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Let $\{W^1\}_{t\geq0}$ and $\{W^2\}_{t\geq0}$ be two Brownian motions with correlation coefficient $\rho \in [0, 1]$, i.e., $\mathbb{E}[(W^1(t)-W^1(s))(W^2(t)-W^2(s))]=\rho(t-s)$ for all $t,s \geq 0$. Show that $<W^1, W^2>_t = \rho t$ for all $t\geq0$.

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Hints:

Show first that $$E[((W^1_t + W^2_t)-(W^1_s + W^2_s))^2] = (2+2\rho)(t-s) $$

Then conclude that

$$ [(2+2\rho)^{-1/2} (W^1 + W^2)]_t =t $$

On the other hand, show (using bilinearity of quadratic covariation) that

$$ [(2+2\rho)^{-1/2} (W^1 + W^2)]_t = [(2+2\rho)^{-1/2} (W^1 + W^2), (2+2\rho)^{-1/2} (W^1 + W^2) ]_t $$ $$ = (1+\rho)^{-1} (t+ [W^1, W^2]_t)$$

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