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Statistics with Baye's theorem and partitions

  1. Feb 6, 2010 #1


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    Well the questions and solutions are in one for some and I will type out the rest.

    http://img249.imageshack.us/img249/2757/47026197.jpg [Broken]

    http://img237.imageshack.us/img237/1802/93774799.jpg [Broken]
    http://img3.imageshack.us/img3/5919/34981698.jpg [Broken]

    http://img19.imageshack.us/img19/7307/25729239.jpg [Broken]

    I need some help doing Q3 and I need some clarification with Q1.

    With Q1, if Y=2, then there are only two possibilities, (1,2) and (1,2). So X can only be equal to 3 thus I am not sure how to write out the distribution table.

    http://img222.imageshack.us/img222/9981/95680883.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 6, 2010 #2
    for Q1, there's no reason not to include (2,2) as well (twice)

    for Q3, three events would be independent if p(A and B and C) = p(A)p(B)p(C), and the three possible pairs are independent (e.g A and C), just a natural extension from two events
  4. Feb 6, 2010 #3


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    Right, but wouldn't then my distribution not all add up to one either way? Σall x P(X=x)≠1

    This I know, but I don't know what numbers I'd write down for P(A) and so on.
  5. Feb 7, 2010 #4

    You should just have Σall x P(X=x | Y=2) =1

    I drew the square with corners at 0, (0,1), (1,1), (1,0) ao P(A) = P(B) =1/2 etc.
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