Discontinuous expression cannot be a Brownian motion? 
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.

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 $\text Corr [X_t ~X_s ]=  \mathbb E [X_t ~X_s ]= s \wedge t$
