$\displaystyle G_{l,m}\left(\mathbf{a}_{l}\right)$ $\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\displaystyle\sum_{k\in\mathcal{K}_{l,m}}\dfrac{1}{a_{l,m_{\beta}\left(k\right)}-a_{l,m_{\alpha}\left(k\right)}}-\dfrac{1}{\left|\mathcal{E}\right|}\sum_{k\in\mathcal{K}_{l,m+1}}\dfrac{1}{a_{l,m_{\beta}\left(k\right)}-a_{l,m_{\alpha}\left(k\right)}}=0.$

$a_{l,m_{\beta}\left(k\right)}$$a_{l,m+1}$$a_{l,m_{\beta}\left(k\right)}$$a_{l,m-1}$$a_{l,m_{\alpha}\left(k\right)}$$a_{l,m_{\alpha}\left(k\right)}$$\mathcal{K}_{l,m}$$\mathcal{K}_{l,m+1}$$a_{l,m}$
Figure A.1: Computation of $G_{l,m}$

In Figure A.1 the terms corresponding to overlapping arrows are equal in the
in the two sums and cancel out.

Let us define the set of edges starting and terminating in $a_{l,m}$:

$\displaystyle\mathcal{K}^{\alpha}_{l,m}=\left\\{k\in\left\\{1,\ldots,\left|\mathcal{E}\right|\right\\}:m_{\alpha}\left(k\right)=m\right\\}$
$\displaystyle\mathcal{K}^{\beta}_{l,m}=\left\\{k\in\left\\{1,\ldots,\left|\mathcal{E}\right|\right\\}:m_{\beta}\left(k\right)=m\right\\}$

We obtain:

$\displaystyle G_{l,m}\left(\mathbf{a}_{l}\right)$ $\displaystyle=$
$\displaystyle\displaystyle-\dfrac{1}{\left|\mathcal{E}\right|}\sum_{k\in\mathcal{K}^{\alpha}_{l,m}}\dfrac{1}{a_{l,m_{\beta}\left(k\right)}-a_{l,m}}+\dfrac{1}{\left|\mathcal{E}\right|}\sum_{k\in\mathcal{K}^{\beta}_{l,m}}\dfrac{1}{a_{l,m}-a_{l,m_{\alpha}\left(k\right)}}$
$\displaystyle=$
$\displaystyle-\dfrac{1}{\left|\mathcal{E}\right|}\sum^{n_{l}}_{m^{\prime}=m+1}\dfrac{n^{l}_{\left(m,m^{\prime}\right)}}{a_{l,m^{\prime}}-a_{l,m}}+\dfrac{1}{\left|\mathcal{E}\right|}\sum^{m-1}_{m^{\prime}=1}\dfrac{n^{l}_{\left(m,m^{\prime}\right)}}{a_{l,m}-a_{l,m^{\prime}}}$
$\displaystyle=$
$\displaystyle-\dfrac{1}{\left|\mathcal{E}\right|}\sum^{n_{l}}_{m^{\prime}=1}\dfrac{n^{l}_{\left(m,m^{\prime}\right)}}{\left|a_{l,m^{\prime}}-a_{l,m}\right|}.$

If we call $\mathbf{F}_{l}$ and $\mathbf{G}_{l}$ the vectors of components
$F_{l,m}$ and $G_{l,m}$ ,$m\in\left\\{1,\ldots,n_{l}\right\\}$ , by (A.3) we
have $\mathbf{G}_{l}=T_{l}\mathbf{F}_{l}$ with $T_{l}$ as the
$\left[n_{l}\times n_{l}\right]$ matrix with the only non zero elements

$\displaystyle\begin{array}[]{ccccc}T_{l\,i,i}=1,&T_{l\,i,i+1}=-1\end{array}$

It is easy to show that $T_{l}$ has full rank $n_{l}$. The two systems are
then equivalent because of the linear independence of rows of $T_{l}$. Finally
by the same argument at the end of the proof of Theorem 2.1 we can get rid of
the $n_{l}$-th equation. ∎

We are ready to present the proof of of Theorem 2.2.

###### Proof.

By Lemma A.2 solutions to $\mathbf{F}_{l}=\mathbf{0}$ are also solution to
$\mathbf{G}_{l}=\mathbf{0}$. We, then, look for a function whose gradient is
equivalent to $\mathbf{G}_{l}$. This requirement can be expressed in the
language of differential forms. A differential form $\boldsymbol{\alpha}$ is
closed if its exterior derivative is zero $d\boldsymbol{\alpha}=0$. A
differential form $\boldsymbol{\alpha}$ is exact if exist a differential form
$\boldsymbol{\beta}$ of degree less than the one of $\boldsymbol{\alpha}$ such
that $\boldsymbol{\alpha}=d\boldsymbol{\beta}$. If the $1$-form:

$\displaystyle\boldsymbol{\gamma}_{l}=\sum^{n_{l}-1}_{m=2}G_{l,m}\left(\mathbf{a}_{l}\right)da_{l,m}$
(81)

is exact in some domain, there exists a $0$-form i.e. a smooth function
$\Psi_{l}$ such that

$\displaystyle d\Psi_{l}=\sum^{n_{l}-1}_{m=2}\dfrac{\partial\Psi_{l}}{\partial
a_{l,m}}\left(\mathbf{a}_{l}\right)da_{l,m}=\boldsymbol{\gamma}_{l}$ (82)

i.e.

$\displaystyle G_{l,m}$ $\displaystyle=$
$\displaystyle\dfrac{\partial\Psi_{l}}{\partial
a_{l,m}}\left(\mathbf{a}_{l}\right),\quad
m\in\left\\{2,\ldots,n_{l}-1\right\\}.$ (83)

Let us pick an open ball $B$ in $\mathbb{R}^{n_{l}-2}$ not including the
points $\left(a_{l,j}=a_{l,i}\right)$, $i,j\in\left\\{1,\ldots,n_{l}\right\\}$
where $\mathbf{G}_{l}$ is not defined. By Poincare lemma if
$\boldsymbol{\gamma}_{l}$ is closed in $B$ is also exact in $B$. Thus if

$\displaystyle
d\boldsymbol{\gamma}_{l}=\sum^{n_{l}-1}_{m,m^{\prime}=2}\left\\{\dfrac{\partial
G_{l,m}}{\partial a_{l,m^{\prime}}}-\dfrac{\partial G_{l,m^{\prime}}}{\partial
a_{l,m}}\right\\}da_{l,m}da_{l,m^{\prime}}=0$

then

$\displaystyle\left\\{\dfrac{\partial G_{l,m}}{\partial
a_{l,m^{\prime}}}-\dfrac{\partial G_{l,m^{\prime}}}{\partial
a_{l,m}}\right\\}=0,$

and there exists $\Psi_{l}$ such that $G_{l,m}$ is its gradient. We have:

$\displaystyle H_{i,i}=\dfrac{\partial G_{l,i}}{\partial a_{l,i}}$
$\displaystyle=$
$\displaystyle\displaystyle\sum^{n_{l}}_{m=1}\dfrac{n^{l}_{\left(i,m\right)}}{\left(a_{l,m}-a_{l,i}\right)^{2}}$
$\displaystyle H_{i,j}=\dfrac{\partial G_{l,i}}{\partial a_{l,j}}$
$\displaystyle=$
$\displaystyle-\displaystyle\dfrac{n^{l}_{\left(i,j\right)}}{\left(a_{l,i}-a_{l,j}\right)^{2}}$

$i\neq j\in\left\\{2,\ldots,n_{l}-1\right\\}$. Since
$n^{l}_{\left(i,j\right)}=n^{l}_{\left(j,i\right)}$ $\boldsymbol{\gamma}_{l}$
is exact whenever $\mathbf{G}_{l}$ is defined and $\Psi_{l}$ exists. After
some algebra we obtain:

$\displaystyle\Psi_{l}\left(\mathbf{a}_{l}\right)$ $\displaystyle=$
$\displaystyle-\dfrac{1}{2\left|\mathcal{E}\right|}\displaystyle\sum^{n_{l}}_{m,m^{\prime}=1}n^{l}_{\left(m,m^{\prime}\right)}\log\left|a_{l,m^{\prime}}-a_{l,m}\right|.$

Since $H$ is a symmetric diagonally dominant matrix, it is everywhere semi-
positive definite, this is equivalent to the convexity of
$\Psi_{l}\left(\mathbf{a}_{l}\right)$. In addition, we are restricting the
domain to the unit hypecube that is a convex domain so the problem is a convex
one.

