On Finite Exchangeability and Conditional Independence

We study the independence structure of finitely exchangeable distributions over random vectors and random networks. In particular, we provide necessary and sufficient conditions for an exchangeable vector so that its elements are completely independent or completely dependent. We also provide a sufficient condition for an exchangeable vector so that its elements are marginally independent. We then generalize these results and conditions for exchangeable random networks. In this case, it is demonstrated that the situation is more complex. We show that the independence structure of exchangeable random networks lies in one of six regimes represented by undirected and bidirected independence graphs in graphical model sense. In addition, under certain additional assumptions, we provide necessary and sufficient conditions for the exchangeable network distributions to be faithful to each of these graphs.


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 [1], 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., [14].
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 [5], 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 [23,7].
However, for finite 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., [6,19,15]. 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. [13]) in order to provide the independence structure of exchangeable distributions.
In particular, we exploit the necessary and sufficient conditions, provided in [22], for faithfulness of probability distributions and graphs. 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 [20,24]. 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, but first show when exchangeable networks are closed under marginalization and conditioning. We then show that, other than complete independence and dependence, there are four other independence structures that can arise for exchangeable networks. This is a consequence of a result in [14]. We then provide assumptions under which intersection and composition properties are necessary and sufficient for the exchangeable random networks to induce the exact same independence structure as that of each of the 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 3, we provide the results for exchangeable random vectors. In Section 4, we provide the results for exchangeable random networks starting with marginalization and conditioning, and then introduce each aforementioned case in a separate subsection. We end with a short discussion on these results in Section 5.

Graph-theoretic concepts
A labeled graph is an ordered pair G = (V, E) consisting of a vertex set V , an edge set E, and a relation that with each edge associates two vertices, called its endpoints. When vertices u and v are the endpoints of an edge, these are adjacent and we write u ∼ v; we denote the corresponding edge as uv.
In this paper, we will restrict the work to simple graphs, i.e. graphs without loops (the endpoints of each edge are distinct) or multiple edges (each pair of vertices are the endpoints of at most one edge). Further, three types of edge, denoted by arrows, arcs (solid lines with two-headed arrows) and lines (solid lines) 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 4.2, we will distinguish only lines and arcs, which respectively form undirected and bidirected graphs.
The labeled 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 ). We omit the term labeled when the context does not give rise to ambiguity.
A subgraph of a graph G = (V G , E G ) is graph F = (V F , E F ) such that V F ⊆ V G and E F ⊆ E G and the assignment of endpoints to edges in F is the same as in G.
The 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 ∼ e 2 if and only if e 1 and e 2 have a common endpoint [27, p. 168]. We will in particular be interested in the line graph of the complete graph, which we will refer to as the incidence graph. Figure 1 displays the incidence graph for V = {1, 2, 3, 4}. The 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 sk(G).

Random networks
Given a finite or countably infinite node set N -representing individuals or actors in a given population of interest -we define a random network over N to be a collection X = (X d , d ∈ D(N )) of binary random variables taking values 0 and 1 indexed by a set D(N ) of dyads. Throughout the paper we take D(N ) to be the set of all unordered pairs ij of nodes in N , and nodes i and j are said to have a tie if the random variable X ij takes the value 1, and no tie otherwise. Thus, a network is a random variable taking value in {0, 1} ( N 2 ) and can therefore be seen as a random simple, undirected, labeled graph with node set N , whereby the ties form the random edges of the graphs. We will write G N for the set of all simple, labeled undirected graphs on N .
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 = (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.

Probabilistic independence models and their properties
An independence model J over a finite set V is a set of triples X, Y | Z (called independence statements), where X, Y , and Z are disjoint subsets of V ; Z may be empty, but ∅, Y | Z and X, ∅ | Z are always included in J . The independence statement X, Y | Z 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 [24].
In order to define probabilistic independence models, consider a set V and a collection of random variables {X α } α∈V with state spaces X α , α ∈ V and joint distribution P . We let X A = {X v } v∈A etc. for each subset A of V . For disjoint subsets A, B, and C of V we use the short notation A ⊥ ⊥ B | C to denote that X A is conditionally independent of X B given X C [4,13], i.e. that for any measurable Ω ⊆ X A and P -almost all x B and x C , We can now induce an independence model J (P ) by letting Similarly we use the notation A ⊥ ⊥ B | C for A, B | C / ∈ J (P ). We say that A and B are completely independent if, for every C ⊆ V \(A∪B), A ⊥ ⊥ B | C. Similarly, we say that A and B are completely dependent if, for any or C has only one member {i}, {j}, or {k}, for better readability, we write i ⊥ ⊥ j | k. We also write A ⊥ ⊥ B when C = ∅, which denotes the marginal independence of A and B.
A probabilistic independence model J (P ) over a set V is always a semigraphoid [20], i.e., it satisfies the four following properties for disjoint subsets A, B, C, and D of V : 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 graphoid ; that is, it also satisfies Furthermore, a graphoid or semi-graphoid for which the reverse implication of the decomposition property holds is said to be compositional, that is, it also satisfies imsart-generic ver. 2014/10/16 file: exch-ind-arxiv.tex date: July 8, 2019 If, for example, P has strictly positive density, the induced probabilistic independence model is always a graphoid; see e.g. Proposition 3.1 in [13]. See also [21] 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, J (P ) is a compositional graphoid; e.g. see [24]. 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 [26]. Another important property that is not necessarily satisfied by probabilistic independence models is singleton-transitivity (also called weak transitivity in [20], where it is shown that for Gaussian and binary distributions P , J (P ) always satisfies it). For i, j, and k, single elements in V , In addition, we have the two following properties: Henceforth, instead of saying that "J(P ) satisfies these properties", we simply say that "P satisfies these properties". First we provide the following known result [22]: For a probability distribution P the following holds: 1. If P satisfies upward-stability then P satisfies composition. 2. If P satisfies downward-stability then P satisfies intersection.

