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Let $W_t$ and $\tilde{W}_t$ be two standard independent Brownian motions and for a constant $-1 \leq \rho \leq 1$, define $X_t := \rho W_t + \sqrt{1-\rho^2} \tilde{W}_t$.

Is $X := (X_t)_{\{t \geq 0\}}$ a Brownian motion?

I know that to show something is a Brownian motion I must check the various criteria, but skipping to one of the later ones it appears that $X$ is discontinuous, e.g. at $\rho = 2$. Does this mean that $X$ cannot be a Brownian motion? It appears to satisfy the other criteria.

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    $\begingroup$ $\rho$ can't be 2, it must be in $[-1,1]$ by your own admission. $\endgroup$
    – Set
    Dec 6, 2015 at 4:18

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It is a Brownian Motion of correlation $\rho$ if $\mid \rho \mid \leq 1$ with $W_t$ and $\sqrt{1-\rho ^2}$ with $\bar W_t$.

One way to show it is by checking is is a martingale with correlation $\text Corr [X_t ~X_s ]= \mathbb E [X_t ~X_s ]= s \wedge t$

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