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Given a probability space $(\Omega,\mathcal{F},P)$, and a random variable $X\colon\Omega\to\Bbb{R}$, we can associate with it its distribution function $F\colon \Bbb{R}\to[0,1]$ defined as \begin{align*} F(x)&=P(X\le x)=P(\{\omega\in\Omega:X(\omega)\le x \}). \end{align*} Further, we can define an induced probability measure $\mu\colon \mathcal{B}\to[0,1]$ by \begin{align*} \mu(A) &= P(X\in A)=P(\{\omega\in\Omega:X(\omega)\in A \}). \end{align*}

I read in this answer that $\mu$ and $F$ uniquely determine eachother.

My question is: can we do this "in reverse", i.e. starting with a probability measure $\mu\colon \mathcal{B}\to[0,1]$, can we then get a unique random variable?


On $(\Bbb{R},\mathcal{B},P)$, I tried to do this starting with $\mu(A)=P(A)$ for all $A\in\mathcal{B}$, where $P(A)=\mathrm{Leb}(A\cap[0,1])$. I believe this yields $X(x)=x$ for all $x\in\Bbb{R}$, and $F(x)=\min(1,\max(0,x))$, so it works in this simple case.

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  • $\begingroup$ $F_X(x)= P(X\leq x) = \mu((-\infty; x])$ for example $\endgroup$ – Slowpoke May 16 '16 at 15:21
  • $\begingroup$ Yes, it works. You start with $(\Omega,\mathcal F,P)$ where $\Omega=\mathbb R$ and $\mathcal F=\mathcal B$ (the Borelsets). Then function $X:\mathbb R\to\mathbb R$ prescribed by $x\mapsto x$ is a random variable. This with $P_X(A):=P(X\in A)=P(A)$ so $P_X=P$. The distribution of $X$ coincides with the probability measure that is part of the probability space. $\endgroup$ – drhab May 16 '16 at 15:26
  • $\begingroup$ @drhab Will the random variable always be the identity? $\endgroup$ – Szmagpie May 16 '16 at 19:39
  • $\begingroup$ No. But if you want $P_X=P$ then you need $X=1_{\mathbb R}$ a.s. My former comment proves actually that for every distribution $P$ on $(\mathbb R,\mathcal B)$ a random variable $X$ can be found having $P$ as its distribution: $P_X=P$. Further it is not a very fruitful subject. If you take $(\mathbb R,\mathcal B)$ as measurable space then you inherit quite some restrictions. Just keep it on the "mysterious" $(\Omega,\mathcal F,P)$ willing to offer anything you want. $\endgroup$ – drhab May 17 '16 at 7:38

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