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Math Help - Prove the probability inequality

  1. #1
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    Prove the probability inequality

    If P[A|B] > P[A|B complement], then show that P[A complement] < P[A complement|B complement].

    What I've got so far:

    0) P[A intersection B]/P[B] > P[A intersection B complement]/P[B complement]

    1) P[A intersection B]*P[B complement] > P[B]*P[A intersection B complement]

    2) P[A intersection B] - P[A intersection B]*P[B] > P[B]*P[A] - P[B]*P[A intersection B]

    3) P[A intersection B] > P[B]*P[A]

    4) P[A] < P[A intersection B]/P[B]

    5) P[A] < P[A|B]

    6) 1 - P[A complement] < P[A|B]

    7) P[A complement] > P[A complement|B]

    I'm stuck at this part. Any help would be greatly appreciated.
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  2. #2
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    What you wrote is correct; after a few lines, you could get the result (replacing P(B) by 1-P(B complement) in the denominator, etc.). However this is not a very direct way to procede... What do you think of the following?:

    P(A|B)>P(A|B^c) implies P(A^c|B)<P(A^c|B^c) (taking complement of boths sides), hence P(A^c)=P(A^c|B)P(B)+P(A^c|B^c)P(B^c)\leq P(A^c|B^c)P(B)+P(A^c|B^c)P(B^c) =P(A^c|B^c). qed.
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  3. #3
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    Quote Originally Posted by Laurent View Post
    hence P(A^c)=P(A^c|B)P(B)+P(A^c|B^c)P(B^c)\leq P(A^c|B^c)P(B)+P(A^c|B^c)P(B^c) =P(A^c|B^c). qed.
    Could you show how you arrived at this? I can't figure it out.
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  4. #4
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    Quote Originally Posted by My Little Pony View Post
    Could you show how you arrived at this? I can't figure it out.
    First equality comes from P(A)=P(A\cap B)+P(A\cap B^c) and definition of conditional expectation. This is sometimes called "law of total expectation" or something alike.
    The inequality is what I proved right before: P(A^c|B)\leq P(A^c|B^c).
    The last equality is P(B)+P(B^c)=1.
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