3 ms·
Eh? (P(X|Y)>P(X) iff P(Y|X)>P(Y)) whenever all the relevant probabilities (P(X), P(Y), P(X and Y), etc.) are strictly positive and strictly less than 1. Now, X
by drdeca 2y ago
Eh? (P(X|Y)>P(X) iff P(Y|X)>P(Y)) whenever all the relevant probabilities (P(X), P(Y), P(X and Y), etc.) are strictly positive and strictly less than 1.
Now, X might be much stronger evidence for Y than Y is for X, but if Y is evidence for X, then (assuming the previously mentioned assumption) then X is evidence for Y, even if only very weak.
- fnordpiglet 2y agoYes but the comment being responded to stated “but there are Y that are not X” which was categorical.
- drdeca 2y agoSorry, I don’t understand this point. What’s the interpretation of your previous message (in light of that) which makes it true?