I have the following definition in probability theory from a paper in game theory.
$\textit{Definition:}$ Suppose that $(X^i)_{i=1}^n$, where $i=\{1,2,\dots,n\}$ and $n<+\infty$, is a collection of finite sets. Define the set $\mathcal{A}=(X,\mu)$, where $X=\Pi_{i=1}^nX_i$ and $\mu$ is the probabolity measure over $X$. When $x\in X$ is drawn according to $\mu$, then every palyer $i$ is informed about the coordinate $x^i$. Even further 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 $\mu(x^i\times X^{-i})$.
I face difficulty in understanding the notation and I have the following questions:
- Is the probability measure $\mu$ defined as follows:
$$\mu:X\to \Delta(X)\subset [0,1]^ n$$
- I strugle to understand the last proposition ''Even further...stands for $\mu(x^i\times X^{-i})$''. What is the intuition and how this transition occurs from $\mu$ to $p$?
If it would be simple to provide an example according to the definition I would appreciate this. For instance lets suppose that n=3, so $X=X^1\times X^2\times X^3= \{a,b\}\times\{z,y\}\times\{u,v\}$, and then $x=(\underbrace{a}_{x^1},\underbrace{z}_{x^1},\underbrace{u}_{x^1})$, what does it mean that $\mu(x^2)>0$ according to the definition above? Can anybody build an example?
Consider $X_1=\{a_1,b_1\}$, $X_2=\{a_2,b_2\}$ and $X_3=\{a_3,b_3\}$. Then $X=\{a_1,b_1\}\times\{a_2,b_2\}\times \{a_3,b_3\}$. Consider following probability on $X$:
Now $X^{-i}=\prod_{j=1, j\neq i}^nX_i$, in our case $X^{-1}=X_2\times X_3$, $X^{-2}=X_1\times X_3$ and $X^{-3}=X_1\times X_2$. This yields the first abuse of notation. For example, by $(a_2,X^{-2})$ we usually denote the set $\{(x_1,a_2,x_3)\in X:\ (x_1,x_3)\in X^{-2}\}$. I hope this meets your understanding so far.
If we now want to consider conditional probabilities, we could think of the probability of $x_1=a_1$ and $x_3=a_3$ if we know $x_2=a_2$. This is indeed $\mu(\{(a_1,x_2,a_3):\ x_2\in X_2\} |\{(x_1,a_2,x_3):\ (x_1,x_3)\in X^{-2}\} )$. Using the definition of conditioned probability yields, that this equals $\mu(\{(a_1,a_2,a_3)\})/\mu((a_2,X^{-2}))=\frac{\frac{1}{5}}{\frac{1}{5}+3\cdot\frac{1}{10}}=\frac{2}{5}$.
If $a=(a_1,a_2,a_3)$, the same notational abuse as above lets us write $a=(a_2,a^{-2})$ with $a^{-2}=(a_1,a_3)$ and by $p(a_2)(a^{-2})$ we obtain a far nicer way of writing the conditional probability we just derived.
I hope those conventions I used make sense for your applications.