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I just had some questions about what the sample space of the following random variables are.

Background: According to Wikipedia, a random variable $X$ is a measurable function from the sample space $\Omega$ to a measurable space $E$, where $\Omega$ belongs to a probability space $(\Omega, \mathcal{F},P)$.

  1. In the literature the following phrase is often encountered: "let $X$ be a Gaussian random variable with zero mean and unit variance, that is, $X \sim \mathcal{N}(0,1)$". Now, in such formulation, mention of the sample space and measurable spaces are typically omitted, I guess for convenience, but if we want to be pedantic, would this be $\Omega = E = \mathbb{R}$?

  2. Let's say I see instead the phrase "Let $X$ be a normalized Haar random vector on $\mathbb{R}^n$". I guess $E$ is the $(n-1)$ unit sphere? What is the sample space $\Omega$?

  3. "Let $U$ be a Haar random distributed unitary on a $d$-dimensional space". What is $\Omega$ and $E$?

  4. Ultimately I am interested in some works on convergence in distribution of products of random matrices $M_n = A_1 A_2\cdots A_n$ as $n \to \infty$, e.g. https://link.springer.com/content/pdf/10.1007%2FBF00532045.pdf . The precise theorem is not important, I am just asking about the mathematical set-up. Here $A_i$ are i.i.d. random square $d \times d$ matrices. My question is, is it obvious that $M_n, M_{n-1}$ are random variables on the same probability space, such that convergence (in distribution) is a meaningful concept? Naively I would have said, the sample space of $M_n$ is larger than the sample space of $M_{n-1}$ because the former involves a longer product of random objects, but I think I am confusing some concepts.

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    $\begingroup$ As in general, it always depends on the context. And this can be different from book to book. Therefore, I would refrain from making a general statement. Hence, it is difficult to make a statement without further sources. And I would be very surprised if someone here could make a more precise statement. $\endgroup$
    – Nils
    Commented Mar 11, 2022 at 1:12
  • $\begingroup$ @Nils I think even an answer about why you say this is imprecise / ambiguous would be helpful. And examples of how it can be different in different contexts. The reason is because, formal probability theory is rooted heavily in the probability triple $(\Omega, \mathcal{F}, P)$. In practice (physics, engineering), we just shut up and calculate: we all intuitively know what a random variable that is e.g. uniform is, or at least think we know how to generate it, without thinking too hard about the underlying probability triple. It would be extremely unsatisfying if there is no middle ground. $\endgroup$
    – nervxxx
    Commented Mar 11, 2022 at 1:24

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