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Psy-Kosh comments on Probability Space & Aumann Agreement - Less Wrong
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Then agent 1 knows that agent 2 knows one of the members of J that have non empty intersection with I(w), and similar for for agent 2.
Presumably they have to tell each other which of their own partitions w is in, right? ie, presumably SOME sort of information sharing happens about each other's conclusions.
And, once that happens, seems like intersection I(w) and J(w) would be their resultant common knowledge.
I'm confused still though what the "meet" operation is.
Unless... the idea is something like this: they exchange probabilities. Then agent 1 reasons "J(w) is a member of J such that it both Intersects I(w) AND would assign that particular probability. So then I can determine the subset of I(w) that intersects with those" and determine a probability from there." And similar for agent 2. Then they exchange probabilities again, and go through an equivalent reasoning process to tighten the spaces a bit more... and the theorem ensures that they'd end up converging on the same probabilities? (each time they state unequal probabilities, they each learn more information and each one then comes up with a set that's a strict subset of the one they were previously considering, but each of their sets always contain the intersection of I(w) and J(w))?
Try a concrete example: Two dice are thrown, and each agent learns one die's value. In addition, each learns whether the other die is in the range 1-3 vs 4-6. Now what can we say about the sum of the dice?
Suppose player 1 sees a 2 and learns that player 2's die is in 1-3. Then he knows that player 2 knows that player 1's die is in 1-3. It is common knowledge that the sum is in 2-6.
You could graph it by drawing a 6x6 grid and circling the information partition of player 1 in one color, and player 2 in another color. You will find that the meet is a partition of 4 elements, each a 3x3 grid in one of the corners.
In general, anything which is common knowledge will limit the meet - that is, the meet partition the world is in will not extend to include world-states which contradict what is common knowledge. If 2 people disagree about global warming, it is probably common knowledge what the current CO2 level is and what the historical record of that level is. They agree on this data and each knows that the other agrees, etc.
The thrust of the theorem though is not what is common knowledge before, but what is common knowledge after. The claim is that it cannot be common knowledge that the two parties disagree.
Not sure... what happens when the ranges are different sizes, or otherwise the type of information learnable by each player is different in non symmetric ways?
Anyways, thanks, upon another reading of your comment, I think I'm starting to get it a bit.
What I don't like about the example you provide is: what player 1 and player 2 know needs to be common knowledge. For instance if player 1 doesn't know whether player 2 knows whether die 1 is in 1-3, then it may not be common knowledge at all that the sum is in 2-6, even if player 1 and player 2 are given the info you said they're given.
This is what I was confused about in the grandparent comment: do we really need I and J to be common knowledge? It seems so to me. But that seems to be another assumption limiting the applicability of the result.