Exchangeability for random vectors and networks
A probability distribution P over a finite vector (X 1 , X 2 , X 3 , . . . , X n ) of random variables is (finitely) exchangeable if for any permutation π ∈ S(n) of the indices 1, 2, 3, . . . , n, the probability distribution of the permuted vector (X π(1) , X π(2) , X π (3), . . . , X π(n) ) is the same as P ; see [1]. 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∈N over a finite nodeset N is said to be (finitely) weakly exchangeable [23,10] if for all permutations π ∈ S(N ) we have that If the matrix X is symmetric -i.e. X ij = X ji , we say it is 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 (adjacency matrix) of a random network and thus the above concepts can be translated into networks. A random network is 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.

Undirected and bidirected graphical models
Graphical models [see, e.g. 13] are statistical models expressing conditional independence statements among a collection of random variables X V = (X v , v ∈ 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 Markov properties.
We say that C separates A and B in an undirected graph G, denoted by A ⊥ u B | C, if every path between A and B has a vertex in C. For a bidirected graph G, we say that C separates A and B, denoted by A ⊥ b B | C, if every path between A and B has a vertex outside C ∪ A ∪ B. Note the obvious duality between this and separation for undirected graphs.
A joint probability distribution P for X V is Markov with respect to an undirected graph [3] or a bidirected graph [2,12]  If, for P and undirected G, A ⊥ u B | C ⇐⇒ A ⊥ ⊥ B | C then we say that P and G are faithful ; Similarly, if, for P and bidirected G, A ⊥ b B | C ⇐⇒ A ⊥ ⊥ B | C then P and G are faithful. Hence, faithfulness implies Markovness, but not the other way around.
For a given probability distribution P , we define the skeleton of P , denoted by sk(P ), to be the undirected graph with vertex set V such that vertices u and v are not adjacent if and only if there is some subset C of V so that u ⊥ ⊥ v | C. Thus, if P is Markov with respect to an undirected graph G then sk(P ) would be a subgraph of G (by considering all remaining vertices to be in C); whereas if P is Markov with respect to a bidirected graph G then sk(P ) is a subgraph of sk(G) (by considering C = ∅).
In general a graph G(P ) is induced by P with skeleton sk(P ). For undirected graphs, let G u (P ) = sk(P ), whereas for bidirected graphs, let G b (P ) be sk(P ) with all edges being bidirected. We shall need the following results from [22]: Proposition 1. Let P be a probability distribution defined over {X α } α∈V . It then holds that P and G u (P ) are faithful if and only if P satisfies intersection, singleton-transitivity, and upward-stability.
Proposition 2. Let P be a probability distribution defined over {X α } α∈V . It then holds that P and G b (P ) are faithful if and only if P satisfies composition, singleton-transitivity, and downward-stability.

Results for vector exchangeability
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∈V . First, notice that exchangeability is closed under marginalization and conditioning: Proof. First we prove closeness under marginalization: For a permutation matrix π of A, let the permutation matrix over V be π * (a) = π(a), for a ∈ A, and π * (b) = b for b ∈ V \ A. The proof then follows from exchangeability of X V . Now we prove closeness under conditioning: By the definition of conditioning, we need to prove that P (x A , x * C ) = P (x π(A) , x * C ) for every π. This is again true by considering π * (a) = π(a), for a ∈ A, and π * (c) = c for c ∈ C.
Notice that the above result implies that the marginal/conditional X A | X C , (i.e., when A ∪ C ⊂ V ) is also exchangeable. We now have the following results: Proposition 4. If P satisfies vector exchangeability then the following holds:

P satisfies upward-stability if and only if it satisfies composition. 2. P satisfies downward-stability if and only if it satisfies intersection.
Proposition 5. If P is exchangeable then it satisfies singleton-transitivity.
We see that the skeleton of an exchangeable distribution can only take a very specific form: Proposition 6. If P is exchangeable then sk(P ) is either an empty or a complete graph.
Proof. If there is any independence statement of form i ⊥ ⊥ j | C then by permutation for all variables, we obtain k ⊥ ⊥ l | C ′ for all k and l. Therefore, sk(P ) is empty. If there is no independence statement of this form then sk(P ) is complete.
Notice that for faithfulness to empty or complete graphs, it is immaterial what type of graph one is discussing as all types of such graphs are Markov equivalent, i.e. they induce the same independence model. Hence, we have the following: Corollary 1. Let P satisfy vector exchangeability. If P is faithful to a graph then the graph is empty or complete.
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:

Theorem 1. If P is exchangeable then P is faithful to either the complete graph or the empty graph if and only if P satisfies the intersection and composition properties.
Proof. The proof follows from Propositions 1, 2, 4, 5, and 6.

Corollary 2. 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.
Corollary 3. 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.
Proof. The proof follows from the fact that a regular Gaussian distribution satisfies the intersection and composition properties. Proposition 7. If P is exchangeable and satisfies intersection (e.g. P has a positive density), and there exists one independence statement induced by P then all variables X v are marginally independent of each other.
Proof. By Proposition 4, P satisfies downward-stability. An independence statement A ⊥ ⊥ B | C, by the use of decomposition implies i ⊥ ⊥ j | C for an arbitrary i ∈ A and j ∈ B. Downward-stability implies i ⊥ ⊥ j. Exchangeability implies all pairwise marginal independences.

