Pearl et al. "Causal Inference in Statistics: A Primer" (2016) p. 26 provides the following definitions of direct cause and cause:

A variable $X$ is a direct cause of a variable $Y$ if $X$ appears in the function that assigns $Y$’s value. $X$ is a cause of $Y$ if it is a direct cause of $Y$, or of any cause of $Y$.

It seems to me the definition of cause is circular (and thus essentially invalid) as it refers back to itself within the definition. Am I misunderstanding it? If not, how could the definition be improved?


It's not circular, it's just recursive. It makes the functional programmer in me smile.

Note that this is actually two definitions - that of a direct cause, and then the more generic definition of (any kind of) cause, using the former as the boundary condition.

It's quite similar to how the set-theoretic definition of natural numbers, if you're familiar with that - there's the base case (empty set for 0 in numbers, direct cause for general causality), and then the recursive part (e.g. 1 is a set of empty sets, and a level-1 generalized cause is the direct cause of a direct cause of Y).

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  • $\begingroup$ I was suspecting something like that, but it is still hard to wrap my head around it... How do I verify whether $X$ is a cause of $Y$ in cases when it is not a direct cause? In such cases, I have to verify that $X$ is a direct cause of a cause of $Y$. But the definition of a cause is not explicit (it is recursive). Could you help me further? $\endgroup$ – Richard Hardy Mar 19 at 13:13
  • $\begingroup$ Y as direct cause of Z: Z = f(Y, u1). X as direct cause of Y: Y = g(X, u2). X as cause of Z: Z = f(g(X, u2), u1), where the 'u's are sets of various other causal factors we're not interested in. If you can only observe X and Z, you may or may not be able to identify causal effects. If you do have Y, it's Bayes and chain rule time. This is a very TL;DR version, you really do need a proper book to cover everything. $\endgroup$ – jkm Mar 19 at 13:35
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    $\begingroup$ I think I got it. I took a walk (before seeing your last comment), and it fell into places. Thank you for the answer! $\endgroup$ – Richard Hardy Mar 19 at 13:58

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