<p>A strategy profile <span class="math-container">$\sigma^*=(\sigma_i^*,\sigma_{-i}^*)$</span> is a Nash equilibrium if for all player <span class="math-container">$i$</span>,
<span class="math-container">\begin{equation}
u_i(s_i,\sigma_{-i}^*)\ge u_i(s_i',\sigma_{-i}^*), \quad \forall s_i\in\mathrm{supp}(\sigma_i^*), \forall s_i'\in S_i.
\end{equation}</span>
Rewriting this condition explicitly in terms of the probabilities over pure strategies:
<span class="math-container">\begin{align}
\sum_{s_{-i}\in S_{-i}}\sigma_{-i}^*(s_{-i})u_i(s_i,s_{-i})
&amp;\ge \sum_{s_{-i}\in S_{-i}}\sigma_{-i}^*(s_{-i})u_i(s_i',s_{-i}), \quad \forall s_i\in\mathrm{supp}(\sigma_i^*), \forall s_i'\in S_i. \tag{1}
\end{align}</span>
Let <span class="math-container">$p_{\sigma^*}(s_i,s_{-i})=\sigma_i^*(s_i)\sigma_{-i}^*(s_{-i})$</span> be the joint distribution implied by the equilibrium strategies. Then condition <span class="math-container">$(1)$</span> can be rewritten as
<span class="math-container">\begin{equation}
\sum_{s_{-i}\in S_{-i}}\underbrace{\sigma_i^*(s_i)\sigma_{-i}^*(s_{-i})}_{p_{\sigma^*}(s_i,s_{-i})}u_i(s_i,s_{-i})
\ge \sum_{s_{-i}\in S_{-i}}\underbrace{\sigma_i^*(s_i)\sigma_{-i}^*(s_{-i})}_{p_{\sigma^*}(s_i,s_{-i})}u_i(s_i',s_{-i}), \quad\forall s_i,s_i'\in S_i.
\end{equation}</span>
This is the definition of correlated equilibrium.</p>

CC BY-SA 4.0
2021-09-07T15:28:57.847Z
I am a bit confused...why we sum only with respect to $\sigma^*_{-i}(s_{-i})$ in the condition $(1)$ and then we chacnge this? I am missing something here....
Hunger Learn
CC BY-SA 4.0
2021-09-07T15:32:27.607Z
Did you mean why $u_i(s_i,\sigma_i)=\sum_{s_{-i}\in S_{-i}} \sigma_{-i}(s_{-i})u_i(s_i,s_{-i})$?
Herr K.
CC BY-SA 4.0
2021-09-07T15:33:46.450Z
Yes! Exactly this is what I mean
Hunger Learn
CC BY-SA 4.0
2021-09-07T15:38:57.600Z
Well, $u_i(s_i,\sigma_{-i})$ is $i$'s expected utility from playing pure strategy $s_i$ (i.e. with probability $\color{red}{1}$) against the mixed strategy profile $\sigma_{-i}$, so his payoff would be $(\color{red}{1})\sigma_{-i}(s_{-i})u_i(s_i,s_{-i})$. If $i$ were to play a mixed strategy $\sigma_i$, then his expected payoff would be $\color{red}{\sigma_i(s_i)}\sigma_{-i}(s_{-i})u_i(s_i,s_{-i})$
Herr K.
CC BY-SA 4.0
2021-09-07T15:43:46.740Z
Ok I understand now. One last thing. Is it $u_i(s_i,\sigma_i)=\sum_{s_{-i}\in S_{-i}}\sigma_{-i}(s_{-i})u_i(s_i,s_{-i})$ or $u_i(s_i,\sigma_{-i})=\sum_{s_{-i}\in S_{-i}}\sigma_{-i}(s_{-i})u_i(s_i,s_{-i})$?
Hunger Learn
CC BY-SA 4.0
2021-09-07T15:50:13.363Z
@KingOdysseus: The latter. The second argument in $u_i$ should be the strategies by players other than $i$.
Herr K.
CC BY-SA 4.0
2021-09-07T15:22:38.467Z
2021-09-07T15:22:38.467Z
Herr K.
<p>This is a classic theorem in game theory, that is left as an excersice in my textbook. Can anybody proove it? I can not thing of anything excpet from the definition of the correlated equilibrium in first place. Here is the theorem and the definition as well.</p>
<p><span class="math-container">$\mathbf{Theorem:}$</span> For every Nash equilibrium <span class="math-container">$\sigma^*$</span>, the probability distribution <span class="math-container">$p_{\sigma^*}$</span> is a correlated equilibrium.</p>
<p><span class="math-container">$\mathbf{Definition:}$</span> A probability distribution <span class="math-container">$p$</span> over the set of action vectors <span class="math-container">$S$</span> is called a
correlated equilibrium if the strategy vector <span class="math-container">$\tau^*$</span> is a Nash equilibrium of the game <span class="math-container">$\Gamma^*(p)$</span>. In other words, for every player <span class="math-container">$i ∈ N$</span>:</p>
<p><span class="math-container">\begin{equation}\Sigma_{s_{-i}\in S_{-i}}p(s_i,s_{-i})u_i(s_i,s_{-i})\geq \Sigma_{s_{-i}\in S_{-i}}p(s_i,s_{-i})u_i(s^{'}_i,s_{-i}),\quad\text{$\forall s_i,s^{'}_i\in S_i$}\end{equation}</span></p>
<p>Every strategy vector <span class="math-container">$\sigma$</span> induces a probability distribution <span class="math-container">$p_{\sigma^*}$</span> over the set of action vector <span class="math-container">$S$</span>.
<span class="math-container">\begin{equation}p_{\sigma^*}(s_1,...,s_n)=\sigma_1(s_1)\times\sigma_2(s_2)\times...\times\sigma_n(s_n)\end{equation}</span></p>
<p><span class="math-container">$\textit{Hint:}$</span> When we relate to a Nash equilibrium <span class="math-container">$\sigma^*$</span> as a correlated equilibrium we mean the probability distribution <span class="math-container">$p_{\sigma^*}$</span> given by the aforementioned equation:</p>

