I'm new to stochastic and trying to do the exercise of finding the variance and mean of $\int_{0}^{t} s^2W_sdW_s$, where $W_s$ is a Brownian motion.
Please, help to understand the way of solving this. This is how I approach it:
For mean: As $s^2W_s$ is a martingale (is it? not really sure how to prove) $ \ \mathbb{E}[s^2W_s]=0$
For variance: $Var(I_t)=\mathbb{E}[I_t^2]-\mathbb{E}[I_t]^2=\mathbb{E}[(\int_{0}^{t} s^2W_sdW_s)^2] - 0$
By applying Itô isometry:
$\mathbb{E}[(\int_{0}^{t} s^2W_sdW_s)^2] = \mathbb{E}[\int_0^t(s^2W_s )^2\, ds]=\int_0^ts^4\mathbb{E}[W_s^2] \,ds =\int_0^t s^5 \, ds = \frac{t^6}{6}$
.
Am I correct?
You are almost there, here is solution filling out some of the gaps. Using the martingale property of the Ito integral (the below variance calculation implies that the relevant square integrability condition is satisfied):
$$\mathbb{E}[\int_{0}^{t} s^2W_sdW_s]=0$$
Using Ito isometry:
$\mathbb{E}[(\int_{0}^{t} s^2W_sdW_s)^2] = \mathbb{E}[\int_0^ts^4W_s^2\, ds]$
Using Fubini (the above can be understood as a Riemann integral) we obtain:
$\int_0^ts^4\mathbb{E}[W_s^2] \,ds =\int_0^t s^5 \, ds = \frac{t^6}{6}$