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\section{Introduction}
The concept of exchangeability has been a natural and convenient assumption to impose in probability theory and for simplifying statistical models. As defined originally for random sequences, it states that any order of a finite number of samples is equally likely. The concept was later generalized for binary random arrays \citep{ald85}, and consequently for random networks in statistical network analysis. In this context, exchangeability is translated into invariance under relabeling
of the nodes of the network, whereby isomorphic graphs have the
same probabilities; see, e.g., \cite{lau17}.

Exchangeability is closely related to the concept of independent and identically distributed random variables. It is an immediate consequence of the definition that independent and identically distributed random variables are exchangeable, but the converse is not true. For infinite sequences, the converse is established by the well-known deFinetti's Theorem \citep{def31}, which implies that in any infinite sequence of exchangeable random variables, the random variables are conditionally independent and identically-distributed given the underlying distributional form. Other versions of deFinetti's Theorem exist for the generalized definitions of exchangeability \citep{sil76,dia08}.

However, for finitely exchangeable random sequences (vectors) and arrays (matrices), the converse does not hold, and the available results basically provide approximations of the infinite case; see, e.g., \cite{dia80,mat95,lau19}. In this paper, we utilize a completely different approach to study the relationship between finite exchangeability and (conditional) independence.  We employ the theory of graphical models (see e.g.\ \cite{lau96}) in order to provide the independence structure of exchangeable distributions.

In particular, we exploit the necessary and sufficient conditions, provided in \cite{sad17}, for faithfulness of probability distributions and graphs, which determine when the conditional independence structure of the distribution is exactly the same as that of a graph in graphical model sense. Thus, we, in practice, work on the induced independence model of an exchangeable probability distribution rather than the distribution itself. The specialization to finitely exchangeable random vectors leads to  necessary and sufficient conditions for an exchangeable vector to be completely independent or completely dependent, meaning that two elements of the vector are conditionally independent or dependent, respectively, given any subset of the remaining elements of the vector. These conditions are namely intersection and composition properties \cite{pea88,stu05}. We also use the results to provide a sufficient condition for an exchangeable vector to be marginally independent.

For random networks, we follow a similar procedure. It was shown in \cite{lau17} that if a distribution over an exchangeable random network could be faithful to a graph then the skeleton of the graph is only one of the four possible types: the empty graph, the complete graph, and the, so-called incidence graph, and its complement (see Proposition \ref{prop:7}). A question is then which graphs that emerge from these skeleta are faithful to the exchangeable distribution. We show that, other than complete independence or dependence, there are four other independence structures that can arise for exchangeable networks. These are faithful to a pair of dual graphs with incidence graph skeleton, and  a pair of dual graphs with the complement of the incidence graph skeleton. We then provide assumptions under which intersection and composition properties are necessary and sufficient for the exchangeable random networks to be faithful to each of these six cases.

The structure of the paper is as follows: In the next section, we provide some definitions and known results needed for this paper from graph theory, random networks, and graphical models. In Section \ref{sec:3}, we provide the results for exchangeable random vectors.  In Section \ref{sec:4}, we provide the results for exchangeable random networks by starting with providing the aforementioned possible cases in Theorem \ref{thm:2}, and then introducing them in separate subsections. We end with a short discussion on these results in Section \ref{sec:5}. We will provide technical definitions and results needed for the proof of Theorem \ref{thm:2}, and its proof, in Appendix \ref{app}.
\section{Definitions and preliminary results}\label{sec:2}
\subsection{Graph-theoretic concepts}
A \emph{(labeled) graph} is an ordered pair $G = (V,E)$ consisting of a
\emph{vertex} set $V$, which is non-empty and finite, an \emph{edge} set $E$, and a relation that with
each edge associates two vertices, called
its \emph{endpoints}. We omit the term labeled in this paper since the context does not give rise to ambiguity.
 When vertices $u$ and $v$ are the endpoints of an
edge, these are
\emph{adjacent} and we write $u\sim v$; we denote the corresponding edge as
$uv$.

In this paper, we will restrict our attention to \emph{simple} graphs, i.e.\ graphs without
loops (this assumption means that the
endpoints of each edge are distinct) or multiple edges (each pair of vertices are the endpoints
of at most one edge). Further, three \emph{types} of edges, denoted by
\emph{arrows}, \emph{arcs} (solid lines with two-headed arrows) and \emph{lines}
(solid lines without arrowheads) have been used in the literature of graphical models.  Arrows can be represented by ordered pairs of vertices, while arcs
and lines by $2$-subsets of the vertex set. However, for our purpose, except for Section \ref{sec:41}, we will distinguish only lines and arcs, which  respectively form
\emph{undirected} and \emph{bidirected} graphs.

The graphs $F=(V_F,E_F)$ and $G=(V_G,E_G)$ are considered equal if and only
if $(V_F,E_F)=(V_G,E_G)$.

A \emph{subgraph} of a graph $G = (V_G,E_G)$ is graph $F = (V_F,E_F)$ such that
$V_F\subseteq V_G$ and $E_F\subseteq E_G$ and the assignment of endpoints to
edges in $F$
is the same as in $G$.

The \emph{line graph} $L(G)$ of a graph $G=(V,E)$ is the intersection graph of
the edge set $E$, i.e.\ its vertex set is $E$ and $e_1\sim e_2$ if and only if
$e_1$ and $e_2$ have a common endpoint \citep[p.\ 168]{wes01}. We will in
particular be interested in the line graph of a complete graph, which we will
refer to as the \emph{incidence graph}. Figure~\ref{fig:2} displays the incidence graph for $V = \{1,2,3,4\}$.
\begin{figure}[htb]
\centering
\begin{tikzpicture}[node distance = 8mm and 8mm, minimum width = 6mm]

    \begin{scope}
      \tikzstyle{every node} = [shape = circle,
      font = \scriptsize,
      minimum height = 6mm,

      inner sep = 2pt,
      draw = black,
      fill = white,
      anchor = center],
      text centered]
      \node(12) at (0,0) {$1,2$};
      \node(13) [below right = of 12] {$1,3$};
      \node(23) [above right = of 12] {$2,3$};
			\node(24) [right = of 23] {$2,4$};
      \node(14) [right = of 13] {$1,4$};
      \node(34) [below right = of 24] {$3,4$};

    \end{scope}

    \begin{scope}
      \draw (12) -- (13);
			\draw (12) -- (14);
			\draw (12) -- (23);
			\draw (12) -- (24);
      \draw (13) -- (23);
			\draw (13) -- (14);
			\draw (13) -- (34);
			\draw (23) -- (24);
			\draw (23) -- (34);
			\draw (14) -- (24);
			\draw (24) -- (34);\draw (14) -- (34);

    \end{scope}

    \end{tikzpicture}
		\caption{\small{The incidence graph for $V=
		    \{1,2,3,4\}$.}
     \label{fig:2}
		\end{figure}
We denote by $L_{-}(n)$ and $L_{\leftrightarrow}(n)$, the undirected incidence graph and the bidirected incidence graph for $n$ nodes, respectively; and by  $L^c_{-}(n)$ and $L^c_{\leftrightarrow}(n)$, the undirected complement of the incidence graph and the bidirected complement of the incidence graph for $n$ nodes, respectively, where the \emph{complement} of graph $G$ refers to a graph with the same node set as $G$, but with the edge set that is the complement set of the edge set of $G$.

The \emph{skeleton} of a graph is the undirected graph where all arrowheads are removed from the graphs, i.e., all edges are replaced by lines. We denote the skeleton of a graph $G$ by $\mathrm{sk}(G)$.

A \emph{walk} $\omega$ is a list $\omega=\langle i_0,e_1,i_1,\dots,e_n,i_n\rangle$ of vertices and edges such that for $1\leq m\leq n$, the edge $e_m$ has endpoints $i_{m-1}$ and $i_m$. When indicating a specific walk, we may skip the edges, and only write the nodes, when the walk we are considering is clear from the context. A \emph{path} is a walk with no repeated nodes.
\subsection{Random networks}
Given a finite node set $\mathcal{N}$ ---
representing \emph{individuals} or \emph{actors} in a given population of
interest ---  we define a
\emph{random network} over $\mathcal{N}$ to be a collection $X = (X_d, d\in
\mathcal{D}(\mathcal{N}) )$
of binary random variables taking values $0$ and $1$ indexed by a set
$\mathcal{D}(\mathcal{N})$, which is a collection of unordered pairs $ij$ of nodes in $\mathcal{N}$. The binary random variables $X_d$ are called \emph{dyads}, and
nodes $i$ and $j$ are said to have
a \emph{tie}  if the random variable $X_{ij}$ takes the value $1$, and no tie
otherwise.
Thus, a random network is a random
variable taking value in $\{0,1\}^{\mathcal{N} \choose 2}$ and can, therefore,
be seen as a random {\it simple, undirected} graph with node set $\mathcal{N}$,
whereby the ties form the random edges of the graphs.

We use the terms
network, node, and tie rather than graph, vertex, and edge to differentiate
from the terminology used in the graphical model sense. Indeed, as we shall
discuss graphical models for networks, we will also consider each dyad $d$ as a
vertex in a graph $G=(\mathcal{D},E)$ representing the dependence structure of the random
variables associated with the dyads, with the edge set of such graph
representing Markov properties of the distribution of $X$.
\subsection{Probabilistic independence models and their properties}\label{sec:propind}
An \emph{independence model} $\mathcal{J}$ over a finite set $V$ is a set of triples $\langle X,Y\,|\, Z\rangle$ (called \emph{independence statements}), where $X$, $Y$, and $Z$ are disjoint subsets of $V$; $Z$
may be empty, but $\langle \varnothing,Y\,|\, Z\rangle$ and $\langle X,\varnothing\,|\, Z\rangle$ are always included in $\mathcal{J}$. The independence statement $\langle X,Y\,|\, Z\rangle$ is read as ``$X$ is independent of $Y$ given $Z$''. Independence models may in  general have a  probabilistic interpretation, but not necessarily. Similarly, not all independence models can be easily represented by graphs. For further discussion on general independence models, see \cite{stu05}.

In order to define probabilistic independence models, consider a set $V$ and a collection of random variables
$\{X_\alpha\}_{\alpha\in V}$ with state spaces $\mathcal{X}_\alpha, \alpha\in V$ and joint distribution $P$. We let $X_A=\{X_v\}_{v\in A}$ etc.\ for each subset $A$ of $V$. For disjoint subsets $A$, $B$, and $C$ of $V$
we use a short notation $A\mbox{\,$\perp\!\!\!\perp$\,} B\,|\, C$ to denote that $X_A$ is \emph{conditionally independent of $X_B$ given $X_C$} \cite{daw79,lau96}, i.e.\ that for any measurable $\Omega\subseteq \mathcal{X}_A$ and $P$-almost all $x_B$ and $x_C$,
$$P(X_A \in \Omega\,|\, X_B=x_B, X_C=x_C)=P(X_A \in \Omega\,|\, X_C=x_C).$$
We can now induce an independence model $\mathcal{J}(P)$ by letting
\begin{displaymath}
\langle A,B\,|\, C\rangle\in \mathcal{J}(P) \text{ if and only if } A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C \text{ w.r.t.\ $P$}.
\end{displaymath}
Similarly we use the notation $A\nolinebreak{\not\hspace{-1.5mm}\ci} B\,|\, C$ for $\langle A,B\,|\, C\rangle\notin \mathcal{J}(P)$.

