= 27d567bc2d48d9f40e9ccc20ea370b15)
PhilGoetz comments on Open Thread, September, 2010-- part 2 - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (858)
I was wrong, and jimrandomh was right. I said:
P(X|A,B) = 1-(1-P(X|A))(1-P(X|B)) if P(A,B) = P(A)P(B)
But P(X|A,B) = 1-P(~X|A,B)
therefore P(~X|A,B) = (1-P(X|A))(1-P(X|B)) = P(~X|A)P(~X|B)
This is equivalent to claiming P(X|A,B) = P(X|A)P(X|B)
And this is wrong for most distributions.