Is deterministic evolution preserved under weak converge of stochastic processes?

Question

Suppose you have a sequence of continuous stochastic processes $X_N$ with $X_N(0)=0$, and that $X_N$ converge weakly on the space of continuous functions, to a stochastic process $X$. Suppose $X_N$ are so that for all $N$ you sample a random variable $Z$ and then you run the process. Furthermore, for all time $t>0$ you know exactly the value of $Z$. That is, the natural filtration $\mathcal F_t$ is trivial for $t=0$ and constant for $t>0$, and $Z$ is measurable with respect to $\mathcal F_t$.

For an example, consider $X_N(t)=Zf_N(t)$ or $X_N(t)=f_t^N(Z)$ where for all $t>0$, $f_t^N$ is invertible over the range of $Z$.

Then is it true that $X$ is also like this? Meaning that the natural filtration is trivial for $t=0$ and $X(t)$ is measurable with respect to $\mathcal F_s$ for $t\geq s$?

Answer 1

Following the discussion in the comments, the statement is not true, even if we stipulate that we only want $X$ to have the desired properties with respect to its own natural filtration.

To see this, let $X_t$ be the process that is identically $0$ up to time $1$ and from then on is equal to either $t - 1$ or $-t + 1$ with probability $\frac{1}{2}$ each. $X$ does not have the property desired since, denoting by $\mathcal X$ the natural filtration of $X$, $\mathcal X_{3/2} \setminus \mathcal X_{1/2}$ is nontrivial, say.

However, now let $Z$ be the random variable that is equal to either $1$ or $-1$ with equal probability, and is independent of $X$. Let $f(t) = \mathbb 1_{t \geq 1} (t - 1)$, and take $X^N := Zf$ for all $N$. We then have weak convergence of $X^N$ to $X$ since indeed their laws on $C[0, T]$ are the same.

Comment: The counterexample given is quite degenerate, in the sense that there is only one “jump” of information in the filtration of $X$, namely between $\mathcal X_1$ and $\mathcal X_{1+}$. However, counterexamples with more complicated filtrations may be constructed without an issue.