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"A strategy cannot be plausibly chosen by a rational player if and only if it is never a best response."

I understand the logic behind neglecting the strategies that are strictly dominated. But why can't there exist a strategy (which is never a best response) which supports a nash equilibrium mixed strategy ?

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Because in a best response mixed strategy, every component pure strategy is again a best response strategy. And in fact, the converse is true too. This is a basic result proved by Nash himself in the early days. A version of the proof is given under proposition 3.1 of this paper on algorithmic game theory.

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The answer related to strict domination by mixed strategies. The following can be proved via Linear Programming duality:

Theorem: In a bimatrix game, the following two statements are equivalent:

  • Pure strategy $i$ of player I is the unique best response to a suitable mixed strategy $q$ of player II.

  • Pure strategy $i$ of player I is not strictly dominated by a mixed strategy.

Proof sketch: I am not going to write out the proof here, but an approach is to write down, in terms of linear equalities, what it means that row $i$ to get expected payoff $u$; next, that all other rows $k$ have smaller payoff than $u$, by introducing a variable that denotes the difference - that variable is then maximized in the linear program; and finally, that the variables describing the mixed strategy $q$ of player II are indeed a probability distribution. Then take the dual of the linear program, and use the fact that the dual has the same optimal objective function value as the primal (in particular when the primal has a positive optimal objective function value).

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