Radon-Nikodym derivative $\frac{d\mathbb{P}_{X}}{d\mathbb{Q}_{X}}$ from $\frac{d \mathbb{P}}{d\mathbb{Q}}$? Assume we have two probability measures $\mathbb{Q}$,$\mathbb{P}$ and assume we know the radon-nikodyn-derivative $\frac{d \mathbb{P}}{d\mathbb{Q}}$.
For some random variable $X$, $\mathbb{P}_{X}(A):= \mathbb{P}(X^{-1}(A))$
Can we say anything about the radon-nikodyn derivative
$$\frac{d\mathbb{P}_{X}}{d\mathbb{Q}_{X}} ?$$
Can we compute it from the given one?
 A: I am not entirely sure if something can be said in general about this Radon-Nikodym derivative, but I will provide a positive answer to your question in the case when $X$ is actually invertible. To this end assume $\mathbb{P}, \mathbb{Q}$ are probability measures on the measurable space $(\Omega, \mathcal{B})$ and denote the aforementioned Radon-Nikodym derivative by $f := \frac{d\mathbb{P}}{d\mathbb{Q}}.$ Also let $X : \Omega \rightarrow \mathbb{R}$ be an invertible random variable. I will denote the pushforwards of the probabilities in the classical manner, i.e. $X_\ast \mathbb{P}$ for example.
By the very definition of the Radon-Nikodym derivative $f,$ we have that for every measurable set $A \subseteq \Omega,$ $\mathbb{P}(A) = \int_A f d \mathbb{Q}.$ Let now $B \in \mathcal{B}(\mathbb{R}).$ We have that $X_\ast \mathbb{P}(B) = \int_B d X_\ast \mathbb{P} = \int_\mathbb{R} \chi_B d X_\ast \mathbb{P} = \int_\Omega \chi_B \circ X d \mathbb{P} = \int_\Omega \chi_B \circ X \cdot f \,\, d\mathbb{Q} = \int_\Omega (\chi_B \cdot f\circ X^{-1}) \circ X \,\, d\mathbb{Q} = \int_\mathbb{R} \chi_B \cdot f \circ X^{-1} \,\, d X_\ast \mathbb{Q} = \int_B f \circ X^{-1} \,\, d X_\ast \mathbb{Q}.$ Since $B$ was arbitrary, it follows that the Radon-Nikodym derivative of $X_\ast \mathbb{P}$ with respect to $X_\ast \mathbb{Q}$ is $\frac{d \, X_\ast \mathbb{P}}{d \, X_\ast \mathbb{Q}} = \frac{d \mathbb{P}}{d \mathbb{Q}} \circ X^{-1},$ which is quite a natural result to expect.
