I am trying to solve following expectation
$E \left [B^2_s \left( \int^t_s B_u dB_u \right)^2 \right]$
with $0 \leq s \leq t \leq T$ and $B_t$ a 1-dim. Brownian motion. Further using $E \left[ . \vert \mathcal{F}_s \right]$ is a hint.
I started by introducing the conditional expectation and seperating the integral
$E \left[ B^2_s E \left[ \left( \int^t_0 B_u dB_u - \int^s_0 B_u dB_u \right)^2 \vert \mathcal{F}_s \right] \right]$
I am not sure about the next step, but I think $\left( \int^t_0 B_u dB_u \right)_t$ are square integrable martingales and I can write
$E \left[ B^2_s E \left[ \left( \int^t_0 B_u dB_u \right)^2 - \left( \int^s_0 B_u dB_u \right)^2 \vert \mathcal{F}_s \right] \right]$
and with $\int^t_0 B_u dB_u = \dfrac{1}{2}(B^2_t - t)$
$E \left[ B^2_s E \left[ \left( \dfrac{1}{2}(B^2_t - t) \right)^2 - \left( \dfrac{1}{2}(B^2_s - s) \right)^2 \vert \mathcal{F}_s \right] \right]$
Any next step of making the equation easier did not have any real result.
Thank you for your help !
Fix $s$ and condition by $\mathcal F_s$, then $M_t=\displaystyle\int_s^tB_sB_u\mathrm dB_u$ defines a martingale with quadratic variation process $\langle M,M\rangle_t=\displaystyle\int_s^t(B_sB_u)^2\mathrm du$ hence the expectation to be computed is $$ (\ast)=E(M_t^2)=E(E(M_t^2\mid\mathcal F_s))=E(\langle M,M\rangle_t)=\int_s^tE(B_s^2B_u^2)\mathrm du. $$ Now, for every $u\geqslant s$, $B_u^2=B_s^2+2B_sW+W^2$ where $W=B_u-B_s$ is independent of $B_s$ with $E(W)=0$ and $E(W^2)=u-s$, hence $$ (\ast)=\int_s^t(E(B_s^4)+(u-s)E(B_s^2))\mathrm du. $$ Plug in $E(B_s^2)=s$ and $E(B_s^4)=3s^2$ in the last integral to conclude that, unless I am mistaken, $$(\ast)=\tfrac12s(t+5s)(t-s).$$