8

I have some understanding of measure theory / real analysis, and some understanding of probability theory, but I'm having some trouble putting the two together.

According to Wikipedia:

Let $(\Omega, \mathcal{F}, P)$ be a probability space and $(E, \mathcal{E})$ a measurable space. Then an $(E, \mathcal{E})$-valued random variable is a function $X : \Omega \to \mathcal{E}$ which is $(\mathcal{F}, \mathcal{E})$-measurable.

Now for example, let's take $X$ be a standard Gaussian random variable, $X \sim \mathcal{N}(0, 1)$.

  • I think $E = \mathbb{R}$ since $X$ takes values in $\mathbb{R}$.
  • Also, we should have $\mathcal{E} = \mathscr{B}(\mathbb{R})$ the Borel $\sigma$-field of $\mathbb{R}$.
  • But, what should $(\Omega, \mathcal{F}, P)$ be?

Furthermore, let's try to calculate $\mathbb{E}[X]$ the mean of $X$. By Wikipedia's definition,

$$\mathbb{E}[X] = \int_\Omega X\, dP = \int_\Omega X(\omega)\, P(d\omega).$$

This raises some questions.

  • How does this relate to the elementary computation: $$\mathbb{E}[X] = \int_{-\infty}^{\infty} x\cdot f_X(x)\, dx$$ How does $f_X : \mathbb{R} \to \mathbb{R}^{\geq 0}$ relate to the measure-theoretic definition of $X$?
  • What is the meaning of $P(d\omega)$? $P$ is a measure so it makes sense to integrate $dP$, but what is $d\omega$?
jcai
  • 638
  • 2
    You don't usually care about the underlying probability space. What matters is the distribution function of your random variable. But if you need to see an example, just take $\Omega = [0,1]$, $P$ the Lebesgue measure and define $X(\omega) = \Phi^{-1}(\omega)$ where $\Phi$ is the distribution function of a standard normal random variable. – Calculon Feb 20 '17 at 12:25
  • Not only you don't care about the underlying probability space, but you don't want to care. Consider $X$ the random variable that gives you the result of a dice roll. What is the underlying probability space? Should it be the space of all possible positions of all atoms in the universe ? Something simplier? something more complex? – Tryss Feb 20 '17 at 13:03
  • As of today, [1] has a description of these five symbols given "in more intuitive terms." [1] https://en.wikipedia.org/wiki/Random_variable#Measure-theoretic_definition – Michael Levy Aug 31 '19 at 23:56

3 Answers3

3

But, what should be $(Ω,F,P)$?

Depends on your application. If want to look at arbitrary random variables, then it is simple arbitrary. If you have a specific example in mind, it is not.

How does this relate to the elementary computation [...]

In this case you assumed that $X \sim \mathcal N[0,1]$. In particular $X$ is a continuous variable. In measure theory we say that the distribution of $X$ is absolutely continuous w.r.t. to Lebesgue measure. The Radon-Nikodym theorem then guarentees the existence of an $f_X$ with the property you have stated, so that we can apply the change of variable formula to make the computation of the expectation easier. Without the change of variable formula, we would have to compute the expectation with the definitions of expectations for indicator functions then simple functions then (?simple integrable then $L_1$ then?) nonnegative functions and then measurable functions.. But, again, this is a particular example, where $X$ is continuous. The measure-theoretic definition of expectation is much more general.

What is the meaning of $P(dω)$?

It doesn't mean anything. It is a notational crutch, like writing $\lim\limits_{n\to \infty} a_n$ instead of $\lim a$. It becomes useful, if you have multiple nested integrals / integrate w.r.t. to product measures.

BCLC
  • 14,197
Stefan Perko
  • 12,867
  • Stefan Perko, suggestion: so that we ... 'can apply the change of variable formula to make the computation of the expectation easier. Without the change of variable formula, we would have to compute the expectation with the definitions of expectations for indicator functions then simple functions then nonnegative functions and then measurable functions. ' ? – BCLC Mar 14 '18 at 08:33
  • Related question? Computing expectation without change of variable formula – BCLC Mar 14 '18 at 08:34
  • @BCLC Knock yourself out. Btw. you can also go 'integrable simple functions' $\to$ '$L_1$-functions' $\to$ 'non-negative functions' if you like. – Stefan Perko Mar 14 '18 at 10:18
  • Thanks, Stefan Perko! ^-^ – BCLC Mar 14 '18 at 11:17
  • Stefan Perko, wait, what's $L_1$? integrable? – BCLC Mar 14 '18 at 11:18
  • @BCLC Yeah, simple integrable functions constitute a dense subspace of $L_1$ (https://en.wikipedia.org/wiki/Lp_space). – Stefan Perko Mar 14 '18 at 11:27
  • integrable comes before nonnegative? I thought integrable was the last step. So the last step is actually measurable and not necessarily integrable? It's been awhile since I took real analysis – BCLC Mar 14 '18 at 12:58
  • If you are okay with proving it just for integrable functions you can stop at the second step. The third step is only necessary if you want to prove the statement for functions taking infinite values by realizing that $f : \Omega \to [0,\infty]$ is s.t. $\sup_{n\in \mathbb N} f\wedge n = f$ and $f\wedge n$ is integrable ($\wedge$ is minimum). - However, now that I look at this more closely I would not recommend this approach. It seems like it can work somehow, but the fact that $f\mapsto f \circ X$ is not necessarily an operator $L_1 \to L_1$ makes it more complicated than I thought. – Stefan Perko Mar 14 '18 at 14:41
0

The probability space is abstract (We know, or suppose it exists but we don't care how it looks). By Radom-Nikodym Theorem, if a measure $\nu$ is absolutely continuous with respect to other $\mu$, this is $\forall A\in\mathcal{F},\mu(A)=0\implies\nu(A)=0$, if both are $\sigma$-finite, there is a measurable function $f$ such that $$\forall A\in\mathcal{F},\nu(A)=\int_Af\mathrm{d}\mu $$ Besides you prove that in this case for all measurable function $g$ we have that $$\int g\mathrm{d}\nu=\int fg\mathrm{d}\mu$$ in this case $\nu(\dot{})=\int_{\dot{}}f_X\mathrm{d}x,\mu=\mathrm{d}x=$lebesgue measure, obviously $f=f_X$ and $g$ is the random variable $X$. $P(dw)$ is just notation.

Julio Maldonado Henríquez
  • 1,318
0

You can use the following derivations:

$\int X dP = \int t dPX^{-1}(dt) = \int t dF = \int t f_{X} dx$ where F is the distribution function for $X$ and $PX^{-1}$ is the pull back measure given by $PX^{-1}(B)=P(X^{-1}(B))$.

Literally an Orange
  • 810