I've been reading stuff about Brownian motions and all that, and I came across the following statement:
On proving that $B_{\frac{s+t}{2}}\sim N(\frac{x+y}{2},\frac{t-s}{4})$ conditionally on $B_s=x, B_t=y$ the following is stated: the random variable defined by
$$ Y=B_{\frac{s+t}{2}}-\frac{1}{2}(B_t+B_s) $$
is Gaussian,and $Y,B_t,B_s$ are jointly normally distributed.
I don't see how, with just given definition of $B$-motion (which are preciesely 1.$ B_0=0$, 2. $B_t-B_s\sim N(0,t-s)$ for any $0\le s<t$, independent of $\sigma(B_u:u\le s$) 3. $B$ has almost surely continuous paths) $Y$ is Gaussian or the three r.v's are jointly normal. I know sum of independent Gaussian is Gaussian and stuff, but we don't have indepeedent one here, do we?