\mu^{\prime},\nu^{\prime})$ described by the
following requirements:

  1. 1.

for $i=0,1,\dots,n-1$, put

$\mu^{\prime}(x_{2i+1},y_{2i+1}):=\ \mu(x_{2i+1},y_{2i+1})-\delta$
$\nu^{\prime}(x_{2i+1},y_{2i+1}):=\ \nu(x_{2i+1},y_{2i+1})+\delta,$

  2. 2.

for $i=1,2,\dots,n$, put

$\mu^{\prime}(x_{2i},y_{2i}):=\ \mu(x_{2i},y_{2i})+\delta$
$\nu^{\prime}(x_{2i},y_{2i}):=\ \nu(x_{2i},y_{2i})-\delta,$

  1. 3.

for $(x,y)\not\in\\{(x_{i},y_{i}):1\leq i\leq 2n\\}$, set

$\mu^{\prime}(x,y)=\mu(x,y),$ $\nu^{\prime}(x,y)=\nu(x,y).$

Note that $\mu$ and $\mu^{\prime}$, as well as $\nu$ and $\nu^{\prime}$, have
the same marginal distributions and hence
$(\mu^{\prime},\nu^{\prime})\in\mathcal{R}$. We also have
$\mu^{\prime}+\nu^{\prime}=\mu+\nu=\eta$ and thus
$(\mu^{\prime},\nu^{\prime})\in\mathcal{R}(\eta)$. This contradicts the
uniqueness of the representation $(\mu,\nu)$ and shows that
$\mathrm{supp}(\eta)$ cannot contain an alternating cycle. By Theorem 1.8,
this ends the proof. ∎

Before the further combinatorial analysis, we need to introduce some useful
auxiliary notation. For $\mu,\nu\in\mathcal{M}([0,1]^{2})$ with
$|\mathrm{supp}(\mu+\nu)|<\infty$, we define a quotient function
$q_{(\mu,\nu)}:\mathrm{supp}(\mu+\nu)\rightarrow[0,1]$ by

$q_{(\mu,\nu)}(x,y)=\frac{\mu(x,y)}{\mu(x,y)+\nu(x,y)}.$

In what follows, we will omit the subscripts and write $q$ for $q_{(\mu,\nu)}$
whenever the choice for $(\mu,\nu)$ is clear from the context.

######  Proposition 3.3.

Let $\mu,\nu\in\mathcal{M}([0,1]^{2})$ and $|\mathrm{supp}(\mu+\nu)|<\infty$.
Then $(\mu,\nu)\in\mathcal{R}$ if and only if the following conditions hold
simultaneously:

  * $\bullet$

for every $x$ satisfying $\mu(\\{x\\}\times[0,1])+\nu(\\{x\\}\times[0,1])>0$,
we have

(3.1) $\sum_{\begin{subarray}{c}y\in[0,1],\\\
(x,y)\in\mathrm{supp}(\mu+\nu)\end{subarray}}q(x,y)\frac{\mu(x,y)+\nu(x,y)}{\mu(\\{x\\}\times[0,1])+\nu(\\{x\\}\times[0,1])}\
=\ x,$

  * $\bullet$

for every $y$ satisfying $\mu([0,1]\times\\{y\\})+\nu([0,1]\times\\{y\\})>0$,
we have

(3.2) $\sum_{\begin{subarray}{c}x\in[0,1],\\\
(x,y)\in\mathrm{supp}(\mu+\nu)\end{subarray}}q(x,y)\frac{\mu(x,y)+\nu(x,y)}{\mu([0,1]\times\\{y\\})+\nu([0,1]\times\\{y\\})}\
=\ y,$

where sums in (3.1) and (3.2) are well defined – in both cases, there is only
a finite number of nonzero summands.

###### Proof.

Due to $|\mathrm{supp}(\mu+\nu)|<\infty$, this is a simple consequence of
Definition 1.4. ∎

Next, we will require an additional distinction between three different types
of points.

######  Definition 3.4.

Let $(\mu,\nu)\in\mathcal{R}$. A point $(x,y)\in\mathrm{supp}(\mu+\nu)$ is
said to be

  1. a)

a lower out point, if $q(x,y)<\min(x,y)$;

  2. b)

an upper out point, if $q(x,y)>\max(x,y)$;

  3. c)

a cut point, if it is not an out point, i.e.

$x\leq q(x,y)\leq y\ \ \ \ \ \text{or}\ \ \ \ \ y\leq q(x,y)\leq x.$

Finally, for the sake of completeness, we include a formal definition of an
axial path.

######  Definition 3.5.

The sequence $\big{(}(x_{i},y_{i})\big{)}_{i=1}^{n}$ with terms in $[0,1]^{2}$
is called an axial path if

  1. $\cdot$

all points $(x_{i},y_{i})$ are distinct;

  2. $\cdot$

we have $x_{i+1}=x_{i}$ or $y_{i+1}=y_{i}$ for all $i$;

  3. $\cdot$

there are at most two points on any horizontal or vertical line.

To develop some intuition, it is convenient to inspect the example given
below.

Figure 2.  Support of a coherent distribution $m$. Purple points (endpoints of
the path) are cut points. Red point represents a mass in $\mathrm{supp}(\mu)$
and is an upper out point. Blue point indicates a mass in $\mathrm{supp}(\nu)$
and it is a lower out point.

######  Example 3.6.

Let $m$ be a probability measure given by

$m\Big{(}\frac{1}{8},\frac{1}{4}\Big{)}=\frac{84}{196},\ \ \
m\Big{(}\frac{1}{2},\frac{1}{4}\Big{)}=\frac{14}{196},\ \ \
m\Big{(}\frac{1}{2},\frac{3}{4}\Big{)}=\frac{14}{196},\ \ \
m\Big{(}\frac{7}{8},\frac{3}{4}\Big{)}=\frac{84}{196}.$

There are five observations, which will be discussed separately.

(i) Consider the decomposition $m=\mu+\nu$, where $(\mu,\nu)$ is determined by
the quotient function

$q\Big{(}\frac{1}{8},\frac{1}{4}\Big{)}=\frac{1}{8},\ \ \
q\Big{(}\frac{1}{2},\frac{1}{4}\Big{)}=1,\ \ \
q\Big{(}\frac{1}{2},\frac{3}{4}\Big{)}=0,\ \ \
q\Big{(}\frac{7}{8},\frac{3}{4}\Big{)}=\frac{7}{8}.$

Using Proposition 3.3, we can check that $(\mu,\nu)\in\mathcal{R}$. For
instance, for $y=\frac{1}{4}$ we get

(3.3) $\frac{q(\frac{1}{8},\frac{1}{4})\cdot
m(\frac{1}{8},\frac{1}{4})+q(\frac{1}{2},\frac{1}{4})\cdot
m(\frac{1}{2},\frac{1}{4})}{m(\frac{1}{8},\frac{1}{4})+m(\frac{1}{2},\frac{1}{4})}\
=\
\frac{\frac{1}{8}\cdot\frac{84}{196}+1\cdot\frac{14}{196}}{\frac{84}{196}+\frac{14}{196}}\
=\ \frac{1}{4},$

which agrees with (3.2). As a direct consequence, by Proposition 1.5, we have
$m\in\mathcal{C}$.

(ii) Observe that $(\frac{1}{8},\frac{1}{4})$ and $(\frac{7}{8},\frac{3}{4})$
are cut points, $(\frac{1}{2},\frac{1}{4})$ is an upper out point and
$(\frac{1}{2},\frac{3}{4})$ is a lower out point. Moreover, $\mathrm{supp}(m)$
is an axial path without cycles – see Figure 2.

(iii) Notably, $(\mu,\nu)$ is a unique representation of $m$. Indeed,
$(\frac{1}{8},\frac{1}{4})$ is the only point in $\mathrm{supp}(m)$ with
$x$-coordinate equal to $\frac{1}{8}$ and hence
$q(\frac{1}{8},\frac{1}{4})=\frac{1}{8}$. Accordingly,
$q(\frac{1}{2},\frac{1}{4})=1$ is now a consequence of (3.3). The derivation
of $q(\frac{1}{2},\frac{3}{4})=0$ and $q(\frac{7}{8},\frac{3}{4})=\frac{7}{8}$
follows from an analogous computation.

