Does the distribution of a process on $\mathbb{R}^{[0,\infty)}$ uniquely define it?

Question

Question: Can I have two different stochastic processes $(A_t)_{t \in [0, \infty)}$, $(B_t)_{t \in [0, \infty)}$ having the same distribution on $\mathbb{R}^{[0, \infty)}$ differ in some ways? Especially, can one of them be a.s. continuous while the other isn't?

Here is where my question originated from. You don't need to read it, but you would help me if you did and maybe spotted my mental mistakes I did:

Prestory: Earlier I had some problems with understanding why two brownian motions have to have the same supremum, because as I understood it, all I knew about two different brownian motions would be, that they share some properties (e.g. the finite dimensional distributions), but for example don't need to share their overall distribution on $\mathbb{R}^{[0,\infty)}$.

Now after doing some research I found out, that from the Kolmogorov theorem it follows that the finite dimensional distributions already uniquly define the distribution of a process on $\mathbb{R}^{[0,\infty)}$ and thus obviously its supremum.

Problem: Now I read further in the script I got that from. It is a method for constructing a brownian motion. At that point we have a consistent family of measures $\mathbb{P}_J$ on $\mathbb{R}^J$, $J \subset [0, \infty)$ finite which are distributed like the finite dimensional distributions of a brownian motion. From that we can construct a uniquely defined measure $\mathbb{P}$ on $\mathbb{R}^{[0, \infty)}$ which has the finite dimensional distributions $\mathbb{P}_J$ of a brownian motion and from that construct a stoch. process with the coordinate projections $\pi_i, i \in [0, \infty)$. The last problem is, that this stoch. process isn't necessarily almost surely continuous. We solve that by showing that there exists a a.s. continuous modification.

Now to my core problem. This modification obviously again shares the same finite dimensional distributions with the original process $\pi$. Thus it also has the same distribution on $\mathbb{R}^{[0, \infty)}$. So the two processes have the exactly same distribution on $\mathbb{R}^{[0, \infty)}$ while one being a.s. continuous and one not?

This bugs me. I thought if two random variables have the same distribution we expect them to behave the same?

Answer 1

Here is a somewhat silly example. Consider the trivial probability measure on the trivial $\sigma$-algebra $\mathcal{X} = \{ \emptyset, \mathbb{R} \}$ on the reals. Then any (non-random) function $f: [0,\infty) \to \mathbb{R}$ gives rise to a stochastic process $A_t = f(t)$ with trivial finite-dimensional distributions. So there are continuous and non-continuous versions of this.

The finite-dimensional distributions determine the product measure uniquely, but that does not mean that they determine properties of sample paths. For some more on this see Stochastic Processes, Indistinguishability and Modifications.

Answer 2

The main point in (A Brownian motion $B$ that is discontinuous at an independent, uniformly distributed random variable $U(0,1)$) is that the finite- dimensional distributions do not determine the distribution of the process. In the continuous case it is true, you can recover the process via the finite-dimensional distributions, (we say that finite dimensional sets form a determining class).

The answer to your question is: No, two processes having the same distributions are identical. The interesting part is to note that two process can have the same finite dimensional distributions $P(X(t)=Y(t))=1$ for every $t$ and they still can be very different processes. The above link is the classical example of this situation.

Usually one proves that $P(X(t)=Y(t))=1$ and from there one would like to conclude that they are indistinguishable $P(X(t)=Y(t), \forall t)=1$. That is not immediate, one needs more conditions e.g. the processes are right-continuous with left limits