most recent 30 from http://mathoverflow.net 2013-05-21T16:24:04Z http://mathoverflow.net/feeds/question/120124 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120124/compute-the-expected-value-of-the-product-between-a-lebesguestieltjes-type-integ Phoebe 2013-01-28T16:24:54Z 2013-01-28T17:44:39Z

<p>Hi, I have the following expected value to compute</p> <p>$E[ \int_{o}^{T} f(t) dt \int_{o}^{T} H(s) dW(s)]$,</p> <p>where $f(t)$ and $H(s)$ are two stochastic processes adapted to the filtration generated by the Brownian motion W.</p> <p>I think that this expected value could be equal to zero, but I really don't know how to give this proof.</p> <p>Thank you in advance for any kind of advice or references.</p> <p>Imma</p>

http://mathoverflow.net/questions/120124/compute-the-expected-value-of-the-product-between-a-lebesguestieltjes-type-integ/120125#120125 Hicham 2013-01-28T16:38:19Z 2013-01-28T17:44:39Z

<p>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 $E(\int_{o}^{T} h^2(s) ds)&lt;+\infty$. </p> <p>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</p> <p>$$\int_0^T(\int_0^Tf(t)dt)H(s)dW_s =\int_0^T(\int_0^sf(t)dt)H(s)dW_s+\int_0^T(\int_s^Tf(t)dt)H(s)dW_s$$</p> <p>then with interverting of the order of integration in the second integral we can write $$\int_0^T(\int_s^Tf(t)dt)H(s)dW_s=\int_0^T(\int_0^tH(s)dW_s)f(t)dt$$</p> <p>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.</p>