Question

Suppose that ?* is a mixed ESS, and let a be an action to which ?* assigns positive probability. Show that if ?* assigns positive probability to an action different from a, then (a, a) is not a pure strategy Nash equilibrium, and hence a is not an ESS.

Answer 1

A dominated strategy is one where the outcome for the player is worse off under all possible situations no matter what the other person does. By Nash's theorem, a mixed strategy is one that assigns probabilities of each strategy being played. A mixed strategy can be strictly dominated, even if none of the pure strategies to which it assigns positive probabilities are strictly dominated.Alternatively, a mixed strategy which dominates a dominated pure strategy can itself be dominated by another mixed strategy with different probability allocations on the pure strategies. So we can show this as follows: The strategy set for a player i is given by S(i,n-i), which depends on the strategy choices of all the other n-i players in the game. Given the strategy choices of the remanining n-i players, whose strategy set is given S(n-i,i). S(i,n-i) is said to be a dominant stategy if S(i,n-i)>S(all i, n-i) given the strategy choices of the other players. Its the best reaction given the strategy choices of the other players. Moreover S(i*,(n-i)*) will be a Nash Equilibrium if S(i*,(n-i)*) does not cause any deviation incentive on the part of any of the players in the game. Now if a mixed strategy S(p,i) (where p is the probaility of choosing strategy i) places a probability p>0 on a dominated startegy S(i^), then in computing the payoffs for that mixed strategy choice, a positive probability will be placed on the dominated pure strategy S(i^). Given this we will always have a mixed strategy S(i%) which places p=0 on the dominated pure strategy S(i^) which results in a higher payoff. So in such a case the S(i%){with p=0 for strategy i}>S(i^){with p>0 for i}, as the payoffs from a dominated strategy will always be less and so will be dominated by a mixed strategy which places 0 probability on such a dominated pure strategy. The computing of the payoffs in playing S(i^) must be borne in mind and this will be lesser for a dominated strategy.