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Proof of a probability inequality
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[QUOTE="Jyan, post: 4671054, member: 471224"] Ah I See. I actually did leave out a few steps in my proof just because writing out in latex is tedious and proofs I read in books seem like they leave out the same kind of stuff I did, but I guess I should be as explicit as possible. As for applying the hint, this is what I have: [tex] B_1 = A_1 [/tex] Through induction, define B_i for all i in N [tex] B_i = A_i \setminus (\bigcup_{i=1}^{i-1}A_i) [/tex] Since B_i is A_i with a set difference, each B_i is a subset (not a strict one though) of A_i, which means that P(B_i) <= P(A_i). Moreover, the union of all the Bs is equal to the union of all the As since repeated elements are not included in the union. (is there a better way to express this?) Moreoever, each B_i removes all common elements with B_(i-1), B_(i-2), ... Therefore, [tex] \bigcap_{i=1}^{\infty}B_i = \emptyset [/tex] So, Because the union of the Bs is equal to the union of the As (first equality), the Bs are disjoint (second equality), and each B_i is a subset of A_i (the inequality): [tex] P(\bigcup_{i=1}^{\infty} A_i) = P(\bigcup_{i=1}^\infty B_i) = \sum_{i=1}^{\infty} P(B_i) \le \sum_{i=1}^{\infty}P(A_i) [/tex] I appreciate the help/comments from everyone, thank you! [/QUOTE]
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Proof of a probability inequality
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