How to get $E[\int_{t}^T2B_sdB_s|\mathcal{F}_t]=0?$

Question

I am computing the following Ito integral $$ E[\int_{0}^T2B_sdB_s|\mathcal{F}_t]=E[\int_{0}^t2B_sdB_s+\int_{t}^T2B_sdB_s|\mathcal{F}_t] $$ $$ =E[\int_{0}^t2B_sdB_s|\mathcal{F}_t]+E[\int_{t}^T2B_sdB_s|\mathcal{F}_t] $$ the first term $\int_{0}^t2B_sdB_s$ is independent of $\mathcal{F}_t$, then $E[\int_{0}^t2B_sdB_s|\mathcal{F}_t]=\int_{0}^t2B_sdB_s$.

But how to get $$ E[\int_{t}^T2B_sdB_s|\mathcal{F}_t]=0? $$

Answer 1

Use Ito on $B_t^2$, and you find out that $d(B_t^2)=2B_tdB_t+dt\implies \int_{[t,T]} 2B_sdB_s=B_T^2-B_t^2-(T-t)$. By using the fact that $E[B_T^2-B_t^2|\mathscr{F}_t]=(T-t)$, the claim follows.