[944] Measurability of a RV
David Itkin
ditkin at andrew.cmu.edu
Sun Nov 5 08:12:34 EST 2017
Hi Richard,
The converse of what you said is true -- that is if [image: \mathcal{F}
\subset \mathcal{G}] and [image: X] is [image: \mathcal{F}]-measurable
then [image: X] is also [image: \mathcal{G}]-measurable. On the other hand
if [image: X] is [image: \mathcal{G}]-measurable then we cannot conclude
anything about it being [image: \mathcal{F}]-measurable since it is
possible that [image: X^{-1}(-\infty,a) \in \mathcal{G}\setminus\matcal{F}] and
so would not be [image: \mathcal{F}]-measurable. The intuition behind this
is that [image: \mathcal{F}] gives us less information than [image:
\mathcal{G}] and so we may not be able to deduce as much from it -- we saw
an example of this in recitation 1 with a two period binomial tree model
where the return of our stock at time 2 was [image: G_2] measurable, but
not [image: G_1] measurable.
I think the easiest way to remember is to have this trivial example in the
back of your mind: take [image: \mathcal{G}] = [image:
\mathcal{P}(\Omega)] (i.e.
the power set) and [image: \mathcal{F} = \{\emptyset, \Omega\}](trivial [image:
\sigma]-algebra). Then all [image: X: \Omega \rightarrow \mathbb{R}]
are [image:
\mathcal{G}]-measurable, but only constant random variables will be [image:
\mathcal{F}]-measurable.
Hope this helps,
David
On Sun, Nov 5, 2017 at 6:36 AM, Richard Rosenbaum <rrosenba at andrew.cmu.edu>
wrote:
> Say G is a sigma algebra and F is a smaller sigma algebra contained in G.
>
> Say X is a RV that is G-measurable.
>
> Why is X not also F-measurable?
>
> This idea was discussed briefly in the TA session on Friday, but I am
> still a bit confused. My intuition would be that since G is larger, and X
> is measurable with respect to G, then X would also be measurable with
> respect to F since it is contained in G.
>
> If anyone could share their thoughts, it would be much appreciated.
>
>
> Regards,
> Rich
>
> --
> Richard Rosenbaum
> Master of Science in Computational Finance Candidate
> Carnegie Mellon University - Class of Dec. 2018
>
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