Bayes theorem: P(A | B) = P(B | A) * P(A) / P(B) Let A = W, B = (X, Y) P(W | X, Y) = P(X, Y | W) * P(W) / P(X, Y) Remembering that in general P(A, B) = P(A | B) * P(B), applying this to P(Y, X) and P(Y, X | W) (Treat the conditioning on W as a specific probability): P(X, Y | W) = P(Y, X | W) = P(Y | X, W) * P(X | W) P(X, Y) = P(Y, X) = P(Y | X) * P(X) Now the right hand side becomes: P(Y | X, W) * P(X | W) * P(W) / (P(Y | X) * P(X)) We're pretty much here! Use the Bayes theorem again on P(X | W) = P(W | X) * P(X) / P(W) Notice! The denominator had a P(X), the numerator had a P(W), so these simplify away, we're left with your statement P(Y | X, W) * P(W | X) / P(Y | X) The trick for proving stuff like this is noticing that it's suspiciously close to bayes theorem already, then refer to what you want to obtain in order to decide what to apply theorems on, things should work out and maybe there's even a simpler way to achieve this, but I used basic probability definitions
Prof. Gal is a genius. Thank you very much MLSS Moscow for bring his lectures to everyone.
Prof. Gal would have been a good doctor writing prescriptions.
I found it annoying that he walked back and forth prob 100 times during the lecture. Apart from that, it was an interesting speech!
P(W|X, Y) = P(Y|X,W,)P(W|X) / P(Y|X)
proof?
may be easy but not able to catch up
it's bayes rule
Bayes theorem: P(A | B) = P(B | A) * P(A) / P(B)
Let A = W, B = (X, Y)
P(W | X, Y) = P(X, Y | W) * P(W) / P(X, Y)
Remembering that in general P(A, B) = P(A | B) * P(B), applying this to P(Y, X) and P(Y, X | W) (Treat the conditioning on W as a specific probability):
P(X, Y | W) = P(Y, X | W) = P(Y | X, W) * P(X | W)
P(X, Y) = P(Y, X) = P(Y | X) * P(X)
Now the right hand side becomes:
P(Y | X, W) * P(X | W) * P(W) / (P(Y | X) * P(X))
We're pretty much here!
Use the Bayes theorem again on P(X | W) = P(W | X) * P(X) / P(W)
Notice! The denominator had a P(X), the numerator had a P(W), so these simplify away, we're left with your statement
P(Y | X, W) * P(W | X) / P(Y | X)
The trick for proving stuff like this is noticing that it's suspiciously close to bayes theorem already, then refer to what you want to obtain in order to decide what to apply theorems on, things should work out and maybe there's even a simpler way to achieve this, but I used basic probability definitions