User:IssaRice/Aumann's agreement theorem

Hal Finney's example

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	1	2	3	4	5	6
1
2

$E=\{\omega \in \Omega :\Pr(A\mid I(\omega ))=q_{1}{\text{ and }}\Pr(A\mid J(\omega ))=q_{2}\}$

One of the assumptions in the agreement theorem is that $E$ is common knowledge. This seems like a pretty strange requirement, since it seems like the posterior probability of $A$ can never change no matter what else the agents condition on in addition to $E$. For example, what if we bring in agent 3 and make the posteriors common knowledge again?

What if we take $E'=\{\omega \in \Omega :\Pr(A\mid I(\omega ))=q_{1}\}$ and say that agent 1 knows $E'$?

In the form of $E$ above, we can change $A$ to be any subset of $\Omega$ and $q_{1},q_{2}$ to be any numbers in $[0,1]$. We can also set the state of the world to be any $\omega \in \Omega$. The agreement theorem says that as we vary these parameters, if we ever find that $(I\wedge J)(\omega )\subset E$, then we must have $q_{1}=q_{2}$.

Define $X((x,y))=x+y$.

$\omega$	$A$	$q_{1}$	$q_{2}$	Explanation
(2, 3)	$2\leq X\leq 6$	1	1	Given these parameters, $E=(I\wedge J)(\omega )$ so $E$ is common knowledge. This satisfies the requirement of the agreement theorem, and indeed 1=1.
(2, 3)	$X=4$	1/3	1/3	Given these parameters, $E=(I\wedge J)(\omega )$ so $E$ is common knowledge. This satisfies the requirement of the agreement theorem, and indeed 1/3=1/3.
(2, 3)	$X=4$	1/3	1/3	Given these parameters, $E=\{2,3\}\times \{2,3\}$, which is not a superset of $(I\wedge J)(\omega )$, so $E$ is not common knowledge. Nonetheless, 1/3=1/3. (Is this a case of mutual knowledge that is not common knowledge?)

Agent 1 knows he rolled a 2 and agent 2 rolled something between 1 and 3. Now, consider that agent 1 is additionally told that agent 2 did not roll a 1. Now agent 1's posterior probability of the event $X=4$ is 1/2. How does this affect the agreement theorem? It seems like agent 1's information partition changes...