Expectation problem of a stochastic integral

Question

I have a question for the following equation: $$ E[B_t^2 \int_{0}^{t} 2s B_s dB_s]= 4 \int_{0}^{t}s^2 ds =4/3t^3 $$ i know about the Ito isometry formula but I don't know what happened with this $B_t^2$. I hope someone can explain me this, thank you.

Answer 1

You have by Ito's formula: \begin{equation} B_t^2 = \int_0^t 2B_sdB_s + t \end{equation} Hence, \begin{align} E[B_t^2\int_0^t 2 sB_sdB_s] &= E[(\int_0^t 2B_sdB_s + t)\int_0^t 2sB_sdB_s]\\ &= E[(\int_0^t 2B_sdB_s\int_0^t 2sB_sdB_s)] + t\underbrace {E[\int_0^t 2sB_sdB_s]}_{=0} \\ &=4\int_0^t sE[B_s^2]ds \quad \text{By isometry} \\ &= 4 \int_0^ts^2ds.\\ \end{align} Note that we have $E[\int_0^t 2B_sdB_s]=0$ (Explain why ?)