Given two not independent Brownian motions, $X$ and $Y$. I was wondering if we can say anything about the quadratic covariation of $X$ and $Y$, $\langle X,Y \rangle\_t$. I know that for two independent Brownian motions, that this quadratic covariation is zero, but does this also hold when we cannot say whether or not the Brownian motions are independent? I'm familiar with the definition of the quadratic (co)variation, but it didn't help me in finding the answer. I'm starting to suspect that we cannot do much with $\langle X,Y \rangle\_t$ when $X$ and $Y$ are dependent.


Any help is appreciated!



Quadratic covariation of two not independent Brownian motions
If the pair $(X,Y)$ is a local martingale wrt. some filtration, and since the quadratic variations are $[X,X]\_t=t$, $[Y,Y]\_t=t$ and $[X,Y]\_t=0$, [Lévy's characterisation](https://en.wikipedia.org/wiki/Brownian_motion#L%C3%A9vy_characterisation) of Brownian motion gives that $(X,Y)$ is a two dimensional Brownian motion, in particular $X,Y$ are independent. 


