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I.S Department, Brooklyn College of CUNY, Brooklyn, NY, 11210. 2C. .... It can be shown nonetheless, that when n individuals communicate probabilities to each ...
Communication, Consensus and Knowledge

Rohit Parikh1,2,3 and Paul Krasucki1,2,4

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C.I.S Department, Brooklyn College of CUNY, Brooklyn, NY, 11210 C.I.S. Department, CUNY Graduate Center, 33 W 42nd St, New York, NY, 10036 3 Department of Mathematics, CUNY Graduate Center, 33 W 42nd St, New York, NY, 10036 4 Current address: Department of Computer Science, Rutgers University, Camden College of Arts and Sciences, Camden, NJ 08102 2

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Running head: Communication and Consensus

Author to whom proofs should be sent: Prof. Rohit Parikh Computer Science, Box 220 City University Graduate Center 33 W 42nd Street New York, NY, 10036-8099

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B1-Parikh, Rohit J. and Krasucki, Paul J.- B2-Communication, Consensus and Knowledge C2-We continue the work of Aumann [1], Geanakoplos and Polemarchakis [4], Cave [3] and Bacharach [2] on common knowledge and consensus, extending the results of [4], [3] and [2] to the case where the number n of communicants is greater than two, but communication is in pairs. When n > 2, then communication in pairs does not lead to common knowledge for the group. We show that, nevertheless, conditional probabilities will converge in any fair protocol for communication, i.e., when none of the participants is blocked from communication. We show that the Cave-Bacharach generalisation of the results of [4] and [1] does not hold for pairwise communication when n > 2, and hence any proof for this case must use methods different from those of [1] and [4]. In particular, our proof uses a convexity condition that is obeyed by conditional probabilities, but is not implied by Cave’s union consistency principle. B6-J. Econ. Theory B7 B8 C4 B4-CUNY Graduate Center (Computer Science department), 33 W 42nd St, New York, NY, 10036; Brooklyn College of CUNY (CIS department), Brooklyn, NY, 11210

Journal of Economic Literature Classification Numbers: 026, 213

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Abstract: We consider the problem of arriving at a consensus in the situation where a group of traders communicate in pairs. We generalise and extend previous results of Geanakoplos and Polemarchakis, Cave and Bacharach. Introduction: In 1976 Robert Aumann proved the following result. Suppose that a group of n individuals share a common prior distribution, but have different private information partitions. If there is common knowledge in the group of the posterior probabilities of some proposition A, then these probabilities must all be equal. Aumann did not address the question of why these posteriors should be common knowledge at all. This question was addressed by Geanakoplos and Polemarchakis who proved that for a group of 2 individuals, even if these posteriors are not common knowledge, but if the members of the group tell each other their respective probabilities, then common knowledge will transpire and the probabilities will all become equal. Subsequently, Cave [3] and Bacharach [2] proved that the results of Aumann, Geanakoplos and Polemarchakis will hold not only for conditional probabilities, but also for any function f from sets of worlds to some domain D, provided only that f has the property that if X, Y are disjoint and f (X) = f (Y ), then f (X) = f (X ∪ Y ). This property is called union consistency by Cave and the sure thing principle by Bacharach. All the authors mentioned above deal with the case n = 2 except for Cave who considers arbitrary finite n but in the context where all members of the group hear all the statements that any of them makes. None of these authors raises the following general questions. 1) What are the states of knowledge created in a group of individuals when communication takes place? and 2) what happens in the case n > 2 when communication is not to the whole group, as in an auction, but pairwise, as happens commonly in commercial transactions? We illustrate the first point by an example. Suppose I telephone you and give you some information, then the fact I have told you is common knowledge between us. If, however, I write you a letter containing the same information, then when you receive the letter, you know the fact (and that you have received the letter) but I do not know that you know, since I do not know that you have received my letter. If you acknowledge the letter and I receive your acknowledgement then I know that you know, but, again, you do not know that I know that you know. In fact, common knowledge cannot be created by letters provided that they are undated and as long as there is some uncertainty in the time of arrival. See [6]. Nonetheless, the fact remains that the entire correspondence, except the very last step, is known to both of us. Moreover, even the last step is known to the one whose turn it is to respond. Thus it can be shown, using the Cave-Bacharach argument that eventual consensus will necessarily transpire between two people, regardless of whether they communicate by phone or by letter. The situation changes if we have more than two people. Of course, they could communicate in a setting where all of them are together. An auction - the case considered by Cave - achieves just that. But of course not all commercial transactions have the setting of auctions. The details of many sales are known only to the parties to the transactions, although each party may participate in many transactions and with many different persons. Thus we can ask what happens when transactions in a group of individuals take place in pairs. What happens in this case is that no 4

