# Independence using reflecting brownian motion

Suppose $X$ and $Y$ are two Brownian motions such that $|X|$ and $|Y|$ are independent. Then it is easy to show that $\langle X,Y \rangle =0$ using the Tanaka formula, for example, and thus $X$ and $Y$ are independent. However this method just applies to Brownian motion.

Now I have two nice processes $X$ and $Y$ with a nice state space (for example, two Brownian motions on a star (walsh brownian motions)) and I know that the associated reflecting Brownian motions (distance to zero) are independent.

Question: How can I deduce that $X$ and $Y$ are independent?

This seems to be true since independent reflecting Brownian motions dont vanish simultaneously after time 0. Could someone help me? Any proof in the case of Brownian motion without using the cross variation sould be a big step!

-
In the second sentence, what are you showing to be 0? –  Brian Rushton Dec 11 '12 at 1:10
The "expectation" notation got interpreted as a tag, so I switched in angle brackets. –  S. Carnahan Dec 11 '12 at 7:02
Wait, if you take 2d Brownian motion conditioned on the event that at time 1 its two coordinates have the same sign, or equivalently if you take two independent Brownian motions $B_1$ and $B_2$ and then set $X=B_1$ and $Y=\pm B_2$ where the sign is chosen so that $X$ and $Y$ have the same sign at time $1$, then $X$ and $Y$ have independent absolute values but are not independent... –  Vincent Beffara May 24 '13 at 11:43

Following up on my comment.

If you take 2d Brownian motion conditioned on the event that at time 1 its two coordinates have the same sign, or equivalently if you take two independent Brownian motions B1 and B2 and then set X=B1 and Y=±B2 where the sign is chosen so that X and Y have the same sign at time 1, then X and Y have independent absolute values but are not independent...

The thing is, proving $\langle X,Y\rangle=0$ is not enough to have independence, you need to assume something more, like the fact that the pair $(X,Y)$ is a Gaussian process, which is not automatic if $X$ and $Y$ are.

So to answer your initial question: without additional hypotheses, you cannot deduce that $X$ and $Y$ are independent, because they do not have to be.

-

If there is a Skorokhod decomposition for both X and Y, it may helps quantifying your intuition about "reflecting Brownian motions don't vanish simultaneously after time 0". This kind of pathwise information may help showing independence.

-