∎

###  A.4 Proof of Theorem 2.3

###### Proof.

In order to apply Radon–Nikodym theorem $\mathbb{P}$ needs to be absolutely
continuous with respect to $\mathbb{Q}$. This assumption amounts to impose the
restrictions

$\displaystyle
a_{l,1}=\min_{k\in\left\\{1,\ldots,\left|\mathcal{E}\right|\right\\}}\alpha_{l,k}=0$

and

$\displaystyle
a_{l,n_{l}}=\max_{k\in\left\\{1,\ldots,\left|\mathcal{E}\right|\right\\}}\beta_{l,k}=1$

(Assumption 2). We have:

$\displaystyle\mathbb{P}\left(u_{l}\in A,k\in B\right)$ $\displaystyle=$
$\displaystyle\sum_{k\in B}\int_{u_{l}\in
A}\dfrac{1}{\left|\mathcal{E}\right|}f\left(u_{l}\right.\left|K=k\right)du_{l}$
$\displaystyle=$ $\displaystyle\sum_{k\in B}\int_{u_{l}\in
A}\dfrac{d\mathbb{P}}{d\mathbb{Q}}\left(u_{l},k\right)\dfrac{1}{\left|\mathcal{E}\right|}\mathbb{I}_{\left[0,1\right]}\left(u_{l}\right)du_{l}$

By the condition on the a’s above, we have
$f\left(u_{l}\right|\left.K=k\right)=f\left(u_{l}\right|\left.K=k\right)\mathbb{I}_{\left[0,1\right]}\left(u_{l}\right)$
and we can read off the derivative :

$\displaystyle\dfrac{d\mathbb{P}}{d\mathbb{Q}}\left(u_{l},k\right)=f\left(u_{l}\right.\left|K=k\right)$
$\displaystyle D_{KL}\left(\mathbb{P}||\mathbb{Q}\right)$ $\displaystyle=$
$\displaystyle\mathbb{E}_{\mathbb{P}}\left[\log\left(\dfrac{d\mathbb{P}}{d\mathbb{Q}}\left(u_{l},k\right)\right)\right]$
$\displaystyle=$
$\displaystyle\displaystyle\sum^{\left|\mathcal{E}\right|}_{k=1}\int^{1}_{0}\log\left(f\left(u_{l}\right.\left|K=k\right)\right)\dfrac{1}{\left|\mathcal{E}\right|}f\left(u_{l}\right.\left|K=k\right)du_{l}$
$\displaystyle=$
$\displaystyle\displaystyle\sum^{\left|\mathcal{E}\right|}_{k=1}\int^{1}_{0}\log\left(\dfrac{1}{\left(\beta_{lk}-\alpha_{lk}\right)}\mathbb{I}_{\left\\{\left[\alpha_{lk},\beta_{lk}\right]\right\\}}\left(u_{l}\right)\right)\dfrac{1}{\left|\mathcal{E}\right|}\dfrac{1}{\left(\beta_{lk}-\alpha_{lk}\right)}\mathbb{I}_{\left\\{\left[\alpha_{lk},\beta_{lk}\right]\right\\}}\left(u_{l}\right)du_{l}$
$\displaystyle=$
$\displaystyle\displaystyle\sum^{\left|\mathcal{E}\right|}_{k=1}\int^{1}_{0}\log\left(\dfrac{1}{\left(\beta_{lk}-\alpha_{lk}\right)}\right)\dfrac{1}{\left|\mathcal{E}\right|}\dfrac{1}{\left(\beta_{lk}-\alpha_{lk}\right)}\mathbb{I}_{\left\\{\left[\alpha_{lk},\beta_{lk}\right]\right\\}}\left(u_{l}\right)du_{l}$
$\displaystyle+$
$\displaystyle\displaystyle\sum^{\left|\mathcal{E}\right|}_{k=1}\int^{1}_{0}\log\left(\mathbb{I}_{\left\\{\left[\alpha_{lk},\beta_{lk}\right]\right\\}}\left(u_{l}\right)\right)\dfrac{1}{\left|\mathcal{E}\right|}\dfrac{1}{\left(\beta_{lk}-\alpha_{lk}\right)}\mathbb{I}_{\left\\{\left[\alpha_{lk},\beta_{lk}\right]\right\\}}\left(u_{l}\right)du_{l}$

As usually done in this kind of computation we assume by convention that
$0\log\left(0\right)=0$, then:

$\displaystyle D_{KL}\left(\mathbb{P}||\mathbb{Q}\right)$ $\displaystyle=$
$\displaystyle\displaystyle\sum^{\left|\mathcal{E}\right|}_{k=1}\int^{1}_{0}\log\left(\dfrac{1}{\left(\beta_{lk}-\alpha_{lk}\right)}\right)\dfrac{1}{\left|\mathcal{E}\right|}\dfrac{1}{\left(\beta_{lk}-\alpha_{lk}\right)}\mathbb{I}_{\left\\{\left[\alpha_{lk},\beta_{lk}\right]\right\\}}\left(u_{l}\right)du_{l}$
$\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\displaystyle\sum^{\left|\mathcal{E}\right|}_{k=1}-\log\left(\left(\beta_{lk}-\alpha_{lk}\right)\right)$
$\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\displaystyle\sum^{\left|\mathcal{E}\right|}_{k=1}-\log\left(\left(a_{l,m_{\beta}\left(k\right)}-a_{l,m_{\alpha}\left(k\right)}\right)\right)$
$\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\displaystyle\sum_{m>m^{\prime}}-n_{\left(m,m^{\prime}\right)}\log\left(a_{l,m}-a_{l,m^{\prime}}\right)$
$\displaystyle=$
$\displaystyle\dfrac{1}{2\left|\mathcal{E}\right|}\displaystyle\sum^{n_{l}}_{m,m^{\prime}=1}-n_{\left(m,m^{\prime}\right)}\log\left|a_{l,m}-a_{l,m^{\prime}}\right|.$

∎

###  A.5 Proof of Lemma 2.2

###### Proof.

Since $\mathbf{U}$ is uniform on $\mathcal{S}^{(x)}$ the random variables
$\left\\{U_{l},l\in\mathcal{D}\right\\}$ are standard uniforms, and since
$\mathbf{Y}$ is a solution to the standard uniform on
$\mathcal{S}^{(y)}=(\mathcal{G}^{(y)},\mathbf{Y})$, the
$\left\\{W_{l},l\in\mathcal{D}\right\\}$ are standard uniform as well.
$W_{l}=\left(y_{lI}x_{lI^{\prime}}+y_{lJ}\left(1-x_{lI^{\prime}}\right)\right)V\\\
+\left(y_{lI}x_{lJ^{\prime}}+y_{lJ}\left(1-x_{lJ^{\prime}}\right)\right)\left(1-V\right)$
and $\mathbf{W}$ and $\left(I^{\prime},J^{\prime}\right)$ are sampled with
uniform probability from $\mathcal{E}^{1}$ ∎

###  A.6 Proof of Lemma2.3

###### Proof.

$\displaystyle\dfrac{d}{2}=\sum^{d}_{l=1}W_{l}$ $\displaystyle=$
$\displaystyle y_{lI}U_{l}+\left(1-U_{l}\right)y_{lJ}$ $\displaystyle=$
$\displaystyle\sum^{d}_{l=1}\left(y_{lI}-y_{lJ}\right)U_{l}+\sum^{d}_{l=1}y_{lJ}$
$\displaystyle=$ $\displaystyle
c_{1}\sum^{d}_{l=1}U_{l}+c_{2}=\dfrac{d}{2}c_{1}+c_{2}$

and $\sum^{d}_{l=1}W_{l}=\dfrac{d}{2}$ if

$\left(1-c_{1}\right)\dfrac{d}{2}=c_{2}.$ (84)

∎

###  A.7 Proof of Lemma 2.4

###### Proof.

The coordinates of the vertexes (i.e. the columns of
$\left(\mathbf{X}^{1},\mathbf{X}^{2}\right)$) inherit the constant sum from
(29) and allow us to define the following subsets of $\mathcal{E}^{t}$

$\displaystyle\mathcal{K}^{1}_{u_{l}}$ $\displaystyle\equiv$
$\displaystyle\left\\{k\in\left\\{1,\ldots,\left|\mathcal{E}^{1}\right|\right\\}:u_{l}\in\left[a_{l,m_{\alpha}\left(k\right)},a_{l,m_{\beta}\left(k\right)}\right]\right\\}$
$\displaystyle\mathcal{K}^{2}_{u_{l}}$ $\displaystyle\equiv$
$\displaystyle\left\\{k\in\left\\{1,\ldots,\left|\mathcal{E}^{1}\right|\right\\}:u_{l}\in\left[a_{l,m_{\alpha}\left(k\right)},a_{l,m_{\beta}\left(k\right)}\right]\right\\}$
$\displaystyle\mathcal{K}_{u_{l}}$ $\displaystyle\equiv$
$\displaystyle\left\\{k\in\left\\{1,\ldots,\left|\mathcal{E}^{1}\sqcup\mathcal{E}^{2}\right|\right\\}:u_{l}\in\left[a_{l,m_{\alpha}\left(k\right)},a_{l,m_{\beta}\left(k\right)}\right]\right\\}$
$\displaystyle=$
$\displaystyle\mathcal{K}^{1}_{u_{l}}\sqcup\mathcal{K}^{2}_{u_{l}}.$

The sufficient conditions for standard marginal uniformity on
$\mathcal{S}^{\prime}$ given in (21) become

$\displaystyle\dfrac{1}{\left|\mathcal{E}^{1}\sqcup\mathcal{E}^{2}\right|}\displaystyle\sum_{k\in\mathcal{K}_{u_{l}}}\dfrac{1}{a_{l,m_{\beta}\left(k\right)}-a_{l,m_{\alpha}\left(k\right)}}=$
$\displaystyle=$
$\displaystyle\dfrac{\left|\mathcal{E}^{1}\right|}{\left|\mathcal{E}^{1}\sqcup\mathcal{E}^{2}\right|}\displaystyle\dfrac{1}{\left|\mathcal{E}^{1}\right|}\sum_{k\in\mathcal{K}^{1}_{u_{l}}}\dfrac{1}{a_{l,m_{\beta}\left(k\right)}-a_{l,m_{\alpha}\left(k\right)}}$
$\displaystyle+$
$\displaystyle\dfrac{\left|\mathcal{E}^{2}\right|}{\left|\mathcal{E}^{1}\sqcup\mathcal{E}^{2}\right|}\displaystyle\dfrac{1}{\left|\mathcal{E}^{2}\right|}\sum_{k\in\mathcal{K}^{2}_{u_{l}}}\dfrac{1}{a_{l,m_{\beta}\left(k\right)}-a_{l,m_{\alpha}\left(k\right)}}$
$\displaystyle=$
$\displaystyle\dfrac{\left|\mathcal{E}^{1}\right|+\left|\mathcal{E}^{2}\right|}{\left|\mathcal{E}^{1}\sqcup\mathcal{E}^{2}\right|}=1.$

