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Let $X:(\Omega, \mathcal F, \mathbb P) \to \mathbb R$ be a random variable. I usually encounter a statement "let $(X_n)_n$ be an i.i.d. sample of $X$". However, I could not imagine how such sequence appears in a specific probability space.

Could you please construct such a sequence from $X$?

Analyst
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  • This is highly non-trivial to prove and not even possible on every probability space. – J. De Ro Mar 30 '22 at 21:19
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    Usually, it means that one constructs a product space from copies of $(\Omega,\mathcal{F},\mathbb{P})$ and defines $X_n(\omega)=X(\omega_n)$. ($\omega$ is a sequence here , i.e., $(\omega_1,\omega_2,\ldots)$.) –  Mar 30 '22 at 21:45
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    This is handled by the Kolmogorov extension theorem: https://en.wikipedia.org/wiki/Kolmogorov_extension_theorem – Mike Earnest Mar 30 '22 at 22:24
  • So this is generally impossible without extending the probability space. Consider $\Omega = {0,1}$ and $P[{0}]=P[{1}]=1/2$. We cannot produce a sequence of i.i.d. $Bern(1/2)$ random variables on this space. – Michael Mar 31 '22 at 00:34

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