Mohammad Cannon

Answered

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)

$\int {\mathrm{\partial}}_{t}\mathrm{log}\left(\frac{{\nu}_{t}(x)}{\nu (x)}\right)d{\nu}_{t}=\int \frac{{\mu}_{t}(x)}{{\nu}_{t}(x)}{\mathrm{\partial}}_{t}\left(\frac{{\nu}_{t}(x)}{\nu (x)}\right)d\nu =-\int {\mathrm{\partial}}_{t}\left(\frac{{\mu}_{t}(x)}{{\nu}_{t}(x)}\right)d{\nu}_{t}.$

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}(x)}{\nu (x)}\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 $\frac{d}{dx}v(x)dx$. And I don't understand the first inequality at all.

I'm also not sure what the notation $\frac{{\nu}_{t}(x)}{\nu (x)}$ is when we're talking about measures - I know it's not simply a fraction but something like a Radon-Nikodym derivative.

Then (according to a paper I'm reading)

$\int {\mathrm{\partial}}_{t}\mathrm{log}\left(\frac{{\nu}_{t}(x)}{\nu (x)}\right)d{\nu}_{t}=\int \frac{{\mu}_{t}(x)}{{\nu}_{t}(x)}{\mathrm{\partial}}_{t}\left(\frac{{\nu}_{t}(x)}{\nu (x)}\right)d\nu =-\int {\mathrm{\partial}}_{t}\left(\frac{{\mu}_{t}(x)}{{\nu}_{t}(x)}\right)d{\nu}_{t}.$

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}(x)}{\nu (x)}\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 $\frac{d}{dx}v(x)dx$. And I don't understand the first inequality at all.

I'm also not sure what the notation $\frac{{\nu}_{t}(x)}{\nu (x)}$ is when we're talking about measures - I know it's not simply a fraction but something like a Radon-Nikodym derivative.

Answer & Explanation

Xzavier Shelton

Expert

2022-06-27Added 26 answers

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paper you are referring to, as there a probably some unchecked typos or even mistakes: since it is on arXiv, it may be not peer reviewed yet.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 ${g}_{t}(x)=\frac{\mathrm{d}{\mu}_{t}}{\mathrm{d}{\nu}_{t}}(x)$ rather that $\frac{{\mu}_{t}(x)}{{\nu}_{t}(x)}$ since here ${\nu}_{t}(x)$ 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 $\lambda =\frac{1}{2}(\mu +\nu )$. For example, it means that

$\frac{\mathrm{\partial}}{\mathrm{\partial}t}(\mathrm{log}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\nu})=\frac{\mathrm{\partial}}{\mathrm{\partial}t}(\mathrm{log}\left(\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}/\frac{\mathrm{d}\nu}{\mathrm{d}\lambda}\right))=\frac{\mathrm{\partial}}{\mathrm{\partial}t}(\mathrm{log}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}-\mathrm{log}\frac{\mathrm{d}\nu}{\mathrm{d}\lambda})=\frac{\frac{\mathrm{\partial}}{\mathrm{\partial}t}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}}{\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}}=\left(\frac{\mathrm{\partial}}{\mathrm{\partial}t}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}\right)\frac{\mathrm{d}\lambda}{\mathrm{d}{\nu}_{t}}.$

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 $f(x)$ it holds that

$\begin{array}{rl}{\int}_{X}f(x)\frac{\mathrm{\partial}}{\mathrm{\partial}t}(\mathrm{log}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\nu}(x)){\nu}_{t}(\mathrm{d}x)& ={\int}_{X}f(x)(\frac{\mathrm{\partial}}{\mathrm{\partial}t}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}(x))\frac{\mathrm{d}\lambda}{\mathrm{d}{\nu}_{t}}(x){\nu}_{t}(\mathrm{d}x)\\ & ={\int}_{X}f(x)(\frac{\mathrm{\partial}}{\mathrm{\partial}t}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}(x))\lambda (\mathrm{d}x)\\ & =\frac{\mathrm{\partial}}{\mathrm{\partial}t}({\int}_{X}f(x)\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}(x)\lambda (\mathrm{d}x))\\ & =\frac{\mathrm{\partial}}{\mathrm{\partial}t}({\int}_{X}f(x){\nu}_{t}(\mathrm{d}x)).\end{array}$

Note also that working with this you have to be accurate in something of the kind ${b}_{t}(x)\frac{\mathrm{\partial}}{\mathrm{\partial}t}{a}_{t}(x)$ which is

${b}_{t}(x){(\frac{\mathrm{\partial}}{\mathrm{\partial}s}{a}_{s}(x))|}_{s=t}$

cause otherwise you may get confused. I hope I didn't.

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 $\lambda =\frac{1}{2}(\mu +\nu )$. For example, it means that

$\frac{\mathrm{\partial}}{\mathrm{\partial}t}(\mathrm{log}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\nu})=\frac{\mathrm{\partial}}{\mathrm{\partial}t}(\mathrm{log}\left(\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}/\frac{\mathrm{d}\nu}{\mathrm{d}\lambda}\right))=\frac{\mathrm{\partial}}{\mathrm{\partial}t}(\mathrm{log}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}-\mathrm{log}\frac{\mathrm{d}\nu}{\mathrm{d}\lambda})=\frac{\frac{\mathrm{\partial}}{\mathrm{\partial}t}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}}{\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}}=\left(\frac{\mathrm{\partial}}{\mathrm{\partial}t}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}\right)\frac{\mathrm{d}\lambda}{\mathrm{d}{\nu}_{t}}.$

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 $f(x)$ it holds that

$\begin{array}{rl}{\int}_{X}f(x)\frac{\mathrm{\partial}}{\mathrm{\partial}t}(\mathrm{log}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\nu}(x)){\nu}_{t}(\mathrm{d}x)& ={\int}_{X}f(x)(\frac{\mathrm{\partial}}{\mathrm{\partial}t}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}(x))\frac{\mathrm{d}\lambda}{\mathrm{d}{\nu}_{t}}(x){\nu}_{t}(\mathrm{d}x)\\ & ={\int}_{X}f(x)(\frac{\mathrm{\partial}}{\mathrm{\partial}t}\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}(x))\lambda (\mathrm{d}x)\\ & =\frac{\mathrm{\partial}}{\mathrm{\partial}t}({\int}_{X}f(x)\frac{\mathrm{d}{\nu}_{t}}{\mathrm{d}\lambda}(x)\lambda (\mathrm{d}x))\\ & =\frac{\mathrm{\partial}}{\mathrm{\partial}t}({\int}_{X}f(x){\nu}_{t}(\mathrm{d}x)).\end{array}$

Note also that working with this you have to be accurate in something of the kind ${b}_{t}(x)\frac{\mathrm{\partial}}{\mathrm{\partial}t}{a}_{t}(x)$ which is

${b}_{t}(x){(\frac{\mathrm{\partial}}{\mathrm{\partial}s}{a}_{s}(x))|}_{s=t}$

cause otherwise you may get confused. I hope I didn't.

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