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

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.

@ZacharySelk Thanks for the reply, I did think of this but I wasn't sure whether an increment had to be defined as $B_{t_k}-B_{t_{k-1}}$ i.e. I wasn't sure whether for example $B_{t_k}-B_{t_{k-2}}$ was allowed as an increment, if you get what i'm trying say? — seraphimk, Apr 20 '18 at 21:05

score 2 · Answer 1 · answered Apr 20 '18 at 20:53

2

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

answered Apr 20 '18 at 20:53

Kroki

are overlapping increments independent i.e. are $B_{t_4}-B_{t_2}$ and $B_{t_3}-B_{t_1}$ independent? – seraphimk Apr 20 '18 at 22:01
1

No they are not independent. – Kroki Apr 20 '18 at 22:06
One more question, for $0 < t_{k-1} < t_k < t_{k+1}$ since $B_{t_{k+1}}=B_{t_{k+1}} - B_0$ does that mean $B_{t_{k}} - B_{t_{k-1}}$ is not independent from $B_{t_{k+1}}$ since they overlap? – seraphimk Apr 21 '18 at 13:17
That is correct. – Kroki Apr 21 '18 at 17:06
Thanks that helped clarify a lot – seraphimk Apr 22 '18 at 15:44

1 Answers1