We have Brownian motion $X_t=at+oB_t$, where $B_t$ represents a standard Brownian motion. To compute expectation, I do:
Since a standard Brownian motion is normally distributed with mean 0 and variance 1, $E[X_t]=at$
Since $E[B_t]=0$, then $Var[X_t]=(at)^2-(a^2t^2)=0$
Which doesn't make sense. Where did I go wrong?