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I have trouble understanding this question,

the first question to my understanding is asking me that for a fixed p , (p,q) is nash equilibrium, prove that all (p,q) are convex.

and for the second, what I understand is that for (p,q) as nash equilibrium. prove that (p,q) is not convex.

If someone could reword the question for me , that would be much appreciated.

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  • $\begingroup$ Is $(p,q)$ the probability distribution players 1 and 2 put on their $m$ and $n$ actions in equilibrium? $\endgroup$ – Pburg Nov 13 '14 at 1:44
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I am assuming we are supposed to think of $(p,q)$ as the probability distribution on actions from two players?

Take the 2x2 pure coordination game as an example. Let pure strategies available to player 1 be $\{A,B\}$ and let the pure strategies available to player 2 be $\{a,b\}$. Then we attach payoffs $u_i(A,a)=u_i(B,b)=1$ for $i=1,2$ and utility is zero from any other outcome.

It should be obvious that $\{A,a\}$ is a Nash equilibrium and the same goes for $\{B,b\}$. Then degenerate mixed strategies $(0,0)$ and $(1,1)$ are in the set of Nash equilibria. Then, we can find a convex combination $(.3,.3)$ that is not a Nash equilibria. If player $1$ chooses $A$ strictly more than $50\%$ of the time, then player $2$ will choose $a$ $100\%$ of the time. So, $(.3,.3)$ can't be a Nash equilibrium. Therefore, the set described in b.) is not convex.

In part a.) we fix the strategy of one player. In this case, the other player's best response will give a convex distribution. For a player to mix over multiple actions, she must be indifferent between each action. So, she is indifferent between all probability distributions of those specified actions in the best response correspondence.

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