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In statistics, it is a common practice to say that "correlation does not mean causation", and mostly the proof for this is given by examples. While that is good for the intuition, it's not rigorous. Ideas such as a third variable which may be causing both is often cited, but again, that's an example.

Can someone give me a Mathematical argument to why this is true? And possibly, give the mathematical condition for when it does hold true? Any leads or answers are much appreciated :) Thanks in advance!

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    $\begingroup$ "correlation" is mathematics, while "causation" is not; so one should not expect a mathematical argument :-) $\endgroup$ – John B Feb 3 '16 at 20:32
  • $\begingroup$ "Cause" is not a mathematical concept but a metaphysical one. $\endgroup$ – leonbloy Feb 3 '16 at 20:35
  • $\begingroup$ I'm sure that the mathematical theory of causation does exist in the domain of formal logic :) ftp.cs.ucla.edu/pub/stat_ser/r338-shrout.pdf $\endgroup$ – spandan madan Feb 3 '16 at 20:37
  • $\begingroup$ Even the "correlation" concept, as used in the sentence, is not equal to the mathematical (statistical) concept of correlation. It shoud rather be "dependence". I'd say: when we find dependent variables we have no right to postulate a direct causation... but we have right to ask for some rational explanation. But the justification of this assertion (if there is one) is not mathematical, nor even physical, but metaphysical. $\endgroup$ – leonbloy Feb 3 '16 at 20:39
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    $\begingroup$ Dependence means: $\exists f,g : E(f(X)g(Y))\ne E(f(X))E(g(Y))$. Linear correlation: $f = g = \operatorname{id}$ special case. One of $X,Y$ doesn′t cause the other means (came upon the spot): there exists $Z$ (with a restriction on $\sigma(Z)$ relative to $\sigma(X,Y)$) such that $X\mid Z$ is independent of $Y\mid Z$. The former doesn't imply the latter and a single counter-example suffices. $\endgroup$ – A.S. Feb 3 '16 at 20:56
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Let's assume that someone thought "correlation does always mean causation". Then a single example to the contrary, that is, a counterexample, does in fact serve as refutation of that statement.

This has the logical form of, for some predicate $P(x)$, ($\exists a: \lnot P(a)) \Rightarrow \lnot (\forall x: P(x))$.

As Einstein said: "No amount of experimentation can ever prove me right; a single experiment can prove me wrong."

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  • $\begingroup$ While disproving by example is scientifically correct, there is always a mathematical reason behind this. I was looking online for this, and found nothing and so I turned to stackexchange :) I hope someone can show the mathematics of this experimental failure :) $\endgroup$ – spandan madan Feb 3 '16 at 20:56

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