Counterexample: Convergence in finite dimensional distributions does not imply weak convergence

Question

I'm working at the following exercise: Give an example of a sequence of stochastic processes $(\mathbb{X}^n)_{n\geq 1}$ such that the finite dimensional distributions converge to the finite dimensional distributions of a process $\mathbb{Y}$ but $\mathbb{X}^n$ does not converge to $\mathbb{Y}$ in distribution. By a theorem we discussed, the sequence can't be tight to achieve this. So I thought about something like $X^n(t) = \delta_{\frac{1}{n}}(t)$. But since this is deterministic I am confused about calculating the finite dimensional distribution, e.g. $P(X^n_{t_1}\leq x_1,\dots,X^n_{t_p}\leq x_p)$ because it seems to me that this depends on the choice of $t_1,\dots,t_n$ but maybe this example just doesn't work. Maybe someone can help me? Cheers, Alex

Answer 1

A classical example (which can be found in Billingsley's book (1968) for example) is the following: we work on $C[0,1]$, the space of continuous functions on the unit interval, and we consider a sequence $(x_n)_{n\geqslant 1}$ of elements of $C[0,1]$ such that $x_n(t)\to 0$ for each $t\in [0,1]$ but the convergence is not uniform. Considering a process $(\mathbb X^n)_{n\geqslant 1}$ such that the distribution of $\mathbf X^n$ is $\delta_{x_n}$, we get a process whose finite dimensional distributions go to $0$ but the process does not converge to $0$.