# Help me elucidate what “the vectors $(B(d):d \in \mathcal {D_n })$ and $(Z_t:t \in \mathcal {D-D_n})$ are independent” mean.

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.

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?

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