# 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.

## 1 Answer

The proof is not correct; the line where you suppose the first-order condition is satisfied with equality is where it fails. To see why, note the affine nature of von-Neumann Morgenstern utility functions means generally the gradient of utility at a maximizer will not be zero.

Consider the prisoner's dilemma.

$$\begin{array}{|c|c|c|} \hline & L & R \\ \hline T & -1,-1 & -3,0 \\ \hline B & 0,-3 & -2,-2 \\ \hline \end{array}$$

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 $p=0$.

Addendum: A nice gut check, if you are looking to generally prove the existence of Nash equilibria, you should be wary of any proof you can come up with that does not explicitly make use of a fixed point theorem. In particular, if one assumes that every finite normal form game possesses a Nash equilibrium as a primitive, one can use this to prove Brouwer's fixed point theorem, so if your proof is not sufficiently 'powerful' to replace the technical machinery needed to prove Brouwer, it likely is missing some subtlety.