3 ms·
No, the other poster is referring to Bayes' rule P(A|B) = P(A)*P(B|A)/P(B). The claim is that it is reasonable to assume P(A) ~ P(B), so that P(A|B) ~ P(B|A).
by nonbel 9y ago
No, the other poster is referring to Bayes' rule P(A|B) = P(A)*P(B|A)/P(B).
The claim is that it is reasonable to assume P(A) ~ P(B), so that P(A|B) ~ P(B|A). I am referring to P(A) and P(B) in the equation.
- Bromskloss 9y agoAh, OK, got it.
- nonbel 9y agoIts hard to tell over the internet, but if you are being sarcastic please let me know. I have no problem with clarifying further.
- Bromskloss 9y agoNo sarcasm! I believe I understand what you were saying now.