An explanation of Aumann's agreement theorem

There's a discussion of practical implications of AAT in my post.

Here is a proof of the equivalence between my definition and Aumann's for "common knowledge". I'm assuming some familiarity with set partitions.

Aumann's definition is in terms of the Kolmogorov model of probability. In particular, a proposition is identified with the set of possible worlds in which the proposition is true.

Let P₁ be that partition of the possible worlds such that two worlds share the same block in P₁ if and only if I condition on the same body of knowledge when computing posterior probabilities in the two worlds*. Let P₂ be the analogous partition of the possible worlds for you. For each world w, let P₁(w) denote the block in my partition containing w, and let P₂(w) be the block in your partition containing w. Let P denote the finest common coarsening of our respective partitions**, and let P(w) be the block of P containing w.

Fix a proposition A. Let p be my posterior probability for A, and let q be yours (in the actual world). Let E be the set of worlds in which I assign posterior probability p to A, while you assign posterior probability q. Formally:

• E = { w : p = prob(A | P₁(w)) and q = prob(A | P₂(w)) }

Let w₀ be the actual world. Aumann's definition says that our respective posterior probabilities are common knowledge (in the actual world) if P(w₀) ⊆ E.

On the other hand, I said that our posterior probabilities are common knowledge when there is a proposition C that is true in the actual world and satisfies the following three conditions:

1. For each world w in C, P₁(w) ⊆ C and P₂(w) ⊆ C.

2. For each world w in C, p = prob(A | P₁(w)).

3. For each world w in C, q = prob(A | P₂(w)).

These definitions are logically equivalent.

For, suppose that our posteriors are common knowledge in Aumann's sense. Then, by setting C = P(w₀), we get that the posteriors are also common knowledge in my sense.

On the other hand, suppose that the posteriors are common knowledge in my sense. It is given that C happened in the actual world, meaning that w₀ ∈ C. By Condition 1, C is a disjoint union of P₁-blocks and a disjoint union of P₂-blocks. This means that C is a block containing w₀ in some common coarsening of P₁ and P₂. Hence, C contains the block P(w₀) containing w₀ in the finest common coarsening of P₁ and P₂. That is, P(w₀) ⊆ C. Conditions 2 and 3 together imply that C ⊆ E. Thus we get that P(w₀) ⊆ E, so our posteriors are also common knowledge in Aumann's sense.

* That this relation induces a partition assumes, in effect, that I know what I don't know — i.e., that there are no unknown unknowns.

** Aumann calls this the meet of P₁ and P₂, because he considers coarsenings to be lower in the partial order of partitions. However, people often use the opposite convention, in which case P would be called the join.

The conditions are logically equivalent. Unfortunately, I don't see a way to show this without using the partition poset terminology that I intended to avoid. Nonetheless, if you unpack the definition of "common knowledge" in Aumann's paper, it is equivalent to what I gave. (ETA: I give the unpacking in this comment.)

Awesome. Thank you. The write-up can now be downloaded here.

Update 2020-05-23: Updated Dropbox link.

I didn't realize that you needed a Scribd login to download the pdf. That is a deal breaker. I will find another place for the document.

IIRC, the standard term in philosophy for the same thing is "mutual belief".

I would like an abstract.

Thanks, added.

IIRC, that's how Aumann's paper was describing it.

No, not as I read it.

My C & E_i correspond to blocks in the partition that Aumann assigns to agent 1. Aumann has agent 1 evaluating the posterior probability of A in world ω by computing p(A | P₁(ω)), where P₁(ω) is the element of 1's partition containing ω. So the agent will never condition on a union P₁(ω₁) ∪ P₁(ω₂) of distinct blocks. (Here I'm following Aumann's notation as closely as possible.) Correspondingly, my agent never conditions on a disjunction [(C & E₁) ∨ (C & E₂)].

But I think that you're right about the generalization being trivial. It should just involve the standard procedure for turning a union A ∪ B into a disjoint union A ∪ (B∖A).