The notions of Bayes Correlated Equilibrium and Bayesian Nash Equilibrium in a game with one player

Question

I would like your help to use the correct terminology to define a Bayes Correlated Equilibrium (BCE) and a Bayesian Nash Equilibrium (BNE) in a "game" with one player. The notion of BCE in an $N$-player game has been introduced in this paper.

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Let me describe the game and give the notion of BCE and BNE in an $N$-player game.

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 $$ \sigma: T\times \Theta\rightarrow \Delta(A) $$

A mixed strategy of player $i$ of the incomplete information game $(G,S)$ is a mapping $$ \beta_i: T_i\rightarrow \Delta(A_i) $$

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 $$ \sum_{a_{-i}, t_{-i}, \theta} \psi(\theta) \pi(t_{-i},t_i| \theta) \sigma(a_{-i}, a_i|t_i, t_{-i}, \theta) u_i(a_i, a_{-i},\theta) $$ $$ \geq \sum_{a_{-i}, t_{-i}, \theta} \psi(\theta) \pi(t_{-i},t_i| \theta) \sigma(a_{-i}, a_i|t_i, t_{-i}, \theta) u_i(\tilde{a}_i, a_{-i},\theta) $$ $\forall \tilde{a}_i\in A_i$.

Definition of BNE: The strategy profile $(\beta_1,...,\beta_N)$ is a BNE for the game $(G,S)$ if, for each $i=1,...,N$, $t_i\in T_i$, and $a_i\in A_i$ such that $\beta(a_i|t_i)>0$, we have $$ \sum_{a_{-i}, t_{-i}, \theta} \psi(\theta) \pi(t_{-i},t_i| \theta) \Pi_{j\neq i} \beta_j(a_j|t_j) u_i(a_i, a_{-i},\theta) $$ $$ \geq \sum_{a_{-i}, t_{-i}, \theta} \psi(\theta) \pi(t_{-i},t_i| \theta) \Pi_{j\neq i} \beta_j(a_j|t_j) u_i(\tilde{a}_i, a_{-i},\theta) $$ $\forall \tilde{a}_i\in A_i$.

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Question: when $N=1$, I've understood from the various versions of the paper I linked that the following terminology may be appropriate. Could you check and provide better suggestions, if any?

• $S$ is called "experiment" in the sense studied by Blackwell (1951; 1953), rather than "information structure".

• $G$ is called "decision problem", rather than "basic game".

• The player is called "decision maker".

• The BCE concept is referred to as "one-player BCE".

• Also, how would you refer to the BNE concept with one player?

• Lastly, are there other crucial papers in the literature introducing the concept of BCE when $N=1$? Or the paper linked is still a valid reference?

Answer 1

• The terms experiment and information structure are used interchangeably by the literature, so both terms are valid regardless of $N$.
• I agree if $N=1$ there is really no game, and it is rather a single-agent decision problem (or decision problem with uncertainty), and yes, you don't have a player, rather a decision maker.
• For the same token, you don't want to say the solution concept if BNE, instead you want to say that the agent maximizes his expected utility given the information she received.
• I was not aware of the term "one-player BCE" but Bergemann and Morris coined the term BCE, so if that is how they call it, it is correct.
• In terms of other references, this is an earlier version of that same paper. However, I think that the paper you linked is the right paper for citing the concept of BCE.

In this later paper, the same two authors made the remarkable connection that the problem of Bayesian Persuasion popularized by Kamenica and Gentzkow could be interpreted in light of the concept of BCE when $N=1$. So perhaps, this later reference of BM can be useful if you want to explore more the case of $N=1$.