# On the conditional distribution of $B_{(s+t)/2}$ conditionally on $(B_t,B_s)$, for Brownian motion $B$

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?

• You forgot one condition in your characterization of Brownian motion. (Note that $X_t=7t$ satisfies 1.-2.-3. but $(X_t)$ is not a Brownian motion.) – Did Jun 7 '16 at 6:56
• Oh, yes, increments should be centered gaussians with variance equal to the length of increment. – user340297 Jun 7 '16 at 7:00
• Which solves your question, right? – Did Jun 7 '16 at 7:03
• Ah, right, because $2Y=B_{(s+t)/2}-B_s - (B_t-B_{(s+t)/2})$ – user340297 Jun 7 '16 at 7:07