In the discrete case this is easy the marginal

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Unformatted text preview: omewhat surprisingly, A and B4 are independent. To check this from (A.6), we note that P (B4 ) = 6/36 and A \ B4 = {(4, 3)} has probability 1/36, so P (A \ B3 ) = 16 1 =· = P (A)P (B3 ) 36 6 36 There are two ways of extending the definition of independence to more than two events. A1 , . . . , An are said to be pairwise independent if for each i 6= j , P (Ai \Aj ) = P (Ai )P (Aj ), that is, each pair is independent. A1 , . . . , An are said to be independent if for any 1 i1 < i2 < . . . < ik n we have P (Ai1 \ . . . \ Aik ) = P (Ai1 ) · · · P (Aik ) If we flip n coins and let Ai = “the ith coin shows Heads,” then the Ai are independent since P (Ai ) = 1/2 and for any choice of indices 1 i1 < i2 < . . . < ik n we have P (Ai1 \ . . . \ Aik ) = 1/2k . Our next example shows that events can be pairwise independent but not independent. Example A.7. Flip three coins. Let A = “the first and second coins are the same,” B = “the second and third coins are the same,” and C = “the third and first coins are the same.” Clearly...
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