Scott Aaronson: Common knowledge and Aumann's agreement theorem
post by gjm · 2015-08-17T08:41:45.179Z · LW · GW · Legacy · 4 commentsThis is a link post for http://www.scottaaronson.com/blog/?p=2410
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The excellent Scott Aaronson has posted on his blog a version of a talk he recently gave at SPARC, about Aumann's agreement theorem and related topics. I think a substantial fraction of LW readers would enjoy it. As well as stating Aumann's theorem and explaining why it's true, the article discusses other instances where the idea of "common knowledge" (the assumption that does a lot of the work in the AAT) is important, and offers some interesting thoughts on the practical applicability (if any) of the AAT.
(Possibly relevant: an earlier LW discussion of AAT.)
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comment by Shmi (shminux) · 2015-08-17T14:58:32.847Z · LW(p) · GW(p)
It includes a great test of whether a given discussion is Aumann-rational:
what rational disagreements should look like: they should follow unbiased random walks, until sooner or later they terminate in common knowledge of complete agreement.
as opposed to
suppose your friend tells you a liberal opinion, then you take it into account, but reply with a more conservative opinion. The friend takes your opinion into account, and replies with a revised opinion. Question: is your friend’s new opinion likelier to be more liberal than yours, or more conservative?
Obviously, more liberal! Yes, maybe your friend now sees some of your points and vice versa, maybe you’ve now drawn a bit closer (ideally!), but you’re not going to suddenly switch sides because of one conversation.
Yet, if you and your friend are Bayesians with common priors, one can prove that that’s not what should happen at all.
comment by GMHowe · 2015-08-23T21:16:41.033Z · LW(p) · GW(p)
Maybe I'm confused, in the 'muddy children puzzle' it seems it would be common knowledge from the start that at least 98 children have muddy foreheads. Each child sees 99 muddy foreheads. Each child could reason that every other child must see at least 98 muddy foreheads. 100 minus their own forehead which they cannot see minus the other child's forehead which the other child cannot see equals 98.
What am I missing?
Replies from: gjm↑ comment by gjm · 2015-12-04T14:46:23.618Z · LW(p) · GW(p)
Common knowledge means I know, and I know that you know, and I know that you know that he knows, and she knows that I know that you know that he knows, and so on -- any number of iterations.
Each child sees 99 muddy foreheads and therefore knows n >= 99. Each child can tell that each other child knows n >= 98. But, e.g., it isn't true that A knows B knows C knows that n >= 98; only that A knows B knows C knows that n>=97: each link in the chain reduces the number by 1. So for no k>0 is it common knowledge that n>=k.
Replies from: GMHowe