<p>I think the reason you're having difficulty is that your definition is actually not equivalent to Bergemann and Morris' definition of BCE, except under specific assumptions which include the ones you state regarding independence. However, in general, we do not wish to make these assumptions. (For example, assuming that <span class="math-container">$t_{-i}$</span> is independent of <span class="math-container">$a_i$</span> conditional on <span class="math-container">$\theta$</span> and <span class="math-container">$t_i$</span> means that the mediator cannot reveal any additional information about other players' types once they have revealed the state to a player. This seems unnecessarily restrictive.)</p>

<p>In the paper, a BCE induces a probability distribution <span class="math-container">$\Pr$</span> that satisfies</p>

<p><span class="math-container">$$ \sum_{a_{-i},t_{-i},\theta}\Pr(a_i,a_{-i},t_i,t_{-i},\theta) u_i(a_i,a_{-i},\theta) \ge \sum_{a_{-i},t_{-i},\theta} \Pr(a_i,a_{-i},t_i,t_{-i},\theta) u_i(a^\prime_i,a_{-i},\theta) $$</span></p>

<p>for all <span class="math-container">$a_i^\prime \in A_i$</span>. If we wish to express this inequality in terms of player <span class="math-container">$i$</span>'s expected utility given their information, then we divide both sides by <span class="math-container">$\Pr(a_i,t_i)$</span>. Similarly, from the inequality </p>

<p><span class="math-container">$$ \sum_{a_{-i},t_{-i},\theta}\Pr(a_{-i},t_{-i},\theta \vert a_i,t_i) u_i(a_i,a_{-i},\theta) \ge \sum_{a_{-i},t_{-i},\theta} \Pr(a_{-i},t_{-i},\theta \vert a_i, t_i) u_i(a^\prime_i,a_{-i},\theta), $$</span></p>

<p>we can multiply both sides by <span class="math-container">$\Pr(a_i,t_i)$</span> to recover the first expression above. To see how this squares with your calculation, notice that</p>

<p><span class="math-container">$$ \begin{align*} \Pr(a_{-i}\vert a_i,t_i,t_{-i},\theta) \Pr(t_{-i}\vert a_i,t_i,\theta)\Pr(\theta\vert a_i,t_i) \Pr(a_i,t_i) &amp;= \Pr(a_{-i},t_{-i},\theta \vert a_i, t_i) \Pr(a_i,t_i) \\
&amp;= \Pr(a_i,a_{-i},t_i,t_{-i},\theta) \\
&amp;= \Pr(a_i,a_{-i}\vert t_i,t_{-i},\theta) \Pr(t_i,t_{-i} \vert \theta) \Pr(\theta)\\
&amp;= \sigma(a_i,a_{-i}\vert t_i,t_{-i},\theta) \pi(t_i,t_{-i} \vert \theta) \psi(\theta).
\end{align*} $$</span></p>

<p>In particular, we have that <span class="math-container">$$ \Pr(a_{-i}\vert a_i,t_i,t_{-i},\theta) \Pr(t_{-i}\vert a_i,t_i,\theta)\Pr(\theta\vert a_i,t_i) = \frac{\sigma(a_i,a_{-i}\vert t_i,t_{-i},\theta) \pi(t_i,t_{-i} \vert \theta) \psi(\theta)}{\Pr(a_i,t_i)}. $$</span> This makes it difficult to get the decomposition you want without making needlessly strong independence assumptions.</p>

<p>When <span class="math-container">$N=1$</span>, there is only one player, so the equilibrium is (in some sense) degenerate. However, the difference between one-player BCE and one-player BNE is that the BCE set is larger, and includes any BNE. (Indeed, this is true even when <span class="math-container">$N&gt;1$</span>.)</p>

<p>For your second question:</p>

<blockquote>
  <p>a BCE is still intended as mapping from state and signal <strong>to a probability distribution over actions</strong>. From reading at p.26 of the linked paper, the authors then say "[...] In that case, the set of BCE correspond to joint distribution of actions <strong>and states</strong> [...]". I'm confused. </p>
</blockquote>