We say that non-empty $A$ and $B$ are \emph{completely independent} if, for every $C\subseteq V\setminus (A\cup B)$, $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C$. Similarly, we say that $A$ and $B$ are \emph{completely dependent} if, for any $C\subseteq V\setminus (A\cup B)$, $A\nolinebreak{\not\hspace{-1.5mm}\ci} B\,|\, C$.

If $A$, $B$, or $C$ has only one member $\{u\}$, $\{v\}$, or $\{w\}$, for better readability, we write $u\mbox{\protect $\: \perp \hspace{-2.3ex v\,|\, w$. We also write $A\mbox{\protect $\: \perp \hspace{-2.3ex B$ when $C=\varnothing$, which denotes the \emph{marginal independence} of $A$ and $B$.

A probabilistic independence model $\mathcal{J}(P)$ over a set $V$ is always a \emph{semi-graphoid} \cite{pea88}, i.e., it satisfies the four following properties for disjoint subsets $A$, $B$, $C$, and $D$ of $V$:
 \begin{enumerate}
    \item $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C$ if and only if $B\mbox{\protect $\: \perp \hspace{-2.3ex A\,|\, C$ (\emph{symmetry});
    \item if $A\mbox{\protect $\: \perp \hspace{-2.3ex B\cup D\,|\, C$ then $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C$ and $A\mbox{\protect $\: \perp \hspace{-2.3ex D\,|\, C$ (\emph{decomposition});
    \item if $A\mbox{\protect $\: \perp \hspace{-2.3ex B\cup D\,|\, C$ then $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C\cup D$ and $A\mbox{\protect $\: \perp \hspace{-2.3ex D\,|\, C\cup B$ (\emph{weak union});
    \item if $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C\cup D$ and $A\mbox{\protect $\: \perp \hspace{-2.3ex D\,|\, C$ then $A\mbox{\protect $\: \perp \hspace{-2.3ex B\cup D\,|\, C$ (\emph{contraction}).
 \end{enumerate}
Notice that the reverse implication of contraction clearly holds by decomposition and weak union. A semi-graphoid for which the reverse implication of the weak union property holds is said to be a \emph{graphoid}; that is, it also satisfies
\begin{itemize}
	\item[5.] if $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C\cup D$ and $A\mbox{\protect $\: \perp \hspace{-2.3ex D\,|\, C\cup B$ then $A\mbox{\protect $\: \perp \hspace{-2.3ex B\cup D\,|\, C$ (\emph{intersection}).
\end{itemize}
Furthermore, a graphoid or semi-graphoid for which the reverse implication of the decomposition property holds is said to be \emph{compositional}, that is, it also satisfies
\begin{itemize}
	\item[6.] if $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C$ and $A\mbox{\protect $\: \perp \hspace{-2.3ex D\,|\, C$ then $A\mbox{\protect $\: \perp \hspace{-2.3ex B\cup D\,|\, C$ (\emph{composition}).
\end{itemize}
If, for example, $P$ has strictly positive density, the induced probabilistic independence model is always a graphoid; see e.g.\ Proposition 3.1 in \cite{lau96}. See also \cite{pet15} for a necessary and sufficient condition for $P$ in order for the intersection property to hold. If the distribution $P$ is a regular multivariate Gaussian distribution, $\mathcal{J}(P)$ is a compositional graphoid; e.g.\ see \cite{stu05}.
Probabilistic independence models with positive densities are not in general compositional; this only holds for special types of multivariate distributions such as, for example,  Gaussian distributions and the symmetric binary distributions used in \cite{wer09}.

Another important property that is not necessarily satisfied by probabilistic independence models is \emph{singleton-transitivity} (also called \emph{weak transitivity} in \cite{pea88}, where it is shown that for Gaussian and binary distributions $P$, $\mathcal{J}(P)$ always satisfies it). For $u$, $v$, and $w$, single elements in $V$,
 \begin{itemize}
	\item[7.] if $u\mbox{\protect $\: \perp \hspace{-2.3ex v\,|\, C$ and $u\mbox{\protect $\: \perp \hspace{-2.3ex v\,|\, C\cup\{w\}$ then $u\mbox{\protect $\: \perp \hspace{-2.3ex w\,|\, C$ or $v\mbox{\protect $\: \perp \hspace{-2.3ex w\,|\, C$ (\emph{singleton-transitivity}).
\end{itemize}
In addition, we have the two following properties:
 \begin{itemize}
\item[8.] if $u\mbox{\protect $\: \perp \hspace{-2.3ex v\,|\, C$ then $u\mbox{\protect $\: \perp \hspace{-2.3ex v\,|\, C\cup\{w\}$ for every $w\in V\setminus \{u,v\}$ (\emph{upward-stability});
\item[9.] if $u\mbox{\protect $\: \perp \hspace{-2.3ex v\,|\, C$ then $u\mbox{\protect $\: \perp \hspace{-2.3ex v\,|\, C\setminus\{w\}$ for every $w\in V\setminus \{u,v\}$ (\emph{downward-stability}).
\end{itemize}
Henceforth, instead of saying that ``$\mathcal{J}(P)$ satisfies these properties'', we simply say that ``$P$ satisfies these properties''. First we provide the following well-known result \cite{sad17}:
\begin{lemma}\label{lem:1}
For a probability distribution $P$ the following holds:
\begin{enumerate}
  \item If $P$ satisfies upward-stability then $P$ satisfies composition.
  \item If $P$ satisfies downward-stability then $P$ satisfies intersection.
\end{enumerate}
\end{lemma}
\subsection{Exchangeability for random vectors and networks}
A probability distribution $P$ over  a finite vector $(X_1, X_2, X_3, \dots, X_n)$ of random variables with the
same shared sample space is \emph{(finitely) exchangeable} if for any permutation $\pi\in S(n)$ of the indices $1, 2, 3, \dots, n$, the probability distribution of the permuted vector $(X_{\pi(1)},X_{\pi(2)},X_\pi{(3)},\dots,X_{\pi(n)})$ is the same as $P$; see \cite{ald85}. We shall for brevity say that the sequence $X$ is exchangeable in the meaning that its distribution is.

We are also concerned with probability distributions on networks that
are finitely exchangeable. A distribution $P$ of a random  matrix $X=(X_{ij})_{i,j\in \mathcal{N}}$  over a finite node set $\mathcal{N}$ with the
same shared sample space is said to be \emph{ (finitely) weakly exchangeable} \citep{sil76,eag78} if  for all permutations $\pi\in S(\mathcal{N})$ we have that
\begin{equation}
    \mathbb{P}\{(X_{ij}=x_{ij})_{i,j\in \mathcal{N}}\}=\mathbb{P}\{(X_{ij}=x_{\pi(i)\pi(j)})_{i,j\in \mathcal{N}}\}.
\end{equation}
If the matrix $X$ is symmetric --- i.e.\ $X_{ij}=X_{ji}$, we say it is \emph{symmetric weakly exchangeable}. Again, we shall for brevity say that $X$ is weakly or symmetric weakly exchangeable in the meaning that its distribution is.

A symmetric binary array with zero diagonal can be interpreted as a matrix of ties (the \emph{adjacency matrix}) of a random network and, thus, the above concepts can be translated into networks. A random network is \emph{exchangeable} if its adjacency matrix is symmetric weakly exchangeable.  Then it is easy to observe that
 a random network is exchangeable if and only if its distribution is invariant
 under relabeling of the nodes of the network.
\subsection{Undirected and bidirected graphical models}\label{sec:23}

Graphical models \citep[see, e.g.][]{lau96} are statistical models expressing
conditional independence statements among a collection of random variables $X_V = (X_v, v
\in V)$ indexed by a finite set $V$. A graphical model is determined by a graph
$G=(V,E)$ over the
indexing set $V$, and the edge set $E$ (which may include edges of undirected, directed
or bidirected type) encodes conditional independence
relations among the variables, or {\it Markov properties}.

We say that $C$ \emph{separates} $A$ and $B$ in an undirected graph $G$, denoted by $A\perp_u B\,|\, C$, if every path between $A$ and $B$ has a vertex in $C$, that is there is no path between $A$ and $B$ outside $C$.  For a bidirected graph $G$, we say that $C$ \emph{separates} $A$ and $B$, denoted by $A\perp_b B\,|\, C$, if every path between $A$ and $B$ has a vertex outside $C\cup A\cup B$, that is there is no path between $A$ and $B$ within $A\cup B\cup C$.  Note the obvious duality between this and separation for undirected graphs. We might skip the subscripts $u$ and $b$ in $\perp_u$ and $\perp_b$ when it is apparent from the context with which separation we are dealing.

A joint probability
distribution $P$ for $X_V$ is \emph{Markovian} with respect to an undirected graph \citep{dar80} with the vertex set $V$ if $A\perp_u B\,|\, C$  implies $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C$. $P$ is \emph{Markovian} with respect to a bidirected graph \citep{cox93,kau96} if $A\perp_b B\,|\, C$  implies $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C$.

For example, in the undirected graph of Figure \ref{fig:001}(a),
the global Markov property implies that $\{u,x\}\mbox{\protect $\: \perp \hspace{-2.3ex w \,|\, v$, whereas in the bidirected graph of Figure \ref{fig:001}(b),
the global Markov property implies that $\{u,x\}\mbox{\protect $\: \perp \hspace{-2.3ex w$.

\begin{figure}[htb]
\centering
\begin{tikzpicture}[node distance = 8mm and 8mm, minimum width = 6mm]

    \begin{scope}
      \tikzstyle{every node} = [shape = circle,
      font = \scriptsize,
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		\caption{{\footnotesize (a) An undirected dependence graph. (b) a bidirected dependence graph.}}\label{fig:001}
		\end{figure}

If, for $P$ and undirected $G$, $A\perp_u B\,|\, C\iff A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C$ then we say that $P$ and $G$ are \emph{faithful}; Similarly, if, for $P$ and bidirected $G$, $A\perp_b B\,|\, C\iff A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C$ then $P$ and $G$ are \emph{faithful}. Hence, faithfulness implies being Markovian, but not the other way around.

