Following that question : When is it true that $\frac{ \partial a } { \partial b} = (\frac{ \partial b } { \partial a} )^{-1} $ I understood how this is true :
$$\frac{ \partial a } { \partial b} = (\frac{ \partial b } { \partial a} )^{-1} $$
Let's complicate it a little bit (perhaps) and take $a$ and $b$ as measures :
$$\frac{ \partial \mathbb P } { \partial \mathbb Q} = (\frac{ \partial \mathbb Q } { \partial \mathbb P} )^{-1} $$ Apparently this is true.
I am thinking about using Radon Nikodym theorem to prove that, when the two measures are equivalent.
But more generally, when is it true? People on the other post wanted to invert the function in a ball around the point of interest using the fact that the function is non zero. However here, the functions are non zero a.s.
Does that create problems I m not thinking about ? What are the solutions?
First of all R-N-derivatives are notated as $\frac{ d \mathbb P } { d \mathbb Q}$ instead of $\frac{ \partial \mathbb P } { \partial \mathbb Q}$
Then: What is meant by $\frac{ d \mathbb P } { d \mathbb Q}$? It's actually defined as the function $$f := \frac{ d \mathbb P } { d \mathbb Q}$$ s.t. $$\mathbb P(A) = \int_A f d\mathbb Q$$ for all measurable $A$. Consider that it follows that for any integrable function $g$ it holds $$\int_A g\; d\mathbb P = \int_A gf \;d\mathbb Q \label{a}\tag{*}$$ If both measures are equivalent we have $f > 0$ $\mathbb P$-a.s. and $\mathbb Q$-a.s and we can consider $$f^{-1} = \frac{1}{f} > 0$$
What's left to show for having $\frac{ d \mathbb Q } { d \mathbb P} = f^{-1}$ is $$\mathbb Q(A) = \int_A f^{-1} d\mathbb P$$ But with $(\ref{a})$ we get $$\int_A f^{-1} d\mathbb P = \int_A f^{-1}f \;d\mathbb Q = \int_A 1 \;d\mathbb Q = \mathbb Q(A)$$ Hence: $$f^{-1} = \frac{ d \mathbb Q } { d \mathbb P}$$