Probability Spaces and Random Variables

Question

Say I have a population $\Omega$, a random variable $X \colon \Omega \to \mathbb{R}$. Assume for example that $X \sim \mathcal{N}(0,1)$. I know that the Probability distribution $\mathbb{P}(X=A)$ ($A\subset \mathbb{R}$) is given by $$\int_A f(x) dx$$ where $f(x)$ is the according density function. Now how does my probability measure space look like? What are the measurable sets on $\Omega$ and what's my measure $\mathbb{P}$ on $\Omega$? I get the intuition behinde random variables and what $\mathbb{P}(X=A)$ is supposed to mean in the real world (it's easier to work with numbers) but when just looking at this abstractly i can't draw an analogy to the Lebesgue measure $\lambda$ on $\mathbb{R}^n$ in the sense that I have some ''reasonable'' set $A \subset \mathbb{R}^n$ with an amount of volume (meaning $A$ is in the $\sigma$-algebra of $\lambda$-measurable sets) and then I can compute that volume by applying $\lambda$ to $A$. Trying to draw the same analogy to $\Omega$ is rather hard since say I have some "reasonable" subset $A\subset \Omega$ and I want to apply $\mathbb{P}$ on it. I can't do that, I am kind of forced to do this through $X$ somehow which makes me think that the $\sigma$-algebra of measurable sets on $\Omega$ is induced by the preimage of $X$ but I don't know.

Answer 1

The truth is that the probability space doesn't really matter. Of course, this is not true at a surface level, but if $(\Omega, \mathcal{F}, \mathbb{P})$ is rich enough for you to be able to define the random variables you care about, then the exact nature of the probability space doesn't really matter.

I think the key here is in your last sentence:

I am kind of forced to do this through $X$ somehow which makes me think that the $\sigma$-algebra of measurable sets on $\Omega$ is induced by the preimage of $X$

I think this captures the notion that, for a random variable $X: \Omega \rightarrow \mathbb{R}$, it is not the probability space $\Omega$ that captures the behaviour of $X$, but rather the law of $X$, which is the pullback measure $\mu_X:=\mathbb{P}\circ X^{-1}$ defined on the measurable space $(\mathbb{R}, \mathcal{B})$.

Now, in the space $(\mathbb{R}, \mathcal{B}, \mu_X)$, the identity function is measurable, i.e. a random variable, with the same law as your original random variable. It is in this sense that I mean that the probability space doesn't matter, because as soon as you have a random variable, you can 'transfer' to a different probability space which more innately contains the behaviour of the random variable.

While there are many probability spaces you can define a uniform random variable on, for example, the law (i.e. the probability meadure) it induces on the space on which it is supported on, $[0,1]$, is unique; its law is the Lebesgue measure.

So to answer your question, if you only know that a random variable $X$ exists on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$, there isn't a lot you can say about what $\Omega$ itself looks like. All you can say is that it carries enough information to carry a random variable like $X$.