Continuity of the densities of a stochastic process 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?)
 A: I do not think sample path continuity suffices.  Here is my alleged counterexample.  The densities are 1 + .5*sin(x/(1-t)), 0 < t < 1 .  As t -> 1 this converges to the uniform by Riemann-Lebesgue, but, of  course, it isn't continuous on [0,1]x[0,1].  To get a stochastic process whose densities these are, let F_t be the cumulant and simple take $X_t(x)  = F_t^{-1}(x)$.   I think the $F_t$'s are continuous enough so that those paths are continuous.$$$$
Two remarks:  1.  your densities have to be weakly continuous in t by the path continuity (a.e convergence implies convergence in distribution,and 2, if it bothers you that I have the discontinuity at an endpoint (t=1), just freeze the process to extend past t=1
A: The truly minimal condition on $X$ that guarantees that the function $(t,x)\mapsto p_t(x):=\chi_t(x)$ is continuous is tautological: $p_t(x)$ is continuous in $(t,x)$ if and only $p_t(x)$ is continuous in $(t,x)$. As far as the minimality is concerned, I don't think you can do much better than this.
However, one can rather easily see that the sample continuity of $X$ is not enough even for the continuity of $p_t(x)$ in $t$ (for fixed $x$). E.g., let
$$p_t(x):=(1+\sin\tfrac xt)f(x)$$
for real $x$ and real $t\ne0$, with $p_0:=f$, where $f$ is the standard normal pdf. Then $p_t$ is a continuous pdf for each $t$ and, by the Riemann–Lebesgue_lemma,
$$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 $x\ne0$.
