ST 380 Probability and Statistics for the Physical Sciences Conditional Probability We always use all available information when we assess the probability of some event. For example, consider this experiment: choose a day at random from the year 2012, and find out the percentage change for that day in Apple’s stock price (AAPL) and in the S&P 500 index (GSPC). Let A be the event that AAPL rises, and B be the event that GSPC rises. 1 / 11 Probability Conditional Probability
ST 380 Probability and Statistics for the Physical Sciences Table : Numbers of days in 2012 GSPC falls ( B ′ ) GSPC rises ( B ) Total AAPL falls ( A ′ ) 75 45 120 AAPL rises ( A ) 43 86 129 Total 118 131 249 2 / 11 Probability Conditional Probability
ST 380 Probability and Statistics for the Physical Sciences AAPL rose on 129 out of 249 days, so P ( A ) = 129 / 249 = 0 . 52. If we feel that current market conditions are similar to those of 2012, we would use that as an estimate of the probability that AAPL rose today. Now suppose that we see that GSPC rose today. Does that change our estimate for AAPL? 3 / 11 Probability Conditional Probability
ST 380 Probability and Statistics for the Physical Sciences Of the 131 days on which GSPC rose, AAPL also rose on 86 days. So given that GSPC rose, the probability that AAPL rose is 86 / 131 = 0 . 66. That is, the information that GSPC rises on a given day changes the probability of a rise in AAPL from 0 . 52 to 0 . 66. We call this the conditional probability of A given that B occurred, and write it as P ( A | B ). 4 / 11 Probability Conditional Probability
ST 380 Probability and Statistics for the Physical Sciences Note that P ( A ∩ B ) = 86 249 and P ( B ) = 131 249 , so P ( A | B ) = 86 131 = P ( A ∩ B ) . P ( B ) This is the general definition of conditional probability: P ( A | B ) = P ( A ∩ B ) . P ( B ) Note the useful consequence: P ( A ∩ B ) = P ( A | B ) × P ( B ) . 5 / 11 Probability Conditional Probability
ST 380 Probability and Statistics for the Physical Sciences Bayes’ Theorem Law of Total Probability Suppose that A 1 , A 2 , . . . , A k are: mutually exclusive ( A i ∩ A j = ∅ , i � = j ) and exhaustive ( A 1 ∪ A 2 ∪ · · · ∪ A k = S ). Then for any event B , P ( B ) = P ( B | A 1 ) P ( A 1 ) + P ( B | A 2 ) P ( A 2 ) + · · · + P ( B | A k ) P ( A k ) k � = P ( B | A i ) P ( A i ) i =1 6 / 11 Probability Conditional Probability
ST 380 Probability and Statistics for the Physical Sciences Bayes’ Theorem P ( A j | B ) = P ( A j ∩ B ) P ( B ) P ( B | A j ) P ( A j ) = � k i =1 P ( B | A i ) P ( A i ) P ( B | A j ) = P ( A j ) × � k i =1 P ( B | A i ) P ( A i ) P ( A j ) is called the prior probability of A j (that is, prior to knowing whether or not B happened); P ( A j | B ) is called the posterior probability of A j (that is, conditional on knowing that B occurred). 7 / 11 Probability Conditional Probability
ST 380 Probability and Statistics for the Physical Sciences Bayes’ Theorem is useful when we know P ( A i ) and P ( B | A i ), i = 1 , 2 , . . . , k . Example 2.31: Incidence of a Rare Disease One in 1,000 adults suffers from a rare disease, for which a less-than-perfect test is available. A 1 = individual has the disease , A 2 = A ′ 1 B = test is positive Specification P ( A 1 ) = 0 . 001 , P ( A 2 ) = 1 − P ( A 1 ) = 0 . 999 P ( B | A 1 ) = 0 . 99 , P ( B | A 2 ) = 0 . 02 8 / 11 Probability Conditional Probability
ST 380 Probability and Statistics for the Physical Sciences Terminology P ( B | A 1 ) is the sensitivity of the test, the probability of correctly identifying someone with the disease, here 0 . 99. P ( B ′ | A 2 ) is the specificity of the test, the probability of correctly identifying someone who does not have the disease, here 1 − 0 . 02 = 0 . 98. Calculation In this case, Bayes’ Theorem gives P ( A 1 | B ) = 0 . 047. Even though the test is quite accurate, a positive result is far from conclusive that the individual suffers from the disease. 9 / 11 Probability Conditional Probability
ST 380 Probability and Statistics for the Physical Sciences Independence Suppose that information about whether B has occurred does not change the probability of A . That is, P ( A | B ) = P ( A ). Then we say that A is independent of B . So P ( A ∩ B ) = P ( A | B ) P ( B ) = P ( A ) P ( B ) . This is symmetric in A and B , so B is also independent of A , and we say simply that A and B are (probabilistically, or statistically) independent. 10 / 11 Probability Independence
ST 380 Probability and Statistics for the Physical Sciences Mutual Independence If more than two events A 1 , A 2 , . . . , A k satisfy P ( A 1 ∩ A 2 ∩ · · · ∩ A k ) = P ( A 1 ) P ( A 2 ) . . . P ( A k ) , they are mutually independent . In the birthday example, because we assumed all outcomes were equally likely, we can show that each person’s birthday is independent of the others. We can often argue that two or more events are defined by physically independent processes, not affected by each other. In this case, we assume that the events are also statistically independent of each other. 11 / 11 Probability Independence
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