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I am studying some concepts about d-separation through the book [1], however, I am not understanding the intuition behind the axiomatization of graphoids. If anyone can help me with a little clarification and knowledge I would be very grateful, it's been 2 days that I have been stuck at this point.

Given $X, Y$ and $Z$ three disjoint sets of nodes from a graph $G$.

1) Symmetry:

$ I(X, Z, Y) \iff\ I(Y, Z, X) $

I don't understand why this bi-implication makes sense in this case. A non-causal relationship between $X$ and $Y$ given $Z$ does not allow one to believe that the same is valid for the reverse. In terms of causality, this does not seem valid to me.

2) Intersection:

$ I(X, Z \cup W, Y) \space and \space I(X, Z \cup Y, W) \implies I(X, Z, W \cup Y)$

I did not understand this property and the reason that makes $X$ independent of $Y \cup W$ given $Z$ in this case. Although the left side of the implication may in fact occur, but we have no guarantees that the right side is true. We cannot assume that all the independence relations indicated in the left side are due only to $Z$.

:)

[1] Modeling and Reasoning with Bayesian Networks - Adnan Dawiche

Edit: There are other axioms, but these are the ones that confuse me.

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1 Answer 1

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For (1), the two sides have exchanged X and Y, while Z is fixed. "X and Y" given Z is the same as "Y and X" given Z (because "X and Y" is the same as "Y and X"). That's all.

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  • $\begingroup$ Thank you so much for your answer! I got it, but my question is: If X is independent of Y given Z why can we believe that Y is independent of X given Z. For me in terms of causal relationship this not is necessarily valid. This is what the property shows. $\endgroup$
    – Jardel Carvalho
    Apr 16, 2020 at 20:25
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    $\begingroup$ The formal definition of independence is $P(X \cap Y | Z) = P(X | Z)P(Y | Z)$. This is clearly symmetric in $X$ and $Y$. Therefore if $X$ is independent of $Y$ (given $Z$), then $Y$ is independent of $X$ (given $Z$). If this property isn't valid for whatever you are studying, then you cannot use this framework. $\endgroup$
    – Ted
    Apr 16, 2020 at 23:05

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