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Math Help - Proving Bonferroni's Inequality

  1. #1
    Bar0n janvdl's Avatar
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    Proving Bonferroni's Inequality

    Bonferroni's Inequality states that:

    P(A \cap B) \geq P(A) + P(B) - 1

    But the first axiom of probability states that:

    P(\Omega ) = 1


    -------------------------


    Am I allowed to do the following?

    P(A \cap B) \geq P(A \cup B) - 1

    P(A \cap B) \geq P(A \cup B) - P(\Omega )

    P(\Omega ) \geq P(A \cup B) - P(A \cap B)

    And the last line must be true because the entire sample size must be at least equal to the union of A and B minus their intersection.


    OR

    Should i rather try to prove it this way:

    Rearrange so that the 1 is isolated...

    1 \geq P(A) + P(B) - P(A \cap B)

    1 \geq P(A \cup B)

    P(\Omega ) \geq P(A \cup B)

    Which is also true...


    Anyone got an idea which one of these is correct?
    Thanks in advance.


    EDIT: The second one might be the correct one seeing as A and B are not disjoint. Am i right?
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  2. #2
    Senior Member tukeywilliams's Avatar
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     \Omega is the sample space. Then  P(A\cup B) \leq P(\Omega) = 1 .

    And  P(A \cup B) = P(A) + P(B) - P(A \cap B) .

    Hence  P(A \cap B) \geq P(A) + P(B) -1 .
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by tukeywilliams View Post
     \Omega is the sample space. Then  P(A\cup B) \leq P(\Omega) = 1 .

    And  P(A \cup B) = P(A) + P(B) - P(A \cap B) .

    Hence  P(A \cap B) \geq P(A) + P(B) -1 .
    Thanks a lot TukeyWilliams +rep+
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