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How do I compute

$E\left(tW_t - \int_0^t W_u du \Big| \mathcal{F}_s \right).$

Given that $W_t$ is standard Brownian motion under the measure $P$ and $\{\mathcal{F}_t, t\ge 0\}$ denotes its standard filtration. could someone guide me on the approach to this problem?

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    @mpiktas: That's often a very good heuristic! I agree with the rest of what you say :)2012-02-21

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The hard part is finding the conditional expectation of the integral $\int_0^t W_u\,du$. It turns out that there is a "Fubini theorem" for conditional expectation:

Proposition. Suppose $X_t$ is a stochastic process, $\mathcal{G}$ is a $\sigma$-field. Recall by Tonelli's theorem that $E \int_a^b |X_u|\,du = \int_a^b E|X_u|\,du$. If this quantity is finite, then $E\left[\left.\int_a^b X_u\,du \;\right|\; \mathcal{G}\right] = \int_a^b E[X_u \mid \mathcal{G}]\,du.$

I'll let you prove this as an exercise. As a hint, use the definition of conditional expectation to show that the right side is the conditional expectation of $\int_a^b X_u\,du$; you'll need to use (classical) Fubini's theorem.

(As a side comment, the same proposition holds, with the same proof, if we replace $[a,b]$ by any other $\sigma$-finite measure space.)

Given this, the rest of the problem should not be too hard.