RCCP (Reichenbach's Common Cause Principle) and confounder definition

RCCP states that if X and Y are statistically dependent, then there exists Z causally influencing both. I've heard a variation of RCCP that states that if X and Y are statistically dependent, one of the following scenarios is possible

1. Z = X
2. Z = Y
3. Z is a confounder that is different from X and Y (Z ≠ Y, Z ≠ X), but causes both X and Y

I think that even Bernhard Schölkopf presented RCCP in this manner.

My question is

Given that X and Y are statistically dependent and Z influences both, Z ≠ Y, Z ≠ X, can we say that Z is a confounder? In other words, are the 2 RCCP formulations written above equivalent?

For example:

If A->B->X and A->Y, is A a confounder? (note that A doesn't directly cause X because of mediator B)

In the case above, what can allow us to define the combination of A and B with Z and state that Z is, indeed, the confounder?

A confounder, by definition, is a variable or set of variables that sets up a backdoor path. So if $$Z$$ causally affects both $$X$$ and $$Y,$$ and you are interested in the causal effect of $$X$$ on $$Y,$$ then $$Z$$ is a confounder. The causal diagram would be this:
Given that $$X$$ and $$Y$$ are statistically dependent and $$Z$$ influences both, $$Z \not= Y, Z \not= X,$$ can we say that $$Z$$ is a confounder?
In your example of $$A\to B\to X$$ and $$A\to Y,$$ you neglect to mention whether there is an arrow $$X\to Y.$$ If there is, and I would assume so, then the variable set $$Z=\{A,B\}$$ is a confounder, because it sets up a backdoor path from $$X$$ to $$Y.$$ The principle that lets us define $$Z$$ in this way is simply the principle of naming things. If we consolidate $$\{A,B\}$$ down to a single variable $$Z,$$ then the causal diagram for your scenario (assuming $$X\to Y$$) would be identical to what's above, because it perfectly captures all the interactions of $$Z$$ with everything that's not $$Z.$$
• @pentavol In the case you outline here in your comment, there can't be any confounding set, because the ONLY way to get to $Y$ is through $X.$ A backdoor path, by definition, has to be a DIFFERENT way to get from $X$ to $Y,$ starting with an arrow into $X.$ If you were to add a direct arrow $A\to Y$ or $B\to Y,$ you'd have some confounders. Aug 16, 2021 at 14:12