Measure theoretic interpretation of sampling a random variable According to this post,  $N$ independent, identical draws from a random variable $Y$ is the same thing as one draw from each of $N$ independent, identically random variables $Y_n$. I am struggling to formulate this observation rigorously from the perspective of measure theory.
I guess sampling from $Y$ n times, can on one hand be thought of as a sequence of events $\omega_n$ from one sample space $\Omega$ or it can be thought of as a realization of $n$ iid random variables. In what sense are they equivalent mathematically?
 A: Let us recall that, from the point of view of measure theory, a probability space is a triple $(\Omega ,\mathcal A,\mu
)$, where $\Omega $ is a set, $\mathcal A$ is a $\sigma $-algebra of subsets of $\Omega $, and $\mu $ is a positive
measure defined on $\mathcal A$, and satisfying $\mu (\Omega )=1$.
On the other hand, a random variable is nothing but a function
$$
  Y:\Omega \to {\mathbb R}
  $$
which is measurable in the sense that $Y^{-1}((-\infty ,a))\in \mathcal A$, for all $a\in {\mathbb R}$.
Incidentally, although Statisticians will likely disagree, there is absolutely nothing random about a random variable!
I any case, given $Y$, as above,
in order to formalize the concept of $n$ independent and identically distributed random variables, one must replace
the probability space $(\Omega ,\mathcal A,\mu )$ with $(\Omega _n,\mathcal A_n,\mu _n)$, where

*

*$
  \Omega _n = \Omega \times \Omega \times \cdots \times \Omega ,
  $


*$\mathcal A_n$ is the $\sigma $-algebra generated by the collection of sets of the form
$E_1\times E_2\times \cdots \times E_n$, where each $E_i$ belongs to $\mathcal A$,


*$\mu _n$ is the product measure, namely the unique measure defined on $\mathcal A_n$ such that
$
  \mu _n(E_1\times E_2\times \cdots \times E_n)= \mu (E_1) \mu (E_2) \ldots \mu (E_n).
  $
Once this is done, we may consider, for each $k=1,2,\ldots ,n$, the random variable $Y_k$ defined by
$$
  Y_k(\omega _1,\omega _2,\ldots , \omega _n) = Y(\omega _k),
  $$
and one may now easily prove that the $Y_k$ are indeed independent, and that they are identically distributed, in the
sense that the push-forward measures on ${\mathbb R}$ given by $(Y_k)_*(\mu _n)$ all coincide with $Y_*(\mu )$.
Recalling that the two alternatives presented by the OP are:

*

*a realization of $n$ iid random variables,


*a sequence of events from the sample space,
I must say that I struggle to make precise mathematical sense of the sentence "a realization of a random variable".  To
me, the only possible meaning of this is "choosing some $\omega $ in $\Omega $ and computing $Y(\omega )$", which
doesn't seem to be an act worthy of such a pompous name.
Nevertheless, according to this logic, (1) should then be formally interpreted as choosing some $\omega =(\omega
_1,\omega _2,\ldots , \omega _n)$ in $\Omega _n$, and computing $Y_1(\omega ),Y_2(\omega ),\ldots ,Y_n(\omega )$.
On the other hand, (2) should be interpreted as preforming $n$ experiments of choosing $\omega $ in $\Omega $, and
computing $Y(\omega )$.
So, if we collect the $n$ chosen $\omega $'s from the paragraph above into an $n$-tuple $(\omega _1,\omega _2,\ldots ,
\omega _n)$, we see that (2) is precisely equivalent to (1).
