If you have two standard brownian motions $B_{t}$ and $C_{t}$ which are independent, then $W_{t}$ = $aB_{t}+C_{t}\sqrt(1-a^2)$ is also a standard brownian motion, for $ -1 < a < 1 $. This can be easily shown, however I am having trouble between finding the covariance between $W_{1}$ and $B_{2}$ for instance. This isn't directly a problem, however I am trying to use this covariance to solve a problem which isn't quite working out.
My attempt:
$Cov(W_{1}, B_{2}) = Cov(aB_{1}+C_{1}\sqrt(1-a^2), B_{2})$
$ = Cov(aB_{1}, B_{2})$ (by linearity of covariance and then independence of $C_{1}$ and $B_{2}$
$ = aCov(B_{1}, B_{2}) = a $
Is there something I'm doing wrong? The reason I am trying to solve this is because I am trying to determine probabilities such as $P(W_{1}<0|B_{2}=-1)$ and want to use their bivariate distribution.
Any help would be appreciated.