<p>We have that <span class="math-container">${\cal I} = ((X^i)_i, \mu)$</span>​ and <span class="math-container">${\cal J} = ((Y^i)_i, \nu)$</span>​ are two information structures.</p>
<p>An Interpretation mapping for player <span class="math-container">$i$</span>​​ is a mapping <span class="math-container">$\phi^i: X^i \to \Delta(Y^i)$</span>​ so it associates with every <span class="math-container">$x^i$</span>​ a distribution over <span class="math-container">$Y^i$</span>​.</p>
<p>Let <span class="math-container">$x^i \in X^i$</span>​. Then <span class="math-container">$\phi^i(x^i)$</span>​ is a distribution over <span class="math-container">$Y^i$</span>​ so <span class="math-container">$\phi(x^i)(y^i)$</span>​ is the probability of this distribution attached to outcome <span class="math-container">$y^i$</span>​.</p>
<p>Then for <span class="math-container">$x = (x^1, \ldots, x^I)$</span> we can define:
<span class="math-container">$$
\phi: \prod_i X^i \to \Delta\left(\prod_i Y^i\right)
$$</span>
Such that for all <span class="math-container">$x = (x^1,\ldots, x^I)$</span> and <span class="math-container">$y = (y^1,\ldots, y^I)$</span> we have the product probability measure:
<span class="math-container">$$
\phi(x)(y) = \phi^1(x^1)(y^1) \cdot \phi^2(x^2)(y^2) \cdot \ldots \cdot \phi^I(x^I)(y^I)
$$</span>
We have that an interpretation <span class="math-container">$\phi= (\phi^i)_i$</span>​ is <strong>consistent</strong> if for every <span class="math-container">$x \in \prod_i X^i$</span> and <span class="math-container">$y \in \prod_i Y^i$</span>:
<span class="math-container">$$
\mathbb{E}_\mu \phi(x)(y^i) = \nu(y),\\
\leftrightarrow \sum_{x \in X} \phi(x)(y) \,\, \mu(x) = \nu(y)
$$</span>
Now define <span class="math-container">$\phi(x)(y) \mu(x) = P(x,y)$</span>​. Then the sum of <span class="math-container">$P(x, y)$</span> over all <span class="math-container">$x$</span> and <span class="math-container">$y$</span> equals 1:
<span class="math-container">$$
\sum_x \sum_y P(x,y) = \sum_x \sum_y \phi(x)(y)\,\, \mu(x) = \sum_y \nu(y) = 1
$$</span>
Also the marginal distributions satisfy:
<span class="math-container">$$
\begin{align*}
&amp;\sum_x P(x,y) = \sum_x \phi(x)(y)\mu(x) = \nu(y),\\
&amp;\sum_y P(x,y) = \sum_y \phi(x)(y) \mu(x) = \mu(x) \sum_y \phi(x)(y) = \mu(x).
\end{align*}
$$</span>
So <span class="math-container">$P(x,y)$</span>​​​ act as a joint probability distribution over <span class="math-container">$\prod_i X^i \times \prod_i Y^i$</span> with marginals <span class="math-container">$\nu$</span> and <span class="math-container">$\mu$</span>. Then <span class="math-container">$\phi(x)(y) = \dfrac{P(x,y)}{\mu(x)}$</span> which  is the same as the conditional probability <span class="math-container">$\phi(x)(y) = P(y|x)$</span>​​.</p>
<p>We can write the marginals <span class="math-container">$P(y) = \nu(y)$</span>​ and <span class="math-container">$P(x) = \mu(x)$</span>​. Also, let <span class="math-container">$P(y^i)$</span> be the marginal distribution of <span class="math-container">$P$</span> with respect to <span class="math-container">$y^i$</span>, <span class="math-container">$P(x^i)$</span> the marginal distribution of <span class="math-container">$P$</span> with respect to <span class="math-container">$x^i$</span> and so on.</p>
<p>Then we can define:
<span class="math-container">$$
q(y^i)(y^{-i}) = P(y^{-i}|y^i) = \dfrac{P(y^i, y^{-i})}{ P(y^i)}.
$$</span>
And:
<span class="math-container">$$
r(x^i)(y^{-i}) = P(y^{-i}|x^i) = \dfrac{ P(x^i, y^{-i})}{ P(x^i)}.