CC BY-SA 4.0
2021-09-07T14:42:19.803Z
@HerrK. how to doublecheck it?
Hunger Learn
CC BY-SA 4.0
2021-09-07T14:52:51.160Z
Is there any typo? Where did you get the definition from? It seems inconsistent with the version given in the Fudenberg-Tirole textbook.
Herr K.
CC BY-SA 4.0
2021-09-07T14:55:22.393Z
well it is from the Maschler, Solan and Zamir Game theory book, but from what I have seen until now, there are some typos in the book in general. However, I do not know abou the this one? @HerrK. can you write the Fundenberg - Tirole version as well to compare it?
Hunger Learn
CC BY-SA 4.0
2021-09-07T15:13:03.393Z
@HerrK. I think that this version is the same with the equivalent version of the definition $2.4B$ in page 57 of Fundenberg - Tirole textbook...please check it if you want! Thank you in advance!
Hunger Learn
CC BY-SA 4.0
2021-09-07T15:23:29.993Z
I now see where the difference is. Sorry for the confusion.
Herr K.
<mathematical-economics><game-theory><nash-equilibrium><real-analysis>
Prove that for every Nash equilibrium $\sigma^*$, the probability distribution $p_{\sigma^*}$ is a correlated equilibrium
Hunger Learn
# Prove that for every Nash equilibrium $\sigma^*$, the probability distribution $p_{\sigma^*}$ is a correlated equilibrium

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

This is a classic theorem in game theory, that is left as an excersice in my textbook. Can anybody proove it? I can not thing of anything excpet from the definition of the correlated equilibrium in first place. Here is the theorem and the definition as well.

$\mathbf{Theorem:}$ For every Nash equilibrium $\sigma^\*$, the probability distribution $p\_{\sigma^\*}$ is a correlated equilibrium.

$\mathbf{Definition:}$ A probability distribution $p$ over the set of action vectors $S$ is called a
correlated equilibrium if the strategy vector $\tau^\*$ is a Nash equilibrium of the game $\Gamma^\*(p)$. In other words, for every player $i ∈ N$:

\begin{equation}\Sigma\_{s\_{-i}\in S\_{-i}}p(s\_i,s\_{-i})u\_i(s\_i,s\_{-i})\geq \Sigma\_{s\_{-i}\in S\_{-i}}p(s\_i,s\_{-i})u\_i(s^{'}\_i,s\_{-i}),\quad\text{$\forall s\_i,s^{'}\_i\in S\_i$}\end{equation}

Every strategy vector $\sigma$ induces a probability distribution $p\_{\sigma^\*}$ over the set of action vector $S$.
\begin{equation}p\_{\sigma^\*}(s\_1,...,s\_n)=\sigma\_1(s\_1)\times\sigma\_2(s\_2)\times...\times\sigma\_n(s\_n)\end{equation}

$\textit{Hint:}$ When we relate to a Nash equilibrium $\sigma^\*$ as a correlated equilibrium we mean the probability distribution $p\_{\sigma^\*}$ given by the aforementioned equation:

