Manfred comments on Approaching Logical Probability - Less Wrong

7 Post author: Manfred 27 February 2014 07:44AM

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Comment author: Manfred 28 February 2014 09:29:13PM 0 points [-]

Ah, I see - is the idea "if we haven't derived probabilities yet, how can we use probabilistic strategies?"

If we use some non-black-box random process, like rolling a die, then I think the problem resolves itself, since we don't have to use probabilities to specify a die, we can just have a symmetry in our information about the sides of the die, or some knowledge of past rolls, etc. Under this picture, the "mixed" in mixed strategy would be externalized to the random process, and it would have the same format as a pure strategy.

Comment author: cousin_it 01 March 2014 09:13:15AM *  0 points [-]

Hmm, no, I was trying to make a different point. Okay, let's back up a little. Can you spell out what you think are the assumptions and conclusions of Savage's theorem with your proposed changes? I have some vague idea of what you might say, and I suspect that the conclusions don't follow from the assumptions because the proof stops working, but by now we seem to misunderstand each other so much that I have to be sure.

Comment author: Manfred 01 March 2014 09:59:42AM *  0 points [-]

I am proposing no changes. My claim is that even though we use english words like "event-space" or "actions" when describing Savage's theorem, the things that actually have the relevant properties in the AMD problem are the strategies.

Cribbing from the paper I linked, the key property of "actions" is that they are functions from the set of "states of the world" (also somewhat mutable) to the set of consequences (the things I have a utility function over). If the state is "I'm at the first intersection" and I take the action (no quotes, actual action) of "go straight," that does return a consequence.

Comment deleted 01 March 2014 02:51:14PM *  [-]
Comment author: Manfred 01 March 2014 10:53:31PM *  0 points [-]

Well, if we're changing what objects are the "actions" in the proof, we're probably also changing which objects are the "states." You only need a strategy once, you don't need a new strategy for each intersection.

If we have a strategy like "go straight with probability p," a sufficient "state" is just the starting position and a description of the game.

Hmm, I'm not sure on what grounds we can actually rule out using the individual intersections as states, though, even though that leads to the wrong answer. Maybe they violate axiom 3, which requires the existence of "constant actions."

Comment author: cousin_it 02 March 2014 12:24:14AM 0 points [-]

Sorry for deleting my comment. I'm still trying to figure out where this approach leads. So now you're saying that "I'm at the first intersection" isn't actually a "state" and shouldn't get a probability?

Comment author: Manfred 02 March 2014 01:21:51AM *  0 points [-]

Right. To quote myself:

P(outcome | do(action)) has no proper place in our agent's decision-making. Savages theorem requires us to use probabilities for the things that determine the outcome; if our action does not determine the outcome, its probability isn't given by Savage's theorem.

And I do think that simultaneously, we can use Cox's theorem to show that the absent-minded driver has some probability P(state | information). It's just not integrated with decision-making in the usual way - we want to obey Savage's theorem for that.

So we'll have a probability due to Cox's theorem. But for decision-making, we won't ever actually need that probability, because it's not a probability of one of the objects Savage's theorem cares about.

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