Using Ito formula, I need to calculate
$$E(\int_0^ts^nB_sd_s)^2.$$
So, I denote $\ X_t \ $ as $ \ \int_0^ts^nB_sd_s \ $ and my $ \ f(x) = x^2, f'(x) = 2x, f''(x)=2. \ $
Ito lemma states that $f(X_t) - f(X_0) = \int_0^tf'(X_s)dX_s + {1\over2}\int_0^tf''(X_s)(dX_s)^2$.
But, actually, I do not understand, how I can compute $dX_s$ and $(dX_s)^2$. Firstly, I thought, it will be $dX_s = s^nB_sds$ and $(dX_s)^2 = s^{2n}B_s^2(ds)^2 = 0$, but I am not sure, whether it is correct.
Then I try to apply Ito lemma like this:
$$(\int_0^ts^nB_sd_s)^2 = f(X_t) = f(X_0) + \int_0^t2X_sdX_s + \int_0^ts^{2n}B_s^2(ds)^2. = 2\int_0^t(s^nB_s)^2ds + 0$$
Is it correct usage of Ito lemma?
Hence, we get $(\int_0^ts^nB_sds)^2 = 2 \int_0^t(s^nB_s)^2ds$.
But I still cannot calculate the expectation of the rhs. Actually, it seems a little bit harder for me. We cannot apply any rules about Ito isometry here, as far as I understand.