I am doing the following exercise:
The thing I am struggling with is the identity given in the hint: $$ E\left( \int^t_0 X_s ds \right)^2 = \int^t_0 \int^t_0 E(X_s X_u)\,ds\, du $$ I am unable to verify this. I have tried various starting points like writing $$ E\left( \int^t_0 X_s ds \right) =\int_{\Omega} \int^t_0 X_s \, ds \, d\mathbb P $$, looking at Fubini's theorem etc, but I seem to be missing the correct starting point. Much appreciated if someone could show me this.
Write the squared integral as $$ \left( \int_0^t X_s \, ds \right)^2 = \left( \int_0^t X_s \, ds \right)\left( \int_0^t X_u \, du \right) = \int_0^t \int_0^t X_s X_u \, ds \, du, $$ where I have applied Fubini's theorem since the intervals are finite and the functions are integrable. Now use the definition of expectation: $$ \mathbb{E}\left( \int_0^t X_s \, ds \right)^2 = \mathbb{E}\left[\int_0^t \int_0^t X_s X_u \, ds \, du \right] = \int_{\Omega} \left( \int_0^t \int_0^t X_s X_u \, ds \, du \right) \, d\mathbb{P} $$ Now all you have to do is to justify using Fubini again, to push the $\mathbb{E}$ integral inside.