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# 531f10BAYES1 - STAT 531 Bayesian Methods HM Kim Department...

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Unformatted text preview: STAT 531: Bayesian Methods HM Kim Department of Mathematics and Statistics University of Calgary Fall 2010 1/47 Bayes’s theorem Bayess theorem relates to the problem of adjudicating between competing hypotheses given observations. Suppose D is an event, i.e. something that either happens or that doesn’t. Suppose C 1 , C 2 , ··· , C J are other events that form a partition. C 1 , C 2 , ··· , C J can be thought of as competing hypotheses to explain the event observed, D . Fall 2010 2/47 In that case, the conditional probability of C j given D is p ( C j | D ) = p ( D | C j ) p ( C j ) p ( D ) To calculate p ( D ), we may need a further result, the Law of Total Probability. , → Law of Total Probability : The overall, or marginal probability of the event, D , can be expressed in terms of the probabilities of C j and the conditional probabilities of D given each of the C j s, as follows: p ( D ) = J ∑ j =1 p ( D | C j ) p ( C j ) . Fall 2010 3/47 , → Example : An item is produced in 3 different factories, C 1 , C 2 , C 3 . The proportions produced in the 3 factories, and the proportions defective in each, are as follows: factory % produced % defective C 1 50 2 C 2 30 3 C 3 20 4 An item is purchased and found to be defective. This is event D . What is the probability that it was from factory C 1 ? Fall 2010 4/47 First, we find the overall probability of a defective, p ( D ), from the Law of Total Probability: p ( D ) = p ( D | C 1 ) p ( C 1 )+ p ( D | C 2 ) p ( C 2 )+ p ( D | C 3 ) p ( C 3 ) = 0 . 02 × . 5+0 . 03 × . 3+0 . 02 × . 2 = 0 . 027 Then, Bayess theorem tells us the probability that the item was from factory C 1 : p ( C 1 | D ) = p ( D | C 1 ) p ( C 1 ) p ( D ) = . 02 × . 5 . 027 = 0 . 37 Fall 2010 5/47 , → Another version : p ( C j | D ) = p ( D | C j ) p ( C j ) p ( D ) = ⇒ p ( C j | D ) ∝ p ( D | C j ) p ( C j ) To implement this, we calculate p ( D | C j ) p ( C j ) for each j , add them up, and then divide by the sum so that they for each add up to 1 (which they have to, because theyre probability of a partition). p ( C 1 | D ) ∝ p ( D | C 1 ) p ( C 1 ) = 0 . 02 × . 5 = 0 . 010 p ( C 2 | D ) ∝ p ( D | C 2 ) p ( C 2 ) = 0 . 03 × . 3 = 0 . 009 p ( C 3 | D ) ∝ p ( D | C 3 ) p ( C 3 ) = 0 . 04 × . 2 = 0 . 008 Then p ( C 1 | D ) = p ( D | C 1 ) p ( C 1 ) . 010+0 . 009+0 . 008 = . 010 . 027 = 0 . 37 Fall 2010 6/47 Another way of looking at this is that C 1 , C 2 and C 3 are the possible states of nature (parameters) D is the data (datum) We then use the data to decide how likely the different states of nature are relative to one another. This is the idea that underlies Bayesian statistics. , → Likelihood, prior and posterior probabilities : p ( D | C j ) is the probability of the data given the state of nature C j ....
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531f10BAYES1 - STAT 531 Bayesian Methods HM Kim Department...

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