- #36
wle
- 336
- 157
stevendaryl said:
- (Factorizability): Assume that ##P(A,B|a,b,\lambda) = P(A|a,b,\lambda) P(B|a,b,\lambda)##.
The derivations I've seen use Bayes' theorem here, which implies $$P(A, B| a, b, \lambda) = P(A | a, b, \lambda) P(B | A, a, b, \lambda) \,.$$ The locality assumption used after this translates to ##P(A | a, b, \lambda) = P(A | a, \lambda)## and ##P(B | A, a, b, \lambda) = P(B | b, \lambda)##.