This story draft by @escholar has not been reviewed by an editor, YET.
Authors:
(1) Nicholas A. G. Johnson ([email protected]);
(2) Theo Diamandis ([email protected]);
(3) Alex Evans ([email protected]);
(4) Henry de Valence ([email protected]);
(5) Guillermo Angeris ([email protected]).
1.1 Symmetric pure strict equilibrium
2 Batched decentralized exchanges
There is a strict, pure equilibrium where all players have equal strategies, given by x = (q/n)1 where q is the optimizer of the following problem:
with variable q ∈ R. We will show some properties of this result first and then show that the pure strategy x = (q/n)1 is, indeed, an equilibrium.
Discussion. It may appear that the condition placed on f is very strong, but in fact, any f not satisfying the above condition has only trivial (or no) equilibria. In particular, since f is concave, if f does not satisfy the above condition, either (a) f is strictly positive everywhere except at f(0) = 0, (b) f is strictly negative everywhere except at f(0) = 0, or (c) f = 0. In the first case, there is no equilibrium as any player can improve their payoff by increasing their strategy. In the second case, any player who plays a nonzero strategy receives negative payoff (whereas playing the zero strategy would give 0 payoff). While, in the third case, any strategy is an equilibrium.
Equilibrium properties. The collection of strategies x = (q/n)1 is clearly pure and symmetric. To see that x = (q/n)1 is a strict equilibrium, note that the best response for any player i, when every other player plays strategy q/n is:
Next, note that q > 0 must satisfy the first order optimality conditions of (4):
This paper is available on arxiv under CC BY 4.0 DEED license.