Q: Can a continuous stochastic process admit a Lebesgue density at some time $t_0$ and no Lebesgue density as some time $t_1$? Let $(X_t)_{t\geq 0}$ be a continuous real-valued stochastic process that is not a linear combination of random variables (so processes of the form $(1-t)Y + t Z$ are excluded for example, we want only processes that remain stochastic conditionally to any countable number of $X_{t_i}$). Given $t_0,t_1 \in \mathbb R$, is it possible that $X_{t_0}$ admits a density w.r.t the Lebesgue measure, but $X_{t_1}$ does not?

A: Yes, let $\left(S_t\right)_{t\geq 0}$ be a geometric brownian motion that solves
$$dS_t = \mu dt + \sigma dW_t \qquad S_0 = x > 0$$
such that $\sigma > 0$ and $\mu < -\sigma^2/2$. Then $S_t \to 0$ almost surely as $t$ goes to infinity. Furthermore, $S_t$ has a density for all $t \geq 0$. See Geometric Brownian Motion on Wikipedia.
Define the real-valued stochastic process $\left(X_t\right)_{t\geq 0}$ as
$$ X_t = S_{\frac{t}{1-t}}, \;\; t< 1; \qquad X_t = 0, \;\; t \geq 1.$$
$X$ is clearly continuous for $t > 1$ and $t < 1$—we take the continuous version of $S$. Continuity at $t = 1$ follows from the fact that $S$ converges almost surely to zero. Furthermore, $X_t$ has a density with respect to the Lebesgue measure for $t \in (0,1)$ due to the fact that $S$ has a density for all time but the law of $X_t$ for $t \geq 1$ is a Dirac mass at zero.
