How to show this game of complete information always has a pure strategy nash equilibrium with positive probability?

Question

Suppose I have the following 3-player binary game with $Y_i\in\{0,1\}$ denote the action of player $i$ and $u_i+\delta_i\sum_{j\neq i}Y_j+\epsilon_i$ is the payoff function of choosing $Y_i=1$ for player $i$ (the payoff function of choosing $Y_i=0$ equals 0), which is common knowledge among all players. $\epsilon_i$ follows a distribution with support $R$, such as a normal distribution. $u_i$ and $\delta_i$ are finite numbers that I specify in my numerical experiment. The players have the following best responses:

$Y_1=\mathbf{1}(u_1+\delta_1(Y_2+Y_3)+\epsilon_1>0),$

$Y_2=\mathbf{1}(u_2+\delta_2(Y_1+Y_3)+\epsilon_2>0),$

$Y_3=\mathbf{1}(u_3+\delta_3(Y_1+Y_2)+\epsilon_3>0),$

where $\mathbf{1}(\cdot)$ is the indicator function.

In general, I guess it's not true that the game always have a pure strategy Nash equilibrium for any value of $(u_i,\delta_i)_{i=1}^3$ and any realization of $(\epsilon_1,\epsilon_2,\epsilon_3)$.

How to show that for any finite values of $(u_i,\delta_i)_{i=1}^3$, this game always have a pure strategy nash equilibrium with strictly positive probability? That is, for any finite value of $(u_i,\delta_i)_{i=1}^3$, there exist a set of $E\subset R^3$ ($E$ might depend on $(u_i,\delta_i)_{i=1}^3$) with strictly positive probability measure such that when the realized value of $(\epsilon_1,\epsilon_2,\epsilon_3)$ falls in $E$, the game has a pure strategy nash equilibrium.

Intuitively, I guess the fact that $(\epsilon_i)_{i=1}^3$ being supported on $R^3$ plays a role, i.e., the $\epsilon_i$ could be large or small enough to rationalize any action. But I don't know how to show it formally.

Answer 1

Given the way the game is defined the payoff of player $i$, $u_i+\delta_i\sum_{j\neq i}Y_j+\epsilon_i$ is not affected by player $i$'s action $Y_i$, so every action of player $i$ is the best response (trivially). Consequently, every possible strategy profile is a Nash equilibrium.

However, if the payoff of player $i$ is

\begin{eqnarray*} (u_i+\delta_i\sum_{j\neq i}Y_j+\epsilon_i)Y_i = \begin{cases} u_i+\delta_i\sum_{j\neq i}Y_j+\epsilon_i & \text{ if } Y_i=1 \\ 0 & \text{if } Y_i=0\end{cases} \end{eqnarray*}

In this case the best response of player $i$ is $\text{BR}_i(Y_{-i}) = \begin{cases} \{0\} &\text{if } u_i+\delta_i\sum_{j\neq i}Y_j+\epsilon_i < 0 \\ \{0,1\} & \text{if } u_i+\delta_i\sum_{j\neq i}Y_j+\epsilon_i= 0 \\ \{1\} & \text{if } u_i+\delta_i\sum_{j\neq i}Y_j+\epsilon_i > 0\end{cases}$

then you can easily show that $(1,1,1)$ is a Nash equilibrium outcome of the above game with positive probability. Assuming all players observe $(u_i,\delta_i,\epsilon_i)_{i=1}^{3}$, $(1,1,1)$ is Nash equilibrium when $\epsilon_i\geq-u_i-2\delta_i$ for all $i$. Given that $\epsilon_i$ are independently normally distributed, the probability that $(1,1,1)$ is a Nash equilibrium is equal to $\Pr(\epsilon_i\geq-u_i-2\delta_i\text{ for all } i)$ $=\Pr(\epsilon_1\geq-u_1-2\delta_1)\Pr(\epsilon_2\geq-u_2-2\delta_2)\Pr(\epsilon_3\geq-u_3-2\delta_3)>0 $ for every $(u_i,\delta_i)_{i=1}^{3}\in\mathbb{R}^6$. So the probability that the Nash equilibrium exists is positive for every $(u_i,\delta_i)_{i=1}^{3}\in\mathbb{R}^6$.