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Id like to prove that $\sigma(X_{1},...,X_{n})$ and $\sigma(X_{n+1},...)$ are independent for independent random variables $X_{i}$.

I've found that $\{X_{1}\in B_{1},...,X_{n}\in B_{n}\}$ generates $\sigma(X_{1},...,X_{n})$ but I can't see why this is the case. Could anyone help me see this?

And what set then generates $\sigma(X_{n+1},...)$?

Help is much appreciated.

Davide Giraudo
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user70267
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  • "Bumped to the homepage by Community ♦ 4 mins ago This question has answers that may be good or bad; the system has marked it active so that they can be reviewed." The answer is correct and adapted to the level of the question. The question itself might have been unsuited to the OP's level of mathematical sophistication (see the comments on the answer), but this is another story. – Did Dec 16 '16 at 10:32

1 Answers1

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Some ideas:

  • first, by an approximation argument, it's enough to show that $\mu(A\cap B)=\mu(A)\mu(B)$ for $A\in\sigma(X_1,\dots,X_n)$ and $B$ in an algebra generating $\sigma(X_{n+1},\dots)$.

  • Then we use the fact that we can express such an algebra, for example the algebra generated by sets of the form $\bigcap_{j\in F}X_j^{-1}(B_j)$, where $B_j$ is a Borel subset of $\Bbb R$ and $F\subset \{n+1,\dots\}$ is finite.

  • Each element of this algebra can be written as an union of elements of the form $\bigcap_{i\in I}X_i^{-1}(B_i)$, where $I\subset \{n+1,\dots\}$ is finite and $B_i$ is a Borel subset of the real line. We can actually express such an union as a disjoint one, and this make the computations easier.

Here, the idea is to go from independence between two finite collections of indexes to an infinite one, and that's why approximations arguments are involved.

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Davide Giraudo
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  • When assuming that $\sigma(X_{1},...X_{n})$ is generated by $A={X_{1}\leq x_{1},...,X_{n}\leq x_{n}}$ and $\sigma(X_{n+1})$ by $B={X_{n+1}\leq x_{n+1},...}$ for example than it is easy to see that $P(A\cap B)=P(A)P(B)$ and thus the $\sigma$-algebras are independent. My problem however is seeing why it is generated by such a set. – user70267 Mar 31 '13 at 20:13
  • You can give a characterization of the algebra generated by a collection. Did you try it in this case? – Davide Giraudo Mar 31 '13 at 21:24
  • No I think that's my problem actually.. That I dont know I can do or how to do that? – user70267 Mar 31 '13 at 21:47
  • @user70267 I've added details. Is it clear? – Davide Giraudo Mar 31 '13 at 21:54
  • If Im completely honest no, it's only confusing me more. I thoughti had a clear idea in my mind as to how to prove how the two are independent and all I did not understand was how that one set was generating the $\sigma$-algebra. Ive been trying to figure this stuff out for days now so I should accept my defeat and drop it, because I'm clearly never going to understand this. I would like to thank you for trying to help me, I do appreciate it. – user70267 Mar 31 '13 at 22:08
  • Well, you can try the following approach: let $\cal F_k$ be the $\sigma$-algebra generated by $X_{n+1},\dots,X_{n+k}$. Then $\mathcal A:=\bigcap_k\cal F_k$ is an algebra which generated $\sigma(X_{n+1},\dots,)$. Now use the approximation argument. – Davide Giraudo Mar 31 '13 at 22:13
  • I do not see how this helps me show that ${X_{1}\in B_{1},...,X_{n}\in B_{n}}$ generates $\sigma(X_{1},...,X_{n})$. – user70267 Mar 31 '13 at 22:20
  • @user70267 $\sigma(X_0, X_1, ..., X_n) = \sigma(\sigma(X_0) \cup \sigma(X_1) \cup ... \cup \sigma(X_n))$ and $\sigma(X_k) = {X_k \in B_k | B_k \in \mathscr{B}}$ ? – BCLC Nov 03 '15 at 19:42