A Non-cooperative Equilibrium for Supergames: A Correction

The purpose of this note is to correct an error in my paper [1]. Under the assumptions of the paper, Proposition 3 is not, in general, true. The point at which the proof goes awry is in the use of the mapping Ω. Ω is treated in the paper as if it were a function (i.e. a point to point mapping); whereas it is in fact a correspondence (a point to set mapping). Ω maps points of the unit simplex into itself, using a subset of the Pareto optimal set to obtain the simplex. A given point in the payoff space may be the image of more than one point in the strategy space. Each strategy, s, which maps into a given point in the Pareto optimal set (i.e. which has a given I associated with it in the simplex), can map into a distinct δ in the simplex. Thus the Brouwer theorem cannot be used. While Ω is surely upper semi-continuous, it need not have convex image sets, ruling out use of the Kakutani fixed point theorem. Furthermore, it need not be lower semi-continuous, ruling out another route to the use of the Brouwer theorem. That route is to use the results of E. Michael [2], which establish that if Ω is lower semi-continuous, then it has a selection which is a continuous function from the unit simplex into itself. The upshot is that the axioms of [1] must be somehow strengthened in one of the ways suggested by the preceding paragraph in order to make the proposition true. One way would be to strengthen A3 so as to make the payoff functions quasi-concave on S. Then the images of Ω would be convex and the Kakutani theorem could be applied. An alternative, which I prefer, is to strengthen A5. Let T = {s | s ∊ S and τ(s) ∊ H*}. T is the subset of strategies in S which map onto the Pareto optimal set, H*. Thus τ−1 is a mapping from H onto S, in general, a correspondence; and, when restricted to H* it is a correspondence from H* onto T. The stronger A5 is: A5′ H* is concave and the mapping τ−1 from H* on to T is lower semi-continuous. A5′ would, of course, be satisfied if the mapping from T to H* were one-to-one.In that case, Ω would be a continuous function and the Brouwer theorem could be directly applied. As A5′ stands, Ω must have a continuous selection to which the latter theorem could be applied. I have seen no way to prove Proposition 3 from the assumptions in [1], nor have I found a more satisfactory method of strengthening them. The only alternatives not mentioned above of which I am aware are essentially minor variations of the two methods.