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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?

Marine Galantin
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1 Answers1

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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: $$f^{-1} = \frac{ d \mathbb Q } { d \mathbb P}$$

Gono
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  • Thank you for the detailled answer. What are the consideration one should be careful about? You re writing 1/f and f^-1, what are the problems that can arise please? I still don't get how a function that could be zero in many different places can be expressed as 1/f – Marine Galantin Feb 24 '20 at 10:42
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    There is nothing to care about. As mentioned above: If both measures are equivalent, $f$ is strictly positive hence cannot be zero. So there is nothing to concern by considering $\frac{1}{f}$. It holds $$\Bbb Q \sim \Bbb P \iff f > 0$$ – Gono Feb 24 '20 at 12:48
  • but it is f > 0 P-as, right ? – Marine Galantin Feb 24 '20 at 13:30
  • It is $f>0$ P-as and Q-as. – Gono Feb 24 '20 at 13:44