# 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?)

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 $$x\ne0$$.
• 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? Aug 26 '20 at 14:00
• @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 $t=0$. Aug 26 '20 at 14:20
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