I need to find the distribution of $B_s + B_t , \forall \ t,s \geq 0$, where $B$ is a standard Brownian motion.
Here's what I've done:
$B_s + B_t = B_t + B_t \sim N(0+0, t+t)=N(0,2t)$
However, the solution combine the $B_t$ and obtain a different variance.
$B_t + B_t = 2B_t \sim N(0,2^2 t)= N(0,4t)$
Shouldn't I be obtaining the same variance regardless of the approach?