Use Itō isometry to calculate $\mathbb{E}X_{t}^{2}$, where $X_{t}= \int_{0}^{t}B_{s}{\boldsymbol d}B_{s}$

Question

Use Itō isometry to calculate $\mathbb{E}X_{t}^{2}$, where $$X_{t}= \int_{0}^{t}B_{s}{\boldsymbol d}B_{s}$$

The process $B_{t}$ itself is an adapted process. Recall that $$\int_{0}^{t}B_{s}{\boldsymbol d}B_{s}= \frac{1}{2}B_{t}^{2}- \frac{t}{2}$$ and $\mathbb{E}\left [ B_{t}^{2} \right ]= t$. Hence $$\mathbb{E}\left [ \int_{0}^{t}B_{s}{\boldsymbol d}B_{s} \right ]= 0$$ How should I continue here ?

Answer 1

I believe that neither the accepted answer nor the beginning of the proof of the OP are acceptable.

The exercise explicitly asked to use Ito Isometry and neither did that. I'd do it like this:

$$ \mathbb{E}\bigg[ \bigg( \int_{0}^{t} B_s dB_s \bigg)^{2} \bigg] =\mathbb{E}\bigg[ \int_{0}^{t} B_{s}^{2} dB_s \bigg] = \int_{0}^{t} \mathbb{E}[B_{s}^{2}] ds = \int_{0}^{t}s ds = \frac{t^2}{2} $$

Answer 2

From the hint, we know that $$ 2X_{t}=2\int_{0}^{t}B_{s}dB_{s}=B_{t}^{2}-t. $$ Squaring both sides yields $$ 4X_{t}^{2}=\left(B_{t}^{2}-t\right)^{2}=B_{t}^{4}-2tB_{t}^{2}+t^{2}. $$ Taking expectations yields $$ 4\mathbb{E}[X_{t}^{2}]=\mathbb{E}[B_{t}^{4}]-2t\mathbb{E}[B_{t}^{2}]+t^{2}. $$ You know that $B_{t}$ is normally distributed with mean zero and variance $t$. Can you finish?