Given that $(B_t)_{t\geq 0}$ is a standard Brownian motion and $0 \leq r < s$ and $x,y \in \mathbb{R}$, how can we show that the three processes $(B_t)_{t \in [0,r)}$, $(B_t)_{t \in (r,s)}$ and $(B_t)_{t \in (s,\infty)}$ are independent conditionally on $B_r=x$ and $B_s =y$.
Since everything is Gaussian I could certainly just calculate the conditional densities and verify that they factorize in the right way to show independence. But is there a more clever way?
What I do know is that for $\lambda \in (0,1)$ the random variable $Y = B_{\lambda r + (1-\lambda)s} - \lambda B_r - (1-\lambda)B_s$ is independent of all $B_t$ with $t \notin (r,s) $ and the variance of $Y$ is $\mathbb{E}[Y^2]=\lambda (1-\lambda) (t-s)$. Clearly $Y$ is related to the "middle segment" of the Brownian motion because $\lambda r +(1-\lambda)s \in (r,s)$, but how do I use the independence here in order to prove the conditional independence of the Brownian motions from above?