这是一份oxford牛津大学作业代写的成功案例
$$
\max _{p}(p \cdot v-e(p)) .
$$
Denote the solution of this problem by $p^{*}(v)$. Assuming that $e$ is differentiable, the first-order condition for this problem implies that
$$
v=e^{\prime}\left(p^{}(v)\right) \text { for all } v . $$
Oxford COURSE NOTES ：
Proof. I first prove part a. By the definition of Nash equilibrium we have $U_{2}\left(\alpha_{1}^{}, \alpha_{2}^{}\right) \geq U_{2}\left(\alpha_{1}^{}, \alpha_{2}\right)$ for every mixed strategy $\alpha_{2}$ of player 2 or, since $U_{2}=-U_{1}$, $U_{1}\left(\alpha_{1}^{}, \alpha_{2}^{}\right) \leq U_{1}\left(\alpha_{1}^{}, \alpha_{2}\right)$ for every mixed strategy $\alpha_{2}$ of player 2
Hence
$$
U_{1}\left(\alpha_{1}^{}, \alpha_{2}^{}\right)=\min {\alpha{2}} U_{1}\left(\alpha_{1}^{*}, \alpha_{2}\right)
$$