$$</span></p>
<blockquote>
<p>I think that <span class="math-container">$q(y^i)(y^{−i})$</span> equals to <span class="math-container">$\frac{\nu(y^{-i},y^i)}{\nu(y^i)}$</span></p>
</blockquote>
<p>I think you are correct. Notice that <span class="math-container">$P(y^i, y^{-i}) = \nu(y^i, y^{-i})$</span> and <span class="math-container">$P(y^i) = \sum_{y^{-i}} P(y^i, y^{-i}) = \sum_{y^{-i}} \nu(y^i, y^{-i}) = \nu(y^i)$</span> so indeed (by Bayes rule):
<span class="math-container">$$
q(y^i)(y^{-i}) = \dfrac{P(y^i, y^{-i})}{P(y^i)} = \dfrac{\nu(y^i, y^{-i})}{\nu(y^i)} = \nu(y^{-i}|y^i)
$$</span></p>
<blockquote>
<p>Also what is the formula of <span class="math-container">$r(x^i)(y^{−i})$</span> and how it ends up to the result that <span class="math-container">$r(x^i)=\mathbb{E}_{p(x^i)}\phi^{-i}(x^{-i})$</span>?</p>
</blockquote>
<p>The comuputation for this expression is a bit more involved. First:
<span class="math-container">$$
\begin{align*}
P(x^i, y^{-i}) &amp;= \sum_{x^{-i}, y^{i}} P(x^i, x^{-i}, y^i, y^{-i}),\\
&amp;= \sum_{x^{-i}, y^{i}} \phi(x^i, x^{-i})(y^i, y^{-i}) \,\, \mu(x^i, x^{-i}),\\
&amp;=\sum_{x^{-i}, y^i} \mu(x^i, x^{-i}) \left[\left(\prod_{j \ne i} \phi^j(x^j)(y^j)\right)\phi^i(x^i)(y^i) \right],\\
&amp;=\sum_{x^{-i}} \mu(x^i, x^{-i}) \left(\prod_{j \ne i}\phi^j(x^j)(y^j)\right) \underbrace{\sum_{y^{i}} \phi^i(x^i)(y^i)}_{=1},\\
&amp;=\sum_{x^{-i}} \mu(x^i,x^{-i})\,\, \phi^{-i}(x^{-i})(y^{-i}),\\
\end{align*}
$$</span>
Next one has,
<span class="math-container">$$
P(x^i) = \sum_{x^{-i}} P(x^i, x^{-i}) =  \sum_{x^{-i}} \mu(x^i, x^{-i}) = \mu(x^i)
$$</span>
Then:
<span class="math-container">$$
\begin{align*}
r(x^i)(y^{-i}) &amp;= \dfrac{P(x^i, y^{-i})}{P(x^i)},\\
&amp;= \sum_{x^{-i}} \frac{\mu(x^i, x^{-i})}{\mu(x^i)} \phi^{-i}(x^{-i})(y^{-i}),\\
&amp;= \sum_{x^{-i}} p(x^i)(x^{-i})\,\, \phi^{-i}(x^{-i})(y^{-i}),\\
&amp;= \mathbb{E}_{p(x^i)} \phi^{-i}(x^{-i})(y^{-i})
\end{align*}
$$</span></p>
<h3>Additional questions and answers</h3>
<blockquote>
<ol>
<li>Does it mean that the product measure is always translated to a conditional probability?</li>
</ol>
</blockquote>
<p>No, that is why the authors impose the defintion that the interpretation should be <strong>consistent</strong>.</p>
<blockquote>
<ol start="2">
<li>Have you seen any textbook that has the same notation as in here?</li>
</ol>
</blockquote>
<p>No. In my opinion, the notation is rather inconvenient to work with ;-)</p>
<blockquote>
<ol start="3">
<li>Why <span class="math-container">$P(y^i, y^{-i}) = \nu(y^i, y^{-})$</span>.</li>
</ol>
</blockquote>
<p>Let <span class="math-container">$y = (y^i, y^{-i})$</span> then:</p>
<p><span class="math-container">$P(y^i, y^{-i}) = P(y)$</span> which is the marginal distribution of <span class="math-container">$P(x,y)$</span> with respect to <span class="math-container">$y$</span>. Condition <span class="math-container">$(1)$</span> shows that this is equal to <span class="math-container">$\nu(y) = \nu(y^i, y^{-i})$</span>.</p>
<blockquote>
<ol start="4">
<li>How did you figure out <span class="math-container">$P(x^i, y^{-i}) = \sum_{x^{-i}, y^i} P(x^i, x^{-i}, y^i, y^{-i})$</span>.</li>
</ol>
</blockquote>
<p>Similar as in point 3, <span class="math-container">$P(x^i, y^{-i})$</span> is the marginal distribution <span class="math-container">$P$</span> with respect to <span class="math-container">$(x^i, y^{-i})$</span>, so you have to 'integrate out' <span class="math-container">$(x^{-i}, y^i)$</span>. As these are finite, this amounts simply to taking the sum.</p>