common knowledge is arrived at by the group: i.e., no new fact becomes common knowledge as a result of communication which was not common knowledge to begin with, see [6], [7]. The reason here is that an individual knows only that portion of the dialogue in which he or she participates. It can be shown nonetheless, that when n individuals communicate probabilities to each other, two at a time, and every individual is repeatedly involved in communication, then their probabilities do become equal. We show that this result generalises to arbitrary functions f which satisfy a certain convexity condition. This condition is properly stronger than Cave’s “union consistency” property, and indeed, union consistency is not strong enough for n > 2. For there does exist a function f , with the union consistency property such that repeated communication of values of f from individual 1 to 2 to 3 to 1 ..., fails to bring about consensus. This last fact shows that the proof of Geanakoplos and Polemarchakis, which uses the union consistency property of conditional probabilities (and only that), does not work for the case where n > 2. We also give a general definition of the formula to be used for updating, which tells us how an individual should revise his set of possible worlds when he receives a communication. Basic Notions and Results : We assume given a space W , the space of states (or of possible worlds) and n participants with finite partitions Pi of W. Let Pi (x) be the equivalence class of world x in partition Pi . Let x ≡ y mean that for all i, Pi (x) = Pi (y). We will say that a subset X of W is closed if x in X and x ≡ y imply that y is also in X. If P + is the coarsest common refinement (join) of the Pi , then X is closed iff it is a union of P + equivalence classes. Note that since the Pi are finite, so is P + . Similarly, we will use the expression i-closed to describe a union of elements of the partition Pi . Note that an i-closed set is closed, but not necessarily vice versa. By a protocol P r we mean a pair of functions s(t), r(t) from the natural numbers (≥ 0) to the set {1,...,n}. Here t stands for time and s(t), r(t) are, respectively, the sender and the recipient of the communication which takes place at time t. Definition : Given a protocol (s(t), r(t)) consider the directed graph whose vertices are the participants {1,...,n} and such that there is an edge from i to j iff there are infinitely many t such that s(t) = i and r(t) = j. Then the protocol is fair if the graph above is strongly connected, i.e. if there is a path of directed edges which passes through every vertex at least once, returning to its origin. If the participants are observing a fair protocol, then not only will each participant be a recipient and a sender infinitely many times, but each participant will receive information from every other, possibly indirectly, infinitely many times. Let f be some function from the set of subsets of the state space W to some domain D (f is the function whose values are communicated). If f is intended to be a conditional probability, then we assume given a probability measure π on W and some event A contained in W . Then f (X) for X ⊆ W is just π(A|X) = π(A ∩ X)/π(X) where π(X) 6= 0

