Cheap talk part 3
I'm going to plunge straight into this. Parts one and two are also available if you are lost.
[ed: as you can see, I end up failing to prove what I meant to prove... also, blogger keeps eating my post.]
Suppose that c = (1-a)/n where n is an integer and 0 <= a < c. In other words, there are n intervals around the unit circle, plus a remainder of a.
(1) Suppose that n is even, and that Senders of type t send a message m = t* =
t modulo c, when t* < a. Now it's a best response for Receiver to randomize equally over possible values of t in response to this message. Proof:
Let's call the possible values of t, t
0 to t
n where t
i = ci + t*. Take any arbitrary t
i. Choose an interval H = (t
i, t
i+1/2] or (t
i, t
i-1/2], such that H does not contain the zero point, and therefore does not contain both t
0 and t
n. (For example, from t
0, go clockwise and from t
n go anticlockwise.)
The interval H now contains n/2 possible values of t. Proof: ts are evenly spaced along the interval, c apart, with the first one at t
i+c or t
i-c (as the interval is open at ti itself). Temporarily normalizing t
i to zero, the interval (0, 1/2] or (0, -1/2] is of length 1/2 and, as 1 = nc+a, 1/2 = (n/2)c + a/2 where 0 <= a/2 < c. Moving from t
i into H therefore moves you towards n/2 equally likely values of t, and away from n/2+1 equally likely values of t. As we are assuming no risk-aversion (Receiver's utility is linear in distance from t), moving away from t
i therefore reduces utility. Similarly, moving away in the opposite direction is moving into an interval of size 1/2, excluding t
i itself, which also must contain the remaining n/2 possible values of t from the set t
j, j ^= i, and away from n/2+1 equally likely values of t.
(Apologies for the HTML notation: ^= means "does not equal".)
Therefore, any a ^= t
i, for all i, is strictly dominated by some pure strategy a = t
i for some i, and any mixed strategies containing anything other than the t
is is strictly dominated by a mixed strategy containing only t
is.
Furthermore, expected distance from t is the same at any t
i (leaving the proof for now, I am fairly sure) and thus the mixed strategy choosing t
i with probability 1/(n+1) is a best strategy.
(2) Given this strategy, Sender's strategy is a best response. Proof: for t = ti, i e {1,2,...,n-1}, the proof is identical to that given above as Sender's ideal point is just the same as to Receiver's ideal point when t=t
i+1. When t = t
n, ...
ah. No wait. When t = t
n Sender's ideal point will be t+c which is strictly greater than t
0 = t+a (we are crossing the zero point). If so, this will surely generate an incentive to deceive Receiver by claiming to be a different type. Blast. Perhaps we can remove the interval [nc, 1) from the set of types that sends an informative message? Well, this is obviously not over yet. More pointless fun awaits.