CC BY-SA 4.0
2021-09-23T14:57:35.803Z
wow...and thanks! I couldn't ask for moore...so let me check if there is anything that i can not understand for a little more...clarification....
Nav89
CC BY-SA 4.0
2021-09-23T15:21:01.597Z
1. You say by the product probability measure we have:

$$\phi(x)(y)=\phi^1(x^1)(y^1)\dot\phi^2(x^2)(y^2)\dot \dots\phi^I(x^I)(y^I)$$
and also $\phi(x)(y)=P(y|x)$ does it mean that the product measure is always translated to a conditional probability?

2. Have you seen any textbook that has the same notation as in here?

3. Why $P(y^i,y^{-i})=\nu(y^i,y^{-i})$? I think that you will say by definition, but I am missing something that makes me not seeing it.

4. How did you figure out $P(x^i,y^{-i})=\sum_{x^{-i}, y^{i}} P(x^i, x^{-i}, y^i, y^{-i})$, $\sum_{y^{i}} \phi^i(x^i)(y^i)=1?$
Nav89
CC BY-SA 4.0
2021-09-23T17:41:27.827Z
@Nav89 I added my answers to my response above. Heres the [wiki](https://en.wikipedia.org/wiki/Marginal_distribution) on how marginal distributions are computed.
tdm
CC BY-SA 4.0
2021-09-24T07:24:12.417Z
So, to sum up,the probability $\phi$ denotes the transition probability of moving from one state of a system $\mathcal{I}$ into another state to $\mathcal{J}$. The extra condition that is needed is the assumption of consistency, right? But how do we translate consistency here?

And my other question is why $\sum_{y^i}\phi^i(x^i)(y^i)=1$, my understanding is teh follwing:

$$\sum_{y^i}\phi(x^i)(y^i)=\phi^1(y^1)(y^1)+\phi^2(y^2)(y^2)+\dots +\phi^I(y^I)(y^I)=\phi^1(y^1|y^1)+\phi^2(y^2|y^2)+\dots+\phi^I(y^I|y^I)=\phi(y|y)=1$$
Nav89
CC BY-SA 4.0
2021-09-24T11:20:35.163Z
@Nav89 we have that $\phi(x^i)$ is a probability distribution over $Y^i$ so if you add up the probababilities, you get 1: $\sum_{y^i \in Y^i} \phi(x^i)(y^i) = 1$.
tdm
CC BY-SA 4.0
2021-09-24T11:38:26.177Z
so what i wrote is wrong i guess :P but i undes=rstand what you say i am ok
Nav89
CC BY-SA 4.0
2021-09-25T10:30:36.243Z
check this one when you have time, you hace a better understanding in this notation than me...it seems so confusing to me

