For a directed graphical model, does D-separation is equivalent to conditional independence. I have got bit confused about this concept from the following statement from this source.
if X and Y are not d-separated given Z in G, then are X and Y are
dependent in all distributions P that factorize over G? This is false,
because a distribution that factorizes over G may yet contain
additional “accidental” independencies not asserted by the graph
structure
Best Answer
D-seperation is not equivalent to conditional independence. The D-seperation of $X$ and $Y$ given $Z$ implies the following conditional independence:
$$P(X,Y|Z) = P(X|Z)P(Y|Z).$$
However D-seperation is a concept that applies specifically to graphical models. You can talk about conditional independence in any context involving random variables.
The statement is talking about something different. What the statement is saying is that just because two nodes $X$ and $Y$ are not D-seperated given some subset of nodes $Z$, that doesn't mean there doesn't exist some probability distribution which factorizes over $G$ for which $X$ and $Y$ are conditionally independent given $Z$. In fact there will always exist such a distribution (see the following paragraph).
Graphical models are only required to not encode any independency which doesn't actually exist in the joint probability distribution, that is to say, they are required to make no unwarranted assumptions. However there's no requirement that they encode every independency that does in fact exist, and often times building a graphical model which captures every existing independency while at the same time avoiding the instantiation of spurious independencies is impossible (such a graph when it does exist is called a perfect I-map). Therefore given a graphical model $G$, any mutually independent probability distribution on the nodes of $G$ trivially factorizes over $G$.