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 !

  • $\begingroup$ I think there is an $H_s$ missing in the definition of $X_t$ $\endgroup$ – Justpassingby Nov 29 '15 at 17:32

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.