(iv) Finally, the representation $(\mu,\nu)$ is minimal; let
$(\tilde{\mu},\tilde{\nu})\in\mathcal{R}$ satisfy $\tilde{\mu}\leq\mu$ and
$\tilde{\nu}\leq\nu$. Suppose that
$(\frac{1}{8},\frac{1}{4})\in\mathrm{supp}(\tilde{\mu}+\tilde{\nu})$. Again,
as $(\frac{1}{8},\frac{1}{4})$ is the only point in $\mathrm{supp}(m)$ with
$x$-coordinate equal to $\frac{1}{8}$, we get
$q_{(\tilde{\mu},\tilde{\nu})}(\frac{1}{8},\frac{1}{4})=\frac{1}{8}$. Next,
assume that
$(\frac{1}{2},\frac{1}{4})\in\mathrm{supp}(\tilde{\mu}+\tilde{\nu})$. As
$\tilde{\nu}(\frac{1}{2},\frac{1}{4})\leq\nu(\frac{1}{2},\frac{1}{4})=0$, we
have $q_{(\tilde{\mu},\tilde{\nu})}(\frac{1}{2},\frac{1}{4})=1$. Likewise, we
can check that

(3.4) $q_{(\tilde{\mu},\tilde{\nu})}(x,y)\ =\ q_{(\mu,\nu)}(x,y)\ \ \ \
\textrm{for all }(x,y)\in\mathrm{supp}(\tilde{\mu}+\tilde{\nu}).$

By Proposition 3.3 and the equation (3.4), we easily obtain that
$\tilde{\mu}+\tilde{\nu}=0$ or
$\mathrm{supp}(\tilde{\mu}+\tilde{\nu})=\mathrm{supp}(m)$. For example,

$\cdot$ if
$(\frac{1}{2},\frac{1}{4})\in\mathrm{supp}(\tilde{\mu}+\tilde{\nu})$, then
(3.1) gives
$(\frac{1}{2},\frac{3}{4})\in\mathrm{supp}(\tilde{\mu}+\tilde{\nu})$;

$\cdot$ if
$(\frac{1}{2},\frac{3}{4})\in\mathrm{supp}(\tilde{\mu}+\tilde{\nu})$, then
(3.2) yields
$(\frac{7}{8},\frac{3}{4})\in\mathrm{supp}(\tilde{\mu}+\tilde{\nu})$.

Therefore, if $\tilde{\mu}+\tilde{\nu}\not=0$, then the measure
$\tilde{\mu}+\tilde{\nu}$ is supported on the same set as $m$ and
$q_{(\tilde{\mu},\tilde{\nu})}\equiv q_{(\mu,\nu)}$. For the same reason, i.e.
using Proposition 3.3 and path structure of $\mathrm{supp}(m)$, it follows
that $\tilde{\mu}+\tilde{\nu}=\alpha\cdot m$ for some $\alpha\in[0,1]$. For
instance, by (3.2) for $y=\frac{1}{4}$, we get

$\frac{\frac{1}{8}\cdot\tilde{m}(\frac{1}{8},\frac{1}{4})+1\cdot\tilde{m}(\frac{1}{2},\frac{1}{4})}{\tilde{m}(\frac{1}{8},\frac{1}{4})+\tilde{m}(\frac{1}{2},\frac{1}{4})}\
=\ \frac{1}{4},$

where $\tilde{m}=\tilde{\mu}+\tilde{\nu}.$ Hence
$\tilde{m}(\frac{1}{8},\frac{1}{4})\tilde{m}(\frac{1}{2},\frac{1}{4})^{-1}=m(\frac{1}{8},\frac{1}{4})m(\frac{1}{2},\frac{1}{4})^{-1}=\frac{84}{14}$.

(v) By the above analysis and Theorem 1.8, we conclude that
$m\in\mathrm{ext}_{f}(\mathcal{C})$.

We are now ready to demonstrate the central result of this section.

######  Theorem 3.7.

If $\eta\in\mathrm{ext}_{f}(\mathcal{C})$, then $\mathrm{supp}(\eta)$ is an
axial path without cycles.

Let us briefly explain the main idea of the proof. For
$\eta\in\mathrm{ext}_{f}(\mathcal{C})$, we inductively construct a special
axial path contained in $\mathrm{supp}(\eta)$, which does not contain any cut
points (apart from the endpoints). We show that axial path obtained in this
process is acyclic and involves all points from $\mathrm{supp}(\eta)$.

###### Proof of Theorem 3.7.

Fix $\eta\in\mathrm{ext}_{f}(\mathcal{C})$ and let $(\mu,\nu)$ be the unique
representation of $\eta$. By $\mathcal{L}(\eta)$ and $\mathcal{U}(\eta)$
denote the sets of lower and upper out points, correspondingly. Choose any
$(x_{0},y_{0})\in\mathrm{supp}(\eta)$. We will consider two separate cases
now:

Case I: $(x_{0},y_{0})$ is an out point. With no loss of generality, we can
assume that $(x_{0},y_{0})\in\mathcal{L}(\eta)$. We then use the following
inductive procedure.

1∘ Suppose we have successfully found $(x_{n},y_{n})\in\mathcal{L}(\eta)$ and
it is the first time we have chosen a point with the $x$-coordinate equal to
$x_{n}$. Since $(x_{n},y_{n})\in\mathcal{L}(\eta)$, we have
$q(x_{n},y_{n})<x_{n}$. By (3.1), there must exist a point
$(x_{n+1},y_{n+1})\in\mathrm{supp}(\eta)$ such that $x_{n+1}=x_{n}$ and
$q(x_{n+1},y_{n+1})>x_{n}$. We pick one such point and add it at the end of
the path. If $(x_{n+1},y_{n+1})$ is a cut point or an axial cycle was just
created, we exit the loop. Otherwise, note that
$(x_{n+1},y_{n+1})\in\mathcal{U}(\eta)$. Go to 2∘.

2∘ Assume we have successfully found $(x_{n},y_{n})\in\mathcal{U}(\eta)$ and
it is the first time we have chosen a point with the $y$-coordinate equal to
$y_{n}$. Since $(x_{n},y_{n})\in\mathcal{U}(\eta)$, we have
$q(x_{n},y_{n})>y_{n}$. By (3.2), there must exist a point
$(x_{n+1},y_{n+1})\in\mathrm{supp}(\eta)$ such that $y_{n+1}=y_{n}$ and
$q(x_{n+1},y_{n+1})<y_{n}$. We pick one such point and add it at the end of
the path. If $(x_{n+1},y_{n+1})$ is a cut point or an axial cycle was just
created, we exit the loop. Otherwise, note that
$(x_{n+1},y_{n+1})\in\mathcal{L}(\eta)$. Go to 1∘.

Figure 3. An example of an axial path constructed by the algorithm. Symbols
$\vee,\wedge$ are placed next to lower ($\vee$) and upper ($\wedge$) out
points. Purple point $(x_{k},y_{k})$ is the endpoint of the path. Red points
represent probability masses in $\mathrm{supp}(\mu)$, while blue points
indicate probability masses in $\mathrm{supp}(\nu)$.

Figure 4. An example of an axial path $\Gamma$ constructed after the second
run of the algorithm. Purple points $(x_{k},y_{k})$ and $(x_{-l},y_{-l})$
(endpoints of $\Gamma$) are cut points. Red points represent probability
masses in $\mathrm{supp}(\mu)$, while blue points indicate probability masses
in $\mathrm{supp}(\nu)$.

As $|\mathrm{supp}(\eta)|<\infty$, the procedure terminates after a finite
number of steps (denote it by $k$) and produces an axial path
$\big{(}(x_{i},y_{i})\big{)}_{i=0}^{k}$ contained in $\mathrm{supp}(\eta)$; to
be more precise, it is also formally possible that $(x_{k},y_{k})$ is a third
point on some horizontal or vertical line (in such a case we have obtained an
axial cycle). By the construction of the loop, point $(x_{k},y_{k})$ is either
an endpoint of an axial cycle or a cut point. Let us show that the first
alternative is impossible. First, we clearly have
$\mathcal{L}(\eta)\subset\mathrm{supp}(\nu)$ and
$\mathcal{U}(\eta)\subset\mathrm{supp}(\mu)$, see Figure 4. Next, assume that
$(x_{k-1},y_{k-1})\in\mathcal{U}(\eta)$. This means that $(x_{k},y_{k})$ was
found in step 2∘ and $q(x_{k},y_{k})<y_{k-1}\leq 1$. Therefore
$(x_{k},y_{k})\in\mathrm{supp}(\nu)$ and there exists an alternating cycle in
$\mathrm{supp}(\eta)$. However, this is not possible because of Proposition
3.2. If $(x_{k-1},y_{k-1})\in\mathcal{L}(\eta)$, the argument is analogous.