Example 1. Consider an exchangeable distribution P over four variables
By exchangeability, all independences of form σ(i) ⊥ ⊥ σ(j) | σ(k) hold for any permutation σ on (i, j, k, l). It is easy to see that none of the semigraphoid axioms can generate new independence statements from these. If intersection holds then, for example, from i ⊥ ⊥ j | k and i ⊥ ⊥ k | j, we obtain i ⊥ ⊥ {j, k}, which by decomposition implies i ⊥ ⊥ j, and hence all marginal independences between singletons. If composition holds then, for example, from i ⊥ ⊥ j | k and i ⊥ ⊥ l | k, we obtain i ⊥ ⊥ {j, l} | k, which by weak union implies i ⊥ ⊥ j | {k, l}, and hence all conditional independences between singletons given the remaining variables.

Marginalization and conditioning for exchangeable random networks
A way to define marginalization and conditioning for random networks is by a more restrictive version where one can only marginalize over and condition on the nodes of the network, not the dyads: For a node set A of the network X, we say that the marginal network X A = (X d , d ∈ D(A)) is the network obtained by marginalization over nodes N \ A of the network, i.e. it is defined by marginalizing over the set of dyads {X ij : i ∈ N \ A, ∀j}. Similarly, we say that the conditional network X B | C is the network obtained by conditioning on nodes C = N \ B of the network, which is defined by conditioning on the set of dyads {X ij : i ∈ C, ∀j}.
Proposition 8. If X is an exchangeable random network then so are the marginal network X A and the conditional network X B | C .
Proof. In a similar fashion to the proof of Proposition 3, the proof follows from defining and using the permutation function π * (a) = π(a), for a ∈ A, and π * (b) = b for b ∈ N \ A for the marginal case, and π * (b) = π(b), for a ∈ B, π * (c) = c for c ∈ C in the conditional case. Notice that {X π * * (a * )π * * (b) : a * ∈ A, ∀b ∈ N \ A} = {X π * (a * )π * (b) : a * ∈ A, ∀b ∈ N \ A} for any permutation π * * as long as π * * (a) = π(a). This result, for example, implies that when in all independence statements in an axiom, we condition on a set C (like for C in composition and intersection), we can simply focus on the conditional probability space given C as long as C = ∪ i∈A {ij, ∀j} for some set of nodes A. However, independence models that only contain statements of from A ⊥ ⊥ B | C, where all A, B, C are all collection of all dyads connected to a node set, are not particularly interesting as their skeleton could only be an empty or a complete graph. Such independence structures act very similarly to those in the vector exchangeability case.
Hence, we henceforth focus on marginalization over and conditioning on arbitrary sets of dyads. 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 π(i)π(j) ) ij∈A = x A ), for any permutation π. However, this is not necessarily equal to P (X A = (x π(i)π(j) ) ij∈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 ∈ A and ik ∈ C). This is because, for a permutation π that maps i to a node other than i, exchangeability of . But, this does not necessarily hold as i is mapped to another node in A but not in C.
However, we have the following: Proposition 9. 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 ∈ A then there is no dyad ik or jk in C for any node k ∈ N . It then holds that the conditional/marginal random network X A | X C is exchangeable.
Proof. Let N (A ∪ C) be the set of all endpoints of dyads in A and C, and define similarly N (A) and N (C). Define the permutation π * ∈ S(N (A ∪ C)) such that π * (i) = π(i) for i ∈ N (A) and π * (k) = k for k ∈ N (C). Notice that this is welldefined since A and C do not share any nodes. Using π * and by exchangeability of X, we conclude that P (x A , x C ) = P ((x π(i)π(j) ) ij∈A , x C ), which, as mentioned before, is equivalent to the exchangeability of X A | X C .

Types of graphs faithful to exchangeable distributions
An analogous result to that of vector exchangeability, concerning the skeleton of an exchangeable probability distribution, was proven in [14]: Proposition 10. If a distribution P over a random network X is exchangeable then sk(P ) is one of the following: 1. the empty graph; 2. the incidence graph; 3. the complement of the incidence graph; 4. the complete graph.
Notice that a pairwise independence statement for an exchangeable P over a random network X is of form ij ⊥ ⊥ kl | C, where i, j, k, l are nodes of X. Depending on the type of sk(P ), these statements take different forms.

