Imagine I have a sequence of two dimensional continuous time processes $X_t^n =(Y_t^n ,Z_t^n)$ and I know $$Y_t^n \to Y_t^\infty$$ and $$Z_t^n \to Z_t^\infty$$ both in distribution. I would like to know if $$(Y_t^n, Z_t^n) \to (Y_t^\infty, Z_t^\infty)$$ jointly in distribution as well.
What conditions does one need to ensure this? Obs: they are not independent! Does it help if they are martingales and/or semimartingales?
Thanks a lot!
This doesn't hold; Note that your parameter $t$ doesn't even enter into your conditions or the final statement. In the most simple case you can set $X_t^n = X_0^n$ for some sequence of random vaiables $X_0^n$. Then your question asks under what conditions the weak convergence of the marginals implies joint weak convergence. However, there is pretty much no answer to this other than "they converge jointly iff they converge jointly".
Consider a sequence of two-dimensional random variables $(Y^n,Z^n)$ like those here, such that $Y^n\xrightarrow[n\to\infty]{D}Y^\infty$ and $Z^n\xrightarrow[n\to\infty]{D}Z^\infty$, but $(Y^n,Z^n)$ does not converge to $(Y^\infty,Z^\infty)$.
Now, for any random variable $Y$ with finite mean you can construct a martingale by setting $Y_t \equiv Y$ for all $t$. Extending the example above to the martingale setting like this, you find an example of a sequence of martingales converging "marginally", but not jointly.
As for sufficient conditions, the independence of $Y^n$ and $Z^n$, or the convergence of one of the sequences to a constant, would be enough to conclude the joint convergence, cf Billingsley: Convergence of Probability Measures, Section 3. Though, I don't think that the martingale structure helps much here.