We have shown that $(x_{k},y_{k})$ is a cut point. Set
$\Gamma_{+}=\bigcup_{i=1}^{k}\\{(x_{i},y_{i})\\}$. Moving on, we can return to
the starting point $(x_{0},y_{0})$ and repeat the above construction in the
reversed direction. By switching the roles of $x$ and $y$-coordinates in steps
1∘ and 2∘, we produce another axial path $(x_{i},y_{i})_{i=0}^{-l}$. Set
$\Gamma_{-}=\bigcup_{i=-1}^{-l}\\{(x_{i},y_{i})\\}$ and

$\Gamma\ =\ \Gamma_{+}\cup\\{(x_{0},y_{0})\\}\cup\Gamma_{-}.$

Repeating the same arguments as before, we show that $(x_{-l},y_{-l})$ is a
cut point and $\Gamma$ is an axial path without cycles, see Figure 4.

It remains to verify that $\mathrm{supp}(\eta)=\Gamma$. This will be
accomplished by showing that there exists
$(\tilde{\mu},\tilde{\nu})\in\mathcal{R}$ with $\tilde{\mu}\leq\mu$,
$\tilde{\nu}\leq\nu$ and $\mathrm{supp}(\tilde{\mu}+\tilde{\nu})=\Gamma$. This
will give the claim: by the minimality of the representation $(\mu,\nu)$, we
will deduce that $\tilde{\mu}+\tilde{\nu}=\alpha\cdot\eta$ for some
$\alpha\in(0,1]$, and hence
$\mathrm{supp}(\tilde{\mu}+\tilde{\nu})=\mathrm{supp}(\eta)$.

We begin with the endpoints of $\Gamma$. As $(x_{k},y_{k})$ is a cut point,
there exists $\gamma\in[0,1]$ such that $q(x_{k},y_{k})=\gamma
x_{k}+(1-\gamma)y_{k}$. We can write

(3.5) $\eta(x_{k},y_{k})\ =\
\eta^{\prime}(x_{k},y_{k})+\eta^{\prime\prime}(x_{k},y_{k}),$

where $\eta^{\prime}(x_{k},y_{k})=\gamma\eta(x_{k},y_{k})$ and
$\eta^{\prime\prime}(x_{k},y_{k})=(1-\gamma)\eta(x_{k},y_{k})$. Set

(3.6) $\mu^{\prime}(x_{k},y_{k})=x_{k}\eta^{\prime}(x_{k},y_{k})\ \ \ \
\text{and}\ \ \ \
\mu^{\prime\prime}(x_{k},y_{k})=y_{k}\eta^{\prime\prime}(x_{k},y_{k}).$

By (3.5) and (3.6), we have

(3.7)
$\mu^{\prime}(x_{k},y_{k})+\mu^{\prime\prime}(x_{k},y_{k})=\Big{(}x_{k}\gamma+y_{k}(1-\gamma)\Big{)}\eta(x_{k},y_{k})=\mu(x_{k},y_{k}).$

Equations (3.5) and (3.7) have a clear and convenient interpretation. Namely,
we can visualize it as ‘cutting’ the point $(x_{k},y_{k})$ into two separate
points: $(x_{k},y_{k})^{\prime}$ with mass $\eta^{\prime}(x_{k},y_{k})$ and
$(x_{k},y_{k})^{\prime\prime}$ with mass $\eta^{\prime\prime}(x_{k},y_{k})$.
Moreover, calculating their quotient functions independently, we get
$q^{\prime}(x_{k},y_{k})=x_{k}$ and $q^{\prime\prime}(x_{k},y_{k})=y_{k}$.
Performing the same ‘cut’ operation on $(x_{-l},y_{-l})$ we can divide this
point into $(x_{-l},y_{-l})^{\prime}$ and $(x_{-l},y_{-l})^{\prime\prime}$
such that $q^{\prime}(x_{-l},y_{-l})=x_{-l}$ and
$q^{\prime\prime}(x_{-l},y_{-l})=y_{-l}$.

Observe that $(x_{k},y_{k})$ and $(x_{k-1},y_{k-1})$ have exactly one common
coordinate, say $y_{k}=y_{k-1}$. Consequently, $(x_{k},y_{k})$ is the only
point in $\Gamma$ with $x$-coordinate equal to $x_{k}$. Additionally, by (3.2)
and $(x_{k-1},y_{k-1})\in\mathcal{U}(\eta)$, this means that
$q(x_{k},y_{k})\not=y_{k}$ and $\gamma>0$. Hence
$\eta^{\prime}(x_{k},y_{k})>0$. Similarly, suppose that $y_{-l}=y_{-l+1}$ (as
presented in Figure 4; for other configurations of endpoints, we proceed by
analogy). Thus, $(x_{-l},y_{-l})$ is the only point in $\Gamma$ with
$x$-coordinate equal to $x_{-l}$. By (3.2) and
$(x_{-l+1},y_{-l+1})\in\mathcal{L}(\eta)$, we have
$\eta^{\prime}(x_{-l},y_{-l})>0$.

Next, consider the following function $\tilde{q}:\Gamma\rightarrow[0,1]$
uniquely determined by the following requirements:

  1. 1.

$\tilde{q}(x_{k},y_{k})=x_{k}$ (if $y_{k}=y_{k-1}$, as we have assumed)
or $\tilde{q}(x_{k},y_{k})=y_{k}$ (in the case when $x_{k}=x_{k-1}$),

  2. 2.

$\tilde{q}(x_{-l},y_{-l})=x_{-l}$ (if $y_{-l}=y_{-l+1}$, as we have assumed)
or $\tilde{q}(x_{-l},y_{-l})=y_{-l}$ (in the case when $x_{-l}=x_{-l+1}$),

  3. 3.

$\tilde{q}(x,y)=0$ for all $(x,y)\in\Gamma\cap\mathcal{L}(\eta)$,

  4. 4.

$\tilde{q}(x,y)=1$ for all $(x,y)\in\Gamma\cap\mathcal{U}(\eta)$.

Set $\delta=\min(a,b,c,d)$, where

$a=\eta^{\prime}(x_{k},y_{k})$ (if $y_{k}=y_{k-1}$) or
$a=\eta^{\prime\prime}(x_{k},y_{k})$ (if $x_{k}=x_{k-1}$),

$b=\eta^{\prime}(x_{-l},y_{-l})$ (if $y_{-l}=y_{-l+1}$) or
$b=\eta^{\prime\prime}(x_{-l},y_{-l})$ (if $x_{-l}=x_{-l+1}$),

$c=\min_{(x,y)\in\Gamma\cap\mathcal{L}(\eta)}\nu(x,y)$,
$d=\min_{(x,y)\in\Gamma\cap\mathcal{U}(\eta)}\mu(x,y).$

Then $\delta>0$, which follows from the previous discussion. Finally, using
the acyclic path structure of $\Gamma$ and Proposition 3.3 (just as in Example
3.6), we are able to find a pair $(\tilde{\mu},\tilde{\nu})\in\mathcal{R}$
with $\mathrm{supp}(\tilde{\mu}+\tilde{\nu})=\Gamma$ and a quotient function
$q_{(\tilde{\mu},\tilde{\nu})}=\tilde{q}$. Letting

$\beta\ =\
\delta\cdot\Big{(}\max_{(x,y)\in\Gamma}(\tilde{\mu}+\tilde{\nu})(x,y)\Big{)}^{-1},$

we see that $\beta\tilde{\mu}\leq\mu$ and $\beta\tilde{\nu}\leq\nu$, as
desired.

Case II: $(x_{0},y_{0})$ is a cut point. Suppose that $x_{0}=y_{0}$ and
$q(x_{0},x_{0})=x_{0}$. Put

$\tilde{\mu}=\mathbbm{1}_{\\{(x_{0},x_{0})\\}}x_{0}\eta(x_{0},y_{0})\ \ \ \
\text{and}\ \ \ \
\tilde{\nu}=\mathbbm{1}_{\\{(x_{0},x_{0})\\}}(1-x_{0})\eta(x_{0},y_{0}).$

We have $(\tilde{\mu},\tilde{\nu})\in\mathcal{R}$ and $\tilde{\mu}\leq\mu$,
$\tilde{\nu}\leq\nu$. Hence $\mathrm{supp}(\eta)=\\{(x_{0},x_{0})\\}$. Next,
assume that $x_{0}\not=y_{0}$. In that case, $q(x_{0},y_{0})$ cannot be equal
to both $x_{0}$ and $y_{0}$ at the same time. This means that we can proceed
just as in Case I (at least in one direction). The only difference is that we
have already located one of the cut points – there is no need to apply the
procedure twice. ∎

From the proof provided, we can deduce yet another significant conclusion.

