I am trying to solve an exercise and I have trouble with the definition of brownian bridge.
"Let (bu , 0 ≤ u ≤ 1) be the Brownian bridge derived by conditioning a one-dimensional Brownian motion (Bu , 0 ≤ u ≤ 1) on B0 = B1 = 0."
Can I just take bu = Bu + u*B1? Or do I have to continue with the more general definition of a brownian bridge (e.i the conditional expectation assuming B1 = 0)
Yes, you can take $b_u = B_u - u B_1$ (note the minus sign). The formulation you have stated is equivalent to the Brownian bridge, in that it has the same distribution. See for example: https://en.wikipedia.org/wiki/Brownian_bridge and Brownian Bridge equivalence of definitions