Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

Thanks!

share|improve this question
1  
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
1  
$\sigma(\mathcal A)$ is "sigma-algebra generated by $\mathcal A$". –  Gerald Edgar Sep 24 '12 at 1:12
add comment

2 Answers

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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
    
Thanks a lot! But I forgot another important condition. Sorry! –  Vladimir Sep 23 '12 at 15:20
1  
To clarify - this answer does not satisfy the updated conditions of the question. –  Vladimir Sep 23 '12 at 17:27
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.