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

Posted 02-02-2009 at 02:12 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.

Suppose A, B and C are independent events. Then

P((A u B) n C) = P((A n C) u (B n C))

* 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

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)so AuB and C are independent also.

= 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),

* 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).Total Comments 0