I am studying the Itô-integral and I am trying to solve the following with the definition of the integral (so $\int_{0}^{t}u_{s}\,dB_{s}$ is the $L^2$-limit of $\sum_{t_j} u^n_{t_j}(B_{t_j}-B_{t_{j-1}})$ for a partition $0=t_1<...< t_n= t$ and some step (or simple) process $u^n$, which is approximating $u_s$).
Let $X_{t}=\int_{0}^{t}e^{s}\,dB_{s}$, and $Y_{t}=\int_{0}^{t}X_{s}\,dB_{s}$. I want to compute $E(Y_{t})$ and $E(Y_{t}^{2})$. I know $E(Y_{t})=0$ since stochastic integrals are centered random variables, but how do I get $E(Y_{t}^{2})$? Can somebody give me a hint?
Using Ito Isometry, we have:
$$\mathbb{E}[X_t^2]=\mathbb{E}\left[\left(\int_0^t e^s dW_s\right)^2\right]=\mathbb{E}\left[\int_0^t (e^s)^2 ds\right]=\frac{1}{2}\int_0^t2e^{2s}ds=0.5(e^{2t}-1)$$
So the variance of $X_t$ is $\mathbb{E}[X_t^2]-\mathbb{E}[X_t]^2=0.5(e^{2t}-1)$ and the expected value of $X_t$ is $\mathbb{E}[X_t]=0$ by the martingale property of Ito Integral. So $X_t$ is normally distributed with mean zero and variance $0.5(e^{2t}-1)$.
Now, again using Ito Isometry:
$$\mathbb{E}[Y_t^2]=\mathbb{E}\left[\left(\int_0^t X_s dW_s\right)^2\right]=\mathbb{E}\left[\int_0^t (X_s)^2 ds\right]=\int_0^t\mathbb{E}[X_s^2]ds=0.5\int_0^t(e^{2s}-1)ds=0.5[0.5e^{2s}-s]_0^t=0.5[0.5e^{2t}-1-t]$$