I have $W_t$ is a Brownian Motion and $$B_t :=W_t-\int_0^t \frac{W_u}{u}du$$ is also a Brownian Motion. I have to show that these two are uncorrelated.
I know for Brownian uncorrelated is equivalent to $E[W_tB_t]=0$.
I have at this point $$ E\left[W_t^2\right]-E\left[W_t\int_0^t\frac{W_u}{u}du\right]=t-....?$$
So i don't know how i compute the second expected value but i know it should be $t$. Can anybody help me?
Ito's formula gives $$ d(B_tW_t)=B_tdW_t+W_tdB_t+dW_tdB_t=(W_t+B_t)dW_t+(1-(W_t^2/t))dt. $$ Therefore, for a local martingale $m_t$, $$ E(B_tW_t)=Em_t+\int_0^t1-(EW_s^2)/s\,ds. $$ The second integral is $0$, it remains to see that $m_t$ is a true martingale. Can you finish from here?