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Re: question 13.9

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  • j1mmy_n3u7120n
    Both follow the normal proof methods .... Note that P(X,Y|e) = P(X,Y,e)/P(e) and P(X|Y,e) = P(X,Y,e)/P(Y,e) and P(Y|e) = P(Y,e)/P(e) The first part is
    Message 1 of 3 , Nov 13, 2004
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      Both follow the normal proof methods ....

      Note that P(X,Y|e) = P(X,Y,e)/P(e)
      and P(X|Y,e) = P(X,Y,e)/P(Y,e)
      and P(Y|e) = P(Y,e)/P(e)

      The first part is immediate from this.

      The second part is almost trivial after this ...
      Note that,
      P(X,Y|e) = P(X|Y,e) P(Y|e)
      also,
      P(X,Y|e) = P(Y|X,e) P(X|e)

      Can u finish it off from here?
      -- AI

      --- In aima-talk@yahoogroups.com, "brandon_pitt_47"
      <brandon_pitt_47@y...> wrote:
      >
      >
      > Hope someone can help me with this one. completely stuck. any
      solutions out there? it's
      > on page 490 about proving the conditionalized version of the general
      product rule.
      >
      > P(X,Y|e) = P(X|Y,e) P(Y|e)
      >
      > part b: prove the conditionalized version of Bayes' Rule in
      Equation (13.10)
      >
      > P(Y|X,e) = P(X|Y,e) P(Y|e)
      > ------------
      > P(X)
      >
      > ANY HELP WOULD BE GREAT.
      >
      > thanks,
      > brandon
    • Brandon Iskey
      yes, i can. thanks again for your help. brandon __________________________________ Do you Yahoo!? Check out the new Yahoo! Front Page. www.yahoo.com
      Message 2 of 3 , Nov 14, 2004
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        yes, i can. thanks again for your help.

        brandon



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