Showing existence of a Nash equilibrium in pure strategy

Question

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.

Answer 1

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.

Answer 2

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.