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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 $\Bbb P^2$?


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.

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  • $\begingroup$ By Kolmogorov extension theorem (see here, here, here or theorem 2.2 of chapter 2 (p. 50) of Karatzas/Shreve). $\endgroup$
    – Sayantan
    Commented Apr 20, 2019 at 3:52
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    $\begingroup$ The question does not make sense unless you define the term distribution of a stochastic process''. One can talk about the distribution if BM as a measure on $C[0,1]$ or on $\mathbb R ^{[0,1]}$ with the product topology. $\endgroup$ Commented Apr 20, 2019 at 5:50

3 Answers 3

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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 $\mathbb{R}^{\infty}$.)

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  • $\begingroup$ You have not mentioned $\mathbb{R}^{[0,\infty)}$. Is there a reason for this? Do you disagree with the proof given by Yuval? Which the counterexample in the most recent answer would support. $\endgroup$
    – feltshire
    Commented Nov 24, 2022 at 16:22
  • $\begingroup$ @feltshire Yuval's proof is correct, it's just that the product $\sigma$-algebra for $\mathbb{R}^{[0,\infty)}$ does not have a special name. Michael is just saying that the product $\sigma$-algebra for $C[0,1]$ is the same as the Borel $\sigma$-algebra generated by the uniform norm, etc. $\endgroup$
    – suncup224
    Commented Jun 6 at 14:13
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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 ${\mathcal C}$.

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  • $\begingroup$ This is apparently contradicted by the short counterexample given in another answer. Would you mind addressing this? $\endgroup$
    – feltshire
    Commented Nov 24, 2022 at 16:14
  • $\begingroup$ @feltshire the "counterexample" is not a counterexample if we use the product $\sigma$-algebra. $X_t^1$ and $X_t^2$ are in fact same in law under the product $\sigma$-algebra $\endgroup$
    – suncup224
    Commented Jun 6 at 14:11
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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.

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  • $\begingroup$ Could you elaborate on what exactly the "laws" here are, in particular which $\sigma$-algebra we are considering on $\{0,1\}^{[0,\infty)}$? Because, if we take the product-$\sigma$-algebra then as far as I understand by the other answer's reasoning the laws should coincide on it as it would be generated by the cylinder sets. So presumably we'd need a larger $\sigma$-algebra, but which one can we use without possibly running into any measurability issues? $\endgroup$
    – hgmath
    Commented Oct 11, 2023 at 20:56

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