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  1. #1
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    Probability question

    Show that for any n events A_1 ,A_2 ,...,A_n in a finite probability space, we have P(A_1 \cup A_2 \cup \cdot \cdot \cdot \cup A_n) \leq P(A_1) + ...+P(A_n)
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  2. #2
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by qwesl View Post
    Show that for any n events A_1 ,A_2 ,...,A_n in a finite probability space, we have P(A_1 \cup A_2 \cup \cdot \cdot \cdot \cup A_n) \leq P(A_1) + ...+P(A_n)
    Have you ever heard of Boole's inequality?

    http://www.andrew.cmu.edu/course/21-228/lec7.pdf
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  3. #3
    MHF Contributor Drexel28's Avatar
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    I really know nothing about probability and so this is more of a question then an answer. Doesn't this follow since the probability measure is by definition countably additive?
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  4. #4
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    Yes but how is this done using only the following axioms (first principles)
    1) P(\emptyset)=0
    2) P(\Omega)=1
    3) P(A_1 \cup \cdot \cdot \cdot \cup A_n)=P(A_1) + ...+ P(A_n) for n disjoint sets A_1 , ...,A_n \subseteq \Omega
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  5. #5
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by qwesl View Post
    Yes but how is this done using only the following axioms (first principles)
    1) P(\emptyset)=0
    2) P(\Omega)=1
    3) P(A_1 \cup \cdot \cdot \cdot \cup A_n)=P(A_1) + ...+ P(A_n) for n disjoint sets A_1 , ...,A_n \subseteq \Omega
    Use google, proofs this common are usually all over the internet.

    http://www.math.ntu.edu.tw/~hchen/te...s/lecture2.pdf

    If you don't like that one there are several others...


    Quote Originally Posted by Drexel28 View Post
    I really know nothing about probability and so this is more of a question then an answer. Doesn't this follow since the probability measure is by definition countably additive?
    Yes, I've seen a proof using this argument, wikipedias in fact.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Anonymous1 View Post


    Yes, I've seen a proof using this argument, wikipedias in fact.
    Oops, probably should have wikid it first! God bless Wikipedia!
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