https://economics.stackexchange.com/questions/47679/mechanisms-of-communication-in-game-theory
Nav89
CC BY-SA 4.0
2021-09-23T13:22:28.547Z
2021-09-23T17:39:46.417Z
2021-09-23T17:39:46.417Z
tdm
<p>The question that I have is a little technical and it has to do with the notation and the combination between some mathamatical properties in the probability theory of information economics.</p>
<p>Say <span class="math-container">$\mathcal{I}=(X,\mu)$</span> and <span class="math-container">$\mathcal{J}=(Y,\nu)$</span> are measurable probability spaces, where <span class="math-container">$\mu$</span> and <span class="math-container">$\nu$</span> denote the probability distributions over <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> respectively. Then, we define <span class="math-container">$\phi:X\to\Delta(Y)$</span>, where <span class="math-container">$\Delta(Y)$</span> is the simplex of <span class="math-container">$Y$</span>. If the image of <span class="math-container">$\mu$</span> by <span class="math-container">$\phi$</span> is <span class="math-container">$\nu$</span> then it holds
<span class="math-container">$$\mathbb{E}_{\mu}\phi(x)(y)=\nu(y),\quad \text{where $\phi(x)(y)$ denotes the conditional probability $\phi(y|x)$}$$</span></p>
<p>Note that <span class="math-container">$\mu$</span> and <span class="math-container">$\nu$</span> are probability measures s.t. for every <span class="math-container">$x^i$</span> such that <span class="math-container">$\mu(x^i)&gt;0$</span>, then <span class="math-container">$p(x^i)\in\Delta(X^{-i})$</span> denotes the conditional probability of <span class="math-container">$\mu$</span> given <span class="math-container">$x^i$</span> over <span class="math-container">$X^{-i}$</span>:
<span class="math-container">$$p(x^i)(x^{-i})=\mu(x^{-i}|x^i)=\frac{\mu(x^{-i},x^{i})}{\mu(x^i)}$$</span></p>
<p>Also, note that <span class="math-container">$\mu(x^i)$</span> stands for <span class="math-container">$\mu(\{x^i\}\times X^{-i})$</span>.</p>
<p>To help you understand the notation, from game theory imagine that <span class="math-container">$i=\{1,2,3\}$</span>, then <span class="math-container">$X=X^1\times X^2\times X^3= \{a^1,b^1\}\times\{a^2,b^2\}\times\{a^3,b^3\}$</span> and <span class="math-container">$x=(\underbrace{a^1}_{x^1},\underbrace{a^2}_{x^2},\underbrace{a^3}_{x^1})$</span></p>
<p>Even furhter, the following probabilities are defined. Let <span class="math-container">$r$</span> denote the ex-ante probability and <span class="math-container">$q$</span> the ex-post probability, they are defined as follows on <span class="math-container">$\Delta(Y^{-i})$</span>:</p>
<p><span class="math-container">$$r(x^i)(y^{−i})= P_{\phi}(y^{−i}|x^i),\quad\text{and}\quad q(y^i)(y^{−i})= P_{\phi}(y^{−i}|y^i)$$</span></p>
<p>for <span class="math-container">$P_{\phi}(x^i,y^i)&gt;0$</span>, where <span class="math-container">$P_{\phi}(x^i,y^i)=\mu(x^i)\phi^{i}(x^i)(y^i)$</span> denotes the probability induced on <span class="math-container">$X\times Y$</span> by <span class="math-container">$\mu$</span> and the transition probability <span class="math-container">$\phi$</span>, then <span class="math-container">$r(x^i)$</span> and <span class="math-container">$q(y^i)$</span> are random vectors with values in <span class="math-container">$\Delta(Y^{-i})$</span> and <span class="math-container">$P_{\phi}(y^i|x^i)=\phi^{i}(x^i)(y^i)$</span> and <span class="math-container">$r(x^i)=\mathbb{E}_{p(x^i)}\phi^{-i}(x^{-i})$</span></p>
<p><span class="math-container">$\textit{Question:}$</span> I struggle to understand how to calculate <span class="math-container">$r()$</span> and <span class="math-container">$q()$</span> with repsect to <span class="math-container">$\mu$</span> and <span class="math-container">$\nu$</span> as well. I can not clarify from the notation how are the ex-post and ex-ante probabilities connected with <span class="math-container">$\phi$</span>, <span class="math-container">$\mu$</span>, <span class="math-container">$\nu$</span> and <span class="math-container">$p$</span>?</p>
<p>I think that <span class="math-container">$q(y^i)(y^{−i})$</span> equals to <span class="math-container">$\frac{\nu(y^{-i},y^i)}{\nu(y^i)}$</span>, but I can not understand how to make the transition from <span class="math-container">$q(y^i)(y^{−i})= P_{\phi}(y^{−i}|y^i)=\dots=\frac{\nu(y^{-i},y^i)}{\nu(y^i)}$</span>. Also what is the formula of <span class="math-container">$r(x^i)(y^{−i})$</span> and how it ends up to the result that <span class="math-container">$r(x^i)=\mathbb{E}_{p(x^i)}\phi^{-i}(x^{-i})$</span>?</p>
<p>Also it seems that <span class="math-container">$p$</span> is the &quot;backwards&quot; version of <span class="math-container">$\phi$</span>. Whereas <span class="math-container">$\phi$</span> tells you the likelihood of each value of <span class="math-container">$Y$</span> given the value of <span class="math-container">$X$</span>, the function <span class="math-container">$p$</span> tells you, given <span class="math-container">$y$</span>, the conditional probability it was produced by each <span class="math-container">$x$</span>. So could I also claim that <span class="math-container">$P_{\phi}(x,y)=\nu(y)p(y)(x)=\mu(x)\phi(x)(y)$</span>?</p>
<p><span class="math-container">$\textit{Hint:}$</span> Keep in mind that the noation <span class="math-container">$f(x)(y)=f(y|x)$</span> in every situation and <span class="math-container">$f(x^i)$</span> is the conditional probability <span class="math-container">$f(x^i)(x^{-i})=f(x^{-i}|x^{i})$</span></p>

