$(B_t)$ is a Brownian motion and i assume that $s&lt;t&lt;u$ we have $$E[B_t |\sigma(B_s,B_u)] = G(B_s,B_u)$$ Does anyone knows the explicit expression of $G$ ? (the calculus is easy but fastidious...). Thank you.
Let $(B_t)_{t\geq 0}$ be a standard Brownian motion in $\mathbb{R}^d$. It is intuitive that, for fixed $s&lt;t&lt;u$  $$\mathbb{E}[B_t\mid \sigma(B_s,B_u)]=B_s+\frac{t-s}{u-s}(B_u-B_s).$$  However, I cannot think of a way to show this rigorously.   If first attempted to take $A\in\sigma(B_s,B_u)$ and show that $\mathbb{E}[1_A B_t]=\mathbb{E}[1_A(B_s+\frac{t-s}{u-s}(B_u-B_s))]$. But I cannot manage to show this equality.   I'd be very thankful for any ideas and suggestions on how to tackle this problem.
Today in a match the setter set the ball too low for me to spike it, so I just dinked it over.  The bottom of the ball was lower than the top of the net at the point where I made contact.  When the defender made contact, again, the bottom of the ball was lower than the top of the net (the top of the ball was over the top of the net though).  I felt that this wasn't technically a block and that he shouldn't have been allowed to hit it a second time, but I wasn't sure so I just let it go.  Was this a legal block or should it have been our point when he hit it a second time in a row?  Edit:  I don't think I was clear about the part where, after I hit the ball, the ball travelled over the net onto the other team's side of the court.  The other player did not reach over the net so there is no issue there.  It's just an issue of whether or not their first contact was a block or a hit.  I believe it was not a block but I can't find a specific ruling to address it. 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