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JGWeissman comments on Causality does not imply correlation - Less Wrong
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Maybe I'm not quite understanding, but it seems to me that your argument relies on a rather broad definition of "causality". B may be dependent on A, but to say that A "causes" B seems to ignore some important connotations of the concept.
I think what bugs me about it is that "causality" implies a directness of the dependency between the two events. At first glance, this example seems like a direct relationship. But I would argue that B is not caused by A alone, but by both A's current and previous states. If you were to transform A so that a given B depended directly on a given A', I think you would indeed see a correlation.
I realize that I'm kind of arguing in a circle here; what I'm ultimately saying is that the term "cause" ought to imply correlation, because that is more useful to us than a synonym for "determine", and because that is more in line (to my mind, at least) with the generally accepted connotations of the word.
This is the right idea. For small epsilon, B(t) should have a weak negative correlation with A(t - epsilon), a weak positive correlation with A(t + epsilon). and a strong positive correlation with the difference A(t + epsilon) - A(t - epsilon).
The function A causes the function B, but the value of A at time t does not cause the value of B at time t. Therefore the lack of correlation between A(t) and B(t) does not contradict causation implying correlation.
Only trivially. Since B = dA/dt, the correlation between B and dA/dt is perfect. Likewise for any other relationship B = F(A): B correlates perfectly with F(A). But you would only compare B and F(A) if you already had some reason to guess they were related, and having done so would observe they were the same and not trouble with correlations at all.
If you do not know that B = dA/dt and have no reason to guess this hypothesis, correlations will tell you nothing, especially if your time series data has too large a time step -- as positively recommended in the linked paper -- to see dA/dt at all.