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Let $B$ be a standard Brownian motion with induced filtration $F$.
Is it true that, for $s<t$, $$ E \left( \int_s^t B_x dx \mid F_s \right) = \int_s^t E \left( B_x \mid F_s\right) dx \;? $$

To prooved this, I would use the dominated convergence theorem.
But I don't see any dominating function.
Viewing the integral as a standard Riemann integral, I think the result follows.

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up vote 7 down vote accepted

Write $B_x=B_x-B_s+B_s$. By linearity of conditional expectation and using the fact that $B_x-B_s$ is independent of $\mathcal F_s$, we are reduced to show that $$E\left(\int_s^t(B_x-B_s)dx\mid\mathcal F_s\right)=0.$$ It's enough to show that $\int_s^t(B_x-B_s)dx$ is independent of $\mathcal F_s$. To see that, write $\int_s^t(B_x-B_s)dx$ as the almost sure limit of $\frac 1n\sum_{j=0}^{n-1}(B_{s+xk/n}-B_s)$. If $X_n$ is independent of $Y$ for all $n$ and $X_n\to X$ almost surely, then $X$ and $Y$ are independent. Indeed if $O_1$ and $O_2$ are open, $$P(X\in O_1,Y\in O_2)=\lim_{n\to +\infty}P(\bigcap_{j\geq n}X_n\in O_1,Y\in O_2)=\lim_{n\to +\infty}P(\bigcap_{j\geq n}X_n\in O_1)P(Y_n\in O_2),$$ and we conclude by a monotone class argument (I should introduce the set of probability one where we have the convergence).

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I rarely see this kind of argument. I enjoyed reading it a lot. Thank you. –  Nicolas Essis-Breton Oct 20 '12 at 15:54

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