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

Question

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.

Answer 1

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 $|\rho|\leqslant1$.