When are the transition densities of an SDE symmetric?

Question

We fix $T>0$. Let $b:[0, T] \times \mathbb{R}^d \rightarrow \mathbb{R}^d$ and $\sigma:[0, T] \times \mathbb{R}^d \rightarrow \mathcal{M}^\text{sym}_{d \times d}(\mathbb{R})$ be measurable and regular enough. Let $(W_t, t \in [0, T])$ be a $d$-dimensional Brownian motion on a probability space $(\Omega, \mathcal A, \mathbb P)$.

For any $(s, x) \in[0, T) \times \mathbb{R}^d$, we consider the SDE $$ d X^x_{s, t} = b (t, X^x_{s, t}) d t+\sigma (t, X^x_{s, t}) d W_t, \quad X^x_{s, s} = x, t \in[s, T]. $$

Let $(p_{s, t})_{0 \leq s<t \leq T}$ be the associated transition densities, i.e., $p_{s, t}(x, \cdot)$ is the density of $X_{s, t}^x$.

Are there some conditions on $b, \sigma$ such that $p_{s, t}$ is symmetric, i.e., $p_{s, t} (x, y)= p_{s, t} (y, x)$? Any reference is greatly appreciated.

Thank you so much for your elaboration!

Answer 1

This question is addressed here. In particular, a diffusion whose transition densities are symmetric is a special case of a $\nu$-symmetric diffusion, and by itself, this symmetry does not uniquely specify the diffusion. For example, the symmetry property holds for any $d$-dimensional diffusion of the form $$ d X_t = \operatorname{div}M(X_t) dt + \sqrt{2} B(X_t) dW_t $$ where $B(x)$ is a $d \times d$ matrix, $M(x) = B(x) B(x)^T$ for all $x \in \mathbb{R}^d$, and $(\operatorname{div}M)_i = \sum_{j=1}^n \partial M_{ij}/ \partial x_j$ is the divergence of a second-order tensor field. To specify the diffusion, it suffices to specify the geometry of the noise, because then the corresponding drift field is uniquely determined. When the density $\nu$ is additionally normalizable i.e. $\int_{\mathbb{R}^d} \nu(x) dx < \infty$, then a related question came up here.