<p>It doesn't matter whether one uses mixed or pure strategies in that place.
If a mixed strategy <span class="math-container">$\tau_i\in\Delta(C_i)$</span> leads to a payoff larger than <span class="math-container">$u_i(\sigma)+\epsilon$</span>, it must put positive probabilities on the set of pure strategies <span class="math-container">$[c_i]$</span> such that <span class="math-container">$u_i(\sigma_{-i},[c_i])&gt;u_i(\sigma)+\epsilon$</span>.</p>

CC BY-SA 4.0
2023-11-29T23:57:44.107Z
Thank you so much! So does it mean that, for the same reason, the Nash equilibrium may as well be defined as follows: $\sigma$ is a Nash equilibrium of $\Gamma$ if and only if $u_i(\sigma) \geq u_i(\sigma_{-i},[c_i]),\quad \forall i \in N,\quad \forall c_i \in C_i$?
Beerus
CC BY-SA 4.0
2023-11-30T06:03:39.417Z
Yes, exactly. $~$
Michael Greinecker
CC BY-SA 4.0
2023-11-29T08:03:15.010Z
2023-11-29T08:03:15.010Z
Michael Greinecker
<p>I am studying infinite strategy sets using Myerson's Game Theory: Analysis of Conflict. On Page 143, he defines an <span class="math-container">$\epsilon$</span>-equilibrium as follows:</p>
<blockquote>
<p><strong>Definition</strong> For any nonnegative number <span class="math-container">$\epsilon$</span>, an <span class="math-container">$\epsilon$</span>-equilibrium of any strategic-form game is a combination of randomized strategies such that no player could expect to gain more than <span class="math-container">$\epsilon$</span> by switching to any of his feasible strategies, instead of following the randomized strategy specified for him. That is, <span class="math-container">$\sigma$</span> is an <span class="math-container">$\epsilon$</span>-equilibrium of <span class="math-container">$\Gamma$</span> if and only if
<span class="math-container">\begin{align*}
u_i(\sigma_{-i},[c_i]) - u_i(\sigma) \leq \epsilon,\quad \forall i \in N,\quad \forall c_i \in C_i.\tag1
\end{align*}</span></p>
</blockquote>
<p>Then, he put:</p>
<blockquote>
<p>when <span class="math-container">$\epsilon=0$</span>, an <span class="math-container">$\epsilon$</span>-equilibrium is just a Nash equilibrium in the usual sense.</p>
</blockquote>
<p>However, I am a bit confused about this expression (1), because of that &quot;<span class="math-container">$[c_i]$</span>&quot; there. Expression (1) means that no player could expect to gain more than <span class="math-container">$\epsilon$</span> by switching to any of his <em><strong>pure</strong></em> strategies (not <em><strong>feasible</strong></em> strategies, because they might as well be mixed strategies <span class="math-container">$\tau_i$</span>). So I feel that the definition should have been stated as:</p>
<blockquote>
<p><span class="math-container">$\sigma$</span> is an <span class="math-container">$\epsilon$</span>-equilibrium of <span class="math-container">$\Gamma$</span> if and only if
<span class="math-container">\begin{align*}
u_i(\sigma_{-i},\tau_i) - u_i(\sigma) \leq \epsilon,\quad \forall i \in N,\quad \forall \tau_i \in \Delta(C_i).\tag2
\end{align*}</span></p>
</blockquote>
<p>I would like to know if my thought is correct. Is this a mistake of the book, or is the amended definition and expression unnecessary? I would really appreciate it if someone could help me check!</p>

<microeconomics><mathematical-economics><game-theory><nash-equilibrium>
About infinite strategy sets and $\epsilon$-equilibrium from Game Theory: Analysis of Confilct by Roger Myerson
Beerus
# About infinite strategy sets and $\epsilon$-equilibrium from Game Theory: Analysis of Confilct by Roger Myerson

(Asked by: username_0 on 2023-11-28)

I am studying infinite strategy sets using Myerson's Game Theory: Analysis of Conflict. On Page 143, he defines an $\epsilon$-equilibrium as follows:

> **Definition** For any nonnegative number $\epsilon$, an $\epsilon$-equilibrium of any strategic-form game is a combination of randomized strategies such that no player could expect to gain more than $\epsilon$ by switching to any of his feasible strategies, instead of following the randomized strategy specified for him. That is, $\sigma$ is an $\epsilon$-equilibrium of $\Gamma$ if and only if
> \begin{align\*}
> u\_i(\sigma\_{-i},[c\_i]) - u\_i(\sigma) \leq \epsilon,\quad \forall i \in N,\quad \forall c\_i \in C\_i.\tag1
> \end{align\*}

Then, he put:

> when $\epsilon=0$, an $\epsilon$-equilibrium is just a Nash equilibrium in the usual sense.

However, I am a bit confused about this expression (1), because of that "$[c\_i]$" there. Expression (1) means that no player could expect to gain more than $\epsilon$ by switching to any of his ***pure*** strategies (not ***feasible*** strategies, because they might as well be mixed strategies $\tau\_i$). So I feel that the definition should have been stated as:

> $\sigma$ is an $\epsilon$-equilibrium of $\Gamma$ if and only if
> \begin{align\*}
> u\_i(\sigma\_{-i},\tau\_i) - u\_i(\sigma) \leq \epsilon,\quad \forall i \in N,\quad \forall \tau\_i \in \Delta(C\_i).\tag2
> \end{align\*}

I would like to know if my thought is correct. Is this a mistake of the book, or is the amended definition and expression unnecessary? I would really appreciate it if someone could help me check!

**Tags:** microeconomics, mathematical-economics, game-theory, nash-equilibrium


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## 1 answers

### Answer by username_1 on 2023-11-29. Score: 2 | Accepted answer

It doesn't matter whether one uses mixed or pure strategies in that place.
If a mixed strategy $\tau\_i\in\Delta(C\_i)$ leads to a payoff larger than $u\_i(\sigma)+\epsilon$, it must put positive probabilities on the set of pure strategies $[c\_i]$ such that $u\_i(\sigma\_{-i},[c\_i])>u\_i(\sigma)+\epsilon$.

> **Comments:**
> **username_0**: Thank you so much! So does it mean that, for the same reason, the Nash equilibrium may as well be defined as follows: $\sigma$ is a Nash equilibrium of $\Gamma$ if and only if $u_i(\sigma) \geq u_i(\sigma_{-i},[c_i]),\quad \forall i \in N,\quad \forall c_i \in C_i$?  
> **username_1**: Yes, exactly. $~$  

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