# Independence of random variables involving Brownian motion

I am reading a book on stochastic analysis and I don't understand the following (i.e. don't know how to prove it rigorously):

Let $B$ be a standard Brownian motion and $\{ \mathcal{F}_t \}$ be the filtration generated by the Brownian motion. Let $s \geq 0$, $t_n > \ldots > t_1 \geq 0$ be fixed real numbers.

Then for any Borel function $f: \mathbb{R}^n \rightarrow \mathbb{R}$, $f(B_{t_1 +s} - B_s, \ldots, B_{t_n +s} - B_s)$ is independent of $\mathcal{F}_s$.

My approach:

I plan to apply the Dynkin's lemma. Let $X_i = B_{t_i +s} - B_s$. I note that \begin{equation} \sigma (f(X_1, X_2, \ldots, X_n)) \subseteq \sigma \bigg( \bigg\{ \{X_1 \leq x_1\} \cap \ldots \cap \{X_n \leq x_n\} \, \bigg|\, x_1, x_2, \ldots, x_n \in \mathbb{R} \bigg\} \bigg) \end{equation} and I know that $\bigg\{ \{X_1 \leq x_1\} \cap \ldots \cap \{X_n \leq x_n\} \, \bigg|\, x_1, x_2, \ldots, x_n \in \mathbb{R} \bigg\}$ is a $\pi-$system. However, I can't show that $$\mathbb{P} ( \{X_1 \leq x_1\} \cap \ldots \cap \{X_n \leq x_n\} \cap A) = \mathbb{P} (\{X_1 \leq x_1\} \cap \ldots \cap \{X_n \leq x_n\}) \mathbb{P} (A), \quad \forall A \in \mathcal{F}_s.$$ Any ideas?

• Do you know that $(B_{t+s})_{t \geq 0}$ and $\mathcal{F}_s$ are independent? – saz Dec 16 '14 at 16:53
• @saz I am actually reading the proof of this, but it uses this fact, so it would be a circular argument if you use this fact. – ashburn Dec 16 '14 at 16:55

Hint: Set $Y_k = B_{t_k + s} - B_{t_{k-1}+s}$ where $t_0 = 0$. Note that $X_k = Y_1 + \dots + Y_k$, so you can express $f(X_1, \dots, X_n)$ as a Borel function of $Y_1, \dots, Y_n$. Now observe that $\{\mathcal{F}_s, \sigma(Y_1), \dots, \sigma(Y_n)\}$ are mutually independent $\sigma$-fields. Use a Dynkin lemma argument to conclude that $\mathcal{F}_s$ and $\sigma(Y_1, \dots, Y_n) = \sigma(\sigma(Y_1), \dots, \sigma(Y_n))$ are independent.