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Let $\mathcal{F}_i^j$ be a collection of sigma-algebras for $i \in \mathbb{N}$ and $j \in \{1,\ldots,n\}$ such that $\mathcal{F}_{i+1}^j \subseteq \mathcal{F}_i^j$ for all $i,j$. I would like to show that

$$\sigma\left(\cup_j\cap_i\mathcal{F}_i^j\right) = \cap_i\sigma\left(\cup_j\mathcal{F}_i^j\right)$$

but I'm not sure how, or whether it is true.


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This is not a research level question. Try mathstackexchange. – Nik Weaver Sep 23 '12 at 15:07
It really was trivial before the edit - sorry. – Vladimir Sep 23 '12 at 15:20
I don't really understand the question. What is $\sigma$? – Joel David Hamkins Sep 23 '12 at 23:22
$\sigma(\mathcal A)$ is "sigma-algebra generated by $\mathcal A$". – Gerald Edgar Sep 24 '12 at 1:12
up vote 2 down vote accepted

$\newcommand{\cF}{\mathcal{F}}$ What you want to show is false. Let $x^j_i$ and be a set of random variables, each taking values in $\{0,1\}$, where $j=1,2$ and $i=1,2,\ldots$. Let $\cF^j_i=\sigma(x^j_i,x^j_{i+1},\ldots)$.

Consider the event $A=$ "$x^1_i=x^2_i$ for all but finitely many $i$".

This event clearly belongs to the right hand side. To see that it does not belong to the left hand side, define the following probability distribution: let the $x^1_i$ be i.i.d uniformly in $\{0,1\}$. Then either let $x^2_i=x^1_i$ for all $i$ or let $x^2_i\ne x^1_i$ for all $i$, with probability $\frac12$ for either option.

Now, we see that the probability of $A$ is $\frac12$. However, the probability of any event in the lhs must be either 0 or 1, due to Kolmogorov's 0-1 law. Hence, $A$ does not belong to the lhs.

This question is related to a question of mine. I guess it is high time that I post the answer to that question too.

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Thank you very much, Ori. I guess another way to see it is that the tail sigma-algebras of each sequence is trivial, while the tail of the sequence of pairs is not. – Vladimir Sep 24 '12 at 5:07
That's right, though I'm not sure why you call it "another way"... – Ori Gurel-Gurevich Sep 24 '12 at 6:47

It's false. Let $n=2$. Suppose $A$ is some $\sigma$-algebra and $B$ is another $\sigma$-algebra strictly containing $A$. Let $\mathcal{F}_1^1=\mathcal{F}_2^2=A$ and $\mathcal{F}^j_i=B$, otherwise.

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Thanks a lot! But I forgot another important condition. Sorry! – Vladimir Sep 23 '12 at 15:20
To clarify - this answer does not satisfy the updated conditions of the question. – Vladimir Sep 23 '12 at 17:27

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