transition probability density of Follmer process

85 Views Asked by At

If Follmer diffusion process $\{X_t\}_{t \in [0,1]}$ satisfies the following SDE $$dX_t= dB_t+\nabla \log P_{1-t}f(X_t)dt, \quad X_0=0$$ where $f=\frac{d\mu}{d\gamma}, \gamma(dx)=(2\pi)^{-\frac{n}{2}}e^{-\frac{|x|^2}{2}}dx$ and $B_t$ is a n-dimensional Brownian motion satisfied $B_0=0$. If we denote heat semigroup $P_t$ given by $$P_th(x)=\mathbb E[h(x+B_t)]$$ for any bounded continuous function $h$.

My Question: How to prove that the diffusion process $\{X_t\}_{t\in [0,1]}$ satisfies the transition probability density $$p_{s,t}(x,y)=\tilde p_{s,t}(x,y)\frac{P_{1-t}f(y)}{P_{1-s}f(x)},$$ where $$\tilde p_{s,t}(x,y)=(2\pi(t-s))^{-\frac n 2} \exp\left(-\frac{|x-y|^2}{2(t-s)} \right).$$