The following is taken from a proof of Lévy's construction of Brownian motion in a book by Peter Mörters and Yuval Peres.

$\mathcal {D_n } := \{\frac {k } {2^n } :1 \le k \le 2^n \} $, the set of dyadic points in $[0,1]$.

I'm not sure how the claim in $(2)$ is defined and thus have a hard time reading the paragraph that follows.

screenshot from book

My first guess was that it mean that each element of the first vector is independent of the second vector seen as a collection of random variables. This seems to be implicated by the last sentence.

But then to conclude that $\frac {B(d-2^{-n }) + B(d+2^{-n })} {2 } + \frac {Z_d } {2^{(n+1)/2 } } $ is independent of $\{Z_t:t \in \mathcal {D-D_n } \} $ I believe we also need $B(d+2^{-n }),B(d+2^{-n })$ and $Z_d $ to be mutually independen. Is this coorect?

Many thanks in advance!

  • $\begingroup$ Maybe what we should use is that if $(X,Y)$ is independent of $Z$ then $X+Y $ is independent of $Z $. $\endgroup$ – MrFranzén May 21 at 12:44

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