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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.

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

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  • $\begingroup$ 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 $(v_t)$, $$E\left(\int_0^tu_sdB_s\cdot\int_0^tv_sdB_s\right)=E\left(\int_0^tu_sv_sds\right)$$ $\endgroup$ – Did Jan 10 at 15:38
  • $\begingroup$ ((This is answering a comment formerly here but now on main, by the OP.)) $\endgroup$ – Did Jan 10 at 15:38
  • $\begingroup$ ah I get it now, thank you! $\endgroup$ – Zugzwangerz Jan 10 at 15:39

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