<p>Formally, a BCE is a probability distribution of actions for each realisation of the state and signals. This probability distribution <em>induces</em> a joint distribution over actions and states. Just integrate out <span class="math-container">$(t_i,t_{-i})$</span> in <span class="math-container">$\Pr(a_i,a_{-i},t_i,t_{-i},\theta)$</span> to get that joint distribution.</p>

<p>Finally, this equilibrium concept is called Bayes Correlated Equilibirum because it extends our definition of correlated equilibrium (defined by Aumann for complete information games) to games of incomplete information, in the same way that Bayes-Nash equilibrium extends Nash equilibrium.</p>

CC BY-SA 4.0
2019-05-22T16:14:38.590Z
I think this answer is better than mine, so I'm just commenting what I think my answer adds to this post:
When $N=1$ it does not reduces to Bayes Nash equilibrium. The reason is that the function $\pi:\Theta\rightarrow \Delta(T_1)$ may contain additional information about the state of the world, $\theta$, for the single player. BNE would be a special case when $\pi$ is uninformative (for example a constant function) in such a case, the player learns nothing about the state of the world after receiving its signal $t_1\in T_1$, and chooses her action based on her prior belief.
Regio
CC BY-SA 4.0
2019-05-22T16:16:15.580Z
Also part of the first question is a duplicate of this one: https://economics.stackexchange.com/questions/29237/bayes-correlated-equilibrium-obedience/29240#29240
Regio
CC BY-SA 4.0
2019-05-22T17:31:52.507Z
@Regio I wanted to say something along the lines you point out here. However, I disagree that BNE is what you get when $\pi$ is uninformative. Instead, you recover BNE when the only information that the player ever gets about $\theta$ is through $\pi$, instead of additionally through $\sigma$ (as it might in an arbitrary BCE).
Theoretical Economist
CC BY-SA 4.0
2019-05-22T17:39:44.490Z
Indeed, there still (generically) exist BCE that are not BNE even if $\pi$ were uninformative. (Say $T$ were a singleton.)
Theoretical Economist
CC BY-SA 4.0
2019-05-22T18:01:49.973Z
Sorry, you're completely right, $\pi$ represents the player's own (prior) information (which could be the same as the common prior $\psi$, if they have no aditional private information). Instead, I wanted to say that BNE coincides with BCE when $\sigma$ is constant over $\Theta$. In which case the recommendation contains no additional information about the state of the world. Feel free to add it to your response :).
Regio
CC BY-SA 4.0
2019-05-23T11:31:26.013Z
Thanks. Questions 1 and 3 are clear. Regarding question 2 I'm still confused about difference between BCE and BNE for the game $(G,S)$ with $N=1$. Actually, I have problems in providing the definition of BNE. Let me start with BCE. A BCE  of the game $(G,S)$ with $N=1$ is a mapping $\sigma: \Theta\times T\rightarrow \Delta(A)$ such that $\forall t \in T$, $\forall a\in A$, $\forall \tilde{a}\neq a$, $\sum_{\theta}\psi(\theta)\pi(t|\theta)\sigma(a|\theta,t)u(a,\theta)\geq \sum_{\theta}\psi(\theta)\pi(t|\theta)\sigma(a|\theta,t)u(\tilde{a},\theta)$. Correct?
Star
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2019-05-23T11:40:21.440Z
Now I try to give the definition of BNE of the game $(G,S)$ when $N=1$. A BNE of the game $(G,S)$ when $N=1$ is a mapping $\beta: T\rightarrow \Delta(A)$ such that $\forall t\in T$, $\forall a\in A$ such that $\beta(a|t)>0$, $\forall \tilde{a}\neq a$, $\sum_{\theta} \psi(\theta) \pi(t|\theta) u(a,\theta)\geq \sum_{\theta} \psi(\theta) \pi(t|\theta) u(\tilde{a},\theta)$. Correct?
Star
CC BY-SA 4.0
2019-05-23T11:43:21.213Z
From your comments above, it seems to me that you are saying something about when it  is the case that the set of BNE is equal to the set of BCE in the game $(G,S)$ under $N=1$. Could you repeat that? I didn't get it. Thanks again
Star
CC BY-SA 4.0
2019-05-26T20:59:43.170Z
@user3285148 both definitions look correct. Why are you so interested in equilibria of one-player games? It seems a little strange, since a one-player game is not a game at all. In any case, I wasn’t actually saying anything about circumstances when the BCE set is the same as the BNE set.
Theoretical Economist
CC BY-SA 4.0
2019-05-26T21:02:02.597Z
However, one instance where the BCE set is the same as the BNE set is when the mediator has no information to reveal to the player. This happens, for example, when $\Theta$ is a singleton. I also think that formalising properly what it means for the mediator to not have information to reveal will show that this is necessary and sufficient for the BCE and BNE sets to coincide (at least when $N=1$), but I’ll want to think about that more carefully to be sure.
Theoretical Economist
CC BY-SA 4.0
2019-05-27T21:45:22.443Z
Thank you. I'm not a theorist. For reasons too long to explain here I'm interested in the BCE's "linear" properties and its robustness to information assumptions when $N=1$. I understand that when $N=1$ we don't have a game anymore. How would you call the "solution concept" corresponding to BCE when $N=1$?
Star
CC BY-SA 4.0
2019-05-22T14:21:45.513Z
2019-05-22T17:44:55.413Z
2019-05-22T17:44:55.413Z
Theoretical Economist
<p>I would like your help to understand the definition of <strong>Bayes Correlated Equilibrium (BCE)</strong> in an incomplete information game at p.7 of <a href="http://www.princeton.edu/~smorris/pdfs/bce.pdf" rel="nofollow noreferrer">this</a> paper.</p>

