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Baye's Theorem w/ Multiple Events

  1. May 17, 2010 #1
    So, Bayes' Theorem states

    P(H|D) = P(D|H) X P(H) / [P(D|H) X P(H) + P(D|~H) x P(~H)], where ~ = negation

    If my hypothesis consists of two events - S and O - would this be

    P(S,O|D) = P(D|S,O) X P(S,O) / [P(D|S,O) X P(S,O) + P(D|~S,~O) x P(~S,~O)]

    or this

    P(S,O|D) = P(D|S,O) X P(S,O) / [P(D|S,O) X P(S,O) + P(D|~S,O) x P(~S,O) + P(D|S,~O) x P(S,~O) + P(D|~S,~O) x P(~S,~O)] ?
     
  2. jcsd
  3. May 18, 2010 #2
    the second formulation looks right.
     
  4. May 20, 2010 #3
    Show how P(D)=P(D|H)P(H)+P(D|1-H)P(1-H)
     
    Last edited: May 20, 2010
  5. May 20, 2010 #4
    HnH' = 1

    D = HnD+ H'nD

    P(D)= P (HnD) + (H'nD)= P(D | H) P(H) + P(D | H') P(H')
     
  6. May 20, 2010 #5
    Can you derive this from Bayes' Theorem?:

    P(D)=P(D|H)P(H)/P(H|D) is equivalent to P(D)=P(D|H)P(H)+P(D|1-H)P(1-H)?

    EDIT: I think it's more clear to use probability notation P(1-H) rather than logical notation P(~H) in what is essentially a conditional probability problem.
     
    Last edited: May 20, 2010
  7. May 25, 2010 #6
    I think you meant

    HuH' = 1

    Given that D = HnD + H'nD can we assume that P(D) = P(HnD) + P(H'nD) ?

    HuH' = 1
    or H' = 1 - H
    HnD = P(D|H)P(H)
    H'nD = P(D|H - 1)(1 - P(H))

    P(D) = P(D|H)P(H) + P(D|H - 1)(1 - P(H))

    but bayes theorem says P(D)=P(D|H)P(H)/P(H|D)

    so i think it's Ok
     
  8. May 25, 2010 #7
    Note:

    D=P(D|H)P(H)+P(D|1-H)P(1-H) is in terms of joint probabilities (terms cancel);

    D=P(D|H)P(H)/P(H|D) is in terms of a conditional probability.

    You can express D either way and you can write a conditional probability in terms of joint probabilities as the OP has done. However, you cannot write D in terms of Bayes Theorem without a conditional probability.
     
  9. May 26, 2010 #8
    Sorry, that's P(D) where D appears in the previous post.
     
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