CC BY-SA 4.0
2021-09-22T16:42:53.417Z
What text does this come from?
Michael Greinecker
CC BY-SA 4.0
2021-09-22T18:22:39.413Z
@MichaelGreinecker this one https://econ-papers.upf.edu/papers/169.pdf
Nav89
CC BY-SA 4.0
2021-09-23T09:16:25.553Z
@MichaelGreinecker more precisely pages 4 and 5 contain all this information.
Nav89
CC BY-SA 4.0
2021-09-24T10:24:09.740Z
@Giskard is it ok if I am going yo edit one or 2 more questions here? I do not want to write a new question, since the new questions will be implied within the same framework
Nav89
CC BY-SA 4.0
2021-09-24T11:22:12.430Z
@nav89 In this case I think you should copy info from the Q, not edit answered Q's.
Giskard
CC BY-SA 4.0
2021-09-24T12:23:38.027Z
`@Giskard ok i will do this right now
Nav89
CC BY-SA 4.0
2021-09-25T10:29:46.027Z
@Giskard check this one when you have time https://economics.stackexchange.com/questions/47679/mechanisms-of-communication-in-game-theory
Nav89
CC BY-SA 4.0
2021-09-25T18:44:03.977Z
@Nav89 I think perhaps you meant to address this to users tdm or Michael Greinecker?
Giskard
CC BY-SA 4.0
2021-09-25T19:33:58.863Z
@Giskard of course but i can not tag them...can you tell me how?
Nav89
CC BY-SA 4.0
2021-09-26T06:04:29.273Z
@Nav89 The same way you tagged me...? (only one tag is allowed per comment)
Giskard
CC BY-SA 4.0
2021-09-26T06:16:52.723Z
@Michael Greinecker check this one when you have time https://economics.stackexchange.com/questions/47679/mechanisms-of-communication-in-game-theory
Nav89
CC BY-SA 4.0
2021-09-26T06:18:29.803Z
@tdm check this one when you have time, you hace a better understanding in this notation than me...it seems so confusing to me https://economics.stackexchange.com/questions/47679/mechanisms-of-communication-in-game-theory?noredirect=1#comment76764_47679
Nav89
<mathematical-economics><game-theory><bayesian-game><probability><information>
Difficulty in understanding the notation related from probability theory with game theory
Nav89
# Difficulty in understanding the notation related from probability theory with game theory

(Asked by: username_0 on 2021-09-22)

The question that I have is a little technical and it has to do with the notation and the combination between some mathamatical properties in the probability theory of information economics.

Say $\mathcal{I}=(X,\mu)$ and $\mathcal{J}=(Y,\nu)$ are measurable probability spaces, where $\mu$ and $\nu$ denote the probability distributions over $X$ and $Y$ respectively. Then, we define $\phi:X\to\Delta(Y)$, where $\Delta(Y)$ is the simplex of $Y$. If the image of $\mu$ by $\phi$ is $\nu$ then it holds
$$\mathbb{E}\_{\mu}\phi(x)(y)=\nu(y),\quad \text{where $\phi(x)(y)$ denotes the conditional probability $\phi(y|x)$}$$

Note that $\mu$ and $\nu$ are probability measures s.t. for every $x^i$ such that $\mu(x^i)>0$, then $p(x^i)\in\Delta(X^{-i})$ denotes the conditional probability of $\mu$ given $x^i$ over $X^{-i}$:
$$p(x^i)(x^{-i})=\mu(x^{-i}|x^i)=\frac{\mu(x^{-i},x^{i})}{\mu(x^i)}$$

Also, note that $\mu(x^i)$ stands for $\mu(\{x^i\}\times X^{-i})$.

