Inferring the Hidden from the observable (Bayes' rule) P(H=h1, h2,,hn,0=01, 02,,On) Conditional prob P(AB)=P(A, B)/P(B P(H=h1,.,hn, O=O1 PO=01,,On) P(H=h,…hn)(O=0n,…On|H=hhn) P(O=01,On) P(O=O1,. ,On) somewhat difficult to calculate But notice P(H=hn,…,hn2O )>P(H=h1,…,hn,O mpliesP(H=h,…,hn|O=0,,O)>P(H=h1,hn|0=01,,On) so can treat P(0=O1, ,0n)as a constantInferring the Hidden from the Observable (Bayes’ Rule) P ( H = h 1, h 2 ,..., hn | O = o 1, o 2 ,..., on ) Conditional Prob: P(A|B) = P(A,B)/P(B) P ( H = h 1,..., hn , O = o 1, ..., on ) = P (O = o 1, ..., on ) P ( H = h 1,..., hn )P (O = o 1 ,..., on | H = h 1,..., hn ) = P (O = o 1, ..., on ) P (O = o 1 ,..., on ) somewhat difficult to calculate But notice: P ( H = h 1, ..., hn , O = o 1,..., on ) > P ( H = h ′1, ..., h ′n , O = o 1, ..., on ) implies P (H = h 1, ..., hn | O = o 1,..., on ) > P (H = h ′1 ,..., h ′n | O = o 1, ..., on ) so can treat P (O = o 1 ,..., on ) as a constant