Change of measure formula for the Föllmer process

Question

While reading a preprint Eldan, Lehec, and Shenfeld - Stability of the logarithmic Sobolev inequality via the Föllmer Process I came across the following SDE in Section 3: $$d X_t=d B_t+\nabla \log P_{1-t} f\left(X_t\right) d t$$ where $B_t$ is the Brownian motion and $P_t$ denotes the heat semigroup.

Given a probability measure $\mu(dx)=f(x)\gamma(dx)$ where $\gamma$ is the gaussian density, one can consider $X_t$ to be the process that is close to a Brownian motion while having the law $\mu$ at time $t=1$.

I would like to understand why the following change of measure formula holds for every test function $u$,

$$\mathbb{E}\left[u\left(X_1\right) \mid \mathcal{F}_t\right]=\frac{P_{1-t}(u f)\left(X_t\right)}{P_{1-t} f\left(X_t\right)}$$

where $\mathcal{F}_t$ denotes the natural filtration associated with the process $X_t$. I looked into Girsanov's theorem in this regard but I don't see any exponential martingales in the above expression, so I am not sure what is going on here.

Answer 1

One can show it using the Feynman-Kac formula. In particular, from Feynman-Kac, we know that $h(x, t) := \mathbb{E}[u(X_1) \mid \mathcal{F}_t ]$ solves the following PDE:

$ \frac{\partial}{\partial t} h(x, t) + \nabla \log P_{1-t} f(X_t)^T \nabla_x h(x, t) + \frac{1}{2} \Delta_x h(x, t) = 0.$

with the terminal condition $h(x, 1) = u(x)$.

Now, note that your proposed function for $h(x, t)$, namely $\frac{P_{1-t} (uf)}{P_{1-t}(f) }$ satisfies the terminal condition, so you just need to plug your proposed function to the above equation to show that it satisfies the PDE above.