0
$\begingroup$

Problem statement

Let us say that we have some set of independent random variables: $X = \{X_i\}^n_{i=1}$ defined over a probability space of $~(\Omega, \mathcal{F}, P)$. I want to understand whether the following holds:

If $\mathcal{F}_X = \sigma\left(\left\{X_1, X_2, \dots, X_n\right\}\right)$, then is it, in general, true that: $\left|\mathcal{F}_X\right| = \sum\limits_{\forall i \in I_n}|\sigma(\{X_i\})|$?

As we know (and as proposed here), as $X$ consists of independent R.V.-s, then we can state that $\forall i,j \in I_n:\sigma(\{X_i\})$ and $\sigma(\{X_j\})$ are independent. But how to proceed from this fact to the split of the cardinality of $\mathcal{F}_X$?

I would appreciate any help, thank you in advance!

Related questions:

  1. Independent random variables and sigma algebras
  2. Cardinality of a sigma algebra of two independent random variables
$\endgroup$
2
  • 1
    $\begingroup$ The comment section to the linked question about the sigma algebra of two independent random variables already contains a counter example. If it doesn't hold for two, then there is no reason to believe that it would hold for general $n$. $\endgroup$ Nov 5 '21 at 23:45
  • $\begingroup$ @LeanderTilstedKristensen oh, that's true, for some reason, I didn't notice it at the first look. It seems, that in the case of mine, the general case can be disproven by a simple counter-example! Thank you for your note! $\endgroup$
    – O.spectrum
    Nov 5 '21 at 23:50
0
$\begingroup$

As correctly mentioned by @LeanderTilstedKristensen, it can be seen, that one of the linked questions already contains a counter-example for the case of two independent random variables. It should not be that hard to show then that in general the proposed cardinality relation does not hold!

$\endgroup$

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.