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Hello,

i came across a remark that states that nash equilibria inside the simplex S_n are unique. Or stated differently: If there is more then one such equilibrium, then they have to lie on the boundary of S_n. Why is this true ?

Thank you

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3 Answers

Here are two true statements like this. I assume we are talking about two player games with finite strategy sets. Let the number of strategies of the two players be $p$ and $q$.

  • For a generic pay off matrix, there is at most one Nash equilibria in the interior of the strategy space. More precisely, this is true on a Zariski open subset of $\mathbb{R}^{2pq}$.

  • For any pay off matrix, the space of Nash equilibria in the interior of the strategy space is connected. In particular, it is impossible to have two isolated interior equilibria.

Let $\Delta^{p-1}$ be the strategy simplex for the first player, and $\Delta^{q-1}$ for the second player. Let $P$ and $Q$ be the payoff functions for each player. Our goal is to understand those points in the interior of $\Delta^{p-1} \times \Delta^{q-1}$ which are Nash equilibria.

Suppose that $(x,y)$ is an interior equilibrium. Now, $Q(x,y)$ is linear in $y$. The only way that a linear function can be optimized at an interior point is that $Q$ is constant as a function of $y$. The condition that $Q$ be constant as a function of $y$ is given by $q-1$ linear equations in the $x$ variables. There is also one more equation: $\sum x_i =1$. Similarly, requiring that $x$ be an interior optimum gives $(p-1)+1$ linear equations in the $y$ variables.

If our payoff matrix is generic, then these equations are independent. We have $q$ equations in $p$ variables and $p$ equations in $q$ variables. If $p \neq q$, then one of these sets of equations will have no solution. If $p=q$, then there is one solution in $\mathbb{R}^p \times \mathbb{R}^q$. If it has positive entries, then it is the unique interior equilibrium. If some of the entries are negative, then there are no interior equilibria.

Even without the assumption that the payoff matrix is generic, the subset of $\Delta^{p-1}$ where some linear equations hold is a polyhedron $K$. Similarly, the subset of $\Delta^{q-1}$ where some linear equations hold is a polyhedron $L$. So the set of interior equilibria always looks like $K \times L$.

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Maybe I am misunderstanding something, but this cannot be true in such a generality. Indeed, if the utility function is constant, then every point is an equilibrium. To have uniqueness, you need some additional property like strict concavity or similar. For instance, you may have a look at this slides http://www.arts.cornell.edu/econ/guerdjikova/teaching/gth2006/ecinf_files/lecture9.pdf just to have a flavor.

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Hello,

Both answers are very helpful. Following Valerios remark i looked for additional properties an found that the author of the text i was reading meant a special class of Nash equilibria called evolutionary stable strategies (ESS) . For those ESS the claim was proved in by Haigh (1975). So, i have to apologize for the false statement in my question.

Thank you for our answers.

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Hello, you don't need to answer, but you can just comment the answers. Moreover, you can use the same account. Finally, in MathOverflow people usually thanks also by voting-up the answers and accepting (the button right below the vote) the answer that they find more helpful. –  Valerio Capraro Feb 24 '12 at 16:42
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