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Let $X_t = \displaystyle\int_{0}^{t}sgn(W_{1s})dW_{1s}$ and $Y_t=W_{2s}$, where $(W_{1s})_s$ and $(W_{2s})_s$ are one-dimensional standard Brownian motions with correlation coefficient $\rho$, that is, $corr(W_{1s},W_{2s})=\rho$ for each $s$. Then, $X$ and $Y$ are clearly standard Brownian motions, because $X$ is a martingale with quadratic variation $[X]_t=t$, for each $t$. Now, I am trying to make sure whether $(X_s,Y_s)_s$ becomes a two-dimensional Brownian motion. My approach was to show whether $(X_t,Y_t)=_d normal$ with covariance varying proportionately with time $t$, in which I am now stuck due in particular to algebra. Any comment will be welcome.

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\begin{align*} \langle W^{1},W^{2} \rangle_{t} &={\mathbb E}[W_{t}^{1}W_{t}^{2}] \\ &={\rm Corr}[W_{t}^{1},W_{t}^{2}]\sqrt[]{{\rm Var}[W_{t}^{1}]}\sqrt[]{{\rm Var}[W_{t}^{2}]} \\ &=\rho t \end{align*} and \begin{align*} \langle X,Y \rangle_{t} &=\langle \int_{0}^{\bullet}{\rm sgn}(W_{s}^{1}){\rm d}W_{s}^{1},\int_{0}^{\bullet}{\rm d}W_{s}^{2} \rangle_{t} \\ &=\int_{0}^{t}{\rm sgn}(W_{s}^{1}){\rm d}\langle W^{1},W^{2} \rangle_{s} \\ &=\int_{0}^{t}{\rm sgn}(W_{s}^{1})\rho{\rm d}s \\ &\not\equiv 0. \end{align*}

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    $\begingroup$ Hello, I got it. Thanks. By the way, since I am now self-studying stochastic calculus, I am stuck in here and there, especially in algebras like the above fifth equality, i.e, a quadratic covariation of two Ito integrals simplifies to an integral whose integrator is just a quadratic covariation of the two martingales concerned. Such advanced books as Springer-Verlag ones becomes infinitely time-consuming blackholes too often. Could you recommend me any book, lecture notes, or anything else that deals with such technical things not too rigorously as well as not too heuristic? Many thanks. $\endgroup$ – metric Feb 15 at 8:42
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    $\begingroup$ Also, my opinion is that the last line would be better if it changes to "Since the quadratic covariation of X and Y is path-dependent, (X,Y) is not a Browniani motion.", instead of just " not zero." Isn't it? $\endgroup$ – metric Feb 15 at 8:47
  • $\begingroup$ Hello, I studied probability theory and stochastic calculus using the book Ikeda--Watanabe (1981) with reference to Karatzas--Shreve (1998). Unfortunately, I cannot recommend you since I don't know the details of the other books. $\endgroup$ – 720773 Feb 15 at 9:59
  • $\begingroup$ If (X,Y) is a two-dimensional Brownian motion, then it satisfies \langle X_{t}Y_{t}\rangle \equiv 0 (cf. page 54, Ikeda--Watanabe (1981)). This implies that its pair is not a two-dimensional Brownian motion since one can prove \langle X_{t}Y_{t}\rangle \not\equiv 0 (As you say, we may need to use the path dependence for the proof). Conversely, if \langle X_{t}Y_{t}\rangle \equiv 0 holds, then its pair is a two-dimensional Brownian motion (cf. page 74, Ikeda-Watanabe (1981)). $\endgroup$ – 720773 Feb 15 at 9:59

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