Let μ t be the law of a random process X t . Let ν...

Mohammad Cannon

Mohammad Cannon

Answered

2022-06-26

Let μ t be the law of a random process X t . Let ν and ν t be arbitrary measures in Ω with the sole restriction that μ t be absolutely continuous w.r.t. ν t , and ν t be absolutely continuous w.r.t. ν.
Then (according to a paper I'm reading)
t log ( ν t ( x ) ν ( x ) ) d ν t = μ t ( x ) ν t ( x ) t ( ν t ( x ) ν ( x ) ) d ν = t ( μ t ( x ) ν t ( x ) ) d ν t .
The second inequality looks like an integration by parts, but I don't know what it means to take " d v" = t ( ν t ( x ) ν ( x ) ) d ν where the " d v" derivative is with respect to a different variable (t) than the integral is taken over (\nu) -- usually " d v" is something like d d x v ( x ) d x. And I don't understand the first inequality at all.
I'm also not sure what the notation ν t ( x ) ν ( 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

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paper you are referring to, as there a probably some unchecked typos or even mistakes: since it is on arXiv, it may be not peer reviewed yet.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 ) = d μ t d ν t ( x ) rather that μ t ( x ) ν t ( x ) since here ν 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 μ , ν λ then
(1) d μ d ν = d μ d λ / d ν d λ .
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 λ, e.g. take λ = 1 2 ( μ + ν ). For example, it means that
t ( log d ν t d ν ) = t ( log ( d ν t d λ / d ν d λ ) ) = t ( log d ν t d λ log d ν d λ ) = t d ν t d λ d ν t d λ = ( t d ν t d λ ) d λ d ν t .
Here we used two facts. First of all, λ should be λ t in general, but since all ν t are said to be dominated by the same measure, we were able to pick a single t-independent dominating measure λ. Moreover,
1 d ν t d λ = d λ d ν t .
Some of these equalities are guaranteed to hold only a.s. but again, everything here is dominated by ν, so that's not a problem. As a result, for every function f ( x ) it holds that
X f ( x ) t ( log d ν t d ν ( x ) ) ν t ( d x ) = X f ( x ) ( t d ν t d λ ( x ) ) d λ d ν t ( x ) ν t ( d x ) = X f ( x ) ( t d ν t d λ ( x ) ) λ ( d x ) = t ( X f ( x ) d ν t d λ ( x ) λ ( d x ) ) = t ( X f ( x ) ν t ( d x ) ) .
Note also that working with this you have to be accurate in something of the kind b t ( x ) t a t ( x ) which is
b t ( x ) ( s a s ( x ) ) | s = t
cause otherwise you may get confused. I hope I didn't.

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