<hr>

<p>Let me summarise the definition provided in the paper. </p>

<p>There are <span class="math-container">$N\in \mathbb{N}$</span> players, with <span class="math-container">$i$</span> denoting a generic player.</p>

<p>There is a finite set of states <span class="math-container">$\Theta$</span>, with <span class="math-container">$\theta$</span> denoting a generic state. </p>

<p>A basic game <span class="math-container">$G$</span> consists of</p>

<ul>
<li><p>for each player <span class="math-container">$i$</span>, a finite set of actions <span class="math-container">$A_i$</span>, where we write <span class="math-container">$A\equiv A_1\times A_2\times ... \times A_N$</span>, and a utility function <span class="math-container">$u_i: A\times \Theta \rightarrow \mathbb{R}$</span>.</p></li>
<li><p>a full support prior <span class="math-container">$\psi\in \Delta(\Theta)$</span>.</p></li>
</ul>

<p>An information structure <span class="math-container">$S$</span> consists of</p>

<ul>
<li><p>for each player <span class="math-container">$i$</span>, a finite set of signals <span class="math-container">$T_i$</span>, where we write <span class="math-container">$T\equiv T_1\times T_2\times ... \times T_N$</span>.</p></li>
<li><p>a signal distribution <span class="math-container">$\pi: \Theta \rightarrow \Delta(T)$</span>.</p></li>
</ul>

<p>A decision rule of the incomplete information game <span class="math-container">$(G,S)$</span> is a mapping 
<span class="math-container">$$
\sigma: T\times \Theta\rightarrow \Delta(A)
$$</span></p>

<p><strong>Definition of BCE:</strong> The decision rule <span class="math-container">$\sigma$</span> is a BCE for the game <span class="math-container">$(G,S)$</span> if, for each <span class="math-container">$i=1,...,N$</span>, <span class="math-container">$t_i\in T_i$</span>, and <span class="math-container">$a_i\in A_i$</span>, we have
<span class="math-container">$$
\sum_{a_{-i}, t_{-i}, \theta} \psi(\theta) \pi(t_{-i}|t_i, \theta) \sigma(a_{-i}| a_i,t_i, t_{-i}, \theta) u_i(a_i, a_{-i},\theta)
$$</span>
<span class="math-container">$$
\geq 
\sum_{a_{-i}, t_{-i}, \theta} \psi(\theta) \pi(t_{-i}|t_i, \theta) \sigma(a_{-i}| a_i,t_i, t_{-i}, \theta) u_i(\tilde{a}_i, a_{-i},\theta)
$$</span>
<span class="math-container">$\forall \tilde{a}_i\in A_i$</span>.</p>

