# Extending Brownian motions

I'm reading through the construction of a Brownian motion on Karatzas and Shreve (Section $$2.3$$).

I'm having trouble filling a detail on the proof of corollary $$3.4$$.

Suppose we have a sequence of probability spaces $$(\Omega_n,\mathcal{F}_n,P_n)$$ for $$n\in \mathbb{N}$$ each with a Brownian motion $$X^{(n)}_t$$ for $$t\in [0,1]$$. Define $$\Omega=\Omega_1\times\Omega_2 \dots$$, $$\mathcal{F}=\mathcal{F}_1 \otimes \mathcal{F}_2 \otimes \dots$$ and $$P=P_1\times P_2\times \dots$$ and we define $$B$$ on $$\Sigma$$ as $$B_t = X_t^{(1)}$$ if $$t\in[0,1]$$ and $$B_t = B_n + X^{(n+1)}_{t-n}$$ if $$t\in [n,n+1]$$.

I'm guessing I can take the natural filtration but with this filtration it doesn't seem obvious that the increments are independent. Is there a way to relate the natural filtration of $$B$$ with the filtrations of each of the Brownian motions $$X^{(n)}$$?

Suppose you have 2 probability spaces and 2 random variables: \begin{align} (\Omega_1, \mathcal F_1, \mathbb P_1), \quad X_1:\Omega_1 \to \mathbb R\phantom{.} \\ (\Omega_2, \mathcal F_2, \mathbb P_2), \quad X_2:\Omega_2 \to \mathbb R. \\ \end{align} When you take product $$\Omega = \Omega_1 \times \Omega_2, \mathcal F= \mathcal F_1 \otimes\mathcal F_2, \mathbb P = \mathbb P_1 \times \mathbb P_2$$, then you can "extend" variables $$X_1,X_2$$ to be defined on $$(\Omega, \mathcal F, \mathbb P)$$ by setting $$\bar X_1(\omega_1, \omega_2) = X_1(\omega_1), \quad \bar X_2(\omega_1, \omega_2) = X_2(\omega_2).$$ This operation gives two independent rvs $$\bar X_1, \bar X_2$$ such that $$\bar X_i \overset D = X_i$$. Similarly for infinite products. The other fact we need is that if $$X_1, Y_1:\Omega_1 \to \mathbb R$$ are independent and $$X_2, Y_2:\Omega_2 \to \mathbb R$$ are independent, then $$\bar X_1, \bar X_2, \bar Y_1, \bar Y_2 :\Omega \to \mathbb R$$ are independent as well. More generally for any number of variables. Both facts can be proven directly by expanding definitions, although general cases are tedious to write.
Now consider Brownian motions $$X_t^{(n)}$$ defined for $$t \in [0, 1]$$ on different probability spaces. You can apply above construction to get independent Brownian motions $$\bar X_t^{(n)}$$ defined on the same probability space. Set $$B_t = \bar X^{(1)}_t$$ for $$t\in [0, 1]$$ and $$B_t = B_n + \bar X^{(n+1)}_{t-n}$$ for $$t \in [n, n+1]$$.
Suppose $$0 \leq t_1 < t_2 < \ldots < t_n$$. Add to this sequence all natural numbers in $$[0, t_n]$$, getting $$0 = u_1 < u_2 < \ldots < u_m, t_i = u_{k_i}$$. Now for any $$k < m$$, $$u_k, u_{k+1} \in [l_k, l_k + 1]$$ for some natural $$l_k$$. By previous facts, $$B_{u_{k+1}} - B_{u_k} = \bar X^{(l_k+1)}_{u_{k+1} - l_k} - \bar X^{(l_k+1)}_{u_{k} - l_k}$$ is family of independent variables, and therefore \begin{align} B_{t_{i+1}} - B_{t_i} &= B_{u_{k_{i+1}}} - B_{u_{k_i}} \\ &= (B_{u_{k_{i+1}}} - B_{u_{k_{i+1} - 1}}) + (B_{u_{k_{i+1} - 1}} - B_{u_{k_{i+1} - 2}}) + \ldots + (B_{u_{k_i+1}}-B_{u_{k_i}}) \end{align} are independent as well.
I don't see how using filtrations would help seeing independence in any case, but you could write $$\sigma(B_s: s \leq t) = \sigma(\sigma(\bar X^{(1)}_s: s \leq 1), \ldots \sigma(\bar X^{(n)}_s: s \leq 1), \sigma(\bar X^{(n+1)}_s: s \leq t - n))$$ for $$t \in [n, n+1]$$.