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Let $B_1$ and $B_2$ be two $\mathbb{R}$-valued Brownian motions with $$\langle B_1,B_2\rangle=\int_0^t\rho_s ds,$$ where $\rho$ is progressively measurable with values in $(-1,1)$. We define $(W_1,W_2)$ by $$W_1(t):=B_1(t)$$ and $$W_2(t):=\int_0^t\frac{1}{\sqrt{1-\rho_s^2}}dB_2(s)-\int_0^t\frac{\rho_s}{\sqrt{1-\rho_s^2}}dB_1(s).$$ Then $(W_1,W_2)$ is a two-dimensional Brownian motion.

To prove this, I should use Levy's charaterization, but I don't even find an ansatz. Can someone help me, please?

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2 Answers 2

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Continuity and the local martingale property should (hopefully) be clear. For the quadratic variation/covariation, look at the quantities $(dW_1)^2$, $(dW_2)^2$, $dW_1dW_2$. The first two should be $dt$, and the third $0$. (Use the fact that $dB_1dB_2=\rho_t dt$). Then integrate to get the required answer.

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One of the characterizations of Multidimensional Brownian Motion $B_i(t)$ is that the covariations satisfy: $$[B_i, B_j]_t = \delta_{ij}t$$

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