<hr>

<p><strong>Question:</strong> </p>

<p><strong>1)</strong> I don't follow how the conditional expectation is computed in the definition. Consider for example the left hand side (LHS) and let me go through each step by using the notation <span class="math-container">$Pr(\cdot)$</span> to generically denote any probability distribution. </p>

<p>The LHS is the expected payoff of player <span class="math-container">$i$</span> where the expectation is computed wrto anything that is unknown to him conditional on what he knows. Hence,
<span class="math-container">$$
\sum_{a_{-i}, \theta, t_{-i}} u_i(a_{-i},a_i, \theta) \times Pr(a_{-i}, \theta, t_{-i}| t_i, a_i)=
$$</span>
<span class="math-container">$$
\sum_{a_{-i}, \theta, t_{-i}} u_i(a_{-i},a_i, \theta) \times Pr(a_{-i}| \theta, t_{-i}, t_i, a_i)\times Pr(\theta, t_{-i}| t_i, a_i)=
$$</span>
<span class="math-container">$$
\sum_{a_{-i}, \theta, t_{-i}} u_i(a_{-i},a_i, \theta) \times \underbrace{Pr(a_{-i}| \theta, t_{-i}, t_i, a_i)}_{\equiv \sigma(a_{-i}| a_i,t_i, t_{-i}, \theta) \text{ [OK!]}}\times \underbrace{Pr(t_{-i}|\theta, t_i, a_i)}_{\equiv \pi(t_{-i}|t_i, \theta)? \text{ Where is $a_i$?} }\times \underbrace{Pr(\theta| t_i, a_i)}_{\equiv \psi(\theta)?\text{ Where are $t_i, a_i$?}}=
$$</span></p>

<p>Are we assuming </p>

<p>1) <span class="math-container">$t_{-i}\perp a_i $</span>, conditional on <span class="math-container">$\theta, t_i$</span></p>

<p>2) <span class="math-container">$\theta \perp t_i, a_i$</span></p>

<p>?</p>

<p><strong>2)</strong> How does the definition of BCE simplify to when <span class="math-container">$N=1$</span>?</p>

<p>From reading at p.25 of the linked paper, it seems that a BCE is still intended as mapping from state and signal <strong>to a probability distribution over actions</strong>. From reading at p.26 of the linked paper, the authors then say "[...] In that case, the set of BCE correspond to joint distribution of actions <strong>and states</strong> [...]". I'm confused. </p>

<p>Also, when <span class="math-container">$N=1$</span>, how does the definition of BCE differ from the definition of Bayesian Nash Equilibrium?</p>

<p><strong>3)</strong> Just as a curiosity, what it is the reason of including the adjective "<strong>correlated</strong>"?</p>

CC BY-SA 4.0
2019-05-22T17:33:21.903Z
Please stop tagging your questions [tag:cooperative-game-theory]. BCE is a non-cooperative equilibrium concept.
Theoretical Economist
<game-theory><bayesian-game>
Clarification on the definition of Bayes Correlated Equilibrium
Star
# Clarification on the definition of Bayes Correlated Equilibrium

(Asked by: username_0 on 2019-05-22)

I would like your help to understand the definition of **Bayes Correlated Equilibrium (BCE)** in an incomplete information game at p.7 of [this](http://www.princeton.edu/~smorris/pdfs/bce.pdf) paper.

---

Let me summarise the definition provided in the paper.

There are $N\in \mathbb{N}$ players, with $i$ denoting a generic player.

There is a finite set of states $\Theta$, with $\theta$ denoting a generic state.

A basic game $G$ consists of

* for each player $i$, a finite set of actions $A\_i$, where we write $A\equiv A\_1\times A\_2\times ... \times A\_N$, and a utility function $u\_i: A\times \Theta \rightarrow \mathbb{R}$.
* a full support prior $\psi\in \Delta(\Theta)$.

