3
$\begingroup$

I am recently reading the proof of the existence of the Nash Equilibrium. As a math student, I do understand the use of Berge's maximum theorem and Kakutani's fixed point theorem, but I am not sure why the mixed-strategy space and the image of the best response correspondence are convex.

For example, if I have two mixed-strategies $\sigma_i,\sigma_i'\in BR_i(\sigma_{-i})$, why is the mix of these two strategies, $\lambda\sigma_i+(1-\lambda)\sigma_i'$ ($0\leqslant\lambda\leqslant1$), also a best response? Since $\sigma_i,\sigma_i'\in BR_i(\sigma_{-i})$ suggests $u_i(\sigma_i,\sigma_{-i})=u_i(\sigma’_i,\sigma_{-i})=\max\{u_i(x_i,\sigma_{-i})\}$, how to prove that $u_i(\lambda\sigma_i+(1-\lambda)\sigma_i',\sigma_{-i})$ also equals to maximum?

$\endgroup$
4
$\begingroup$

I would have thought that the if the opponent strategy $\sigma_{-i}$ is given, even if it is a mixed strategy, then $$u_i(\lambda\sigma_i+(1-\lambda)\sigma_i', \sigma_{-i})= \lambda u_i(\sigma_i, \sigma_{-i})+(1-\lambda)u_i(\sigma_i', \sigma_{-i})$$ and so if $u_i(\sigma_i, \sigma_{-i})=u_i(\sigma_i', \sigma_{-i})$ then $u_i(\lambda\sigma_i+(1-\lambda)\sigma_i', \sigma_{-i})$ is equal to both of them. If they are best responses then so too is the linear combination strategy

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.