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Let $(X_t)_{t\geq0}$ and $(Y_t)_{t \geq0}$ denote two continuous stochastic processes on a probability space $(\Omega, \mathcal{F}, P)$ and let $C$ denote space of continuous functions from $[0,\infty)$ to $\mathbb{R}$.

Let $\pi_x\colon C \rightarrow \mathbb{R}$ given by $\pi_x(f)=f(x)$ and equip $C$ with the sigma algebra $\mathcal{E} = \sigma(\{\pi_x : x\geq 0\})$.

Consider the now the $\mathcal{F}$-$\mathcal{E} $-measurable mappings $\mathbb{X}, \mathbb{Y} \colon \Omega \rightarrow C$ given by $$ \omega \mapsto X_{(\cdot)}(\omega) \quad \text{and} \quad \omega \mapsto Y_{(\cdot)}(\omega). $$ I want to prove that if $(X_t)_{t\geq0}$ and $(Y_t)_{t \geq0}$ have the same marginal distributions, then $P_\mathbb{X}=P_\mathbb{Y}$.


My progress: Supposing that $(X_t)_{t\geq0}$ and $(Y_t)_{t \geq0}$ have the same marginal distributions, then for $t_1,\dots, t_n, A_1,\dots,A_n$ $$ P(X_{t_1}\in A_{1},\dots, X_{t_n} \in A_{n})=P(Y_{t_1}\in A_{1},\dots, Y_{t_n} \in A_{n}) $$ by definition. So working from this, we get that $$ \bigcap_{k=1}^n \{X_{t_k} \in A_k\} = \bigcap_{k=1}^n \{\pi_{t_k} \circ\mathbb{X} \in A_k\} = \bigcap_{k=1}^n \{\mathbb{X} \in \pi_{t_k}^{-1}(A_k)\} = \{\mathbb{X} \in \bigcap_{k=1}^n \pi_{t_k}^{-1}(A_k) \} $$ but this is where I become stuck. We want to show $P(\mathbb{X} \in B) =P(\mathbb{Y} \in B) $ where $B \in \mathcal{B}$, an intersection stable generator set of $\mathcal{E}$. So the question is if $$ \{\bigcap_{k=1}^n \pi_{t_k}^{-1}(A_k) \colon n \in \mathbb{N}, t_1,\dots,t_n \in \mathbb{R}, A_1,\dots, A_n \in \mathcal{B}(\mathbb{R}) \} $$ is an intersection stable generating set of $\mathcal{E}$. Can anyone help me? Am I on the right track? Also my textbook claims this is only true if the processes are continuous but I can't seem to find anywhere where I would use this continuity...

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  • $\begingroup$ Since $π_x\not\in C$, why is $\mathscr E=σ(\{π_x\mid x\geqslant0\})$ a σ-algebra on $C$? $\endgroup$ Dec 30, 2021 at 2:46
  • $\begingroup$ $\mathcal{E} $ is the smallest $\sigma$-algebra that makes the $\pi_x$'s measurable. Hence it is a collection of sets of continuous functions. $\endgroup$
    – Jacobiman
    Dec 30, 2021 at 10:19
  • $\begingroup$ That's a very standard result you can check George Lowther's blog "Almost sure" the question is adressed for example in lemma 1 here unless mistaken : almostsuremath.com/2009/11/03/… $\endgroup$
    – TheBridge
    Dec 30, 2021 at 14:52

1 Answer 1

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From the hypothesis and Dynkin's $\pi-\lambda$ theorem [1], you can deduce that the two processes have the same distributions when restricted to rational times. Then continuity gives the desired conclusion.

Without continuity, the following is a well known counterexample: Let $\{X_t\}$ be standard Brownian motion, let $U$ be uniformly distributed in $[0,1]$ and let $\{Y_t\}$ be obtained from $\{X_t\}$ by setting $Y_U=X_U+1$ and $Y_t=X_t$ for all $t \ne U$.

[1] https://en.wikipedia.org/wiki/Dynkin_system

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