Given the joint distribution, how to find the transformation beween the marginals?

Question

Assume p(x,y) is the joint density of two random variables X and Y. Is there any way to construct the transformation Y=f(X) which corresponds to the joint density p(x,y)?

Answer 1

Such an $f$ exists iff $Y$ is $\sigma(X)$-measurable (you can show this using a standard machinery proof). If this is the case, we have $$Y = E[Y|X] = \int_{\mathbb{R}} yp_{Y|X}(y|X)dy,$$ where $p_{Y|X}p(y|x) = p(x, y)/p_X(x)$ and $p_X(x) = \int_{\mathbb{R}} p(x, y)dy.$

$f$ must be unique up to a $\mu_X$-null set, where $\mu_X$ is the distribution of $X$ (i.e., the measure $\mu_X(A) = \int_A p_X(x)dx).$