Logical Uncertainty: an addendum

And I forgot to mention one thing in the last post which is relevant. Gaifman, in his paper, states that if in \mathcal A  we have that A\rightarrow B then P(B|\mathcal A) \geq P(A|\mathcal A). I’ll quickly show that that’s a theorem of my approach, and, indeed, any similar approach.

Suppose that ``A\rightarrow B" \in X. In that case, then, my approach has that P(B|AX) = 1, because the agent knows B is logically implied by A. If that’s the case, then:

\begin{aligned} P(B|X) &= P(B|AX)P(A|X)+P(B|\bar AX)P(\bar A|X) \\  &=P(A|X)+P(B|\bar AX)P(\bar A|X)\\  &\geq P(A|X)\end{aligned}

With equality if and only if either P(A|X) = 1 (A is logically certain given X) or P(B|\bar AX) = 0 (B is also impossible when A is false, which means it’s logically equivalent to A). So maybe this was fairly obvious to you, but if it wasn’t, now you have that proof in your background list of proofs and theorems!

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2 Responses to Logical Uncertainty: an addendum

  1. Pingback: Bayes’ Theorem | An Aspiring Rationalist's Ramble

  2. Pingback: Logical Uncertainty | An Aspiring Rationalist's Ramble

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