On page 2 in Karatzas and Shreve: Brownian Motion and Stochastic Calculus it is said that a stochastic process $Y$ is a modification of $X$ if for all $t$:
$P(X_t=Y_t)=1$.
If both are stochastic processes into $\mathbb{R}^d$ it is said that they have the same finite dimensional distribution if for every $n$ and every $A \in \mathcal{B}(\mathbb{R}^{nd})$, and any real numbers $0 \le t_1<t_2< \ldots<t_n <\infty$ we have
$P((x_{t_1},x_{t_2},\ldots,x_{t_n})\in A)=P((y_{t_1},y_{t_2},\ldots,y_{t_n})\in A)$
Is it trivial that the first condition gives the second condition? How do we prove it? I think that in order to show this, you are not supposed to used the facts learned in elementary probability and statistics classes, but use product measures of some kind? Does it in some way follow from the theory that concerns the extensions of measures and product measures?
Best Answer
If $(y_t)$ is a modification of $(x_t),$ then the right hand side of (1) equals zero: \begin{eqnarray*}\Bigl|\mathbb{P}((x_{t_1},\ldots,x_{t_n})\in A)-\mathbb{P}((y_{t_1},\ldots,y_{t_n})\in A)\Bigr|&\leq& \mathbb{P}\Bigl((x_{t_1},\dots,x_{t_n})\neq (y_{t_1},\dots,y_{t_n})\Bigr)\\ &\leq&\sum_{i=1}^n \mathbb{P}(x_{t_i}\neq y_{t_i}).\tag1 \end{eqnarray*}