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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:

  • $cov(W_t,B_t)=E[W_t\int_0^tsgn(W_s)dW_s]=E[\int_0^t1dW_s\int_0^tsgn(W_s)dW_s]=E[\int_0^tsgn(W_s)ds]=\int_0^tE[sgn(W_s)]ds=\int_0^t0ds=0$

  • $cov(W_t^2,B_t)=E[(\int_0^t2W_sdW_s+t)\int_0^tsgn(W_s)dW_s]\\=E[2\int_0^tW_ssgn(W_s)ds]+tE[\int_0^tsgn(W_s)dW_s]=2\int_0^tE[W_ssgn(W_s)]ds\\=2\int_0^tE[W_s|W_s>0]-E[W_s|W_s<0]ds\\=4\int_0^tE[W_s|W_s>0]ds\\ =4\int_0^t 1/\pi ds=4t/pi$

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    $\begingroup$ You are missing the point that a pair of 1D Brownian motion needs to be jointly gaussian to be a 2D Brownian motion. Once you will have realized that, note that you are simply supposed to use repeatedly the polarization formula $$E\left(\int_0^tX_sdW_s\cdot\int_0^tY_sdW_s\right)=E\left(\int_0^tX_sY_sds\right)$$ for suitable processes $X$ and $Y$. $\endgroup$ – Did Jan 14 '18 at 13:00
  • $\begingroup$ Thank you for your suggestion. I edited my question with the computations. I understand why $cov(W_t,B_t)=0$ is not equal to being a 2D Brownian Motion. But i still don't get the point behind determining $cov(W_t^2,B_t)$ since it doesn't seem to me like a contradiction to being 2D gaussian. $\endgroup$ – Andrew Jan 14 '18 at 13:50
  • $\begingroup$ And yet it is: if $(X,Y)$ is jointly gaussian, uncorrelated, and centered, what is $E(X^2Y)$ already? (FYI, there are mistakes in your computation of the covariance of $W_t^2$ and $B_t$.) $\endgroup$ – Did Jan 14 '18 at 14:05
  • $\begingroup$ Where did i do mistakes? I don't get what $E(X^2Y)$ is supposed to be in that case (but is it correct that is not equal to 0?). $\endgroup$ – Andrew Jan 14 '18 at 14:15

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