In a discrete time process, how is a filtration generated by random variables?

Question

Say I have random variables $X_1, X_2, ...X_n$ and someone said that these functions generate the filtration, ie $F_n = \sigma(X_1,...X_n)$ does that mean: $F_n = \sigma(X_1) \otimes \sigma(X_2) \otimes ...\sigma(X_n)$?

I don't think so because I am also told that filtration $F_a$ is a strict subset of $F_b$ for $b>a$ but I don't think you can say $F_n = \sigma(X_1) \otimes \sigma(X_2) \otimes ...\sigma(X_a)$ is a subset of $F_n = \sigma(X_1) \otimes \sigma(X_2) \otimes ...\sigma(X_b)$ because one is $a$ dimensions and the other is $b$ dimensions.

So how do i understand $\sigma(X_1,...X_n)$?

Answer 1

$\sigma (X_1,X_2,..,X_n)$ is the smallest sigma field on $\Omega$ w.r.t. which each $X_i$ is measurable. It is equal to the collection of all sets of the form $((X_1,X_2,..,X_n)^{-1}(E)$ where $E$ is a Borel set in $\mathbb R^{n}$.

Answer 2

What is meant by this is that

$F_n=\sigma(X_1,\ldots,X_n)$,

Is the smallest sigma algebra such that the first $n$ random variables are measurable functions. Then for $m>n$ you have that $F_n\subseteq F_m$ since the first $n$ random variables are measurable with respect to $F_m$ (by definition, it is the smallest sigma algebra such that the first $m>n$ are measurable with respect to it) and hence $F_n$, being the smallest such sigma algebra with this property, necessarily is contained in $F_m$.