covariance of integral of Brownian 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!
 A: $$\mathbb E(X(t)X(s))=\int_0^t\int_0^s\mathbb E(B(u)B(v))\,\mathrm dv\,\mathrm du=\int_0^t\int_0^s\min\{u,v\}\,\mathrm dv\,\mathrm du$$
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.
A: As @Did explained in his answer, the covariance of the integrated Brownian motion is obtained by integrating the covariance of the usual Brownian motion. However, he refrained from computing said integral, which led to some confusion in the comments.
To evaluate the integral, we make use of the identity $\min\{u,v\}=u-1[v<u](u-v)$. Recall that we are assuming $0<s<t<\infty$. Therefore
$$
\int_0^s\int_0^t \min\{u,v\}\ dv \ du=\int_0^s\int_0^t u\ dv \ du-\int_0^s\int_0^u u-v\ dv \ du=\frac{ts^2}{2}-\frac{s^3}{6}.
$$
Note that @martin's calculation in the comments is incorrect: it consists of just the first term above, and comes from assuming that $u\leq v$ throughout the region of integration (which is false).
A: $$E[x(s)x(t)] = \int_0^t \int_0^s E\bigg(B(u)B(v)\bigg)dvdu$$
holds true for the linearity of expection. It seems Fubini's theorem in the arguement of @ did comes to treat expectation operator as integral which different from the theory in the context of measure theory which proposes
$$E(X)=E_{P1}\big(E_{P2}(X)\big)=E_{P2}\big(E_{P1}(X)\big)$$
if $X$ is jointly measurable and either nonnegative or jointly integrable.