For a given probability distribution $P$, we define the \emph{skeleton} of
$P$, denoted by $\mathrm{sk}(P)$, to be the undirected graph with the vertex set $V$ such
that vertices $u$ and $v$ are not adjacent if and only if there is \emph{some} subset $C$
of $V$ so that $u\mbox{\,$\perp\!\!\!\perp$\,} v\,|\, C$.  Thus, if $P$ is Markovian with respect to an
undirected graph $G$ then $\mathrm{sk}(P)$ would be a subgraph of $G$ (since for every missing edge $ij$ in $G$, $i\mbox{\protect $\: \perp \hspace{-2.3ex j\,|\, V\setminus\{i,j\}$); and if $P$ is
Markovian with respect to a bidirected graph $G$ then $\mathrm{sk}(P)$ is a subgraph of $\mathrm{sk}(G)$ (since for every missing edge $ij$ in $G$, $i\mbox{\protect $\: \perp \hspace{-2.3ex j$).

In general, a graph $G(P)$ is induced by $P$ with skeleton $\mathrm{sk}(P)$. For undirected graphs, let $G_u(P)=\mathrm{sk}(P)$, whereas for bidirected graphs, let $G_b(P)$  be $\mathrm{sk}(P)$ with all edges being bidirected. We shall need the following results from \cite{sad17} (where the first part was first shown in \cite{pea85}):

\begin{prop}\label{prop:und}
Let $P$ be a probability distribution defined over $\{X_\alpha\}_{\alpha\in V}$. It then holds that
\begin{enumerate}
  \item $P$ and $G_u(P)$ are faithful if and only if $P$ satisfies intersection, singleton-transitivity, and upward-stability.
  \item $P$ and $G_b(P)$ are faithful if and only if $P$ satisfies composition, singleton-transitivity, and downward-stability.
\end{enumerate}
\end{prop}
\subsection{Duality in independence models and graphs}\label{sec:dual}
The \emph{dual} of an independence model $\mathcal{J}$ (defined in \cite{mat92} under the name of \emph{dual relation}) is the independence model defined by $\mathcal{J}^d=\{\langle A,B\,|\, V\setminus (A\cup B\cup C)\rangle:\nn\langle A,B\,|\, C\rangle\in \mathcal{J}\}$; see also \cite{drt10}. We will need the following lemma regarding the duality of independence models:
\begin{lemma}\label{lem:dual}
For an independence model  $\mathcal{J}$ and its dual $\mathcal{J}^d$,
\begin{enumerate}
  \item $\mathcal{J}$ is semi-graphoid if and only if $\mathcal{J}^d$ is semi-graphoid;
  \item $\mathcal{J}$ satisfies intersection  if and only if $\mathcal{J}^d$ satisfies composition; and vice versa;
  \item $\mathcal{J}$ satisfies singleton-transitivity if and only if $\mathcal{J}^d$ satisfies singleton-transitivity;
  \item  $\mathcal{J}$ satisfies upward-stability  if and only if $\mathcal{J}^d$ satisfies downward-stability; and vice versa.
\end{enumerate}
\end{lemma}
\begin{proof}
1., 2., and 3.\ are proven in \cite{lne07}, which showed that if  $\mathcal{J}$ is singleton-transitive compositional graphoid, so is the dual couple of $\mathcal{J}$ (although this is proven based on the so-called \emph{Gaussoid} formulation). An alternative proof for these,  with the same formulation as in this paper, is found in \cite{mal13}. (In fact the statement in the mentioned article was proven for a generalization of singleton-transitivity, called \emph{dual decomposable transitivity}).

In order to prove 4., suppose that $\mathcal{J}$ satisfies upward-stability, and assume that $\langle i,j\,|\, C\rangle\in \mathcal{J}^d$ and $k\in C$. Therefore, $\langle i,j\,|\, V\setminus (\{i,j\}\cup C)\rangle\in \mathcal{J}$. Hence, $\langle i,j\,|\, V\setminus (\{i,j\}\cup C)\cup\{k\}\rangle\in \mathcal{J}$ because of upward-stability. Therefore  $\langle i,j\,|\, C\setminus\{k\}\rangle\in \mathcal{J}^d$, which implies downward-stability of $\mathcal{J}^d$. The other direction is similar.
\end{proof}
As an additional statement to the lemma, it can also be shown that $\mathcal{J}$ is closed under \emph{marginalization} if and only if $\mathcal{J}^d$ is closed under \emph{conditioning}; and vice versa; but this result is not needed in this paper.

In addition, for separation in graphs, we will use the following lemma related to duality:
\begin{lemma}\label{lem:5}
Let $G_u$ and $G_b$ be an undirected and a bidirected graph such that $\mathrm{sk}(G_u)=\mathrm{sk}(G_b)$. Then, the independence model $\mathcal{J}(G_b)$ induced by $G_b$ is $\mathcal{J}^d(G_u)$ and vice versa.
\end{lemma}
\begin{proof}
Since the separation satisfies the composition property, it is sufficient to prove the statement for singletons. Suppose that there is a connecting path $\omega$ between $i$ and $j$ given $C$ in $G_u$. This means that no inner vertex of $\omega$ is in $C$; thus they are all in $V\setminus (\{i,j\}\cup C)$. Therefore, in $G_b$, $i$ and $j$ are connecting given $V\setminus (\{i,j\}\cup C)$. The other direction (where $i\nolinebreak{\not\hspace{-1.5mm}\dse} j\,|\, C$ in $G_b\Rightarrow i\nolinebreak{\not\hspace{-1.5mm}\dse} j\,|\, V\setminus (A\cup B\cup C)$ in $G_u$) is proven in a similar way.
\end{proof}

In fact, the second part of Proposition \ref{prop:und}, could be implied by the first part, and vice versa, using Lemmas \ref{lem:dual} and \ref{lem:5}; and the second part of Lemma \ref{lem:1}, could be implied by the first part, and vice versa, using Lemma \ref{lem:dual}.
\section{Results for vector exchangeability}\label{sec:3}
We shall study the relationship between vector exchangeability and conditional independence by using the definitions and results in the previous section. In the entire section, we assume that $P$ is a probability distribution defined over the vector $(X_v)_{v\in V}$. First, notice that exchangeability is closed under marginalization and conditioning:
\begin{prop}\label{prop:3n}
If $X_V$ is an exchangeable random vector then so are the marginal vectors $X_A$, for $A\subseteq V$, and the conditional vectors $X_A\,|\, X_C=x^*_C$, for disjoint $A,C$ where $V=A\cup C$, if they exist.
\end{prop}
\begin{proof}
First we prove closedness under marginalization: For a permutation matrix $\pi$ of $A$, let the permutation matrix over $V$ be $\pi^*(a)=\pi(a)$, for $a\in A$, and $\pi^*(b)=b$ for $b\in V\setminus A$. The proof then follows from exchangeability of $X_V$.

Now we prove closedness under conditioning: By the definition of conditioning, we need to prove that $P(x_A,x^*_C)=P(x_{\pi(A)},x^*_C)$ for every $\pi$. This is again true by considering $\pi^*(a)=\pi(a)$, for $a\in A$, and $\pi^*(c)=c$ for $c\in C$.
\end{proof}
Notice that the above result implies that the marginal/conditional  $X_A\,|\, X_C$, (i.e., when $A\cup C\subset V$) is also exchangeable. We now have the following results:
\begin{prop}\label{prop:3}
If $P$ satisfies vector exchangeability then the following holds:
\begin{enumerate}
  \item $P$ satisfies upward-stability if and only if it satisfies composition.
  \item $P$ satisfies downward-stability if and only if it satisfies intersection.
\end{enumerate}
\end{prop}
\begin{proof}
1. ($\Rightarrow$) follows from Lemma \ref{lem:1}. To prove ($\Leftarrow$), let $i\mbox{\protect $\: \perp \hspace{-2.3ex j\,|\, C$. By exchangeability, for an arbitrary $k\notin C\cup\{i,j\}$, we have $i\mbox{\protect $\: \perp \hspace{-2.3ex k\,|\, C$. Composition implies $i\mbox{\protect $\: \perp \hspace{-2.3ex j\cup k\,|\, C$. Weak union implies $i\mbox{\protect $\: \perp \hspace{-2.3ex j\,|\, C\cup\{k\}$.

2. follows from 1.\ and the consequence of duality provided in Lemma \ref{lem:dual}
\end{proof}

\begin{prop}\label{prop:4}
If $P$ is exchangeable then it satisfies singleton-transitivity.
\end{prop}
\begin{proof}
If $i\mbox{\protect $\: \perp \hspace{-2.3ex j\,|\, C$ and $k\notin C\cup\{i,j\}$ then clearly $i\mbox{\protect $\: \perp \hspace{-2.3ex k\,|\, C$ and $k\mbox{\protect $\: \perp \hspace{-2.3ex j\,|\, C$.
\end{proof}
We see that the skeleton of an exchangeable distribution can only take a very specific form:
\begin{prop}\label{prop:5}
If $P$ is exchangeable then $\mathrm{sk}(P)$ is either an empty or a complete graph.
\end{prop}
\begin{proof}
If there is any independence statement of form $i\mbox{\protect $\: \perp \hspace{-2.3ex j\,|\, C$ then by permutation for all variables, we obtain $k\mbox{\protect $\: \perp \hspace{-2.3ex l\,|\, C'$  for all $k$ and $l$. Therefore, $\mathrm{sk}(P)$ is empty. If there is no independence statement of this form then $\mathrm{sk}(P)$ is complete.
\end{proof}

