<p>Given the way the game is defined the payoff of player <span class="math-container">$i$</span>, <span class="math-container">$u_i+\delta_i\sum_{j\neq i}Y_j+\epsilon_i$</span> is not affected by player <span class="math-container">$i$</span>'s action <span class="math-container">$Y_i$</span>, so every action of player <span class="math-container">$i$</span> is the best response (trivially). Consequently, every possible strategy profile is a Nash equilibrium.</p>
<p>However, if the payoff of player <span class="math-container">$i$</span> is</p>
<p><span class="math-container">\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 &amp; \text{ if } Y_i=1 \\ 0 &amp; \text{if } Y_i=0\end{cases} \end{eqnarray*}</span></p>
<p>In this case the best response of player <span class="math-container">$i$</span> is
<span class="math-container">$\text{BR}_i(Y_{-i}) = \begin{cases} \{0\} &amp;\text{if } u_i+\delta_i\sum_{j\neq i}Y_j+\epsilon_i &lt; 0 \\ \{0,1\} &amp; \text{if } u_i+\delta_i\sum_{j\neq i}Y_j+\epsilon_i= 0 \\ \{1\} &amp; \text{if } u_i+\delta_i\sum_{j\neq i}Y_j+\epsilon_i &gt; 0\end{cases}$</span></p>
<p>then you can easily show that <span class="math-container">$(1,1,1)$</span> is a Nash equilibrium outcome of the above game with positive probability. Assuming all players observe <span class="math-container">$(u_i,\delta_i,\epsilon_i)_{i=1}^{3}$</span>, <span class="math-container">$(1,1,1)$</span> is Nash equilibrium when <span class="math-container">$\epsilon_i\geq-u_i-2\delta_i$</span> for all <span class="math-container">$i$</span>. Given that <span class="math-container">$\epsilon_i$</span> are independently normally distributed, the probability that <span class="math-container">$(1,1,1)$</span> is a Nash equilibrium is equal to
<span class="math-container">$\Pr(\epsilon_i\geq-u_i-2\delta_i\text{ for all } i)$</span>
<span class="math-container">$=\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)&gt;0 $</span>
for every <span class="math-container">$(u_i,\delta_i)_{i=1}^{3}\in\mathbb{R}^6$</span>. So the probability that the Nash equilibrium exists is positive for every <span class="math-container">$(u_i,\delta_i)_{i=1}^{3}\in\mathbb{R}^6$</span>.</p>

CC BY-SA 4.0
2024-08-13T08:06:16.480Z
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?
ExcitedSnail
CC BY-SA 4.0
2024-08-13T08:12:59.653Z
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
CC BY-SA 4.0
2024-08-13T06:16:34.970Z
2024-08-13T07:12:09.853Z
2024-08-13T07:12:09.853Z
Amit
<p>Suppose I have the following 3-player binary game with <span class="math-container">$Y_i\in\{0,1\}$</span> denote the action of player <span class="math-container">$i$</span> and <span class="math-container">$u_i+\delta_i\sum_{j\neq i}Y_j+\epsilon_i$</span> is the payoff function of choosing <span class="math-container">$Y_i=1$</span> for player <span class="math-container">$i$</span> (the payoff function of choosing <span class="math-container">$Y_i=0$</span> equals 0), which is common knowledge among all players. <span class="math-container">$\epsilon_i$</span> follows a distribution with support <span class="math-container">$R$</span>, such as a normal distribution. <span class="math-container">$u_i$</span> and <span class="math-container">$\delta_i$</span> are finite numbers that I specify in my numerical experiment. The players have the following best responses:</p>
<p><span class="math-container">$Y_1=\mathbf{1}(u_1+\delta_1(Y_2+Y_3)+\epsilon_1&gt;0),$</span></p>
<p><span class="math-container">$Y_2=\mathbf{1}(u_2+\delta_2(Y_1+Y_3)+\epsilon_2&gt;0),$</span></p>
<p><span class="math-container">$Y_3=\mathbf{1}(u_3+\delta_3(Y_1+Y_2)+\epsilon_3&gt;0),$</span></p>
<p>where <span class="math-container">$\mathbf{1}(\cdot)$</span> is the indicator function.</p>
<p>In general, I guess it's not true that the game always have a pure strategy Nash equilibrium for any value of <span class="math-container">$(u_i,\delta_i)_{i=1}^3$</span> and any realization of <span class="math-container">$(\epsilon_1,\epsilon_2,\epsilon_3)$</span>.</p>
<p>How to show that for any finite values of <span class="math-container">$(u_i,\delta_i)_{i=1}^3$</span>, this game always have a pure strategy nash equilibrium  with strictly positive probability? That is, for any finite value of <span class="math-container">$(u_i,\delta_i)_{i=1}^3$</span>, there exist a set of <span class="math-container">$E\subset R^3$</span> (<span class="math-container">$E$</span> might depend on <span class="math-container">$(u_i,\delta_i)_{i=1}^3$</span>) with strictly positive probability measure such that when the realized value of <span class="math-container">$(\epsilon_1,\epsilon_2,\epsilon_3)$</span> falls in <span class="math-container">$E$</span>, the game has a pure strategy nash equilibrium.</p>
<p>Intuitively, I guess the fact that <span class="math-container">$(\epsilon_i)_{i=1}^3$</span> being supported on <span class="math-container">$R^3$</span> plays a role, i.e., the <span class="math-container">$\epsilon_i$</span> could be large or small enough to rationalize any action. But I don't know how to show it formally.</p>

CC BY-SA 4.0
2024-08-13T04:00:21.750Z
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 $0$)?
Herr K.
CC BY-SA 4.0
2024-08-13T07:59:48.560Z
@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.
ExcitedSnail
<game-theory><nash-equilibrium><proof>
How to show this game of complete information always has a pure strategy nash equilibrium with positive probability?
ExcitedSnail
# How to show this game of complete information always has a pure strategy nash equilibrium with positive probability?

(Asked by: username_0 on 2024-08-13)

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.

**Tags:** game-theory, nash-equilibrium, proof

> **Comments:**
> **username_1**: 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 $0$)?  
> **username_0**: @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.  

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## 1 answers

### Answer by username_2 on 2024-08-13. Score: 3 | Accepted answer

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 $(u\_i,\delta\_i)\_{i=1}^{3}\in\mathbb{R}^6$.

> **Comments:**
> **username_0**: 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?  
> **username_2**: 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.  

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