# Is this a valid proof for a Mixed Nash Equilibria ( and also a Pure N.E)?

Let $G = \{ \bar{S}_1 ,\bar{S}_2 ; u_1 , u_2 \}$ be a normal-form two player game. A mixed strategy $\bar{\bf{p}}_i = (p(s_i))_{s_i \in \bar{S}_i}$ for player $i$ is a probability distribution where, $(p(s_i)) \geq 0$ and $\Sigma_{s_i \in \bar{S}_i} p(s_i) = 1$.

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The expected payoff for player $1$ according to the mixed strategy $\bar{\bf{p}}_1$ is denoted: $$u_1(p_1,p_2) = \Sigma_{s_1 \in \bar{S}_1} [p_1(s_1) p_2(s_2)] u_i(s_1,s_2) = \sum_{n = 1}^N \sum_{m=1}^M p_{1n} p_{2m} u_1(s_1,s_2) = \langle p_1 , A_1 p_2\rangle$$ where $A_1$ is the payoff matrix for player $1$.

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So player $1$ wishes to maximise $u_1(p_1,\bar{p}_2)$ w.r.t. $p_1$. That is; $$\exists \ \bar{p}_1 : \frac{d}{dp_1} \langle p_1 , A_1 \bar{p}_2\rangle |_{p_1 = \bar{p}_1} = 0$$

Hence, $$\exists \ \bar{p}_1 : u_1(\bar{p}_1,\bar{p}_2) \geq u_1(p_1,\bar{p}_2)$$ ie. a Mixed Nash Equilibria.

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For a Pure Nash Equilibria, set $p(s_j) = 1$ and $p(s_i)_{i \ne j} = 0$ such that the payoff is maximised in this fashion.

$$\begin{array}{|c|c|c|} \hline & L & R \\ \hline T & -1,-1 & -3,0 \\ \hline B & 0,-3 & -2,-2 \\ \hline \end{array}$$
In particular, there is a unique equilibrium, $\{B,R\}$, which is trivially a mixed strategy equilibrium (though not, for example 'totally mixed'). Fix the column agent's strategy at $R$ and consider $$p = \mathbb{P}(\textrm{Row plays }T)$$ Then in particular, $$U_1(p) = -3(p) - 2(1-p) = - 1 - p$$ which has non-zero derivative everywhere, but also clearly is maximized at $p=0$.