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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 !!!

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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,

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  • $\begingroup$ Thank you very much for your answer !! :) Are you sure that a 2d BM can have components with non-zero correlation ? If we have a (2d) Gaussian vector, this is surely true, but I think that, by definition, a d-dimensional BM has independent components... $\endgroup$ – Fred G. Dec 7 '15 at 12:12
  • $\begingroup$ One moment...how do you know that $[B,X]_t=\rho t$ ? If we prove it rigorously, then we can say the correlation coefficient is not zero and therefore the two processes not indepedent ! $\endgroup$ – Fred G. Dec 7 '15 at 14:11
  • $\begingroup$ I checked some reference, I think you are right, standard definition of multi dimensional BM assumes independence across components. So you only have to show that the quadratic covariation is non zero. $\endgroup$ – Jay.H Dec 7 '15 at 14:28
  • $\begingroup$ Modified my answer. $\endgroup$ – Jay.H Dec 7 '15 at 14:32
  • $\begingroup$ Perfect ! Thank you !! The conclusion follows from the fact that the quadratic covariation of two independent processes is zero, right ? How would you prove this last statement ? $\endgroup$ – Fred G. Dec 7 '15 at 14:32

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