Densities of strong solutions to SDE's

Question

I have the following problem for which I could not find appropriate references: Suppose we have $V\in C^2(\mathbb{R}^n)$ with $\mathrm{Hess}(V)>0$ (or a bit more restrictive, I could assume $\mathrm{Hess}(V)\geq R$ for some $R>0$). Then consider the following SDE ($(B_t)_{t\geq 0}$ being an $n$-dimensional Brownian motion): $$dX_t = \sqrt{2}dB_t - \nabla V(X_t) dt$$ I would like to know the following:

Does at any time $t > 0$ the distribution of a strong solution to this SDE have a density (wrt. to the Lebesgue measure)?

I know that at least in the Gaussian case the answer to the question is positive, since in this case the SDE is solved by the Ornstein-Uhlenbeck process, which is normally distributed, and strong uniqueness holds for the SDE so any two solutions have the same distribution. I am not sure about the general case though when $V$ differs from the Gaussian potential.

Thank you very much for any help!

What do you mean by "general case"? What is $(B_t)_{t \geq 0}$ (in the general case)? — saz, Jun 10 '18 at 17:43
Sorry, with the general case I mean when $V$ is not the Gaussian potential but the general function defined in the beginning (i.e. $V\in C^2$ etc.). — Corram, Jun 11 '18 at 18:58
If $V$ is smooth, Hormander's theorem guarantees smooth transition densities: see section 7 of this nice set of notes. However, I think that a shorter and more elementary use of Girsanov and Novikov can do the trick here as well, but the details may get a bit annoying, see this question and perhaps try to adapt the tricks used there. — shalin, Jun 11 '18 at 23:57
thank you shalop for the nice reference. so it definitely works at least when $V\in C^\infty$, which is very good to know. — Corram, Jun 12 '18 at 17:52

Densities of strong solutions to SDE's

0 Answers0