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Let $(\Omega, \mathcal{A}, P)$ be a probability space with a sequence $X_i, i \geq 1$ of random variables. Let $\mathcal{F}_{n,m}:=\sigma(X_n, \dotsc, X_m)$ for $n \leq m$.

Consider $\mathcal{F}_{n,m} \cap \mathcal{F}_{n,p}$.

Is it true that this is equal to $\mathcal{F}_{n, \min(m,p)}$?

Now consider $\mathcal{F}_{n,m} \cup \mathcal{F}_{n',m'}$, where $n'<n<m<m'$.

Is it true that this is equal to $\mathcal{F}_{n',m'}$, because $\mathcal{F}_{n,m} \subseteq\mathcal{F}_{n',m'}$?

In general, if we have two sigma-algebras $\mathcal{C}$ and $\mathcal{D}$. Then $\mathcal{C}\cup \mathcal{D}$ need not be a sigma-algebra again. We use $\sigma(\mathcal{C} \cup \mathcal{D})$ to denote the smallest sigma-algebra containing these two?

Intuitively, the intersection of a sigma-algebra is the sigma-algebra that makes both sigma-algebras measurable (how would one express that intuitively?). This means that an event that is only measurable with respect to one sigma-algebra does not occur in the intersection. The intersection only contains information contained in both sigma-algebras?

The union of a sigma-algebra would intend to make all the events in one of these sigma-algebra measurable, so it contains any information contained in any of these two sets. But since this need not be closed, we need to "close" it, in order to again be meaningful? Is a union of increasing sigma-algebras always a sigma-algebra?

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    $\begingroup$ "Is a union of increasing sigma-algebras always a sigma-algebra?" Finitely many, yes, trivially. Infinitely many, no. $\endgroup$ – Did Oct 30 '15 at 15:25
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In general, if $A\subset B$, then $A\cap B=A$. So indeed $$\mathcal F_{n,m}\cap \mathcal F_{n,p} = \mathcal F_{n,m\wedge p}$$ (where $\wedge$ denotes $\min$). It is also true that if $n'<n<m<m'$ then $$\mathcal F_{n,m}\cup \mathcal F_{n',m'} = \mathcal F_{n',m'}, $$ as $A\subset B$ implies $A\cup B=B$. Note that these relations follow from basic set theory, and don't have anything to do with $\sigma$-algebras in particular.

However, to be precise as to what $\mathcal F_{n,m}$ means here; for a generic random variable $X$ we define $\sigma(X)$ to be the intersection of sub-$\sigma$-algebras $\mathcal F$ of $\mathcal A$ such that $X$ is $\mathcal F$-measurable. It turns out that $$\sigma(X) = \{ X^{-1}(B) : B\in\mathcal B\}, $$ where $\mathcal B$ is the Borel $\sigma$-algebra. This is straightforward to prove by showing that the bracketed set is a $\sigma$-algebra, using the fact that $$\mathcal B=\sigma(\{(-\infty, x] : x\in\mathbb R\}).$$ In general, for a sequence of random variables $\{X_i\}$, we define $$\sigma(X_1, \ldots, X_n)=\sigma\left(\bigcup_{i=1}^n \sigma(X_i)\right),$$ and $$\sigma(X_1, X_2, \ldots) = \sigma\left(\bigcup_{i=1}^\infty \sigma(X_i)\right). $$ So in this case, $$\mathcal F_{n,m} = \sigma(X_n,\ldots, X_m) = \sigma\left(\bigcup_{i=n}^m \sigma(X_i)\right).$$ To answer your last question, if $\{\mathcal F_i\}$ is an increasing sequence of $\sigma$-algebras, then $$\mathcal F_n = \bigcup_{i=1}^n \mathcal F_i,$$ and $\mathcal F_n$ is a $\sigma$-algebra, from which we conclude.

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