[Economics] Showing existence of a Nash equilibrium in pure strategy

Solution 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.

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Solution 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.