An information structure $S$ consists of

* for each player $i$, a finite set of signals $T\_i$, where we write $T\equiv T\_1\times T\_2\times ... \times T\_N$.
* a signal distribution $\pi: \Theta \rightarrow \Delta(T)$.

A decision rule of the incomplete information game $(G,S)$ is a mapping
$$
\sigma: T\times \Theta\rightarrow \Delta(A)
$$

**Definition of BCE:** The decision rule $\sigma$ is a BCE for the game $(G,S)$ if, for each $i=1,...,N$, $t\_i\in T\_i$, and $a\_i\in A\_i$, we have
$$
\sum\_{a\_{-i}, t\_{-i}, \theta} \psi(\theta) \pi(t\_{-i}|t\_i, \theta) \sigma(a\_{-i}| a\_i,t\_i, t\_{-i}, \theta) u\_i(a\_i, a\_{-i},\theta)
$$
$$
\geq
\sum\_{a\_{-i}, t\_{-i}, \theta} \psi(\theta) \pi(t\_{-i}|t\_i, \theta) \sigma(a\_{-i}| a\_i,t\_i, t\_{-i}, \theta) u\_i(\tilde{a}\_i, a\_{-i},\theta)
$$
$\forall \tilde{a}\_i\in A\_i$.

---

**Question:**

**1)** I don't follow how the conditional expectation is computed in the definition. Consider for example the left hand side (LHS) and let me go through each step by using the notation $Pr(\cdot)$ to generically denote any probability distribution.

The LHS is the expected payoff of player $i$ where the expectation is computed wrto anything that is unknown to him conditional on what he knows. Hence,
$$
\sum\_{a\_{-i}, \theta, t\_{-i}} u\_i(a\_{-i},a\_i, \theta) \times Pr(a\_{-i}, \theta, t\_{-i}| t\_i, a\_i)=
$$
$$
\sum\_{a\_{-i}, \theta, t\_{-i}} u\_i(a\_{-i},a\_i, \theta) \times Pr(a\_{-i}| \theta, t\_{-i}, t\_i, a\_i)\times Pr(\theta, t\_{-i}| t\_i, a\_i)=
$$
$$
\sum\_{a\_{-i}, \theta, t\_{-i}} u\_i(a\_{-i},a\_i, \theta) \times \underbrace{Pr(a\_{-i}| \theta, t\_{-i}, t\_i, a\_i)}\_{\equiv \sigma(a\_{-i}| a\_i,t\_i, t\_{-i}, \theta) \text{ [OK!]}}\times \underbrace{Pr(t\_{-i}|\theta, t\_i, a\_i)}\_{\equiv \pi(t\_{-i}|t\_i, \theta)? \text{ Where is $a\_i$?} }\times \underbrace{Pr(\theta| t\_i, a\_i)}\_{\equiv \psi(\theta)?\text{ Where are $t\_i, a\_i$?}}=
$$

Are we assuming

1) $t\_{-i}\perp a\_i $, conditional on $\theta, t\_i$

2) $\theta \perp t\_i, a\_i$

?

**2)** How does the definition of BCE simplify to when $N=1$?

From reading at p.25 of the linked paper, it seems that a BCE is still intended as mapping from state and signal **to a probability distribution over actions**. From reading at p.26 of the linked paper, the authors then say "[...] In that case, the set of BCE correspond to joint distribution of actions **and states** [...]". I'm confused.

Also, when $N=1$, how does the definition of BCE differ from the definition of Bayesian Nash Equilibrium?

**3)** Just as a curiosity, what it is the reason of including the adjective "**correlated**"?

**Tags:** game-theory, bayesian-game

> **Comments:**
> **username_1**: Please stop tagging your questions [tag:cooperative-game-theory]. BCE is a non-cooperative equilibrium concept.  

