Mechanisms of communication in game theory

Question

In the spirit of the previous question that I have done, here considering the paper here I am trying to make the matching definition $2.2$ here.

I will give two definitions and I would like to clarify how are they connected

$\textit{Definition of a communication mechanism:}$ A communication mechanism is a triple $\mathcal{C}=((T^i)_i, (Y^i)_i , l )$, where $T^i$ is $i's$ finite set of messages, $Y^i$ is $i's$ finite set of signals, and $l: T=(\Pi_{i\in I} T^i)\to \Delta(Y)$ is the signal function. When $t$ is the profile of messages sent by the players to the mechanism, $y\in Y$ is drawn according to $l(t)$ and player $i$ is informed of $y_i$. Furthermore, $\mathcal{T}_i=\Delta(T_i)$ represents the set of mixed messages for player $i$ and $l$ is extended to $\mathcal{T}$ by

$$\underbrace{l(\tau)( y)}_{l(y|\tau)}=\mathbb{E}_{\tau} l(t)( y)=\mathbb{E}_{\tau}l(y|t)=\sum_{\tau\in \mathcal{T}}l(y|t)\times \tau(t)$$.

In this point I clarify that, $$l(\tau)(y)=l(y|\tau),\quad\text{some different way of defining the conditional expectation}$$ and hence since $l(\tau)(y)$ is a function of $y$ we could write for sipmlicity

$$\nu(y):=\sum_{\tau\in \mathcal{T}}l(y|t)\times \tau(t)=l(\tau)(y)$$

$\textit{Q1:}$ Is it right to redifine $l(\tau)(y)$ to $\nu(y)$ and how could I change the $\sum$ to $\int$? I mean could it be right to ew-write $\nu(y)=\int_{\mathcal{T}}l(t|y)d(\tau(t))?$

the next definition says

$\textit{Definition of a game with a communication mechanism:}$ Given a compact game $G$ and a communication mechanism $\mathcal{C}$, then if $\Gamma(\mathcal{C}, G)$ is the game $G$ extended by $\mathcal{C}$, it unfolds as follows:

• each player $i$ sends a message $t^i$ to the mechanism
• $y\in Y$ is drawn according to $l(t)$ and each player $i$ is informed of $y^i$
• each player $i$ chooses $\sigma^i\in \Sigma^i$ according to $y^i$
• the vector payoff is $g(\sigma)$ A strategy for player $i$ is given by a mixed message $\tau^i\in \mathcal{T}^i$ and by a mapping $F^i: Y^i\to \Sigma^i$. The payoff function is given by $$g_{\mathcal{C}}(\tau, F) =\mathbb{E}_{l(\tau)}g(F(y))$$

$\textit{Q1:}$ I assume that since y is drawn according to l(\tau) then the index in the last payoff function should be $\mathbb{E}_{l(\tau)(y)}$ instead of $\mathbb{E}_{l(\tau)}$ why is this not the case?

$\textit{Q3:}$ By writing donw the the payoff function in terms of sums or integrals (instead of terms in expextations) if the probability disrtibution is continuous, what is going to be the formuation?

Answer 1

First, $l(\tau)(y)$ is a function of both $\tau$ and $y$, and the dependence on $\tau$ is essential; this is how communication happens. Since $T$ and $Y$ are finite, there is no point in introducing integrals.

$\mathbb{E}_{l(\tau)}g(F(y))$ is the expectation over the function when the random value $y$ is distributed according to $l(\tau)$. That is, $$\mathbb{E}_{l(\tau)}g(F(y))=\sum_{y\in Y}\Big(g(F(y))\cdot l(\tau)(y)\Big).$$ Here, $l(\tau)$ is a distribution over $Y$ that assigns probability $l(\tau)(y)$ to $y$.