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?

  • 1
    $\begingroup$ in this point $\mathbb{E}_{l(\tau)}l(t)( y)$ you should write $\mathbb{E}_{\tau}l(y| t)$....i think $\endgroup$ Sep 25, 2021 at 11:53
  • 1
    $\begingroup$ I corrected this... $\endgroup$
    – Nav89
    Sep 25, 2021 at 11:58

1 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 $y$.

  • $\begingroup$ The point is that the index $l(\tau)$ as it is written is not convenient to work. it says $\mathbb{E}_{l(\tau)}$ but instead the probability measure is $l(\tau)(y)$. I'd prefer the notation to be $\mathbb{E}_{l(y|\tau)}$. Even further, $l(\tau)(y)=l(y|\tau)$, so by conditioning on $\tau$ it means that $\tau$ is "known" and the only "unkown" parameter here is $y$, isn't it? $\endgroup$
    – Nav89
    Sep 28, 2021 at 7:13
  • $\begingroup$ As for the integrals, I am trying to think how could someone write all these in terms of integrals...this does not have to do if they are finite or not. How could the notation become? Say taht $T$ and $Y$ are some intervals that are subsets of real numbers... $\endgroup$
    – Nav89
    Sep 28, 2021 at 7:18
  • $\begingroup$ $l(τ)(y)$ is a number, not a measure. $\endgroup$ Sep 28, 2021 at 10:52
  • $\begingroup$ Wait....what? if $P$ is the probability measure then $\mathbb{E}_{P}$ represents the expectation operator over $P$. Furthermore $l$ denotes a pdf if you read the definition carefully and since this is the case it is a probability measure. If you understand something else, tell me what am I missing here.... $\endgroup$
    – Nav89
    Sep 28, 2021 at 12:46
  • $\begingroup$ Well, $l(\tau)$ is a probability measure on $Y$ and it assigns the value $l(\tau)(y)$ to the point $y$. $\endgroup$ Sep 28, 2021 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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