Conditional expectation of an integral depending of a Brownian Motion Hello there. I would like to find $$\mathbb{E} [B_1 | \int_0^2 B_s ds ] $$ where $(B_t)_{t \geq 0}$ is a standard Brownian motion.
I tried the following but I'm pretty sure this is utterly false :
"If $(B_1,\int_0^2 B_s ds)$ is a Gaussian vector, we would have
$$   \mathbb{E} [B_1 | \int_0^2 B_s ds ] = a \int_0^2 B_s ds$$ for some $a\in \mathbb{R}$.
It follow that $$\frac{3}{2} = \mathbb{E}[B_1\int_0^2 B_s ds] = \mathbb{E} [(\int_0^2 B_s ds) \mathbb{E} [B_1 | \int_0^2 B_s ds ]] = a \mathbb{E}[\int_0^2 B_s ds] = a\frac{4}{3} $$
And I would conclude that $a=\frac{9}{8}$."
Problem is $(B_1,\int_0^2 B_s ds)$ is probably not a Gaussian vector... so any idea ?

$(B_1,\int_0^{2} B_s ds)$ is indeed jointly Gaussian and this implies that the conditional expectation is just a  constant $c$  times $\int_0^{2} B_s ds$. The constant can be evaluted using $E[B_1\int_0^{2} B_s ds]=E[\int_0^{2} B_s ds)]^{2}$. Can you do this (simple) evaluation? [LHS is $\int_0^{2} EB_1B_s ds$].
