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Simpler article (with a proof): http://oyhus.no/CorrelationAndCausation.html http://oyhus.no/CorrelationAndCausation.html Edit: also regarding the last sentenc

• infinity0 10y ago
  I think the point of "correlation does not imply causation", refers to the literal prepositional logic sense of "=>". Yes, correlation suggests causation, i.e. P(causation|correlation) > P(causation) from a Bayes perspective. That doesn't mean you should discount the possibility of ¬causation, merely that its probability is smaller. And "how much smaller" could be very close to 0, so it would still be hasty to say "implies", which linguistically implies "=>". A better phrasing would be "correlation suggests but does not imply causation". (edit: e.g. as per that xkcd comic, mentioned by other posters. edit2: I mixed up the proof with the OP article. the proof uses "evidence of" which is also good.) But yes, nice proof nonetheless. I like how causation is basically defined as P(c|a) = 1, showing how most complex philosophical issues are actually irrelevant (for this particular result).
  • dragonwriter 10y ago
    Actually, as I see it, the main point of "correlation does not imply causation" is mostly "correlation does not imply a particular causal relationship". While coincidence is possible, as well, its mostly about the fact that you can't conclude A causes B from a correlation between A and B alone, because the correlation may be due to the fact that B causes A, or the fact that A and B are both caused by C. That's why you need the correlation, plus an explanatory theory of the causation, plus evidence to reject alternative causal relationships, to have a decently strong basis for concluding a particular causal relationship.
    • infinity0 10y ago
      I think, what I said is the same as what you said, just with some more maths. > you can't conclude A causes B right, this is what I referred to as "=>" > plus an explanatory theory of the causation, plus evidence yes, this all works together to build up the "how much smaller". An explanatory theory basically allows you to make predictions and run tests to collect more data to pump into the application of Bayes' theorem as used by that proof, improving your confidence of the difference between P(a|c) and P(a).
      • jsprogrammer 10y ago
        Belief in "correlation implies causation" admits the Law of Excluded Middle fallacy. Just because you make an observation consistent with your beliefs, does not mean that you can claim all other explanations (complement of your beliefs) are invalid (primarily because you do not know what they are or could be).
      • JadeNB 10y ago
        > > you can't conclude A causes B > right, this is what I referred to as "=>" Except that it isn't quite, since you were careful to clarify that your \implies (i.e., '=>' or '⇒') was the \implies of propositional logic, which explicitly disclaims any causal relationship. \implies in that context says precisely and only that the antecedent is false, or the consequent is true. In this sense, 2 + 2 = 4 \implies Barack Obama is currently the president of the US, and 2 + 2 = 5 \implies George Bush is currently the president of the US, even though there is no causal relationship in either case.
        • infinity0 10y ago
          "is" in language can often mean "is a subset of", I was using the term that way. Whilst you are right that "=>" disclaims any philosophical relationship, the proof covers all definitions of "cause" that one might reasonably come up with. So "you can't conclude A => B" implies (with probability 1) that "you can't conclude A causes B": The proof only defines "causation" as some event "a" for which "P(c|a) = 1". This is the same property that "=>" has in propositional logic, and there is no implication of philosophical causation here either. But the proof still works, as a consequence of its definitions. So in other words, the proof says: if causation causes correlation then P(a|c) > P(a) (i.e. correlation is evidence of causation) but we can't say causation is definitely true (P(a) = 1), however you want to define "causes" as long as it has the property that P(c|a) = 1.
  • deleted 10y ago
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  • jerf 10y ago
    Shorter: Correlation does not Aristotelian-imply causation. Correlation probabilistically-implies causation. Aristotelian logic is a necessary step towards understanding logic, but on its own it's not actually that useful because in reality we just don't have enough things that we can safely approximate as 100% true for it to work reliably. Incidentally, this also means that the classic lists of "argument fallacies" often contain a few fallacies themselves, as an argument being Aristotelian-fallacious does not mean that it's practically- or probabilistically-fallacious. But I will agree that this is generally a distinction without a difference; I only rarely see someone accuse someone else of committing a fallacy where the accusation is Aristotelian-correct but not probabilistically-correct. But, rolling back around to the original point, most of them are indeed thoughtless recitations of the "correlation does not imply causation" mantra when, probabilistically, the correlation being observed can be reasonably interpreted as evidence.
• mhermher 10y ago
  P(c|a) is not 1. If we assume correlation is Pearson, then the "shape" of the causation has to be a linear effect for P(c|a) = 1. A U-shaped effect will probably give you a Pearson 0, etc. I do love the work of Pearl, et. al. But even they do not admit Pearson (or any other type) of correlation, but rather the nebulous "statistical dependence". So the proof only works if your statistical dependence tool of choice matches the nature of your causal effect perfectly.
  • xtacy 10y ago
    > But even they do not admit Pearson (or any other type) of correlation, but rather the nebulous "statistical dependence". The notion of "Statistical dependence" is not nebulous. X and Y are independent if the joint distribution factorises as p(X, Y) = p(X) p(Y) > even they do not admit Pearson (or any other type) of correlation, Precisely. They operate purely in probabilistic dependence/independence terminology, from a theoretical point of view.
    • mhermher 10y ago
      I understand, but from an inference point of view, it is nebulous because the distributions are unknown. Statistical dependence can be determined easily if you know the distributions, but not if you only have samples. The "c" in the proof, I assume means "observed correlation". Because we are in fact talking about "[Observed] correlation does not imply [Unobserved] causation", right?
      • xtacy 10y ago
        I partly agree with you, but I wouldn't go to the extent of using "easily" in this phrase -- > Statistical dependence can be determined easily if you know the distributions, Even if you know the distribution, a statistical test will make Type-I/II errors that you would have to take care of. Actually, I find the text linked above hard to understand, without properly defining 'c.' What's the sample space? In general, my sentiments are with the xkcd comic strip, but nothing more. Pearl's theories lay a firm foundation for communicating a causal hypothesis and manipulating it algebraically, but the true tests of causal hypothesis are: - Experimental evidence - The predictions it makes, in cases where experiments are hard to perform (e.g., in physics, when we make certain causal conjectures about how the universe works).
• xiler 10y ago
  I'm no expert on this subject but the proof in the first article seems a bit dodgy. Bayesian inference is a type of induction [1]. So using it to prove induction merely begs the question? [1] http://plato.stanford.edu/entries/induction-problem/#BaySub http://plato.stanford.edu/entries/induction-problem/#BaySub
  • dustintran 10y ago
    The author doesn't really apply Bayesian inference. He applies Bayes' rule which is a mathematical property.