User:IssaRice/Aumann's agreement theorem

Hal Finney's example

$\overline{)\begin{array}{ccccccc}& 1& 2& 3& 4& 5& 6\\ 1& 2& 3& 4& 5& 6& 7\\ 2& 3& 4& 5& 6& 7& 8\\ 3& 4& 5& 6& 7& 8& 9\\ 4& 5& 6& 7& 8& 9& 10\\ 5& 6& 7& 8& 9& 10& 11\\ 6& 7& 8& 9& 10& 11& 12\end{array}}$

	1	2	3	4	5	6
1
2

$E=\{\omega \in \mathrm{\Omega }:Pr(A\mid I(\omega ))=q_1\text{ and }Pr(A\mid J(\omega ))=q_2\}$

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 $E^{\prime }=\{\omega \in \mathrm{\Omega }:Pr(A\mid I(\omega ))=q_1\}$ and say that agent 1 knows $E^{\prime }$?

In the form of $E$ above, we can change $A$ to be any subset of $\mathrm{\Omega }$ and $q_1,q_2$ to be any numbers in $[0,1]$. We can also set the state of the world to be any $\omega \in \mathrm{\Omega }$. 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)	$2\le X\le 6$	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?)