Do finite dimensional distributions determine the law of a stochastic process?

Question

For $i=1,2$, let $\{X_t^i\}_{t\geq 0}$ be $\Bbb R^d$-valued stochastic processes adapted to $\{\mathscr F^i_t\}_{t\geq 0}$ on the probability space $(\Omega^i,\mathscr F^i,\Bbb P^i)$. Suppose the two processes have the same finite dimensional distributions, i.e. for any $n\in\Bbb N$ and $t_1<\cdots<t_n$ and $A\in \mathscr B(\Bbb R^{dn})$ we have $$\Bbb P^1[(X^1_{t_1},\cdots,X^1_{t_n})\in A]=\Bbb P^2[(X^2_{t_1},\cdots,X^2_{t_n})\in A]$$ Is it true that $\{X_t^1\}_{t\geq 0}$ under $\Bbb P^1$ has the same law as $\{X_t^2\}_{t\geq 0}$ under $\Bbb P^2$?

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In the context where the $X^i$ are solutions to a SDE, the book Brownian Motion and Stochastic Calculus (Chapter 5, Proposition 3.10) proves the equivalence in law by showing that the finite dimensional distributions are the same. However, it is not clear to me how the equivalence in law follows.

Answer 1

Yes, for both discrete-time and continuous-time processes.

For a discrete time process (time series), its law is a probability measure on $\mathbb{R}^{\infty}$, with the Borel $\sigma$-algebra generated by the product topology.

For a continuous-time process with continuous sample-paths, its law is a probability measure on $C[0,1]$, with the Borel $\sigma$-algebra generated by the uniform norm.

For a continuous-time process with cadlag sample-paths, its law is a probability measure on $D[0,1]$, with the Borel $\sigma$-algebra generated by the Skorohod tpology.

In all three cases, the finite dimensional cylinder sets form a determining class, i.e. two measures are the same is they agree on the finite dimensional cylinder sets.

(However, only in the discrete-time case do the the finite dimensional cylinder sets form a convergence determining class on $\mathbb{R}^{\infty}$.)

Answer 2

I understand the question is that is two process have the same "finite dimensional distribution" then they are the same. Notice a stochastic process induces a measure on ${\mathbb R}^{[0, \infty)}$, the space of all maps $[0, \infty)\to {\mathbb R}$, thus we just need to show $P_1=P_2$, where $P_i$ is the measure on ${\mathbb R}^{[0, \infty)}$ induced by $X^i$, where $i=1, 2$.

This can be done by "Dynkin $\pi$-$\lambda$ theorem", see p49 of the book you mentioned, or see https://en.wikipedia.org/wiki/Dynkin_system

Briefly, consider the collection ${\mathcal F}$ of subsets $A\in{\mathbb R}^{[0, \infty)}$ so that $P_1(A)=P_2(A)$. Now ${\mathcal F}$ contains the collection ${\mathcal C}$ of all "cylinder sets", i.e. those of the form $\{X\,|\,(X_{t_1}, ...,X_{t_n})\in B\}$, where $B$ is Borel in ${\mathbb R}^n$. The collection ${\mathcal C}$ forms a "$\pi$-system", i.e. if $A, B\in {\mathcal C}$ then $A\cap B\in {\mathcal C}$. Then notice ${\mathcal F}$ is a "Dynkin system", i.e. if $A_1,A_2, ...$ is a disjoint sequence in ${\mathcal F}$, then $\cup_i A_i\in{\mathcal F}$, and if $A\in {\mathcal F}$ then ${\mathbb R}^{[0, \infty)}-A\in {\mathcal F}$. Now apply the Dynkin π-λ theorem, we see ${\mathcal F}$ contains the $\sigma$-algebra generated by ${\mathcal C}$, and this says $P_1=P_2$ since the canonical $\sigma$-algebra on ${\mathbb R}^{[0, \infty)}$ is just the one generated by ${\mathcal C}$.

Answer 3

Here's an example showing why some condition such as cadlag is needed. Let $U$ be a random variable distributed uniformly on the interval $[0,1]$. Consider the $\{0,1\}$-valued processes $X^1_t = 0$ and $X^2_t = \mathbf{1}(t-U \in \mathbb{Q})$, where $\mathbb{Q}$ is the set of rational numbers. These processes have the same finite-dimensional distributions but they are not the same in law.