To help you understand the notation, from game theory imagine that $i=\{1,2,3\}$, then $X=X^1\times X^2\times X^3= \{a^1,b^1\}\times\{a^2,b^2\}\times\{a^3,b^3\}$ and $x=(\underbrace{a^1}\_{x^1},\underbrace{a^2}\_{x^2},\underbrace{a^3}\_{x^1})$

Even furhter, the following probabilities are defined. Let $r$ denote the ex-ante probability and $q$ the ex-post probability, they are defined as follows on $\Delta(Y^{-i})$:

$$r(x^i)(y^{−i})= P\_{\phi}(y^{−i}|x^i),\quad\text{and}\quad q(y^i)(y^{−i})= P\_{\phi}(y^{−i}|y^i)$$

for $P\_{\phi}(x^i,y^i)>0$, where $P\_{\phi}(x^i,y^i)=\mu(x^i)\phi^{i}(x^i)(y^i)$ denotes the probability induced on $X\times Y$ by $\mu$ and the transition probability $\phi$, then $r(x^i)$ and $q(y^i)$ are random vectors with values in $\Delta(Y^{-i})$ and $P\_{\phi}(y^i|x^i)=\phi^{i}(x^i)(y^i)$ and $r(x^i)=\mathbb{E}\_{p(x^i)}\phi^{-i}(x^{-i})$

$\textit{Question:}$ I struggle to understand how to calculate $r()$ and $q()$ with repsect to $\mu$ and $\nu$ as well. I can not clarify from the notation how are the ex-post and ex-ante probabilities connected with $\phi$, $\mu$, $\nu$ and $p$?

I think that $q(y^i)(y^{−i})$ equals to $\frac{\nu(y^{-i},y^i)}{\nu(y^i)}$, but I can not understand how to make the transition from $q(y^i)(y^{−i})= P\_{\phi}(y^{−i}|y^i)=\dots=\frac{\nu(y^{-i},y^i)}{\nu(y^i)}$. Also what is the formula of $r(x^i)(y^{−i})$ and how it ends up to the result that $r(x^i)=\mathbb{E}\_{p(x^i)}\phi^{-i}(x^{-i})$?

Also it seems that $p$ is the "backwards" version of $\phi$. Whereas $\phi$ tells you the likelihood of each value of $Y$ given the value of $X$, the function $p$ tells you, given $y$, the conditional probability it was produced by each $x$. So could I also claim that $P\_{\phi}(x,y)=\nu(y)p(y)(x)=\mu(x)\phi(x)(y)$?

$\textit{Hint:}$ Keep in mind that the noation $f(x)(y)=f(y|x)$ in every situation and $f(x^i)$ is the conditional probability $f(x^i)(x^{-i})=f(x^{-i}|x^{i})$

**Tags:** mathematical-economics, game-theory, bayesian-game, probability, information
> **Comments:**
> **username_1**: What text does this come from?  
> **username_0**: @MichaelGreinecker this one https://econ-papers.upf.edu/papers/169.pdf  
> **username_0**: @MichaelGreinecker more precisely pages 4 and 5 contain all this information.  
> **username_0**: @Giskard is it ok if I am going yo edit one or 2 more questions here? I do not want to write a new question, since the new questions will be implied within the same framework  
> **username_2**: @nav89 In this case I think you should copy info from the Q, not edit answered Q's.  
> **username_0**: `@Giskard ok i will do this right now  
> **username_0**: @Giskard check this one when you have time https://economics.stackexchange.com/questions/47679/mechanisms-of-communication-in-game-theory  
> **username_2**: @Nav89 I think perhaps you meant to address this to users tdm or Michael Greinecker?  
> **username_0**: @Giskard of course but i can not tag them...can you tell me how?  
> **username_2**: @Nav89 The same way you tagged me...? (only one tag is allowed per comment)  
> **username_0**: @Michael Greinecker check this one when you have time https://economics.stackexchange.com/questions/47679/mechanisms-of-communication-in-game-theory  
> **username_0**: @tdm check this one when you have time, you hace a better understanding in this notation than me...it seems so confusing to me https://economics.stackexchange.com/questions/47679/mechanisms-of-communication-in-game-theory?noredirect=1#comment76764_47679  

---

## 1 answers

### Answer by username_3 on 2021-09-23. Score: 6 | Accepted answer

We have that ${\cal I} = ((X^i)\_i, \mu)$​ and ${\cal J} = ((Y^i)\_i, \nu)$​ are two information structures.

An Interpretation mapping for player $i$​​ is a mapping $\phi^i: X^i \to \Delta(Y^i)$​ so it associates with every $x^i$​ a distribution over $Y^i$​.

Let $x^i \in X^i$​. Then $\phi^i(x^i)$​ is a distribution over $Y^i$​ so $\phi(x^i)(y^i)$​ is the probability of this distribution attached to outcome $y^i$​.

