Finding the expected value of martingale increments on conditioned on brownian bridge I want to show that $\mathbb{E}[W_t - W_s | \mathcal{G}_s] = \frac{t-s}{1-s}(W_1 - W_s)$ for $0 \leq s \leq t$, where $W_t$ is the Weiner process, and $\mathcal{G}_s = \sigma(\mathcal{F}_t^W, \sigma(W_1))$, where $\mathcal{F}_t^W$ is the filtration of $W_t$. 
This is driving me crazy, I just keep getting the wrong answer. 
Using the linearity of expectation, we have $\mathbb{E}[W_t - W_s | \mathcal{G}_s] 
 = \mathbb{E}[W_t| \mathcal{G}_s] - \mathbb{E}[W_s| \mathcal{G}_s]$.  It seems to me that since $\mathcal{F}_s^W \subset \mathcal{G}_s$, we should have $\mathbb{E}[W_s| \mathcal{G}_s] = W_s$ (is this part wrong?)
Now we are left with $\mathbb{E}[W_t| \mathcal{G}_s]$. I already know from this post: conditional expected value of a brownian motion that $\mathbb{E}[W_t|W_s] = \frac{t}{s}W_t$ for $0 <t <s$, so since $\sigma(W_1) \subset \mathcal{G}_s$, we have $\mathbb{E}[W_t| \mathcal{G}_s] = tW_1$. But then this gives $\mathbb{E}[W_t - W_s | \mathcal{G}_s] = tW_1 - W_s$, which is wrong. 
How do I properly deal with the fact that the sigma algebra gives us information about the future and past? Am I supposed to somehow use the fact that the question is only asking about the difference? Hints would be appreciated!
Thanks very much in advance. 
 A: 
Geometric Intuition

As $\mathcal{G}_s$ and $W_1$ are given, the values of $W_s$ and $W_1$ are known. Thus
$$
\mathbb{E}\left(W_t-W_s|\mathcal{G}_s\right)
$$
is the expectation of a bridge $X_t=\left(W_t-W_s\right)|\mathcal{G}_s$, whose trajectory passes through $0$ when $t=s$ and through $W_1-W_s$ when $t=1$. Intuitively, the expectation of this $X_t$, namely $\mathbb{E}X_t$, would be the linear interpolation between $\left(t,X_t\right)|_{t=s}=\left(s,0\right)$ and $\left(t,X_t\right)|_{t=1}=\left(1,W_1-W_s\right)$, i.e.,
$$
\mathbb{E}X_t=X_s+\frac{\left(W_1-W_s\right)-0}{1-s}\left(t-s\right)=\frac{t-s}{1-s}\left(W_1-W_s\right).
$$
This intuition can be used to guess what the conditional expectation is (if it is not given). Upon this guess, a possible proof with mathematical rigor could be as follows.

Rigorous Proof

Just need to show these two facts, provided that $0<s<t<1$:


*

*$\left(W_t-W_s\right)-\frac{t-s}{1-s}\left(W_1-W_s\right)$ is independent from $W_s$, and hence independent from $\mathcal{F}_s^W$;

*$\left(W_t-W_s\right)-\frac{t-s}{1-s}\left(W_1-W_s\right)$ is independent from $W_1$.


Since $\left(W_t-W_s\right)-\frac{t-s}{1-s}\left(W_1-W_s\right)$, $W_s$ and $W_1$ are all Gaussian with fixed $s$ and $t$, it suffices to check if
$$
\text{Cov}\left(\left(W_t-W_s\right)-\frac{t-s}{1-s}\left(W_1-W_s\right),W_u\right)=0
$$
for $u=s,1$.
Note that
\begin{align}
&\text{Cov}\left(\left(W_t-W_s\right)-\frac{t-s}{1-s}\left(W_1-W_s\right),W_u\right)\\
&=\text{Cov}\left(W_t-\frac{1-t}{1-s}W_s-\frac{t-s}{1-s}W_1,W_u\right)\\
&=\text{Cov}\left(W_t,W_u\right)-\frac{1-t}{1-s}\text{Cov}\left(W_s,W_u\right)-\frac{t-s}{1-s}\text{Cov}\left(W_1,W_u\right)\\
&=t\wedge u-\frac{1-t}{1-s}s\wedge u-\frac{t-s}{1-s}1\wedge u,
\end{align}
where $a\wedge b=\min\left\{a,b\right\}$.
Recall that $0<s<t<1$. When $u=s$,
$$
t\wedge s-\frac{1-t}{1-s}s\wedge s-\frac{t-s}{1-s}1\wedge s=s-\frac{1-t}{1-s}s-\frac{t-s}{1-s}s=0;
$$
when $u=1$,
$$
t\wedge 1-\frac{1-t}{1-s}s\wedge 1-\frac{t-s}{1-s}1\wedge 1=t-\frac{1-t}{1-s}s-\frac{t-s}{1-s}1=0.
$$
Therefore, $\left(W_t-W_s\right)-\frac{t-s}{1-s}\left(W_1-W_s\right)$ is independent from $W_s$ (and hence independent from $\mathcal{F}_s^W$) and $W_1$, and hence independent from $\mathcal{G}_s$. As a result,
$$
\mathbb{E}\left[\left(W_t-W_s\right)-\frac{t-s}{1-s}\left(W_1-W_s\right)\Bigg|\mathcal{G}_s\right]=\mathbb{E}\left(\left(W_t-W_s\right)-\frac{t-s}{1-s}\left(W_1-W_s\right)\right)=0,
$$
which immediately leads to
$$
\mathbb{E}\left(W_t-W_s|\mathcal{G}_s\right)=\mathbb{E}\left[\frac{t-s}{1-s}\left(W_1-W_s\right)\Bigg|\mathcal{G}_s\right]=\frac{t-s}{1-s}\left(W_1-W_s\right).
$$
More details could be found in this reference, especially sections 4 and 6 therein.