Lemma 2. An independence statement ij ⊥
⊥ kl | C, i = j, k = l, for an exchangeable P over a random network X is in one of the following forms depending on the type of sk(P ): 1. empty graph: no constraints on i, j, k, l; 2. incidence graph: i = l, k and j = l, k; 3. complement of the incidence graph: i = k or i = l or j = k or j = l; 4. complete graph: no possible i, j, k, l, i.e., no such statement.
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 ⊥ ⊥ kl | C.
It is clear that if 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, but 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 require some definitions and results for the more general types of graphs, which we only need for the results in this subsection. We provide these here before providing the result. If the reader is only interested in the main result (Theorem 2) and not the proof, we suggest that they skip the material leading to this theorem.
In the remainder of this section, by a graph, we mean a graph with simultaneous undirected, directed, or bidirected edges. We also only focus on maximal graphs, which are graphs where a missing edge between vertices u and v imply that there exists a separation statement of form u ⊥ v | C, for some C. A walk ω is a list ω = i 0 , e 1 , i 1 , . . . , e n , i n of vertices and edges such that for 1 ≤ m ≤ n, the edge e m has endpoints i m−1 and i m . A section ρ of a walk is a maximal subwalk consisting only of lines, meaning that there is no other subwalk that only consists of lines and includes ρ. Thus, any walk decomposes uniquely into sections; these are not necessarily edge-disjoint and sections may also be single vertices. A section ρ on a walk ω is called a collider section if one of the following walks is a subwalk of ω: i ≻ρ≺ j, i≺ ≻ρ≺ j, i≺ ≻ρ≺ ≻ j. All other sections on ω are called non-collider sections.
A trisection is a walk i, ρ, j , where ρ is a section. If in the trisections, i and j are distinct and not adjacent then the trisection is called unshielded. We say that a trisection is collider or non-collider if its section ρ is collider or non-collider respectively. We do not intend to define the separation criterion for the general type of graphs here -the reader can refer to [16]. We only provide the following immediate consequence of the definition: for an unshielded collider trisection with endpoints i and j in a graph, and for each statement of from i ⊥ j | C, it holds that, for every t ∈ ρ, t / ∈ C; in an unshielded non-collider trisection with endpoints i and j in a graph, for each i ⊥ j | C, it holds that, for some t ∈ ρ, t ∈ C.
As mentioned before, two graphs are called Markov equivalent if they induce the same independence model. Proof. Because of maximality, G and H have the same skeleton. An unshielded trisection in these graphs cannot be a collider in one and a non-collider in the other. This is because if that is the case (say an unshielded trisection between i and j and separation i ⊥ j | C), by Markov equivalence, it implies that an inner vertex t of the corresponding section is not in C in one, but in C in the other, which is a contradiction.
Lemma 4. Suppose that a distribution P over an exchangeable random network with nodes 1, . . . , n is faithful to a graph G. Then P is faithful to the graph H obtained by any permutation of {1, . . . , n} on n 2 vertices of G; hence G and H are Markov equivalent.
Proof. Let J σ (P ) be the independence model induced by permuting the nodes of a random network with probability distribution P by the map σ. Because of exchangeability, this is the same as J (P ). We know that J (P ) = J (G). These imply that J σ (P ) = J (G). Notice that J (G) is also invariant under σ, i.e. J (G) = J σ (G). Clearly, this is the independence model induced by the graph G after its vertices are mapped by the permutation σ of the nodes of the network. Hence, P is faithful to G σ , which is G with mapped vertices by σ.
Theorem 2. If a distribution P over an exchangeable random network with n nodes is faithful to an anterial graph G then G is Markov equivalent to one of the following graphs: 1. the empty graph; 2. the undirected incidence graph L − (n); 3. the bidirected incidence graph L ↔ (n); 4. the undirected complement of the incidence graph L c − (n); 5. the bidirected complement of the incidence graph L c ↔ (n); 6. the complete graph.
Proof. Cases 1 and 6 are trivial. Thus first assume that case 2 of Proposition 10 holds. If there are no unshielded collider trisections in G then G is Markov equivalent to L − (n). Thus, suppose that there is an edge 12, 13, on which there is an arrowhead at vertex 13, and which is in an unshielded collider trisection. Notice that the other endpoint of the collider trisection does not contain 1 or 2. Now by a permutation that only swaps 2 and 3, and by Lemma 4, we conclude that there is a Markov equivalent graph to G in which there is an arrowhead at 12 on the edge 12, 13, and 12, 13 is on an unshielded collider trisection, where 13 is an endpoint. This, by Lemma 3, implies that there must be arrowheads at both 12 and 13 on the edge 12, 13. Therefore, there is an arrowhead at every vertex in G since any edge ij, ik can be mapped to 12, 13 by a permutation. Now assume that case 3 of Proposition 10 holds. Again, if there is no unshielded colliders in G then G is Markov equivalent to L c − (n). Thus, suppose that there is an edge 12, 34, on which there is an arrowhead at vertex 34, and which is in an unshielded collider trisection. Notice that the other endpoint of the collider trisection contains 1 or 2. Now by a permutation that only swaps 1 and 3, and 2 and 4, and by Lemma 4, we conclude that there is a Markov equivalent graph to G in which there is an arrowhead at 12 on the edge 12, 34, and 12, 34 is on an unshielded collider trisection, where 34 is an endpoint. This, by Lemma 3, implies that there must be arrowheads at both 12 and 34 on the imsart-generic ver. 2014/10/16 file: exch-ind-arxiv.tex date: July 8, 2019 edge 12, 34. Therefore, there is an arrowhead at every vertex in G since any edge ij, kl can be mapped to 12, 34 by a permutation.
Hence, for exchangeable random networks, there are six regimes available. Although the following method is not unique, here we provide a simple test to decide in which regime a given exchangeable distribution lies: for arbitrary fixed nodes i, j, k, l, m of the network, test the following: • ij ⊥ ⊥ kl | C, for some C, and ij ⊥ ⊥ ik | C ′ , for some C ′ ⇒ Empty graph; • ij ⊥ ⊥ kl | C, for some C, and ij ⊥ ⊥ ik | C ′ , for any C ′ ⇒ L(n): • ij ⊥ ⊥ kl | C, for any C, and ij ⊥ ⊥ ik | C ′ , for some C ′ ⇒ L c (n): ⊥ kl | C, for any C, and ij ⊥ ⊥ ik | C ′ , for any C ′ ⇒ Complete graph.
Of course, there may be cases where this test is not in a sense consistent: for example, consider a permissible case where ij ⊥ ⊥ kl | C 1 and ij ⊥ ⊥ kl | C 2 , but ik ∈ C 1 and ik / ∈ C 2 . Notice, however, that in such cases P cannot be faithful to any of the graphs of Theorem 2 anyway.
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 2. However, in special cases this holds. We detail this below separately for each regime mentioned above.

The incidence graph case
In this section, we assume that sk(P ) = L(n). The following example shows that, in principle, intersection and composition are not sufficient for faithfulness of P and L − (n).

