Is the Brownian motion $B_{t_i}$ independent from the increments $B_{t_k}-B_{t_{k-1}}$ (where $t_i <t_{k-1}<t_k$)

Question

Hi I know that by definition a Brownian motion $B_t$ has independent increments, i.e.

$B_{t_1},B_{t_2}-B_{t_1},...,B_{t_k}-B_{t_{k-1}}$ are independent for all $0\le t_1 < t_2 <...< t_{k-1} < t_k$

However is the Brownian motion say $B_{t_i}$ independent of the increment $B_{t_k}-B_{t_{k-1}}$? (Where $t_i < t_{k-1}$)

I'm just a little confused here as I don't view $B_{t_i}$ as an increment.

Answer 1

Yes. Indeed, $B_{t_i}, B_{t_{k-1}} - B_{t_i}, B_{t_k} - B_{t_{k-1}}$ are independent.