Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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).

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.