I am stuck with the following problem :
I have a Brownian motion $B_t$ and an Ito process $$X_t:=\int_0^t sgn(B_s)\ d B_s,$$ where $sgn(x)=1$ when $x \geq 0$ and $sgn(x)=-1$ when $x<0$.
I have to say whether the couple $(B_t, X_t)$ is a 2-dimensional Brownian motion or not.
I have a hint saying "you may show that the components of the vector are not independent".
Therefore, I guess the answer is NO.
My idea to prove that $X_t$ and $B_t$ are not independent is to show that the expectation of the product is different from the product of the expectations, which is zero.
To compute $\mathbb{E} [X_t B_t]$, I would compute the distribution of $X_t B_t$ via Ito's lemma (integration by parts, actually) and then consider the expectation.
Is this feasible ? Once I have the distribution of $X_t B_t$ I am not so sure about how to procede. Could you help me ?
Thank you very much for any help !!!
One way of disproof is to check the quadratic variation: If $B_t, X_t$ are jointly Brownian, then the quadratic variation should be
$[B,X]_t = 0$
But by the condition, we actually have
$[B,X]_t = \int_0^t sign(B_s) ds$ , which is not even deterministic,