# Proving that if $(X_t)_{t\geq0}$ and $(Y_t)_{t \geq0}$ are continuous and have the same marginal distributions, then $P_\mathbb{X}=P_\mathbb{Y}$.

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...

• Since $π_x\not\in C$, why is $\mathscr E=σ(\{π_x\mid x\geqslant0\})$ a σ-algebra on $C$? Dec 30, 2021 at 2:46
• $\mathcal{E}$ is the smallest $\sigma$-algebra that makes the $\pi_x$'s measurable. Hence it is a collection of sets of continuous functions. Dec 30, 2021 at 10:19
• 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/… Dec 30, 2021 at 14:52

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 $$t \ne U$$.