This is somewhat of a stupid question, but I am stuck on coming up with an airtight justification. Let $X_1, \ldots, X_n$ be defined on a probability space $(\Omega, \mathcal{F}, P)$. There are several equivalent definitions for the sigma-field generated by $X_1, \ldots, X_n$. One particular one is:

$$\mathcal{F}_n = \{ B = \{\omega: (X_1(\omega), \ldots, X_n(\omega)) \in A\},\; A \in B(\mathbb{R}^n) \}$$

Where $B(\mathbb{R}^n)$ is the Borel sigma-field on $\mathbb{R}^n$. This is clearly a sigma-algebra since $B(\mathbb{R}^n)$ is a sigma-field, and furthermore is a sub-sigma-field of $\mathcal{F}$ since the transformation is measurable.

My question is: can $\mathcal{F}_n$ be re-written as:

$$\mathcal{F}_n = \{ \cap_{i=1}^n \{\omega: X_i(\omega) \in B_i\}, \;B_i \in B(\mathbb{R}) \}$$

In other words, the set of all sets that can be written as intersections of inverse images of Borel sets under $X_1, \ldots, X_n$. It seems obvious, because $B(\mathbb{R}^n) = B(\mathbb{R}) \times \ldots \times B(\mathbb{R})$, but I feel like I may be missing something. Can anyone confirm or disprove this?

Question answered: Did pointed out (in the comments) where I had gotten confused. Many thanks to everyone. The correct formulation should be: $$\mathcal{F}_n = \sigma \Big(\cap_{i=1}^n \{\omega: X_i(\omega) \in B_i\}, \;B_i \in B(\mathbb{R}) \Big)$$

  • 1
    $\begingroup$ Have you checked whether your $\mathcal F_n$ is a $\sigma$-algebra? $\endgroup$ – Davide Giraudo Mar 3 '13 at 16:07
  • $\begingroup$ I thought it was obvious from the above justification, but I haven't rigorously verified it... I will take a look in a bit. Thanks! $\endgroup$ – gogurt Mar 3 '13 at 16:24
  • 1
    $\begingroup$ The assertion that $B(\mathbb R^n)$ is $B(\mathbb R)\times\cdots\times B(\mathbb R)$ is either wrong or usually written, to avoid ambiguities, rather as $B(\mathbb R^n)=B(\mathbb R)\otimes\cdots\otimes B(\mathbb R)$. $\endgroup$ – Did Mar 3 '13 at 16:24
  • 1
    $\begingroup$ Did you're absolutely right. I got my notation mixed up, which is why I got confused. So essentially, the second "version" of $\mathcal{F}_n$ I gave should be the sigma-algebra generated by what I wrote. Many thanks. $\endgroup$ – gogurt Mar 3 '13 at 16:38

Either the assertion that $B(\mathbb R^n)$ is $B(\mathbb R)\times\cdots\times B(\mathbb R)$ is wrong or it is more usually (and, to avoid ambiguities, more preferably) written as $B(\mathbb R^n)=B(\mathbb R)\otimes\cdots\otimes B(\mathbb R)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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