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., $p_{s, t} (x, y)= p_{s, t} (y, x)$?

Thank you so much for your elaboration!

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!