4 ms·

> > 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

by JadeNB 11y 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 11y 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.
- dragonwriter 11y ago
  That's not entirely true, at least, using the normal definition of "causes" that is of interest in correlation vs. causation discussions, which certainly includes "causes" which are contributors to the occurrence of an effect but do not alone guarantee it (e.g., smoking causes cancer, but it is not true that smoking implies cancer in the propositional logic sense.)

by JadeNB 11y ago

infinity0 11y 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.

dragonwriter 11y ago
That's not entirely true, at least, using the normal definition of "causes" that is of interest in correlation vs. causation discussions, which certainly includes "causes" which are contributors to the occurrence of an effect but do not alone guarantee it (e.g., smoking causes cancer, but it is not true that smoking implies cancer in the propositional logic sense.)