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Unformatted text preview: Chapter 3 Independence Suppose that it is known that event A has occurred. We update the probability of the event B from P(B) to P(B  A) = P(BA) P(A) P(B  A) can be larger, smaller, or the same as P(B) If P(B  A) = P(B), then the occurrence of A does not change our estimate of the probability of B P(B  A) = P(B) P(B c  A) = P(B c ) The occurrence of A does not change the odds favoring B P(B):P(B c ) = P(B  A):P(B c  A) Equivalently, the occurrence of B does not change the odds favoring A In such cases, A and B are said to be independent events The formal definition of independence uses A and B in more symmetric fashion If P(B  A) = P(AB)/P(A) = P(B), then P(AB) = P(A)P(B) Definition: Events A and B are said to be (stochastically) independent if P(AB) = P(A)P(B) Do not confuse the notions of independent events and mutually exclusive events P(AB) = P(A)P(B) 0 for independent events, so they cannot be mutually exclusive P(AB) = 0 P(A)P(B) for mutually exclusive events, so they cannot be independent Magic mantra to memorize Independent events cannot be mutually exclusive Mutually exclusive events cannot be independent Note: There are trivial uninteresting exceptions when either P(A) or P(B) equals 0 If A and B are independent events, then so are A and B c , A c and B, and A c and B c P(AB c ) = P(A) P(AB) = P(A) P(A)P(B) = P(A)[(1 P(B)] = P(A)P(B c ) The other two results can be proved similarly (Try them!) Independence of events is of great help in calculations; P(AB) is just P(A)P(B) for independent events A and B For example, P(A B) = P(A) + P(B) P(AB) = P(A) + P(B) P(A)P(B) P(A B c ) = P(A)+P(B c ) P(AB c ) = P(A) + P(B c ) P(A)P(B c ) Even though independence is a great help in calculations, it cannot and should not be used indiscriminately (e.g., whenever you are stuck in solving a problem and cant figure out a way to proceed) DO NOT ASSUME that events are independent unless the problem explicitly says so On homework and exams, if you are asked to determine whether or not A and B are independent events, just find P(AB) and compare it to the product P(A)P(B) But this is not the way the concept of independence is used in probability theory Physical independence versus stochastic independence It might be reasonable to assume on the basis of study and preliminary analysis of the physical phenomenon under consideration that two events are physically independent Physical independence is an assumption that we justify based purely on physical considerations We conclude from physical principles that occurrence (or nonoccurrence) of one event seems to have no influence on the occurrence of another event Example: two successive tosses of a fair coin. Does the occurrence of a head on the first toss influence the result of the second toss?...
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This test prep was uploaded on 09/28/2007 for the course BTRY 4080 taught by Professor Schwager during the Fall '06 term at Cornell University (Engineering School).
 Fall '06
 SCHWAGER

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