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.
Using the polarization formula as suggested:
$cov(W_t^2,B_t)=E[(\int_0^t2W_sdW_s+t)\int_0^tsgn(W_s)dW_s]\\=E[2\int_0^tW_ssgn(W_s)ds]+tE[\int_0^tsgn(W_s)dW_s]=2\int_0^tE[W_ssgn(W_s)]ds\\=2\int_0^tE[W_s|W_s>0]-E[W_s|W_s<0]ds\\=4\int_0^tE[W_s|W_s>0]ds\\ =4\int_0^t 1/\pi ds=4t/pi$