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My problem is the following : let $B_t$ be a standard Brownian motion and $H_t$ a progressive measurable process such that $\mathbb{E}\left(\int_0^{+\infty} H_t^2\ dt \right)<+\infty$. Denote $X_t=\int_0^t H_s\ d Bs$ and $X_{\infty}= \int_0^{\infty} H_s\ d B_s$. Compute $\mathbb{E} \left(X_{\infty} | \mathcal{F}_t\right)$.

My idea would be to write $X_{\infty}=\lim_{t \to \infty} X_t$ (this should be true a.s.) and exploit the fact that $X_t$ is a martingale (I have prove it). Can I swap limit and conditional expectation ? The expectation is basically an integral and therefore we could use, for example, Lebesgue dominated convergence theorem, but I do not think that the assumptions are fulfilled...it does not seem to me that $X_t$ is bounded, even if its limit exists in $L^2$.

Thank you very much for any help you can provide !

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  • $\begingroup$ I think there is an $H_s$ missing in the definition of $X_t$ $\endgroup$ – Justpassingby Nov 29 '15 at 17:32
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I would decompose the integration interval defining $X_\infty$ into the part that is measurable with respect to $\cal F_t$, i.e., $X_t,$ and the rest, which is $\int_t^\infty H_s\ dB_s.$ Since Brownian motion is a martingale the second term has zero conditional expectation w.r.t. $\cal F_t$ so the first term is the conditional expectation.

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