Brownian Motion Independent Increments

Question

I want to prove the following result but I'm not sure how to go about it:

Let $X=\left\{ X_t \,\vert\,0\leq t<\infty\right\}$ be a stochastic process for which $X_0,X_{t_1}-X_{t_0},\ldots,X_{t_n}-X_{t_{n-1}}$ are independent random variables, for every integer $n\geq 1$ and indices $0=t_0 < t_1 < \cdots < t_n < \infty$. Then for any fixed $0\leq s < t <\infty$, the increment $X_t -X_s$ is independent of $\mathscr F_s^X$.

The statement means trivially that $X_t - X_s$ is independent of $\sigma(X_k)$ for all $0 \leq k \leq s$ and is therefore independent of $\bigcup\limits_{0 \leq k \leq s} \sigma(X_k)$.

However, this is not necessarily a $\pi$ system so I can't necessarily apply Dynkin's theorem to state that $\mathcal{F}_s^X \equiv \sigma\left(\bigcup\limits_{0 \leq k \leq s} \sigma(X_k)\right)$ is independent of $X_t - X_s$.

Does anyone know any generating $\pi$ systems for which $X_t-X_s$ is independent?

Answer 1

check out 'Karatzas and Shreve 98' page 38-39, solution of 1.8, where you are further lead to 'Doob 53' page 604. There you are told, that

$\mathcal{F}_s^X=\left\{ \left\{(X_{t_1}, X_{t_2}, ... )\in B\right\}\:|\: 0\leq t_1 < t_2 < ... \leq s,\: B\in \mathcal{B}(\mathbf{R})\otimes \mathcal{B}(\mathbf{R})\otimes ... \right\}$,

sometimes a handy tool when you are in a natural filtration setting. Therefore, I suppose that the countable cylindersets of the form

$\left\{ X_{t_1}\in B_1, X_{t_2}\in B_2, ...\:|\: 0\leq t_1 < t_2 < ... \leq s,\: B_n\in \mathcal{B}(\mathbf{R})\right\}$

is the generator that you are looking for.