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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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    I've tried to interpret what exactly do you want. Check whether I succeded. This question though belongs to math.SE, flagging accordingly.2012-02-21
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    The Fs (filteration) comes within the brackets though.2012-02-21
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    It makes no sense then, since one random variable is conditioned, but another not.2012-02-21
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    @mpiktas: I don't understand these comments about the notation. Shouldn't it be pretty clear the intention is $\mathbb E(t W_t - \int_0^t W_u \>\mathrm d u \mid \mathcal F_s )$?2012-02-21
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    **Hints**: What do you know about $W_t$ as a process, particularly about $\mathbb E( W_t \mid \mathcal F_s)$? Now, note that the integral is just a *Riemann* integral, it's not a stochastic integral. Does that help?2012-02-21
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    @cardinal, you are right, my mistake.2012-02-21
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    @mpiktas: I was just unsure. I thought I must be missing something as you're always very sharp and on top of these things. :)2012-02-21
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    @cardinal, I follow simple rule, if it's not pretty, it's not right. Sometimes it does not work :) Also I was thinking about making distinction between $E(X+Y|\Omega)$ and $EX+E(Y|\Omega)$, but for conditional expectations the former is always interpreted as $EX|\Omega+EY|\Omega$, and the latter is pretty uncommon.2012-02-21
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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.