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I have this exercise:

Let $(B_t^1,B_t^1)_{t \in (0,\infty)}$ be a standard two-dimensional brownian motion. Prove that the process $((X_t^1, X_t^2))_{t \in (0,\infty)}$ defined by $$ X_t^1 = \int_0^t cos(B_s^1)dB_s^1 - \int_0^t sin(B_s^1)dB_s^2$$ $$X_t^2 = \int_0^t sin(B_s^1)dB_s^1 + \int_0^t cos(B_s^1)dB_s^2$$ is a standard two-dimensional brownian motion.

I was thinking to solve it by using a two dimensional levy characterization of brownian motion. However, I just now the 1 dimensional version, so in general I would check that that given a martingale $M_t$, it is a standard brownian motion if $\langle M \rangle_t = t$. Is it the right way of proving this ? Maybe by considering a covariation ?

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1 Answer 1

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Yes, you can prove it using the multidimensional Levy theorem. This is, checking that the process $((X_t^1, X_t^2))_{t \in (0,\infty)}$ is a 2-dimensional local martingale and that

  • $[X^{1}]_t=[X^{2}]_t = t;$
  • $[X^{1},X^{2}]_t = 0.$
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  • $\begingroup$ Thank you !!! just to check I don't misunderstand, the quantities in the squared brackets are quadratic variation (point 1) and quadratic covariation (point 2)? $\endgroup$
    – m120p
    Commented Nov 12, 2020 at 17:13
  • $\begingroup$ @120548: yes, that's right. $\endgroup$
    – UBM
    Commented Nov 12, 2020 at 17:15

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