Mixed strategy nash equilbrium

Question

In a mixed strategy Nash equilibrium it is always the case that:

a) for each player, each pure strategy that is played with negative probability yields the same expected payoff as the equilibrium mixed strategy itself.

b) for each player, each pure strategy yields the same expected payoff as the equilibrium mixed strategy itself.

c) Each player strictly prefers their mixed strategy to any pure strategy, given that others conform to the equilibrium.

d) For each player, each pure strategy that is played with positive probability yields the same expected payoff as the equilibrium mixed strategy itself.

e) none of the above

Answer 1

• a) makes no sense (you can't put negative probability on an action)
• b) is incorrect. Pure actions that are not played with positive probability in the mix may have arbitrarily bad payoffs since those payoffs never result in any case.
• c) is incorrect. Suppose you did prefer a mix over $a_1,a_2$ (with strictly positive weights on each) to a given pure strategy, $a_1$. Then it must be that the mix gives a higher payoff than $a_1$. But then $a_2$ must give higher payoffs than the mix, as the mix is just a weighted average of $a_1$ and $a_2$.

• d) is correct. To see why, suppose it were not true. Then there would be some pure strategy $a^*$ that you were putting positive probability on in the mix that either gave higher or lower expected payoff than the mix. In the first case (higher), you would then not be willing to put positive weight on other actions -- you would just play $a^*$. In the latter case (lower), you would not be willing to put positive weight on $a^*$.

• e) is incorrect because d) is true.

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