Change of measure in probability theory

Question

When we look at the change of measure in probability theory, are the R-N derivative always computed on the level of probability density (defined as measures on $\mathbb{R}$)? Does it make sense to look at any R-N derivative of abstract probability measures on the abstract probability space in a change of measure?

I also have some confusion illustrated in the following example.

Given an abstract probability abstract space $(\mathbf{\Omega}, \mathbf{P})$. Look at random variables

• $X\sim \text{Exp}(1)$

• $Y \sim \text{Exp}(2)$ where $Y\perp X$, and

• $Y' \sim \text{Exp}(2)$ where $Y'=\frac{1}{2} X$.

  1. If we realize these on $\Omega = \mathbb{R}$ with three (separate) probability measures $\mathbb{P}$, $\mathbb{Q}$, and $\mathbb{Q'}$, then I think $\frac{d\mathbb{Q}}{d\mathbb{P}}$ and $\frac{d\mathbb{Q}'}{d\mathbb{P}}$ are the same. Thus the $Y$ and $Y'$ makes no difference here, and for any bounded continuous function $f$ $$\mathbb{E}_{\mathbb{P}} [f(X)] = \mathbb{E}_{\mathbb{Q}} \Big[f(Y)\frac{d\mathbb{Q}}{d\mathbb{P}}\Big] = \mathbb{E}_{\mathbb{Q'}} \Big[f(Y')\frac{d\mathbb{Q'}}{d\mathbb{P}}\Big].$$

  2. On the other hand, what if we realize everything on $\Omega = \mathbb{R}^3$ with a joint density $p(x,y,y')$. Now we have three marginals $\mathbb{P}$, $\mathbb{Q}$, and $\mathbb{Q'}$. Does it make sense to perform a change of measure from $\mathbb{P}$ to $\mathbb{Q}$ or $\mathbb{Q'}$? And I think the R-N derivatives $\frac{d\mathbb{Q}}{d\mathbb{P}}$ and $\frac{d\mathbb{Q}'}{d\mathbb{P}}$ are no longer the same.

So between (1) and (2), which one is the more natural approach in a change of measure in probability theory?

Answer 1

I'm not exactly sure if this answers your question, but the idea is to always take Radon-Nikodym derivatives wrt to Lebesgue measure. To make sense of this, we push the probability measure $\mathbb{P}$ into $\mathbb{R}$ by considering the law $\mu_X(A):=\mathbb{P}(X\in A)$ (where $A$ is a Borel subset of the real line). This is helpful, because it replaces the abstract space $X$ with the familiar space $\mathbb{R}$. When $\mu_X$ is absolutely continuous wrt Lebesgue measure, it has a Radon-Nikodym derivative, and we have $$ \mu_X(A)=\int_A \frac{d\mu_X}{dm}dm $$ where $m$ is the Lebesgue measure on $\mathbb{R}$.