Thank you. q technically can be ANY probability distribution. However, only few actually can minimize the second term. If P(z|x) is analytical you let q(z)=P(z|x), otherwise, you would probably have to do some Variational Bayes (will do a video on this on a later date).
3:40 wait what? Which laws are you applying to get ln a - ln b - ln c + ln b = ln (a / b) - ln (c / b) ? Or probably rather ln a - ln b - ln c + ln b = ln (a) / b + ( - ln (b) / c) which I don't follow either. a is P(x, z) b is q(z) c is P(z | x) You can do of course ln a - ln b - ln c + ln b = ln (a / b) + ln (b / c), but your rewriting I don't understand.
by far the best and simplest explanation of EM
After watching this video finally got some clarity about EM Algorithm thank you very much.....
Thank you. q technically can be ANY probability distribution. However, only few actually can minimize the second term. If P(z|x) is analytical you let q(z)=P(z|x), otherwise, you would probably have to do some Variational Bayes (will do a video on this on a later date).
Is there more than one way to define KL? I have seen examples where p and q is the other way.
Excellent work!
Great video, and one of the shortest on youtube
best explanation i have seen. thanks!
Nice video. I am just wondring how we do select the function q.
thanks, this video is pretty straight and clear ,AND SOOOO HELPFUL ;D
3:40 wait what? Which laws are you applying to get ln a - ln b - ln c + ln b = ln (a / b) - ln (c / b) ? Or probably rather ln a - ln b - ln c + ln b = ln (a) / b + ( - ln (b) / c) which I don't follow either.
a is P(x, z)
b is q(z)
c is P(z | x)
You can do of course ln a - ln b - ln c + ln b = ln (a / b) + ln (b / c), but your rewriting I don't understand.
- ln c + ln b = -(ln c - ln b) = - ln(c/b)
ah ok, and then you also *do* mean that the entire fraction is in scope of the ln rather than only the numerator as it looks in your video?
yeah sorry. I really should have put in brackets
Ok, thanks a lot for clarifying.
Whoops you are absolutely correct, will fix it
Thank you!
good