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More generally, let the domain D be the reals. We say that the function f is convex if for all subsets X,Y of W , if X and Y are disjoint, and X,Y are closed then there are positive reals a,b such that a + b = 1 and f (X ∪ Y ) = af (X) + bf (Y ). In other words, f (X ∪ Y ) lies in the open interval between f (X) and f (Y ). Note that we will get f (X1 ∪ X2 ∪ ...Xk ) = a1 f (X1 ) + ... + ak f (Xk ) provided that the Xi are pairwise disjoint closed sets and a1 , ...ak will be positive reals such that a1 + ... + ak = 1. Note also that if the f (Xi ) are all equal, then f (X1 ∪ ... ∪ Xk ) = f (X1 ). Hence convexity implies the union consistency property,5 at least for closed sets, the only subsets of W that we will need to be concerned with. Suppose now that P r = (s(t), r(t)) is some protocol. We define by induction on t the message m(x, t) sent at time t, and C(x, i, t), the set of possible states for i at time t, given that the real state is x. C(x, i, 0) = Pi (x) If i = r(t), then C(x, i, t + 1) = C(x, i, t) ∩ {y|m(y, t) = m(x, t)} where m(x, t) = f (C(x, s(t), t)). and otherwise, C(x, i, t + 1) = C(x, i, t). We let v(x, i, t) = f (C(x, i, t)) so that m(x, t) is just v(x, s(t), t). We assume that all the communications are instantaneous, but this assumption is not essential. Lemma 1: If x ≡ y then for all i, t, m(x, t) = m(y, t) and C(x, i, t) = C(y, i, t). Proof : Straightforward, by induction on t. Theorem 1: There is a t0 such that for all x, i, and all t, t0 > t0 , C(x, i, t) = C(x, i, t0 ). Hence for all x, i, v(x, i, t) has a limiting value v(x, i, ∞). Moreover, if f is convex, and the protocol is fair, then this limiting value does not depend on i, i.e. v(x, i, ∞)) = v(x, j, ∞)) for all i, j. Corollary: If W is a probability space such that all nonempty elements of P + have nonzero probability, and A is some event, then the limiting probability of A is the same for all i, j, provided that the protocol is fair. Proof of corollary: We note that in this case P (u, i, t), the probability of A for i at time t, where the real world is u, is π(A|C(u, i, t)) and π(A|X) is a convex function of X. We now turn to the proof of the theorem itself. For all x, i, C(x, i, t) is a decreasing function of t. Also it is a nonempty union of P + equivalence classes. Since P + is finite, it is eventually constant. Let t(x, i) be the least t such that for t0 > t, the value of C(x, i, t0 ) is the same. Then t(x, i) depends only on the P + equivalence class of x. Since P + has finitely many equivalence classes, and there are finitely many i, there is a T such that C(x, i, t) is constant for all t0 > T regardless of x, i. T does of course depend on the choice of the protocol (s(t), r(t)). It follows also that m(x, t) = f (C(x, s(t), t)) depends only on x and s(t) if t > T . We write C(x, i, ∞) to indicate the limiting value of C(x, i, t). 5

f is union consistent if whenever sets X and Y are disjoint and f (X) = f (Y ), then f (X) = f (Y ) = f (X ∪ Y ).

