<p>Rewrite <span class="math-container">$i$</span>'s payoff as
<span class="math-container">\begin{align}
u_i(a_i,a_{-i})&amp;=\sum_{j\in N}a_j-c_ia_i\\
&amp;=(a_1+\cdots+\color{red}{a_i}+\cdots+a_n)-c_ia_i\\
&amp;=\underbrace{(a_1+\cdots+a_n)}_{\text{without $a_i$}}+\color{red}{a_i}-c_ia_i\\
&amp;=\sum_{j\in N\setminus\{i\}}a_j+(1-c_i)a_i.
\end{align}</span>
Thus, <span class="math-container">$i$</span>'s best response only depends on the relative magnitude of <span class="math-container">$c_i$</span>.</p>

CC BY-SA 4.0
2021-09-07T17:57:18.640Z
Ah I see, it was only a rewrite of the original payoff. I thought we had to calculate from there. Now I understand. Thank you so much again!
Maybeline Lee
CC BY-SA 4.0
2021-09-07T17:55:12.360Z
2021-09-07T17:55:12.360Z
Herr K.
<p>Consider the following strategic game with complete information. Set
of players is <span class="math-container">$N = \{1, . . . , n\}$</span>, where <span class="math-container">$n ≥ 2$</span>. Each player <span class="math-container">$i ∈ N$</span> chooses action
<span class="math-container">$a$</span> <span class="math-container">$i ∈ \{0, 1\}$</span>. Payoff of each <span class="math-container">$i ∈ N$</span> from action profile <span class="math-container">$ (a_i)_{i∈N}$</span> is <span class="math-container">$\sum_{
j∈N} a_j−c_ia_i,$</span> where <span class="math-container">$c_i &gt; 0$</span> is a parameter. Assume <span class="math-container">$c_i$</span> is not an integer for each <span class="math-container">$i ∈ N$</span>. The interpretation of the game is as follows. Each person in a society
either takes an action that benefits everyone in the society but is costly for
her, or not.</p>
<p>The solution it gives is as follows: Payoff of each <span class="math-container">$i ∈ N$</span> from action profile <span class="math-container">$ (a_i)_{i∈N}$</span> equals <span class="math-container">$\sum_{j∈N/\{i\}} a_j+(1-c_i)a_i,$</span> Hence there exists unique NE in which <span class="math-container">$a_i = 0$</span> for any player <span class="math-container">$i ∈ N$</span> with <span class="math-container">$1 − c_i &lt; 0$</span> and <span class="math-container">$a_i = 1$</span> for any player <span class="math-container">$i∈ N$</span> with <span class="math-container">$1 − c_i &gt; 0.$</span></p>
<p>But I'm not understanding where this payoff came from</p>

<microeconomics><game-theory>
Equilibrium in strategic game with complete information
Maybeline Lee
# Equilibrium in strategic game with complete information

(Asked by: username_0 on 2021-09-07)

Consider the following strategic game with complete information. Set
of players is $N = \{1, . . . , n\}$, where $n ≥ 2$. Each player $i ∈ N$ chooses action
$a$ $i ∈ \{0, 1\}$. Payoff of each $i ∈ N$ from action profile $ (a\_i)\_{i∈N}$ is $\sum\_{
j∈N} a\_j−c\_ia\_i,$ where $c\_i > 0$ is a parameter. Assume $c\_i$ is not an integer for each $i ∈ N$. The interpretation of the game is as follows. Each person in a society
either takes an action that benefits everyone in the society but is costly for
her, or not.

The solution it gives is as follows: Payoff of each $i ∈ N$ from action profile $ (a\_i)\_{i∈N}$ equals $\sum\_{j∈N/\{i\}} a\_j+(1-c\_i)a\_i,$ Hence there exists unique NE in which $a\_i = 0$ for any player $i ∈ N$ with $1 − c\_i < 0$ and $a\_i = 1$ for any player $i∈ N$ with $1 − c\_i > 0.$

But I'm not understanding where this payoff came from

**Tags:** microeconomics, game-theory


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## 1 answers

### Answer by username_1 on 2021-09-07. Score: 3 | Accepted answer

Rewrite $i$'s payoff as
\begin{align}
u\_i(a\_i,a\_{-i})&=\sum\_{j\in N}a\_j-c\_ia\_i\\
&=(a\_1+\cdots+\color{red}{a\_i}+\cdots+a\_n)-c\_ia\_i\\
&=\underbrace{(a\_1+\cdots+a\_n)}\_{\text{without $a\_i$}}+\color{red}{a\_i}-c\_ia\_i\\
&=\sum\_{j\in N\setminus\{i\}}a\_j+(1-c\_i)a\_i.
\end{align}
Thus, $i$'s best response only depends on the relative magnitude of $c\_i$.

> **Comments:**
> **username_0**: Ah I see, it was only a rewrite of the original payoff. I thought we had to calculate from there. Now I understand. Thank you so much again!  

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