∎

###  A.8 Proof of Lemma 2.5

###### Proof.

The random vector $\mathbf{W}\in[0,1]^{d}$ with elements

$\displaystyle W_{1}$ $\displaystyle=$ $\displaystyle
x_{{\pi\left(1\right)}I}V+\left(1-V\right)x_{{\pi\left(1\right)}J}$ (85)
$\displaystyle\vdots$ $\displaystyle W_{d}$ $\displaystyle=$ $\displaystyle
x_{{\pi\left(d\right)}I}V+\left(1-V\right)x_{{\pi\left(d\right)}J}$

defines a permutation of $\mathbf{U}$. If $\mathbf{U}$ is strict $d$-CTM on
$\mathcal{S}=(\mathcal{G},\mathbf{X})$ then the permuted vector is strict
$d$-CTM on $\mathcal{S}^{\prime}=(\mathcal{G},P_{\pi}\mathbf{X})$, because the
sum is preserved by permutations. ∎

###  A.9 Proof of Corollary 1

###### Proof.

Follows the same lines as the proof of Theorem 2.1 by replacing the common $V$
by the coordinate-specific $V_{l}\sim\mathcal{U}[0,1]$. ∎

###  A.10 Proof of Theorem 3.1

###### Proof.

Let us define the random variables

$\displaystyle V_{l,k}$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ccc}V_{l}&&\hbox{if
}l\in\mathcal{L}^{+}_{k}\\\ &&\\\ 1-V_{l}&&\hbox{if
}l\in\mathcal{L}^{-}_{k}\end{array}\right.,$ (89)

the event $\left\\{U_{l}\leq u_{l}\left|K=k\right.\right\\}$ can be rewritten

$\displaystyle\left\\{U_{l}\leq u_{l}\left|K=k\right.\right\\}$
$\displaystyle=$
$\displaystyle\left\\{x_{li\left(k\right)}V_{l}+\left(1-V_{l}\right)x_{lj\left(k\right)}\leq
u_{l}\left|K=k\right.\right\\}$ $\displaystyle=$
$\displaystyle\left\\{V_{l,k}\leq
F_{U_{l}\left|K\right.}\left(u_{l};k\right)\left|K=k\right.\right\\}$

Let $v_{l,k}=F_{U_{l}\left|K\right.}\left(u_{l};k\right)$ for $l=1,\ldots,d$,
$k=1,\ldots,\left|\mathcal{E}\right|$ then:

$\displaystyle\mathbb{P}\left(U_{1}\leq u_{1},\ldots,U_{d}\leq
u_{d}\left|K=k\right.\right)=\mathbb{P}\left(V_{1,k}\leq
v_{1,k},\ldots,V_{d,k}\leq v_{d,k}\left|K=k\right.\right)$ $\displaystyle=$
$\displaystyle\mathbb{P}\left(\left.\left(\bigcap_{l\in\mathcal{L}^{-}_{k}}\left\\{\left(1-V_{l}\right)\leq
v_{l,k}\right\\}\right)\cap\left(\bigcap_{l\in\mathcal{L}^{+}_{k}}\left\\{V_{l}\leq
v_{l,k}\right\\}\right)\right|K=k\right)$ $\displaystyle=$
$\displaystyle\mathbb{E}\left[\prod_{l\in\mathcal{L}^{-}_{k}}\mathbb{I}_{\left(1-v_{l,k},1\right]}\left(V_{l}\right)\prod_{l\in\mathcal{D}\setminus\mathcal{L}^{-}_{k}}\mathbb{I}_{\left[0,v_{l,k}\right]}\left(V_{l}\right)\right]$
$\displaystyle=$
$\displaystyle\mathbb{P}\left(R_{\mathcal{L}^{-}_{k},\frac{1}{2}}(\mathbf{V})\leq\mathbf{v}_{k}\right).$

∎

###  A.11 Proof of Corollary 3

###### Proof.

Assuming $V_{l}=V$ for $l=1,\ldots,d$ in the proof of Theorem 3.1, it follows
that

$\displaystyle\mathbb{P}\left(U_{1}\leq u_{1},\ldots,U_{d}\leq
u_{d}\left|K=k\right.\right)=\mathbb{P}\left(V_{1,k}\leq
v_{1,k},\ldots,V_{d,k}\leq v_{d,k}\left|K=k\right.\right)$ $\displaystyle=$
$\displaystyle\mathbb{P}\left(\left.\left(\bigcap_{l\in\mathcal{L}^{-}_{k}}\left\\{\left(1-V\right)\leq
v_{l,k}\right\\}\right)\cap\left(\bigcap_{l\in\mathcal{L}^{+}_{k}}\left\\{V\leq
v_{l,k}\right\\}\right)\right|K=k\right)$ $\displaystyle=$
$\displaystyle\mathbb{P}\left(1-v^{-}_{k}<V\leq
v^{+}_{k}\left|K=k\right.\right)$ $\displaystyle=$
$\displaystyle\mathbb{I}_{\left[0,v^{+}_{k}\right]}\left(1-v^{-}_{k}\right)\int^{v^{+}_{k}}_{1-v^{-}_{k}}dv$
$\displaystyle=$ $\displaystyle\max\left(v^{+}_{k}+v^{-}_{k}-1,0\right),$

where in (A.11) we used the identity
$\displaystyle\max_{l\in\mathcal{L}^{-}_{k}}\left(1-v_{l,k}\right)=1-\displaystyle\min_{l\in\mathcal{L}^{-}_{k}}v_{l,k}$

The expression for the bivariate conditional distributions is obtained by
substituting $u_{l^{\prime\prime}}=1$ in
$v_{l^{\prime\prime},k}=F_{U_{l^{\prime\prime}}\left|K\right.}\left(u_{l^{\prime\prime}};k\right)$
for all $l^{\prime\prime}\neq l^{\prime}$ and $l^{\prime\prime}\neq l$ where
$l^{\prime\prime},l^{\prime},l\in\mathcal{D}$, thus obtaining:

$\displaystyle
v^{+}_{k}=\min\left(v_{l,k},v_{l^{\prime},k}\right),v^{-}_{k}=1$
$\displaystyle\hbox{ if }l^{\prime},l\in\mathcal{L}^{+}_{k}$ $\displaystyle
v^{-}_{k}=\min\left(v_{l,k},v_{l^{\prime},k}\right),v^{+}_{k}=1$
$\displaystyle\hbox{ if }l^{\prime},l\in\mathcal{L}^{-}_{k}$ $\displaystyle
v^{+}_{k}=v_{l,k},v^{-}_{k}=v_{l^{\prime},k}$ $\displaystyle\hbox{ if
}l^{\prime}\in\mathcal{L}^{-}_{k},l\in\mathcal{L}^{+}_{k}$ $\displaystyle
v^{-}_{k}=v_{l,k},v^{+}_{k}=v_{l^{\prime},k}$ $\displaystyle\hbox{ if
}l^{\prime}\in\mathcal{L}^{+}_{k},l\in\mathcal{L}^{-}_{k}.$

∎

###  A.12 Proof of Proposition 3.1

###### Proof.

By corollary 2 we obtain:

$\displaystyle\mathbb{P}\left(\mathbf{U}\leq\mathbf{W}\right)=\dfrac{1}{\left|\mathcal{E}\right|}\sum^{\left|\mathcal{E}\right|}_{k_{\mathbf{U}}=1}\mathbb{P}\left(R_{\mathbf{V},\mathcal{L}^{-}_{k_{\mathbf{U}}}}\left(\mathbf{V}\right)\leq\mathbf{Y}_{k_{\mathbf{U}}}\right)$

with
$\mathbf{Y}_{k_{\mathbf{U}}}=\left(F_{U_{1}\left|K_{\mathbf{U}}\right.}\left(W_{1};k_{\mathbf{U}}\right),\ldots,F_{U_{d}\left|K_{\mathbf{U}}\right.}\left(W_{d};k_{\mathbf{U}}\right)\right)$.

∎

###  A.13 Proof of Proposition 3.2

###### Proof.

Using the definition of the reflection
$R_{\mathcal{L},\frac{1}{2}}\left(\mathbf{V}\right)$, we obtain:

$\displaystyle\rho\left(F_{\mathbf{V},\mathcal{D}\setminus\mathcal{L}}\right)$
$\displaystyle=$
$\displaystyle\dfrac{2^{d}\left(d+1\right)}{2^{d}-\left(d+1\right)}\left(\mathbb{E}\left[\prod_{l\in\mathcal{L}}V_{l}\prod_{l\in\mathcal{D}\setminus\mathcal{L}}\left(1-V_{l}\right)\right]-\dfrac{1}{2^{d}}\right).$

For our general representation in (36) the expected value of the product is:

$\displaystyle\mathbb{E}\left[\prod^{d}_{l=1}U_{l}\right]$ $\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\sum^{\left|\mathcal{E}\right|}_{k=1}\mathbb{E}\left[\left.\prod^{d}_{l=1}U_{l}\right|K=k\right]=\dfrac{1}{\left|\mathcal{E}\right|}\sum^{\left|\mathcal{E}\right|}_{k=1}\mathbb{E}\left[\left.\prod^{d}_{l=1}U_{l}\right|K=k\right]$
$\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\sum^{\left|\mathcal{E}\right|}_{k=1}\mathbb{E}\left[\prod^{d}_{l=1}\left(x_{l,i\left(k\right)}V_{l}+x_{l,j\left(k\right)}\left(1-V_{l}\right)\right)\right]$
$\displaystyle=$
$\displaystyle\sum^{d}_{m=0}\sum_{\tiny\begin{array}[]{c}\mathcal{L}_{m}\subseteq\mathcal{D}\\\
\left|\mathcal{L}_{m}\right|=m\end{array}}\xi_{\mathcal{L}_{m}}\mathbb{E}\left[\prod_{l\in\mathcal{L}_{m}}V_{l}\prod_{l\in\mathcal{D}\setminus\mathcal{L}_{m}}\left(1-V_{l}\right)\right],$

with

$\xi_{\mathcal{L}_{m}}=\dfrac{1}{\left|\mathcal{E}\right|}\sum^{\left|\mathcal{E}\right|}_{k=1}\left(\prod_{l\in\mathcal{L}_{m}}x_{l,i\left(k\right)}\prod_{l\in\mathcal{D}\setminus\mathcal{L}_{m}}x_{l,j\left(k\right)}\right).$

