A squared integrals

Question

Is it true that $$ \left(\int_{0}^{t}f(u)du\right)^{2}=\left(\int_{0}^{t}f(u)du\right)\cdot\left(\int_{0}^{t}f(v)dv\right)=\int_{0}^{t}\int_{0}^{t}f(u)f(v)dudv\;? $$ I have tried to prove it with Cauchy-Bunyakovsky-Schwarz inequality, but I couldn't. The original problem is to determine $$ \mathbb{D}^{2}\left(\int_{0}^{t}W(s)ds\right) $$ where $W(t)$ is a Wiener process. I know $$ \mathbb{D}^{2}\left(\int_{0}^{t}W(s)ds\right)=\mathbb{E}\left(\left(\int_{0}^{t}W(s)ds\right)^{2}\right)-\underbrace{\mathbb{E}^{2}\left(\int_{0}^{t}W(s)ds\right)}_{0}=\mathbb{E}\left(\left(\int_{0}^{t}W(s)ds\right)\cdot\left(\int_{0}^{t}W(u)du\right)\right). $$ Our teacher has written it is $$ \mathbb{E}\left(\left(\int_{0}^{t}W(s)ds\right)\cdot\left(\int_{0}^{t}W(u)du\right)\right)=\mathbb{E}\left(\int_{0}^{t}\int_{0}^{t}W(s)W(u)duds\right), $$ but I don't really see why it is true.

Answer 1

You can always pull out a constant from an integral.

$\displaystyle \int_0^t \alpha f(u)du=\alpha \int_0^t f(u)du\quad$ for $\quad\displaystyle\alpha=\int_0^t f(v)dv$.

This is because $t$ is independent from $u,v$ so both integrals are just constants in regard to the other.

This would not be true, if the bound was not independent, see this result:

• Is the integral squared equal to two times the integral from $a$ to $b$ and from $x$ to $b$?