This is my first question and I hope it is ok :)
Reading in a book I came along this answer to a question I did not understand and I would like to understand this very much:
Problem: Calculate the mean and variance of the Itô integral $\int_0^tW_s^2dWs$ with $W_s$ being an Brownian motion.
The solution says that $I_t=\int_0^tW_s^2dW_s$, that Itô integrals are martingales and that therefore $E(I_t)=I_0=0$, I get this part. I also understand that $$Var(I_t)=E(I_t^2)=E(\int_0^tW_s^2ds)$$ by Itô isometry and that $E(W_s^4)=3s^2$ and therefore $Var(I_t)=t^3$. I do not understand that last part unfortunatly. How to we get to the variance?
Also how is it that we can get from the $E(W_s^4)$ to $E(\int_0^tW_s^2ds)$ and how did we use the facts $E(I_t)=I_0=0$ in the beginning?
The process $I=\{I_t: t \geq 0\}$ is a martingale and thus (check out the properties of a martingale) the expected value remains constant, i.e., $$\forall t \geq 0, \ \ \mathbb{E}[I_t]=\mathbb{E}[I_0]=\mathbb{E}[0]=0 $$ The variance of $I_t$ is then given by $$Var(I_t)=\mathbb{E}[I_t^2]-\mathbb{E}[I_t]^2=\mathbb{E}[(\int_0^tW_s^2 \, ds)^2]-0=\int_0^t\mathbb{E}[W_s^4] \, ds =\int_0^t 3s^2 \, ds = t^3$$ where we used Itô's isometry which states that $$\mathbb{E}[(\int_0^tY_s \, ds)^2]=\int_0^t\mathbb{E}[Y_s^2] \, ds$$