The Spearman’s $\rho\left(F_{\mathbf{U}}\right)$ can be expressed as a
function of the $\rho$s on the different
$F_{\mathbf{V},\mathcal{D}\setminus\mathcal{L}_{m}}$ :

$\displaystyle\rho\left(F_{\mathbf{U}}\right)$ $\displaystyle=$
$\displaystyle\sum^{d}_{m=0}\sum_{\tiny\begin{array}[]{c}\mathcal{L}_{m}\subseteq\mathcal{D}\\\
\left|\mathcal{L}_{m}\right|=m\end{array}}\xi_{\mathcal{L}_{m}}\rho\left(F_{\mathbf{V},\mathcal{D}\setminus\mathcal{L}_{m}}\right)+\dfrac{2^{d}\left(d+1\right)}{2^{d}-\left(d+1\right)}\dfrac{1}{2^{d}}\left(\xi^{*}-1\right)$
(94)

with

$\displaystyle\xi^{*}$ $\displaystyle=$
$\displaystyle\sum^{d}_{m=0}\sum_{\tiny\begin{array}[]{c}\mathcal{L}_{m}\subseteq\mathcal{D}\\\
\left|\mathcal{L}_{m}\right|=m\end{array}}\xi_{\mathcal{L}_{m}}=\dfrac{1}{\left|\mathcal{E}\right|}\sum^{\left|\mathcal{E}\right|}_{k=1}\prod^{d}_{l=1}\left(x_{l,i\left(k\right)}+x_{l,j\left(k\right)}\right).$

If $\mathbf{V}$ is reflection invariant then
$\bar{F}_{\mathbf{V},\mathcal{D}\setminus\mathcal{L}_{m}}=\bar{F}_{\mathbf{V}}$
in eq. (A.13). The condition $0<\xi^{*}\leq 1$ implies
$\underline{\rho}\leq\rho\left(F_{\mathbf{U}}\right)<\bar{\rho}$ where
$\underline{\rho}=\min\\{\rho\left(F_{\mathbf{V}}\right),-(d+1)/(2^{d}-\left(d+1\right))\\}$
and
$\bar{\rho}=\max\\{\rho\left(F_{\mathbf{V}}\right),-(d+1)/(2^{d}-\left(d+1\right))\\}$.
Since $\rho_{\min}\leq\rho\left(F_{\mathbf{V}}\right)$ if we are able to show
that $-(d+1)/(2^{d}-(d+1))\leq\rho_{\min}$ we obtain the result. Let
$\mathbf{V}^{\ast}$ such that $\rho(F_{\mathbf{V}^{\ast}})=\rho_{\min}$

$\displaystyle-\dfrac{\left(d+1\right)}{2^{d}-\left(d+1\right)}\leq\dfrac{2^{d}\left(d+1\right)}{2^{d}-\left(d+1\right)}\left(\mathbb{E}\left[\prod^{d}_{l=1}V^{\ast}_{l}\right]-\dfrac{1}{2^{d}}\right)$
(96) $\displaystyle\Longleftrightarrow
2^{d}\mathbb{E}\left[\prod^{d}_{l=1}V^{\ast}_{l}\right]\geq 0$ (97)

which is satisfied since $V_{l}^{\ast}$ is a.s. positive for all
$l=1,\ldots,d$. ∎

###  A.14 Proof of Proposition 4.1

###### Proof.

$U_{1}\sim\mathcal{U}[0,1]$ and

$\displaystyle U_{l}$ $\displaystyle=$
$\displaystyle\gamma_{l}-\dfrac{U_{1}}{d-1},\quad\gamma_{l}=\dfrac{l-1}{d-1}$

or

$\displaystyle U_{l}$ $\displaystyle=$
$\displaystyle\dfrac{l}{d-1}U_{1}+\dfrac{l-1}{d-1}\left(1-U_{1}\right)\sim\mathcal{U}\left[\dfrac{l-1}{d-1},\dfrac{l}{d-1}\right].$

It could be verified that $U_{l}$ has the same distribution of extracting on
the edge $\left(d,1\right)\in C_{d}\left(\left\\{1\right\\}\right)$ . Random
permutation permute simultaneously the coordinate of the vertexes that remain
one the first circular permutation of the other , The exchangeable version
then has uniform support on the edges that start from a permutation of the
first vertex and link the vertex resulting from an addition first circular
permutation. This is exactly the support of the the exchangeable version of
CCV with dependence graph $C_{d}\left(\left\\{1\right\\}\right)$ ∎

###  A.15 Proof of Proposition 4.2

###### Proof.

Let $m-1=\left\lfloor dU\right\rfloor\in\left[0,\ldots,d-1\right]$ and
$V=dU-\left\lfloor
dU\right\rfloor=\left(dU\right)\mathrm{mod}\,1\sim\mathcal{U}[0,1]$ . We can
write (1.1) as:

$\displaystyle U_{l}$ $\displaystyle=$
$\displaystyle\left(\dfrac{l-1+m-1+V}{d}\right)\mathrm{mod}\,1,$
$\displaystyle=$
$\displaystyle\left(\left(\dfrac{l+m-2}{d}\right)\mathrm{mod}\,1+\dfrac{V}{d}\right)\mathrm{mod}\,1$
$\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ccc}\dfrac{l+m-2}{d}+\dfrac{V}{d}&\hbox{if}&m<d+2-l\\\
\dfrac{l+m-2-d}{d}+\dfrac{V}{d}&\hbox{if}&m\geq d+2-l\end{array}\right.$
$\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ccc}\dfrac{l+m-1}{d}V+\dfrac{l+m-2}{d}\left(1-V\right)&\hbox{if}&m<d+2-l\\\
\dfrac{l+m-1-d}{d}V+\dfrac{l+m-2-d}{d}\left(1-V\right)&\hbox{if}&m\geq
d+2-l\end{array}\right.$

∎

###  A.16 Proof of Corollary 6

###### Proof.

The unique values for coordinates over the $l$-th dimension are:

$\displaystyle n_{l}$ $\displaystyle=$ $\displaystyle d+1$
$\displaystyle\mathbf{a}_{l}$ $\displaystyle=$
$\displaystyle\dfrac{1}{d}\left(0,\ldots,d\right)^{T}$

In the projected graph, only nearby vertexes are connected by edges, the
difference in projected coordinates joined is constant and the number of edges
is $d$. We have:

$\displaystyle n^{l}_{\left(m,m^{\prime}\right)}$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ccc}1&&\left|m-m^{\prime}\right|=1\\\
&&\\\ 0&&\hbox{otherwise}\end{array}\right.$ $\displaystyle\mathcal{K}_{l,m}$
$\displaystyle=$ $\displaystyle\left(m-1,m\right)$ $\displaystyle
a_{lm}-a_{l,m-1}$ $\displaystyle=$ $\displaystyle\dfrac{1}{d}$

with $m,m^{\prime}=1,\ldots,n_{l}$. Given those preliminary computation we are
ready to show standard marginal uniformity using the conditions in Theorem
2.1:

$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\displaystyle\sum_{k\in\mathcal{K}_{l,m}}\dfrac{1}{a_{l,m_{\beta}\left(k\right)}-a_{l,m_{\alpha}\left(k\right)}}=\dfrac{1}{\left|\mathcal{E}\right|}\dfrac{n^{l}_{\left(m,m^{\prime}\right)}}{a_{lm}-a_{l,m-1}}=\frac{d}{d}=1$

Concerning the constant sum constraint, it is sufficent to sum over $l$ ,
$x_{l,m}$.

$\displaystyle\sum^{d}_{l=1}x_{l,m}$ $\displaystyle=$
$\displaystyle\sum^{d+2-m-1}_{l=1}\dfrac{l+m-1}{d}+\sum^{d}_{l=d+2-m}\dfrac{l+m-1-d}{d}$
$\displaystyle=$ $\displaystyle\dfrac{d+1}{2}.$

∎

###  A.17 Proof of Proposition 4.3

###### Proof.

In the $3$-dimensional case the proposal of Arvidsen and Johnsson, (1982) is

$\displaystyle U_{1}$ $\displaystyle=$ $\displaystyle U\qquad
U\sim\mathcal{U}[0,1]$ $\displaystyle U_{2}$ $\displaystyle=$
$\displaystyle\left\\{U+1/2\right\\}$ $\displaystyle U_{3}$ $\displaystyle=$
$\displaystyle 1-\left\\{2U\right\\}$

Case $U\in\left[0,1/2\right]$

$\displaystyle U_{1}$ $\displaystyle=$ $\displaystyle U$ $\displaystyle U_{2}$
$\displaystyle=$ $\displaystyle U+1/2$ $\displaystyle U_{3}$ $\displaystyle=$
$\displaystyle 1-2U$

Case $U\in\left[1/2,1\right]$

$\displaystyle U_{1}$ $\displaystyle=$ $\displaystyle U$ $\displaystyle U_{2}$
$\displaystyle=$ $\displaystyle U-1/2$ $\displaystyle U_{3}$ $\displaystyle=$
$\displaystyle 1-\left(2U-1\right)$

