Let $B$ be a standard Brownian motion on $[0,1]$ and further let $0\leq s \leq t \leq 1$ with $\mathcal{G}_{s} = \sigma( B_{r}: 0 \leq r \leq s) \lor \sigma(B_{1})$
Show that $$\mathbb E [B_{t} - B_{s} \lvert \mathcal{G}_{s}]=\frac{t-s}{1-s} (B_{1}-B_{s})$$
using the properties of Gaussian processes.
My attempt:
$$\mathbb E [B_{t} - B_{s} \lvert \mathcal{G}_{s}]=\mathbb E [B_{t} -B_{1}+ B_{1}-B_{s} \lvert \mathcal{G}_{s}]=\mathbb E [B_{t} -B_{1}\lvert \mathcal{G}_{s}]+(B_{1}-B_{s})$$
So now how am I supposed to use the fact that the process is Gaussian to compute $\mathbb E [B_{t} -B_{1}\lvert \mathcal{G}_{s}]$?
My remarks: I am confused how the properties of Gaussian random variables help us here. Any ideas?
Consider the process $$ X_t=B_{s+t}-B_s-\frac{t}{1-s}(B_1-B_s), $$ $0\le t\le 1-s$. Since $X_t$ is independent of $\{B_t:t\in [0,1]\setminus (s,1)\}$ (see, e.g., Lemma 4 here), \begin{align} \mathsf{E}[B_{s+t}-B_s\mid\mathcal{G}_s]&=\mathsf{E} X_t+\frac{t}{1-s}(B_1-B_s) \\ &=\frac{t}{1-s}(B_1-B_s). \end{align} because $X_t$ is a Brownian bridge with $\mathsf{E} X_t=0$.