Notice that for faithfulness to empty or complete graphs, it is immaterial
whether one considers undirected or bidirected interpretation of graphs.
\begin{coro}
Let $P$ satisfy vector exchangeability. If $P$ is faithful to a graph (both under undirected interpretation and under bidirected
interpretation) then the graph is empty or complete.
\end{coro}
Hence, there are two regimes available: if there is no independence statement implied by $P$ then we are in the complete graph regime; and if there is at least one conditional independence statement implied by $P$ then we are in the empty graph regime. The following also provides conditions for the opposite direction of the above result:
\begin{theorem}
If $P$ is exchangeable then $P$ is faithful to a graph (both under undirected interpretation and under bidirected
interpretation) if and only if $P$ satisfies the intersection and composition properties. The graph must then be either empty or complete.
\end{theorem}
\begin{proof}
The first result follows from Propositions \ref{prop:und}, \ref{prop:3}, and \ref{prop:4}. The second follows from Proposition \ref{prop:5}.
\end{proof}
Indeed other exchangeable distributions may exist but they are not faithful to a graph. We then have the following corollaries:
\begin{coro}
 Let $P$ be exchangeable and there exists an independence statement induced by $P$. It then holds that  all variables $X_v$ are completely independent of each other if and only if $P$ satisfies intersection and composition.
\end{coro}
\begin{coro}
Let $P$ be a regular exchangeable Gaussian distribution. If there is a zero element in its covariance matrix then all $X_v$ are completely independent; and otherwise they are completely dependent.
\end{coro}
\begin{proof}
The proof follows from the fact that a regular Gaussian distribution satisfies the intersection and composition properties.
\end{proof}
Notice that the above statement could be shown otherwise since a zero off-diagonal entry of the covariance matrix can be permuted by exchangeability to every other off-diagonal entry, making the covariance matrix diagonal.
\begin{prop}
If $P$ is exchangeable, satisfies intersection (this holds when $P$ has a positive density), and if there exists one independence statement induced by $P$ then all variables $X_v$ are marginally independent of each other.
\end{prop}
\begin{proof}
By Proposition \ref{prop:3}, $P$ satisfies downward-stability. An independence statement $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C$, by the use of decomposition implies $i\mbox{\protect $\: \perp \hspace{-2.3ex j\,|\, C$ for an arbitrary $i\in A$ and $j\in B$.  Downward-stability implies $i\mbox{\protect $\: \perp \hspace{-2.3ex j$. Exchangeability implies all pairwise marginal independences.
\end{proof}
\begin{example}
Consider an exchangeable distribution $P$ over four variables $(i,j,k,l)$ with $i\mbox{\protect $\: \perp \hspace{-2.3ex j\,|\, k$. By exchangeability, all independences of form $\pi(i)\mbox{\protect $\: \perp \hspace{-2.3ex \pi(j)\,|\, \pi(k)$ hold for any permutation $\pi$ on  $(i,j,k,l)$. It is easy to see that none of the semi-graphoid axioms can generate new independence statements from these. If intersection holds then, for example, from $i\mbox{\protect $\: \perp \hspace{-2.3ex j\,|\, k$ and $i\mbox{\protect $\: \perp \hspace{-2.3ex k\,|\, j$, we obtain $i\mbox{\protect $\: \perp \hspace{-2.3ex \{j,k\}$, which by decomposition implies $i\mbox{\protect $\: \perp \hspace{-2.3ex j$, and hence all marginal independences between singletons.  If composition holds then, for example, from $i\mbox{\protect $\: \perp \hspace{-2.3ex j\,|\, k$ and $i\mbox{\protect $\: \perp \hspace{-2.3ex l\,|\, k$, we obtain $i\mbox{\protect $\: \perp \hspace{-2.3ex \{j,l\}\,|\, k$, which by weak union implies $i\mbox{\protect $\: \perp \hspace{-2.3ex j\,|\, \{k,l\}$, and hence all conditional independences between singletons given the remaining variables.
\end{example}

\section{Results for exchangeability for random networks}\label{sec:4}
Henceforth, in the context of random networks, we consider vectors whose components are indexed by dyads (i.e.\ two-element subsets of the node set $\mathcal{N}$). Thus,
for a conditional independence statement of the form $A\mbox{\protect $\: \perp \hspace{-2.3ex B \,|\, C$, $A,B,C \subset \mathcal{D}(\mathcal{N})$ are
pairwise disjoint subsets of dyads. (Later on, we particularly write the conditioning sets as simply $C$, which should be considered a subset of dyads.) Notice that we simply use the notation $ij$ for a dyad whose endpoints are nodes $i$ and $j$. This is different from the notation $\{i,j\}$, which indicates the set of two nodes $i$ and $j$ as used in the previous section.
\subsection{Marginalization and conditioning for exchangeable random networks}
Here we focus on marginalization over and conditioning on arbitrary sets of dyads. Notice that by \emph{marginalizing over} a set $M$, we mean we marginalize the set $M$ out, which results in a distribution over $\mathcal{D}(\mathcal{N})\setminus M$.
However, exchangeable networks are not always closed under marginalization over or conditioning on an arbitrary set of dyads:

For a marginal network $X_A$, where $A$ is a subset of dyads, exchangeability and summing up all probabilities over values of the dyads that are marginalized over imply that $P(X_A=x_A)=P((X_{\pi(i)\pi(j)})_{ij\in A}=x_A)$, for any permutation $\pi$. However, this is not necessarily equal to $P(X_A=(x_{\pi(i)\pi(j)})_{ij\in A})$, which is what we need for exchangeability of the marginal to hold. For conditioning, in fact, $X_A\,|\, X_C$ is not exchangeable if a node appears in a dyad in $A$ and a dyad in $C$ (e.g.\ $i$ appearing in $ij\in A$ and $ik\in C$). This is because, for a permutation $\pi$ that maps $i$ to a node other than $i$, exchangeability of $X_A\,|\, X_C$ is equivalent to $P(x_A\,|\, x_C)=P((x_{\pi(i)\pi(j)})_{ij\in A}\,|\, x_C)$, which itself is equivalent to $P(x_A,x_C)=P((x_{\pi(i)\pi(j)})_{ij\in A},x_C)$. But, this does not necessarily hold as $i$ is mapped to another node in $A$ but not in $C$.

However, we have the following:
\begin{prop}
Let $A$ and $C$  be disjoint subsets of dyads of an exchangeable random network $X$ such that $A$ and $C$ do not share any nodes, i.e.\ if $ij\in A$ then there is no dyad $ik$ or $jk$ in $C$ for any node $k\in\mathcal{N}$. It then holds that the conditional/marginal random network $X_A\,|\, X_C$ is exchangeable.
\end{prop}
\begin{proof}
Let $\mathcal{N}(A\cup C)$ be the set of all endpoints of dyads in $A$ and $C$, and define similarly $\mathcal{N}(A)$ and $\mathcal{N}(C)$. Define the permutation $\pi^*\in S(\mathcal{N}(A\cup C))$ such that $\pi^*(i)=\pi(i)$ for $i\in \mathcal{N}(A)$ and $\pi^*(k)=k$ for $k\in \mathcal{N}(C)$. Notice that this is well-defined since $A$ and $C$ do not share any nodes. Using $\pi^*$ and by exchangeability of $X$, we conclude that $P(x_A,x_C)=P((x_{\pi(i)\pi(j)})_{ij\in A},x_C)$, which, as mentioned before, is equivalent to the exchangeability of $X_A\,|\, X_C$.
\end{proof}

\subsection{Types of graphs faithful to exchangeable distributions}\label{sec:41}
An analogous result to that of vector exchangeability, concerning the skeleton of an exchangeable probability distribution, was proven in \cite{lau17}:
\begin{prop}\label{prop:7}
If a distribution $P$ over a random network $X$ is exchangeable then $\mathrm{sk}(P)$ is one of the following:
\begin{enumerate}
  \item the empty graph;
  \item the incidence graph;
  \item the complement of the incidence graph;
  \item the complete graph.
\end{enumerate}
\end{prop}
Notice that a pairwise independence statement for an exchangeable $P$ over a random network $X$ is of form $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$, where $i,j,k,l$ are nodes of $X$. Depending on the type of $\mathrm{sk}(P)$, these statements take different forms.
\begin{lemma}\label{lem:2}
Suppose that there exists an independence statement of form $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$, $i\neq j,k\neq l$, for an exchangeable $P$ over a random network $X$. It is then in one  of the following forms depending on the type of $\mathrm{sk}(P)$:
\begin{enumerate}
  \item empty graph $\Rightarrow$ no constraints on $i,j,k,l$;
  \item incidence graph $\Rightarrow$ $i\neq l,k$ and $j\neq l,k$;
  \item complement of the incidence graph $\Rightarrow$ $i=k$ or $i=l$ or $j=k$ or $j=l$;
\end{enumerate}
if $\mathrm{sk}(P)$ is the complete graph then it is not possible to have such a conditional independence statement.
\end{lemma}
\begin{proof}
The proof follows from the fact if there is an edge between $ij$ and $kl$ in $G(P)$ then there is no statement of form $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$.
\end{proof}
It is clear that if $\mathrm{sk}(P)$ is the complete graph then every dyad is completely dependent on every other dyad. Thus we consider the cases of the incidence graph skeleton and the complement of the incidence graph skeleton separately. However, before this, we show that among the known graphical models, only undirected and bidirected graphs can be faithful to an exchangeable distribution.

In order to do so, we can start off by considering any class of \emph{mixed graphs}, i.e.\ graphs with simultaneous undirected, directed, or bidirected edges that use the (unifying) separation criterion introduced in \cite{sadl16}. To the best of our knowledge, the largest class of such graphs is the class of \emph{chain mixed graphs} \cite{sadl16}, which includes the classes of \emph{ancestral graphs} \cite{ric02}, \emph{LWF} \cite{lau89} and \emph{regression chain graphs} \cite{wer94}, and several others; see \cite{sadl16}. We require some definitions and results, including the formulation of the separation criterion, which we only need for the results in this subsection. We provide these together with the proof of the main result (Theorem \ref{thm:2}) in Appendix \ref{app}. If the reader is only interested in the statement of the theorem and not the proof, we suggest that they skip the material in the appendix.

Two graphs are called \emph{Markov equivalent} if they induce the same independence model.
\begin{theorem}\label{thm:2}
If a distribution $P$  over an exchangeable random network with $n$ nodes is faithful to a chain mixed graph $G$ then $G$ is Markov equivalent to one of the following graphs:
\begin{enumerate}
  \item the empty graph;
  \item  the undirected incidence graph, $L_{-}(n)$;
  \item  the bidirected incidence graph, $L_{\leftrightarrow}(n)$;
  \item   the undirected complement of the incidence graph, $L^c_{-}(n)$;
  \item  the bidirected complement of the incidence graph, $L^c_{\leftrightarrow}(n)$;
  \item the complete graph.
\end{enumerate}
\end{theorem}
Hence, for exchangeable random networks, there are six regimes available, where these are three pairs that are complement of each other. Within the two non-trivial pairs, the two undirected and bidirected cases act as dual of each other as described in Section \ref{sec:dual}. We will make use of this duality to simplify the results and proofs.

Although the following method is not unique, here we provide a simple test to decide in which regime a given exchangeable distribution lies:
\begin{alg}\label{alg:1}
For arbitrary fixed nodes $i,j,k,l,m$ of a given exchangeable network, test the following:
\begin{itemize}
 \item $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$, for some $C$, and $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C'$, for some $C'\Rightarrow$ Empty graph;
 \item $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$, for some $C$, and $ij\nolinebreak{\not\hspace{-1.5mm}\ci} ik\,|\, C'$, for any $C'\Rightarrow L(n)$:
 \begin{itemize}
 \item $ik\in C\Rightarrow L_{-}(n)$;
 \item $ik\notin C\Rightarrow L_{\leftrightarrow}(n)$;
\end{itemize}
 \item $ij\nolinebreak{\not\hspace{-1.5mm}\ci} kl\,|\, C$, for any $C$, and $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C'$, for some $C'\Rightarrow L^c(n)$:
 \begin{itemize}
 \item $lm\in C'\Rightarrow L^c_{-}(n)$;
 \item $lm\notin C'\Rightarrow L^c_{\leftrightarrow}(n)$;
\end{itemize}
 \item $ij\nolinebreak{\not\hspace{-1.5mm}\ci} kl\,|\, C$, for any $C$, and $ij\nolinebreak{\not\hspace{-1.5mm}\ci} ik\,|\, C'$, for any $C'\Rightarrow$ Complete graph.
\end{itemize}
\end{alg}
\begin{prop}
If a distribution $P$  over an exchangeable random network is faithful to a graph $G$ then Algorithm \ref{alg:1} determines the Markov equivalence class of $G$.
\end{prop}
\begin{proof}
First, we show that this test covers all the possible cases: The first level tests ($ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$ and $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C'$) clearly cover all the cases concerning the skeleton of the graph. But, the second level test ($ik\in C$ or $lm\in C'$), which only concerns the non-trivial skeletons, might not be consistent: For example, first consider the $L(n)$ case, and assume that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C_1$ and $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C_2$, for some $C_1,C_2$, but $ik\in C_1$ and $ik\notin C_2$. However, in such cases, it is easy to show that
$P$ cannot be faithful to either of the two undirected or bidirected graphs. The case of $L^c(n)$ is similar.

