16.322 Stochastic Estimation and Control Professor Vander Velde 80%chance of passing and a bad component has 30% chance of passing. On the average, we think that the process produces 60% good components, 40% bad ones. If a component passes the test, what is the probability that it is a Note that whether a certain component is good or bad is really not probabilistic-it is either good or bad. But we dont know what it is. Our knowledge based on all prior experience, intuition, or whatever is expressed in the prior probabilities P(good)=P(G)=A=0.6 P("bad")=P(B)=A2=04 P(P|G)=0.8 P(P|B)=03 P(GP) P(GP(PIG) 06(0.8)048 P(G)P(P|G)+P(B)P(P|B)0608)+04(0.3)06008 Suppose we have n-l observations P(A4|E1…E-1) Take one more observation, e P(AE1…En) P(4E-En)=P(E1、E) P(AE1…,En ∑P(AE1-En) P(ELEn-DP(A EEnP(En) >P(EE)P(A IEEn-DP(E, IA.. P(AKIELE-P(E IA) P(AIEEM-IP(E, IA) This is of the same form as the relation first written down, with P(AkIE-En-) in place of P(Ak). This says that Bayes'rule can be applied repetitively to account for any number of observations if the observations are conditional independent-conditioned on the alternatives Lecture16.322 Stochastic Estimation and Control Professor Vander Velde Lecture 2 80% chance of passing and a bad component has 30% chance of passing. On the average, we think that the process produces 60% good components, 40% bad ones. If a component passes the test, what is the probability that it is a good one? Note that whether a certain component is good or bad is really not probabilistic – it is either good or bad. But we don’t know what it is. Our knowledge based on all prior experience, intuition, or whatever is expressed in the prior probabilities. 1 2 ("good") ( ) 0.6 ("bad") ( ) 0.4 ( | ) 0.8 ( | ) 0.3 ( ) ( | ) 0.6(0.8) 0.48 ( | ) 0.8 ( ) ( | ) ( ) ( | ) 0.6(0.8) 0.4(0.3) 0.60 P PG A P PB A PPG PP B PGPP G PG P PGPP G PBPP B = == = == = = = = == + + Suppose we have n −1 observations 1 1 ( | ... ) PA E E k n− Take one more observation, En 1 1 1 1 1 1 1 1 11 1 1 11 11 1 1 1 1 ( ... ) ( | ... ) ( ... ) ( ... ) ( ... ) ( ... ) ( | ... ) ( | ... ) ( ... ) ( | ... ) ( | ... ) ( | ... ) ( | ) ( | ... ) ( | ) k n k n n k n i n i n k nn n n i n ni n i k n nk i n ni i P AE E PA E E PE E P AE E P AE E P E E P A E E P E AE E P E E P A E E P E AE E PA E E PE A PA E E PE A − − − − − − = = = = ∑ ∑ ∑ This is of the same form as the relation first written down, with ( ) 1 1 | ... PA E E k n− in place of ( ) P Ak . This says that Bayes’ rule can be applied repetitively to account for any number of observations if the observations are conditionally independent – conditioned on the alternatives