Then for $x = (x^1, \ldots, x^I)$ we can define:
$$
\phi: \prod\_i X^i \to \Delta\left(\prod\_i Y^i\right)
$$
Such that for all $x = (x^1,\ldots, x^I)$ and $y = (y^1,\ldots, y^I)$ we have the product probability measure:
$$
\phi(x)(y) = \phi^1(x^1)(y^1) \cdot \phi^2(x^2)(y^2) \cdot \ldots \cdot \phi^I(x^I)(y^I)
$$
We have that an interpretation $\phi= (\phi^i)\_i$​ is **consistent** if for every $x \in \prod\_i X^i$ and $y \in \prod\_i Y^i$:
$$
\mathbb{E}\_\mu \phi(x)(y^i) = \nu(y),\\
\leftrightarrow \sum\_{x \in X} \phi(x)(y) \,\, \mu(x) = \nu(y)
$$
Now define $\phi(x)(y) \mu(x) = P(x,y)$​. Then the sum of $P(x, y)$ over all $x$ and $y$ equals 1:
$$
\sum\_x \sum\_y P(x,y) = \sum\_x \sum\_y \phi(x)(y)\,\, \mu(x) = \sum\_y \nu(y) = 1
$$
Also the marginal distributions satisfy:
$$
\begin{align\*}
&\sum\_x P(x,y) = \sum\_x \phi(x)(y)\mu(x) = \nu(y),\\
&\sum\_y P(x,y) = \sum\_y \phi(x)(y) \mu(x) = \mu(x) \sum\_y \phi(x)(y) = \mu(x).
\end{align\*}
$$
So $P(x,y)$​​​ act as a joint probability distribution over $\prod\_i X^i \times \prod\_i Y^i$ with marginals $\nu$ and $\mu$. Then $\phi(x)(y) = \dfrac{P(x,y)}{\mu(x)}$ which is the same as the conditional probability $\phi(x)(y) = P(y|x)$​​.

We can write the marginals $P(y) = \nu(y)$​ and $P(x) = \mu(x)$​. Also, let $P(y^i)$ be the marginal distribution of $P$ with respect to $y^i$, $P(x^i)$ the marginal distribution of $P$ with respect to $x^i$ and so on.

Then we can define:
$$
q(y^i)(y^{-i}) = P(y^{-i}|y^i) = \dfrac{P(y^i, y^{-i})}{ P(y^i)}.
$$
And:
$$
r(x^i)(y^{-i}) = P(y^{-i}|x^i) = \dfrac{ P(x^i, y^{-i})}{ P(x^i)}.
$$

> I think that $q(y^i)(y^{−i})$ equals to $\frac{\nu(y^{-i},y^i)}{\nu(y^i)}$

I think you are correct. Notice that $P(y^i, y^{-i}) = \nu(y^i, y^{-i})$ and $P(y^i) = \sum\_{y^{-i}} P(y^i, y^{-i}) = \sum\_{y^{-i}} \nu(y^i, y^{-i}) = \nu(y^i)$ so indeed (by Bayes rule):
$$
q(y^i)(y^{-i}) = \dfrac{P(y^i, y^{-i})}{P(y^i)} = \dfrac{\nu(y^i, y^{-i})}{\nu(y^i)} = \nu(y^{-i}|y^i)
$$

> Also what is the formula of $r(x^i)(y^{−i})$ and how it ends up to the result that $r(x^i)=\mathbb{E}\_{p(x^i)}\phi^{-i}(x^{-i})$?