---

## 1 answers

### Answer by username_1 on 2019-05-22. Score: 3 | Accepted answer

I think the reason you're having difficulty is that your definition is actually not equivalent to Bergemann and Morris' definition of BCE, except under specific assumptions which include the ones you state regarding independence. However, in general, we do not wish to make these assumptions. (For example, assuming that $t\_{-i}$ is independent of $a\_i$ conditional on $\theta$ and $t\_i$ means that the mediator cannot reveal any additional information about other players' types once they have revealed the state to a player. This seems unnecessarily restrictive.)

In the paper, a BCE induces a probability distribution $\Pr$ that satisfies

$$ \sum\_{a\_{-i},t\_{-i},\theta}\Pr(a\_i,a\_{-i},t\_i,t\_{-i},\theta) u\_i(a\_i,a\_{-i},\theta) \ge \sum\_{a\_{-i},t\_{-i},\theta} \Pr(a\_i,a\_{-i},t\_i,t\_{-i},\theta) u\_i(a^\prime\_i,a\_{-i},\theta) $$

for all $a\_i^\prime \in A\_i$. If we wish to express this inequality in terms of player $i$'s expected utility given their information, then we divide both sides by $\Pr(a\_i,t\_i)$. Similarly, from the inequality

$$ \sum\_{a\_{-i},t\_{-i},\theta}\Pr(a\_{-i},t\_{-i},\theta \vert a\_i,t\_i) u\_i(a\_i,a\_{-i},\theta) \ge \sum\_{a\_{-i},t\_{-i},\theta} \Pr(a\_{-i},t\_{-i},\theta \vert a\_i, t\_i) u\_i(a^\prime\_i,a\_{-i},\theta), $$

we can multiply both sides by $\Pr(a\_i,t\_i)$ to recover the first expression above. To see how this squares with your calculation, notice that

$$ \begin{align\*} \Pr(a\_{-i}\vert a\_i,t\_i,t\_{-i},\theta) \Pr(t\_{-i}\vert a\_i,t\_i,\theta)\Pr(\theta\vert a\_i,t\_i) \Pr(a\_i,t\_i) &= \Pr(a\_{-i},t\_{-i},\theta \vert a\_i, t\_i) \Pr(a\_i,t\_i) \\
&= \Pr(a\_i,a\_{-i},t\_i,t\_{-i},\theta) \\
&= \Pr(a\_i,a\_{-i}\vert t\_i,t\_{-i},\theta) \Pr(t\_i,t\_{-i} \vert \theta) \Pr(\theta)\\
&= \sigma(a\_i,a\_{-i}\vert t\_i,t\_{-i},\theta) \pi(t\_i,t\_{-i} \vert \theta) \psi(\theta).
\end{align\*} $$

In particular, we have that $$ \Pr(a\_{-i}\vert a\_i,t\_i,t\_{-i},\theta) \Pr(t\_{-i}\vert a\_i,t\_i,\theta)\Pr(\theta\vert a\_i,t\_i) = \frac{\sigma(a\_i,a\_{-i}\vert t\_i,t\_{-i},\theta) \pi(t\_i,t\_{-i} \vert \theta) \psi(\theta)}{\Pr(a\_i,t\_i)}. $$ This makes it difficult to get the decomposition you want without making needlessly strong independence assumptions.

When $N=1$, there is only one player, so the equilibrium is (in some sense) degenerate. However, the difference between one-player BCE and one-player BNE is that the BCE set is larger, and includes any BNE. (Indeed, this is true even when $N>1$.)

For your second question:

> a BCE is still intended as mapping from state and signal **to a probability distribution over actions**. From reading at p.26 of the linked paper, the authors then say "[...] In that case, the set of BCE correspond to joint distribution of actions **and states** [...]". I'm confused.

Formally, a BCE is a probability distribution of actions for each realisation of the state and signals. This probability distribution *induces* a joint distribution over actions and states. Just integrate out $(t\_i,t\_{-i})$ in $\Pr(a\_i,a\_{-i},t\_i,t\_{-i},\theta)$ to get that joint distribution.

Finally, this equilibrium concept is called Bayes Correlated Equilibirum because it extends our definition of correlated equilibrium (defined by Aumann for complete information games) to games of incomplete information, in the same way that Bayes-Nash equilibrium extends Nash equilibrium.

