paulfchristiano comments on Reflection in Probabilistic Logic - Less Wrong

63 Post author: Eliezer_Yudkowsky 24 March 2013 04:37PM

You are viewing a comment permalink. View the original post to see all comments and the full post content.

Comments (171)

You are viewing a single comment's thread. Show more comments above.

Comment author: paulfchristiano 23 March 2013 02:49:09AM *  17 points [-]

Is there anything blocking you from setting P("a<P('A')<b")=1 for all cases, and therefore satisfying the stated reflection principle trivially?

Yes, the theory is logically coherent, so it can't have P("a < P('A')") = 1 and P("a > P('A')") = 1.

For example, the following disquotational principle follows from the reflection principle (by taking contrapositives):

P( x <= P(A) <= y) > 0 ----> x <= P(A) <= y

The unsatisfying thing is that one direction has "<=" while the other has "<". But this corresponds to a situation where you can make arbitrarily precise statements about P, you just can't make exact statements. So you can say "P is within 0.00001 of 70%" but you can't say "P is exactly 70%."

I would prefer be able to make exact statements, but I'm pretty happy to accept this limitation, which seems modest in the scheme of things---after all, when I'm writing code I never count on exact comparisons of floats anyway!

This was intended as a preliminary technical report, and we'll include much more discussion of these philosophical issues in future write-ups (also much more mathematics).

Comment author: abramdemski 23 March 2013 08:11:49AM 10 points [-]

Ok, I see!

To put it a different way (which may help people who had the same confusion I did):

We can't set all statements about probabilities to 1, because P is encoded as a function symbol within the language, so we can't make inconsistent statements about what value it takes on. ("Making a statement" simply means setting a value of P to 1.)

I'm very pleased with the simplicity of the paper; short is good in this case.

Comment author: abramdemski 23 March 2013 10:10:11PM *  1 point [-]

Actually, we can use coherence to derive a much more symmetric disquotation principle:

P(x>P(A)>y)=1 => x>P(A)>y.

Suppose P(x>P(A)>y)=1. For contradiction, suppose P(A) is outside this range. Then we would also have P(w>P(A)>z)=1 for some (w,z) mutually exclusive with (x,y), contradicting coherence.

Right?

Comment author: paulfchristiano 24 March 2013 03:35:32AM 2 points [-]

Not quite---if P(A) = x or P(A) = y, then they aren't in any interval (w, z) which is non-overlapping (x, y).

We can obtain P(x > P(A) > y) =1 ---> x >= P(A) >= y by this argument. We can also obtain P(x >= P(A) >= y) > 0 ---> x >= P(A) >= y.

Comment author: abramdemski 24 March 2013 02:55:17PM *  1 point [-]

So, herein lies the "glut" of the theory: we will have more > statements than are strictly true. > will behave as >= should: if we see > as a conclusion in the system, we have to think >= with respect to the "true" P.

A "gap" theory of similar kind would instead report too few inequalities...

Comment author: paulfchristiano 24 March 2013 05:18:20PM 5 points [-]

Yes, there is an infinitesimal glut/gap; similarly, the system reports fewer >= statements than are true. This seems like another way at looking at the trick that makes it work---if you have too many 'True' statements on both sides you have contradictions, if you have too few you have gaps, but if you have too many > statements and too few >= statements they can fit together right.

Comment author: abramdemski 24 March 2013 07:56:57AM 1 point [-]

Ah, right, good!

Comment author: abramdemski 23 March 2013 09:28:39PM 1 point [-]

For example, the following disquotational principle follows from the reflection principle (by taking contrapositives):

P( x <= P(A) <= y) > 0 ----> x <= P(A) <= y

The unsatisfying thing is that one direction has "<=" while the other has "<".

The negated statements become 'or', so we get x <= P(A) or P(A) <= y, right?

To me, the strangest thing about this is the >0 condition... if the probability of this type of statement is above 0, it is true!

Comment author: JeremyHahn 23 March 2013 10:44:59PM 4 points [-]

I agree that the derivation of (4) from (3) in the paper is unclear. The negation of a<b<c is not a>=b>=c.

Comment author: abramdemski 23 March 2013 11:51:35PM 1 point [-]

Ah, so there are already revisions... (I didn't have a (4) in the version I read).