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

Question

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.

Answer 1

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?