covariance of integral of Brownian

Question

What is the covariance of the process $X(t) = \int_0^t B(u)\,du$ where $B$ is a standard Brownian motion? i.e., I wish to find $E[X(t)X(s)]$, for $0<s<t<\infty$. Any ideas?

Thanks you very much for your help!

Answer 1

$$\mathbb E(X(t)X(s))=\int_0^t\int_0^s\mathbb E(B(u)B(v))\,\mathrm dv\,\mathrm du=\int_0^t\int_0^s\min\{u,v\}\,\mathrm dv\,\mathrm du$$ Edit: As @TheBridge noted in a comment, the exchange of the order of integration is valid by Fubini theorem, since $\mathbb E(|B(u)B(v)|)\leqslant\mathbb E(B(u)^2)^{1/2}\mathbb E(B(v)^2)^{1/2}=\sqrt{uv}$, which is uniformly bounded on the domain $[0,t]\times[0,s]$ hence integrable on this domain.