Part 1 seems to have little to do with how I remember the theorem. Here is the abstract of Aumann’s paper.
Two people, 1 and 2, are said to have common knowledge of an event $E$ if both know it, 1 knows that 2 knows it, 2 knows that 1 knows is, 1 knows that 2 knows that 1 knows it, and so on. THEOREM. If two people have the same priors, and their posteriors for an event $A$ are common knowledge, then these posteriors are equal.
Your 2 implies the following claim:
Agent 1 knows that agent 2 knows E if and only if agent 2 knows that agent 1 knows E.
This claim is obviously false.
Here is another corollary of your definition:
Suppose P1(w)={w,v}, P2(w)={w,u}. Then at w, I know E={w,v,u}, but at v, I do not know E! So I can distinguish between w and v by checking my knowledge, even though I cannot distinguish between w and v!
Part 3 is correct. Indeed, common knowledge of honesty is a requirement.
Agent 1 knows that agent 2 knows E if and only if agent 2 knows that agent 1 knows E.
How does it imply that? (It well might, within the context of the agreement theorem. My recollection is that you assume from the start that A1 and A2 have common knowledge of E.)
This claim is obviously false.
Why? If knowledge means “justified true belief”, then for agent 1 to know that agent 2 know E, agent 1 must also know E, and vice-versa. This doesn’t prove the claim that you say I am making, but goes most of the way towards proving it.
Suppose P1(w)={w,v}, P2(w)={w,u}. Then at w, I know E={w,v,u}, but at v, I do not know E! So I can distinguish between w and v by checking my knowledge, even though I cannot distinguish between w and v!
This is true, except that P1 and P2 should range over events, not over world states. This is a step that the theorem relies on. Are you claiming that this is false?
Terms: P1 means what Aumann calls P-superscript1; N1 means what Aumann calls cursive-P superscript 1. P1(E) = {w,v} means that, after observing event E, A1 knows that the world is in one of the states {w, v}. N1 is the set that describes the range of P1. E is an event, meaning a set of possible world states. Aumann doesn’t define what an ‘event’ is, other than implicitly in how he uses the variable E, so I hope I’m getting that right.
I’m constructing this partly from memory—sorry, this is a complex proof, and Aumann’s paper is skimpy on definitions, several of which (like “meet” and “join”) are left undefined and hard to find defined anywhere else even with Google. I really can’t do this justice without more free time than I have in the next several months.
What I think Aumann is saying is that, if A1 knows E, and knows that A2 knows E, then for every state x in P1(E), for every event D such that x is in P2(D) , P2(D) is a subset of E. Saying this allows Aumann to go on and show that A1 and A2 can iteratively rule out possibilities until they converge on believing the same thing.
This requires knowing more than what we mean when we say “A1 knows that A2 knows E”. When we say that, we mean that A1 knows the world is in one of the states in E, and knows that A2 knows the world is in one of the states in E. But it is possible that there is some state x, that the world is not in, but that is a member of E and of P1(E), but not of P2(E).
My recollection is that this is the problem: If you only consider conditions involving 3 possible world states, like the w, u, and v in the above example, then you can show that these things are equivalent. The agents can always use their mutual knowledge to iteratively eliminate possible states until they agree. For instance, if the situation is that P1({w,v,u})={w,v}, P2({w,v,u})={w,u}, then P1 and P2 can use their common knowledge to conclude w, and thus agree. But if you consider conditions where P1, P2, and E contain more than 3 different states between them, you can find situations that have multiple possible solutions, which the agents cannot choose between; and so cannot converge.
The definition you gave was symmetric. If I misread it, my apologies.
Why? If knowledge means “justified true belief”, then for agent 1 to know that agent 2 know E, agent 1 must also know E, and vice-versa. This doesn’t prove the claim that you say I am making, but goes most of the way towards proving it.
True, but it’s impossible to go the rest of the way. If you see a dog and I see both you and the dog through a one-way mirror, then I know that you know that there’s a dog there but you don’t know that I know that there is a dog.
I am having trouble matching up your notation with the notation I’m used to.
There are two operations, which I am used to calling P and K. They also have a number attached to them.
P takes sets to bigger sets or else to themselves. P1({w}) is what A1 thinks is possible when w is true. P1(S) for any set S is what A1 might think is possible given that something in S is true.
K takes sets to smaller sets or else to themselves. K1(S) is the set of possible states of the world where A1 knows that S is true.
What I think Aumann is saying is that, if A1 knows E, and knows that A2 knows E, then for every state x in P1(E), for every event D such that x is in P2(D) , P2(D) is a subset of E. Saying this allows Aumann to go on and show that A1 and A2 can iteratively rule out possibilities until they converge on believing the same thing.
That seems to translate to the statement:
Whenever E, A1 knows that A2 knows that E.
which is stronger than just:
In the current state w, A1 knows that A2 knows that E.
Unless your P is my K in which case it translates to “E is the whole space” because all x are in K(the whole space).
This requires knowing more than what we mean when we say “A1 knows that A2 knows E”.
Aumann’s theorem is based on common knowledge, which is the very strong statement that A1 knows E, and A2 knows that, and A1 knows that, and so on.
