Density for Translated Process

Question

Let $M$ be a (compact) Riemannian manifold. Let $v$ be a smooth vector field on $M$ with flow $\Theta_t$. Let $L$ be an elliptic second order differential operator on $M$ that generates the Ito process $X_t$.

Define the process $Y_t := \Theta_t^* X_t$. I read that the generator of $Y_t$ is the time-dependent differential operator $$L_s = \Theta_s^* L - v$$ I do not really know how to prove this though, and there is no explanation in the article I read. Is the underlying probability measure of $Y_t$ absolutely continuous w.r.t. $X_t$? If yes, what is its density?

I know the Girsanov formula, which states e.g. that the process generated by $L-v$ has the density $$\exp\left( \int_0^t v(B_s) \mathrm{d} B_s - \int_0^t |v(B_s)|^2 \mathrm{d} s\right)$$ However, in this situation, also the second-order part is altered and I don't know how to deal with this.

Answer 1

The distribution of the process $Y$ is locally equivalent to the distribution of $X$ if and only if the flow $\Theta$ preserves the principal symbol of $L$.

If $\Theta$ does not preserve the principal symbol of $L$, then the two processes $X$ and $Y$ do not have the same quadratic variation and thus their distributions can not be locally equivalent.

If $\Theta$ preserves the principal symbol of $L$, then the distributions are locally equivalent and we can use Girsanov theorem to compute the density.