Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?


    Aldebaran
     
  2. jcsd
  3. Sep 13, 2007 #2

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

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

    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).

    :rofl:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Probability of n hypothesis
  1. Probability rank-n (Replies: 6)

Loading...