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

CC BY-SA 4.0
2019-12-01T05:59:57.657Z
2019-12-01T05:59:57.657Z
Henry
<p>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.</p>

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

<game-theory><nash-equilibrium>
Confusion about the convexity of the best response correspondence
justadzr
# Confusion about the convexity of the best response correspondence

(Asked by: username_0 on 2019-12-01)

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?

**Tags:** game-theory, nash-equilibrium

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## 1 answers

### Answer by username_1 on 2019-12-01. Score: 4 | Accepted answer

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


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