######  Corollary 3.8.

If $\eta\in\mathrm{ext}_{f}(\mathcal{C})$, then $q(x,y)=0$ for all
$(x,y)\in\mathcal{L}(\eta)$ and $q(x,y)=1$ for all
$(x,y)\in\mathcal{U}(\eta)$. Except for the endpoints of this axial path
(which are cut points), $\mathrm{supp}(\eta)$ consists of lower and upper out
points, appearing alternately.

###### Proof.

Note that $\mathcal{L}(\eta)$ and $\mathcal{U}(\eta)$ are well defined as the
representation of $\eta$ is unique. The statement follows directly from the
proof of Theorem 3.7. ∎

##  4\. Asymptotic estimate

Equipped with the machinery developed in the previous sections, we are ready
to establish the asymptotic estimate (1.3). We need to clarify how the
properties of $\mathrm{ext}_{f}(\mathcal{C})$ covered in the preceding part
apply to this problem. Referring to the prior notation, we will write

$(X,Y)\in\mathcal{C}_{f}\ \ \ \ \ \text{or}\ \ \ \ \
(X,Y)\in\mathrm{ext}_{f}(\mathcal{C}),$

to indicate that the distribution of a random vector $(X,Y)$ is a coherent (or
an extremal coherent) measure with finite support.

######  Proposition 4.1.

For any $\alpha>0$, we have

$\sup_{(X,Y)\in\mathcal{C}}\mathbb{E}|X-Y|^{\alpha}\ \ =\ \
\sup_{(X,Y)\in\mathcal{C}_{f}}\mathbb{E}|X-Y|^{\alpha}.$

###### Proof.

Fix any $(X,Y)\in\mathcal{C}$. As shown in [5, 7], there exists a sequence
$(X_{n},Y_{n})\in\mathcal{C}_{f}$ such that

(4.1) $\max\Big{\\{}|X-X_{n}|,|Y-Y_{n}|\Big{\\}}\ \leq\ \frac{1}{n},\ \ \ \ \
\ \mbox{for all}\ n=1,\,2,\,\ldots$

almost surely. Consequently, by dominated convergence and (4.1), we obtain

$\mathbb{E}|X-Y|^{\alpha}\ \ =\ \
\lim_{n\rightarrow\infty}\mathbb{E}|X_{n}-Y_{n}|^{\alpha},$

and thus

$\mathbb{E}|X-Y|^{\alpha}\ \ \leq\ \ \sup_{n\in\mathbb{N}}\
\mathbb{E}|X_{n}-Y_{n}|^{\alpha}\ \ \leq\ \
\sup_{(X,Y)\in\mathcal{C}_{f}}\mathbb{E}|X-Y|^{\alpha}.$

This proves the ‘$\leq$’-inequality, while in the reversed direction it is
evident. ∎

Next, we will apply the celebrated Krein–Milman theorem, see [18].

######  Theorem 4.2 (Krein–Milman).

A compact convex subset of a Hausdorff locally convex topological vector space
is equal to the closed convex hull of its extreme points.

The above statement enables us to restrict the analysis of the estimate (1.3)
to extremal measures. Precisely, we have the following statement.

######  Proposition 4.3.

For any $\alpha>0$, we have

$\sup_{(X,Y)\in\mathcal{C}_{f}}\mathbb{E}|X-Y|^{\alpha}\ \ =\ \
\sup_{(X,Y)\in\mathrm{ext}_{f}(\mathcal{C})}\mathbb{E}|X-Y|^{\alpha}.$

###### Proof.

Let $Z=C([0,1]^{2},\mathbb{R})$; then $Z^{*}$ is the space of finite signed
Borel measures with the total variation norm $\|\cdot\|_{\mathrm{TV}}$. Let us
equip $Z^{*}$ with the topology of weak∗ convergence. Under this topology,
$Z^{*}$ is a Hausdorff and a locally convex space. For a fixed
$m\in\mathcal{C}_{f}$, let

$\mathcal{C}_{m}\ =\ \\{m^{\prime}\in\mathcal{C}_{f}:\
\mathrm{supp}(m^{\prime})\subseteq\mathrm{supp}(m)\\}$

denote the family of coherent distributions supported on the subsets of
$\mathrm{supp}(m)$. Firstly, observe that $\mathcal{C}_{m}$ is convex.
Secondly, we can easily verify that
$\mathrm{ext}(\mathcal{C}_{m})=\mathcal{C}_{m}\cap\mathrm{ext}_{f}(\mathcal{C})$.
Plainly, if $m^{\prime}\in\mathcal{C}_{m}$ and $m^{\prime}=\alpha\cdot
m_{1}+(1-\alpha)\cdot m_{2}$ for some $\alpha\in(0,1)$ and
$m_{1},m_{2}\in\mathcal{C}$, then
$\mathrm{supp}(m^{\prime})=\mathrm{supp}(m_{1})\cup\mathrm{supp}(m_{2})$ and
we must have $m_{1},m_{2}\in\mathcal{C}_{m}$. Hence
$\mathrm{ext}(\mathcal{C}_{m})\subset\mathrm{ext}_{f}(\mathcal{C})$, whereas
$\mathrm{ext}_{f}(\mathcal{C})\cap\mathcal{C}_{m}\subset\mathrm{ext}(\mathcal{C}_{m})$
is obvious.

Moreover, we claim that $\mathcal{C}_{m}$ is compact in the weak∗ topology.
Indeed, by the Banach–Alaoglu theorem,

$B_{Z^{*}}\ =\ \\{\mu\in Z^{*}:\ \|\mu\|_{\mathrm{TV}}\leq 1\\}$

is weak∗ compact. As $\mathcal{C}_{m}\subset B_{Z^{*}}$, it remains to check
that $\mathcal{C}_{m}$ is weak∗ closed. We can write
$\mathcal{C}_{m}=\mathcal{C}\cap\mathcal{P}_{m}$, where $\mathcal{P}_{m}$
stands for the set of all probability measures supported on the subsets of
$\mathrm{supp}(m)$. Note that $\mathcal{P}_{m}$ is clearly weak∗ closed.
Lastly, coherent distributions on $[0,1]^{2}$ are also weak∗ closed, as
demonstrated in [6].

Thus, by Krein–Milman theorem, there exists a sequence
$(m_{n})_{n=1}^{\infty}$ with values in $\mathcal{C}_{m}$, satisfying

(4.2) $m_{n}\ =\
\beta_{1}^{(n)}\eta_{1}^{(n)}+\beta_{2}^{(n)}\eta_{2}^{(n)}+\dots+\beta_{k_{n}}^{(n)}\eta_{k_{n}}^{(n)},$

where
$\eta_{1}^{(n)},\dots,\eta_{k_{n}}^{(n)}\in\mathrm{ext}(\mathcal{C}_{m})$ and
$\beta_{1}^{(n)},\dots,\beta_{k_{n}}^{(n)}$ are positive numbers summing up to
$1$, such that

(4.3) $\int_{[0,1]^{2}}f\ \mathrm{d}m_{n}\ \longrightarrow\ \int_{[0,1]^{2}}f\
\mathrm{d}m,$

for all bounded, continuous functions $f:[0,1]^{2}\rightarrow\mathbb{R}$. Put
$f(x,y)=|x-y|^{\alpha}$. By (4.3) and (4.2), we have

$\displaystyle\int_{[0,1]^{2}}|x-y|^{\alpha}\ \mathrm{d}m\ \ \ $
$\displaystyle\leq\ \ \ \sup_{n\in\mathbb{N}}\int_{[0,1]^{2}}|x-y|^{\alpha}\
\mathrm{d}m_{n}$ $\displaystyle\leq\ \ \
\sup_{\begin{subarray}{c}n\in\mathbb{N},\\\ 1\leq i\leq
k_{n}\end{subarray}}\int_{[0,1]^{2}}|x-y|^{\alpha}\ \mathrm{d}\eta_{i}^{(n)}$
$\displaystyle\leq\ \ \
\sup_{\eta\in\mathrm{ext}_{f}(\mathcal{C})}\int_{[0,1]^{2}}|x-y|^{\alpha}\
\mathrm{d}\eta,$

and hence

$\sup_{(X,Y)\in\mathcal{C}_{f}}\mathbb{E}|X-Y|^{\alpha}\ \ \leq\ \
\sup_{(X,Y)\in\mathrm{ext}_{f}(\mathcal{C})}\mathbb{E}|X-Y|^{\alpha}.$

The reverse inequality is clear. ∎

Now, we have the following significant reduction. Denote by $\mathcal{S}$ the
family of all finite sequences $\textbf{z}=(z_{0},z_{1},\dots,z_{n+1})$,
$n\in\mathbb{N}$, with $z_{0}=z_{n+1}=0$, $\sum_{i=1}^{n}z_{i}=1$ and
$z_{i}>0$ for $i=1,2,\dots,n$. We emphasize that $n=n(\textbf{z})$, the length
of z, is also allowed to vary. In what follows, we will write $n$ instead of
$n(\textbf{z})$; this should not lead to any confusion.