> **Comments:**
> **username_2**: I think this answer is better than mine, so I'm just commenting what I think my answer adds to this post: When $N=1$ it does not reduces to Bayes Nash equilibrium. The reason is that the function $\pi:\Theta\rightarrow \Delta(T_1)$ may contain additional information about the state of the world, $\theta$, for the single player. BNE would be a special case when $\pi$ is uninformative (for example a constant function) in such a case, the player learns nothing about the state of the world after receiving its signal $t_1\in T_1$, and chooses her action based on her prior belief.  
> **username_2**: Also part of the first question is a duplicate of this one: https://economics.stackexchange.com/questions/29237/bayes-correlated-equilibrium-obedience/29240#29240  
> **username_1**: @username_2 I wanted to say something along the lines you point out here. However, I disagree that BNE is what you get when $\pi$ is uninformative. Instead, you recover BNE when the only information that the player ever gets about $\theta$ is through $\pi$, instead of additionally through $\sigma$ (as it might in an arbitrary BCE).  
> **username_1**: Indeed, there still (generically) exist BCE that are not BNE even if $\pi$ were uninformative. (Say $T$ were a singleton.)  
> **username_2**: Sorry, you're completely right, $\pi$ represents the player's own (prior) information (which could be the same as the common prior $\psi$, if they have no aditional private information). Instead, I wanted to say that BNE coincides with BCE when $\sigma$ is constant over $\Theta$. In which case the recommendation contains no additional information about the state of the world. Feel free to add it to your response :).  
> **username_0**: Thanks. Questions 1 and 3 are clear. Regarding question 2 I'm still confused about difference between BCE and BNE for the game $(G,S)$ with $N=1$. Actually, I have problems in providing the definition of BNE. Let me start with BCE. A BCE  of the game $(G,S)$ with $N=1$ is a mapping $\sigma: \Theta\times T\rightarrow \Delta(A)$ such that $\forall t \in T$, $\forall a\in A$, $\forall \tilde{a}\neq a$, $\sum_{\theta}\psi(\theta)\pi(t|\theta)\sigma(a|\theta,t)u(a,\theta)\geq \sum_{\theta}\psi(\theta)\pi(t|\theta)\sigma(a|\theta,t)u(\tilde{a},\theta)$. Correct?  
> **username_0**: Now I try to give the definition of BNE of the game $(G,S)$ when $N=1$. A BNE of the game $(G,S)$ when $N=1$ is a mapping $\beta: T\rightarrow \Delta(A)$ such that $\forall t\in T$, $\forall a\in A$ such that $\beta(a|t)>0$, $\forall \tilde{a}\neq a$, $\sum_{\theta} \psi(\theta) \pi(t|\theta) u(a,\theta)\geq \sum_{\theta} \psi(\theta) \pi(t|\theta) u(\tilde{a},\theta)$. Correct?  
> **username_0**: From your comments above, it seems to me that you are saying something about when it  is the case that the set of BNE is equal to the set of BCE in the game $(G,S)$ under $N=1$. Could you repeat that? I didn't get it. Thanks again  
> **username_1**: @user3285148 both definitions look correct. Why are you so interested in equilibria of one-player games? It seems a little strange, since a one-player game is not a game at all. In any case, I wasn’t actually saying anything about circumstances when the BCE set is the same as the BNE set.  
> **username_1**: However, one instance where the BCE set is the same as the BNE set is when the mediator has no information to reveal to the player. This happens, for example, when $\Theta$ is a singleton. I also think that formalising properly what it means for the mediator to not have information to reveal will show that this is necessary and sufficient for the BCE and BNE sets to coincide (at least when $N=1$), but I’ll want to think about that more carefully to be sure.  
> **username_0**: Thank you. I'm not a theorist. For reasons too long to explain here I'm interested in the BCE's "linear" properties and its robustness to information assumptions when $N=1$. I understand that when $N=1$ we don't have a game anymore. How would you call the "solution concept" corresponding to BCE when $N=1$?  

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