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Why does a sigma-algebra represent the information available at a given time? I understand the idea of filtration and stopping-time, given that each sigma-algebra represent the info we have at a specific time, but why is that?

For instance in a game of dice rolls (or anyone you want), what would be total universe and the available information in forms of sigma-algebra, at the n-th turn? thanks

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In a game of dice rolls, the total universe $\Omega$ would be irrelevant (as it nearly always is in probabilistic modeling, as long as it is large enough) and the sigma-algebra after the $n$th roll $X_n$ would be $\mathcal F_n=\sigma(X_k;1\leqslant k\leqslant n)$. The global sigma-algebra $\mathcal F$ on $\Omega$ may be any sigma-algebra containing $\mathcal F_\infty=\sigma(X_k;k\geqslant 1)$ since one wants each function $X_n:\Omega\to\{0,1\}$ to be a random variable on $(\Omega,\mathcal F)$, that is, to be measurable with respect to $\mathcal F$.

A time-scale interpretation might be helpful here. Imagine that the $n$th throw happens at time $n$. Then, at time $n$, the results $X_k$ for $k\geqslant n+1$ are not available yet, hence one can combine the values $X_k$ for $k\leqslant n$ in any measurable way and stay in the realm of the random variables measurable with respect to $\mathcal F_n$, but not any value $X_k$ for $k\geqslant n+1$ since these throws did not happen yet.

Note that the global sigma-algebra $\mathcal F$ may contain some extra information not in $\mathcal F_\infty$, for example the temperature $T_n$ of the room where the $n$th throw occurs, and/or the age $A_n$ of the operator throwing the $n$th dice, and so on.

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    In a sequence of die-rolls, one sigma-algebra that is sometimes of interest other than $\mathcal{F}_n$ is the tail sigma-algebra.2012-06-13
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The Doob-Dynkin lemma relates them in an intuitive way for most of the standard applications in probability theory.

Suppose you have a probability space $(\Omega,\Sigma,\mu)$ and two random variables $f:\Omega\to\mathbb{R}$ and $g:\Omega\to\mathbb{R}$. That $g$ only depends on $f$ can be interpreted as saying that you know the value of $g$ whenever you know the value of $f$. This means that you can find a function $h:\mathbb{R}\to\mathbb{R}$ such that $g(\omega)=h(f(\omega))$. In other words, $g=h\circ f$. Now the Doob-Dynkin emma says that the following are equivalent:

  1. There is a measurable function $h:\mathbb{R}\to\mathbb{R}$ such that $g=h\circ f$.
  2. The random variable $g$ is measurable with respect to the $\sigma$-algebra generated by $f$, that is the $\sigma$-algebra $\{f^{-1}(B):B\textrm{ is a Borel set}\}$.

Most naturally occuring $\sigma$-algebras are of the form $\{f^{-1}(B):B\textrm{ is a Borel set}\}$ for some random variable $f$. This is equivalent to the $\sigma$-algebra being countably generated.

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Say you know the values of $X_1+\cdots+X_n$ and $X_1^2+\cdots+X_n^2$. Then you can find the values of $\bar X = (X_1+\cdots+X_n)/n$ and $S^2 = ((X_1-\bar X)^2+\cdots+(X_n-\bar X)^2)/(n-1)$, and likewise if you know the values of those latter two quantities, then you can find the first two. So they are in a sense equivalent. Saying they are equivalent is the same as saying they generate the same sigma-algebra. Hence conditioning on them is the same as conditioning on the sigma-algebra that they generate. The details of the particular choice of which of these pairs don't matter, you you speak of conditioning on a sigma-algebra.