On the independence and stationarity of supremum of increments of a Brownian Motion In between an exercise I did, I have the following affirmation that is meant to be used without proof: given a standard Brownian Motion $B$, the rvs defined as $Z_n=\sup_{0\leq u\leq 1}|B_{u+n}-B_n|$ are i.i.d. This is intuitively clear but I would like to prove it and I don't know how to properly do it.
As for the independence I think I should prove first that $\sigma\left(\left|B_{u+n}-B_n\right| :{0\leq u\leq 1}\right)$ is independent of $\sigma(B_t:t\leq n)$ and from there just use that $Z_n$ is measurable with respect to the first sigma algebra. But I am not sure how to do this either. 
As for the "identically distributed" part I believe there should be a charaterization for the distribution of the supremum of rvs in terms of the distributions of the rvs (all of them in this case would have the so called folded distribution, i.e., the distribution of the absolute value of a normal). But again, I'm not sure how to formalize this.
Any ideas? This comes from a course on stochastic calculus with rigurous foundations on measure theory.
Thank you!
 A: You can notice that the $\sigma$-algebra generated by $Z_n$ is (up to sets of measure $0$) generated the union of 
$\mathcal F^{(n)}_j:=\sigma\left( \left|B_{n+k2^{-j}}-B_n\right|,1\leqslant k\leqslant 2^j\right)$, $j\geqslant 1$. This follows from the continuity of paths of Brownian motion. 
Now, using a monotone class argument, it suffices to prove that for a fixed $j\geqslant 1$, the sequence $\left(\mathcal F^{(n)}_j\right)_{n\geqslant 1}$ is independent. This is a consequence of the following two facts:


*

*$\mathcal F^{(n)}_j=\sigma\left( \left|B_{n+(k+1)2^{-j}}-B_{nk2^{-j}}\right|,0\leqslant k\leqslant 2^j-1\right)$ and 

*the independence of the increments of Brownian motion.


For the identical distribution, note that for each $t$, 
$$\mathbb P\left\{\sup_{0\leqslant u\leqslant 1}\left|B_{u+n}-B_n\right|\leqslant t\right\}=\lim_{j\to +\infty}\mathbb P\left\{\max_{1\leqslant k\leqslant 2^j} \left|B_{n+k2^{-j}}-B_n\right|\leqslant t\right\},$$
and use the fact that the vectors 
$\left(B_{n+k2^{-j}}-B_n\right)_{1\leqslant k\leqslant 2^j}$ and $\left(B_{k2^{-j}}\right)_{1\leqslant k\leqslant 2^j}$
have the same distribution.
