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 $q_1=q_2$.