5
$\begingroup$

Well, I have some simple, maybe silly question about random variables, but there is something that I can not undestand when we define them. Suppese that, we have some random variable $X$, that is defined in a standard probability space $(\Omega,\mathcal{F}_s,\mathbb{P})$, where $\mathbb{P}:\mathcal{F}_s\rightarrow [0,1]$. I struggle to undestand which is the space , that this random variable is defined. Specifically, the random variable $X$ is a function $X:\Omega\rightarrow R$, where R is some arbitrary space and probabyly the real line. Can we claim that $X\in L^2(\Omega,\mathcal{F}_s,\mathbb{P})$ or this is wrong? How can we know indeed where this random variable belongs to?

Maybe my whole skeptic is wrong, so forgive me in advance, but I am a begginer, who wants to understand, this mathematical theory!

$\endgroup$
2
  • 3
    $\begingroup$ It is $X:\Omega\rightarrow\mathbb{R}$. $\endgroup$
    – Michael
    Mar 20, 2020 at 14:52
  • $\begingroup$ @Michael You are right! $\endgroup$
    – Hunger Learn
    Mar 20, 2020 at 15:26

1 Answer 1

Reset to default
3
$\begingroup$

Here's a few of the basic definitions.

Given your setup, a random variable is a function $X : \Omega \to \mathbb R$, that is, it is a function whose domain is $\Omega$.

But not just any old function from $\Omega$ to $\mathbb R$ is allowed as a random variable. There is a requirement, namely that for every open interval $(a,b) \subset \mathbb R$, the set $X^{-1}(a,b) = \{\omega \in \Omega \mid X(\omega) \in (a,b)\}$ is an element of the $\sigma$-algebra $\mathcal F_s$. In other words, $X^{-1}(a,b)$ is required to be an event.

Because of this requirement, we can define the expectation of a random variable $X$ as an integral, namely $$E(X) = \int_\Omega X \, d \mathbb P $$ Now as I have stated this so far, that integral might be infinite. So, one might wish to impose one more requirement on the definition of a random variable, namely that $E(X)$ is finite. This would be equivalent to requiring that $X \in L^1(\Omega, \mathcal F_s, \mathbb P)$.

$\endgroup$
5
  • $\begingroup$ thank you for your answear, it is very helpful, but I have something in mind that I want to make it clear a little further. Usually, I think that we define random variables $X(\omega,t)$ where we say that if we fix time, then it is a finction of the sample space $\Omega$, we can have in mind a Brownian motion, or if we fix it as function of time $t$, where as we know $t\in [0+\infty)$. Is this completely different from my question in the beginning? $\endgroup$
    – Hunger Learn
    Mar 20, 2020 at 15:03
  • 1
    $\begingroup$ It looks like you are thinking about a stochastic process, which is a deeper concept that a random variable. But that's another question. $\endgroup$
    – Lee Mosher
    Mar 20, 2020 at 15:21
  • $\begingroup$ yes you are right, I just figured out the same think! Well in such a case stochastic process is some $X\in L^2(\Omega,\mathcal{F}_s,\mathbb{P})$ $\endgroup$
    – Hunger Learn
    Mar 20, 2020 at 15:23
  • $\begingroup$ so to understand the concept in depth, if a function is a random variable and it has a finiete integral then, we can say that it holds $X\in L^1(\Omega,\mathcal{F}_s,\mathbb{P})$. In addition to this, if the random variable is a square integrable function, then it holds that it also has a finite variance and as a consequence $X\in L^2(\Omega,\mathcal{F}_s,\mathbb{P})$ or $\mathbb{V}ar(X)+\int_{A}(X-\mathbb{E}(X))f(X)dX<\infty$, $A\in\Omega$. Am I right? $\endgroup$
    – Hunger Learn
    Mar 23, 2020 at 9:38
  • 1
    $\begingroup$ That sounds about right. $\endgroup$
    – Lee Mosher
    Mar 23, 2020 at 14:04

You must log in to answer this question.

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