Example 2.
Suppose that there is an exchangeable P that induces ij ⊥ ⊥ kl | C ij,kl , where C ij,kl = {ik, il, jk, jl}. Exchangeability implies that ij ⊥ ⊥ kl | C ij,kl for all i, j, k, l. Suppose, in addition, that P satisfies upward-stability. It is easy to see that sk(P ) = L − (n). Moreover, by Lemma 1, P satisfies composition.

Moreover, P satisfies intersection: Notice that none of the semi-graphoid axioms plus upward-stability and composition imply an independence statement of form
both contain a node i. In addition, it can be seen that these axioms imply that if there is A ⊥ ⊥ B | C then for every ij ∈ A and kl ∈ B, it holds that C ij,kl ⊆ C. Let C A,B = ij∈A,kl∈B C ij,kl . By these two observations, we conclude that if A ⊥ ⊥ B | C ∪ D and A ⊥ ⊥ D | C ∪ B then C A,D ⊆ C. By upward stability all statements of form ij ⊥ ⊥ kl | C where C ij,kl ⊆ C hold. Hence, by composition, we have that A ⊥ ⊥ B | C. Hence, by contraction, A ⊥ ⊥ B ∪ D | C. However, P does not satisfy singleton-transitivity: Consider ij ⊥ ⊥ kl | C ij,kl and ij ⊥ ⊥ kl | C ij,kl ∪{km}. If, for contradiction, singleton-transitivity holds then, because sk(P ) = L(n), we have that ij ⊥ ⊥ 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 ⊥ ⊥ kl | C ij,kl }. By Proposition 1, it is implied that, although intersection and composition are satisfied, P is not faithful to L − (n).
However, under certain assumptions, intersection and composition are sufficient for faithfulness. In order to proceed, we need to assume two additional regimes within the incidence graph case related to the cases of Theorem 2: the undirected incidence graph case and the bidirected incidence graph case. As in Example 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 ⊥ ⊥ kl | C implies that C ij,kl ⊆ 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 ⊥ ⊥ kl | C implies that C ij,kl ∩C = ∅.

The undirected incidence graph case
Let also C ij = {ir, jr : ∀r = i, j}. This is a minimal separator w.r.t. its size in L − (n): Proof. 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 / ∈ C then at least km and lm must be in C. The same vertices km and lm appear if im / ∈ 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 | ≤ |C|.
To prove the third statement, suppose, for contradiction, that C = C ij and C = 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 ∪ C kl in a C. Hence, C ⊂ C ij ∪ C kl . If n > 5 and, say, im, jm / ∈ C but km, lm ∈ C then consider the vertices ih, jh, kh, lh. Without loss of generality, assume that ih, jh ∈ C but kh, lh / ∈ C. Then the path kl, kh, mh, jm, ij connects kl and ij, a contradiction. The cases where n = 3, 4, 5 are easy to check.
However, not all minimal separators of L − (n) are of the form above. For example, in L − (6), consider the set C = {ik, il, im, jk, jl, jm, nk, nl, mh}. It holds that ij ⊥ kl | C. In addition, C is minimal in the sense that if we remove any vertex from C, the separation does not hold. However, C is not of the form C pq for any pair p, q ∈ {i, j, k, l, m, h}.
Proposition 12. Let a distribution P be defined over an exchangeable random network and sk(P ) = L(n). Suppose that for some i, j, k, l, and every minimal C such that ij ⊥ ⊥ kl | C, it holds that C ij,kl ⊆ C and C is invariant under swapping imsart-generic ver. 2014/10/16 file: exch-ind-arxiv.tex date: July 8, 2019 k and m and l and h for every m, h. It then holds that if P satisfies composition then it satisfies upward-stability and singleton-transitivity.
Proof. First, we prove upward-stability. By Lemma 2, we know that i, j, k, l are all different. Suppose that ij ⊥ ⊥ 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 / ∈ C ∪ {ij, kl}, we prove that ij ⊥ ⊥ kl | C ∪ {mh} by induction on |C|. Notice that, by assumption, mh / ∈ C ij,kl . In addition, without loss of generality, we can assume that m, h = 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 / ∈ C ij,kl , we have that ij ⊥ ⊥ mh | C. By composition ij ⊥ ⊥ {kl, mh} | C, which, by weak union, implies ij ⊥ ⊥ kl | C ∪ {mh}. Now suppose that ij ⊥ ⊥ kl | C, C is not minimal, and, for every C ′ such that , for every op. Consider the independence statement ij ⊥ ⊥ kl | C 0 such that C 0 ⊂ C and C 0 is minimal. We have that ij ⊥ ⊥ mh | C 0 . By induction hypothesis, we can add vertices to the conditioning set in order to obtain ij ⊥ ⊥ mh | C. Now, again composition and weak union imply the result.
Singleton-transitivity also follows from the above argument. We need to show (Notice that because of upward-stability ij ⊥ ⊥ kl | C ∪{mh} is immaterial.) Again without loss of generality, we can assume that m, h = i, j, and the above argument showed that ij ⊥ ⊥ mh | C.
Theorem 3. Let a distribution P be defined over an exchangeable random network and sk(P ) = L(n). Suppose that for some i, j, k, l, and every minimal C such that ij ⊥ ⊥ kl | C, it holds that C ij,kl ⊆ C and C is invariant under swapping k and m and l and h for every m, h. Then P is faithful to L − (n) if and only if P satisfies the intersection and composition properties.
Proof. The proof follows from Propositions 12 and 1.

