<p><strong>Question 1</strong></p>

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

<p>This is true even though there are other ways to represent fully informative information structures. Think of <span class="math-container">$T$</span> as labels. A fully informative information structure should use each element of <span class="math-container">$T$</span> to label only one state of the world. That way, when the DM observes the realized label, <span class="math-container">$t$</span>, they know which is the state. By assuming that <span class="math-container">$T=V$</span>, a natural way to label each state with a different label is to assign each state their own label, i.e. <span class="math-container">$P_{T|V}(t|v)=1$</span> if and only if <span class="math-container">$t=v$</span>. 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 <span class="math-container">$S^c$</span>, 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). </p>

<p><strong>Question 2</strong></p>

<p>the answer is "Not necessarily" for both questions. </p>

<p>1) If the DM is choosing a mixed strategy (as you correctly point out this is possible even with complete information) then <span class="math-container">$1&gt;P^c_{Y|V}(y|v)&gt;0$</span> for the actions that the DM is mixing when they learn that the state is <span class="math-container">$v$</span>. </p>

<p>2) If there is an action that is optimal for more than one state of the world  then <span class="math-container">$1&gt;P^c_{V|Y}(v|y)&gt;0$</span> for the states for which <span class="math-container">$y$</span> is optimal. </p>

<p>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.</p>

<p><strong>Question 3</strong></p>

<p>Yes it is true:</p>

<p>Claim: a BCE that satisfies that <span class="math-container">$P_{Y,V}(y,v)&gt;0$</span> for all <span class="math-container">$(y,v)\in Y\times V$</span> cannot be induced by a completely informative signal. </p>

<p>Proof: Proceed by contradiction. Suppose a BCE, <span class="math-container">$P_{Y,V}^*$</span>, satisfies the restriction and is induced by a completely informative information structure. </p>

<p>Consider some state, <span class="math-container">$v_0$</span>, for which action <span class="math-container">$y_0$</span> 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. </p>

<p>Now, without loss of generality, we can assume that the completely informative structure is the signal defined as <span class="math-container">$S^c$</span>. </p>

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

<p>We conclude that <span class="math-container">$P_{Y,V}^*(y_0,v_0)=0$</span>, i.e. a contradiction!</p>

CC BY-SA 4.0
2020-02-18T22:35:03.223Z
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.
Star
CC BY-SA 4.0
2020-02-19T13:02:19.610Z
Actually: it should be for SOME $k\neq y$, rather than FOR EACH $k\neq y$ as I erroneously wrote.
Star
CC BY-SA 4.0
2020-02-19T19:29:32.313Z
I'll edit my answer to prove the statement via contradiction.
Regio
CC BY-SA 4.0
2020-02-18T21:59:23.437Z
2020-02-19T19:49:49.763Z
2020-02-19T19:49:49.763Z
Regio
<p><strong>Model</strong></p>

<p>Consider a game where a decision maker (DM) has to choose action <span class="math-container">$y\in \mathcal{Y}$</span> possibly without being fully aware of the state of the world. </p>

<p>The state of the world has support <span class="math-container">$\mathcal{V}$</span>. </p>

<p>When DM chooses action <span class="math-container">$y\in \mathcal{Y}$</span> and the state of the world is <span class="math-container">$v\in \mathcal{V}$</span>, she receives the payoff <span class="math-container">$u(y,v)$</span>. </p>

<p>Let <span class="math-container">$P_V\in \Delta(\mathcal{V})$</span> be the DM's prior. </p>

<p>The DM also processes some signal <span class="math-container">$T$</span> with support <span class="math-container">$\mathcal{T}$</span> distribution <span class="math-container">$P_{T|V}$</span>  to refine his prior and get a posterior on <span class="math-container">$V$</span>, denoted by <span class="math-container">$P_{V|T}$</span>, via the Bayes rule. </p>

<p>Let <span class="math-container">$S\equiv \{\mathcal{T}, P_{T|V}\}$</span> be called "information structure". </p>

