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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?

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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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