Limits of integration to find $\int_0^tB_s^2\text{d}s$

Question

I am currently attempting to find the variance of $\int_0^tB_s^2\text{d}s$. In particular, I am confused about the integral of $\mathbb{E}\left(\int_0^tB_s^2\text{d}s\right)^2$. I know that repeated applications of Fubini's theorem leads us to $\mathbb{E}\left(\int_0^tB_s^2\text{d}s\right)^2=\int_0^t\int_0^t\mathbb{E}(B_r^2B_s^2)^2\text{d}r\text{d}s$. I also know that $\mathbb{E}(B_r^2B_s^2)^2=2s^2+rs$. What I don't understand is why we cannot integrate this directly, i.e. $$\int_0^t\int_0^t\mathbb{E}(B_r^2B_s^2)^2\text{d}r\text{d}s=\int_0^t\int_0^t2s^2+rs\text{d}r\text{d}s=\int_0^t\frac{2t^3}3+\frac{rt^2}2\text{d}r=\frac{11t^4}{12}.$$ I have read from these links here, here and here that I should split the integral when $r<s$ and $s<r$, and use the symmetry of the integrand to obtain $$\int_0^t\int_0^t\mathbb{E}(B_r^2B_s^2)^2\text{d}r\text{d}s=2\int_0^t\int_0^s2s^2+rs\text{d}r\text{d}s=\frac{7t^4}{12}.$$ Perhaps I have overlooked something from my elementary multivariable calculus, but I really can't think of any reason why. Help is appreciated, thanks!

Answer 1

It is not true that $$\mathbb{E}(B_r^2 B_s^2) = 2s^2 + rs$$ You can see this because the expression on the left of the equality sign is symmetric in $r$ and $s$ while the expression on the right is not.

Rather, this equality holds for $s \leq r$. I.e.

$$\mathbb{E}(B_r^2 B_s^2) = \begin{cases}2s^ 2 + rs & s \leq r \\ 2r^ 2 + rs & r \leq s\end{cases}$$

Now, it should be obvious why you have to split the limits of integration.