# Features of the set of Bayes Correlated Equilibrium in a single-agent decision problem with uncertainty

I would like your help to better understand the notion of Bayes Correlated Equilibrium (BCE) in a single-agent decision problem with uncertainty The notion of BCE is provided in this paper for a generic $$N$$-player game.

Consider the following single-agent choice problem under uncertainty.

Let $$V$$ be the state of the world with support $$\mathcal{V}$$ and probability distribution $$P_V\in \Delta(\mathcal{V})$$. First, let nature draw a realisation $$v$$ of $$V$$ from $$P_V$$. Then, let the decision maker choose an action $$y\in \mathcal{Y}$$, with $$\mathcal{Y}$$ finite, without observing $$v$$. Upon the decision has been made, the decision maker gets a payoff $$u(y,v)$$.

Before choosing an action, the decision maker can receive some signal to refine her prior about the state of the world.

Now, I want to use Theorem 1 in Bergemann and Morris (2016) to characterise the set of optimal strategies under minimal assumptions on the amount of information that is processed by the decision maker. To do that, I introduce the notion of one-player BCE.

A one-player BCE of the game described is a probability distribution $$P_{Y,V}\in \Delta(\mathcal{Y}\times \mathcal{V})$$ such that:

1) $$\forall v \in \mathcal{V}$$ $$\sum_{y\in \mathcal{Y}}P_{Y,V}(y,v)=P_V(v)$$

2) $$\forall y\in \mathcal{Y}$$ and $$\forall \tilde{y}\neq y$$ $$\sum_{v\in \mathcal{V}} [u(y,v)-u(\tilde{y},v)] \times P_{Y,V}(y,v)\geq 0$$

Let $$\mathcal{P}$$ be the set of BCE. Let $$\mathcal{Q}\equiv \{P_Y\in \Delta(\mathcal{Y}): P_Y(y)\equiv \sum_{v\in \mathcal{V}}P_{Y,V}(y,v) \text{ }\forall y \in \mathcal{Y}, P_{Y,V}\in \mathcal{P}\}$$

Questions:

1) Is $$\mathcal{Q}$$ convex? I guess it is because $$\mathcal{P}$$ is convex. Could you confirm?

2) Let $$\mathcal{B}\equiv \{b\in \mathbb{R}^{|\mathcal{Y}|}: b^Tb\leq 1\}$$ be the unit ball in $$\mathbb{R}^{|\mathcal{Y}|}$$. Let us stack in a $$|\mathcal{Y}|\times 1$$ vector the image set of every probability distribution $$P_{Y}\in \mathcal{Q}$$. With some abuse of notation, let us still denote any such a vector by $$P_{Y}\in \mathcal{Q}$$. Does $$\max_{P_Y\in \mathcal{Q}} b^T P_Y$$ exists $$\forall b\in \mathcal{B}$$? I guess the answer to this question depends on the features of $$\mathcal{Q}$$. For example, if $$\mathcal{Q}$$ is compact, then the maximum above exists. Could you help?

• Some notations are undefined---$v_y$, $v_{\tilde{y}}$, $\mathcal{P}$, etc. For 1), if $\mathcal{P}$ is convex, than $\mathcal{Q}$ is also, since it's the image of a convex set under a linear map. (The map of taking marginal distribution). For 2), if $\mathcal{Q}$ is convex, then it's a linear program with convex constraint, which always have a solution. – Michael Aug 11 '19 at 2:32
• Thanks. Notation on $v_y, v_{\tilde{y}}$ fixed. $\mathcal{P}$ is the set of BCE. – user3285148 Aug 11 '19 at 11:34
• From your definition of $\mathcal{P}$, you can see that indeed it is convex. It is the intersection of two convex sets, one from each condition. – Michael Aug 12 '19 at 3:19
• @Michael. I dont think that your claim about 2) is correct. If convexity was sufficient than $\underset{x \in B_1(0)}{max} 1 \times x$ would have a solution as the unit open ball around zero($B_1(0)$) is convex. You need closedness at some level. – Grada Gukovic Aug 12 '19 at 8:39

Yes, $$Q$$ is convex and closed precisely because $$P$$ is and Q is its image under a continuous function. Therefore, $$Q$$ is compact.