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

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.

• What does the argument in the indicator function mean? As it currently stands, the argument is just a number. What condition(s) must that number satisfy in order for the indicator function to produce a $1$ (or $0$)? Commented Aug 13 at 4:00
• @HerrK. Thanks for pointing that out. I missed comparing the expression to 0 inside the indicator function. I also forgot to specify that the expression is the payoff when player chooses 1 and the payoff of choosing 0 is 0. Have changed my question. Commented Aug 13 at 7:59

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 $$(u_i,\delta_i)_{i=1}^{3}\in\mathbb{R}^6$$.

• Thanks! I think this works and also made me realized some typos in my question. I've revised the question. Also, following your argument, it seems that any $(Y^*_1,Y^*_2,Y^*_3)$ can be supported as Nash equilibrium in some regions of $(\epsilon_1,\epsilon_2,\epsilon_3)$, right? Commented Aug 13 at 8:06
• Yes, for every $(Y_1^*,Y_2^*,Y_3^*)$, we can find a region of $(\epsilon_1,\epsilon_2,\epsilon_3)$ where it is a Nash equilibrium.
– Amit
Commented Aug 13 at 8:12