I'm trying to calculate this: $\mathbb{E} \left[\displaystyle \left(\int_0^t B_s^2 ds\right)^2\right]$, where $B_s$ is Brownian process with zero mean and variance $s$. I tried Ito isometry, but that's not useful here.
An idea would be calculating $\displaystyle \int_0^t B_s^2 ds$, raising to power 2 and then taking expectation, but I don't know how to calculate $\displaystyle \int_0^t B_s^2 ds$, because we can't use integration by part here.
Is there any hint?
$$ \begin{split} E\left(\int_0^t B^2_sds\times \int_0^t B^2_udu\right) &= 2E\left(\int_0^t \left(\int_0^uB^2_sds\right) B^2_udu\right) \\ &= 2\int_0^t\left(\int_0^uE(B^2_sB^2_{u})\right)dsdu \\ &= 2\int_0^t\left(\int_0^uE(B^2_s(B_{s}+B'_{u-s})^2\right)dsdu \\ &=2\int_0^t\left(\int_0^uE(B^4_s+2B^3_sB'_{u-s}+B^2_sB'^2_{u-s})\right)dsdu\\ &=2\int_0^t\left(\int_0^u(3s^2+0+s(u-s))\right)dsdu \\ &=2\int_0^t\left(\int_0^u(2s^2+su)\right)dsdu \\ &= \frac{7}{12}t^4. \end{split} $$