Let $X=(X\_t)\_{t\in I}$ ($I\subset\mathbb{R}$ an interval) be a stochastic process with continuous sample paths and such that $X\_t$ admits a continuous Lebesgue density $\chi\_t\in C(\mathbb{R}^d)$ for each $t\in I$.


Are you aware of (minimal) conditions on $X$ which guarantee that the function $(t,x)\mapsto \chi\_t(x)$ is continuous on $I\times\mathbb{R}^d$?


(Could it be that the sample-continuity of $X$ already suffices?)



Continuity of the densities of a stochastic process
The truly minimal condition on $X$ that guarantees that the function $(t,x)\mapsto p\_t(x):=\chi\_t(x)$ is continuous is tautological: $p\_t(x)$ is continuous in $(t,x)$ if and only $p\_t(x)$ is continuous in $(t,x)$. As far as the minimality is concerned, I don't think you can do much better than this.


However, one can rather easily see that the sample continuity of $X$ is not enough even for the continuity of $p\_t(x)$ in $t$ (for fixed $x$). E.g., let
$$p\_t(x):=(1+\sin\tfrac xt)f(x)$$
for real $x$ and real $t\ne0$, with $p\_0:=f$, where $f$ is the standard normal pdf. Then $p\_t$ is a continuous pdf for each $t$ and, by the [Riemann–Lebesgue\_lemma](https://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma),
$$F\_t(x):=\int\_{-\infty}^x p\_t(u)\,du$$
is continuous in real $t$ for each real $x$. Moreover, $F\_0$ is continuous and strictly increasing (in fact, $F\_t$ is so for each real $t$). Hence, the process $(X\_t)$ defined by the formula
$$X\_t:=F\_t^{-1}(U),$$
where $U$ is a random variable uniformly distributed on the interval $(0,1)$, has continuous paths. Also, $p\_t$ is the pdf of $X\_t$, for each $t$. However, $p\_t(x)$ is not continuous in $t$ for any real $x\ne0$.