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To proceed with the proof of the theorem, we assume for simplicity that n = 3, and that for all t, s(t) = t mod 3 and r(t) = t + 1 mod 3. This protocol is evidently fair (we will call this protocol a “round-robin” protocol). The reader should note that our arguments will also apply to any n and to any fair protocol. We will say that p is a possible limiting value for i if there is an x such that p = f ((C(x, i, ∞)). Let the possible limiting values for individual 1 be p1 , ..., pk with p1 < p2 < ... < pk . Similarly (q1 , ..., qr ) are the values for 2 (in order) and (r1 , ..., rm ) are the limiting values for 3. We want to show that for all x, the limiting values generated by x do not depend on i. Let p be a possible limiting value for 1 and let E(p) be the set of all x such that f ((C(x, 1, ∞)) = p. Then E(p) is a disjoint union of those sets C(x, 1, ∞) such that f (C(x, 1, ∞)) = p. For if y is in C(x, 1, ∞), then y is compatible with all the information received by 1 when the actual world is x. Hence y would have generated the same information for 1 that x did and so C(y, 1, ∞) = C(x, 1, ∞). Thus f (E(p)) = p by the union consistency principle, which, as we remarked, is implied by convexity. Call p and q compatible if there is an x such that f (C(x, 1, ∞)) = p and f (C(x, 2, ∞)) = q. We now note that for such an x, for large t with t = 1 mod 3, 1 sends the value p to 2 but C(x, 2, t + 1) = C(x, 2, t) ∩ {y|f (C(y, 1, t)) = p} = C(x, 2, t) = C(x, 2, ∞) and hence {y|f (C(y, 1, t) = p} must contain C(x, 2, ∞). Thus E(p) contains C(x, 2, ∞). Thus E(p) is a union of sets of the form C(u, 2, ∞) and hence, by convexity, p must be an average of those values qi which are compatible with it. Consider p1 , the smallest of the p’s. The set of values q compatible with it must contain at least one element ≤ p1 and hence q1 ≤ p1 . By symmetry r1 ≤ q1 and p1 ≤ r1 . Hence p1 = q1 = r1 . But now note that only q1 can be compatible with p1 since the other qi are strictly greater and the average of q1 with any other qi would be strictly greater than q1 . Similarly, only r1 is compatible with q1 and only p1 with r1 . Thus, after time T , if any of 1,2,3 is sending p1 , they all are, and if one of them isn’t, then none of them is. Thus in particular, p2 can be compatible only with qi for i ≥ 2, q2 with ri for i ≥ 2 and r2 with pi for i ≥ 2. We can now repeat our argument and show that p2 = q2 = r2 and any of them forces the others. By induction we get, for all i, pi = qi = ri and all force each other. What happens if n > 3 or if we are using a fair protocol other than round robin? We note now that our argument only used the fact that we are able to find a chain t1 < t2 < ...., < tp , all greater than T and such that (a) s(t1 ) = 1 (b) the sender at tj+1 is the recipient at tj and (c) the chain passes through all participants, finally returning to 1. (1 may occur in between also). But this is implied by the fact that the protocol is fair. This proves theorem 1. Example 1: Let’s consider the case of statistical inference by Bayesian individuals. Suppose there are two dice, d1 is fair but the other, d2 is fair only between { 1,2,3,4,5}, never showing 6. d1 is worth 1 dollar and d2 is worth a hundred dollars. The two dice are put in a box which is 7

shaken, and then one of the dice is taken out and tossed, showing a 2. Of two gamblers A,B, A is told that the result was under 4, while B is told that it is even. A now evaluates the probability that the die is d2 as 6/11, and hence his expected value for the die will be $55. However, B thinks that the probability that the die is d2 is only 4/9 and his expected value for the offered die will be $45. Thus if the die is offered at $50, clearly A would buy it, and B would not. What will happen if they communicate their values of the die to each other? Will these values converge? To put this problem abstractly, Suppose that the random variable X takes values in a set W . X is distributed according to one of a finite number of probability distributions W1 , W2 , ..., Wk . We also have n individuals each of whom has his information partition Pi of the set W . The individuals share a utility function u:{ W1 , W2 , ..., Wk } → R, say u(Wi )=ui . The result w of an experiment performed (to measure X) is (partially) communicated to each person by specifying Pi (w). The individuals communicate values of f : 2W → R where for A ⊆ W , f (A) = ki=1 p(Wi |A)·ui . In order for f to be defined for all i-closed subsets of W , we need to assume that for A ∈ P + , p(A) > 0. It is easy to check then that for disjoint, non-empty subsets A1 ,A2 of W, if p(A1 ∪ A2 ) 6= 0, then f (A1 ∪ A2 ) = f (A1 ) · (p(A1 )/p(A1 ∪ A2 )) + f (A2 ) · (p(A2 )/p(A1 ∪ A2 )). Hence f is convex so that when values of f are exchanged, then by theorem 1, consensus will be reached. The fact that there are only two gamblers is of course not essential. P

