Expectation of a squared Ito integral

Question

I want to evaluate the following expectation: $$ \mathbb{E}\left[ \left(\int_0^T W_t^2 d W_t \right)^2\right], $$ where $W_{t}$ is a Brownian motion. For now, I have $$ \left(\int_{0}^{T}W^{2}_{t}\, d W_{t}\right)^2=\frac {1}{9}W^{6}_{T}+\left(\int_{0}^{T}W_{t}\, d t\right)^2-\frac {2}{3}W^{3}_{T}\int_{0}^{T}W_{t}\, d t. $$ I think I can deal with the $\frac {1}{9}W^{6}_{T}$ part. There are two issues I don't know how to approach. One is that I don't know the variance of $\int_{0}^{T}W_{t}\, d t$. The other one is about how to compute the expectation of $\frac {2}{3}W^{3}_{T}\int_{0}^{T}W_{t}\, d t$.

Answer 1

$\newcommand{E}{\mathbb E}$

For the variance-term, use $$ \left(\int_0^T W_t dt\right)^2 = \int_0^T \int_0^T W_tW_sdtds $$ whence $$ \mathbb E\left(\int_0^T W_t dt\right)^2 = \int_0^T \int_0^T \mathbb E[W_tW_s]dtds = \int_0^T \int_0^T (t \wedge s) dtds $$

Where the Fubini theorem for exchanging the expectation and double integral applies since $\int_0^T \int_0^T \mathbb E|W_tW_s|dtds < \infty$ (for that, see the argument here).

For the other , imagine we had a function $f(t)$ in place of $W_t$. The resulting integral would look like $\frac 23 f(T)^3 \int_0^T f(t)dt$. Now, $f(T)^3$ is just a constant, therefore one can slip it inside and write $\frac 23 \int_0^T f(T)^3f(t)dt$. Since the integral of a Brownian motion evaluated at a sample point $\omega$ is just the integral of some continuous function, the same logic holds on an $\omega$-by-$\omega$ basis and therefore $$ \frac 23 W_T^3 \int_0^T W_tdt = \frac 23 \int_0^T W_T^3W_tdt $$ whence $$ \mathbb E\left[\frac 23 W_T^3 \int_0^T W_tdt\right] = \frac 23 \mathbb E\left[\int_0^T W_T^3W_tdt\right] = \frac 23\int_0^T \E[W_T^3W_t]dt $$ where we can exchange the expectation and integral using the Fubini theorem once again , because $\int_0^T \mathbb E|W_T^3W_t| dt<\infty$. To see this, use an argument very similar to the one earlier: $$ |W_T^3W_t| = |(W_t + (W_T- W_t))^3W_t| \\ \leq |W_t^4| + 3|W_t^3||W_T-W_t| +3|W_t^2||(W_T-W_t)^2| + |W_t||(W_T-W_t)^3| $$

and the expectation of the RHS is some polynomial in $t$ , which is obviously integrable on $[0,T]$. Once this is done, you can use the fact that $$ \E[W_T^3W_t] = \E[(W_t +(W_T-W_t))^3W_t] \\ = \E[W_t^4] + 3\E[W_t^3]\E[W_T-W_t] + 3\E[W_t^2]\E[(W_T-W_t)^2] + \E[W_t]\E[(W_T-W_t)^3] $$

using the independence of increments. Using the usual formulas, this comes to $3t^2 + 3t(T-t)$, which you can integrate over $[0,T]$ and multiply by $\frac 23$ to get the value of that respective term.