User:IssaRice/Aumann's agreement theorem

Information partitions

Actually, can't the information states overlap? What if, when Bob rolls a 1, Alice is told "1 or 3" and when Bob rolls a 2 is told "2 or 3" (with the rest of the situation remaining the same)? Then it seems like $I((2,1))=\{(2,1),(2,3)\}$ and $I((2,3))=\{(2,2),(2,3)\}$, which overlap.

Similarly in Geanakoplos's example on p. 261, in his example the agent is conveniently told whether the number is even or odd, giving a partition. But what if the agent is told whether the number is 1-7 or 2-8? It seems like this information states setup restricts the examples we can formalize, and the examples people consider seem to conveniently be formalizable.

In this case, if the real number happens to be 2-7, then the agent is told both "1-7" and "2-8", so he can deduce it is 2-7. Similarly, if he is told "1-7", then he can tell the number is 1, for if it wasn't 1, he would have also been told "2-8". So the information partition ends up being $\{\{1\},\{2,\dots ,7\},\{8\}\}$. So a rational agent can make their possible information states into a partition by reasoning about which clues are communicated. A general version of this is probably some basic theorem in game theory or a related field, but I am highly ignorant.

Actually, the above operation seems to be to "refine" the "partition" until it becomes an actual partition. The coarsest partition such that each element of the partition is a subset of the original set.

Join and meet

Given two partitions of a set, the join is the coarsest common refinement, and the meet is the finest common coarsening.

TODO give more formal definition

TODO give imperative definition

General process for reaching agreement

In the idealized agreement theorem setup, the process for two agents, Alice and Bob, agreeing on posteriors of some event $A$ will look like this:

1. Alice and Bob each have an information partition.
2. They each make some observation, updating their posteriors. At this point, their posteriors for $A$ might be the same, but might not.
3. Alice and Bob tell each other their posteriors.
4. After hearing the other's posterior, each modifies their own information partition. (The underlying sample space remains the same, but the partitioning changes, so that the posterior can change.)
5. Repeat (1)–(4) until $E=\{\omega \in \mathrm{\Omega }:Pr(A\mid I(\omega ))=q_1\text{ and }Pr(A\mid J(\omega ))=q_2\}$ becomes common knowledge for some constants $q_1,q_2\in [0,1]$. In other words, keep exchanging posteriors until they become common knowledge. In step (2), after the first round the observation becomes the other's professed posterior.
6. At this point, Aumann's agreement theorem applies, and we conclude $q_1=q_2$. Alice and Bob agree!

What are the actual mechanics of observing the other's posterior and updating on it? I think what happens is that Bob gives Alice some set of omegas, and Alice gives Bob some other set of Omegas, and each intersects it with their current information set. And this resulting set becomes the new information set. Specifically suppose $q_2=Pr(A\mid J(\omega ))$. Then Bob gives Alice the set $\{\omega \in \mathrm{\Omega }:Pr(A\mid J(\omega ))=q_2\}$. This way, Alice can carry out updates without knowing what $J(\omega )$ is or even what $q_2$ is.

Maybe knowing $J$ and $q_2$ is equivalent to knowing the set above, in which case Alice can receive $J$ and $q_2$ and compute the set on her own.

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).

Given the above description, what are the possible states of knowledge of Alice?

$I=\{\{1\}\times \{1,2,3\},\dots ,\{6\}\times \{1,2,3\},\{1\}\times \{4,5,6\},\dots ,\{6\}\times \{4,5,6\}\}$

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

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

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

Now let $A\subset \mathrm{\Omega }$ be an arbitrary event, and let $q_1,q_2\in [0,1]$ be constants. We now define the event

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

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 $q_1=q_2$.

Let $X:\mathrm{\Omega }\to \mathbf{R}$ be a random variable defined by $X(x,y)=x+y$. In other words, $X$ is the sum of the values of the two dice.

$\omega$	$A$	$q_1$	$q_2$	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)	$X=5$	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?)
(2, 3)	$X=6$	0	1/3	Given these parameters, $E=\{1,2\}\times \{3\}$, which is not a superset of $(I\wedge J)(\omega )$, so $E$ is not common knowledge.

Alice knows she rolled a 2 and Bob rolled something between 1 and 3. Now, consider that Alice is additionally told that Bob did not roll a 1. Now Alice's posterior probability of the event $X=4$ is 1/2. How does this affect the agreement theorem? It seems like Alice's information partition changes; in particular, Bob's roll gets divided into $\{\{1\},\{2,3\},\{4,5,6\}\}$. Bob's information partition remains the same, so now $(I\wedge J)((2,3))=\{1,2,3\}\times \{2,3\}$. Now for the event $X=4$, $Pr(X=4\mid I((2,3)))=1/2$ and $Pr(X=4\mid J((2,3)))=1/3$ and the set $E=\{(1,2),(1,3),(2,2),(2,3)\}$ is not a superset of $(I\wedge J)((2,3))$ so the agreement theorem doesn't apply. And indeed, that's good because $1/2\ne 1/3$.

Another thing: can't Bob deduce that Alice must have rolled a 1 or 2 by knowing her posterior that X=4 is 1/2? If that's the case, it seems like the meet reduces to {(1,2), (1,3), (2,2), (2,3)}, which is the same as E, so the agreement theorem applies again, with the posterior of X=4 being 1/2. Actually, this might be the whole point of Aumanning, because otherwise the theorem just doesn't apply in most cases.

Aumann's coin flip example

let's hope i get this right. let A be the event "heads on the second toss", H be the event "heads on first toss" and T be the event "tails on first toss". B is the random variable representing the bias of the coin, and b is a specific value. B is distributed uniformly in $[0,1]$.

$\begin{array}{rr}Pr(A\mid H)& ={\int }_{b\in [0,1]}Pr(A\mid H,B=b)\cdot Pr(B=b\mid H)\\ & ={\int }_{b\in [0,1]}b\cdot \frac{Pr(H\mid B=b)\cdot Pr(B=b)}{Pr(H)}\\ & =\underset{0}{\overset{1}{\int }}b\cdot 2bdb\\ & =2/3\end{array}$

and

$\begin{array}{rr}Pr(A\mid T)& ={\int }_{b\in [0,1]}Pr(A\mid T,B=b)\cdot Pr(B=b\mid T)\\ & ={\int }_{b\in [0,1]}b\cdot \frac{Pr(T\mid B=b)\cdot Pr(B=b)}{Pr(T)}\\ & =\underset{0}{\overset{1}{\int }}b\cdot 2(1-b)db\\ & =1/3\end{array}$