<p>A strategy for the DM is <span class="math-container">$P_{Y|T}$</span>. Such a strategy is optimal if it maximises his expected payoff, where the expectation is computed using the posterior. </p>

<p>Let us now define the concept of 1-player Bayesian Correlated Equilibrium provided in Bergemann and Morris (2013,2016,etc.).</p>

<p><span class="math-container">$P_{Y,V}\in  \Delta(\mathcal{Y}\times \mathcal{V})$</span> is a 1 player Bayesian Correlated Equilibrium if</p>

<p>1) <span class="math-container">$\sum_{y\in \mathcal{Y}}P_{Y,V}(y,v)=P_V(v)$</span> for each <span class="math-container">$v\in \mathcal{V}$</span></p>

<p>2) <span class="math-container">$\sum_{v\in \mathcal{V}}u(y,v) P_{Y,V}(y,v)\geq \sum_{v\in \mathcal{V}}u(k,v) P_{Y,V}(y,v)$</span> for each <span class="math-container">$y$</span> and <span class="math-container">$k\neq y$</span>.</p>

<hr>

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

<hr>

<p><strong>Question 1:</strong> </p>

<p><strong>How does the 1-player Bayesian Correlated Equilibrium induced by the complete information structure look like?</strong></p>

<p>This is my attempt to answer. </p>

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

<p>Hence, from (<span class="math-container">$\star$</span>) and for each <span class="math-container">$(y,v)$</span>
<span class="math-container">$$
P^{c}_{Y,V}(y,v)=\sum_{t\in \mathcal{T}}P_{Y|T}(y|t)P_{T|V}(t|v)P_V(v)=
\sum_{t\in \mathcal{V}}P_{Y|T}(y|t)P_{T|V}(t|v)P_V(v)=
P_{Y|T}(y|v)P_V(v)
$$</span></p>

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

<hr>

<p><strong>Question 2:</strong> </p>

<p>Is it true that, for each <span class="math-container">$v\in \mathcal{V}$</span>, <span class="math-container">$P^{c}_{Y|V}(y|v)\equiv \frac{P^{c}_{Y,V}(y,v)}{P_V(v)}$</span> should be equal to <span class="math-container">$1$</span> for a <span class="math-container">$y\in \mathcal{Y}$</span> and zero otherwise? </p>

<p>Is it true that, for each <span class="math-container">$y\in \mathcal{Y}$</span>,  <span class="math-container">$P^{c}_{V|Y}(v|y)\equiv \frac{P^{c}_{Y,V}(y,v)}{\sum_{v\in \mathcal{V}}P^c_{Y,V}(y,v)}$</span> should be equal to <span class="math-container">$1$</span> for a <span class="math-container">$v\in \mathcal{V}$</span> and zero otherwise? </p>

<hr>

<p><strong>Question 3:</strong>
Is it true that by adding the constraint <span class="math-container">$P_{Y,V}(y,v)&gt;0$</span>  (<strong>strictly</strong>) for all <span class="math-container">$(y,v)\in \mathcal{Y}\times \mathcal{V}$</span> in the definition of 1-player Bayes Correlated Equilibrium above  <strong>we exclude <span class="math-container">$P^c_{Y,V}$</span>? Why?</strong></p>

<game-theory><bayesian-game>
Bayes Correlated Equilibrium with complete information
Star
# Bayes Correlated Equilibrium with complete information

(Asked by: username_0 on 2020-02-18)

**Model**

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?**

**Tags:** game-theory, bayesian-game

---

## 1 answers

### Answer by username_1 on 2020-02-18. Score: 1 | Accepted answer

**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!

> **Comments:**
> **username_0**: 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.  
> **username_0**: Actually: it should be for SOME $k\neq y$, rather than FOR EACH $k\neq y$ as I erroneously wrote.  
> **username_1**: I'll edit my answer to prove the statement via contradiction.  

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