gRR comments on What a reduction of "could" could look like - Less Wrong

53 Post author: cousin_it 12 August 2010 05:41PM

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Comment author: gRR 18 February 2012 05:41:31PM *  0 points [-]

Thanks! This site is rather large :) And the most interesting (for me) stuff is rarely linked from the sequences. That is, it seems so... Is there a more efficient way to get up to date? [I would SO like to have found this site 5 years ago!]

The axioms were not proposed seriously as a complete solution, of course. The (Ax2) can be thought of as a cached consequence of a long chain of reasoning. The main point of the post was about the word "could", after all.

(Ax3) is just an axiomatic restatement of the "utility maximization" goal. I don't see why an agent can't know about itself that it's a utility maximizer.

A contradiction can be inferred from your axioms in this way: (4) says A=2, which implies [A=1 => W=0], which by (Ax3) together with [A=2 => W=1000000] implies that A=1, which is a contradiction and a decision to do whatever (I think you've also mixed up A=1 and A=2 while going from (Ax2) to (1) and (2)).

Yes, A=1 and A=2 were mixed up in the (Ax2), fixed it now. But I don't see your contradiction. (Ax3) + [A=1 => W=0] + [A=2 => W=1000000] does not imply A=1. It implies NOT(A=1)

Comment author: Vladimir_Nesov 18 February 2012 07:21:42PM *  0 points [-]

Is there a more efficient way to get up to date?

See the posts linked to from UDT, ADT, TDT, section "Decision theory" in this post.

I don't see why an agent can't know about itself that it's a utility maximizer.

It can, it's just a consideration it can't use in figuring out how the outcome depends on its decision. While looking for what is influenced by a decision, the decision itself should remain unknown (given that the decision is determined by what this dependence turns out to be, it's not exactly self-handicapping).

Yes, A=1 and A=2 were mixed up in the (Ax2), fixed it now. But I don't see your contradiction. (Ax3) + [A=1 => W=0] + [A=2 => W=1000000] does not imply A=1. It implies NOT(A=1)

It seems the mixing up of A=1 and A=2 didn't spare this argument, but it's easy enough to fix. From A=2 we have NOT(A=1), so [A=1 => W=1,000,000,000] and together with [A=2 => W=1,000,000] by (Ax3) we have A=1, contradiction.

Comment author: gRR 18 February 2012 11:56:07PM 1 point [-]

I thought about a different axiomatization, which would not have the same consistency problems. Not sure whether this is the right place, but:

Let X be a variable ranging over all possible agents.

(World Axioms)
forall X [Decides(X)=1 => Receives(X)=1000000]
forall X [Decides(X)=2 => Receives(X)=1000]

(Agent Axioms - utility maximization)
BestX = argmax{X} Receives(X)
Decision = Decides(BestX)

Then Decision=1 is easily provable, regardless of whether any specific BestX is known.
Does not seem to lead to contradictions.

Comment author: gRR 18 February 2012 08:16:52PM 1 point [-]

Yes, I see now, thanks.

If I attempted to fix this, I would try to change (Ax3) to something like:

forall a1,a2,w1,w2 ((A=a1 => W=w1) AND (A=a2 => W=w2) AND (w1>w2)) => NOT (FinalA=a2)

where FinalA is the actual decision. But then, why should the world's axiom (Ax2) be defined in terms of A and not FinalA? This seems conceptually wrong...

Ok, I'll go read up on the posts :)

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