The bidirected incidence graph case
The dual couple of an independence model J is the independence model defined by [18]; see also [9]. [17] showed that if J is singleton-transitive compositional graphoid, so is the dual couple of J (although this is proven based on the so-called Gaussoid formulation). Here, we use the following lemma: Lemma 5. Let G u and G b be an undirected and a bidirected graph such that sk(G u ) = sk(G b ). Then, the independence model Proof. Since the separation satisfies the composition property, it is sufficient to prove the statement for singletons. Suppose that there is a connecting path π between i and j given C in G u . This means that no inner vertex of π is in C; thus they are all in V \({i, j}∪C). Therefore, in G b , i and j are connecting given We show that C c ij \ {kl} is a maximal separator w.r.t. the size: The proof follows from Proposition 11 and Lemma 5, and considering |C c ij \ {kl}| = |C c ij | − 1. Again, similar to the L − (n) case, not all maximal separators of L ↔ (n) are of the form above. For example, in L ↔ (6), and for C = {ih, jh, mk, ml}, it holds that ij ⊥ kl | C. In addition, C is maximal.
Proposition 14. Let a distribution P be defined over an exchangeable random network and sk(P ) = L(n). Suppose that for some i, j, k, l, and every maximal C such that ij ⊥ ⊥ kl | C ∪ {mh}, for every m, h, it holds that C ij,kl ∩ (C ∪ {mh}) = ∅ and C is invariant under swapping k and m and l and h. It then holds that if P satisfies intersection then it satisfies downward-stability and singletontransitivity.
Proof. First, we prove downward-stability. By Lemma 2, we know that i, j, k, l are all different. Suppose that ij ⊥ ⊥ kl | C ∪ {mh}. Notice that, because of exchangeability, the assumptions of the statement hold for every i, j, k, l, m, h. We prove that ij ⊥ ⊥ kl | C by reverse induction on |C|. Notice that, by assumption, mh / ∈ C ij,kl . In addition, without loss of generality, we can assume that m, h = 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 maximal. Because of exchangeability, by the invariance of C under the aforementioned swaps, and since mh / ∈ C ij,kl , we have that ij ⊥ ⊥ mh | C ∪ {kl}. By intersection ij ⊥ ⊥ {kl, mh} | C, which, by decomposition, implies ij ⊥ ⊥ kl | C. Now suppose that ij ⊥ ⊥ kl | C ∪ {mh}, C is not maximal, and, for every By induction hypothesis, we can remove vertices from the conditioning set in order to obtain ij ⊥ ⊥ mh | C ∪ {kl}. Now, again intersection and decomposition imply the result.
Singleton-transitivity also follows from the above argument. We need to show that ij ⊥ ⊥ kl | C and ij ⊥ ⊥ kl | C ∪ {mh} imply ij ⊥ ⊥ mh | C or mh ⊥ ⊥ kl | C. (Notice that because of downward-stability ij ⊥ ⊥ kl | C is immaterial.) Again without loss of generality, we can assume that m, h = i, j, and the above argument showed that ij ⊥ ⊥ {kl, mh} | C, which by decomposition, implies ij ⊥ ⊥ mh | C.
Theorem 4. Let a distribution P be defined over an exchangeable random network and sk(P ) = L(n). Suppose that for some i, j, k, l, and every maximal C such that ij ⊥ ⊥ kl | C ∪{mh}, for every m, h, it holds that C ij,kl ∩(C ∪{mh}) = ∅ and C is invariant under swapping k and m and l and h. Then P is faithful to L ↔ (n) if and only if P satisfies the intersection and composition properties.
Proof. The proof follows from Propositions 14 and 2.

The compliment of the incidence graph case
In this section, we assume that sk(P ) = L c (n). We show again that, under certain assumptions, intersection and composition are sufficient for faithfulness. We consider two additional regimes in the compliment of the incidence graph case related to the cases of Theorem 2: the undirected complement of the incidence graph case and the bidirected complement of incidence graph case. Let C c ijk = {lm : ∀l, m / ∈ {i, j, k}}. For the faithfulness results to the undirected case, an assumption that is used is that for some (and because of exchangeability for all) i, j, k, ij ⊥ ⊥ ik | C implies that C c ijk ⊆ C. For the faithfulness results to the bidirected case, one assumption is that for some (and because of exchangeability for all) i, j, k, ij ⊥ ⊥ ik | C implies that C c ijk ∩ C = ∅.

