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For Ito SDE $$ dx = f(x,t)dt + g(x,t)dw$$

It transforms initial distribution $p(x_0)$ to distribution $p(x_T)$ at time $T$. My question is whether or not there exists a "reverse" sde $dx = f'(x,t)dt + g'(x,t)dw$ such that if the initial distribution is $q(x_0)=p(x_T)$, the transformed distribution at time $T$ under the sde is $q(x_T)=p(x_0)$. Under which assumption, such "reverse" sde exists and what are $f',g'$?

My naive guess is $$ dx = [f(x,T-t) + g(x,T-t)g^T(x,T-t)\nabla log q(x_T|x_0)]dt + g(x,T-t)dw $$ But I am not sure about correctness and under what assumptions such "reverse" sde exists.

Update: My previous question only requires end marginals. It should be $ q(x_{T-t})=p(x_t) $.

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  • $\begingroup$ That would mean to reverse entropy. Both processes are dissipative, both widen the distribution. $\endgroup$ Commented Mar 7, 2021 at 18:58
  • $\begingroup$ Notation is crappy. What you likely mean is it transforms an initial distribution, say $p_{0}$, to some distribution $p_{T}$ at time $T$. Now, regarding you question I suggest you dig into literaturem focussing on key word: Doob's transform, h-transform, bridges. $\endgroup$
    – Tobsn
    Commented Mar 7, 2021 at 19:39

1 Answer 1

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I think your question was motivated by the recent wave of (denoising) diffusion models in generative modelling in ML. Here are some links I found useful for the math background of the reverse time SDE.

https://ludwigwinkler.github.io/blog/ReverseTimeAnderson/

https://www.sciencedirect.com/science/article/pii/0304414982900515

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  • $\begingroup$ This is a link only answer. $\endgroup$
    – Nobody
    Commented May 11, 2022 at 7:26
  • $\begingroup$ Thanks for the link. The first blog solves my question! $\endgroup$ Commented May 11, 2022 at 15:54

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