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I'm trying to figure out how to determine if these two models are I-equivalent. Google didn't properly give me a solid answer so far. Any idea on to determine it? Thank you. Image of two models

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  • $\begingroup$ equivalent in which sense? their joint distributions are different. $\endgroup$
    – gunes
    Commented Mar 17, 2019 at 19:44
  • $\begingroup$ How do you generally theorize if they the two directed probabilistic graphical models are equivalent or not? $\endgroup$
    – RinW
    Commented Mar 17, 2019 at 20:02
  • $\begingroup$ @RinW Where did you find this problem? Is it online somewhere with solutions? $\endgroup$
    – Abraham
    Commented Mar 16, 2020 at 5:20
  • $\begingroup$ @Abraham it was online, but I can't remember the source of it since it was almost a year ago. Sorry $\endgroup$
    – RinW
    Commented Mar 16, 2020 at 18:38

2 Answers 2

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Equivalence definition matters here. But, I-equivalence is a common query in Bayesian Networks, in which I just assumed you're asking for that, but you need to specify in your question as well. For two BNs to be I-equivalent, they need to have the same skeleton, and immoralities. Check the lecture here. Immorality is defined as a V-structure that a node $X$, having parents $Y,Z$ without an edge in between them.

In your graphs, the skeletons are clearly the same. And, the only V-structure in both graphs is the $B,C,D$ triple. So, the two graphs have the same V-structures and they are I-equivalent.

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  • $\begingroup$ Thank you, I'm trying to find its equivalence by same set of independence assumptions. If the structure changes, C->B and D->B would the same equivalence exist since the skeleton changes? $\endgroup$
    – RinW
    Commented Mar 17, 2019 at 20:47
  • $\begingroup$ Yes, this is I-equivalence: check this also: ai.stanford.edu/~moises/tutorial/sld019.htm $\endgroup$
    – gunes
    Commented Mar 17, 2019 at 20:49
  • $\begingroup$ So according to the second link, the second network on it is not i-equivalence because it does not share the same V-structure, right? The link between D & E. $\endgroup$
    – RinW
    Commented Mar 17, 2019 at 20:56
  • $\begingroup$ Yes, on the right figure S,E,D create a new V-structure, that's not present on the left one. Similarly for C,E,D. $\endgroup$
    – gunes
    Commented Mar 17, 2019 at 20:57
  • $\begingroup$ I think the same logic applies when imgur.com/YzA9Toc is considered as well. Since the V-structures changes for B between the two. $\endgroup$
    – RinW
    Commented Mar 17, 2019 at 21:16
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You can read section 9 of this article and this piece of information would be helpful:

I-equivalence can be determined by looking at the skeleton graphs, i.e. the graphs formed by converting directed edges to undirected edges. Two graphs are I-equivalent if they have the same skeleton and the same set of v-structures

In your case, you just compare the v-structures in each graph(both contain only $B\rightarrow C \leftarrow D$) and remove the direction to compare the skeleton(the same), so they are I-equivalent.

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