# Compute the covariance of $W_t$ and $B_t=\int_0^t\mathrm{sgn}(W)dW$, for a Brownian motion $W$

by Andrew   Last Updated January 14, 2018 13:20 PM

From the Levy charaterization we know that $B_t$ is a Brownian Motion, as well as that $(B_t,W_t)$ is a two-dimensional Brownian Motion, if the cross-variation $<W,B>_t=0$.

First I am supposed to compute $cov(W_t,B_t)$. After that I should disapprove that $(B_t,W_t)$ is a two-dimensional Brownian Motion with the hint to compute $cov(W_t^2,B_t).$ My thoughts so far:

• I must say that I am a bit confused. Isn't it true that two Brownian Motions form a two-dimensional Brownian Motion if their covariance is $0$, since they are gaussian?
• I tried calculating the covariance with the tower property $cov(W_t,B_t)=E[W_tB_t]=E[W_tE[\int_0^t sgn(W_t)dW_t|W_t]]$, but I don't understand the dependence of $\int_0^t sgn(W_t)dW_t$ and $W_t$

• Furthermore I ignored the hint since I don't know what to do with it (always good I know) and computed $<W,B>_t=<\int_0^. 1 dW_t,\int_0^tsgn(W_t) dW_t>=\int_0^tsgn(W_t)ds.$ Doesn't seem to me like $0$ but unsure how to prove it.

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