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!


1 Answer 1


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 $u_i(\sigma_{-i},[c_i])>u_i(\sigma)+\epsilon$.

  • $\begingroup$ 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 $u_i(\sigma) \geq u_i(\sigma_{-i},[c_i]),\quad \forall i \in N,\quad \forall c_i \in C_i$? $\endgroup$
    – Beerus
    Nov 29, 2023 at 23:57
  • 1
    $\begingroup$ Yes, exactly. $~$ $\endgroup$ Nov 30, 2023 at 6:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.