conditional expectation of brownian motion

Solution 1:

Actually, it just might be that OP is the next Paul Lévy, as the result holds!

Unfortunately, I am far from being as bright as Lévy, so I'll apply the same kind of tricks that I used in this answer.

Set $$ Z_t=B_t-\frac{u-t}{u-s}B_s-\frac{t-s}{u-s}B_u, $$ which you can check is a Gaussian random variable independent of $B_s$ and $B_u$.

It follows that \begin{align*} E[B_t\,|\,\sigma(B_s,B_u)] &=E[Z_t\,|\,\sigma(B_s,B_u)]+\frac{u-t}{u-s}B_s+\frac{t-s}{u-s}B_u\\ &=E[Z_t]+\frac{u-t}{u-s}B_s+\frac{t-s}{u-s}B_u\\ &=\frac{u-t}{u-s}B_s+\frac{t-s}{u-s}B_u\\ &=B_s+\frac{t-s}{u-s}\left(B_u-B_s\right). \end{align*}

Solution 2:

As a side note for further reading, this property of the Brownian motion is called the harness property. Processes satisfying this property have been studied in the literature, for instance in the paper

Harnesses, Levy bridges and Monsieur Jourdain