Whether two one-dimensional Brownian motions yield a two-dimensional Brownian motion

Question

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.

Answer 1

\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*}