The undirected compliment of the incidence graph case
Let C j = {jr : ∀r = j}. Recall also that C c ij = {lm : ∀l, m / ∈ {i, j}}. C c ij is a minimal separator w.r.t. its size in L c − (n): Proposition 15. In L c − (n), n > 4, it holds that ij ⊥ ik | C c ij , and if ij ⊥ ik | C then |C c ij | ≤ |C|. In fact, if C = C c ij and C = C c ik then |C c ij | < |C|. Proof. 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 c ij . To prove the second statement, consider the sets C c ij \ C c ik = C k \ {ik, jk} and C c ik \ C c ij = C j \ {ij, jk}, i.e., the neighbours of ij and ik with the joint neighbours removed. For every subset S of C k \ {ik, jk}, clearly there are at least the same number of vertices in C j \ {ij, jk} adjacent to members of S. Hence, the Hall's marriage theorem [11] implies the result.
To prove the third statement, suppose, for contradiction, that C = C c ij and C = C c ik and |C| = |C c ij |. Using the fact that, for every missing member of C c ij in C, there is at least a member of C c ik that should be in C, there cannot be any vertex outside C c ij ∪ C c ik in a C. Hence, C ⊂ C c ij ∪ C c ik . If n > 4 and, km, jo / ∈ C, where o = m, the path ik, jo, km, ij connects ik and ij, a contradiction. Thus, say, km, jm / ∈ C (which implies jl, kl ∈ C). Then the path ik, jm, il, km, ij connects ik and ij, a contradiction.
However, not all minimal separators of 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 ⊥ ik | C. In addition, C is minimal. However, C is not of the form in the above proposition.
Proposition 16. Let a distribution P be defined over an exchangeable random network and sk(P ) = L c (n). Suppose that for some i, j, k, and every minimal C such that ij ⊥ ⊥ ik | C, C c ijk ⊆ C and C is invariant under swapping k and m for every m. It then holds that if P satisfies composition then it satisfies upward-stability and singleton-transitivity.
Proof. First, we prove upward-stability. By Lemma 2, we know that the form of independencies for sk(P ) = L c (n) is ij ⊥ ⊥ ik | C, as provided in the statement of the proposition. Suppose that ij ⊥ ⊥ ik | C. Notice that, because of exchangeability, the assumptions of the statement hold for every i, j, k, l, m. For a variable lm / ∈ C ∪ {ij, ik}, we prove that ij ⊥ ⊥ ik | C ∪ {lm} by induction on |C|. Notice that, by assumption, lm ∈ C ijk . Without loss of generality, we can assume that l ∈ {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 / ∈ C c ijk , we have that ij ⊥ ⊥ im | C. By composition ij ⊥ ⊥ {ik, im} | C, which, by weak union, implies ij ⊥ ⊥ ik | C ∪{im}. If l = k then by swapping k and i, we have that jk ⊥ ⊥ ik | C. Now, notice that the statement ij ⊥ ⊥ 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 ⊥ ⊥ ik | C. By this, ij ⊥ ⊥ ik | C, and the use of composition, we obtain {ij, km} ⊥ ⊥ ik | C, which, by weak union, implies ij ⊥ ⊥ ik | C ∪ {km}. Now suppose that ij ⊥ ⊥ ik | C, C is not minimal, and, for every C ′ such that , for every op. Consider the independence statement ij ⊥ ⊥ ik | C 0 such that C 0 ⊂ C and C 0 is minimal. We have that ij ⊥ ⊥ im | C 0 or km ⊥ ⊥ ik | C 0 . By induction hypothesis, we can add vertices to the conditioning set in order to obtain ij ⊥ ⊥ im | C or km ⊥ ⊥ 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 ⊥ ⊥ ik | C and ij ⊥ ⊥ ik | C ∪ {lm} imply ij ⊥ ⊥ lm | C or lm ⊥ ⊥ ik | C. (Notice that because of upward-stability ij ⊥ ⊥ ik | C ∪{lm} is immaterial.) Again without loss of generality, we can assume that l = i or l = k, and the above argument showed that ij ⊥ ⊥ im | C or km ⊥ ⊥ ik | C.
Theorem 5. Let a distribution P be defined over an exchangeable random network and sk(P ) = L c (n). Suppose that for some i, j, k, and every minimal C such that ij ⊥ ⊥ ik | C, C c ijk ⊆ C and C is invariant under swapping k and m for every m. Then P is faithful to L c − (n) if and only if P satisfies the intersection and composition properties.
Proof. The proof follows from Propositions 16 and 1.

The bidirected compliment of the incidence graph case
C ij \ {ik} is a maximal separator w.r.t. its size in L c ↔ (n): The proof follows from Proposition 15 and Lemma 5, and considering |C ij \ {ik}| = |C ij | − 1.
As is the case for L c − (n), not all minimal separators of L c ↔ (n) are of the form above. For example, in L c ↔ (5), consider the set C = {kl, il, jk, jl}. It holds that ij ⊥ ik | C, and C is maximal.
Proposition 18. Suppose that for some i, j, k, and every maximal C such that ij ⊥ ⊥ ik | C ∪ {lm}, for every l, m, C c ijk ∩ (C ∪ {lm}) = ∅ and C is invariant under swapping k and m. It then holds that if P satisfies intersection then it satisfies downward-stability and singleton-transitivity.
Proof. First, we prove downward-stability. By Lemma 2, we know that the form of independencies for sk(P ) = L c (n) is ij ⊥ ⊥ ik | C, as provided in the statement of the proposition. Suppose that ij ⊥ ⊥ ik | C ∪ {lm}. Notice that, because of exchangeability, the assumptions of the statement hold for every i, j, k, l, m. We prove that ij ⊥ ⊥ ik | C by reverse induction on |C|. Notice that, by assumption, lm ∈ C ijk . Without loss of generality, we can assume that l ∈ {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 maximal. We have two cases: If l = i then because of exchangeability, by the invariance of C under swapping k and m, and since im / ∈ C c ijk , we have that ij ⊥ ⊥ im | C ∪ {ik}. By intersection ij ⊥ ⊥ {ik, im} | C, which, by decomposition, implies ij ⊥ ⊥ ik | C. If l = k then by swapping k and i, we have that jk ⊥ ⊥ ik | C ∪ {im}. Now, notice that the statement ij ⊥ ⊥ 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 ⊥ ⊥ ik | C ∪ {ij}. By this, ij ⊥ ⊥ ik | C ∪{km}, and the use of intersection, we obtain {ij, km} ⊥ ⊥ ik | C, which, by decomposition, implies ij ⊥ ⊥ ik | C. Now suppose that ij ⊥ ⊥ ik | C ∪ {lm}, C is not maximal, and, for every , for every op. Consider the independence statement ij ⊥ ⊥ ik | C 0 ∪ {lm} such that C ⊂ C 0 and C 0 is maximal. We have that ij ⊥ ⊥ lm | C 0 ∪ {ik} or lm ⊥ ⊥ ik | C 0 ∪ {ij} depending on whether l = i or l = k. By induction hypothesis, we can remove vertices from the conditioning set in order to obtain ij ⊥ ⊥ lm | C ∪ {ik} or lm ⊥ ⊥ ik | C ∪ {ij}. Now, again intersection and decomposition imply the result.
Theorem 6. Let a distribution P be defined over an exchangeable random network and sk(P ) = L c (n). Suppose that for some i, j, k, and every maximal C such that ij ⊥ ⊥ ik | C ∪ {lm}, for every l, m, C c ijk ∩ (C ∪ {lm}) = ∅ and C is invariant under swapping k and m. Then P is faithful to L c ↔ (n) if and only if P satisfies the intersection and composition properties.
Proof. The proof follows from Propositions 18 and 2.

