# Each diffusion SDE is associated to a *unique* family of transition kernels

I consider an SDE of the form $$dX_t=b(X_t) \, dt + \sigma(X_t) \, dW_t$$, with $$b$$ and $$\sigma$$ globally Lipschitz on $$\mathbb{R}^n$$.

How can I prove that there exists a unique family of transition kernels $$\mathscr{K}=\{k(dx_t,t\mid x_s,s) \mid s\leq t\}$$ with the property that a stochastic process $$(X_t)_{t\geq 0}$$ is a solution to the SDE if and only if it is a Markov process with transition kernels $$\mathscr{K}$$?

It should be somehow obvious because under the stated assumptions on the coefficients we have uniqueness of solutions for any fixed initial data and each solution is Markov process. However, I'm failing to see how these two things imply the statement above since, a-priori, two solutions $$(X_t)_{t\geq 0}$$ and $$(Y_t)_{t\geq 0}$$ with different initial data $$X_0\neq Y_0$$ could have different transition kernels. How do we rule this out?

EDIT: I want to sum up what I have found about this questions after doing more research. There are essentially three ways of proving uniqueness of the transition kernels.

1. The most elementary one is the proof of the Markov property (Theorem $$7.1.2$$) in Oskendal, which holds unchanged for any initial distribution on $$X_0$$ and which shows that all kernels are equals by giving an explicit expression for them.

2. Another way is to do what Figalli does in the paper mentioned by Thomas Kojar (Proposition $$4.1$$), where he proves uniqueness of solutions to the Kolmogorov forward equation. This, together with the fact that all transition kernels $$k_{\mu_0}(dx,t|y,s)$$ (relative to the solution to the SDE with $$X_0\sim \mu_0$$) solve the same Kolmogorov forward equation (which is independent of $$\mu_0$$, as shown in here), gives an alternative way to answer my question.

3. A third way is contained in the proof of Corollary 264 in these notes from CMU, which uses that the transition kernels of a Feller process are determined by its generator (which in turn follows from the Hille-Yosida theorem).

• not really, even in the case of Brownian motion i.e. $p_{t}=\frac{1}{2}\Delta p$, one has to sit down and do energy estimates to get smoothness. (if one doesn't write down the explicit solution). Commented Aug 2, 2023 at 20:27
• Then what does it mean to say that $p$ solves the PDE if it is not regular enough for the PDE to make sense (possibly not even in the weak sense)? Commented Aug 2, 2023 at 20:29