这是一份anu澳大利亚国立大学ECON2141作业代写的成功案例

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 $U_{2}=-U_{1}$, $$ U_{1}\left(\alpha_{1}^{}, \alpha_{2}^{}\right) \leq U_{1}\left(\alpha_{1}^{}, \alpha_{2}\right) \text { for every mixed strategy } \alpha_{2} \text { of player } 2 .
$$
Hence
$$
U_{1}\left(\alpha_{1}^{}, \alpha_{2}^{}\right)=\min {\alpha{2}} U_{1}\left(\alpha_{1}^{}, \alpha_{2}\right) . $$ Now, the function on the right hand side of this equality is evaluated at the specific strategy $\alpha_{1}^{}$, so that its value is not more than the maximum as we vary $\alpha_{1}$, namely $\max {a{1}} \min {a{2}} U_{1}\left(\alpha_{1}, \alpha_{2}\right)$. Thus we conclude that
$$
U_{1}\left(\alpha_{1}^{}, \alpha_{2}^{}\right) \leq \max {\alpha{1}} \min {\alpha{2}} U_{1}\left(\alpha_{1}, \alpha_{2}\right) .
$$

ECON2141 COURSE NOTES ：

$U_{1}\left(\alpha_{1}^{}, \alpha_{2}\right) \geq U_{1}\left(\alpha_{1}^{}, \alpha_{2}^{}\right)$ for every mixed strategy $\alpha_{2}$ of player 2 , 342 or $U_{2}\left(\alpha_{1}^{}, \alpha_{2}\right) \leq U_{2}\left(\alpha_{1}^{}, \alpha_{2}^{}\right)$ for every mixed strategy $\alpha_{2}$ of player 2 .
Similarly,
$$
U_{2}\left(\alpha_{1}, \alpha_{2}^{}\right) \geq U_{2}\left(\alpha_{1}^{}, \alpha_{2}^{}\right) \text { for every mixed strategy } \alpha_{1} \text { of player } 1 \text {, } $$ or $$ U_{1}\left(\alpha_{1}, \alpha_{2}^{}\right) \leq U_{1}\left(\alpha_{1}^{}, \alpha_{2}^{}\right) \text { for every mixed strategy } \alpha_{1} \text { of player } 1 \text {, }
$$
so that $\left(\alpha_{1}^{}, \alpha_{2}^{}\right)$ is a Nash equilibrium of the game.