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Suppose that a distribution F is given. Does there always exist a sequence of independent random variables following F? Moreover, given two distributions F and G and a random variable X following F, does there exist a random variable Y following G, such that X and Y are independent?

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    I don't understand the phrase "independent random variables following F" even after a search. Would you define it?2012-10-18
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    @Ross Millikan I am sure the OP means "independent random variables having same CDF $F$"2012-10-18

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The answer to the first question is YES by a consequence of the famous Kolmogorov extension theorem. Answer to the second is NO. See the comments below.

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    The answer to the first question is YES. The answer to the second question might be NO if the probability space used for F is too small (consider F Bernoulli, Omega a two-points set, and G anything not Dirac nor Bernoulli). Kolmogorov extension theorem ensures that **there exists Omega** and some random variables X and Y on Omega with F and G as distributions (likewise for an i.i.d. sequence with distribution F).2012-10-18
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    @did: I do not understand. I think the answer to the second one is also YES. Kolmogorov extension theorem doesn't care about what $F$ and $G$ are. Given $F$ and $G$, it always takes $\Omega$ to be $\mathbb{R}^2$ and $\mathcal{F}$ as $\mathcal{B}(\mathbb{R}^2)$ and defines $(X,Y)$ on $\Omega$ as $(X,Y)(\omega_1,\omega_2)=(\omega_1,\omega_2)$. Now, if we define $P(X\le x, Y\le y)=F(x)G(y)$ we will have $X$ and $Y$ independent.2012-10-18
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    The OP states *given a random variable X following F*. Nothing says that X is realized on a specific Omega, for example R^2. If F and G are all that you are given then yes, you may choose Omega as you want, for example R^2 (or another probability space since R^2 is by no means the only possibility in Kolmogorov extension theorem).2012-10-18
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    Yes, you are right. I'm mistaken. Sorry.2012-10-18
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    I've edited the answer accordingly. Thank you.2012-10-18