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Independent Probability

  1. Oct 22, 2009 #1

    x^2

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    Independent Probability....

    Hello,

    Could someone point me in the right direction of how to prove that P(A|B) = P(A) if and only if P(B|A) = P(B)? I think I understand the program and I can't formulate any contradictions, but I'm having difficulty showing this property with a formal proof.

    Thanks,
    x^2
     
  2. jcsd
  3. Oct 23, 2009 #2

    lanedance

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    Re: Independent Probability....

    can you start with the definition

    [tex] P(A|B) = \frac{P(A \cap B)}{P(B)} [/tex]
     
  4. Oct 23, 2009 #3

    x^2

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    Re: Independent Probability....

    [tex]P(A|B) = \frac{P(A \cup B)}{P(B)} = P(A)[/tex]

    [tex]P(B|A) = \frac{P(A \cup B)}{P(A)} = P(B)[/tex]

    [tex]P(B|A) = \frac{P(A \cup B)}{P(B)} = P(A) [/tex]

    [tex]P(A|B) = \frac{P(A \cup B)}{P(A)} = \frac{P(A \cup B)}{P(A)} = P(B|A)[/tex]

    I think that is right..... Thank you for the hint!
    x^2
     
  5. Oct 23, 2009 #4

    lanedance

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    Re: Independent Probability....

    i think you mean intersections, not unions

    a better way to show it would be to start with
    [tex]P(A|B) = P(A) [/tex]

    and use the definition and simplify to show P(B|A) = P(B)
     
    Last edited: Oct 23, 2009
  6. Oct 23, 2009 #5

    x^2

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    Re: Independent Probability....

    Yes, sorry, you are correct; they should be intersections and not unions.

    Thank you for the help!
    x^2
     
  7. Oct 23, 2009 #6

    honestrosewater

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    Re: Independent Probability....

    What is going on in this proof? It looks like you are assuming P(A|B) = P(A) and P(B|A) = P(B) = P(A), and then you don't prove what you set out to prove.

    What does P(A|B) = P(A) mean? It means that B's occurring has no effect on the probability of A occurring, i.e., A is independent of B, yes? You need to show that A being independent of B also makes B independent of A. So there is a hint: Do you have a rule that has a different version for independent events?

    To prove an equivalence, you can prove the implication both ways: Assume P(A|B) = P(A) and use this to derive P(B|A) = P(B). Then assume P(B|A) = P(B) and use this to derive P(A|B) = P(A).

    So the first line in your proof should be

    1] P(A|B) = P(A)​

    What can you say given also

    1] P(A|B) = P(A)
    2] P(A ∩ B) = P(B) * P(A|B)​

    Remember that you are trying to get to P(B|A) = P(B). So, as a general rule, P(B|A) = ??

    Note that A and B are arbitrary events, so the proof in the other direction will be the same.
     
  8. Oct 23, 2009 #7

    lanedance

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    Re: Independent Probability....

    sorry yeah corected post
     
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