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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?)

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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 $x\ne0$.

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  • $\begingroup$ Many thanks for your answer, Iosif Pinelis. I accept your counterexample (and thus also the one that mike gave in his answer). I would hope though that, maybe, continuity of $(t,x)\mapsto p_t(x)$ could be achieved by imposing sufficient growth conditions on the differences $\mathbb{E}[|X_t-X_s|^\alpha]$ (akin to Kolmogorov's continuity theorem). Do you have any ideas how sufficient `uniformity' conditions of this sort might look like? $\endgroup$
    – fsp-b
    Commented Aug 26, 2020 at 14:00
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    $\begingroup$ @fsp-b : I don't think Kolmogorov-type conditions will help. If needed, you can make $X_t$ even closer to $X_0$ (for small $t$) by replacing $\sin\frac xt$ by something like $\sin(xe^{1/t})$ or $\sin(x\exp(e^{1/t}))$, with very high frequencies near $t=0$. $\endgroup$
    – Iosif Pinelis
    Commented Aug 26, 2020 at 14:20
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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

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  • $\begingroup$ Thanks for your comment, mike. I agree that sample-path continuity alone is insufficient. $\endgroup$
    – fsp-b
    Commented Aug 26, 2020 at 14:02

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