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


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.

  • $\begingroup$ in this case, are the Xk random variables or just numbers? $\endgroup$ – lezebulon Jun 13 '12 at 9:24
  • $\begingroup$ ??? Random variables, of course. $\endgroup$ – Did Jun 13 '12 at 9:26
  • $\begingroup$ so what I don't understand here (and that's the root of my problem imo) is why don't we know X(n+1) at time n? I'd say that all the X(k) are uniforms on [1;6], so I don't see how playing turns changes this knowledge $\endgroup$ – lezebulon Jun 13 '12 at 10:00
  • $\begingroup$ To say that we do not know X(n+1) at time n means that X(n+1) is not measurable with respect to F(n). To wit, F(n) is precisely the collection of the events that are measurable at time n, that is, the events that depend only on the outcomes X(k) for k at most n. For example the event A that X(1)+X(2)+X(3) is even is in F(3) but not in F(2): if someone gives you the values X(1)(omega) and X(2)(omega), you are not able to determine if omega is in A or not. But if someone gives you the values X(1)(omega), X(2)(omega) and X(3)(omega), you are. $\endgroup$ – Did Jun 13 '12 at 10:39
  • $\begingroup$ So if I understand correctly, F(n) does not represent the elements I am able to measure at time n (because I can measure A anytime : it has probability 0.5). But F(n) represents the events that I can tell at time n if they are realized for my outcome so far $\endgroup$ – lezebulon Jun 13 '12 at 16:33

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.


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.


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