$(B_t)$ is a Brownian motion and i assume that $s<t<u$ 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.
2026-03-29 13:46:43.1774792003
conditional expectation of the Brownian motion
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