The algorithm also outputs the correct regimes: The tests of $ij\nolinebreak{\not\hspace{-1.5mm}\ci} kl\,|\, C$ and $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C'$ clearly determine the skeleton $\mathrm{sk}(P)$. The test $ik\in C$ then determines the type of edges since, in $\langle ij,ik,kl\rangle$, if $ik\notin C$ then $ij$ and $kl$ cannot be separated in the undirected graph; and if $ik\in C$ then $ij$ and $kl$ cannot be separated in the bidirected graph. The test $lm\in C'$ can be proven similarly.
\end{proof}
Unlike the vector exchangeability case, the intersection and composition properties are not in general sufficient for faithfulness of exchangeable network distributions to the graphs provided in Theorem \ref{thm:2}. However, in special cases this holds. We detail this below for each regime mentioned above.
\subsection{The incidence graph case}
In this section, we assume that $\mathrm{sk}(P)=L_{-}(n)$. The following example shows that, in principle, intersection and composition are not sufficient for faithfulness of $P$ and $L_{-}(n)$ (and $P$ and the bidirected incidence graph $L_{\leftrightarrow}(n)$).
\begin{example}\label{ex:2}
Suppose that there is an exchangeable $P$ that induces $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C_{ij,kl}$, where  $C_{ij,kl}=\{ik,il,jk,jl\}$. Exchangeability implies that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C_{ij,kl}$ for all $i,j,k,l$. Suppose, in addition, that $P$ satisfies upward-stability. Notice that here we do not show that such a probability distribution necessarily exists -- one can treat this defined independence model as an example of ``a network-exchangeable semi-graphoid".

It is easy to see that $\mathrm{sk}(P)= L_{-}(n)$. Moreover, by Lemma \ref{lem:1}, $P$ satisfies composition.
 In addition, $P$ satisfies intersection: Notice that none of the semi-graphoid axioms plus upward-stability and composition imply an independence statement of form $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C$ where $A,B$ both contain a node $i$. In addition, it can be seen that these axioms imply that if there is $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C$ then for every $ij\in A$ and $kl\in B$, it holds that $C_{ij,kl}\subseteq C$. Let $C_{A,B}=\bigcup_{ij\in A,kl\in B}C_{ij,kl}$. By these two observations, we conclude that if $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C\cup D$ and $A\mbox{\protect $\: \perp \hspace{-2.3ex D\,|\, C\cup B$ then $C_{A,D}\subseteq C$. By upward-stability all statements of form  $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$ where $C_{ij,kl}\subseteq C$ hold. Hence, by composition, we have that  $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C$. Hence, by contraction,  $A\mbox{\protect $\: \perp \hspace{-2.3ex B\cup D\,|\, C$.

However, $P$ does not satisfy singleton-transitivity: Consider $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C_{ij,kl}$ and $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C_{ij,kl}\cup\{km\}$. If, for contradiction, singleton-transitivity holds then, because $\mathrm{sk}(P)=L_{-}(n)$, we have that $ij\mbox{\protect $\: \perp \hspace{-2.3ex km\,|\, C_{ij,kl}$. But, it can be seen that this statement is not in the independence model since no compositional graphoid axioms can generate this statement from $\{ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C_{ij,kl}\}$.

By Proposition \ref{prop:und} (1.), it is implied that, although intersection and composition are satisfied, $P$ is not faithful to $L_{-}(n)$.

If we now suppose that $P$ induces $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C^d_{ij,kl}$, where $C^d_{ij,kl}=V\setminus (C_{ij,kl}\cup\{ij,kl\})$, $P$ is exchangeable, and it satisfies downward-stability then, by using the duality (Lemmas \ref{lem:dual} and \ref{lem:5}), we conclude that although intersection and composition are satisfied, $P$ is not faithful to $L_{\leftrightarrow}(n)$.
\end{example}
However, under certain assumptions, intersection and composition are sufficient for faithfulness. We will consider the two dual regimes within the incidence graph case related to the cases of Theorem \ref{thm:2}: the undirected incidence graph case and the bidirected incidence graph case. We make use of the duality between these to immediately extend the results in the undirected case to the bidirected case.

As in Example \ref{ex:2}, let $C_{ij,kl}=\{ik,il,jk,jl\}$. For the faithfulness results to the undirected case, one assumption that is used is that for some (and because of exchangeability for all) $i,j,k,l$, $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$ implies that $C_{ij,kl}\subseteq C$. For the faithfulness results to the bidirected case, one assumption is that for some (and because of exchangeability for all) $i,j,k,l$, $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$ implies that $C_{ij,kl}\cap C=\varnothing$.

For a separation statement $A\,\mbox{$\perp$}\, B\,|\, C$, we define $C$ to be a \emph{minimal} separator in the case that if we remove any vertex from $C$, the separation does not hold; we define a \emph{maximal} separator similarly.  Let also $C_{ij}=\{ir,jr:\forall r\neq i,j\}$ and  $C^d_{ij}=V\setminus (C_{ij}\cup \{ij\})=\{lm:\forall l,m\notin\{i,j\}\}$. It holds that $C_{ij}$ is a minimal separator of $ij,kl$ in $L_{-}(n)$ and $C^d_{ij}\setminus\{kl\}$ is a maximal separator  in $L_{\leftrightarrow}(n)$:
\begin{prop}\label{prop:12}

\begin{enumerate}
  \item In $L_{-}(n)$, it holds that $ij\,\mbox{$\perp$}\, kl\,|\, C_{ij}$, and if $ij\,\mbox{$\perp$}\, kl\,|\, C$  then $|C_{ij}|\leq |C|$. In fact, if $C\neq C_{ij}$ and $C\neq C_{kl}$ then $|C_{ij}|< |C|$.
  \item In $L_{\leftrightarrow}(n)$, it holds that $ij\,\mbox{$\perp$}\, kl\,|\, C^d_{ij}\setminus\{kl\}$, and if $ij\,\mbox{$\perp$}\, kl\,|\, C$  then $|C^d_{ij}|\geq |C|+1$. In fact, if $C\neq C^d_{ij}\setminus\{kl\}$ and $C\neq C^d_{kl}\setminus\{ij\}$ then $|C^d_{ij}|> |C|+1$.
\end{enumerate}
\end{prop}
\begin{proof}
1.\ The first claim is straightforward to prove since every path from $kl$ to $ij$ must pass through an adjacent vertex of $ij$, which contains $i$ or $j$. To prove the second statement, notice that if $im\notin C$ then at least $km$ and $lm$ must be in $C$. The same vertices $km$ and $lm$ appear if $jm\notin C$ too, but for no other vertices of $C_{ij}$ missing in $C$. Thus, for every missing member of $C_{ij}$ in $C$, there is at least a member of $C_{kl}$ that should be in $C$. Hence, $|C_{ij}|\leq |C|$.

To prove the third statement, suppose, for contradiction, that $C\neq C_{ij}$ and $C\neq C_{kl}$ and $|C|=|C_{ij}|$. Using the fact that, for every missing member of $C_{ij}$ in $C$, there is at least a member of $C_{kl}$ that should be in $C$, there cannot be any vertex outside $C_{ij}\cup C_{kl}$ in a $C$. Hence, $C\subset C_{ij}\cup C_{kl}$. If $n>5$ and, say, $im,jm\notin C$ but $km,lm\in C$ then consider the vertices $ih,jh,kh,lh$. Without loss of generality, assume that $ih,jh\in C$ but $kh,lh\notin C$. Then the path $\langle kl,kh,mh,jm,ij\rangle$ connects $kl$ and $ij$, a contradiction. The cases where $n=3,4,5$ are easy to check.

2.\ The proof follows from 1.\ by using the duality (Lemma \ref{lem:5}), and observing that $|C^d_{ij}\setminus\{kl\}|=|C^d_{ij}|-1$.
\end{proof}
However, not all minimal separators of $ij,kl$ in $L_{-}(n)$ are of the form above. For example, in $L_{-}(6)$, consider the set $C=\{ik,il,im,jk,jl,jm,hk,hl,mh\}$. It holds that $ij\,\mbox{$\perp$}\, kl\,|\, C$. However, $C$ is not of the form $C_{pq}$ for any pair $p,q\in\{i,j,k,l,m,h\}$. The same can be said about $L_{-}(n)$: Consider $C^d=V\setminus (C\cup \{ij,kl\})$. By Lemma \ref{lem:5}, we have that $ij\,\mbox{$\perp$}\, kl\,|\, C^d$ in $L_{\leftrightarrow}(6)$, and, in addition, $C^d$ is maximal.
\begin{prop}\label{prop:3n0}
Let a distribution $P$ be defined over an exchangeable random network and $\mathrm{sk}(P)=L_{-}(n)$, and consider some nodes $i,j,k,l$.
\begin{enumerate}
  \item Suppose that for every minimal $C$ such that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$, it holds that $C_{ij,kl}\subseteq C$ and $C$ is invariant under swapping $k$ and $m$, and $l$ and $h$, for every $m,h$, $m\neq h$, $mh\notin C\cup\{ij,kl\}$. It then holds that if $P$ satisfies composition then it satisfies upward-stability and singleton-transitivity.
  \item Suppose that for every maximal $C$ such that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$, it holds that $C_{ij,kl}\cap C=\varnothing$ and $C$ is invariant under swapping $k$ and $m$, and $l$ and $h$, for every $m,h$, $m\neq h$, $mh\in C$. It then holds that if $P$ satisfies intersection then it satisfies downward-stability and singleton-transitivity.
\end{enumerate}
\end{prop}
\begin{proof}
By Lemma \ref{lem:2}, we know that $i,j,k,l$ are all different.

