# Clarification on the definition of Bayes Correlated Equilibrium

I would like your help to understand the definition of Bayes Correlated Equilibrium (BCE) in an incomplete information game at p.7 of this 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"?

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 $$N>1$$.)

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.
• 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 May 22 '19 at 16:14
• @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 May 22 '19 at 17:31
• Indeed, there still (generically) exist BCE that are not BNE even if $\pi$ were uninformative. (Say $T$ were a singleton.) – Theoretical Economist May 22 '19 at 17:39
• 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 May 22 '19 at 18:01