I have been trying to understand the graphoid axioms, one of which–Decomposition–claims that
$$ (X \bot Y,W |Z) \implies (X \bot Y|Z).$$
In Pearl's book Causality, he writes $\bot$ with two vertical lines, and omits the comma. The list is just after equation 1.29 in that text.
Axiom C4 in this lecture by Steffen Lauritzen, uses the comma. Wikipedia seems to claim that it is a logical and uses a different notation:
$$I(X,Z,Y)\ \text{ means that }\ (X \bot Y |Z).$$
There are several answers to questions like this, but as far as I can tell none of them actually give a definition of the comma. Neither do any writings by Pearl or any other online sources I have read, except where they contradict each other. I can't prove the statement if I can't tell what it is saying!
Now before you just say, "It's an and" or some such, I will point out that multiple people have claimed that it is a set intersection, a set union, a logical and–meaning that it isn't one statement but rather two to begin with? and I don't know what else. I tried my best guess and found a counterexample, so clearly I don't know what it is trying to say.
I want to know what is the definition of the claim. Exactly. No hand waving. If you can carefully distinguish sets from random variables from probabilities that would also be a big help.
If I ever figure this out I promise to share my understanding as clearly as possible with others.
Best Answer
Concatenation means and in Pearl's independence axioms. So Pearl's Decomposition $(X\bot YW \;| Z)$ should be read as "$X$ is independent of $(Y\&W)$ given $Z$."
Pearl uses comma for and in probability statements. This seems clear given (1.28)and (1.29), which are pretty standard defintions of conditional independence and regular probabilistic independence.
With this understanding, you can derive Decomposition from the definition of conditional independence in (1.28) [p. 11 in 2nd ed.].
Intersection (the set-theoretic operation, not the graph axiom) is the set-theoretic version of and. They're equivalent because you can convert statements to sets in which they hold, and vice versa.
Comma is and in the probability statements in Wikipedia. In $I(X,Y,Z)$ the commas distinguish arguments to the predicate $I$, which means "$X$ is independent of $Y$ given $Z$" as stated on the page.
Lauritzen seems to use comma for and, though I haven't checked carefully. That looks right, though, and he uses $\cup$ (union, or) in other places.