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Given two probability spaces $(E,\mathcal{B},\mu)$ and $(E,\mathcal{B},\nu)$, can we find two random variables $X$ and $Y$ and the same probability space $(\Omega,\mathcal{F},\mathbb{P})$, such that $X:(\Omega,\mathcal{F},\mathbb{P})\rightarrow(E,\mathcal{B})$ and $Y:(\Omega,\mathcal{F},\mathbb{P})\rightarrow(E,\mathcal{B})$ and $X$, $Y$ are both measurable and $\mu$ and $\nu$ are the corresponding distribution measures ?

More generally, the two probability spaces can be different? To be specific, Given two probability spaces $(E,\mathcal{B},\mu)$ and $(G,\mathcal{H},\nu)$, can we find two random variables $X$ and $Y$ and the same probability space $(\Omega,\mathcal{F},\mathbb{P})$, such that $X:(\Omega,\mathcal{F},\mathbb{P})\rightarrow(E,\mathcal{B})$ and $Y:(\Omega,\mathcal{F},\mathbb{P})\rightarrow(G,\mathcal{H})$ and $X$, $Y$ are both measurable and $\mu$ and $\nu$ are the corresponding distribution measures ?

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  • $\begingroup$ What is a random variable? Can it take values in arbitrary probability spaces? $\endgroup$ – Justpassingby Dec 14 '15 at 17:12
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    $\begingroup$ Sure. Let $\Omega=E\times G$ and $\mathbb{P}=\mu\times \nu.$ $\endgroup$ – user940 Dec 14 '15 at 17:12
  • $\begingroup$ @ByronSchmuland I got it! Thanks so much! $\endgroup$ – Michael Dec 14 '15 at 17:37
  • $\begingroup$ @Justpassingby Yes, it need not to be a real-valued function. en.wikipedia.org/wiki/Random_variable $\endgroup$ – Michael Dec 14 '15 at 17:38
  • $\begingroup$ @ByronSchmuland If constructed in this way, $X$ and $Y$ are independent. Can we construct $X$ and $Y$ such that they are not independent? $\endgroup$ – Michael Dec 14 '15 at 17:44

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