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In an interview I have been asked to solve: $$\mathbb{E}\left[\left(\dfrac{1}{T}\int^T_0W_tdt\right)^2\right]$$

$\textbf{Attempt:}$

$$ \begin{align} \mathbb{E}\left[\left(\dfrac{1}{T}\int^T_0W_tdt\right)^2\right] &= \dfrac{1}{T^2}\mathbb{E}\left[\left(\int^T_0W_tdt\right)\left(\int^T_0W_udu\right)\right]\\ &= \dfrac{1}{T^2}\mathbb{E}\left[\int^T_0\int^T_0W_tW_udtdu\right]\\ &= \dfrac{1}{T^2}\int^T_0\int^T_0\mathbb{E}\left[W_tW_u\right]dtdu \\ &= \dfrac{1}{T^2}\int^T_0\int^T_0min(u,t)\,dtdu \end{align} $$

Then, I couldn't continue. What would be the next steps?

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  • 2
    $\begingroup$ Assuming the first integral is done on $u$, split the second integral as an integral on $[0,u]$ and one on $[u,T]$ $\endgroup$ – dan_fulea Apr 7 at 18:01
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Thanks to @dan_fulea hint, here are the next steps to the solution:

$$ \begin{align} \mathbb{E}\left[\left(\dfrac{1}{T}\int^T_0W_tdt\right)^2\right] &= \dfrac{1}{T^2}\int^T_0\int^T_0min(u,t)\,dtdu \\ &= \dfrac{1}{T^2}\int^T_0\left(\int^u_0min(u,t)\,dt+\int^T_umin(u,t)\,dt\right)du \\ &= \dfrac{1}{T^2}\int^T_0\left(\int^u_0t\,dt+\int^T_uu\,dt\right)du\\ &= \dfrac{1}{T^2}\int^T_0\left(\dfrac{u^2}{2}+u(T-u)\right)du\\ &= \dfrac{1}{T^2}\int^T_0\left(uT-\dfrac{u^2}{2}\right)du\\ &= \dfrac{1}{T^2}\times \left(\dfrac{T^3}{2}-\dfrac{T^3}{6}\right)\\ &= \dfrac{T}{3} \end{align} $$

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