1.\ First, we prove upward-stability. Suppose that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$. Notice that, because of exchangeability, the assumptions of the statement hold for every $i,j,k,l,m,h$. For a variable $mh\notin C\cup\{ij,kl\}$, we prove that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C\cup\{mh\}$ by induction on $|C|$. Notice that, by assumption, $mh\notin C_{ij,kl}$. In addition, without loss of generality, we can assume that $m,h\neq i,j$ since otherwise we can swap $i$ and $k$ and $j$ and $l$ and proceed as follows, and finally swap them back

The base case is when $C$ is minimal. Because of exchangeability, by the invariance of $C$ under the aforementioned swaps, and since $mh\notin C_{ij,kl}$, we have that $ij\mbox{\protect $\: \perp \hspace{-2.3ex mh\,|\, C$. By composition $ij\mbox{\protect $\: \perp \hspace{-2.3ex \{kl,mh\}\,|\, C$, which, by weak union, implies $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C\cup\{mh\}$.

Now suppose that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$, $C$ is not minimal, and, for every  $C'$ such that $|C'|<|C|$,  $[ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C'\Rightarrow ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C'\cup\{op\}]$, for every $op$. Consider the independence statement $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C_0$ such that $C_0\subset C$ and $C_0$ is minimal. We have that $ij\mbox{\protect $\: \perp \hspace{-2.3ex mh\,|\, C_0$. By induction hypothesis, we can add vertices to the conditioning set in order to obtain $ij\mbox{\protect $\: \perp \hspace{-2.3ex mh\,|\, C$.
Now, again composition and weak union imply the result.

Singleton-transitivity also follows from the above argument. We need to show that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$ and $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C\cup\{mh\}$ imply $ij\mbox{\protect $\: \perp \hspace{-2.3ex mh\,|\, C$ or $mh\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$. (Notice that because of upward-stability $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C\cup\{mh\}$ is immaterial.) Again without loss of generality, we can assume that $m,h\neq i,j$, and the above argument showed that $ij\mbox{\protect $\: \perp \hspace{-2.3ex mh\,|\, C$.

2.\ The proof follows from part 1.\ and the duality (Lemma \ref{lem:dual}).
\end{proof}

\begin{theorem}
Let a distribution $P$ be defined over an exchangeable random network and $\mathrm{sk}(P)=L_{-}(n)$, and consider some nodes $i,j,k,l$.
\begin{enumerate}
  \item Suppose that for every minimal $C$ such that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$, it holds that $C_{ij,kl}\subseteq C$ and $C$ is invariant under swapping $k$ and $m$, and $l$ and $h$, for every $m,h$, $m\neq h$, $mh\notin C\cup\{ij,kl\}$. Then $P$ is faithful to $L_{-}(n)$ if and only if $P$ satisfies the intersection and composition properties.
  \item  Suppose that for  every maximal $C$ such that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$, it holds that $C_{ij,kl}\cap C=\varnothing$ and $C$ is invariant under swapping $k$ and $m$, and $l$ and $h$, for every $m,h$, $m\neq h$, $mh\in C$. Then $P$ is faithful to $L_{\leftrightarrow}(n)$ if and only if $P$ satisfies the intersection and composition properties.
\end{enumerate}
\end{theorem}
\begin{proof}
The proof follows from Propositions \ref{prop:3n0} and \ref{prop:und}.
\end{proof}

\subsection{The complement of the incidence graph case}
In this section, we assume that $\mathrm{sk}(P)=L^c_{-}(n)$. We show again that, under certain assumptions, intersection and composition are sufficient for faithfulness. We again utilize the duality of the two additional regimes in the complement of the incidence graph case related to the cases of Theorem \ref{thm:2}: the undirected complement of the incidence graph case and the bidirected complement of incidence graph case. Let $C^d_{ijk}=\{lm:\forall l,m\notin\{i,j,k\}\}$. For the faithfulness results to the, respectively, undirected and bidirected case, an assumption that is used is that for some (and because of exchangeability for all) $i,j,k$, $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$ implies that $C^d_{ijk}\subseteq C$ for undirected case and  $C^d_{ijk}\cap C=\varnothing$ for the bidirected case.

Let $C_j=\{jr:\forall r\neq j\}$. Recall also that $C_{ij}=\{ir,jr:\forall r\neq i,j\}$ and $C^d_{ij}=\{lm:\forall l,m\notin\{i,j\}\}$. $C^d_{ij}$ is a minimal separator of $ij,ik$ in $L^c_{-}(n)$, and $C_{ij}\setminus\{ik\}$ is a maximal separator in $L^c_{\leftrightarrow}(n)$:
\begin{prop}\label{prop:12n}
Let $n>4$.
\begin{enumerate}
  \item In $L^c_{-}(n)$, it holds that $ij\,\mbox{$\perp$}\, ik\,|\, C^d_{ij}$, and if $ij\,\mbox{$\perp$}\, ik\,|\, C$  then $|C^d_{ij}|\leq |C|$. In fact, if $C\neq C^d_{ij}$ and $C\neq C^d_{ik}$ then $|C^d_{ij}|< |C|$.
  \item In $L^c_{\leftrightarrow}(n)$, $n>4$, it holds that $ij\,\mbox{$\perp$}\, ik\,|\, C_{ij}\setminus\{ik\}$, and if $ij\,\mbox{$\perp$}\, ik\,|\, C$  then $|C_{ij}|\geq |C|+1$. In fact, if $C\neq C_{ij}\setminus\{ik\}$ and $C\neq C^d_{ik}\setminus\{ij\}$ then $|C^d_{ij}|> |C|+1$.
\end{enumerate}
\end{prop}
\begin{proof}
1.\ The first claim is straightforward to prove since every path from $ik$ to $ij$ must pass through an adjacent vertex of $ij$, which does not contain $i$ or $j$. This set is $C^d_{ij}$. To prove the second statement, consider the sets $C^d_{ij}\setminus C^d_{ik}=C_k\setminus\{ik,jk\}$ and $C^d_{ik}\setminus C^d_{ij}=C_j\setminus\{ij,jk\}$, i.e., the neighbours of $ij$ and $ik$ with the joint neighbours removed. For every subset $S$ of $C_k\setminus\{ik,jk\}$, clearly there are at least the same number of vertices in $C_j\setminus\{ij,jk\}$ adjacent to members of $S$. Hence, the Hall's marriage theorem \cite{hal35} implies the result.

To prove the third statement, suppose, for contradiction, that $C\neq C^d_{ij}$ and $C\neq C^d_{ik}$ and $|C|=|C^d_{ij}|$. Using the fact that, for every missing member of $C^d_{ij}$ in $C$, there is at least a member of $C^d_{ik}$ that should be in $C$, there cannot be any vertex outside $C^d_{ij}\cup C^d_{ik}$ in a $C$. Hence, $C\subset C^d_{ij}\cup C^d_{ik}$. If $n>4$ and, $km,jo\notin C$, where $o\neq m$,  the path $\langle ik,jo,km,ij\rangle$ connects $ik$ and $ij$, a contradiction. Thus, say, $km,jm\notin C$ (which implies $jl,kl\in C$). Then the path $\langle ik,jm,il,km,ij\rangle$ connects $ik$ and $ij$, a contradiction.

2.\ The proof follows from the previous part and Lemma \ref{lem:5}, and observing that $|C_{ij}\setminus\{ik\}|=|C_{ij}|-1$.
\end{proof}
However, not all minimal separators of $ij,ik$ in $L^c_{-}(n)$ are of the form above. For example, in $L^c_{-}(5)$, consider the set $C=\{kl,jl,il,lm\}$. It holds that $ij\,\mbox{$\perp$}\, ik\,|\, C$. In addition, $C$ is minimal. However, $C$ is not of the form in the above proposition. In $L^c_{\leftrightarrow}(5)$, for the set $C^d=V\setminus (C\cup\{ij,ik\})$, by using Lemma \ref{lem:5}, we see that $ij\,\mbox{$\perp$}\, ik\,|\, C^d$, and $C^d$ is maximal.
\begin{prop}\label{prop:3nn0}
Let a distribution $P$ be defined over an exchangeable random network and $\mathrm{sk}(P)=L_{-}^c(n)$, and consider some nodes $i,j,k$.
\begin{enumerate}
  \item Suppose that for every minimal $C$ such that $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$,  $C^d_{ijk}\subseteq C$ and $C$ is invariant under swapping $k$ and $m$ for every $m$ with an $l$ such that $lm\notin C$. It then holds that if $P$ satisfies composition then it satisfies upward-stability and singleton-transitivity.
  \item Suppose that for  every maximal $C$ such that $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$, $C^d_{ijk}\cap C=\varnothing$ and $C$ is invariant under swapping $k$ and $m$  for every $m$ with an $l$ such that $lm\in C$. It then holds that if $P$ satisfies intersection then it satisfies downward-stability and singleton-transitivity.
\end{enumerate}
\end{prop}
\begin{proof}
By Lemma \ref{lem:2}, we know that the form of independencies for $\mathrm{sk}(P)=L^c_{-}(n)$ is $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$, as provided in the statement of the proposition.

1.\ First, we prove upward-stability. Suppose that $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$. Notice that, because of exchangeability, the assumptions of the statement hold for every $i,j,k,l,m$. For a variable $lm\notin C\cup\{ij,ik\}$, we prove that $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C\cup\{lm\}$ by induction on $|C|$. Notice that, by assumption, $lm\in C_{ijk}$.
Without loss of generality, we can assume that $l\in\{i,j,k\}$, and further, $l=i$ or $l=k$ since otherwise we can swap $j$ and $k$ and proceed as follows, and finally swap them back.

The base case is when $C$ is minimal. We have two cases: If $l=i$ then because of exchangeability, by the invariance of $C$ under swapping $k$ and $m$, and since $im\notin C^d_{ijk}$, we have that $ij\mbox{\protect $\: \perp \hspace{-2.3ex im\,|\, C$. By composition $ij\mbox{\protect $\: \perp \hspace{-2.3ex \{ik,im\}\,|\, C$, which, by weak union, implies $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C\cup\{im\}$.  If $l=k$ then by swapping $k$ and $i$, we have that $jk\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$. Now, notice that the statement $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$ does not change under the $kj$-swap, which implies that $C$ is also invariant under swapping $j$ and $m$. By this swap, we have $km\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$. By this, $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$, and the use of composition, we obtain $\{ij,km\}\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$, which, by weak union, implies $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C\cup\{km\}$.

Now suppose that $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$, $C$ is not minimal, and, for every  $C'$ such that $|C'|<|C|$,  $[ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C'\Rightarrow ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C'\cup\{op\}]$, for every $op$. Consider the independence statement $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C_0$ such that $C_0\subset C$ and $C_0$ is minimal. We have that $ij\mbox{\protect $\: \perp \hspace{-2.3ex im\,|\, C_0$ or $km\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C_0$. By induction hypothesis, we can add vertices to the conditioning set in order to obtain $ij\mbox{\protect $\: \perp \hspace{-2.3ex im\,|\, C$ or $km\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$.
Now, again composition and weak union imply the result.

