I don’t think that the math in Aumann’s agreement theorem says what Aumann’s paper says that it says. The math may be right, but the translation into English isn’t.
Let N1 and N2 be partitions of Omega … Ni is the information partition of i; that is, if the true state of the world is w [an element of Omega], then i is informed of that element Pi(w) of Ni that contains w.
Given w in Omega, an event E is called common knowledge at w if E includes that member of meet(N1, N2) that contains w.
Let A be an event, and let Qi denote the posterior probability p(A|Ni) of A given i’s information; i.e., Qi(w) [ = p(A | Pi(w)) ] = p(A ^ Pi(w) / p(Pi(w)).
Proposition: Let w be in Omega … If it is common knowledge at w that Q1 = q1 and Q2 = q2, then q1 = q2.
Proof: Let P be the member of meet(N1, N2) that contains w. Write P = union over all j of Pj, where the Pj are disjoint members of P1. Since Q1 = q1 throughout P, we have p(A ^ Pj) / p(Pj) = q1 for all j; hence p(A ^ Pj) = q1p(Pj), and so by summing over j we get p(A^P) = q1p(P). Similarly p(A^P) = q2p(P), and so q1=q2.
meet(N1, N2) is not an intersection; it’s a very aggressive union of the subsets in the partitions N1 and N2 of Omega. It’s generated this way:
M = {w}, Used = {}
while (M neq Used)
Take an element m from M \ Used.
Find H1 in N1 and H2 in N2 containing m.
M = union(M, H1, H2).
Used = Used u {m}
Note in particular that P is a member of meet(N1, N2) that contains elements of Omega taken from H2, that are not in H1. To say that Q1 = q1 throughout P means that, for every x in P, Q1(x) = p(A | P1(x)) = q1. This is used to infer that p(A ^ Pj) = p(Pj) q1 for every Pj in N1.
This is a very strange thing to believe, given the initial conditions. The justification (as Robin Hanson pointed out to me) is that “common knowledge at w that Q1=q1” is defined to mean just that: Q1(x) = q1 for all x in the member P of the meet(N1,N2) containing w.
Now comes the translation into English. Aumann says that this technical definition of “common knowledge at w of the posteriors” means the same as “agent 1 and agent 2 both know both of their posteriors”. And the justification for that is this: “Suppose now that w is the true state of the world, P1 = P1(w), and E is an event. To say that 1 ‘knows’ E means that E includes P1. To say that 1 knows that 2 knows E means that E includes all P2 in N2 that intersect P1. …” et cetera, to closure.
And this, I think, is wrong. If 1 knows that 2 knows E, 1 knows that E includes P1 union some P2 that intersects with P1, not that E includes P1 union all P2 that intersect with P1. So the “common knowledge” used in the theorem doesn’t mean the same thing at all that we mean in English when we say they “know each others’ posteriors”.
Also, Aumann adds after the proof that it implicitly assumes that the agents know each others’ complete partition functions over all possible worlds. Which is several orders of magnitude of outlandish; so the theorem can never be applied to the real world.
I don’t think that the math in Aumann’s agreement theorem says what Aumann’s paper says that it says. The math may be right, but the translation into English isn’t.
Aumann’s agreement theorem says:
Let N1 and N2 be partitions of Omega … Ni is the information partition of i; that is, if the true state of the world is w [an element of Omega], then i is informed of that element Pi(w) of Ni that contains w.
Given w in Omega, an event E is called common knowledge at w if E includes that member of meet(N1, N2) that contains w.
Let A be an event, and let Qi denote the posterior probability p(A|Ni) of A given i’s information; i.e., Qi(w) [ = p(A | Pi(w)) ] = p(A ^ Pi(w) / p(Pi(w)).
Proposition: Let w be in Omega … If it is common knowledge at w that Q1 = q1 and Q2 = q2, then q1 = q2.
Proof: Let P be the member of meet(N1, N2) that contains w. Write P = union over all j of Pj, where the Pj are disjoint members of P1. Since Q1 = q1 throughout P, we have p(A ^ Pj) / p(Pj) = q1 for all j; hence p(A ^ Pj) = q1p(Pj), and so by summing over j we get p(A^P) = q1p(P). Similarly p(A^P) = q2p(P), and so q1=q2.
meet(N1, N2) is not an intersection; it’s a very aggressive union of the subsets in the partitions N1 and N2 of Omega. It’s generated this way:
M = {w}, Used = {}
while (M neq Used)
Note in particular that P is a member of meet(N1, N2) that contains elements of Omega taken from H2, that are not in H1. To say that Q1 = q1 throughout P means that, for every x in P, Q1(x) = p(A | P1(x)) = q1. This is used to infer that p(A ^ Pj) = p(Pj) q1 for every Pj in N1.
This is a very strange thing to believe, given the initial conditions. The justification (as Robin Hanson pointed out to me) is that “common knowledge at w that Q1=q1” is defined to mean just that: Q1(x) = q1 for all x in the member P of the meet(N1,N2) containing w.
Now comes the translation into English. Aumann says that this technical definition of “common knowledge at w of the posteriors” means the same as “agent 1 and agent 2 both know both of their posteriors”. And the justification for that is this: “Suppose now that w is the true state of the world, P1 = P1(w), and E is an event. To say that 1 ‘knows’ E means that E includes P1. To say that 1 knows that 2 knows E means that E includes all P2 in N2 that intersect P1. …” et cetera, to closure.
And this, I think, is wrong. If 1 knows that 2 knows E, 1 knows that E includes P1 union some P2 that intersects with P1, not that E includes P1 union all P2 that intersect with P1. So the “common knowledge” used in the theorem doesn’t mean the same thing at all that we mean in English when we say they “know each others’ posteriors”.
Also, Aumann adds after the proof that it implicitly assumes that the agents know each others’ complete partition functions over all possible worlds. Which is several orders of magnitude of outlandish; so the theorem can never be applied to the real world.