**Tags:** mathematical-economics, game-theory, nash-equilibrium, real-analysis

> **Comments:**
> **username_0**: @HerrK. how to doublecheck it?  
> **username_1**: Is there any typo? Where did you get the definition from? It seems inconsistent with the version given in the Fudenberg-Tirole textbook.  
> **username_0**: well it is from the Maschler, Solan and Zamir Game theory book, but from what I have seen until now, there are some typos in the book in general. However, I do not know abou the this one? @HerrK. can you write the Fundenberg - Tirole version as well to compare it?  
> **username_0**: @HerrK. I think that this version is the same with the equivalent version of the definition $2.4B$ in page 57 of Fundenberg - Tirole textbook...please check it if you want! Thank you in advance!  
> **username_1**: I now see where the difference is. Sorry for the confusion.  

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## 1 answers

### Answer by username_1 on 2021-09-07. Score: 9 | Accepted answer

A strategy profile $\sigma^\*=(\sigma\_i^\*,\sigma\_{-i}^\*)$ is a Nash equilibrium if for all player $i$,
\begin{equation}
u\_i(s\_i,\sigma\_{-i}^\*)\ge u\_i(s\_i',\sigma\_{-i}^\*), \quad \forall s\_i\in\mathrm{supp}(\sigma\_i^\*), \forall s\_i'\in S\_i.
\end{equation}
Rewriting this condition explicitly in terms of the probabilities over pure strategies:
\begin{align}
\sum\_{s\_{-i}\in S\_{-i}}\sigma\_{-i}^\*(s\_{-i})u\_i(s\_i,s\_{-i})
&\ge \sum\_{s\_{-i}\in S\_{-i}}\sigma\_{-i}^\*(s\_{-i})u\_i(s\_i',s\_{-i}), \quad \forall s\_i\in\mathrm{supp}(\sigma\_i^\*), \forall s\_i'\in S\_i. \tag{1}
\end{align}
Let $p\_{\sigma^\*}(s\_i,s\_{-i})=\sigma\_i^\*(s\_i)\sigma\_{-i}^\*(s\_{-i})$ be the joint distribution implied by the equilibrium strategies. Then condition $(1)$ can be rewritten as
\begin{equation}
\sum\_{s\_{-i}\in S\_{-i}}\underbrace{\sigma\_i^\*(s\_i)\sigma\_{-i}^\*(s\_{-i})}\_{p\_{\sigma^\*}(s\_i,s\_{-i})}u\_i(s\_i,s\_{-i})
\ge \sum\_{s\_{-i}\in S\_{-i}}\underbrace{\sigma\_i^\*(s\_i)\sigma\_{-i}^\*(s\_{-i})}\_{p\_{\sigma^\*}(s\_i,s\_{-i})}u\_i(s\_i',s\_{-i}), \quad\forall s\_i,s\_i'\in S\_i.
\end{equation}
This is the definition of correlated equilibrium.

> **Comments:**
> **username_0**: I am a bit confused...why we sum only with respect to $\sigma^*_{-i}(s_{-i})$ in the condition $(1)$ and then we chacnge this? I am missing something here....  
> **username_1**: Did you mean why $u_i(s_i,\sigma_i)=\sum_{s_{-i}\in S_{-i}} \sigma_{-i}(s_{-i})u_i(s_i,s_{-i})$?  
> **username_0**: Yes! Exactly this is what I mean  
> **username_1**: Well, $u_i(s_i,\sigma_{-i})$ is $i$'s expected utility from playing pure strategy $s_i$ (i.e. with probability $\color{red}{1}$) against the mixed strategy profile $\sigma_{-i}$, so his payoff would be $(\color{red}{1})\sigma_{-i}(s_{-i})u_i(s_i,s_{-i})$. If $i$ were to play a mixed strategy $\sigma_i$, then his expected payoff would be $\color{red}{\sigma_i(s_i)}\sigma_{-i}(s_{-i})u_i(s_i,s_{-i})$  
> **username_0**: Ok I understand now. One last thing. Is it $u_i(s_i,\sigma_i)=\sum_{s_{-i}\in S_{-i}}\sigma_{-i}(s_{-i})u_i(s_i,s_{-i})$ or $u_i(s_i,\sigma_{-i})=\sum_{s_{-i}\in S_{-i}}\sigma_{-i}(s_{-i})u_i(s_i,s_{-i})$?  
> **username_1**: @KingOdysseus: The latter. The second argument in $u_i$ should be the strategies by players other than $i$.  

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