Example 2: We now give an example to show that Cave’s union consistency property does not imply consensus for n > 2. Let the space W equal the set {1,2,3,4,5,6,7,8}. The function f satisfying the union consistency property takes integer values and is defined as follows: let the subsets of W be numbered X1 ,X2 ,... and let num(Xi ) be i. Now we let: f ({1, 2}) = f ({3, 4}) = f ({1, 2, 3, 4}) = 1 f ({5, 6}) = f ({7, 8}) = f ({5, 6, 7, 8}) = 2 f ({1, 3}) = f ({5, 7}) = f ({1, 3, 5, 7}) = 3 f ({2, 4}) = f ({6, 8}) = f ({2, 4, 6, 8}) = 4 f ({1, 5}) = f ({2, 6}) = f ({1, 2, 5, 6}) = 5 f ({3, 7}) = f ({4, 8}) = f ({3, 4, 7, 8}) = 6 and for all other subsets X of W , f (X) = num(X) + 6. This ensures that f has the union consistency property. Now the participants are A,B,C and their partitions PA , PB , PC are defined by PA = {{1, 2}, {3, 4}, {5, 6}, {7, 8}} PB = {{1, 3}, {2, 4}, {5, 7}, {6, 8}} PC = {{1, 5}, {2, 6}, {3, 7}, {4, 8}}. The real world is 1. Now notice that A sends a 1 to B so B knows that the real world is in {1,2,3,4}. But he learns nothing since he knew anyway that it was in {1,3}. Similarly, B sends a 3 to C who learns nothing thereby and C sends a 5 to A who also learns nothing. Since the three participants have learned nothing in the first round, they will repeat their signals ad infinitum and consensus will not take place.

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The following, rather colorful example is due to Aumann. Suppose there are three detectives, Adams, Brown and Cox. In detective school, they learn that adults are more reliable than teenagers (maturity), teenagers than children (accuracy), and children than adults (disingenuousness). There are three material witnesses in the case: an adult, a teenager, and a child. It is common knowledge that Adams interviews the adult and the teenager, Brown the child and the adult, and Cox the teenager and the child. Predictably, Adams believes the adult, Brown the child, and Cox the teenager. Adams now tells Brown whom he is going to arrest, which Brown knew, since he also interviewed the adult, but predictably, Brown prefers the testimony of the child. He, in turn, tells Cox who he is going to arrest, and so on. Nobody gains any information from what he is told, and nobody’s mind is changed by what he is told. If, however, there were only two detectives, then Cave’s union consistency theorem would apply, and they could not differ on whom they were going to arrest. We have shown above that the union consistency property doesn’t lead to a consensus in case of pairwise communication in a group of more then 2 people. Still, it is a natural question whether we can weaken the convexity condition so that consensus will still be guaranteed. We will say that f is weakly convex if for disjoint, closed A,B, f (A ∪ B) lies in the closed interval between f (A) and f (B). The difference now is that even when f (A) and f (B) are different, f (A∪B) may equal one of f (A) and f (B). With simple convexity, this would be prohibited. Clearly weak convexity implies the union consistency property and is implied by convexity. Example 3: We give an example of a weakly convex f such that when values of f are communicated by 4 persons using the “round-robin” protocol then no consensus is reached. Let the space W be the set { a,b,c,d,e} . Let the function f : 2W → { 0, 1} be defined as follows: f (a) = f (b) = 1 f (c) = f (d) = f (e) = 0 f (a ∪ c) = f (a ∪ d) = f (a ∪ e) = f (b ∪ c) = f (b ∪ d) = f (b ∪ e) = f (a ∪ b) = 1. f (c ∪ d) = f (c ∪ e) = f (d ∪ e) = 0 f (a ∪ c ∪ d) = f (b ∪ d ∪ e) = f (c ∪ d ∪ e) = 0 f (a ∪ d ∪ e) = f (b ∪ c ∪ d) = f (a ∪ c ∪ e) = f (b ∪ c ∪ e) = f (a ∪ b ∪ c) = f (a ∪ b ∪ d) = f (a ∪ b ∪ e) = 1 f (a ∪ c ∪ d ∪ e) = f (b ∪ c ∪ d ∪ e) = 0 f (a ∪ b ∪ c ∪ d) = f (a ∪ b ∪ d ∪ e) = f (a ∪ b ∪ c ∪ e) = 1 f (a ∪ b ∪ c ∪ d ∪ e) = 1 It can be easily checked that f is weakly convex. Intuitively, say that A dominates B if f (A) and f (B) are different, A and B are disjoint and f (A ∪ B) equals f (A). Then in our example, either of a and b dominates over any one of c, d, e, and even over any two of c, d, e, except that c, d together dominate a, and d, e together dominate b. However a, b together dominate over any combination of c, d, e.

