# Covariation of Wiener processes, $\langle W_1,W_2\rangle_t = \rho t$.

I'm wondering why this is true: $\langle W_1,W_2\rangle_t = \rho t$.

Where $W_1$ and $W_2$ are standard Brownian Motion.

I know that $\langle W_1,W_2\rangle_t = 0.5\big[ \langle W_1 + W_2 \rangle - \langle W_1 \rangle_t - \langle W_2 \rangle_t \big]$.

But I'm not sure how to progress from here.

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Are $W_1$, $W_2$ independent...? What is $\rho$? –  saz Nov 24 '12 at 9:44
@saz No. $\rho$ is the correlation coefficient. The equality $\langle W_1,W_2 \rangle_t = \rho t$ is provided in the (ungraded) homework question's preamble where $\rho \in [-1,1]$. –  Jase Nov 24 '12 at 9:45

This does not have to be true, this is a (very strong) hypothesis about the joint distribution of two Brownian motions $(W_1(t))_{t\geqslant0}$ and $(W_2(t))_{t\geqslant0}$ defined on a common probability space, which may be true or not, hence, trying to prove it is pointless.
At most, one can note that $\rho t=\mathbb E(W_1(t)W_2(t))$ and, since $\mathbb E(W_1(t)^2)=\mathbb E(W_2(t)^2)=t$, the variance-covariance inequality imposes that indeed $|\rho|\leqslant1$.