The construction in Eq (2.4) of Rüschendorf and Uckelmann, (2002) is obtained
by setting $W_{l}=-1+2U_{l}$. ∎

###  A.18 Proof of Proposition 4.4

###### Proof.

Let $V=\left\\{b^{d-2}U_{1}\right\\}$ and $m=\left\lfloor
b^{d-2}U_{1}\right\rfloor$. Then $V$ is a standard uniform random variable and
$m$ takes values on the integers $\left\\{0,\ldots,d-1\right\\}$ with equal
probabilities. We have:

$\displaystyle U_{1}$ $\displaystyle=$
$\displaystyle\dfrac{V}{b^{d-2}}+\dfrac{m}{b^{d-2}}=V\dfrac{m+1}{b^{d-2}}+\left(1-V\right)\dfrac{m}{b^{d-2}}$
$\displaystyle=$ $\displaystyle Vz_{1,m+1}+\left(1-V\right)y_{1,m+1}.$

For $i=2,\ldots,d-1$:

$\displaystyle U_{i}$ $\displaystyle=$
$\displaystyle\left\\{b^{i-2}\left(V\dfrac{m+1}{b^{d-2}}+\left(1-V\right)\dfrac{m}{b^{d-2}}\right)+1/2+1/2V-1/2V\right\\}$
$\displaystyle=$
$\displaystyle\left\\{V\left(b^{i-2}\dfrac{m+1}{b^{d-2}}+1/2\right)+\left(1-V\right)\left(b^{i-2}\dfrac{m}{b^{d-2}}+1/2\right)\right\\}$
$\displaystyle=$ $\displaystyle\left\\{V\left(\left\lfloor
b^{i-2}\dfrac{m}{b^{d-2}}+1/2\right\rfloor+\left\\{b^{i-2}\dfrac{m}{b^{d-2}}+1/2\right\\}+\dfrac{b^{i-2}}{b^{d-2}}\right)\right.$
$\displaystyle+$ $\displaystyle\left.\left(1-V\right)\left(\left\lfloor
b^{i-2}\dfrac{m}{b^{d-2}}+1/2\right\rfloor+\left\\{b^{i-2}\dfrac{m}{b^{d-2}}+1/2\right\\}\right)\right\\}$
$\displaystyle=$
$\displaystyle\left\\{V\left(\left\\{b^{i-2}\dfrac{m}{b^{d-2}}+1/2\right\\}+\dfrac{b^{i-2}}{b^{d-2}}\right)+\left(1-V\right)\left(\left\\{b^{i-2}\dfrac{m}{b^{d-2}}+1/2\right\\}\right)\right\\}$
$\displaystyle=$ $\displaystyle Vz_{i,m+1}+\left(1-V\right)y_{i,m+1},\quad
i=1,\ldots,d-1$ $\displaystyle U_{d}$ $\displaystyle=$ $\displaystyle
1-\left\\{\left(V+m\right)\right\\}=1-V=Vz_{d,m+1}+\left(1-V\right)y_{d,m+1}$

The final result for $U_{i}$ follows from $z_{i,m+1}\leq 1$. Then the convex
combination of $z_{i,m+1}$ and $y_{i,m+1}$ is in $\left[0,1\right]$. ∎

###  A.19 Proof of Corollary 6

###### Proof.

We now show that the base-$b$ Ardvisen and Johnson random vector has standard
uniform marginals for each $b\in\mathbb{N}$ using conditions in Theorem 2.1.
This would also showcase the use of our notation in a non trivial example. Let
us derive another expression for the $\mathbf{y}$’s that would simplify the
coming computations

$\displaystyle y_{1,m}$ $\displaystyle=$ $\displaystyle\dfrac{m-1}{b^{d-2}}$
$\displaystyle y_{k,m}$ $\displaystyle=$
$\displaystyle\left(b^{k-2}\dfrac{m-1}{b^{d-2}}+\dfrac{1}{b}\right)\mathrm{mod}\,1=\left(\dfrac{m-1+b^{d-k-1}}{b^{d-k}}\right)\mathrm{mod}\,1$
$\displaystyle=$
$\displaystyle\dfrac{\left(m-1+b^{d-k-1}\right)\mathrm{mod}\,b^{d-k}}{b^{d-k}},\quad
k\in\left\\{2,\ldots,d-1\right\\}$ $\displaystyle y_{d,m}$ $\displaystyle=$
$\displaystyle 1$

for $m\in\\{1,\ldots,b^{d-2}\\}$, where the last line follows from the fact
that $m-1+b^{d-k-1}$ is an integer, the inclusion of the case $k=d$ from he
fact that $a\left(\hbox{mod }1\right)=0$ for $a$ integer. In particular this
representation allow an easy computation of key quantities. First of all ,
coordinates of projected vertexes i.e. the vector of ordered unique values in
$\mathbf{y}_{l}$ and $\mathbf{z}_{l}$, $l\in\left\\{1,\ldots,d\right\\}$ :

$\displaystyle n_{l}$ $\displaystyle=$ $\displaystyle
b^{d-\max\left(l,2\right)}+1$ $\displaystyle\mathbf{a}_{l}$ $\displaystyle=$
$\displaystyle\dfrac{1}{n_{l}-1}\left(0,\ldots,n_{l}-1\right)^{T}$

Then, because of the mod function, values of $y_{k,m}$ repeat themselves after
$b^{d-\max\left(l,2\right)}$ positions. The multiplicity of each unique value
in $\mathbf{y}_{l}$ is the the length of the vector divided by the period,
i.e.: $\frac{b^{d-2}}{b^{d-\max\left(l,2\right)}}=b^{\max\left(l,2\right)-2}$.
Considering the fact that all values with the exception of $0$ are also
repeated in the same way in $\mathbf{z}_{l}$, we get:

$\displaystyle\left|\mathcal{M}_{l,1}\right|$ $\displaystyle=$
$\displaystyle\left|\mathcal{M}_{l,n_{l}}\right|=b^{\max\left(l,2\right)-2}$
$\displaystyle\left|\mathcal{M}_{l,j}\right|$ $\displaystyle=$ $\displaystyle
2b^{\max\left(l,2\right)-2},j=2,\ldots,n_{l}-1$

Finally, in the projected graph only nearby vertexes are connected by edges,
the difference in projected coordinates joined is constant and the number of
edges correspond to the number of replicas in $\mathbf{y}_{l}$ of such a
vertex. We have:

$\displaystyle n^{l}_{\left(m,m^{\prime}\right)}$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ccc}b^{\max\left(l,2\right)-2}&&\left|m-m^{\prime}\right|=1\\\
&&\\\ 0&&\hbox{otherwise}\end{array}\right.$ $\displaystyle\mathcal{K}_{l,m}$
$\displaystyle=$ $\displaystyle\left(m-1,m\right)$ $\displaystyle
a_{lm}-a_{l,m-1}$ $\displaystyle=$
$\displaystyle\dfrac{1}{b^{d-\max\left(l,2\right)}}$

with $m,m^{\prime}=1,\ldots,n_{l}$. Given those preliminary computation we are
ready to show standard marginal uniformity using the conditions in Theorem
2.1:

$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\displaystyle\sum_{k\in\mathcal{K}_{l,m}}\dfrac{1}{a_{l,m_{\beta}\left(k\right)}-a_{l,m_{\alpha}\left(k\right)}}=\dfrac{1}{\left|\mathcal{E}\right|}\dfrac{n^{l}_{\left(m,m^{\prime}\right)}}{a_{lm}-a_{l,m-1}}=\frac{b^{\max\left(l,2\right)-2}b^{d-\max\left(l,2\right)}}{b^{d-2}}=1$

Concerning the constant sum constraint, it is known since Arvidsen and
Johnsson, (1982) (see also Craiu and Meng, (2005)) that the case $b=2$
satisfies $d$-CTM. We show that the first vertex ($m=1$) sum to $\frac{d}{2}$
only in the case $b=2$, implying that $b\neq 2$ cases are not $d$-CTM.

$\displaystyle y_{1,1}$ $\displaystyle=$ $\displaystyle\dfrac{0}{b^{d-2}}=0$
$\displaystyle y_{k,1}$ $\displaystyle=$
$\displaystyle\left\\{b^{k-2}\dfrac{0}{b^{d-2}}+\dfrac{1}{b}\right\\}=\dfrac{1}{b}$
$\displaystyle k$ $\displaystyle\in$
$\displaystyle\left\\{2,\ldots,d-1\right\\}$ $\displaystyle y_{d,1}$
$\displaystyle=$ $\displaystyle 1$

then we should solve for $b$ the following equation

$\displaystyle\sum^{d}_{k=1}y_{k,1}=\dfrac{d-2}{b}+1=\dfrac{d}{2}$

which has as unique solution $b=2$. ∎

###  A.20 Proof of Proposition 4.5

###### Proof.

For each $l\in\mathcal{D}$

$\displaystyle U_{l,t}$ $\displaystyle=$
$\displaystyle\dfrac{1}{d}\left(\pi_{k}\left(l\right)+U_{l,t-1}\right)$
$\displaystyle=$
$\displaystyle\dfrac{1}{d}\left(\pi_{k}\left(l\right)+U_{l,t-1}\right)+\dfrac{\pi_{k}\left(l\right)}{d}U_{l,t-1}-\dfrac{\pi_{k}\left(l\right)}{d}U_{l,t-1}$
$\displaystyle=$
$\displaystyle\dfrac{\pi_{k}\left(l\right)+1}{d}U_{l,t-1}+\dfrac{\pi_{k}\left(l\right)}{d}\left(1-U_{l,t-1}\right)$

∎

###  A.21 Proof of Proposition 4.6

###### Proof.

For each t $-1\leq X_{it}\leq 1$ let us define
$\mathbf{U}_{t}=\dfrac{\mathbf{X}_{t}+1}{2}$, we have:

