I'm on Linux, as if you didn't know. Bah. I'm sick of having an operating system built by volunteers. I want a professional product that just works. My idealism has run out.
Now, on to more amusing things. That cheap-talk game. (You don't find abstract game theory interesting? Shame on you.)
Although it is true that there is no equilibrium in which types within different intervals along the circle play different messages, there may be an equilibrium with an infinity of messages. For example, suppose that c = 1/3 (recall that c is the difference between Sender's and Receiver's ideal points on the circle). Then, there is an equilibrium in which the message space M is [0, 1/3) and Senders send a message equal to their type, modulo 1/3. For example, senders with types 1/6, 1/2 and 5/6 would all send m = 1/6.
Receiver's best response is then any mixed strategy having as its support only actions with the possible values of Sender's type t. For example, if m = 1/6, Receiver can play any combination of 1/6, 1/2 and 5/6. The expected distance from sender's type of any one of these actions is (1/3 * 0 + 1/3*1/3 + 1/3*1/3) = 2/9. The expected distance from sender's type of any other action a is greater than this. The simplest way to see this is to imagine the three possible values of t. By moving Receiver's action a away from one of these values, two of the values get farther away, while one of them gets closer:
Now we consider Sender's best response to Receiver's strategy. First note that if Receiver plays, say "2/3" with probability 1 on receiving m = 0, but randomizes over all 3 possible types for e.g. m = 1/10, then when t = 1/3+1/10 it will be more sensible to send m = 0 (guaranteeing a response close to her bliss point) than the assumed strategy of m = 1/10. So we focus on the case when Receiver randomizes with equal probability over all 3 types, for all messages m in [0,1/3).
In this case, the logic above also holds true for Sender's utility. By sending a message reflecting her true type, she elicits a best response of one of her three possible types. The logic for Sender is just as for Receiver: this has a one in three chance of being at her bliss point (at t+c, where t+c is one of the three possible types) and a 2/3 chance of being 1/3 away; any alternative message would generate a greater expected distance from t+c. Thus, honesty is a best response to Receiver's strategy, and we have our Bayesian Nash equilibrium.
The same logic holds whenever c = 1/n, and n is odd. Then, any message m in [0,1/n) where m corresponds to t modulo 1/n will generate a best response of choosing one of the possible types: any move away from a possible type moves (n+1)/2 of the possible types away from you, and (n-1) towards you, all by the same distance.
On the other hand, when n is even, the equilibrium remains technically but is much less plausible. Now all Receiver's strategies are equally good, because moving action a away from a possible type brings n/2 possible types closer while n/2 move farther away by the same distance. The case with n=4 is shown below:
The equilibrium remains, technically, because choosing a equal to some possible value of t is still a best response. But there would be no incentive, in the long term, for Receiver to play this equilibrium. When n is odd, there are plenty of best responses that don't lead to equilibrium (such as mixed strategies where Receiver doesn't always randomize equally between points) but in the long term one might expect Receiver to play "nicely" in order to keep the flow of useful information going. (I don't know if anyone has toyed with repeated cheap-talk games.)
Before moving on to the general case, where c does not "fit neatly" into the circle, it is worth pointing out that here, as c decreases, the potential equilibria are getting less informative rather than more, in the following two senses:
- The message space for e.g. c = 1/5 is [0,1/5), which is a proper subset of the message space for c = 1/3, [0,1/3).
- The expected distance from Receiver's (and Sender's) ideal point, when c=1/n, is (n^2-1)/4(n^2). This increases as n increases, with a limit of 1/4. For example, expected distance when c = 1/3 is just 2/9, when c = 1/5 it is 6/25.
That's all for now. Next post, I'll have a shot at the general case. I suspect this will also have an equilibrium of some kind.