######  Proposition 4.4.

For any $\alpha\geq 1$, we have

(4.4) $\sup_{(X,Y)\in\mathrm{ext}_{f}(\mathcal{C})}\mathbb{E}|X-Y|^{\alpha}\ \
=\ \ \sup_{{\textbf{z}\in\mathcal{S}}}\
\sum_{i=1}^{n}z_{i}\Big{|}\frac{z_{i}}{z_{i-1}+z_{i}}-\frac{z_{i}}{z_{i}+z_{i+1}}\Big{|}^{\alpha}.$

###### Proof.

Consider an arbitrary $\eta\in\mathrm{ext}_{f}(\mathcal{C})$ and let
$(\mu,\nu)$ be its unique representation. Recall, based on Theorem 3.7, that
$\mathrm{supp}(\eta)$ is an axial path without cycles. Set
$\mathrm{supp}(\eta)=\\{(x_{i},y_{i})\\}_{i=1}^{n}$ and let
$q:\mathrm{supp}(\eta)\rightarrow[0,1]$ be the quotient function associated
with $(\mu,\nu)$. In this setup, by (3.1) and (3.2), we can write

(4.5) $\int_{[0,1]^{2}}|x-y|^{\alpha}\ \mathrm{d}\eta\ \ =\ \
\sum_{i=1}^{n}z_{i}\Big{|}\frac{q_{i-1}z_{i-1}+q_{i}z_{i}}{z_{i-1}+z_{i}}-\frac{q_{i}z_{i}+q_{i+1}z_{i+1}}{z_{i}+z_{i+1}}\Big{|}^{\alpha},$

where $z_{0}=z_{n+1}=0$, $q_{0}=q_{n+1}=0$,

$q_{i}=q(x_{i},y_{i})$ and $z_{i}=\eta(x_{i},y_{i})$, for all
$i=1,\,2,\,\dots,n$.

Note that if $n=1$, then both sides of (4.5) are equal to zero; hence $\eta$
does not bring any contribution to (4.4). Hence, from now on, we will assume
that $n\geq 2$. Notice that by Corollary 3.8, the sequence
$(q_{1},q_{2},\dots,q_{n})$ is given by

$(q_{1},0,1,0,1,\dots,q_{n})$ or $(q_{1},1,0,1,0,\dots,q_{n})$

– except for $q_{1}$ and $q_{n}$, $(q_{2},\dots,q_{n-1})$ is simply an
alternating binary sequence. Furthermore, the right-hand side of (4.5) is the
sum of

(4.6) $P(q_{1}):=\
z_{1}\Big{|}q_{1}-\frac{q_{1}z_{1}+q_{2}z_{2}}{z_{1}+z_{2}}\Big{|}^{\alpha}\
+\
z_{2}\Big{|}\frac{q_{1}z_{1}+q_{2}z_{2}}{z_{1}+z_{2}}-\frac{q_{2}z_{2}+q_{3}z_{3}}{z_{2}+z_{3}}\Big{|}^{\alpha}$

and some other terms not involving $q_{1}$. Since $\alpha\geq 1$, $P$ is a
convex function on $[0,1]$ and hence it is maximized by some
$q_{1}^{\prime}\in\\{0,1\\}$; in the case of $P(0)=P(1)$, we choose
$q_{1}^{\prime}$ arbitrarily. Depending on $q_{1}^{\prime}$, we shall now
perform one of the following transformations
$(q,z)\mapsto(\tilde{q},\tilde{z})$:

a. If $q_{1}^{\prime}\not=q_{2}$, we let $\tilde{n}=n$,
$\tilde{q}_{1}=q_{1}^{\prime}$ and $\tilde{q}_{i}=q_{i}$ for
$i\in\\{0\\}\cup\\{2,3,\dots,n+1\\}$, $\tilde{z}_{i}=z_{i}$ for
$i\in\\{0,1,\dots,n+1\\}$. This operation only changes $q_{1}$ into
$q_{1}^{\prime}$ – we increase the right-hand side of (4.5) by “correcting”
the quotient function on the first atom.

b. If $q_{1}^{\prime}=q_{2}$, we take $\tilde{n}=n-1$, $\tilde{q}_{0}=0$,
$\tilde{z}_{0}=0$ and

$\tilde{q}_{i}=q_{i+1},\ \
\qquad\tilde{z}_{i}=\frac{z_{i+1}}{z_{2}+z_{3}+\ldots+z_{n}}\ \qquad\ \mbox{
for }\ \ \ i\in\\{1,2,\dots,\tilde{n}+1\\}.$

This modification removes the first atom and rescales the remaining ones. It
is easy to see that for the transformed sequences $(\tilde{q},\tilde{z})$, the
right-hand side of (4.5) does not decrease.

Performing a similar transformation for the last summand in (4.5) (depending
on $q_{n}^{\prime}$ and $q_{n-1}$) we obtain a pair of sequences
$(\tilde{q},\tilde{z})$, such that
$(\tilde{q}_{1},\dots,\tilde{q}_{\tilde{n}})$ is an alternating binary
sequence and

$\displaystyle\int_{[0,1]^{2}}|x-y|^{\alpha}\ \mathrm{d}\eta\ \ \ $
$\displaystyle\leq\ \ \
\sum_{i=1}^{\tilde{n}}\tilde{z}_{i}\Big{|}\frac{\tilde{q}_{i-1}\tilde{z}_{i-1}+\tilde{q}_{i}\tilde{z}_{i}}{\tilde{z}_{i-1}+\tilde{z}_{i}}-\frac{\tilde{q}_{i}\tilde{z}_{i}+\tilde{q}_{i+1}\tilde{z}_{i+1}}{\tilde{z}_{i}+\tilde{z}_{i+1}}\Big{|}^{\alpha}$
$\displaystyle=\ \ \
\sum_{i=1}^{\tilde{n}}\tilde{z}_{i}\Big{|}\frac{\tilde{z}_{i}}{\tilde{z}_{i-1}+\tilde{z}_{i}}-\frac{\tilde{z}_{i}}{\tilde{z}_{i}+\tilde{z}_{i+1}}\Big{|}^{\alpha}$
$\displaystyle\leq\ \ \sup_{\tilde{\textbf{z}}}\
\sum_{i=1}^{n}z_{i}\Big{|}\frac{z_{i}}{z_{i-1}+z_{i}}-\frac{z_{i}}{z_{i}+z_{i+1}}\Big{|}^{\alpha},$

which proves the inequality ‘$\leq$’ in (4.4). The reverse bound follows by a
straightforward construction, involving measures with quotient functions equal
to $0$ or $1$ (see (4.5)). ∎

We require some further notation. Given $\alpha>0$, let
$\Phi_{\alpha}:\mathcal{S}\rightarrow[0,1]$ be defined by

$\Phi_{\alpha}(z)\ \ =\ \
\sum_{i=1}^{n}z_{i}\Big{|}\frac{z_{i}}{z_{i-1}+z_{i}}-\frac{z_{i}}{z_{i}+z_{i+1}}\Big{|}^{\alpha}.$

By the preceding discussion, for $\alpha\geq 1$ we have

$\sup_{(X,Y)\in\mathcal{C}}\mathbb{E}|X-Y|^{\alpha}\ =\
\sup_{z\in\mathcal{S}}\Phi_{\alpha}(z),$

and our main problem amounts to the identification of

(4.7)
$\limsup_{\alpha\to\infty}\Big{[}\alpha\cdot\sup_{z\in\mathcal{S}}\Phi_{\alpha}(z)\Big{]}.$

It will later become clear that $\limsup$ in (4.7) can be replaced by an
ordinary limit. We begin by making some introductory observations.

######  Definition 4.5.

Fix $\alpha\geq 1$ and let $\textbf{z}=(z_{0},z_{1},\dots,z_{n+1})$ be a
generic element of $\mathcal{S}$. For $1\leq i\leq n$, we say that the term
(component) $z_{i}$ of z is _significant_ if

$\sqrt{\alpha}\cdot z_{i-1}<z_{i}$ and $\sqrt{\alpha}\cdot z_{i}<z_{i+1}$,

or

$z_{i-1}>\sqrt{\alpha}\cdot z_{i}$ and $z_{i}>\sqrt{\alpha}\cdot z_{i+1}$.

