Conditional Probability We always use all available information when - - PowerPoint PPT Presentation

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.

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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

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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

• ur estimate for AAPL?

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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).

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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).

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ST 380 Probability and Statistics for the Physical Sciences

Bayes’ Theorem

Law of Total Probability Suppose that A1, A2, . . . , Ak are: mutually exclusive (Ai ∩ Aj = ∅, i = j) and exhaustive (A1 ∪ A2 ∪ · · · ∪ Ak = S). Then for any event B, P(B) = P(B|A1)P(A1) + P(B|A2)P(A2) + · · · + P(B|Ak)P(Ak) =

k

• i=1

P(B|Ai)P(Ai)

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ST 380 Probability and Statistics for the Physical Sciences

Bayes’ Theorem P(Aj|B) = P(Aj ∩ B) P(B) = P(B|Aj)P(Aj) k

i=1 P(B|Ai)P(Ai)

= P(Aj) × P(B|Aj) k

i=1 P(B|Ai)P(Ai)

P(Aj) is called the prior probability of Aj (that is, prior to knowing whether or not B happened); P(Aj|B) is called the posterior probability of Aj (that is, conditional

• n knowing that B occurred).

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ST 380 Probability and Statistics for the Physical Sciences

Bayes’ Theorem is useful when we know P(Ai) and P(B|Ai), 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. A1 = individual has the disease, A2 = A′

1

B = test is positive Specification P(A1) = 0.001, P(A2) = 1 − P(A1) = 0.999 P(B|A1) = 0.99, P(B|A2) = 0.02

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ST 380 Probability and Statistics for the Physical Sciences

Terminology P(B|A1) is the sensitivity of the test, the probability of correctly identifying someone with the disease, here 0.99. P(B′|A2) 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(A1|B) = 0.047. Even though the test is quite accurate, a positive result is far from conclusive that the individual suffers from the disease.

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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.

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ST 380 Probability and Statistics for the Physical Sciences

Mutual Independence If more than two events A1, A2, . . . , Ak satisfy P(A1 ∩ A2 ∩ · · · ∩ Ak) = P(A1)P(A2) . . . P(Ak), 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

• f 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.

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