For stochastic process $X_t$ with marginals $\mu_t$, is it true that the sample-path continuity of $X_t$ implies $\mu_t$ is weakly continuous in $t$?

Question

I need to prove or disprove that for a stochastic process $(X_t)_{t \in [0,1]}$ with marginals $(\mu_t)_{t \in [0,1]}$ on $\mathbb{R}$, if the sample paths of $(X_t)_{t \in [0,1]}$ are continuous, then $(\mu_t)_{t \in [0,1]}$ is weakly continuous in $t$. I see a similar question Continuity of the densities of a stochastic process, but it's for the discontinuity of densities of $(X_t)_{t \in [0,1]}$. Any help is greatly appreciated.

Answer 1

Yes, under mild assumptions.

If the state space $E$ is Polish (including $E = \mathbb{R}^n$ in particular), then the space $\mathcal{P}(E)$ of Borel probability measures on $E$, with the weak topology, is metrizable, and so it suffices to show weak sequential continuity. That is, for every sequence of times $t_n \to t$, that we have $\mu_{t_n} \to \mu_t$ weakly.

Let $f : E \to \mathbb{R}$ be continuous and bounded. We must show $\int_E f\,d\mu_{t_n} \to \int_E f\,d\mu_t$, which is to say that $\mathbb{E}[f(X_{t_n})] \to \mathbb{E}[f(X_t)]$. Now we have $X_{t_n} \to X_t$ a.s., by sample path continuity. Hence $f(X_{t_n}) \to f(X_t)$ a.s., because $f$ is continuous. Hence $\mathbb{E}[f(X_{t_n})] \to \mathbb{E}[f(X_t)]$ by dominated convergence, because $f$ is bounded.