2022-06-26

Let ${\mu }_{t}$ be the law of a random process ${X}_{t}$. Let $\nu$ and ${\nu }_{t}$ be arbitrary measures in $\mathrm{\Omega }$ with the sole restriction that ${\mu }_{t}$ be absolutely continuous w.r.t. ${\nu }_{t}$, and ${\nu }_{t}$ be absolutely continuous w.r.t. $\nu$.
Then (according to a paper I'm reading)
$\int {\mathrm{\partial }}_{t}\mathrm{log}\left(\frac{{\nu }_{t}\left(x\right)}{\nu \left(x\right)}\right)d{\nu }_{t}=\int \frac{{\mu }_{t}\left(x\right)}{{\nu }_{t}\left(x\right)}{\mathrm{\partial }}_{t}\left(\frac{{\nu }_{t}\left(x\right)}{\nu \left(x\right)}\right)d\nu =-\int {\mathrm{\partial }}_{t}\left(\frac{{\mu }_{t}\left(x\right)}{{\nu }_{t}\left(x\right)}\right)d{\nu }_{t}.$
The second inequality looks like an integration by parts, but I don't know what it means to take "$dv$" = ${\mathrm{\partial }}_{t}\left(\frac{{\nu }_{t}\left(x\right)}{\nu \left(x\right)}\right)d\nu$ where the "$dv$" derivative is with respect to a different variable (t) than the integral is taken over (\nu) -- usually "$dv$" is something like $\frac{d}{dx}v\left(x\right)dx$. And I don't understand the first inequality at all.
I'm also not sure what the notation $\frac{{\nu }_{t}\left(x\right)}{\nu \left(x\right)}$ is when we're talking about measures - I know it's not simply a fraction but something like a Radon-Nikodym derivative.

Xzavier Shelton

Expert

As I have mentioned, I would not trust every single part of the paper you are referring to, as there a probably some unchecked typos or even mistakes: since it is on arXiv, it may be not peer reviewed yet. But some things I can help you with clarifying. I think it is more formally correct to write ${g}_{t}\left(x\right)=\frac{\mathrm{d}{\mu }_{t}}{\mathrm{d}{\nu }_{t}}\left(x\right)$ rather that $\frac{{\mu }_{t}\left(x\right)}{{\nu }_{t}\left(x\right)}$ since here ${\nu }_{t}\left(x\right)$ is not even defined, as a measure it is a function of sets, not of points.
One thing that makes is easier to work with RN derivatives is the following: if $\mu ,\nu \ll \lambda$ then
$\begin{array}{}\text{(1)}& \frac{\mathrm{d}\mu }{\mathrm{d}\nu }=\frac{\mathrm{d}\mu }{\mathrm{d}\lambda }/\frac{\mathrm{d}\nu }{\mathrm{d}\lambda }.\end{array}$
Note that it makes total sense symbolically, but still on the left hand side of (1) you have a RN derivative, not any true ratio, whereas on the right hand side you have indeed a usual ratio of (density) functions. You can always find this $\lambda$, e.g. take $\lambda =\frac{1}{2}\left(\mu +\nu \right)$. For example, it means that
$\frac{\mathrm{\partial }}{\mathrm{\partial }t}\left(\mathrm{log}\frac{\mathrm{d}{\nu }_{t}}{\mathrm{d}\nu }\right)=\frac{\mathrm{\partial }}{\mathrm{\partial }t}\left(\mathrm{log}\left(\frac{\mathrm{d}{\nu }_{t}}{\mathrm{d}\lambda }/\frac{\mathrm{d}\nu }{\mathrm{d}\lambda }\right)\right)=\frac{\mathrm{\partial }}{\mathrm{\partial }t}\left(\mathrm{log}\frac{\mathrm{d}{\nu }_{t}}{\mathrm{d}\lambda }-\mathrm{log}\frac{\mathrm{d}\nu }{\mathrm{d}\lambda }\right)=\frac{\frac{\mathrm{\partial }}{\mathrm{\partial }t}\frac{\mathrm{d}{\nu }_{t}}{\mathrm{d}\lambda }}{\frac{\mathrm{d}{\nu }_{t}}{\mathrm{d}\lambda }}=\left(\frac{\mathrm{\partial }}{\mathrm{\partial }t}\frac{\mathrm{d}{\nu }_{t}}{\mathrm{d}\lambda }\right)\frac{\mathrm{d}\lambda }{\mathrm{d}{\nu }_{t}}.$
Here we used two facts. First of all, $\lambda$ should be ${\lambda }_{t}$ in general, but since all ${\nu }_{t}$ are said to be dominated by the same measure, we were able to pick a single $t$-independent dominating measure $\lambda$. Moreover,
$\frac{1}{\frac{\mathrm{d}{\nu }_{t}}{\mathrm{d}\lambda }}=\frac{\mathrm{d}\lambda }{\mathrm{d}{\nu }_{t}}.$
Some of these equalities are guaranteed to hold only a.s. but again, everything here is dominated by $\nu$, so that's not a problem. As a result, for every function $f\left(x\right)$ it holds that
$\begin{array}{rl}{\int }_{X}f\left(x\right)\frac{\mathrm{\partial }}{\mathrm{\partial }t}\left(\mathrm{log}\frac{\mathrm{d}{\nu }_{t}}{\mathrm{d}\nu }\left(x\right)\right){\nu }_{t}\left(\mathrm{d}x\right)& ={\int }_{X}f\left(x\right)\left(\frac{\mathrm{\partial }}{\mathrm{\partial }t}\frac{\mathrm{d}{\nu }_{t}}{\mathrm{d}\lambda }\left(x\right)\right)\frac{\mathrm{d}\lambda }{\mathrm{d}{\nu }_{t}}\left(x\right){\nu }_{t}\left(\mathrm{d}x\right)\\ & ={\int }_{X}f\left(x\right)\left(\frac{\mathrm{\partial }}{\mathrm{\partial }t}\frac{\mathrm{d}{\nu }_{t}}{\mathrm{d}\lambda }\left(x\right)\right)\lambda \left(\mathrm{d}x\right)\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }t}\left({\int }_{X}f\left(x\right)\frac{\mathrm{d}{\nu }_{t}}{\mathrm{d}\lambda }\left(x\right)\lambda \left(\mathrm{d}x\right)\right)\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }t}\left({\int }_{X}f\left(x\right){\nu }_{t}\left(\mathrm{d}x\right)\right).\end{array}$
Note also that working with this you have to be accurate in something of the kind ${b}_{t}\left(x\right)\frac{\mathrm{\partial }}{\mathrm{\partial }t}{a}_{t}\left(x\right)$ which is
${b}_{t}\left(x\right){\left(\frac{\mathrm{\partial }}{\mathrm{\partial }s}{a}_{s}\left(x\right)\right)|}_{s=t}$
cause otherwise you may get confused. I hope I didn't.

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