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I wanted to ask whether the general method of computing pure strategy Nash Equilibria as used in this thread Finding mixed Nash equilibria in continuous games gives you ALL possible Nash equilibria.

I have twice differentiable payoff functions $f1$ and $f2$ for which

$ \begin{align} & \frac{ \partial^{2} f_{1}}{ \partial p_1^{2}}< 0, \\ & \frac{ \partial^{2} f_{2}}{ \partial p_2^{2}}<0 \end{align} $

holds, thus the conditions for maxima are given.

When solving the system of equations

$ \begin{align} & \frac{ \partial f_{1}}{ \partial p_1} = 0, \\ & \frac{ \partial f_{2}}{ \partial p_2} = 0 \end{align} $

for $p_1$ and $p_2$, I get a unique solution for the equilibrium prices. What I wanted to know however, is whether this is the ONLY possible pure strategy Nash Equilibrium for this game or just ONE NE? How can one prove the uniquness of the NE?

mlc
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Rudolf Deichmann
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1 Answers1

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The conditions in your OP are not sufficient to ensure uniqueness. Here is a counterexample.

Suppose $$f_1(s_1,s_2) = s_1 s_2 - \frac{s_1^2}{2}$$ and $$f_2(s_1,s_2) = s_1 s_2 - \frac{s_2^2}{2}$$ Each payoff function $f_i$ is strictly concave in $s_i$. The best reply functions are $B_1(s_2) = s_2$ and $B_2 (s_1) = s_1$ so any strategy profile $(s_1,s_2) = (x,x)$ for $x$ in $[0,1]$ is a pure strategy Nash equilibrium.

A well-known sufficient condition for uniqueness is called diagonal strictly concavity. See J.B. Rosen, "Existence and uniqueness of equilibrium point for concave $n$-person games", Econometrica 33, 520-534.

mlc
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  • Thank you for your answer. In the case you are showing there are infinetly many solutions to the system of equations $f_1'= 0$ and $f_2'=0$ . However, I was asking whether any NE can exist which are NOT solutions to this system of equations. Do all NE necessarily have to be solutions to the system $f_1'= 0$ ,$f_2'=0$ ? – Rudolf Deichmann Aug 28 '17 at 23:56