A question about martingale and Brownian motion It is well known that the Brownian motion $B=(B_t)_{t\ge 0}$ is a martingale with respect to its natural filtration $\mathscr{F}_t$ and the fixed probability measure $\mathbb{P}$, i.e.
$$\mathbb{E}(B_t|\mathscr{F}_s)=B_s,\quad s\ge t$$
Now we limit the $t$ in the interval $[0,1]$ and enlarge the filtration with $\sigma(B_1)$ being added into each  $\mathscr{F}_t$,   $t\in[0,1]$, i.e.$$\tilde{\mathscr{F}_t}:=\sigma(\mathscr{F}_t \cup \sigma(B_1)) ,\quad t\in[0,1].$$
My question is how to calculte $\mathbb{E}(B_t|\tilde{\mathscr{F}}_s)$ $~$ for $~$ $0\le s\le t <1$ ?
Someone says the result is $\mathbb{E}(B_t|\tilde{\mathscr{F}}_s)=\frac{1-t}{1-s}B_s + \frac{t-s}{1-s} B_1$ but I don't know why and cannot  verify it.
I had been stuck by this question for two days long and had no idea.
If you know how to solve it , please do not hesitate to help me.
I am looking forward to your answers!
 A: In the other answer, I put the result using "well-known stuff". No let us try to prove it elementarily.
The usual idea in such proofs is to decompose things in parts in $\widetilde{\mathcal F_s}$ and independent parts.
We have $B_t = B_s + (B_t - B_s)$. The first term is in $\widetilde{\mathcal F_s}$, the other one is not completely independent. Since our variables form a Gaussian process, we can use covariances to check independence, and can use Gram-Schmidt orthogonalization to separate independence from dependence.
The projection of $(B_t - B_s)$ on $(B_1 - B_s)$ is given by $\dfrac {\mathrm{cov}((B_t - B_s),(B_1 - B_s))}{\mathrm{cov}((B_1 - B_s),(B_1 - B_s))}(B_1 - B_s) = \dfrac {t-s}{1-s}(B_1 - B_s)$.
So $B_t - B_s = \dfrac {t-s}{1-s}(B_1 - B_s) + R$ where the remainder $R$ is defined as $B_t - B_s - \dfrac {t-s}{1-s}(B_1 - B_s)$, and is independent from $B_1-B_s$ (by construction, but you can now check it manually if you want, the covariance is zero).
Also, $R$ is independent from $\mathcal F_s$, you can check it manually with covariances, or see that it is a function of the process $B_{s+\cdot}-B_s$.
Because you have a gaussian process, covariances are enough to check independence, so $R$ is independent of $\widetilde{\mathcal F_s}$.
Finally $B_t = B_s + \dfrac {t-s}{1-s}(B_1 - B_s) + R$, where the first two terms are in $\widetilde{\mathcal F_s}$, the other one is independent from it. Hence the result by taking conditional expectations on both sides. ($E[R] = 0$, everything is centered)
A: That's a cool question.
Intuitively, if you know your Brownian up to time $s$ and at time $1$, then conditional on what you know, the gap between $s$ and $1$ is filled by a Brownian bridge (see "general case" in https://en.wikipedia.org/wiki/Brownian_bridge ).
So $E(B_t \mid \widetilde {\mathcal F}_s) = B_s + \frac{t-s}{1-s}(B_1 - B_S)$.
