1
$\begingroup$

Why do we define random variables as functions $X: (\Omega, \mathcal{F}) \rightarrow (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ when we can define them as $X: (\Omega, \mathcal{F}) \rightarrow (\mathbb{R}, \mathcal{P}(\mathbb{R}))$ ?

Don't we want to eventually be able to talk about the probability of $\{X \in A\}$ for all $A \subseteq \mathbb{R}$?

$\endgroup$
2
  • $\begingroup$ Writing both $X\in A$ and $A\in\Bbb R$ in the same formula cannot be right; if you are going to do probability or measure theory (and even if not), understanding the difference between elements and subsets is crucial. Also, like all such operators (think $\sum$ and such), $\forall$ should come before the expression where the variable it introduces is used. $\endgroup$ Commented Jun 12, 2021 at 9:44
  • $\begingroup$ Small error thanks $\endgroup$
    – Darby Bond
    Commented Jun 12, 2021 at 11:04

1 Answer 1

1
$\begingroup$

I think what you are proposing is what probabilists did for a long time, before probability theory had the rigorous mathematical foundation it has today. So what's the problem?

The issues are non-measurable sets, sets which don't have a measure. An example is given by the Vitali-Set.

So while we would ideally want to talk about the probability of $\{X\in A\}$ for all $A \subseteq \Bbb{R}$, this shows that this is too much to ask for. This is the reason we construct the Borel $\sigma-$Algebra, where we get rid of these problematic sets.

This may be a bit of a intuitive explanation. All these notions are usually made very precise in a course on measure theory. Does this answer your question?

Edit (in response to the comment):

Let's say you had a random Variable $X \sim \mathcal{U([0,1])}$, a random variable with continuous uniform distribution on the interval $[0,1]$. Let's say that we work on the $\sigma-$Algebra $\mathcal{P}(\Bbb{R})$ and take $V$ to be the Vitali set (as linked above). What would be $\Bbb{P}(X \in V)$? By definition we would have

$$\Bbb{P}(X \in V) = \int_V \unicode{x1D7D9}_{[0,1]}(x)dx = \int_{V \cap [0,1]}1dx. $$

But this doesn't make sense, since we are trying to integrate over a non-measurable set. So it is important to make sure to work only with sets that have a well defined (Lebesgue-)measure.

$\endgroup$
3
  • $\begingroup$ So I am doing a course on measure-theoretic probability. To me, this set $\{ X \in A \}$. where $A \in \mathbb{R}$ is "measurable" because the measure $\mathbb{P}$ is applied to sets $A \in \mathcal{F}$. So as long as $\{ X \in A \} \in \mathcal{F}$, it will have a well-defined probability right? $\endgroup$
    – Darby Bond
    Commented Jun 12, 2021 at 7:55
  • $\begingroup$ I added something in my answer, does this make sense to you? $\endgroup$ Commented Jun 12, 2021 at 8:06
  • 1
    $\begingroup$ This is a good answer. I'll accept this $\endgroup$
    – Darby Bond
    Commented Jun 12, 2021 at 8:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .