Ismor Fischer, 5/29/2012 3.3-1 3.3 Bayes’ Formula Suppose that, for a certain population of individuals, we are interested in comparing sleep disorders – in particular, the occurrence of event A = “Apnea” – between M = Males and F = Females. S = Adults under 50 M F A A ∩ M A ∩ F Also assume that we know the following information: P(M) = 0.4 P(A | M) = 0.8 (80% of males have apnea) prior probabilities P(F) = 0.6 P(A | F) = 0.3 (30% of females have apnea) Given here are the conditional probabilities of having apnea within each respective gender, but these are not necessarily the probabilities of interest. We actually wish to calculate the probability of each gender, given A. That is, the posterior probabilities P(M | A) and P(F | A). To do this, we first need to reconstruct P(A) itself from the given information. P(A | M) P(A ∩ M) = P(A | M) P(M) P(M) P(Ac | M) c c P(A ∩ M) = P(A | M) P(M) P(A) = P(A | M) P(M) + P(A | F) P(F) P(A | F) P(A ∩ F) = P(A | F) P(F) P(F) P(Ac | F) c c P(A ∩ F) = P(A | F) P(F) Ismor Fischer, 5/29/2012 3.3-2 So, given A… P(M ∩ A) P(A | M) P(M) P(M | A) = P(A) = P(A | M) P(M) + P(A | F) P(F) (0.8)(0.4) 0.32 = (0.8)(0.4) + (0.3)(0.6) = 0.50 = 0.64 and posterior P(F ∩ A) P(A | F) P(F) P(F | A) = = probabilities P(A) P(A | M) P(M) + P(A | F) P(F) (0.3)(0.6) 0.18 = (0.8)(0.4) + (0.3)(0.6) = 0.50 = 0.36 S Thus, the additional information that a M F randomly selected individual has apnea (an A event with probability 50% – why?) increases the likelihood of being male from a prior probability of 40% to a posterior probability 0.32 0.18 of 64%, and likewise, decreases the likelihood of being female from a prior probability of 60% to a posterior probability of 36%. That is, knowledge of event A can alter a prior 0.08 0.42 probability P(B) to a posterior probability P(B | A), of some other event B. Exercise: Calculate and interpret the posterior probabilities P(M | Ac) and P(F | Ac) as above, using the prior probabilities (and conditional probabilities) given. More formally, consider any event A, and two complementary events B1 and B2, (e.g., M and F) in a sample space S. How do we express the posterior probabilities P(B1 | A) and P(B2 | A) in terms of the conditional probabilities P(A | B1) and P(A | B2), and the prior probabilities P(B1) and P(B2)? Bayes’ Formula for posterior probabilities P(Bi | A) in terms of prior probabilities P(Bi), i = 1, 2 PB()i ∩ A P( A|Bii )() P B P(Bi | A) = = PA() PA|B(11 )( PB ) +PA|B ( 22 )( PB ) Ismor Fischer, 5/29/2012 3.3-3 In general, consider an event A, and events B1, B2, …, Bn, disjoint and exhaustive. S B1 B2 . Bn A A ∩ B1 A ∩ B2 . A ∩ Bn Prior Probabilities P(A | B1) P(B1) P(A ∩ B1) c P(A | B1) c P(A ∩ B1) Law of Total Probability P(A | B2) P(B2) P(A ∩ B2) c P(A ∩ Bj) P(A | B2) c P(A ∩ B2) n P(A | B3) P(B3) P(A ∩ B3) P(A) = ∑ P(A | Bj) P(Bj) c j=1 P(A | B3) c P(A ∩ B3) . P(A | Bn) P(Bn) P(A ∩ Bn) c P(A | Bn) c P(A ∩ Bn) Bayes’ Formula (general version) For i = 1, 2, …, n, the posterior probabilities are… PB()i ∩ A P( A|Bii )() P B P(Bi | A) = = n . PA() ∑ P( A|Bjj )( P B ) j=1 Reverend Thomas Bayes 1702 - 1761 .