Skip to main content

Timeline for answer to Intuition for random variable being $\sigma$-algebra measurable? by Bjørn Kjos-Hanssen

Current License: CC BY-SA 3.0

Post Revisions

10 events
when toggle format what by license comment
Sep 5, 2021 at 9:30 comment added roi_saumon ...You say "we're given which of the Borel sets $\omega$ belongs to" but if you know for instance $\omega\in (\omega-\epsilon,\omega+\epsilon)$ then we also have $\omega\in (\omega-\epsilon/2,\omega+\epsilon/2)$ or $\omega\in (\omega-\epsilon/3,\omega+\epsilon/3)$ or whatever. But how do you define $E[X|\mathcal{B}]$? Indeed, $W[X|(\omega-\epsilon,\omega+\epsilon)]$, $W[X|(\omega-\epsilon/2,\omega+\epsilon/2)]$ or $W[X|(\omega-\epsilon/3,\omega+\epsilon/3)]$ are different. I just don't understand how point 6) works in this case.
Sep 5, 2021 at 9:30 comment added roi_saumon @BjørnKjos-Hanssen, Maybe there is something I am not getting right. I consider $([0,1],\mathcal{B},P)$ where $\mathcal{B}$ are the borel sets and P the Lebesgue measure on $[0,1]$. A partition of $[0,1]$ with borel sets (which doesn't generate the borelians since by the linked answer this is not possible) would be $[0,1]= \sqcup_{\omega\in[0,1]}\{\omega\}$. However we know that $\omega\in \{\omega\}$ and by analogy with your answer we would like to have $E[X|\mathcal{B}]=E[X|\{\omega\}]= E[X:{\omega}]/P(\{\omega\})$ but $P(\{x\})=0$ so this is not well defined...
Sep 5, 2021 at 7:04 comment added Bjørn Kjos-Hanssen @roi_saumon That's interesting. But really, isn't it the same... we're given which of the Borel sets $\omega$ belongs to?
Sep 4, 2021 at 21:43 comment added roi_saumon @BjørnKjos-Hanssen, I was thinking for instance the borelians on $[0,1]$ as it is explained in this link.
Sep 4, 2021 at 13:59 comment added roi_saumon @BjørnKjos-Hanssen, I didn't understand how this works for $\sigma$-algebras that are not generated from a partition of sets
May 31, 2018 at 9:17 comment added BCLC Bjørn Kjos-Hanssen, in between 5 and 6, could we perhaps have a countably infinite definition (then 6 would be uncountable) ?
May 23, 2018 at 22:55 comment added Bjørn Kjos-Hanssen Thanks, @BCLC!!
May 23, 2018 at 14:01 comment added BCLC Just like the standard machine for Lebesgue integration. Brilliant!
Mar 5, 2014 at 3:07 vote accept BoZenKhaa
Mar 5, 2014 at 3:07
Feb 26, 2014 at 7:14 history answered Bjørn Kjos-Hanssen CC BY-SA 3.0