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nshepperd comments on Inherited Improbabilities: Transferring the Burden of Proof - Less Wrong
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It isn't!
In general it is not true that P(A|B) = P(B|A). P(A|Guilt) depends on the prior probabilities of A and Guilt, as well as P(Guilt|A). For example, say we have four possible proofs A, B, C, D, and P(Guilt|A or B or C) = 1, and P(Guilt|D) = 0. Our prior is all four are equally likely: P(A) = P(B) = P(C) = P(D) = 0.25. P(Guilt) is then 0.75 = P(Guilt|A)P(A) + P(Guilt|B)P(B)...
Given this, we have:
P(A|Guilt) isn't 1. But it's 33%, which is still higher than the prior %25: that is, Guilt is evidence for A.
By the way I think it might help if you avoid talking in proofs and implication and 100% certainty. In hypothetical examples it's useful to set things to P(X) = 1, but in the real world evidence is always probabilistic; nothing's ever 100%.
Ahhh, that helps clear things up. For some reason I'd been understanding you as saying that, given P(Guilt|A) = 1, P(A|Guilt) was also 1. It looks like what you meant was just that Guilt is evidence for, but not necessarily 100% proof of, A. Am I getting that all correct?
Yes.
P(Guilt|A) = P(A|Guilt) only when P(A) = P(Guilt). In which case it would be 100% proof. But that is a rare situation.
Nitpick: the two conditionals also be equal if A and Guilt were mutually exclusive. (in that case, of course, the two conditionals would be both zero)