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.

The state of the world has support $\mathcal{V}$.

When DM chooses action $y\in \mathcal{Y}$ and the state of the world is $v\in \mathcal{V}$, she receives the payoff $u(y,v)$.

Let $P_V\in \Delta(\mathcal{V})$ be the DM's prior.

The DM also processes some signal $T$ with support $\mathcal{T}$ distribution $P_{T|V}$ to refine his prior and get a posterior on $V$, denoted by $P_{V|T}$, via the Bayes rule.

Let $S\equiv \{\mathcal{T}, P_{T|V}\}$ be called "information structure".

A strategy for the DM is $P_{Y|T}$. Such a strategy is optimal if it maximises his expected payoff, where the expectation is computed using the posterior.

Let us now define the concept 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) $\sum_{v\in \mathcal{V}}u(y,v) P_{Y,V}(y,v)\geq \sum_{v\in \mathcal{V}}u(k,v) P_{Y,V}(y,v)$ for each $y$ and $k\neq y$.

Theorem 1 in Bergemann and Morris (2016) claims that $P_{Y,V}$ is a 1-player Bayesian Correlated Equilibrium if and only if there exists an information structure $S\equiv \{\mathcal{T}, P_{T|V}\}$ and an optimal strategy $P_{Y|T}$ for the DM such that $P_{Y,V}$ is induced by $P_{Y|T}$, i.e., for each $(y,v)\in \mathcal{Y}\times \mathcal{V}$ $$ (\star) \hspace{1cm}P_{Y,V}(y,v)=\sum_{t\in \mathcal{T}}P_{Y|T}(y|t)P_{T|V}(t|v)P_V(v) $$ [for simplicity, I have assumed that $\mathcal{T}$ is finite]

Question 1:

How does the 1-player Bayesian Correlated Equilibrium induced by the complete information structure look like?

This is my attempt to answer.

The way in which I represent the complete information structure is $$ S^{c}\equiv \{\mathcal{T}\equiv \mathcal{V}, P_{T|V}(t|v)=1\text{ if $t=v$ and $0$ otherwise}\} $$ Under $S^c$, $P_{Y|T}$ is an optimal strategy if for each $t\in \mathcal{T}$ and for each $y\in \mathcal{Y}$ such that $P_{Y|T}(y|t)>0$ we have that $$ u(y,t)\geq u(k,t) \text{ }\forall k\neq y $$ [Note that, even under the complete information structure, the optimal strategy may be mixed, if two actions lead to the same payoff $u$.]

Hence, from ($\star$) and for each $(y,v)$ $$ P^{c}_{Y,V}(y,v)=\sum_{t\in \mathcal{T}}P_{Y|T}(y|t)P_{T|V}(t|v)P_V(v)= \sum_{t\in \mathcal{V}}P_{Y|T}(y|t)P_{T|V}(t|v)P_V(v)= P_{Y|T}(y|v)P_V(v) $$

For example, suppose $\mathcal{Y}\equiv \{1,2,3\}$, $\mathcal{V}\equiv \{1,2,3\}$, $P_V(1)=P_V(2)=P_V(3)=1/3$, and and $$ u(1,1)=2, u(1,2)=4, u(1,3)=3\\ u(2,1)=2, u(2,2)=3, u(2,3)=3\\ u(3,1)=1, u(3,2)=3, u(3,3)=3\\ $$ Then, a possible optimal $P_{Y|T}$ under $S^c$ is $$ P_{Y|T}(1|1)=1/2, P_{Y|T}(1|2)=0, P_{Y|T}(1|3)=1/3\\ P_{Y|T}(2|1)=1/2, P_{Y|T}(2|2)=1/3, P_{Y|T}(2|3)=1/3\\ P_{Y|T}(3|1)=0, P_{Y|T}(3|2)=1/3, P_{Y|T}(3|3)=1/3\\ $$ and the corresponding 1-player Bays Correlated Equilibrium is $$ P^c_{Y,V}(1,1)=1/6, P^c_{Y,V}(1,2)=0, P^c_{Y,V}(1,3)=1/9\\ P^c_{Y,V}(2,1)=1/6, P^c_{Y,V}(2,2)=1/9, P^c_{Y,V}(2,3)=1/9\\ P^c_{Y,V}(3,1)=0, P^c_{Y,V}(3,2)=2/9, P^c_{Y,V}(3,3)=1/9\\ $$

Question 2:

Is it true that, for each $v\in \mathcal{V}$, $P^{c}_{Y|V}(y|v)\equiv \frac{P^{c}_{Y,V}(y,v)}{P_V(v)}$ should be equal to $1$ for a $y\in \mathcal{Y}$ and zero otherwise?