$\displaystyle\mathbf{U}_{t}=\dfrac{1}{3}\mathbf{U}_{t-1}+\dfrac{2}{3}\dfrac{\mathbf{V}_{k}+1}{2}$
(102)

and $\mathbf{V}_{k}+1$ is a permutation of $\left\\{0,1,2\right\\}$, so that
the generalization to $d$ dimension is straightforward:

$\displaystyle\mathbf{U}_{t}=\dfrac{1}{d}\mathbf{U}_{t-1}+\left(1-\dfrac{1}{d}\right)\dfrac{\mathcal{D}^{\pi}_{t}}{d-1}=\dfrac{1}{d}\left(\mathcal{D}^{\pi}_{t}+\mathbf{U}_{t-1}\right)$
(103)

∎

###  A.22 Proof of Corollary 7

###### Proof.

With the line segment representation, of proposition 4.6 the unique values are
$a_{l,m}=\dfrac{m}{d}$, $m=0,\ldots,d$ and the segments extremes are
$\beta_{l,k}=\frac{\pi_{k}\left(l\right)+1}{d}$ and
$\alpha_{l,k}=\frac{\pi_{k}\left(l\right)}{d}$. Then, since
$m_{\beta}\left(k\right)=\pi_{k}\left(l\right)+1$ and
$m_{\alpha}\left(k\right)=\pi_{k}\left(l\right)$, we obtain:

$\displaystyle a_{l,m_{\beta}\left(k\right)}-a_{l,m_{\alpha}\left(k\right)}$
$\displaystyle=$ $\displaystyle\dfrac{1}{d}$

and

$\displaystyle\mathcal{K}_{l,m}$ $\displaystyle=$
$\displaystyle\left\\{k\in\left\\{1,\ldots,\left|\mathcal{E}\right|\right\\}:\left(m_{\alpha}\left(k\right)\leq
m-1\right)\cap\left(m_{\beta}\left(k\right)\geq m\right)\right\\}$
$\displaystyle=$
$\displaystyle\left\\{k\in\left\\{1,\ldots,\left|\mathcal{E}\right|\right\\}:\pi_{k}\left(l\right)=m-1\right\\}$

leading to $\left|\mathcal{K}_{l,m}\right|$ being equal to the number of
permutations with the l-th element equal to $m-1$. We obtain then
$\left|\mathcal{K}_{l,m}\right|=\left(d-1\right)!$, which implies for each
$m=2,\ldots,d$:

$\displaystyle F_{l,m}\left(\mathbf{a}_{l}\right)$ $\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\displaystyle\sum_{k\in\mathcal{K}_{l,m}}\dfrac{1}{a_{l,m_{\beta}\left(k\right)}-a_{l,m_{\alpha}\left(k\right)}}$
$\displaystyle=$
$\displaystyle\dfrac{1}{d!}\displaystyle\sum_{k\in\mathcal{K}_{l,m}}d=\dfrac{d}{d!}\left|\mathcal{K}_{l,m}\right|=1$

we now investigate if the constant sum is preserved over iterations. As
previously shown, the line segment representation of ILH satisfies the first
constraint in equations (2.3) with $c_{1}=\dfrac{1}{d}$. The second constraint
is:

$\sum^{d}_{l=1}\dfrac{\pi_{k}\left(l\right)}{d}=\sum^{d-1}_{l=0}\dfrac{l}{d}=\dfrac{1}{d}\left(d-1\right)\dfrac{d}{2}=\left(1-\dfrac{1}{d}\right)\dfrac{d}{2}=\left(1-c_{1}\right)\dfrac{d}{2}$

and if $U_{l,t-1}$ has constant sum then $U_{l,t}$ in (67) has constant sum by
Lemma 2.3. ∎

###  A.23 Proof of Proposition 4.7

###### Proof.

In the ILH($T$) case we can apply the deterministic composition. By ILH
construction the $\mathbf{V}$ vector has independent components, and in the
segments belong to non-intersecting hypercubes of smaller size. Let’s consider
a generic construction with the latter characteristics.

$\displaystyle\mathbb{P}\left(\mathbf{U}\leq\mathbf{W}\right)=$
$\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\sum^{\left|\mathcal{E}\right|}_{k_{\mathbf{U}}=1}\mathbb{P}\left(R_{\mathbf{V},\mathcal{L}^{-}_{k_{\mathbf{U}}}}\left(\mathbf{V}\right)\leq\mathbf{Y}_{k_{\mathbf{U}}}\right)$
$\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\sum^{\left|\mathcal{E}\right|}_{k_{\mathbf{U}}=1}\mathbb{P}\left(\mathbf{V}\leq\mathbf{Y}_{k_{\mathbf{U}}}\right)$
$\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\sum^{\left|\mathcal{E}\right|}_{k_{\mathbf{U}}=1}\mathbb{E}_{\mathbf{W}}\left[\prod^{d}_{l=1}\dfrac{\min\left(\max\left(W_{l},\alpha_{l,k_{\mathbf{U}}}\right),\beta_{l,k_{\mathbf{U}}}\right)-\alpha_{l,k_{\mathbf{U}}}}{\beta_{l,k_{\mathbf{U}}}-\alpha_{l,k_{\mathbf{U}}}}\right]$
$\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\sum^{\left|\mathcal{E}\right|}_{k_{\mathbf{U}}=1}\dfrac{1}{\left|\mathcal{E}\right|}\sum^{\left|\mathcal{E}\right|}_{k_{\mathbf{W}}=1}\mathbb{E}_{\mathbf{V}}\left[\prod^{d}_{l=1}\dfrac{\min\left(\max\left(\alpha_{l,k_{\mathbf{W}}}+\left(\beta_{l,k_{\mathbf{W}}}-\alpha_{l,k_{\mathbf{W}}}\right)V_{l},\alpha_{l,k_{\mathbf{U}}}\right),\beta_{l,k_{\mathbf{U}}}\right)-\alpha_{l,k_{\mathbf{U}}}}{\beta_{l,k_{\mathbf{U}}}-\alpha_{l,k_{\mathbf{U}}}}\right]$
$\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\sum^{\left|\mathcal{E}\right|}_{k_{\mathbf{U}}=1}\dfrac{1}{\left|\mathcal{E}\right|}\sum^{\left|\mathcal{E}\right|}_{k_{\mathbf{W}}=1}\prod^{d}_{l=1}\dfrac{\mathbb{E}_{V_{l}}\left[\min\left(\max\left(\alpha_{l,k_{\mathbf{W}}}+\left(\beta_{l,k_{\mathbf{W}}}-\alpha_{l,k_{\mathbf{W}}}\right)V_{l},\alpha_{l,k_{\mathbf{U}}}\right),\beta_{l,k_{\mathbf{U}}}\right)\right]-\alpha_{l,k_{\mathbf{U}}}}{\beta_{l,k_{\mathbf{U}}}-\alpha_{l,k_{\mathbf{U}}}}$

Since the boxes with different segments do not overlap in the sum for
$k_{\mathbf{W}}$ the only contributions come from
$k_{\mathbf{U}}=k_{\mathbf{W}}$:

$\displaystyle\mathbb{P}\left(\mathbf{U}\leq\mathbf{W}\right)$
$\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|^{2}}\sum^{\left|\mathcal{E}\right|}_{k_{\mathbf{U}}=1}\prod^{d}_{l=1}\dfrac{\mathbb{E}_{V_{l}}\left[\alpha_{l,k_{\mathbf{U}}}+\left(\beta_{l,k_{\mathbf{U}}}-\alpha_{l,k_{\mathbf{U}}}\right)V_{l}\right]-\alpha_{l,k_{\mathbf{U}}}}{\beta_{l,k_{\mathbf{U}}}-\alpha_{l,k_{\mathbf{U}}}}$
$\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|^{2}}\sum^{\left|\mathcal{E}\right|}_{k_{\mathbf{U}}=1}\prod^{d}_{l=1}\dfrac{\left(\beta_{l,k_{\mathbf{U}}}-\alpha_{l,k_{\mathbf{U}}}\right)/2}{\beta_{l,k_{\mathbf{U}}}-\alpha_{l,k_{\mathbf{U}}}}$
$\displaystyle=$
$\displaystyle\dfrac{1}{\left|\mathcal{E}\right|}\dfrac{1}{2^{d}}$

The number of vertexes in the ILH($T$) case is $\left(d!\right)^{T}$ and the
result for $\tau$ follows.

