Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space (if necessary we can assume to work with the unit interval with the Lebsegue measure) and let $X: \Omega \to \mathbb{R}$ be a random variable with law $\mu$ (meaning that $X_\sharp \mathbb{P}=\mu$). Suppose it is given a probability measure $\gamma$ on $\mathbb{R}^2$ with first marginal $\mu$ (meaning that $\pi^1_\sharp \gamma = \mu$, where $\pi^1(x,y)=x$ is the projection on the first factor from $\mathbb{R}^2$ to $\mathbb{R}$). Can we find a random variable $Y:\Omega \to \mathbb{R}$ s.t. $(X,Y)_\sharp \mathbb{P}=\gamma$?
I think that the problem is equivalent to the following: given the disintegration of $\gamma$ w.r.t. $\mu$, call it $\{\mu_x\}_{x \in \mathbb{R}}$, can we find a random variable $Y$ s.t.
$$\mathbb{P}(Y \in B \mid X=x) = \mu_x(B)$$ for $\mu$-a.e. $x \in \mathbb{R}$ and every $B \in \mathcal{B}(\mathbb{R})$?
Edit: I add the following comment that may be helpful. It seems (see Bogachev 10.7.7 Corollary) that a sufficient condition is the existence of a random variable $Z: \Omega \to [0,1]$ with uniform distribution independent from the given $X$. Is it always possible if the probability space is the unit interval with the Lebesgue measure?
Best Answer
Even when $(\Omega,\mathcal F, \Bbb P)=([0,1],\mathcal B_{[0,1]},\text{d}x)$, there does not always exist $Y$ for which $(X,Y)_\sharp \Bbb P=\gamma$.
Finally, let $X$ be the random variable on $([0,1],\mathcal B_{[0,1]},\text{d}x)$ defined by $X(\omega)=\omega$. Obviously, $\sigma(X)=\mathcal B_{[0,1]}$. In the case that $\gamma=\mu\times \nu$ is a product measure, where $\nu$ is not concentrated on a single point, $(X,Y)_\sharp \Bbb P=\gamma$ would imply $Y$ is a non-constant random variable is independent of $X$. This contradicts our lemma.