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?


1 Answer 1


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:

enter image description here

So the answer to your question

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?

would be "Yes."

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 $Z.$

  • $\begingroup$ In the above example, if A->(allows)->B->(causes)->X and A->(causes)->X and X->(causes)->Y, {A,B} is still the confounder set? I'm asking this because A does not "cause" B, but just allows for B, which causes X $\endgroup$
    – pentavol
    Aug 16, 2021 at 6:25
  • $\begingroup$ @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. $\endgroup$ Aug 16, 2021 at 14:12

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