Denoting $\xi^{*}_{t}$, the case for $t\leq T$ iterations we compute:

$\displaystyle\xi^{*}_{1}$ $\displaystyle=$
$\displaystyle\dfrac{1}{d!}\sum^{d!}_{k_{1}=1}\prod^{d}_{l=1}\dfrac{\left(2\pi_{k_{1}}\left(l\right)+1\right)}{d}$
$\displaystyle\xi^{*}_{2}$ $\displaystyle=$
$\displaystyle\dfrac{1}{d!}\sum^{d!}_{k_{2}=1}\dfrac{1}{d!}\sum^{d!}_{k_{1}=1}\prod^{d}_{l=1}\left(\dfrac{\pi_{k_{1}}\left(l\right)}{d^{2}}+\dfrac{\pi_{k_{2}}\left(l\right)}{d}+\dfrac{\pi_{k_{1}}\left(l\right)+1}{d^{2}}+\dfrac{\pi_{k_{2}}\left(l\right)}{d}\right)$
$\displaystyle=$
$\displaystyle\dfrac{1}{d!}\sum^{d!}_{k_{2}=1}\dfrac{1}{d!}\sum^{d!}_{k_{1}=1}\prod^{d}_{l=1}\left(\sum^{2}_{t=1}\dfrac{2\pi_{k_{t}}\left(l\right)}{d^{2-t+1}}+\dfrac{1}{d^{2}}\right)$
$\displaystyle\vdots$ $\displaystyle\xi^{*}_{T}$ $\displaystyle=$
$\displaystyle\dfrac{1}{d!}\sum^{d!}_{k_{T}=1}\cdots\dfrac{1}{d!}\sum^{d!}_{k_{1}=1}\prod^{d}_{l=1}\left(\sum^{T}_{t=1}\dfrac{2\pi_{k_{t}}\left(l\right)}{d^{T-t+1}}+\dfrac{1}{d^{T}}\right)$

We simplify the expression for $T=2$ by expanding the product

$\displaystyle\xi^{*}_{2}$ $\displaystyle=$
$\displaystyle\dfrac{1}{d!}\sum^{d!}_{k_{2}=1}\dfrac{1}{d!}\sum^{d!}_{k_{1}=1}\sum^{d}_{m_{2}=0}\sum_{{\tiny\begin{array}[]{c}\mathcal{L}_{m_{2}}\subseteq\mathcal{D}\\\
\left|\mathcal{L}_{m_{2}}\right|=m_{2}\end{array}}}\left\\{\prod_{l\in\mathcal{L}_{m_{2}}}\dfrac{2\pi_{k_{2}}\left(l\right)}{d}\right\\}\left\\{\prod_{l\in\mathcal{D}\setminus\mathcal{L}_{m_{2}}}\dfrac{2\pi_{k_{1}}\left(l\right)+1}{d^{2}}\right\\}$
$\displaystyle=$
$\displaystyle\sum^{d}_{m_{2}=0}\sum_{{\tiny\begin{array}[]{c}\mathcal{L}_{m_{2}}\subseteq\mathcal{D}\\\
\left|\mathcal{L}_{m_{2}}\right|=m_{2}\end{array}}}\dfrac{1}{d!}\sum^{d!}_{k_{2}=1}\left\\{\prod_{l\in\mathcal{L}_{m_{2}}}\dfrac{2\pi_{k_{2}}\left(l\right)}{d}\right\\}\dfrac{1}{d!}\sum^{d!}_{k_{1}=1}\left\\{\prod_{l\in\mathcal{D}\setminus\mathcal{L}_{m_{2}}}\dfrac{2\pi_{k_{1}}\left(l\right)+1}{d^{2}}\right\\}$

The sums over permutations can be reduced. Consider the first sum, the product
is independent from $\pi_{k_{2}}\left(l\right)$ for
$l\in\mathcal{D}\setminus\mathcal{L}_{m_{2}}$ and invariant for permutations
of the product itself. Then the product is the same for
$\left(d-m_{2}\right)!m_{2}!$ permutations. We consider only the ordered
products multiplied by $\left(d-m_{2}\right)!m_{2}!$. This amounts to summing
over the set of combinations of $\mathcal{D}$ with $m_{2}$ elements that we
denote by $\mathcal{C}_{\mathcal{D},m_{2}}$:

$\displaystyle\dfrac{1}{d!}\sum^{d!}_{k_{2}=1}\prod_{l\in\mathcal{L}_{m_{2}}}\dfrac{2\pi_{k_{2}}\left(l\right)}{d}$
$\displaystyle=$ $\displaystyle{d\choose
m_{2}}^{-1}\sum_{{\tiny\left\\{i_{1},\ldots,i_{m_{2}}\right\\}\in\mathcal{C}_{\mathcal{D},m_{2}}}}\prod^{m_{2}}_{l=1}\dfrac{2\left(i_{l}-1\right)}{d}$
$\displaystyle\xi^{*}_{2}$ $\displaystyle=$

$\displaystyle\sum^{d}_{m_{2}=0}{d\choose m_{2},d-m_{2}}\left\\{{d\choose
m_{2}}^{-1}\sum_{{\tiny\left\\{i_{1},\ldots,i_{m_{2}}\right\\}\in{\mathcal{D}\choose
m_{2}}}}\prod^{m_{2}}_{l=1}\dfrac{2\left(i_{l}-1\right)}{d}\right\\}\left\\{{d\choose
d-m_{2}}^{-1}\sum_{{\tiny\left\\{i_{1},\ldots,i_{d-m_{2}}\right\\}\in\mathcal{C}_{\mathcal{D},d-m_{2}}}}\prod^{d-m_{2}}_{l=1}\dfrac{2\left(i_{l}-1\right)+1}{d^{2}}\right\\}$

$\displaystyle=$

$\displaystyle\sum_{m_{2}+m_{1}+m_{0}=d}{d\choose
m_{2},m_{1},m_{0}}\prod^{2}_{t=1}\left\\{{d\choose
m_{t}}^{-1}\sum_{{\tiny\left\\{i_{1},\ldots,i_{m_{t}}\right\\}\in\mathcal{C}_{\mathcal{D},m_{t}}}}\prod^{m_{t}}_{l=1}\dfrac{2\left(i_{l}-1\right)}{d^{2-t+1}}\right\\}\left(\dfrac{1}{d^{2}}\right)^{m_{0}}$

The same line of reasoning leads to:

$\displaystyle\xi^{*}_{T}$ $\displaystyle=$

$\displaystyle\sum_{m_{0}+m_{1}+\ldots+m_{T}=d}{d\choose
m_{0},m_{1},\dots,m_{T}}\prod^{T}_{t=1}\left\\{{d\choose
m_{t}}^{-1}\sum_{{\tiny\left\\{i_{1},\ldots,i_{m_{t}}\right\\}\in\mathcal{C}_{\mathcal{D},m_{2}}}}\prod^{m_{t}}_{l=1}\dfrac{2\left(i_{l}-1\right)}{d^{T-t+1}}\right\\}\left(\dfrac{1}{d^{T}}\right)^{m_{0}}$

∎

###  A.24 Proof of Lemma 5.1

###### Proof.

Let $\mathbf{U}_{l}=\left(U_{l}^{1},\ldots,U_{l}^{p}\right),l\in\mathcal{D}$,
$\mathbf{U}_{l}$ is a vector of standard uniform independent random variables.
We define
$\mathbf{m}_{l}=\left\\{m^{1}_{l},\ldots,m^{p}_{l}\right\\},m^{i}_{l}=1,\ldots,n_{l}$,
$A^{\mathbf{m}_{l}}=A_{m^{1}_{l}}\times\ldots\times A_{m^{p}_{l}}$,
$A_{m^{l}_{1}}=\left[a_{l,m^{i}_{l}-1},a_{l,m^{i}_{l}}\right]$ ,
$A^{\mathbf{m}}=A_{\mathbf{m}_{1}}\times\ldots\times A_{\mathbf{m}_{d}}$. By
Corollary 1 we have $\cup_{\mathbf{m}}A^{\mathbf{m}}=\left[0,1\right]^{dp}$
and $A^{\mathbf{m}}\cap A^{\mathbf{m}^{\prime}}=\emptyset$ if
$\mathbf{m}\neq\mathbf{m}^{\prime}$. We introduce also a notation for the
marginal boxes, given a subset $\mathcal{R}_{s}\subset\mathcal{D}$,
$\left|\mathcal{R}_{s}\right|=s\leq d$ we call
$A_{\left\\{m^{i}_{l}\right\\}_{l\in\mathcal{R}_{s}}}$ the box
$A_{m^{i}_{t_{1}}}\times\ldots\times A_{m^{i}_{t_{s}}}$ and
$A_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}$ the box
$A_{\mathbf{m}_{t_{1}}}\times\ldots\times A_{\mathbf{m}_{t_{s}}}$ . The
multinomial theorem implies:

$\displaystyle\mathbb{E}_{a-LH}\left[\left(\dfrac{1}{\sqrt{d}}\sum^{d}_{l=1}f\left(\mathbf{U}_{l}\right)\right)^{r}\right]-\mathbb{E}_{b-LH}\left[\left(\dfrac{1}{\sqrt{d}}\sum^{d}_{l=1}f\left(\mathbf{U}_{l}\right)\right)^{r}\right]$
$\displaystyle=$ $\displaystyle\displaystyle
d^{-r/2}\sum_{r_{1}+\ldots+r_{d}=r}\dfrac{r!}{r_{1}!\ldots
r_{d}!}\left\\{\mathbb{E}_{a-LH}\left[\prod^{d}_{l=1}f\left(\mathbf{U}_{l}\right)^{r_{l}}\right]-\mathbb{E}_{b-LH}\left[\prod^{d}_{l=1}f\left(\mathbf{U}_{l}\right)^{r_{l}}\right]\right\\}$

For each expectation inside the sum, since $\sum^{d}_{l=1}r_{l}=r$ does not
vary with $d$, in the limit a different subset $r_{t_{1}},\ldots,r_{t_{s}}$,
$t_{1},\ldots,t_{s}\in\mathcal{R}_{s}\subset\mathcal{D}$, $s\leq r$ , of
exponents will be different from zero. Then for $c=a,b$ we rewrite:

$\displaystyle\lim_{d\rightarrow\infty}\mathbb{E}_{c-LH}\left[\prod^{d}_{l=1}f\left(\mathbf{U}_{l}\right)^{r_{l}}\right]=\lim_{d\rightarrow\infty}\mathbb{E}_{c-LH}\left[\prod^{s}_{l=1}f\left(\mathbf{U}_{t_{l}}\right)^{r_{t_{l}}}\right]$

For $c=a,b$ we obtain:

