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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!

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    $\begingroup$ Of course if $X$ and $Y$ are dependent, their quadratic covariation need not be zero. Just take $X = Y$... $\endgroup$ – Mateusz Kwaśnicki Apr 14 at 14:25
  • $\begingroup$ @MateuszKwaśnicki Hah, you're right. Silly that I didn't consider that. Thanks! $\endgroup$ – S. Crim Apr 14 at 15:32
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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 of Brownian motion gives that $(X,Y)$ is a two dimensional Brownian motion, in particular $X,Y$ are independent.

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If you know that the two Brownian motions $X$ and $Y$ have correlation $\rho$, then their quadratic covariation is

\begin{equation*} [X, Y]_t = \rho t \end{equation*}

Check out this blog post for a nice proof.

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