The set of all significant components of $z$ will be denoted by
$\phi_{\alpha}(z)$. Whenever a component $z_{i}$ of z ($1\leq i\leq n$) is not
significant, we say that $z_{i}$ is _negligible_. The terms $z_{0}$ and
$z_{n+1}$ will be treated as neither significant nor negligible.

Now we will show that the contribution of all negligible terms of $z$ to the
total sum $\Phi_{\alpha}(z)$ vanishes in the limit $\alpha\to\infty$.
Precisely, we have the following.

######  Proposition 4.6.

For $\alpha\geq 1$ and $z\in\mathcal{S}$, we have

$\Phi_{\alpha}(z)\ \ \leq\ \ \Psi_{\alpha}(z)\ +\
\Big{|}1-\frac{1}{1+\sqrt{\alpha}}\Big{|}^{\alpha},$

where $\Psi_{\alpha}:\mathcal{S}\rightarrow[0,1]$ is defined by

$\Psi_{\alpha}(z)\ \ =\ \
\sum_{z_{i}\in\phi_{\alpha}(z)}z_{i}\Big{|}\frac{z_{i}}{z_{i-1}+z_{i}}-\frac{z_{i}}{z_{i}+z_{i+1}}\Big{|}^{\alpha}.$

###### Proof.

Since $z_{1}+z_{2}+\dots+z_{n}=1$, it is sufficient to show that

(4.8) $\Big{|}\frac{z_{i}}{z_{i-1}+z_{i}}-\frac{z_{i}}{z_{i}+z_{i+1}}\Big{|}\
\ \leq\ \ \Big{|}1-\frac{1}{1+\sqrt{\alpha}}\Big{|},$

for all negligible components $z_{i}$. Assume that (4.8) does not hold. Since
the ratios $z_{i}/(z_{i-1}+z_{i})$ and $z_{i+1}/(z_{i}+z_{i+1})$ take values
in $[0,1]$, we must have

(4.9)
$\min\Big{\\{}\frac{z_{i}}{z_{i-1}+z_{i}},\frac{z_{i}}{z_{i}+z_{i+1}}\Big{\\}}\
\ <\ \ \frac{1}{1+\sqrt{\alpha}}$

and

(4.10)
$\max\Big{\\{}\frac{z_{i}}{z_{i-1}+z_{i}},\frac{z_{i}}{z_{i}+z_{i+1}}\Big{\\}}\
\ >\ \ \frac{\sqrt{\alpha}}{1+\sqrt{\alpha}}.$

It remains to note that component $z_{i}$ fulfilling (4.9) and (4.10) is
significant. ∎

It is also useful to consider some special arrangements consisting of three
successive components $(z_{i-1},z_{i},z_{i+1})$ of the generic sequence
$z\in\mathcal{S}$.

######  Definition 4.7.

Let $\textbf{z}=(z_{0},z_{1},\dots,z_{n+1})$ be an element of $\mathcal{S}$.
For $1\leq i\leq n$, we say that a subsequence $(z_{i-1},z_{i},z_{i+1})$ of z
is

  * $\bullet$

a _split_ , if $z_{i-1}>z_{i}<z_{i+1}$,

  * $\bullet$

a _peak_ , if $z_{i-1}<z_{i}>z_{i+1}$.

In what follows, let $\mathcal{S}^{\prime}$ be the subset of all those
$z\in\mathcal{S}$, which satisfy:

  1. 1.

$z_{i-1}\not=z_{i}$ for all $i\in\\{1,2,\dots,n+1\\}$,

  2. 2.

there are no split subsequences in $z$,

  3. 3.

there is exactly one peak in $z$,

  4. 4.

there is exactly one negligible component $z_{j_{0}}$ in $z$, and $z_{j_{0}}$
is the center of the unique peak $(z_{j_{0}-1},z_{j_{0}},z_{j_{0}+1})$.

######  Proposition 4.8.

For $\alpha\geq 1$, we have

$\sup_{z\in\mathcal{S}}\Psi_{\alpha}(z)\ \ \leq\ \
\sup_{z\in\mathcal{S}^{\prime}}\Psi_{\alpha}(z).$

###### Proof.

Let us start by outlining the structure of the proof. Pick an arbitrary
$z\in\mathcal{S}$. We will gradually improve $z$ by a series of subsequent
combinatorial reductions

$z\longrightarrow z^{(1)}\longrightarrow z^{(2)}\longrightarrow
z^{(3)}\longrightarrow z^{(4)},$

such that

$\Psi_{\alpha}(z)\ \leq\ \Psi_{\alpha}(z^{(i)})\ \leq\ \Psi_{\alpha}(z^{(j)})$
for $1\leq i\leq j\leq 4$,

and $z^{(i)}$ will satisfy the requirements from $1.$ to $i.$ in the
definition of $\mathcal{S}^{\prime}$. This will give
$\Psi_{\alpha}(z)\leq\Psi_{\alpha}(z^{(4)})$ for some
$z^{(4)}\in\mathcal{S}^{\prime}$ and the claim will be proved.

1\. $z\rightarrow z^{(1)}$. Put $z=(z_{0},z_{1},\dots,z_{n+1})$. If
$z_{i-1}\not=z_{i}$ for all $i\in\\{1,2,\dots,n+1\\}$, then we are done.
Otherwise, let $i_{0}$ be the smallest index without this property. As
$z_{0}=0$ and $z_{1}$ is strictly positive, we must have $i_{0}>1$.
Analogously, we have $i_{0}<n+1$. Consequently, observe that $z_{i_{0}-1}$ and
$z_{i_{0}}$ are negligible. Examine the transformation $z\mapsto\tilde{z}$,

(4.11) $(\dots,z_{i_{0}-1},z_{i_{0}},z_{i_{0}+1},\dots)\ \longrightarrow\
w^{-1}\cdot(\dots,z_{i_{0}-1},z_{i_{0}+1},\dots),$ $w\ =\ 1-z_{i_{0}},$

which removes $z_{i_{0}}$ and rescales the remaining elements. If
$z_{i_{0}+1}\in\phi_{\alpha}(z)$, then $w^{-1}z_{i_{0}+1}$ will remain a
significant component of $\tilde{z}$. The contribution of $z_{i_{0}+1}$ (and
all the other significant components of $z$) to the overall sum will grow by a
factor of $w^{-1}>1$. The contribution of $z_{i_{0}-1}$ to $\Psi_{\alpha}(z)$
is zero and it can only increase if $z_{i_{0}-1}$ becomes significant.
Therefore $\Psi_{\alpha}(z)\leq\Psi_{\alpha}(\tilde{z})$. After a finite
number of such operations, we obtain a sequence $z^{(1)}$ for which $1.$
holds.

2\. $z^{(1)}\rightarrow z^{(2)}$. Set $z^{(1)}=(z_{i}^{(1)})_{i=0}^{n+1}$ and
suppose that $(z_{i_{0}-1}^{(1)},z_{i_{0}}^{(1)},z_{i_{0}+1}^{(1)})$ is a
split for some $i_{0}\in\\{2,3,\dots,n-1\\}$ – by the definition of split
configuration, $i_{0}$ must be greater than $1$ and smaller than $n$.
Accordingly, note that $z_{i_{0}}^{(1)}$ is negligible and consider the
preliminary modification $z^{(1)}\mapsto\hat{z}^{(1)}$ given by

$(\dots,z_{i_{0}-1}^{(1)},z_{i_{0}}^{(1)},z_{i_{0}+1}^{(1)},\dots)\
\longrightarrow\ (\dots,z_{i_{0}-1}^{(1)},0,z_{i_{0}+1}^{(1)},\dots),$

which changes $z_{i_{0}}^{(1)}$ into $0$ (so
$\hat{z}^{(1)}\not\in\mathcal{S}$: we will handle this later). As
$z_{i_{0}-1}^{(1)}>z_{i_{0}}^{(1)}$, we have

(4.12)
$\Bigg{|}\frac{z_{i_{0}-1}^{(1)}}{z_{i_{0}-2}^{(1)}+z_{i_{0}-1}^{(1)}}-\frac{z_{i_{0}-1}^{(1)}}{z_{i_{0}-1}^{(1)}+z_{i_{0}}^{(1)}}\Bigg{|}\
\ \ <\ \ \
\Bigg{|}\frac{z_{i_{0}-1}^{(1)}}{z_{i_{0}-2}^{(1)}+z_{i_{0}-1}^{(1)}}-1\Bigg{|},$

if only ${z_{i_{0}-1}^{(1)}\in\phi_{\alpha}(z^{(1)})}$. Similarly, as
$z_{i_{0}}^{(1)}<z_{i_{0}+1}^{(1)}$, we get

