# Showing existence of a Nash equilibrium in pure strategy

Consider a game with $$N$$ players, each indexed by $$i=1,...,N$$. Every player $$i$$ has to choose a $$J\times 1$$ vector of actions $$a_i\equiv (a_{i,1},...,a_{i,J})$$ where each $$a_{i,j}$$ can be zero or one. For each player $$i$$, let $$a_{-i}$$ denote the actions of the other players.

The payoff of each player $$i$$ is $$u_i(a_i, a_{-i}; \theta)+v_i(a_i; \delta)$$ where $$u_i$$ and $$v_i$$ are some parametric functions of the parameters $$\theta$$ and $$\delta$$. Moreover $$v_i$$ is monotone decreasing in $$\sum_{j=1}^J a_{i,j}$$.

A pure strategy Nash equilibrium (PSNE) of the game is $$a^*\equiv (a_1^*,...,a_N^*)$$ solving $$a_i^*\in \arg\max_{a_i\in \{0,1\}^J} u_i(a_i, a^*_{-i}; \theta)+v_i(a_i; \delta) \quad \forall i=1,...,N$$

Question: Suppose that I'm able to show that a PSNE exists for $$\theta=\theta_0$$ and $$\delta=\delta_0$$ where $$\theta_0$$ and $$\delta_0$$ are some specific real values of the parameters $$\theta$$ and $$\delta$$. Can I conclude that a PSNE exists for $$\theta=\theta_0$$ and any value of $$\delta$$?

In particular, here, I'm wondering whether the claim may follow from the fact that $$v_i(a_i; \delta)$$ enters additively, does not depend on $$a_{-i}$$, and it is monotone.

• What does it mean that $v$ is monotone decreasing in $\sum_{j=1}^J a_{i,j}$? This is not a variable of $v$. May 12 '21 at 20:20

No. Consider the matching pennies game $$\begin{array}{c |c} & H & T & \\ \hline H & (1,-1) & (-1,1)\\ T & (-1,1) & (1,-1)\\ \end{array}$$ which we know no PSNE exists.

Define the function $$v_i(\cdot,\cdot)$$ by $$v_i(H,0) = 5 \\ v_i(x,y) = 0 \quad \forall (x,y) \not = (H,0)$$

At $$\delta = 0$$, we have the game $$\begin{array}{c |c} & H & T & \\ \hline H & (5,4) & (-4,1)\\ T & (-1,5) & (1,-1)\\ \end{array}$$ which has a PSNE of $$(H,H)$$.

For any other value of $$\delta$$, we're back in the original game so no PSNE exists.

Redefine $$H = 0, T = 1$$ and you have your monotone decreasing requirement.

Seems like this class of games is very general. (Or I don't get the definition.) Note that $$\theta$$ is not even used, it is just some parameter that has value $$\theta_0$$ throughout.

As long as there exist two games $$G_{\delta}$$ and $$G_{\delta_0}$$, were both have the same number of players $$J$$, each player has two strategies in both $$G_{\delta}$$ and $$G_{\delta_0}$$, and $$G_{\delta}$$ has a Nash-equilibrium but $$G_{\delta_0}$$ does not, then one can get negation by defining $$v(a; \delta_0) := u_{\delta_0}(a) - u_{\delta}(a)$$ where $$u_{\delta}$$ is the vector of payoffs in $$G_{\delta}$$, and $$u_{\delta_0}$$ is the vector of payoffs in $$G_{\delta_0}$$ given strategy profile $$a$$.

There is still the possibility that either all or no games have a Nash-equilibria, but this is not true on the class of these two strategies per player games, as shown by Walrasian Auctioneer's answer.