However it is easy to see where this can come from. For instance, if I say “I think that the sky is blue” then it’s essentially common knowledge that I said “I think that the sky is blue”
Is that the source of your confusion?
For instance, if the situation is that P1({w,v,u})={w,v}, P2({w,v,u})={w,u}, then P1 and P2 can use their common knowledge to conclude w, and thus agree.
You have P1 and P2 taking big things to small things which means that they are K.
But if you consider conditions where P1, P2, and E contain more than 3 different states between them, you can find situations that have multiple possible solutions, which the agents cannot choose between; and so cannot converge.
However they will be able to agree that it is one of those states. Moreover neither of them will have any greater information than that it’s one of those states. Argument occurs when I believe “A, not B” and you believe “B, not A”, not if we both believe “A or B”.
Part 1 seems to have little to do with how I remember the theorem. Here is the abstract of Aumann’s paper.
Your 2 implies the following claim:
Agent 1 knows that agent 2 knows E if and only if agent 2 knows that agent 1 knows E.
This claim is obviously false.
Here is another corollary of your definition:
Suppose P1(w)={w,v}, P2(w)={w,u}. Then at w, I know E={w,v,u}, but at v, I do not know E! So I can distinguish between w and v by checking my knowledge, even though I cannot distinguish between w and v!
Part 3 is correct. Indeed, common knowledge of honesty is a requirement.
How does it imply that? (It well might, within the context of the agreement theorem. My recollection is that you assume from the start that A1 and A2 have common knowledge of E.)
Why? If knowledge means “justified true belief”, then for agent 1 to know that agent 2 know E, agent 1 must also know E, and vice-versa. This doesn’t prove the claim that you say I am making, but goes most of the way towards proving it.
This is true, except that P1 and P2 should range over events, not over world states. This is a step that the theorem relies on. Are you claiming that this is false?
Terms: P1 means what Aumann calls P-superscript1; N1 means what Aumann calls cursive-P superscript 1. P1(E) = {w,v} means that, after observing event E, A1 knows that the world is in one of the states {w, v}. N1 is the set that describes the range of P1. E is an event, meaning a set of possible world states. Aumann doesn’t define what an ‘event’ is, other than implicitly in how he uses the variable E, so I hope I’m getting that right.
I’m constructing this partly from memory—sorry, this is a complex proof, and Aumann’s paper is skimpy on definitions, several of which (like “meet” and “join”) are left undefined and hard to find defined anywhere else even with Google. I really can’t do this justice without more free time than I have in the next several months.
What I think Aumann is saying is that, if A1 knows E, and knows that A2 knows E, then for every state x in P1(E), for every event D such that x is in P2(D) , P2(D) is a subset of E. Saying this allows Aumann to go on and show that A1 and A2 can iteratively rule out possibilities until they converge on believing the same thing.
This requires knowing more than what we mean when we say “A1 knows that A2 knows E”. When we say that, we mean that A1 knows the world is in one of the states in E, and knows that A2 knows the world is in one of the states in E. But it is possible that there is some state x, that the world is not in, but that is a member of E and of P1(E), but not of P2(E).
My recollection is that this is the problem: If you only consider conditions involving 3 possible world states, like the w, u, and v in the above example, then you can show that these things are equivalent. The agents can always use their mutual knowledge to iteratively eliminate possible states until they agree. For instance, if the situation is that P1({w,v,u})={w,v}, P2({w,v,u})={w,u}, then P1 and P2 can use their common knowledge to conclude w, and thus agree. But if you consider conditions where P1, P2, and E contain more than 3 different states between them, you can find situations that have multiple possible solutions, which the agents cannot choose between; and so cannot converge.
The definition you gave was symmetric. If I misread it, my apologies.
True, but it’s impossible to go the rest of the way. If you see a dog and I see both you and the dog through a one-way mirror, then I know that you know that there’s a dog there but you don’t know that I know that there is a dog.
I am having trouble matching up your notation with the notation I’m used to.
There are two operations, which I am used to calling P and K. They also have a number attached to them.
P takes sets to bigger sets or else to themselves. P1({w}) is what A1 thinks is possible when w is true. P1(S) for any set S is what A1 might think is possible given that something in S is true.
K takes sets to smaller sets or else to themselves. K1(S) is the set of possible states of the world where A1 knows that S is true.
That seems to translate to the statement:
Whenever E, A1 knows that A2 knows that E.
which is stronger than just:
In the current state w, A1 knows that A2 knows that E.
Unless your P is my K in which case it translates to “E is the whole space” because all x are in K(the whole space).
Aumann’s theorem is based on common knowledge, which is the very strong statement that A1 knows E, and A2 knows that, and A1 knows that, and so on.
However it is easy to see where this can come from. For instance, if I say “I think that the sky is blue” then it’s essentially common knowledge that I said “I think that the sky is blue”
Is that the source of your confusion?
You have P1 and P2 taking big things to small things which means that they are K.
However they will be able to agree that it is one of those states. Moreover neither of them will have any greater information than that it’s one of those states. Argument occurs when I believe “A, not B” and you believe “B, not A”, not if we both believe “A or B”.