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gRR comments on Logical Uncertainty as Probability - Less Wrong Discussion

3 Post author: gRR 29 April 2012 10:26PM

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Tags: probability logical_uncertainty

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Comment author: gRR 30 April 2012 12:52:40AM 0 points [-]

But given any finite set (X) of true statements in Peano arithmetic I invent statement in Peano arithmetic that isn't in that set but that you would instantly assign p=1.

It is entirely possible for P(S|X) to be < 1, although P(S|X∪{S}) = 1. There's nothing inconsistent about updating on new evidence.

Comment author: Jack 30 April 2012 01:44:55AM *  1 point [-]

So then it isn't true that P(S|X) = 1 if and only if S∈OLC(X).

Comment author: gRR 30 April 2012 02:18:39AM *  0 points [-]

If X = { "2*2=4" }, then P("2*2=4 ∨ 0=1" | X) = 1, because "2*2=4 ∨ 0=1" ∈ OLC(X).
However, P("2*3=6" | X) < 1, although P("2*3=6" | X ∪ {"2*3=6"}) = 1.

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