3
$\begingroup$

Consider a game where a decision maker (DM) has to choose action $y\in \mathcal{Y}$ possibly without being fully aware of the state of the world $V$. The state of the world has support $\mathcal{V}$. The DM receives the payoff $u(y,v)$ depending on the chosen action $y$ the realisation $v$ of $V$. Let $P_V\in \Delta(\mathcal{V})$ be the DM's prior.

Is the following the correct definition of 1 player Bayesian Correlated Equilibrium provided in Bergemann and Morris (2013,2016,etc.)?

$P_{Y,V}\in \Delta(\mathcal{Y}\times \mathcal{V})$ is a 1 player Bayesian Correlated Equilibrium if

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

2) $\sum_{v\in \mathcal{V}}u(y,v)P_{Y,V}(y,v)\geq \sum_{v\in \mathcal{V}}u(\tilde{y},v)P_{Y,V}(y,v)$ for each $y$ and $\tilde{y}\neq y$.

In particular, I have doubts about $2)$: what if there is a $y$ such that $P_{Y,V}(y,v)=0$ for each $v\in \mathcal{V}$? Am I missing something?

$\endgroup$
  • 3
    $\begingroup$ Shouldn't 1 be a summation over $y$? As for 2, if your notation is correct, then if $P_{Y,V}(y,v) = 0$ for each $v$, both left and right hand sides are $0$? I do not see the problem? $\endgroup$ – Walrasian Auctioneer Nov 21 at 0:31
  • $\begingroup$ Yes, changed it sorry $\endgroup$ – user3285148 Nov 21 at 10:19
2
$\begingroup$

The concept of the BCE from their 2016 paper is similar to what you have. I think Bergemann and Morris' intuitive explanation is valuable so I'll paraphrase it here.

Each player in the game has a decision rule that chooses an action, $y$, dependent on the state of the world $V$, and the player's information set, which we'll call $S$. This information set includes both a finite set of signals for each player, $T_i$, and a signal distribution, $\pi: \mathcal{V} \rightarrow \Delta T$. As you wrote your example, you assume the set of signals is a singleton, leaving us with only a player's prior. This is a possible information structure, but is not necessary.

We can thus write the the decision rule as a mapping, $\sigma$,

\begin{align*} \sigma : S \times V \rightarrow \Delta Y \end{align*}

The lone criteria for a CBE in this setting is that each players' decision rule is ``obedient''. By obedient we simply mean that the action, $y$, chosen by the decision rule must be the optimal action for the player. Thus, a player will always follow the action chosen by their decision rule.

I believe you are confusing the information structure and the decision rule. My information set is not a function of the action I choose in this setting, so $P_{V,Y}(y,v)$ does not have any meaning. Thus, you don't need to be concerned about the existence of a $y$ such that $P_{V,Y}(y,v)=0$ for all $v$.

It is possible to be in this setting that there exists an action $y$ such that $\sigma(y_i|t_i)=0$ for all signals, $t$. But this would simply mean the player never chooses that action in equilibrium.

Is it possible that there exists a signal $t$ such that $\sigma(y_i|t_i)=0$ for all actions, $y$? No, and it would follow for the basic Nash existence proof, given certain constraints on $u(\cdot),$ $\mathcal{Y}$ and $\mathcal{V}$.

$\endgroup$
1
$\begingroup$

You have specialized the definition of BCE in two dimensions: there is only one player, and the player has no private information. If you want to allow for private information you can let the player have some signal $\pi:\mathcal{V}\rightarrow\Delta(T_i)$

And let the decision rule $P_{\mathcal{Y},\mathcal{T},\mathcal{V}}\in\Delta(\mathcal{Y}\times \mathcal{T}\times \mathcal{V})$ be a single-player BCE if

  1. $\sum_{y\in Y}P_{\mathcal{Y},\mathcal{T},\mathcal{V}}(y,t,v)=\pi(t|v)P_{\mathcal{V}}(v)$

  2. For each $t\in \mathcal{T}$, and $y\in \mathcal{Y}$: $$\sum_{v\in \mathcal{V}}u(y,v)P_{\mathcal{Y},\mathcal{T},\mathcal{V}}(y,t,v)\geq\sum_{v\in \mathcal{V}}u(\tilde y,v)P_{\mathcal{Y},\mathcal{T},\mathcal{V}}(y,t,v)$$ for all $\tilde y\neq y$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.