User:IssaRice/Aumann's agreement theorem

Hal Finney's example

Let Alice and Bob be two agents. Each rolls a die, and knows what they rolled. In addition to this, each knows whether the other rolled something in the range 1–3 versus 4–6. As an example, suppose Alice rolls a 2 and Bob rolls a 3. Then Alice knows that the outcome is one of (2,1), (2,2), or (2,3), and Bob knows that the outcome is one of (1,3), (2,3), or (3,3).

In the possible worlds notation, if we let $I$ be Alice's information partition and $J$ be Bob's information partition, we would write:

• $\omega =(2,3)$
• $I(\omega )=\{(2,1),(2,2),(2,3)\}$
• $J(\omega )=\{(1,3),(2,3),(3,3)\}$
• $(I\wedge J)(\omega )=\{1,2,3\}\times \{1,2,3\}$

${\begin{array}{c|cccccc}&1&2&3&4&5&6\\\hline 1&2&3&4&5&6&7\\2&3&4&5&6&7&8\\3&4&5&6&7&8&9\\4&5&6&7&8&9&10\\5&6&7&8&9&10&11\\6&7&8&9&10&11&12\end{array}}$

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

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...