Let $T = [0, \infty)$. We use $\mathcal{B}_1$ to denote the Borel algebra on $\mathbb{R}^1$. Let $\{ X_t \}_{t \in T}$ and $\{ Y_t \}_{t \in T}$ be real-valued continuous stochastic processes. Suppose that $\{ X_t \}_{t \in T}$ and $\{ Y_t \}_{t \in T}$ are equivalent in law, i.e.,
\[
P(X_{t_1} \in E_1 \land \dots \land X_{t_n} \in E_n) = P(Y_{t_1} \in E_1 \land \dots \land Y_{t_n} \in E_n)
\]
for $n \in \mathbb{N}$, $t_1, \dots, t_n \in T$ and $E_1, \dots, E_n \in \mathcal{B}_1$. Let $\pi_t \colon \mathbb{R}^T \to \mathbb{R}$ be a projection defined by $\pi_t(f) = f(t)$. Let
\[
\mathcal{B} _K (\mathbb{R}^T) = \bigvee _{t \in T} \pi _t^{-1} (\mathcal{B} _1).
\]
We want to prove that
\[
P(X \in E) = P(Y \in E)
\]
for $E \in \mathcal{B} _K (\mathbb{R}^T)$.
I can't see how to prove this proposition. Could you tell me a proof?
The same question is found here, but they are discussing measurableness.
Actually, I am not sure this proposition is true or not. I am thinking of continuous Gaussian processes and I feel that it would be great if this proposition follows.
Best Answer
The equation holds when $E$ is a cylinder set, i.e., a set of the form $\{(x_{t_1},x_{t_2},...,x_{t_n}) \in A\}$ where $A$ is a Borel set in $\mathbb R^{n}$. By the $\pi-\lambda$ theorem it holds for all $E$ is the $\sigma-$algebra generated by cylinder sets, but this is nothing but $\mathcal B_K (\mathbb R^{T})$