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Lecture4

# Lecture4 - 1 INDEPENDENCE Lecture 4 ORIE3500/5500...

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Unformatted text preview: 1 INDEPENDENCE Lecture 4 ORIE3500/5500 Summer2009 Chen Independence and Conditional Independence 1 Independence Definition. Two events A and B are said to be independent if P ( A ∩ B ) = P ( A ) P ( B ) . This can be written in terms of conditional probability as well. Note that the definition is equivalent to saying P ( A | B ) = P ( A ) when P ( B ) > 0. If P ( B ) = 0 then A and B are always independent. Example Does there exist a set A such that A is independent of itself? We need P ( A ∩ A ) = P ( A ) 2 which means P ( A )(1- P ( A )) = 0. Therefore, we only need P ( A ) = 0 or 1. For example, A = φ or A = Ω. It would be observed that the concepts of disjoint sets and independence are different. Independence is defined with the probability of that set. Three events A,B and C are said to be independent if P ( A ∩ B ) = P ( A ) P ( B ) P ( A ∩ C ) = P ( A ) P ( C ) P ( B ∩ C ) = P ( B ) P ( C ) and P ( A ∩ B ∩ C ) = P ( A ) P ( B ) P ( C ) . If only the first 3 equations holds then the sets are said to be pairwise inde- pendent. Notice, P ( A ∩ B ∩ C ) = P ( A ) P ( B ) P ( C ) is not enough for independence. Example Consider two independent rolls of a fair six-sided die, and the following events: A = { 1st roll is1 , 2 , or 3 } , 1 1 INDEPENDENCE B = { 1st roll is3 , 4 , or 5 } , C = { the sum of the two rolls is 9 } . We have P ( A ∩ B ) = 1 6 6 = 1 2 · 1 2 = P ( A ) P ( B ) , P ( A ∩ C ) = 1 36 6 = 1 2 · 4 36 = P ( A ) P ( C ) , P ( B ∩ C ) = 1 12 6 = 1 2 · 4 36 = P ( B ) P ( C ) . However, P ( A ∩ B ∩ C ) = 1 36 = 1 2 · 1 2 · 4 36 = P ( A ) P ( B ) P ( C ) . A collection of N events A 1 ,A 2 ,...,A n are said to be independent if any selection of r (1 < r ≤ n ) events A k 1 ,A k 2 ,...,A k r (1 ≤ k 1 < k 2 < ··· < k r ≤ n ) from the collection of n events, satisfies P ( A k 1...
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Lecture4 - 1 INDEPENDENCE Lecture 4 ORIE3500/5500...

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