Why does $\int_0^1 \int_0^1 \text E \left(W^2(s) W^2(t) \right) \,ds\,dt = 2 \int_0^1 \int_0^t \text E \left(W^2(s) W^2(t) \right) \,ds \,dt$

Question

Let $W(t)$ denote the standard Wiener Process, i.e. $W(t) \sim \text N(0, t)$ for $t \geq 0$.

I have shown that

$$\text E \left(W^2(s) W^2(t) \right) = st + 2\left[\min(s,\, t)\right]^2$$ and I used it to show that

\begin{align} \int_0^1 \int_0^1 \text E \left(W^2(s) W^2(t) \right) \,ds \,dt &= \int_0^1 \left[ \int_0^t \text E \left(W^2(s) W^2(t) \right) \,ds +\int_t^1 \text E \left(W^2(s) W^2(t) \right) \,ds \right] \,dt \\ &= \int_0^1 \left[ \int_0^t \left( st + 2s^2 \right) \,ds +\int_t^1 \left( st+2t^2 \right) \,ds \right] \,dt \\ &= \frac{7}{12} \end{align}

It turns out that

\begin{align} 2 \int_0^1 \int_0^t \text E \left(W^2(s) W^2(t) \right) \,ds \,dt &= 2 \int_0^1 \int_0^t \left( st + 2s^2 \right) \,ds \,dt \\ &= \frac{7}{12} \end{align}

Question:

I am hoping someone could provide some insight as to why $\int_0^1 \int_0^1 \text E \left(W^2(s) W^2(t) \right) \,ds \,dt = 2 \int_0^1 \int_0^t \text E \left(W^2(s) W^2(t) \right) \,ds \,dt$

Answer 1

The integrand is symmetric in $s$ and $t$.

The second integral only integrates over the region where $s<t$