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Suppose that ($\Omega$, $\mathcal{F}$,$P$) where $\mathcal{F}$ is the $\sigma$-algebra of Lebesgue measurable subsets of $\Omega\equiv[0,1]$ and $P$ is the Lebesgue measure. Let $G:\mathbb{R}\to[0,1]$ be an arbitrary distribution function. My questions:

  1. Can we always construct $X:\Omega\to\mathbb{R}$ so that $X$ has $G$ as its CDF?
  2. If the answer is 'Yes' to the above, what are other triplets besides the particular $(\Omega,\mathcal{F},P)$ given above that will yield the same answer?

If $G$ is continuous and has range $[0,1]$, $X(\omega)=G^{-1}(\omega)$ for $\omega\in[0,1]$ works: $$ \Pr[X\leq x]=P[\omega:X(\omega)\leq x]=P[\omega:\omega\leq G(x)]=P([0,G(x)])=G(x). $$ But I don't know how to deal with more general $G$.

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    $\begingroup$ I believe the answer to 1. is yes, and that right-continuous (or left-continuous) inverses are a possible method of construction, see here: people.math.ethz.ch/~embrecht/ftp/generalized_inverse.pdf also searching for the term "quantile function" might also be relevant, I think (i.e. I'm not 100% here) $\endgroup$ – Chill2Macht Jan 8 '17 at 17:17
  • $\begingroup$ Even if $G$ is continuous, it may not be 1-1, so you need to give some thought to what you mean by $G^{-1}$. The key is to design $X(\omega)$ so that $\{\omega: X(\omega)\le x\} = [0,G(x)]$ (or $[0,G(x))$). $\endgroup$ – John Dawkins Jan 8 '17 at 17:45
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A necessary and sufficient condition is that $(\Omega,\mathcal{F},P)$ is an atomless probability space. An atom in a probability space is a set $E\in\mathcal{F}$such that $P(E)>0$ and for all $F\subseteq E$ with $F\in\mathcal{F}$ either $P(F)=0$ or $P(F)=P(E)$. The proof of sufficiency is somewhat messy and naturally proceeds by constructing a uniformly distributed random variable with values in $[0,1]$ .

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I nice reference for the sufficient conditions for 1. to be possible is Lemma 2.3 in Berkes and Philipp (1979): Approximation Thorems for Independent and Weakly Dependent Random Vectors

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