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Probability of n hypothesis

  1. Sep 13, 2007 #1
    Ciao all,

    I have a conjunction of n non-independant hypothesis:

    h1 and h2 and .... and hn.

    I want to calculate the probability of such conjunction. To do so, I used the product rule p(h1 and h2) =p(h1).p(h2|h1)

    p(h1 and h2 and .... and hn) = p(h1).p(h2|h1).p(h3|h1 and h2 )....

    Do you think that this is correct?

  2. jcsd
  3. Sep 13, 2007 #2


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    I think it is.
  4. Sep 14, 2007 #3

    Yes the proof was pretty simple:

    p(a1 and a2 and... and an)
    = p(an | a1 and a2 and... and an-1). p(a1 and a2 and... and an-1).
    = p(an | a1 and a2 and... and an-1). p(an-1 | a1 and a2 and... and an-2).p(a1 and a2 and... and an-2).

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