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Let $X:\Omega\longrightarrow \mathbb{R}$ be a random variable. Suppose we take $n$ "observations" (aka. "samples") from $X$, and we put them into an $n$-tuple $(a_1, .., a_n)$, where $a_i$ is the $i$-th observation.

Now suppose we take an $n$-tuple $(X_, ..., X)$, made of $n$ copies of our random variable $X$, and we sample it once, resulting in an $n$-tuple $(b_1, ..., b_n)$.

  1. Are these 2 processes mathematically equivalent?
  2. If equivalent, are there any mathematical advantages of taking one viewpoint over the other?
  3. Does it make sense to build vectors of random variables? Can we even call them vectors? Do they form a vector space?
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closed as off-topic by Did, user91500, José Carlos Santos, Namaste, Glorfindel Jul 21 '17 at 19:07

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    $\begingroup$ Your "Now suppose..." is unclear. If by "$n$ copies of our random variable $X$" you mean (as user4646336 understands) you mean $X(\omega),X(\omega),\ldots)$ the answer is "No". But if you mean construct a new probability space by taking the $n$-fold product of the old one, and use $X(\omega_1), X(\omega_2)\ldots)$ the answer is "Yes". For questions 2 and 3, the answer then becomes Yes. The advantage is sometimes a considerable notational simplification. Look out, however, for "Big" vector spaces (such as spaces of continuous functions), where questions of measurability arise $\endgroup$ – kimchi lover Jul 19 '17 at 12:35
  • $\begingroup$ Continuing to reinvent the wheel about very basic definitions of probability theory puts you at risk of receiving phony answers and believing them, as happened with your previous probabilistic question (but not here). To get a reliable text on the subject, to study it, and then to ask questions on the points remaining unclear to you, would be much preferable. But I am curious: would you imagine asking similarly uninformed questions about étale cohomology, or is this an option only when probability is involved? $\endgroup$ – Did Jul 21 '17 at 14:49
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Random variables are just functions of the space of states.

Sampling is just evaluating them.

Having $X(w_1), X(w_2), ..., X(w_n)$ is the same as having $(X(w_1), X(w_2), ..., X(w_n))$.

What is not the same is doing things like $(X(w), X(w), ..., X(w))$.

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