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We have 4 participants and their partitions are defined as follows: P1 = {{e}, {a, c, d}, {b}} P2 = {{c}, {a, d, e}, {b}} P3 = {{c}, {b, d, e}, {a}} P4 = {{e}, {b, c, d}, {a}} If the real world is d then 1 sends value 0 to 2, but 2 learns nothing since P2 (d) = {a, d, e} ⊆ {w|f (P1 (w)) = 0} = {a, c, d, e} and 2 sends value 1 to 3. Similarly 3 learns nothing upon receiving 1 from 2 and sends value 0 to 4, 4 learns nothing and sends 1 to 1 and 1 also doesn’t learn anything. This will now continue with no consensus ever being reached. We have shown that weak convexity does not work for n = 4 (and in fact for n ≥ 4, whereas the union consistency property is strong enough for n = 2. Thus the question arises what happens if n = 3. Theorem 2: If f is weakly convex and the protocol P r is fair then if 3 participants communicate values of f according to P r then consensus on the value of f must be reached. Proof : Without loss of generality we can restrict ourselves to the “round-robin” protocol. As in the proof of theorem 1 we can define sets of possible limiting values for 1,2,3. In order to prove our theorem observe that the following lemmas hold: Lemma 2: If there is a space W , finite partitions P1 ,...,Pn of W and a weakly convex function f s.t. for some fair protocol P r, the sets of possible limiting values are P 1 ,...,P n , then there exist finite partitions P10 ,..., Pn0 of W and protocol P r0 with the same graph as P r such that executing P r0 , we get the same set of limiting values, but no one gains any knowledge during the execution of P r0 . I.e. C(w, i, 1) = C(w, i, ∞) for all i,w. Proof: Simply take Pi0 (x) to be C(x, i, ∞) and let P r0 (t) = (s(t + T ), r(t + T )) where P r(t) = (s(t), r(t)) and T is such that stability (no further change in the C’s) is always reached by time T . In particular, if no consensus was reached in the first case, then in the second case also there will be no consensus, and moreover, no learning will ever take place. Lemma 3: Suppose there is a space W , a world w ∈ W and a weakly convex function f such that when w is the actual world, and values of f are communicated according to a fair protocol, then no consensus takes place. Then there is another weakly convex function f 0 , taking values in {0,1} such that no consensus is reached when values of f 0 are communicated according to the same protocol. To see that this lemma is true, assume by lemma 2, that the sets C(x, i, ∞) are the same as the sets Pi (x). Let x, i, j be such that i, j differ on the limiting value when the real world is x, and say v(x, i, ∞) = f (Pi (x)) < v(x, j, ∞) = f (Pj (x)). Let α be such that f (Pi (x)) < α < f (Pj (x)). and α is not a value of f . Now let, for all closed X: f 0 (X) = 0 if f (X) < α and f 0 (X) = 1 if f (X) > α. The reader can easily check that f 0 is weakly convex and consensus does not take place. In fact,

