Given two real-valued random variables $X, Y$ with distributions $\mu_X, \mu_Y$. Suppose $\mu_X<\!<\mu_Y$, then the Radon-Nikodym derivative $\frac{d\mu_Y}{d\mu_X}(\cdot)$ exists $\mu_X$-a.e. on $\mathbb{R}$.
It seems that joint distribution $(X, Y)$ does not affect their Radon-Nikodym derivative at all? Since the measure-theoretic definition of Radon-Nikodym derivative as a limit is given by
$$\frac{d\mu_Y}{d\mu_X}(\cdot) = \lim_{r \rightarrow 0} \frac{\mu_Y(B(r,\cdot))}{\mu_X(B(r,\cdot))}.$$
I am just a bit unsure since the "change of measure" type of argument is ubiquitous in probability theory and yet it isn't affected by the joint law. (Unless there is some other type of change of measure argument in probability theory.)
Best Answer
In the setting of the Radon-Nikodym theorem, the two measures $\mu_X$ and $\mu_Y$ are measures on the same measurable space $(\Omega, \mathcal{F})$. When you have two random variables, and you want to talk about the Radon-Nikodym derivative, you first need $\mu_X$ and $\mu_Y$ to be on the same measurable space (in probability, usually $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$. But after this, any information about the joint distribution of $X$ and $Y$ is irrelevant. If you take a step back and ignore the probabilistic context, you will see that the Radon-Nikodym derivative concerns two measures on the same measurable space, and not really two marginal measures on some larger space like $\mathbb{R} \times \mathbb{R}$ or $\Omega \times \Omega$.