Is it true that, for each $y\in \mathcal{Y}$, $P^{c}_{V|Y}(v|y)\equiv \frac{P^{c}_{Y,V}(y,v)}{\sum_{v\in \mathcal{V}}P^c_{Y,V}(y,v)}$ should be equal to $1$ for a $v\in \mathcal{V}$ and zero otherwise?

Question 3: Is it true that by adding the constraint $P_{Y,V}(y,v)>0$ (strictly) for all $(y,v)\in \mathcal{Y}\times \mathcal{V}$ in the definition of 1-player Bayes Correlated Equilibrium above we exclude $P^c_{Y,V}$? Why?


1 Answer 1


Question 1

Yes, the BCE induced by a completely informative information structure will look like this.

This is true even though there are other ways to represent fully informative information structures. Think of $T$ as labels. A fully informative information structure should use each element of $T$ to label only one state of the world. That way, when the DM observes the realized label, $t$, they know which is the state. By assuming that $T=V$, a natural way to label each state with a different label is to assign each state their own label, i.e. $P_{T|V}(t|v)=1$ if and only if $t=v$. However, reshuffling the labels is as informative. For example if the label "low" is sent with probability 1 when the state is "high" and the label "high" is sent with probability 1 when the state is "low" the information structure is not in $S^c$, but is also perfectly informative because after receiving the signal (or label) "low" the DM learns that the state is "high" for sure. (There are ways to represent this formally using permutations, or other ways, but I thought it will be more clear with words).

Question 2

the answer is "Not necessarily" for both questions.

1) If the DM is choosing a mixed strategy (as you correctly point out this is possible even with complete information) then $1>P^c_{Y|V}(y|v)>0$ for the actions that the DM is mixing when they learn that the state is $v$.

2) If there is an action that is optimal for more than one state of the world then $1>P^c_{V|Y}(v|y)>0$ for the states for which $y$ is optimal.

In many papers, people assume that each action is strictly better for one and only one state of the world. In that case, your two statements are true.

Question 3

Yes it is true:

Claim: a BCE that satisfies that $P_{Y,V}(y,v)>0$ for all $(y,v)\in Y\times V$ cannot be induced by a completely informative signal.

Proof: Proceed by contradiction. Suppose a BCE, $P_{Y,V}^*$, satisfies the restriction and is induced by a completely informative information structure.

Consider some state, $v_0$, for which action $y_0$ is not optimal. Note that if there is no such a state, then the decision problem would be trivial since all actions would be optimal for all states of the world. I assume that such a pair exists.

Now, without loss of generality, we can assume that the completely informative structure is the signal defined as $S^c$.

By assumption $P_{Y,V}^*(y_0,v_0)=\sum_{t\in T}P_{Y|T}(y_0|t)P_{T|V}^c(t|v_0)P_V(v_0)>0$, therefore at least one of the summands must be strictly positive. However, $P_{T|V}^c(t|v)=0$ for all $t\neq v_0$. Further, for $t=v_0$, it must be that $P_{Y|T}(y|v_0)=0$ since $y_0$ is not optimal given $v_0$.

We conclude that $P_{Y,V}^*(y_0,v_0)=0$, i.e. a contradiction!

  • $\begingroup$ Thanks a lot: regarding question 3, can you use the claim to answer my question? My attempt: if there exists $(y,v)\in \mathcal{Y}\times \mathcal{V}$ such that $u(y,v)< u(k,y)$ for each $k\neq y$, then by imposing $P^c_{Y,V}(y,v)>0$ for each $(y,v)\in \mathcal{Y}\times \mathcal{V}$ we are able to exclude $P^c_{Y,V}$ from the set of 1-player Bayes Correlated Equilibria. $\endgroup$
    – Star
    Commented Feb 18, 2020 at 22:35
  • $\begingroup$ Actually: it should be for SOME $k\neq y$, rather than FOR EACH $k\neq y$ as I erroneously wrote. $\endgroup$
    – Star
    Commented Feb 19, 2020 at 13:02
  • 1
    $\begingroup$ I'll edit my answer to prove the statement via contradiction. $\endgroup$
    – Regio
    Commented Feb 19, 2020 at 19:29

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