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$\newcommand{\ci}{\perp\!\!\!\perp}$ Although a probabilistic directed acyclic graph (DAG) can only be inferred from conditional independence (CI) properties of the variables up to a Markov equivalence class, are there CI properties which cannot be represented as a DAG?

Maybe answering my question, in the paper "Beware of the DAG!", NIPS 2008, from AP Dawid, the author says:

"It is important to note that, for given variable set $\mathscr{V}$, the collection of CI properties $\mathscr{C}$ that can be represented by a DAG is very special. Thus with $\mathscr{V} =\{X,Y,Z\}$, the pair of properties $\{X \ci Y,X\ci Y\ |\ Z\}$ has no DAG representation."

Is that so? What about a DAG without edges, wouldn't it satisfy those properties?

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    $\begingroup$ Since the D in DAG means directed, a 'DAG' without arrows is not a DAG. $\endgroup$ – Juho Kokkala Jan 23 '15 at 16:35
  • $\begingroup$ Well, you could put an arrow between X and Z or Y and Z, if you'd like... In any case, one could argue the trivial graph is a directed acyclic graph, since there are no cycles and every edge is directed (it is also undirected, but that's the beauty of zero). $\endgroup$ – Leo Azevedo Jan 23 '15 at 16:40
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    $\begingroup$ By 'a DAG without arrows' in the final sentence, do you mean a DAG without edges at all? $\endgroup$ – Juho Kokkala Jan 23 '15 at 16:55
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    $\begingroup$ Corrected 'DAG without arrows' to 'DAG without edges' $\endgroup$ – Leo Azevedo Jan 23 '15 at 17:32
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    $\begingroup$ @Alexis: I agree. My previous comments stems from the original wording of the question, 'DAG without arrows'. I misinterpreted that to mean 'a graph with undirected edges'. $\endgroup$ – Juho Kokkala Jan 23 '15 at 19:28
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$\newcommand{\ci}{\perp\!\!\!\perp}$ Based on comments, I assume by 'a DAG without arrows' you mean the DAG with no edges.

The graph with no edges indeed has the conditional independence properties $X \ci Y$ and $X\ci Y\ |\ Z$. However, the point is that there is no DAG that implies exactly those conditional independence properties. The trivial DAG also implies other conditional independence properties, for example $Y \ci Z$, and thus it is not a DAG representation of the pair of properties $\{X \ci Y,X\ci Y\ |\ Z\}$.

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  • $\begingroup$ I see, thx. Completing the question, are there complete sets of CI conditions, i.e., for which we know the CI condition between every subset of variables, but it is still not possible to represent it in a DAG? I've heard something about neighboring colliders and CI implying cyclic graphs. $\endgroup$ – Leo Azevedo Jan 23 '15 at 17:42

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