(4.13)
$\Bigg{|}\frac{z_{i_{0}+1}^{(1)}}{z_{i_{0}}^{(1)}+z_{i_{0}+1}^{(1)}}-\frac{z_{i_{0}+1}^{(1)}}{z_{i_{0}+1}^{(1)}+z_{i_{0}+2}^{(1)}}\Bigg{|}\
\ <\ \ \
\Bigg{|}1-\frac{z_{i_{0}+1}^{(1)}}{z_{i_{0}+1}^{(1)}+z_{i_{0}+2}^{(1)}}\Bigg{|},$

as long as ${z_{i_{0}+1}^{(1)}\in\phi_{\alpha}(z^{(1)})}$. By (4.12) and
(4.13), with a slight abuse of notation (the domain of $\Psi_{\alpha}$
formally does not contain $\hat{z}^{(1)}$, but we may extend the definition
for $\Psi_{\alpha}(\hat{z}^{(1)})$ in a straightforward way), we can write
$\Psi_{\alpha}(z^{(1)})\leq\Psi_{\alpha}(\hat{z}^{(1)})$. Now, let us denote

$\hat{z}^{(1,\leftarrow)}\ =\
(0,\hat{z}_{1}^{(1)},\dots,\hat{z}_{i_{0}-1}^{(1)},0)$

and

$\hat{z}^{(1,\rightarrow)}\ =\
(0,\hat{z}_{i_{0}+1}^{(1)},\dots,\hat{z}_{n}^{(1)},0).$

In other words, sequences $\hat{z}^{(1,\leftarrow)}$ and
$\hat{z}^{(1,\rightarrow)}$ are two consecutive parts of $\hat{z}^{(1)}$ and
we can restore $\hat{z}^{(1)}$ by glueing their corresponding zeroes together.
Moreover, after normalizing them by the weights

$w^{(1,\leftarrow)}=\sum_{i=1}^{i_{0}-1}\hat{z}_{i}^{(1)}$ and
$w^{(1,\rightarrow)}=\sum_{i=i_{0}+1}^{n}\hat{z}_{i}^{(1)}$,

we get $(w^{(1,\leftarrow)})^{-1}\hat{z}^{(1,\leftarrow)},\
(w^{(1,\rightarrow)})^{-1}\hat{z}^{(1,\rightarrow)}\in\mathcal{S}$. Next, in
this setup, we are left with

$\displaystyle\Psi_{\alpha}(\hat{z}^{(1)})\ \ \ =$ $\displaystyle\ \ \
w^{(1,\leftarrow)}\cdot\Psi_{\alpha}\left(\frac{\hat{z}^{(1,\leftarrow)}}{w^{(1,\leftarrow)}}\right)$
$\displaystyle+$ $\displaystyle\ \ \
w^{(1,\rightarrow)}\cdot\Psi_{\alpha}\left(\frac{\hat{z}^{(1,\rightarrow)}}{w^{(1,\rightarrow)}}\right)$
$\displaystyle\leq$ $\displaystyle\ \ \
\max\left\\{\Psi_{\alpha}\left(\frac{\hat{z}^{(1,\leftarrow)}}{w^{(1,\leftarrow)}}\right),\Psi_{\alpha}\left(\frac{\hat{z}^{(1,\rightarrow)}}{w^{(1,\rightarrow)}}\right)\right\\},$

where we have used $w^{(1,\leftarrow)}+w^{(1,\rightarrow)}=1$. Let

$\tilde{z}^{(1)}\ =\ \arg\max\left\\{\Psi_{\alpha}(z):\
z\in\left\\{\frac{\hat{z}^{(1,\leftarrow)}}{w^{(1,\leftarrow)}},\frac{\hat{z}^{(1,\rightarrow)}}{w^{(1,\rightarrow)}}\right\\}\right\\}.$

By the construction, we have
$\Psi_{\alpha}(z^{(1)})\leq\Psi_{\alpha}(\tilde{z}^{(1)})$, the new sequence
$\tilde{z}^{(1)}$ is shorter than $z^{(1)}$ and $\tilde{z}^{(1)}$ contains
less split configurations than $z^{(1)}$. After repeating this procedure
($z^{(1)}\mapsto\tilde{z}^{(1)}$) multiple times, we acquire a new sequence
$z^{(2)}$ obeying $1.$ and $2.$

3\. $z^{(2)}\rightarrow z^{(3)}$. Surprisingly, it is enough to put
$z^{(3)}=z^{(2)}$. Indeed, we can show that sequence $z^{(2)}$ already
satisfies the third condition. First, suppose that
$(z_{j_{0}-1}^{(2)},z_{j_{0}}^{(2)},z_{j_{0}+1}^{(2)})$ and
$(z_{j_{1}-1}^{(2)},z_{j_{1}}^{(2)},z_{j_{1}+1}^{(2)})$ are two different
peaks with indices $j_{0}<j_{1}$. Hence, as
$z_{j_{0}}^{(2)}>z_{j_{0}+1}^{(2)}$ and $z_{j_{1}-1}^{(2)}<z_{j_{1}}^{(2)}$,
there is at least one point $i_{0}\in\\{j_{0}+1,\dots,j_{1}-1\\}$ at which we
are forced to “flip” the direction of the previous inequality sign:

$z_{j_{0}-1}^{(2)}<z_{j_{0}}^{(2)}>z_{j_{0}+1}^{(2)}>\dots>z_{i_{0}}^{(2)}<\dots<z_{j_{1}-1}^{(2)}<z_{j_{1}}^{(2)}>z_{j_{1}+1}^{(2)}.$

Equivalently, this means that
$(z_{i_{0}-1}^{(2)},z_{i_{0}}^{(2)},z_{i_{0}+1}^{(2)})$ is a split
configuration. This contradicts our initial assumptions about $z^{(2)}$ (the
requirement $2.$ is not met) and proves that there is at most one peak in
$z^{(2)}$. Second, we have

$0=z_{0}^{(2)}<z_{1}^{(2)}$ and $z_{n}^{(2)}>z_{n+1}^{(2)}=0$,

so there exists a point $j_{0}$ at which the direction of the inequalities
must be changed from ‘$<$’ to ‘$>$’. Thus, there is at least one peak in
$z^{(2)}$.

4\. $z^{(3)}\rightarrow z^{(4)}$. Let $z^{(3)}=(z_{i}^{(3)})_{i=0}^{n+1}$ and
assume that $(z_{j_{0}-1}^{(3)},z_{j_{0}}^{(3)},z_{j_{0}+1}^{(3)})$ is the
unique peak of $z^{(3)}$:

(4.14)
$0<z_{1}^{(3)}<\dots<z_{j_{0}-1}^{(3)}<z_{j_{0}}^{(3)}>z_{j_{0}+1}^{(3)}>\dots>z_{n}^{(3)}>0.$

Further reasoning is similar to the previous ones (from points $1.$ and $2.$),
so we will just sketch it. If the requirement $4.$ is not satisfied, pick a
negligible component $z_{i_{0}}^{(3)}$ with $i_{0}\not=j_{0}$. Next, apply the
transformation $z^{(3)}\mapsto\tilde{z}^{(3)}$ defined by (4.11), i.e. remove
$z_{i_{0}}^{(3)}$ and rescale the remaining components. Thanks to the ‘single
peak structure’ (4.14), all the significant components of $z^{(3)}$ remain
significant for $\tilde{z}^{(3)}$. The terms associated with components
$z^{(3)}_{i}\in\phi_{\alpha}(z^{(3)})\setminus\\{z_{i_{0}-1}^{(3)},z_{i_{0}+1}^{(3)}\\}$
are not changed (and their contribution grows after the rescaling). The
summands corresponding to $z_{i_{0}-1}^{(3)}$ and $z_{i_{0}+1}^{(3)}$ can only
increase, just as in (4.12) and (4.13). Therefore
$\Psi_{\alpha}(z^{(3)})\leq\Psi_{\alpha}(\tilde{z}^{(3)})$. After several
repetitions and discarding of all unnecessary negligible components (beyond
the central $z_{j_{0}}$), we finally obtain the desired sequence
$z^{(4)}\in\mathcal{S}^{\prime}$. ∎

We proceed to the proof of our main result.

