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In bayesian epistemology, we usually find the bayesian theorem expressed in relation with three proposition: H, hypoyesis, E, evidence, K, background knowledge.

In particular, in "Bayes or Bust?" by Earman, we find it expressed, on page 33, in this form:

(1) P(H/E&K) = [P(H/K)*P(E/H&K)] /P(E/K)

But the usual formulation, for just P(H/E), should be:

(2) P(H/E) = [P(H)*P(E/H)] / P(E)

Where does (1) come from?

Secondly, on page 64, Earman explains how, if: a) {H,K} implies E b) 0< P(H/K) <1 c)0< P(E/K) <1

Then, from bayes theorem, we have that

P(H/E&K) = P(H/K)/P(E/K)

and from this, applying (b) and (c), we get that P(H/E&K) > P(H/K)

I do not understand what is the role played by (b) and (c).

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  • Clauses b) and c) that you quote from page 64 are identical.
    – Dave
    Feb 18 at 15:47
  • Sorry, it is an error. I am correcting it now
    – PwNzDust
    Feb 18 at 15:57
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For the first question, he's just conditioning everything on the "background knowledge" -- this is a way of making explicit the idea that in formulating this model there are one or more assumptions being made.

For part 2, item a) can be written in terms of probabilities as $P(E \vert H,K) =1$. The other ones are to avoid divide by zero problems, and trivial solutions when either of the probabilities are 1.

Formatted equation

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  • Thanks for the answer. The author goes on and conclude that then it is always the case that P(H/E&K)>P(H/K). This should be because P(E/K) and P(H/K) are always rational numbers between 0 and 1, and so P(H/K) = P(H/E&K) / rational, and so P(H/E&K)>P(H/K), right?
    – PwNzDust
    Feb 19 at 8:25

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