# Correlation device that induces a specific transition probability

Taking a look at this paper of Forges and Vida the authors define a correlation device in page $$102$$, that is a standard probability space $$\left(\Omega,\mathcal{B},\mu\right)$$, They assume that the probability space represents the extraneous events, select in $$\left(\Omega,\mathcal{B}\right)$$ according to $$\mu$$ and are idepedent to the game. Well one fisrt question here is, what changes if the extraneous events are not indepedent to the game?

The correlation device is added to the extended game by cheap talk phase, but I can not understand the reason. Finally in page $$103$$ in the sketch or proof parapgraph, the following senteces is quoted. Let $$q$$ be a communication equilibrium of $$\Gamma$$ (it is the basic game). We gradually construct a set of messages $$M$$, a correlation device $$\mu$$ for $$ext_M\Gamma$$, and equilibrium strategies in $$(ext_M\Gamma)^{\mu}$$ that induce the transition probability $$q$$"

Note that the transition probability is a mapping that assigns the profile of types send by the players to the communication mechanism $$q$$ to the set of actions.

$$\textbf{Question:}$$ What does it mean that they construct a correlation device $$\mu$$ for the extended by cheap talk phase game $$ext_M\Gamma$$, and equilibrium strategies in $$(ext_M\Gamma)^{\mu}$$ that induce the transition probability $$q$$? What does the term induce denotes? Are $$\mu$$ and $$q$$ connected mathematically somehow? How are tese two types of equilibria, the communication equilibria with a mediator and the equilibrium with the correlation device are connected?

The common prior $$p\in\Delta L$$ and the transition probability $$q:L\to\Delta A$$ induce a joint distribution on $$L\times A$$ in which the pair $$(l,a)$$ is selected with probability $$p_l\cdot q_l(a)$$. You can also recover $$q$$ outside (strategically irrelevant) type profiles $$l$$ such that $$p_l$$ by the usual formula for conditional probabilities.
Now, an equilibrium of the extended game leads to eventually (possibly "at infinity", this is long cheap talk) both players choosing an action. You can then calculate the joint distribution of types and chosen actions. If it is the same as the distribution above, we can say that the equilibrium of the extended game induces $$q$$.
If the randomization device would not be independent of the type distribution, you would need to actually define a joint distribution on $$\Omega\times L$$. If this distribution is not independent, a player may learn about the other player's type than what they learn from their own type alone. In that case, $$\Omega$$ would be a signal of types and not just a correlation device used to implement correlated equilibria of the extended game.
• So, you mean that by adding the state space $(\Omega,\mathcal{B},\mu)$ it is like we are adding one more variable to our model, that randomizes somehow the game. Thus, instead of having $p(l)q(a|l)$, we have a new factor added and this becomes, $\mu(\omega)p(l)q(a|l)$? What if $l$ was based on $\omega$ in the sense that given $\omega\in\Omega$ then $l$ is drawn according to $p:\omega\to\Delta(l)$ and hence $(\omega,l,a)$ were selected according to $\mu(\omega) p(l_i,l_{-i}|\omega)q(a|l_i,l_{-i},\omega)$? This type or randomiation still hilds right? Commented Dec 11, 2021 at 8:48
• In the model, the correlation between $\omega$ and $l$ comes from the strategies of the players. Why should a player's type be drawn according to a correlation device? This makes little sense for the interpretation. Commented Dec 11, 2021 at 9:28
• The correlation comes between $\omega$ and $l$ comes from the strategies, but the authors have not seen the connection somehow in the maths. It's just an assumption, but where does this help? In the correlated equilibrium of the cheap talk game, the mediator (or device) does not receive any information from the players. He makes recommendations on how to exchange messages, but remains fully ignorant of the players’ types. (see in page $98$). But from the $(ext_M\Gamma)^{\mu}$ game $q$ is induced. How is this so? How can one probability measure $\mu$ induce another $q$ without any dependence? Commented Dec 11, 2021 at 10:32
• With the caononical game, they take recommendations according to $q$ which in essence means that this $q$ is the correlated strategy of the extended game by the device when the players are sincere and obednient in equilibrium. But what is the connection between $\mu$ and $q$ in cheap talk game? Intuitively and in terms of maths... I don't know if I am clear....in my question...in the previous comment.... Commented Dec 11, 2021 at 10:36