Actuarial Outpost If A,B,C are independent, then AnB (resp., AuB) and C are independent
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# If A,B,C are independent, then AnB (resp., AuB) and C are independent

Posted 02-02-2009 at 01:12 AM by Marid Audran

(Using n for intersection and u for union, since the built-in TeX is terrible with those symbols.)

Oddly enough, Rosenthal's probability book doesn't seem to even point this out, and my other probability book (Hogg/Craig, from several years back) doesn't see fit to prove it or make a big deal of it. I think it's a big deal. If I've got it right, then I think the proof helps explain why "independence" is defined as a distinct phenomenon from "pairwise independence."* If you only have {A,B},{B,C}, and {A,C} independent, then {AuB,C} might not be independent.

Suppose A, B and C are independent events. Then
P((A n B) n C) = P(A n B n C) = P(A)P(B)P(C) = P(A n B) P(C),
so that AnB and C are independent. Also, applying some basic set theory, we have

P((A u B) n C) = P((A n C) u (B n C))
= P(A n C) + P(B n C) - P(A n B n C)
= P(A)P(C) + P(B)P(C) - P(A)P(B)P(C)
= (P(A)+P(B)-P(A n B))P(C) = P(A u B)P(C),
so AuB and C are independent also.

* In other words: In order for more than two events to be independent, it's not enough that any two of them be independent in the P(AnB)=P(A)P(B) sense. Instead, the probability of the intersection must equal the product of the probabilities for any finite subcollection of events, e.g.: P(AnBnC) = P(A)P(B)P(C).
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