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since f 0 supplies less information than f , there can be no learning with f 0 either. We note that for 2-valued functions, weak convexity is equivalent to the union consistency property. The reason theorem 2 does not hold for the union consistency property is that lemma 2 does not hold, so that the reduction from the general case to the 2-valued case cannot be carried out. To prove the theorem, assume that the values of a weakly convex function f are communicated between 1,2,3 using the “round-robin” protocol and consensus is not reached. By lemmas 1 and 2 we can assume that f is { 0,1} -valued and no knowledge is gained during the execution of the protocol. Call a closed set X unstable iff for some persons i,j in a world w from X, i and j have different S values. Let U = {X|X is unstable}. U 6= ∅ since we assumed that consensus is not reached.We can now express U in two ways: (i) as the union of all C(w, i, ∞) s.t. f (C(w, i, ∞)) = 0 and f (C(w, i 1, ∞)) = 1 and (ii) as the union of all C(w, i, ∞) s.t. f (C(w, i, ∞)) = 1 and f (C(w, i 1, ∞)) = 0 (where x y = x − y mod 3). It is not hard to see that each union is disjoint. But then the first representation forces us to assign f (U ) = 0 while the second representation requires f (U ) = 1 so we have a contradiction. This proves theorem 2. Common Knowledge is not Necessary : If a function f is convex, or f is weakly convex and n ≤ 3 then by theorems 1 and 2 a consensus on the value of f in a fair protocol will eventually be reached, i.e. the values of f will become the same. We now point out that there need be no common knowledge in this situation. Suppose that individuals a,b,c etc. do not have watches, but at time t with t = 1 mod 3, a receives a signal from outside that causes him to send a value to b. Similarly with b and c. In that case, a does not know when b communicates with c, b does not know when c communicates with a, etc. The argument for convergence is not affected. However, we can now show that if some proposition P is common knowledge at time t+1, then it was common knowledge at time t. For if P is common knowledge, so is the fact that P is common knowledge. (This is a property of common knowledge). Suppose that t = 2 mod 3. Then it is not known to a that b communicated with c. If P is common knowledge at t + 1, then letting C(P ) denote the fact that P is common knowledge, C(P ) is true and so is Ka (C(P)), i.e. a knows C(P ). But a’s knowledge did not change between t and t + 1 so Ka (C(P )) was true at t, and hence so was C(P ) since a cannot know a false proposition. Applying backward induction on t, we see that C(P ) was true at t = 0, before communication started. However, the consensus value is not known in general before communication starts. Hence it was not common knowledge and never will be common knowledge, though it will be known to everyone [6]. This point is of some importance. It is shown in [7] that knowledge can occur at various levels, and while common knowledge is the highest of these levels, it would be wasteful to demand it when a lower level will do, and is attainable. It is important in a society that everyone know (at least approximately) the prices of commodities. However, it does not make sense to demand that everyone knows the details of every commercial transaction. 11

Acknowledgements: We thank R. Aumann, A. Chandra, H. Gaifman, J. Halpern, S. Lipschutz, A. Mate and H. Polemarchakis for fruitful discussions. In particular, the example with detectives was given to us by Aumann, example 3 is joint with Chandra, and example 1 is joint with Gaifman. References

1. R. Aumann, Agreeing to Disagree, Annals of Statistics, 4 (1976), 1236-1239. 2. M. Bacharach, Some Extensions of a Claim of Aumann in an Axiomatic Model of Knowledge, J. Economic Theory, 37 (1985), 167-190. 3. J.A.K. Cave, Learning to Agree, Economics Letters 12 (1983), 147-152. 4. M. Chandy and J. Misra, How Processes Learn, Proceedings of 4th ACM Conference on Principles of Distributed Computing (1985) pp 204-214. 5. J. Geanakoplos and H. Polemarchakis, We Can’t Disagree Forever, J. Economic Theory, 28 (1982), 192-200. 6. Halpern and Moses, Knowledge and Common Knowledge in a Distributed Environment, Proc. 3rd ACM Conf. on Principles of Distributed Computing (1984), 50-61. (To appear in Jour. Assoc. Comp. Mach.) 7. R. Parikh, Levels of Knowledge in Distributed Computing, IEEE Symposium on Logic in Computer Science, Boston 1986, 322-331. 8. R. Parikh and R. Ramanujam, Distributed Processes and the Logic of Knowledge, Logics of Programs, Springer Lecture Notes in Computer Science, 193, (1985), 256-268.

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List of symbols used



“belongs to” symbol



empty set symbol



union of two sets symbol

S

union of a family of sets



intersection of two sets symbol



inclusion symbol



“infinity” symbol



arrow (right)



equivalence relation symbol

6=

not equal



less than or equal



greater than or equal

π

greek letter pi

Σ

greek letter capital sigma

α

greek letter alpha

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