= 27d567bc2d48d9f40e9ccc20ea370b15)
Vladimir_Nesov comments on An explanation of Aumann's agreement theorem - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (18)
It seems from your presentation like first agent can only hold one of E1 or E2, etc., where E1, E2, etc. are disjoint subsets of the sample space. But the same result holds if the agent can hold the state of knowledge that is a union of such sets, so the first agent could believe E1+E2, etc. There seems to be no reason to not make this generalization. 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).