Let B be a standard brownian motion.
NOTE: in the exercise is suggeste to use that if X is$ N(o,\sigma^2)$ then$ E[X^4]=3\sigma^4$, so i would like to use this result.
I would like to compute $$E\bigg[B_s^2\big(\int_s^tB_udB_u\big)^2\bigg]$$
I started in the following way:
$$E\bigg[B_s^2\big(\int_s^tB_udB_u\big)^2\bigg]=E\Bigg[B_s^2E\bigg[\big(\int_s^tB_udB_u\big)^2\bigg|\mathcal F_s\bigg]\Bigg]=\frac14E\Bigg[B_s^2E\bigg[\big(B_t^2-t-(B_s^2-s) \ \big)^2\bigg|\mathcal F_s\bigg]\Bigg]=\frac14E\Bigg[B_s^2E\bigg[(B_t^2-t)^2-(B_s^2-s)^2\bigg|\mathcal F_s\bigg]\Bigg]$$
since by Ito $$\int_0^tB_udB_u- \int_0^sB_udB_u=\frac12[B_t^2-t-(B_s^2-s)]$$ and the last equality follows by property of the square integrable martingale
how could i proceed then?
Best Answer
Proceed as follows,
$$I=E\bigg[B_s^2\big(\int_s^tB_udB_u\big)^2\bigg]=E\Bigg[B_s^2E_s\bigg[\big(\int_s^tB_udB_u\big)^2\bigg]\Bigg]=E\Bigg[B_s^2E_s\bigg[\int_s^tB_u^2du\bigg]\Bigg]$$
Use Ito's rule
$$d(B_u^2)=2B_udB_u + du,\implies B_u^2=B_s^2+2\int_s^uB_\tau dB_\tau+(u-s)$$
Then,
$$E_s\bigg[\int_s^tB_u^2du\bigg]=\int_s^tB_s^2du+\int_s^t(u-s)du=B_s^2(t-s)+\frac12(t-s)^2$$
Thus, $$I=E\Bigg[B_s^2[B_s^2(t-s)+\frac12(t-s)^2]\Bigg]=3s^2(t-s)+\frac12s(t-s)^2$$ where $E[B_s^4] = 3s^2$ and $E[B_s^2] =s$ are used.