###### Proof of Theorem 1.3.

We start with the lower estimate, for which the argument is simpler. By
Proposition 4.4 and reformulation (4.7), for $\alpha>2$ we have

$\displaystyle\alpha\cdot\sup_{(X,Y)\in\mathcal{C}}\mathbb{E}|X-Y|^{\alpha}\ \
\ =$ $\displaystyle\ \ \ \alpha\cdot\sup_{z\in\mathcal{S}}\Phi_{\alpha}(z)$
$\displaystyle\geq$ $\displaystyle\ \ \
\alpha\cdot\Phi_{\alpha}\left(0,\frac{1}{\alpha},\frac{\alpha-2}{\alpha},\frac{1}{\alpha},0\right)$
$\displaystyle=$ $\displaystyle\ \ \
\alpha\cdot\frac{2}{\alpha}\left|1-\frac{1}{\alpha-1}\right|^{\alpha}\
\xrightarrow{\alpha\rightarrow\infty}\ \ \frac{2}{e}.$

Now we turn our attention to the upper estimate. By Propositions 4.6 and 4.8,
we get

$\alpha\cdot\sup_{(X,Y)\in\mathcal{C}}\mathbb{E}|X-Y|^{\alpha}\ \ \ \leq\ \ \
\alpha\cdot\Bigg{(}\Big{|}1-\frac{1}{1+\sqrt{\alpha}}\Big{|}^{\alpha}\ +\
\sup_{z\in\mathcal{S}^{\prime}}\Psi_{\alpha}(z)\Bigg{)}.$

Next, because of

$\lim_{\alpha\to\infty}\alpha\cdot\Big{|}1-\frac{1}{1+\sqrt{\alpha}}\Big{|}^{\alpha}\
=\ 0,$

it is enough to provide an asymptotic estimate for
$\alpha\cdot\sup_{z\in\mathcal{S}^{\prime}}\Psi_{\alpha}(z)$. Fix an arbitrary
$z=(z_{0},z_{1},\dots,z_{n+1})\in\mathcal{S}^{\prime}$ and let $z_{j_{0}}$ be
the center of the unique peak contained in $z$:

$0<z_{1}<\dots<z_{j_{0}-1}<z_{j_{0}}>z_{j_{0}+1}>\dots>z_{n}>0.$

As $z_{j_{0}}$ is the only negligible component contained in $z$, we have

$\sqrt{\alpha}\cdot z_{i}<z_{i+1}$ for $1\leq i\leq j_{0}-1$,

and

$z_{i-1}>\sqrt{\alpha}\cdot z_{i}$ for $j_{0}+1\leq i\leq n$.

In particular, we get $0\leq z_{j_{0}-1},z_{j_{0}+1}<1/\sqrt{\alpha}$.
Consequently, we can write $\Psi_{\alpha}(z)=A+B+C$, where

$A\ =\
\sum_{|i-j_{0}|>2}z_{i}\Big{|}\frac{z_{i}}{z_{i-1}+z_{i}}-\frac{z_{i}}{z_{i}+z_{i+1}}\Big{|}^{\alpha},$
$B\ =\
z_{i_{0}-2}\Big{|}\frac{z_{i_{0}-2}}{z_{i_{0}-3}+z_{i_{0}-2}}-\frac{z_{i_{0}-2}}{z_{i_{0}-2}+z_{i_{0}-1}}\Big{|}^{\alpha}\
+\
z_{i_{0}+2}\Big{|}\frac{z_{i_{0}+2}}{z_{i_{0}+1}+z_{i_{0}+2}}-\frac{z_{i_{0}+2}}{z_{i_{0}+2}+z_{i_{0}+3}}\Big{|}^{\alpha}$

and

$C\ =\
z_{i_{0}-1}\Big{|}\frac{z_{i_{0}-1}}{z_{i_{0}-2}+z_{i_{0}-1}}-\frac{z_{i_{0}-1}}{z_{i_{0}-1}+z_{i_{0}}}\Big{|}^{\alpha}\
+\
z_{i_{0}+1}\Big{|}\frac{z_{i_{0}+1}}{z_{i_{0}}+z_{i_{0}+1}}-\frac{z_{i_{0}+1}}{z_{i_{0}+1}+z_{i_{0}+2}}\Big{|}^{\alpha}.$

We will examine these three parts separately.

_The term $A$._ Since $z_{i}/(z_{i-1}+z_{i})$ and $z_{i}/(z_{i}+z_{i+1})$
belong to $[0,1]$, we may write

$\displaystyle A\ \ \ \leq$ $\displaystyle\ \ \ \sum_{i=1}^{j_{0}-3}z_{i}\ \
+\ \ \sum_{i=j_{0}+3}^{n}z_{i}$ $\displaystyle<$ $\displaystyle\ \ \
z_{j_{0}-3}\cdot\sum_{i=0}^{j_{0}-4}\left(\frac{1}{\sqrt{\alpha}}\right)^{i}\
\ +\ \
z_{j_{0}+3}\cdot\sum_{i=0}^{n-j_{0}-3}\left(\frac{1}{\sqrt{\alpha}}\right)^{i}$
$\displaystyle<$ $\displaystyle\ \ \
(z_{j_{0}-1}+z_{j_{0}+1})\cdot\frac{1}{\alpha}\cdot\sum_{i=0}^{\infty}\left(\frac{1}{\sqrt{\alpha}}\right)^{i}$
$\displaystyle<$ $\displaystyle\ \ \
\frac{2}{\alpha\sqrt{\alpha}}\cdot\sum_{i=0}^{\infty}\left(\frac{1}{\sqrt{\alpha}}\right)^{i}\
\ =\ \ \frac{2}{\alpha(\sqrt{\alpha}-1)}$

and hence

$\alpha\cdot A\ <\ \frac{2}{\sqrt{\alpha}-1}\
\xrightarrow{\alpha\rightarrow\infty}\ 0.$

_The term $B$._ We have

$\displaystyle B\ \ \ \leq$ $\displaystyle\ \ \
z_{i_{0}-2}\Big{|}1-\frac{z_{i_{0}-2}}{z_{i_{0}-2}+z_{i_{0}-1}}\Big{|}^{\alpha}\
\ +\ \
z_{i_{0}+2}\Big{|}\frac{z_{i_{0}+2}}{z_{i_{0}+1}+z_{i_{0}+2}}-1\Big{|}^{\alpha}$
$\displaystyle<$ $\displaystyle\ \
z_{i_{0}-2}\left|1-\frac{z_{i_{0}-2}}{z_{i_{0}-2}+\frac{1}{\sqrt{\alpha}}}\right|^{\alpha}\
\ +\ \
z_{i_{0}+2}\left|\frac{z_{i_{0}+2}}{\frac{1}{\sqrt{\alpha}}+z_{i_{0}+2}}-1\right|^{\alpha}$
$\displaystyle\leq$ $\displaystyle\ \ \
2\cdot\sup_{x\in[0,1]}x\left|1-\frac{x}{x+\frac{1}{\sqrt{\alpha}}}\right|^{\alpha}\
\ =\ \
\frac{2}{\sqrt{\alpha}(\alpha-1)}\cdot\left(1-\frac{1}{\alpha}\right)^{\alpha}.$

This yields

$\alpha\cdot B\ <\
\frac{2\sqrt{\alpha}}{\alpha-1}\cdot\left(1-\frac{1}{\alpha}\right)^{\alpha}\
\xrightarrow{\alpha\rightarrow\infty}\ 0.$

_The term $C$._ Finally, we observe that

$\displaystyle C\ \ \ \leq$ $\displaystyle\ \ \
z_{i_{0}-1}\Big{|}1-\frac{z_{i_{0}-1}}{z_{i_{0}-1}+z_{i_{0}}}\Big{|}^{\alpha}\
\ +\ \
z_{i_{0}+1}\Big{|}\frac{z_{i_{0}+1}}{z_{i_{0}}+z_{i_{0}+1}}-1\Big{|}^{\alpha}$
$\displaystyle\leq$ $\displaystyle\ \
z_{i_{0}-1}\left|1-z_{i_{0}-1}\right|^{\alpha}\ \ +\ \
z_{i_{0}+1}\left|z_{i_{0}+1}-1\right|^{\alpha}$ $\displaystyle\leq$
$\displaystyle\ \ \ 2\cdot\sup_{x\in[0,1]}x\left|1-x\right|^{\alpha}\ \ =\ \
\frac{2}{\alpha+1}\cdot\left(1-\frac{1}{\alpha+1}\right)^{\alpha}.$

Consequently, we obtain

$\alpha\cdot C\ \leq\
\frac{2\alpha}{\alpha+1}\cdot\left(1-\frac{1}{\alpha+1}\right)^{\alpha}\
\xrightarrow{\alpha\rightarrow\infty}\ \frac{2}{e}.$

The estimates for $A$, $B$ and $C$ give the desired upper bound. The proof is
complete. ∎

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