Independence of two random variables derived from a Brownian motion

Question

If $X = B_1 + B_3 - B_2$ and $Y = B_1 - B_3 + B_2$ Where $B_t$ is Brownian Motion for $t \geq 0$ And I want to state with certainty whether $X$ and $Y$ are indep or not,

do I simply just $\text{Cov}(X,Y) = \text{Cov}(B_1 + B_3 - B_2, B_1 - B_3 + B_2)$ which is non-zero, and thus the two R.v's are not independent?

Answer 1

Yeah, that would be a way to do it.

May I suggest the following way of ordering the terms in the covariance calculation?

$$\text{cov}(X,Y) = \text{cov}(B_1+(B_3-B_2),B_1-(B_3-B_2))$$ $$=\text{cov}(B_1,B_1)-\text{cov}(B_1,B_3-B_2)+\text{cov}(B_3-B_2,B_1)-\text{cov}(B_3-B_2,B_3-B_2)$$ $$=\text{var}(B_1)-\text{var}(B_3-B_2)$$ $$=1-1=0.$$

Underway I use that $B_1$ and $B_3-B_2$ are independent, and thus that their covariance is $0$.

You may know that even though the covariance of two random variables is $0$, the random variables may still be dependent. In this case, however, a covariance of $0$ does imply independence (why?)