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Math Help - Bonferroni's Inequality Proof

  1. #1
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    Acolman, Mexico
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    Bonferroni's Inequality Proof

    Hello,

    I am confused with the following question.

    Suppose E_1, E_2,..., E_n are events, then

    \sum_{i=1}^{n} P(E_i) - \sum_{i<j}P(E_i E_j) \leq P(\cup_{i=1}^{n}E_i) \leq \sum_{i=1}^{n} P(E_i)

    The right hand side is Boole's inequality. Prove the left hand side.

    I am trying to prove this using induction, but I am not sure if I am doing it correctly.
    For n=2,
    Let E_1 and E_2 be events,
    then, \sum_{i=1}^{2} P(E_i) - \sum_{i<j}P(E_i E_j) = P(E_1) + P(E_2) - P(E_1 E_2) = P(E_1 \cup E_2).
    Thus, this holds true for n=2.

    Am I going on the right track?
    Any suggestion for the inductive step?

    Thanks in advance,
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  2. #2
    Member
    Joined
    Mar 2008
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    Acolman, Mexico
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    If I prove the Inclusion-Exclusion formula

    \sum_{i=1}^{n}P(E_i)-\sum_{i<j}P(E_i E_j) + \sum_{i<j<k}P(E_i E_j E_k) - ... +(-1)^{n-1}P(\cap_{i=1}^{n} E_i)= P(\cup_{i=1}^{n} E_i)

    how can I get the following inequality?

    \sum_{i=1}^{n} P(E_i) - \sum_{i<j}P(E_i E_j) \leq P(\cup_{i=1}^{n}E_i)
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  3. #3
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    have u solved the problem?

    Interested to know how to do this too
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