$\displaystyle\mathbb{E}_{c-LH}\left[\mathbb{E}_{c-LH}\left[\prod^{s}_{l=1}f\left(\mathbf{U}_{t_{l}}\right)^{r_{t_{l}}}\left|\left\\{\mathbf{U}_{l}\in
A^{\mathbf{m}_{l}}\right\\}_{l\in\mathcal{R}_{s}}\right.\right]\right]$
$\displaystyle=$
$\displaystyle\sum_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}\mu_{c-LH}\left(A_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}\right)\mathbb{E}_{c-LH}\left[\prod^{s}_{l=1}f\left(\mathbf{U}_{t_{l}}\right)^{r_{t_{l}}}\left|\left\\{\mathbf{U}_{l}\in
A^{\mathbf{m}_{l}}\right\\}_{l\in\mathcal{R}_{s}}\right.\right]$

$\displaystyle\mu_{c-LH}\left(A_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}\right)=\prod^{p}_{i=1}\mathbb{P}_{c-LH}\left(\left\\{U^{i}_{l}\in\left(a_{l,m^{i}_{l}-1},a_{l,m^{i}_{l}}\right]\right\\}_{l\in\mathcal{R}_{s}}\right)$
$\displaystyle=$
$\displaystyle\prod^{p}_{i=1}\dfrac{1}{\left|\mathcal{E}\right|}\sum_{k_{i}\in\mathcal{E}}\mathbb{P}_{c-LH}\left(\left.\left\\{U^{i}_{l}\in\left(a_{l,m^{i}_{l}-1},a_{l,m^{i}_{l}}\right]\right\\}_{l\in\mathcal{R}_{s}}\right|K_{i}=k_{i}\right)$
$\displaystyle=$

$\displaystyle\prod^{p}_{i=1}\dfrac{1}{\left|\mathcal{E}\right|}\sum_{k_{i}\in\mathcal{E}}\left[\prod^{s}_{l=1}\mathbb{I}_{\mathcal{K}_{m^{i}_{t_{l}}}}\left(k_{i}\right)\right]\mathbb{E}_{c}\left[\prod^{s}_{l=1}\mathbb{I}_{\left(\dfrac{a_{l,m^{i}_{t_{l}}-1}-\alpha_{t_{l},k_{i}}}{\beta_{t_{l},k_{i}}-\alpha_{t_{l},k_{i}}},\dfrac{a_{t_{l},m^{i}_{l}}-\alpha_{t_{l},k_{i}}}{\beta_{t_{l},k_{i}}-\alpha_{t_{l},k_{i}}}\right]}\left(V^{i}_{t_{l},k_{i}}\right)\right]$

$\displaystyle=$
$\displaystyle\prod^{p}_{i=1}\dfrac{1}{\left|\mathcal{E}\right|}\sum_{k_{i}\in\bigcap^{s}_{l=1}\mathcal{K}_{m^{i}_{t_{l}}}}\mathbb{E}_{c}\left[\prod^{s}_{l=1}\mathbb{I}_{\left(\dfrac{\alpha_{t_{l},k_{i}}-\alpha_{t_{l},k_{i}}}{\beta_{t_{l},k_{i}}-\alpha_{t_{l},k_{i}}},\dfrac{\beta_{t_{l},k_{i}}-\alpha_{t_{l},k_{i}}}{\beta_{t_{l},k_{i}}-\alpha_{t_{l},k_{i}}}\right]}\left(V^{i}_{t_{l},k_{i}}\right)\right]$
$\displaystyle=$
$\displaystyle\prod^{p}_{i=1}\dfrac{1}{\left|\mathcal{E}\right|}\sum_{k_{i}\in\bigcap^{s}_{l=1}\mathcal{K}_{m^{i}_{t_{l}}}}\mathbb{E}_{c}\left[\prod^{s}_{l=1}\mathbb{I}_{\left(0,1\right]}\left(V^{i}_{t_{l},k_{i}}\right)\right]$
$\displaystyle=$
$\displaystyle\prod^{p}_{i=1}\dfrac{1}{\left|\mathcal{E}\right|}\sum_{k_{i}\in\bigcap^{s}_{l=1}\mathcal{K}_{m^{i}_{t_{l}}}}\mu_{c}\left(\left(0,1\right]^{s}\times\left(0,1\right]^{d-s}\right)$
$\displaystyle=$
$\displaystyle\prod^{p}_{i=1}\dfrac{\left|\bigcap^{s}_{l=1}\mathcal{K}_{m^{i}_{t_{l}}}\right|}{\left|\mathcal{E}\right|}=\left\\{\begin{array}[]{ccc}\left(\dfrac{\left(d-s\right)!}{d!}\right)^{p}&\hbox{if}&m^{i}_{t_{1}}\neq
m^{i}_{t_{s}}\hbox{ for each }i=1,\ldots,p\\\ 0&&\hbox{ otherwise
}\end{array}\right.$

where
$\left|\bigcap^{d}_{l=1}\mathcal{K}_{m^{i}_{l}}\right|=\left(d-s\right)!$ if
$m^{i}_{t_{1}}\neq\ldots\neq m^{i}_{t_{s}}$ and $0$ otherwise because, in the
LH case, it represents the number of permutation of the $d$-dimensional vector
$\left(0,\ldots,d-1\right)$ when a number $s$ of the components are held fixed
and there is no permutation with two equal indexes.

We showed that
$\mu_{c-LH}\left(A_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}\right)$
is independent from the measure of $\mathbf{V}^{i}$, then if

(107)

we obtain the statement of the Lemma.

We denote as $B_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}$ the
circumscribed ball of
$A_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}$. The measure of
$B_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}$ is bounded above
by the measure of the union of the boxes adjacent to
$A_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}$ and
$A_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}$ itself. In the
worst case when none of the adjacent boxes has two or more $m$s equal, we
obtain:

$\displaystyle\mu_{c-LH}\left(A_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}\right)\leq\mu_{c-LH}\left(B_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}\right)\leq\left(2s+1\right)\mu_{c-LH}\left(A_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}\right)$

The result follows from Lebesgue-Besicovitch Differentiation Theorem
(Besicovitch,, 1945). We refer to section 1.7 of Evans and Gariepy, (1991) for
a contemporary treatment. In particular, we write, for $c=a,b$ and
$\left(\mathbf{u}^{\star}_{1},\ldots,\mathbf{u}^{\star}_{d}\right)$, the
center of the box $A_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}$:

$\displaystyle\lim_{d\rightarrow\infty}\left|\mathbb{E}_{c-LH}\left[\prod^{s}_{l=1}f\left(\mathbf{U}_{t_{l}}\right)^{r_{t_{l}}}\left|\left\\{\mathbf{U}_{l}\in
A^{\mathbf{m}_{l}}\right\\}_{l\in\mathcal{R}_{s}}\right.\right]-\prod^{s}_{l=1}f\left(\mathbf{u}^{\star}_{l}\right)^{r_{t_{l}}}\right|$
$\displaystyle\leq$
$\displaystyle\lim_{d\rightarrow\infty}\mathbb{E}_{c-LH}\left[\left|\prod^{s}_{l=1}f\left(\mathbf{U}_{t_{l}}\right)^{r_{t_{l}}}-\prod^{s}_{l=1}f\left(\mathbf{u}^{\star}_{l}\right)^{r_{t_{l}}}\right|\left|\left\\{\mathbf{U}_{l}\in
A^{\mathbf{m}_{l}}\right\\}_{l\in\mathcal{R}_{s}}\right.\right]$
$\displaystyle\leq$
$\displaystyle\lim_{d\rightarrow\infty}\dfrac{\displaystyle\int_{A_{\left\\{m_{l}\right\\}_{l\in\mathcal{R}_{s}}}}\left|\prod^{s}_{l=1}f\left(\mathbf{U}_{t_{l}}\right)^{r_{t_{l}}}-\prod^{s}_{l=1}f\left(\mathbf{u}^{\star}_{l}\right)^{r_{t_{l}}}\right|d\mu_{c-LH}}{\mu_{c-LH}\left(A_{\left\\{m_{l}\right\\}_{l\in\mathcal{R}_{s}}}\right)}$
$\displaystyle\leq$
$\displaystyle\lim_{d\rightarrow\infty}\dfrac{\left(2s+1\right)}{\mu_{c-LH}\left(B_{\left\\{m_{l}\right\\}_{l\in\mathcal{R}_{s}}}\right)}\int_{A_{\left\\{m_{l}\right\\}_{l\in\mathcal{R}_{s}}}}\left|\prod^{s}_{l=1}f\left(\mathbf{U}_{t_{l}}\right)^{r_{t_{l}}}-\prod^{s}_{l=1}f\left(\mathbf{u}^{\star}_{l}\right)^{r_{t_{l}}}\right|d\mu_{c-LH}$
$\displaystyle\leq$
$\displaystyle\lim_{d\rightarrow\infty}\dfrac{\left(2s+1\right)}{\mu_{c-LH}\left(B_{\left\\{m_{l}\right\\}_{l\in\mathcal{R}_{s}}}\right)}\int_{B_{\left\\{m_{l}\right\\}_{l\in\mathcal{R}_{s}}}}\left|\prod^{s}_{l=1}f\left(\mathbf{U}_{t_{l}}\right)^{r_{t_{l}}}-\prod^{s}_{l=1}f\left(\mathbf{u}^{\star}_{l}\right)^{r_{t_{l}}}\right|d\mu_{c-LH}\leq
0$

The last passage follows from
$\lim_{d\rightarrow\infty}\left(2s+1\right)\mu_{c-LH}\left(A_{\left\\{\mathbf{m}_{l}\right\\}_{l\in\mathcal{R}_{s}}}\right)=0$
and corollary 1 in section 1.7 of Evans and Gariepy, (1991). Because the limit
is independent from the measure of $\mathbf{V}^{i}$, equation (107) is
verified. ∎

###  A.25 Proof of Theorem 5.1

###### Proof.

We apply lemma 5.1 substituting by the method of moments the measure $\mu$
with the independence measure obtaining an ordinary hypercube sample. Since
$f$ is bounded we can apply Theorem 1 in Owen, (1992) . ∎