Singleton-transitivity also follows from the above argument. We need to show that $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$ and $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C\cup\{lm\}$ imply $ij\mbox{\protect $\: \perp \hspace{-2.3ex lm\,|\, C$ or $lm\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$. (Notice that because of upward-stability $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C\cup\{lm\}$ is immaterial.) Again without loss of generality,  we can assume that $l=i$ or $l=k$, and the above argument showed that $ij\mbox{\protect $\: \perp \hspace{-2.3ex im\,|\, C$ or $km\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$.

2.\ The proof follows from part 1.\ and the duality (Lemma \ref{lem:dual}).
\end{proof}
\begin{theorem}
Let a distribution $P$ be defined over an exchangeable random network and $\mathrm{sk}(P)=L^c_{-}(n)$, and consider some nodes $i,j,k$.
\begin{enumerate}
  \item Suppose that for every minimal $C$ such that $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$,  $C^d_{ijk}\subseteq C$ and $C$ is invariant under swapping $k$ and $m$ for every $m$ with an $l$ such that $lm\notin C$. Then $P$ is faithful to $L^c_{-}(n)$ if and only if $P$ satisfies the intersection and composition properties.
  \item  Suppose that for  every maximal $C$ such that $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$, $C^d_{ijk}\cap C=\varnothing$ and $C$ is invariant under swapping $k$ and $m$  for every $m$ with an $l$ such that $lm\in C$. Then $P$ is faithful to $L^c_{\leftrightarrow}(n)$ if and only if $P$ satisfies the intersection and composition properties.
\end{enumerate}
\end{theorem}
\begin{proof}
The proof follows from Propositions \ref{prop:3nn0} and \ref{prop:und}.
\end{proof}

\subsection{The empty graph case}
Clearly every minimal separator in the empty graph is the empty set, and the maximal separator is all the remaining vertices. In the case where $\mathrm{sk}(P)$ is the empty graph, we have the following
\begin{prop}\label{prop:3nnnn0}
Let a distribution $P$ be defined over an exchangeable random network, and $\mathrm{sk}(P)$ be the empty graph, and consider some nodes $i,j,k,l$.
\begin{enumerate}
  \item Suppose that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl$ and $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik$ hold. It then holds that if $P$ satisfies composition then it satisfies upward-stability and singleton-transitivity.
  \item Suppose that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, V\setminus\{ij,kl\}$ and $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, V\setminus\{ij,ik\}$ hold. It then holds that if $P$ satisfies intersection then it satisfies downward-stability and singleton-transitivity.
\end{enumerate}
\end{prop}
\begin{proof}
By Lemma \ref{lem:2}, we know that $i,j,k,l$ are disjoint.

1.\ First, we prove upward-stability. Suppose that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$ or $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$. Notice that, because of exchangeability, the assumptions of the statement hold for $i,j,k,l$ that are all different. We prove that $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C\cup\{mh\}$ or $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C\cup\{mh\}$, for every $mh\notin C\cup\{ij,kl\}$ or $mh\notin C\cup\{ij,ik\}$, respectively, by induction on $|C|$.

The base case is when $C=\varnothing$. First, consider the case where $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl$. If $mh\notin C_{ij,kl}$ then by swapping $k$ and $m$ and $l$ and $h$, we obtain $ij\mbox{\protect $\: \perp \hspace{-2.3ex mh$. If $mh\in C_{ij,kl}$ then say $mh=jl$. We use $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik$ and first swap $i$ and $j$ to obtain $ij\mbox{\protect $\: \perp \hspace{-2.3ex jk$. Now we swap $k$ and $l$ to obtain $ij\mbox{\protect $\: \perp \hspace{-2.3ex jl$. (The other three cases of $mh$ are similar.) Now composition and weak-union imply the result.

Now, consider the case where $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik$. First suppose that $mh\in C_{ij,kl}$. If $mh=il$ then by swapping $j$ and $l$, we obtain $il\mbox{\protect $\: \perp \hspace{-2.3ex ik$. If $mh=jk$ then by swapping $i$ and $k$, we obtain $jk\mbox{\protect $\: \perp \hspace{-2.3ex ik$. If $mh=jl$ then by swapping $i$ and $j$ and then $k$ and $l$, we obtain $ij\mbox{\protect $\: \perp \hspace{-2.3ex jl$. If $mh\notin C_{ij,kl}$ then use $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl$ and swap $k$ and $m$ and $l$ and $h$ to obtain $ij\mbox{\protect $\: \perp \hspace{-2.3ex mh$. Now composition and weak-union imply the result.

The inductive step is similar to the inductive step of Proposition \ref{prop:3n0} or Proposition \ref{prop:3nn0} depending on the form $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$ or $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$.

Singleton-transitivity also follows from the above argument.

2.\ The proof follows from the first part and the duality (Lemma \ref{lem:dual})
\end{proof}
\begin{theorem}
Let a distribution $P$ be defined over an exchangeable random network, and $\mathrm{sk}(P)$ be empty. Suppose also that one of the following cases holds for $i,j,k,l$ that are all different:
\begin{itemize}
  \item[a)] $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl$ and $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik$;
  \item[b)] $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, V\setminus\{ij,kl\}$ and $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, V\setminus\{ij,ik\}$.
\end{itemize}
Then $P$ is faithful to the empty graph if and only if $P$ satisfies the intersection and composition properties.
\end{theorem}
\begin{proof}
The proof follows from Propositions \ref{prop:3nnnn0} and \ref{prop:und}.
\end{proof}
\section{Summary and discussion}\label{sec:5}
Our results concern the conditional independence models induced by exchangeable distributions for random vectors and random networks. We have shown that exchangeable random vectors are completely independent of each other or completely dependent of each other if they satisfy intersection and composition properties. In addition, they are marginally independent if there exists at least one independence statement and the intersection property is satisfied. The intersection property is well-understood, and we know that a positive joint density is a sufficient condition for it to hold; thus it is particularly important to study the composition property for exchangeable random vectors, which, in this case, is simplified to $A\mbox{\protect $\: \perp \hspace{-2.3ex B\,|\, C\Rightarrow A\mbox{\protect $\: \perp \hspace{-2.3ex (B\cup D)\,|\, C$, for every disjoint $D$.

For exchangeable random networks, as it turned out, the situation is much more complicated. As an important extension of our results in \cite{lau17}, we showed that the independence structures of exchangeable random networks that can be represented by a graph in graphical model sense are one of the six possible cases: completely dyadic-independent, faithful to the undirected or bidirected incidence graphs, faithful to the undirected or bidirected complement of the incidence graph, or completely dyadic-dependent.

The undirected and bidirected versions of the incidence graph and its complement are in fact dual to each other, so in a sense there are four regimes available. In other words, with exchangeability and duality factored in, one passes from the skeleton to the graphical Markov equivalence class. Exploiting this duality, all the results for the undirected case can be extended to the bidirected case. In addition, the remaining four cases are just two cases modulo graph complement as well. Although we failed to do so, it would
be especially nice if this duality could be understood, so that one can simply present the results with two-fold duality arguments.

We have provided a simple test to decide in which of the six regimes an exchangeable random network lies in cases when it has a ``structured" independence structure. The main two elements of the four ``non-trivial" cases is whether an independence is of form $ij\mbox{\protect $\: \perp \hspace{-2.3ex kl\,|\, C$ or $ij\mbox{\protect $\: \perp \hspace{-2.3ex ik\,|\, C$; and whether $C_{ij,kl}=\{ik,il,jk,jl\}$ is in $C$ or is disjoint from $C$.

We, in fact, do not have ``necessary" and sufficient conditions for whether an exchangeable random network is structured, but rather sufficient conditions that, in addition to the expected intersection and composition properties, are mainly based on whether a minimal (maximal) separator is invariant under the mentioned node swaps of the network. For testing purposes, it is important to stress that the only conditioning set that needs to be tested are the minimal ones, which would significantly improve the computational complexity of any relevant algorithms.

Indeed, in practice, it is more important to understand the independence structures of exchangeable statistical network models for  random networks with (in most situations) binary dyads. One point is that binary distributions always satisfy singleton-transitivity \cite{drts09}, a necessary condition for faithfulness,  although under our sufficient assumptions this condition is automatically satisfied. In general, however, it would be useful to study which actual exchangeable models for networks (such as exchangeable exponential random graph models \cite{was96}) satisfy the provided conditions, both when we deal with binary random networks or weighted ones.

redpajama_set_name: RedPajamaArXiv


Tailoring our services to the needs of each nation we serve, Sonrie volunteers work together with other organizations in the U.S. to improve care provided by the Centro de Rehabilitación in Santo Domingo, Dominican Republic. On-going instructional training, assistance with fitting of difficult-to-fit patients, and program evaluation are some of the ways we have joined together to help amputees in the Dominican Republic. Partners in this endeavor are: Physicians for Peace and World Rehabilitation Fund.
Electrocution victim, Emeterio was difficult to fit due to extreme limb scarring.
After suffering a tragic fall from a train in the Dominican Republic, triple amputee Wilkin, age 17 was brought to the U.S. for prosthetic care. He now receives follow-up care provided by Sonrie volunteers in his home country. Wilkin has returned to school and is studying computers. His smile is evidence of his new outlook on life.
Some patients also receive assistance devices to improve the quality of their lives.


text:
Design With Nature curated by Mario Cucinella, Salone del Mobile.Milano 2022
Taking a sustainable approach to social, environmental and economic issues is a value that the Salone del Mobile.Milano has always considered important and, now more than ever, absolutely paramount. Accordingly, it has drawn up and finessed its sustainability policy, in a bid to integrate sustainability into both company strategy and management and that of the various events it organises, thus helping to make the entire trade fair system more competitive, user-friendly and fit for the future.
The policy ratifies and sets out the Salone's belief that creating sustainable and long-lasting value, through the management of trade fairs and exhibitions, must be achieved with respect for the ethical principles of integrity and honesty, inclusion and transparency, in pursuit of ongoing improvement, operational efficiency and economical management. The document sets out the concrete commitments and objectives the Salone has identified as drivers for its own business model on its path towards inclusive, responsible and sustainable development.
Download the Sustainability Policy
What this actually means is that the Salone del Mobile undertakes to organise all the activities in such a way as to be geared to the efficient use of resources, cutting waste and environmental impact, and protecting the human rights, health and safety of those taking part in the event and those workers involved in its organisation, and assuring benefits for all the companies taking part and the local community.
Tangible proof of its commitment to coming up with solutions and activating virtuous practices is, first and foremost, its membership of the United Nations Global Compact – the leading initiative at global level for business sustainability, which brings together over 15,000 companies from more than 160 world countries. It constitutes a very real appeal to companies all over the world to support and apply the values of sustainability long-term. The 10 principles of the UNGC promote the protection of human rights, a more responsible approach to the environment and the fight against corruption in all its forms.
SUSTAINABLE INSTALLATION GUIDELINES
The Salone is also conscious that business sustainability initiatives are much more effective and far-reaching when planned with the active participation of all their stakeholders, which is why it has distributed the Green Guidelines to all the exhibiting companies, so as to provide them with the greatest possible support when it comes to designing, planning and carrying out their activities at the fair.
Download the Green Guidelines
Thanks to the efforts of all those involved, Supersalone was awarded ISO 14064-1:2019 certification in 2021 – a first sign that it is not just our duty, but that it is also possible, to engrain a change of course. Now, the process of obtaining ISO 20121 certification for sustainable events management is already up and running.

