Thursday, 8 September 2005

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, t0 to tn where ti = ci + t*. Take any arbitrary ti. Choose an interval H = (ti, ti+1/2] or (ti, ti-1/2], such that H does not contain the zero point, and therefore does not contain both t0 and tn. (For example, from t0, go clockwise and from tn 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 ti+c or ti-c (as the interval is open at ti itself). Temporarily normalizing ti 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 ti 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 ti therefore reduces utility. Similarly, moving away in the opposite direction is moving into an interval of size 1/2, excluding ti itself, which also must contain the remaining n/2 possible values of t from the set tj, 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 ^= ti, for all i, is strictly dominated by some pure strategy a = ti for some i, and any mixed strategies containing anything other than the tis is strictly dominated by a mixed strategy containing only tis.

Furthermore, expected distance from t is the same at any ti (leaving the proof for now, I am fairly sure) and thus the mixed strategy choosing ti 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=ti+1. When t = tn, ...

ah. No wait. When t = tn Sender's ideal point will be t+c which is strictly greater than t0 = 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.