The set of Nash equilibrium is convex?

Question

In our class professor said

[0,0   1,0
 0,1   2,2]

this game has 2 NE. (0,0) and (2,2). Then he said "well this serves as an example that set of NE is not convex". Can anybody clarify this statement?

Answer 1

I guess he was comparing it to zero-sum games.

In zero-sum games, the set of optimal strategies is convex, so if $a$ is one optimal strategy and $b$ is another, choosing $a$ with probability $\alpha\in(0,1)$ and $b$ w.p. $1-\alpha$ is also optimal.

As you can see from the example, this is not true in non-zero sum games. A convex combination of equilibria is not an equilibrium by itself.

Note that the comparison is just, as in zero-sum games any pair of optimal strategies is a Nash equilibrium.

Answer 2

I would perhaps interpret this as a statement contrasting the set of Nash equilibria with the structure of the best-reply correspondences for finite normal-form games. An integral part of the proof of Nash's theorem is that for any agent $i$ and any tuple of mixed strategies of all the other agents $\sigma_{-i}$, the best-response for $i$, that is: $$ BR_i(\sigma_{-i}) = \underset{\sigma_i}{\arg\max} u_i(\sigma_i, \sigma_{-i}), $$ is convex-valued, and hence so too is $BR(\sigma) = \big(BR_1(\sigma_{-1}), \ldots, BR_N(\sigma_{-N})\big)$. Even in spite of this latter correspondence having convex values, however, its set of fixed points, i.e. the Nash equilibria of the game, are generally non-convex.