# Compute $E\left((B_t−1)^2\int ^t_0(B_s+1)^2 dB_s\right)$, where $(B_t)$ is a standard Brownian motion

Compute $$E((B_t−1)^2\int ^t_0(B_s+1)^2 dB_s)$$ for $$t≥0$$ given that $$(B_t)_{t≥0}$$ is a Standard Brownian Motion.

Presume we will need to compute $$E((B_t+B_s)-(B_s-1))^2$$ to get some independent terms but really stuck on what to do with the integral part. Thanks for any help with this question.

• @Did could you clarify why $E(B_{t}X_{t})=E(\int_0^tY_{s}ds)$. – Zugzwangerz Jan 10 at 15:35

Consider the processes $$X_t=\int ^t_0Y_s dB_s\qquad Y_t=(B_t+1)^2$$ By repeated applications of Itô isometry, one gets:
• $$E(X_t)=0$$
• $$B_tX_t=\displaystyle\int_0^tdB_s\cdot\int ^t_0Y_sdB_s$$ hence $$E(B_tX_t)=E\left(\int_0^tY_s ds\right)=\int_0^tE(Y_s)ds$$
• $$(B_t^2-t)X_t=\displaystyle\int_0^t2B_sdB_s\cdot\int ^t_0Y_sdB_s$$ hence $$E((B_t^2-t)X_t)=E\left(\int_0^t2B_sY_s ds\right)=2\int_0^tE(B_sY_s)ds$$
Finally, if one can compute $$E(Y_t)$$ and $$E(B_tY_t)$$, the proof is complete. Can you?
• Like at every other step, use Itô isometry, based on the fact that $$d\langle B,B\rangle_t=dt$$ hence, for every suitable processes $(u_t)$ and $(v_t)$, $$E\left(\int_0^tu_sdB_s\cdot\int_0^tv_sdB_s\right)=E\left(\int_0^tu_sv_sds\right)$$ – Did Jan 10 at 15:38