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Why Bayes' Rule is useful

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  • Jim Davies
    P(H|E) = P(H and E) / P(E) is not as useful as Bayes Rule P(H|E) = (P(E|H)*P(H))/P(E) But I m not sure why. Is it because the former requires P(H and E)
    Message 1 of 3 , Jan 31, 2011
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      P(H|E) = P(H and E) / P(E)

      is not as useful as Bayes' Rule

      P(H|E) = (P(E|H)*P(H))/P(E)

      But I'm not sure why. Is it because the former requires P(H and E) assumes independence, which we might not have?

      JimDavies
    • j1mmy_n3u7120n
      ... This is the definition of Conditional Probability. ... Bayes Rule is derived from the definition of conditional probability. Hence, all assumptions made
      Message 2 of 3 , Jan 31, 2011
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        > P(H|E) = P(H and E) / P(E)
        This is the definition of Conditional Probability.

        > P(H|E) = (P(E|H)*P(H))/P(E)
        Bayes' Rule is derived from the definition of conditional probability.
        Hence, all assumptions made in the definition of conditional probability
        will be true even for Bayes' Rule

        > Is it because the former requires P(H and E) assumes independence,
        P(H and E) does not assume independence (or dependence for that matter).
        It just represents joint distribution.

        > Why is the definition of conditional probability not as useful as the
        Bayes' Rule?
        That is because in most applications, when you want to predict P(H|E),
        the most estimable quantity that you have with you is P(E|H). For
        example, if you want to predict what disease is afflicting a person
        based on his symptoms i.e P(Disease | Symptoms), the data that you will
        have with you is which diseases have what all symptoms, from which you
        can estimate P(Symptoms | Disease).

        Regards.

        --- In aima-talk@yahoogroups.com, "Jim Davies" <jim.davies@...> wrote:
        >
        > P(H|E) = P(H and E) / P(E)
        >
        > is not as useful as Bayes' Rule
        >
        > P(H|E) = (P(E|H)*P(H))/P(E)
        >
        > But I'm not sure why. Is it because the former requires P(H and E)
        assumes independence, which we might not have?
        >
        > JimDavies
        >
      • YKY (Yan King Yin, 甄景贤)
        A good way to memorize Bayes rule is this: P(A,B) = P(A|B)P(B) = P(B|A)P(A) ^ probability of A and B This is always true. You can easily get Bayes rule
        Message 3 of 3 , Mar 7 8:25 PM
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          A good way to memorize Bayes rule is this:

          P(A,B)  = P(A|B)P(B)    = P(B|A)P(A)
          ^ probability of "A and B"

          This is always true.

          You can easily get Bayes rule by moving one factor under the other side.

          Bayes rule allows one to calculate "B given A" when we are given "A given B", ie, it allows to reverse the direction of a conditional probability.

          KY
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