55 sometimes we are given conditional probabilities

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Sometimes we are given conditional probabilities and we need to compute the probability that several events occur simultaneously. Using the definition of conditional probability we can always write P A B P A | B P B so that, given P B and the conditional probability P A | B , we can find P A B . 56
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We can extend this to many events by applying this representation repeatedly: P A k A k 1 A 1 P A k | A k 1 A 1 P A k 1 A 1 P A k | A k 1 A 1 P A k 1 | A k 2 A 1 P A k 2 A 1 and so on, to get P A k A k 1 A 1 P A k | A k 1 A 1 P A k 1 | A k 2 A 1 P A 2 | A 1 P A 1 57
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By simply reversing the order of events we can also write the relationship – which is useful for sequential experiments – as P A 1 A 2 A k P A 1 P A 2 | A 1  P A k 1 | A k 2 A 1 P A k | A k 1 A 1 58
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EXAMPLE : An urn contains 4 balls: 3 white and 1 red. We draw the balls from the urn (without replacing balls) until the red ball is drawn. What is the probability of getting the red ball on the third draw? To get the red ball on the third draw we must draw two white balls followed by a red ball. As a shorthand, we want P white , white , red . 59
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Now P white , white , red P white P white | white P red | white , white 3 4 2 3 1 2 1 4 Notice that, at each draw, the probability of drawing a particular color ball is simply the number of balls of that color over the total number of balls in the urn. 60
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As a special case of the general result, suppose A j : j 1,. .., k is monotonically decreasing, that is, A k A k 1 ... A 2 A 1 .(So A k is the least likely event and A 1 is the most likely. If A j occurs then we know A j 1 had to occur.) Then P A k P A k | A k 1 P A k 1 | A k 2  P A 2 | A 1 P A 1 This simple expression has lots of applications in econometrics, including to so-called duration models. 61
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5 . 1 Bayes Rule For two events A and B , each with positive probability, start with the definition of conditional probability, twice: P A | B P A B P B P B | A P B A P A P A B P A From the second equation, P A B P B | A  P A ; plugging into the expression for P A | B , we get P A | B P B | A P A P B 62
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Recall that B B A B A c and the two sets in the union are disjoint. Therefore, P B P B A P B A c P B | A  P A P B | A c P A c This formula itself is useful. Rewriting it slightly gives
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55 Sometimes we are given conditional probabilities and we...

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