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Hi, I have the following expected value to compute

$E[ \int_{o}^{T} f(t) dt \int_{o}^{T} H(s) dW(s)]$,

where $f(t)$ and $H(s)$ are two stochastic processes adapted to the filtration generated by the Brownian motion W.

I think that this expected value could be equal to zero, but I really don't know how to give this proof.

Thank you in advance for any kind of advice or references.


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up vote 0 down vote accepted

You can write $h(s)=\int_0^Tf(t)dtH(s)$ then your expectation could be written as $E[ \int_{o}^{T} h(s) dW(s)]$. This integral is $0$ if you can prove $(\int_{o}^{t} h(s) dW(s)) $ to be a martingale, for example $E(\int_{o}^{T} h^2(s) ds)<+\infty$.

As you noticed in your comment, this procedure is correct if we can suppose that the process $h(s)$ is adapted, which is the case if $f$ is deterministic. Otherwise we can the decompose the integral as the in the following

$$\int_0^T(\int_0^Tf(t)dt)H(s)dW_s =\int_0^T(\int_0^sf(t)dt)H(s)dW_s+\int_0^T(\int_s^Tf(t)dt)H(s)dW_s$$

then with interverting of the order of integration in the second integral we can write $$\int_0^T(\int_s^Tf(t)dt)H(s)dW_s=\int_0^T(\int_0^tH(s)dW_s)f(t)dt$$

The process apperaing in the integrals are now adapted, and you can add the condition so that Fubini works and to justify the martangality of the integrals. Hope thsi help.

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If $f(t)$ is a deterministic function this procedure works. In this case, I'm not completely sure! Because I don't know how to prove that $h(s)$ is adapted to the filtration generated by the Brownian motion. Thank you for your help!! – Phoebe Jan 28 '13 at 17:14
I edited the first answer to take into account your comment – Hicham Jan 28 '13 at 17:45
Thank you.....I think this is the right way to proceed. Definitely, this expected value cannot be zero. – Phoebe Jan 29 '13 at 11:07

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