The empty graph case
Clearly the minimal separator in the empty graph is the empty set, and the maximal separator is all the remaining vertices. In the case where sk(P ) is the empty set, we have the following conditions for intersection and composition being sufficient for faithfulness: Proof. First, we prove upward-and downward-stability in (a) and (b). By Lemma 2, we know that sk(P ) is an empty graph. a) Suppose that ij ⊥ ⊥ kl | C or ij ⊥ ⊥ ik | C. Notice that, because of exchangeability, the assumptions of the statement hold for every disjoint i, j, k, l. We prove that ij ⊥ ⊥ kl | C ∪ {mh} or ij ⊥ ⊥ ik | C ∪ {mh}, for every mh / ∈ C ∪ {ij, kl} or mh / ∈ C ∪ {ij, ik}, respectively, by induction on |C|. The base case is when C = ∅. First, consider the case where ij ⊥ ⊥ kl. If mh / ∈ C ij,kl then by swapping k and m and l and h, we obtain ij ⊥ ⊥ mh. If mh ∈ C ij,kl then say mh = jl. We use ij ⊥ ⊥ ik and first swap i and j to obtain ij ⊥ ⊥ jk. Now we swap k and l to obtain ij ⊥ ⊥ jl. (The other three cases of mh are similar.) Now composition and weak-union imply the result. Now, consider the case where ij ⊥ ⊥ ik. First suppose that mh ∈ C ij,kl . If mh = il then by swapping j and l, we obtain il ⊥ ⊥ ik. If mh = jk then by swapping i and k, we obtain jk ⊥ ⊥ ik. If mh = jl then by swapping i and j and then k and l, we obtain ij ⊥ ⊥ jl. If mh / ∈ C ij,kl then use ij ⊥ ⊥ kl and swap k and m and l and h to obtain ij ⊥ ⊥ mh. Now composition and weak-union imply the result.
The inductive step is similar to the inductive step of Proposition 12 or Proposition 16 depending on the form ij ⊥ ⊥ kl | C or ij ⊥ ⊥ ik | C. b) Suppose that ij ⊥ ⊥ kl | C or ij ⊥ ⊥ ik | C. Notice that, because of exchangeability, the assumptions of the statement hold for every disjoint i, j, k, l. We prove that ij ⊥ ⊥ kl | C \ {mh} or ij ⊥ ⊥ ik | C \ {mh}, for every mh ∈ C, by reverse induction on |C|.
The base case is when C = V \ {ij, kl} or C = V \ {ij, ik}. First, consider the case where ij ⊥ ⊥ kl | V \ {ij, kl}. If mh / ∈ C ij,kl then by swapping k and m and l and h, we obtain ij ⊥ ⊥ mh | V \ {ij, mh}. If mh ∈ C ij,kl then say mh = jl. We use ij ⊥ ⊥ ik | V \ {ij, ik} and first swap i and j to obtain ij ⊥ ⊥ jk | V \ {ij, jk}.
Now we swap k and l to obtain ij ⊥ ⊥ jl | V \ {ij, jl}. (The other three cases of mh are similar.) Intersection and decomposition imply the result. Now, consider the case where ij ⊥ ⊥ ik | V \ {ij, ik}. If mh ∈ C ij,kl then for il swap j and l, for jk swap i and j, for jl swap i and j and then k and l, and apply intersection and decomposition. If mh / ∈ C ij,kl then use ij ⊥ ⊥ kl | V \ {ij, kl}, and swap k and m and l and h, and again apply intersection and decomposition.
The inductive step is similar to the inductive step of Proposition 14 or Proposition 18 depending on the form ij ⊥ ⊥ kl | C or ij ⊥ ⊥ ik | C. Singleton-transitivity also follows from the above argument, since we showed that (the immaterial) ij ⊥ ⊥ ik | C and ij ⊥ ⊥ ik | C ∪{lm} imply ij ⊥ ⊥ lm | C ∪{ik} or lm ⊥ ⊥ ik | C ∪ {ij}. Now, again intersection and decomposition imply the result.
Theorem 7. Let a distribution P be defined over an exchangeable random network and sk(P ) is empty. Suppose also that one of the following cases holds for disjoint i, j, k, l: Then P is faithful to the empty graph if and only if P satisfies the intersection and composition properties.

Summary and discussion
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 ⊥ ⊥ B | C ⇒ A ⊥ ⊥ (B∪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 [14], we showed that exchangeable random networks' independence structures that are wellstructured -in the sense that they can be represented by a graph in graphical model sense-are of one of the six possible cases: completely dyadic-independent, faithful to the undirected or bidiredted incidence graphs, faithful to the undirected or bidiredted compliment of the incidence graph, or completely dyadicdependent.
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 ⊥ ⊥ kl | C or ij ⊥ ⊥ 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 statistical network models for exchangeable random networks with (in most situations) binary dyads. One point is that binary distributions always satisfy singleton-transitivity [8], 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 [25]) satisfy the provided conditions, both when we deal with binary random networks or weighted ones.