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Let $G = (V,E)$ be a directed acyclic graph.

Let $i \rightarrow j$ be an edge such that the parents of $j$ are exactly:

  • the parents of $i$,
  • and $i$.

Let $L\left(G \right)$ be the set defined by {a vertex and its parents, for all vertices in $G$}

Let $G' = (V,F)$ where $F$ is $E$ with $j \rightarrow i$ instead of $i \rightarrow j$.

Given a set of random variables $(X_{v})_{v \in V}$ for which the joint density can be factorized over $L\left(G \right)$, I would like to show that the joint density could also be factorized over $L\left(G' \right)$.


Let $P$ be the parents of $i$ in the graph $G$, thus the parents of $j$ in the graph $G'$.

So I would like to show that there exists $g_{i}$ and $g_{j}$ such that: $g_{i}(x_{i},x_{j},x_{P}) \times g_{j}(x_{j},x_{P}) = f_{i}(x_{i},x_{P}) \times f_{j}(x_{i},x_{j},x_{P})$.

I am unsure whether I could just consider that there was a swap of $i$ and $j$, and I could take $g_{i}=f_{j}$ and $g_{j}=f_{i}$. Or maybe, there is something to do with conditional probabilities and Bayes formula.

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1 Answer 1

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What you are looking for is termed a covered edge reversal, which preserves Markov equivalence of directed acyclic graphs. A proof of equivalence is given in Lemma 1 of A Transformational Characterization of Equivalent Bayesian Network Structures, by Max Chickering. The relationship between Markov properties and factorisation is well established in the theory of graphical models, for example Theorem 5.14 of Probabilistic Networks and Expert Systems.

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    $\begingroup$ This is the perfect answer, I realize now that I was getting lost with my own proof. :) $\endgroup$
    – Wok
    Nov 14, 2010 at 8:42

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