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A definition in my book states

$ \sigma^* \in \Delta$ is a Nash equilibrium for $\Gamma$ if for all $i \in I$ and all $\mu_i \in \Delta_i, \; U_i (\sigma^*) \geq U_i(\sigma^* \backslash \mu_i)$. The set of Nash equilibria for $\Gamma$ is denoted $Eq(\Gamma)$

What does the $\backslash$ here represent, divide or element not included?

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  • $\begingroup$ What do you know about the definition of the Nash equilibrium in informal words, rather than the formal mathematical definition? $\endgroup$ – EnergyNumbers Jul 18 '16 at 16:33
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    $\begingroup$ Nothing, this is the first I've heard about it. $\endgroup$ – Sunhwa Jul 18 '16 at 16:35
  • $\begingroup$ I've skimmed the Wikipedia page, they use the notation: $$ \forall i, x_i \in S: f_i (x_i^*, x_{-1}^*) \geq f_i (x_i, x_{-1}^*) $$ Where I think the $x_{-1}^*$ stands for the nash equilibrium actions (best result by not changing) of all the other players (without player $i$). But still, I don't see how the notation in my question says this. $\endgroup$ – Sunhwa Jul 18 '16 at 16:56
  • $\begingroup$ The definition is rather odd. What's the reference? Never seen it before. In your previous comment it should probably read $-i$. $\endgroup$ – clueless Jul 18 '16 at 18:31
  • $\begingroup$ Concerning the wiki entry, you may define $x_i^* \in \arg\max_{x_i \in S_i}f_i(x_i,x_{-i}^*)$. $\endgroup$ – clueless Jul 18 '16 at 18:37
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Informally, a (possibly mixed) set of strategy choices is a Nash equilibrium if no player can expect to benefit by changing strategies while the other players keep their strategies unchanged.

Here I would interpret your definition as $\Delta$ being the set of all possible strategy choices across all players, $\sigma^*$ being a set of strategy choices which will be a Nash equilibrium if it satisfies the conditions, $i$ being a player, and $\sigma^* \backslash \mu_i$ being a change in player $i$'s strategy while the other players' strategies stay the same. So I would take $\backslash$ having a type of "not included" interpretation.

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    $\begingroup$ This makes sense if we define $\sigma^* \setminus \mu_i:=(\ldots,\sigma^*_{i-1}, \mu_i,\sigma^*_{i+1},\ldots)$. However, it's more conventional and intuitive if you write $U_i(\sigma^*_i, \sigma_{-i}^*) \geq U_i(\mu_i, \sigma_{-i}^*) \forall \mu_i \in \Delta_i$, where $\sigma^*_{-i}:=(\ldots,\sigma^*_{i-1}, \sigma^*_{i+1},\ldots) \in \Delta \setminus \Delta_i$. $\endgroup$ – clueless Jul 20 '16 at 12:06

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