# Calculus of conditional expectation wrt $\sigma(B_1-B_s)$

On a classic probability space with $B$ a brownian motion. How to prove for : $$0\leq s<t\leq 1,\ \ \ \mathbb E (B_t-B_s\mid B_1-B_s)=\frac{t-s}{1-s}(B_1-B_s)$$

What I have done: I am sure that we have the brownian filtration : $\mathcal F_s$ independant to $\sigma (B_1-B_s)$ and also $\mathcal F_t\vee \sigma (B_1) = \mathcal F_t\vee \sigma (B_1-B_s)$. So then we can write : $$\mathbb E (B_t-B_s\mid B_1-B_s)=\mathbb E (B_t-B_s\mid \mathcal F_s\vee \sigma (B_1))$$

Define $X=B_t-B_s$ and $Y=B_1-B_s$. Now since $(X,Y)$ are jointly normaly distributed, with mean $0$ and variance $\Sigma$ you have $$E(X|Y=y)=E(X)+\frac{Cov(X,Y)}{Var(Y)}(y-E(Y))$$ which shall translate into your desired results.
Note $Cov(X,Y)=Cos(B_t-B_s,B_1-B_s)=t-s$ since $t\leq1$, and $Var(Y)=1-s$.
• Do you have a proof of that (even an internet link) ? And how do you assert that $(X,Y)$ are jointly normaly distributed (they are not independant) ? – Al Bundy Nov 14 '16 at 8:41
• For $f$ a borelian function we can write : $B_1-B_s=f(B_1-B_t, B_t-B_s)$. Is it independant to $B_t-B_s$ ? I don't think so. – Al Bundy Nov 14 '16 at 8:49