The comuputation for this expression is a bit more involved. First:
$$
\begin{align\*}
P(x^i, y^{-i}) &= \sum\_{x^{-i}, y^{i}} P(x^i, x^{-i}, y^i, y^{-i}),\\
&= \sum\_{x^{-i}, y^{i}} \phi(x^i, x^{-i})(y^i, y^{-i}) \,\, \mu(x^i, x^{-i}),\\
&=\sum\_{x^{-i}, y^i} \mu(x^i, x^{-i}) \left[\left(\prod\_{j \ne i} \phi^j(x^j)(y^j)\right)\phi^i(x^i)(y^i) \right],\\
&=\sum\_{x^{-i}} \mu(x^i, x^{-i}) \left(\prod\_{j \ne i}\phi^j(x^j)(y^j)\right) \underbrace{\sum\_{y^{i}} \phi^i(x^i)(y^i)}\_{=1},\\
&=\sum\_{x^{-i}} \mu(x^i,x^{-i})\,\, \phi^{-i}(x^{-i})(y^{-i}),\\
\end{align\*}
$$
Next one has,
$$
P(x^i) = \sum\_{x^{-i}} P(x^i, x^{-i}) = \sum\_{x^{-i}} \mu(x^i, x^{-i}) = \mu(x^i)
$$
Then:
$$
\begin{align\*}
r(x^i)(y^{-i}) &= \dfrac{P(x^i, y^{-i})}{P(x^i)},\\
&= \sum\_{x^{-i}} \frac{\mu(x^i, x^{-i})}{\mu(x^i)} \phi^{-i}(x^{-i})(y^{-i}),\\
&= \sum\_{x^{-i}} p(x^i)(x^{-i})\,\, \phi^{-i}(x^{-i})(y^{-i}),\\
&= \mathbb{E}\_{p(x^i)} \phi^{-i}(x^{-i})(y^{-i})
\end{align\*}
$$

### Additional questions and answers

> 1. Does it mean that the product measure is always translated to a conditional probability?

No, that is why the authors impose the defintion that the interpretation should be **consistent**.

> 2. Have you seen any textbook that has the same notation as in here?

No. In my opinion, the notation is rather inconvenient to work with ;-)

> 3. Why $P(y^i, y^{-i}) = \nu(y^i, y^{-})$.

Let $y = (y^i, y^{-i})$ then:

$P(y^i, y^{-i}) = P(y)$ which is the marginal distribution of $P(x,y)$ with respect to $y$. Condition $(1)$ shows that this is equal to $\nu(y) = \nu(y^i, y^{-i})$.

> 4. How did you figure out $P(x^i, y^{-i}) = \sum\_{x^{-i}, y^i} P(x^i, x^{-i}, y^i, y^{-i})$.

Similar as in point 3, $P(x^i, y^{-i})$ is the marginal distribution $P$ with respect to $(x^i, y^{-i})$, so you have to 'integrate out' $(x^{-i}, y^i)$. As these are finite, this amounts simply to taking the sum.

> **Comments:**
> **username_0**: wow...and thanks! I couldn't ask for moore...so let me check if there is anything that i can not understand for a little more...clarification....  
> **username_0**: 1. You say by the product probability measure we have:  $$\phi(x)(y)=\phi^1(x^1)(y^1)\dot\phi^2(x^2)(y^2)\dot \dots\phi^I(x^I)(y^I)$$ and also $\phi(x)(y)=P(y|x)$ does it mean that the product measure is always translated to a conditional probability?  2. Have you seen any textbook that has the same notation as in here?  3. Why $P(y^i,y^{-i})=\nu(y^i,y^{-i})$? I think that you will say by definition, but I am missing something that makes me not seeing it.  4. How did you figure out $P(x^i,y^{-i})=\sum_{x^{-i}, y^{i}} P(x^i, x^{-i}, y^i, y^{-i})$, $\sum_{y^{i}} \phi^i(x^i)(y^i)=1?$  
> **username_3**: @Nav89 I added my answers to my response above. Heres the [wiki](https://en.wikipedia.org/wiki/Marginal_distribution) on how marginal distributions are computed.  
> **username_0**: So, to sum up,the probability $\phi$ denotes the transition probability of moving from one state of a system $\mathcal{I}$ into another state to $\mathcal{J}$. The extra condition that is needed is the assumption of consistency, right? But how do we translate consistency here?  And my other question is why $\sum_{y^i}\phi^i(x^i)(y^i)=1$, my understanding is teh follwing:  $$\sum_{y^i}\phi(x^i)(y^i)=\phi^1(y^1)(y^1)+\phi^2(y^2)(y^2)+\dots +\phi^I(y^I)(y^I)=\phi^1(y^1|y^1)+\phi^2(y^2|y^2)+\dots+\phi^I(y^I|y^I)=\phi(y|y)=1$$  
> **username_3**: @Nav89 we have that $\phi(x^i)$ is a probability distribution over $Y^i$ so if you add up the probababilities, you get 1: $\sum_{y^i \in Y^i} \phi(x^i)(y^i) = 1$.  
> **username_0**: so what i wrote is wrong i guess :P but i undes=rstand what you say i am ok  
> **username_0**: check this one when you have time, you hace a better understanding in this notation than me...it seems so confusing to me  https://economics.stackexchange.com/questions/47679/mechanisms-of-communication-in-game-theory  

---

