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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.

Roger Crook
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  • Since you're talking about measures, then it's not strictly accurate to say that the probability function is defined for every subset of real numbers. In particular, depending on what measure you put on the space, some subsets of $\Bbb R$ may not be measurable and therefore their probability will be undefined. – Addem Jul 25 '23 at 22:19
  • This may be of interest to you. – Mittens Jul 25 '23 at 22:20
  • You mean $P[X \in A]$, not $P[X=A]$. – Michael Jul 25 '23 at 23:42

1 Answers1

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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 $X$.

Julius
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