redpajama_set_name: RedPajamaCommonCrawl


text:
Early Success Won't Change Milwaukee Brewers Rebuilding Plans
Image via jsonline.com
The Milwaukee Brewers have been one of the biggest surprises in baseball this season. Heading into the season, the Brew Crew was expected to continue making incremental progress in a long rebuilding process after finishing 16 games under .500 last year. But nearly two months into the 2017 season, the Brewers have far surpassed expectations. Milwaukee is currently six games over .500 and in first place in the NL Central. However, Brewers GM David Stearns says that the team's success early in 2017 doesn't mean the club will expedite the rebuilding process.
Since taking over as GM prior to the 2016 season, Stearns has had a plan in place to acquire young talent and build from the ground up. So far, he's done just that, as the Brewers boast a rather impressive farm system following a series of trades made over the past year. However, with the addition of major league players like Eric Thames and Travis Shaw over the winter, the Brewers have received a short-term boost, enabling them to compete in what appears to be a wide open NL Central. But Stearns doesn't want that success to alter the team's long-term plans.
"The plan is to consistently win major-league games over a period of years," says Stearns. "If that process begins a little bit sooner than people expected, that's great. And we'll react accordingly. But it doesn't change our strategy; it doesn't change our plan. Our goal is to create an organization that's going to compete in this division for a lengthy period of time. If that starts a little bit sooner, that's great."
Milwaukee's surprising start to the season has some already speculating that the Brewers could be buyers on the trade market this summer. Last summer, Stearns traded away all-star catcher Jonathan Lucroy and closer Jeremy Jeffress for a package of prospects that have helped to improve the club's farm system. Is it possible for the Brewers to be on the other side of such a transaction this season? Could the Brewers turn around and trade some of those prospects for an impact player who could help them secure a division title or wild card spot in 2017?
Stearns indicates that he's open to just about anything. But at the same time, he's not about to sell the farm so the Brewers can take a shot at something special in 2017.
"There's no skipping steps," Stearns reiterates. "But that's true whether it's 2017 or 2020. We are never going to be an organization that skips steps. We're always going to have to balance near term and long term. There are other organizations that have done that exceptionally well. We think we can walk that balance as well."
Some of Milwaukee's best prospects, including outfielders Lewis Brinson and Brett Phillips and pitchers Josh Hader, Luis Ortiz, and Brandon Woodruff ,should be ready to contribute at the big league level in 2018, if not later this season. That means a year from now, the Brewers could have several more pieces of the puzzle in place and be in an even better position to compete at the top of the NL Central than they are right now.
The promise of that future could keep Stearns from making any major moves this summer, even if the Brewers remain in contention for a playoff spot deep into the season. This would be a sound approach that keeps the Brewers in a good position to compete in 2018 and beyond.
However, there's another side of the coin. Milwaukee has played well thus far, but the Brewers may need help down the line if they're going to remain ahead of the Cubs, the defending world champions, and the Cardinals, a perennial contender. Not making any moves to improve the team this summer could be frustrating for the Brewers, including manager Craig Counsell, who says he manages every game to win, even if the team is still in rebuilding mode.
"I don't think you should ever put limits on a baseball season," says Counsell. "David has talked about this, too. You only get so many seasons, I don't care who you are. So, certainly at the start, why put limits on it? I think we'll always think like that."
Milwaukee's early season success has put Stearns in a tough spot between trying to compete right now and still keeping an eye on the future. Few thought Stearns and the Brewers would be in this position quite so soon, but it's happened. Despite the challenges it may present later this season, Stearns isn't about to complain about it.
"The overall plan here is to win major-league games," says Stearns. "That's what we're trying to do and that's what we're building toward."


text:
Government launches £500m fund to help struggling startups
Posted on 21st April, 2020
Government will sink £250m into the fund and expects remaining half to be matched by private investors
Chancellor of the Exchequer Rishi Sunak has launched a £500m fund to help startups
Chancellor of the Exchequer Rishi Sunak has launched a £500m 'Future Fund' to assist UK-based start-ups struggling as a result of the ongoing pandemic.
The money is part of a £1.25bn package designed to buoy struggling young companies; with the £500m assisting venture-backed startups and the remaining £750m supporting SMBs involved in research and development.
The scheme – run in partnership with the British Business Bank – is designed to help those companies that are not eligible for the government's Coronavirus Business Interruption Scheme (CBILS) and will run from May to September.
The government has already agreed an initial £250m for the fund, with private investors expected to make up the remaining half.
Loans will range from £125,000 to £5m, providing a company can find private investment to match, and will carry an eight per cent interest rate, payable after a three-year period.
"Our start-ups and businesses driving research and development are one of our great economic strengths, and will help power our growth out of the coronavirus crisis," Sunak said in a press statement.
"This new, world-leading fund will mean they can access the capital they need at this difficult time, ensuring dynamic, fast-growing firms across all sectors will be able to continue to create new ideas and spread prosperity."
To be eligible to apply to the fund, a firm must be UK-based, has previously raised at least £250,000 in equity from third-party investors in the last five years and can attract match funding from private investors and institutions. Full eligibility criteria are pending.
James Tucker, CEO of mortgage technology provider Twenty7Tec, welcomed the news: "This has to be a welcome move for both tech and research-led businesses. Many of them are loss-making in their first years and so need firm backing from business angels, early investors and venture capitalists.
"The challenge for tech businesses is that there has been less appetite for risk over recent weeks, for obvious reasons. But this move by the chancellor using convertible loans looks set to plug a gap in the market and give a real boost to the companies that will emerge as the tech stars of the future.
"If you've got a track record of raising capital and want to shore up confidence in the business, the Future Fund looks like a good way of doing so. For many start-ups, this will be a massive lifeline. We're waiting to see the full details on how the fund will work, but at first blush, the terms seem reasonable – 36-month maximum term, convertible loan that must be equally or better matched by private funding."
However, some greeted the news with some scepticism. Mark Brownridge, director-general of the Enterprise Investment Scheme Association (EISA), said that private investors are hesitant to invest in anything at the moment and need "advantageous tax treatment" than matched funding.
"Positioning the additional support as a convertible loan, carrying an eight per cent interest rate and a 20 per cent discount on conversion, all in favour of the government, will be of very little encouragement for private investors to invest alongside," he stated.
"We need to be focusing on encouraging new investors, and the only way to do that is through more advantageous tax treatment for them, which is what the EISA, along with other investment funds, has been lobbying for.
"Over the coming weeks, we need to be working with the government in finalising the terms of the new Future Fund, to ensure both that the terms are as good as the CBILS scheme, and that we do all we can to encourage the additional investment appetite from private investors, which must include more favourable tax treatment for an interim period whilst the uncertainties caused by the crisis continue."
SOURCE: https://www.channelweb.co.uk/news/4014168/government-launches-gbp500m-fund-help-struggling-startups
UK start-up companies receive funding boost from government
UK start-up companies receive funding boost from government The government's…
Mortgage searches continuing to increase
The volume of searches for purchase mortgages was up 2.87%…


The original 13mm 'HUFLON' plectrums !
This 13mm plectrum is the result of 2 hours of work! 'HUFLON' is an advanced polymer which contains a proprietary filler. This unique material has a very low coefficient of friction, is self-lubricant, offers excellent frictional behaviour and completely insane wear resistance and of course a fantastic grip!


"Shambles", "utter disgrace" and "worst event I've ever been to" were just a few of the comments made after Wednesday's AU welcome party.
After being a heavily publicised and sold-out event, the AU welcome party ended early due to "safety concerns". These 'concerns' came about by the lack of organisation of the new security team in place.
The queue for the entrance comprised of two checkpoints: an ID check and a bag check/metal detector. The first checkpoint moved relatively quickly. The security in the second checkpoint asked that people be scanned in one by one. This led to huge bottlenecking issues as people were ushered in quickly only to be stopped at the detectors. People crowded around the entrance and yet more people tried to slip in from the sides. With no more space to move, the security team stationed near the detectors shouted at everyone to 'move back' – all whilst people from the other part of the queue were still being moved forward.
This caused even more chaos as the crowd was pushed forward and back, barriers were knocked over and people were hurt.
In a surreal turn of events, a fight then broke out amongst people dressed as Peter Pan (hockey), Ninja Turtles (Futsal) and people dressed as 'average joes' in yellow (dodgeball). There were also accounts of people punching the security and spitting – these people were later escorted off the premises. The confusion, violence and mismanagement further hindered people from being let in. An estimated 100 people were let in the building last night (although the security team disagree, saying about 500 people were inside the venue). Many people left the queue, choosing to go home or to wait out the chaos in Shakespeare's Head down the road. The AU tried to rectify the poor management by suggesting people go to Zoo instead. This received many angry comments as people felt 'cheated' for paying £9 for an event they were not allowed into, only to pay another £3 for Zoo entrance. The fun of the Heroes and Villains night seemed to have descended into a more serious situation altogether.
What has been said about the night?
People have called the new security team 'unprofessional and aggressive'. One student, employed by Tuns, told the Beaver: "The new security team were extremely rude and couldn't handle a student event. [They] wouldn't let a friend of mine get his bag and then squared up to him even though he does work at Tuns. I went over to vouch for him and they started having a go at me too. Such a shambles". When interviewed yesterday by The Beaver, the student said that they both took up their grievances with the security team with the Student Union. It reportedly turned into a "'shouting match"' resulting in their termination of employment at Tuns.
Another student posted on the Facebook event calling it "probably the worst queue I've been in". There were many accounts of students feeling unsafe and wanting to leave.
The Beaver received reports from several organisers of the event, who said there was a lack of respect and communication between the event organisers and the security team. Reports say that the security team were calling LSE students in the queue "animals". When event organisers tried to negotiate a different plan of action for the night, reportedly some security members replied "stop telling us what to do" – again, some felt this entailed a lack of respect from the hired companies towards the student body.
The security were interviewed and said that they "followed the book" and have defended themselves by saying they have run "bigger events than this" before. Students and organisers have criticised that the decision to shut the event down and little communication between the two checkpoints showed "very little intuition on the part of the security team".
They are also trying to secure refunds for those affected and will release a statement by next Wednesday of the outcome of this request.
Investigations are still ongoing and the AU exec are yet to release their reports about the incident.
A Whirlwind 48 Hour Stop in Barcelona.