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Decompostion of conditional probabilities

  1. Mar 24, 2012 #1
    So If B=B1UB2U.....UBn would Pr(A|B)=Pr(A|B1)Pr(B1)+Pr(A|B2)Pr(B2)+.....+Pr(A|Bn)Pr(Bn)?

    I haven't found a formula for this but it makes intuitive sense.
     
  2. jcsd
  3. Mar 24, 2012 #2

    chiro

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    Hey torquerotates.

    Before I give my take on the answer, do you assume that all the B1, B2, etc are disjoint? In other words is Bi AND Bj = NullSet for i != j?
     
  4. Mar 24, 2012 #3
    Yes they are disjoint.
     
  5. Mar 24, 2012 #4

    chiro

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    For two events I get the following:

    P(A|B1UB2) = P(A and (B1UB2))/P(B1UB2) = P((A and B1) U (A and B2))/P(B1UB2) = P(A and B1)/P(B1UB2) + P(A and B2)/P(B1UB2) - P(A and B1 and B2)/P(B1UB2)

    = P(A and B1)/(P(B1) + P(B2)) + P(A and B1)/(P(B1) + P(B2)) - 0

    = 1/P(B)[P(A and B1) + P(A and B2)]

    = 1/P(B)[P(A and B1]P(B1)/P(B1) + P(A and B2)P(B2)/P(B2)]

    = 1/P(B)[P(A|B1)P(B1) + P(A|B2)P(B2)] if we assume P(Bi) > 0 and is a valid probability.

    If the above is right then your answer is wrong in general but right when B is the universal probability space.

    Because your probability P(B) may not be one, it means that the above needs to be corrected using this information. Just for an example consider P(B1) = P(B2) = 1/4 and P(A|B1) = P(A|B2). But since 1/4[P(A|B1) + P(A|B2)] = 1/2P(A|B) for this example since P(B1) = P(B2) and since B1 and B2 are disjoint we have:

    Then under these conditions we have:

    1/P(B) x [P(A|B1)P(B1) + P(A|B2)P(B2)]
    = 1/(1/2) x [P(A|B1)x1/4 + P(A|B1)x1/4]
    =(1/4)/(1/2) x [P(A|B1) + P(A|B2)]
    = 1/2 x [(1/2)/(1/4)]xP(A|B)
    = 1/2 x 2 x P(A|B)
    = P(A|B) for this example.

    If we did not 'normalize' by P(B) we would not have gotten the right answer.

    While I think the above is right, I would like others to check if they could just to make sure, but I'm only using a few identities which are basically distribution over sets and multiplication by 'x/x' terms for P(Bi)/P(Bi).
     
  6. Mar 25, 2012 #5

    Stephen Tashi

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    Did you do any examples?

    Let
    A = the roll of a fair die produces an even number
    B = the face that comes up is a 1 or a 2
    B1 = the face that comes up is a 1
    B2 = the face that comes up is a 2
     
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