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 $f(x^i)(x^{-i})=f(x^{-i}|x^{i})$

  • $\begingroup$ What text does this come from? $\endgroup$ Commented Sep 22, 2021 at 16:42
  • $\begingroup$ @MichaelGreinecker this one econ-papers.upf.edu/papers/169.pdf $\endgroup$
    – Nav89
    Commented Sep 22, 2021 at 18:22
  • $\begingroup$ @MichaelGreinecker more precisely pages 4 and 5 contain all this information. $\endgroup$
    – Nav89
    Commented Sep 23, 2021 at 9:16
  • $\begingroup$ @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 $\endgroup$
    – Nav89
    Commented Sep 24, 2021 at 10:24
  • 1
    $\begingroup$ @Nav89 I think perhaps you meant to address this to users tdm or Michael Greinecker? $\endgroup$
    – Giskard
    Commented Sep 25, 2021 at 18:44

1 Answer 1


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.

  1. 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 ;-)

  1. 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 $\nu(y) = \nu(y^i, y^{-i})$.

  1. 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.

  • $\begingroup$ 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.... $\endgroup$
    – Nav89
    Commented Sep 23, 2021 at 14:57
  • $\begingroup$ 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 $P(x^i,y^{-i})=\sum_{x^{-i}, y^{i}} P(x^i, x^{-i}, y^i, y^{-i})$, $\sum_{y^{i}} \phi^i(x^i)(y^i)=1?$ $\endgroup$
    – Nav89
    Commented Sep 23, 2021 at 15:21
  • 1
    $\begingroup$ @Nav89 I added my answers to my response above. Heres the wiki on how marginal distributions are computed. $\endgroup$
    – tdm
    Commented Sep 23, 2021 at 17:41
  • $\begingroup$ 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: $$\sum_{y^i}\phi(x^i)(y^i)=\phi^1(y^1)(y^1)+\phi^2(y^2)(y^2)+\dots +\phi^I(y^I)(y^I)=\phi^1(y^1|y^1)+\phi^2(y^2|y^2)+\dots+\phi^I(y^I|y^I)=\phi(y|y)=1$$ $\endgroup$
    – Nav89
    Commented Sep 24, 2021 at 7:24
  • $\begingroup$ @Nav89 we have that $\phi(x^i)$ is a probability distribution over $Y^i$ so if you add up the probababilities, you get 1: $\sum_{y^i \in Y^i} \phi(x^i)(y^i) = 1$. $\endgroup$
    – tdm
    Commented Sep 24, 2021 at 11:20

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