# 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!

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$$\mathbb E(X(t)X(s))=\int_0^t\int_0^s\mathbb E(B(u)B(v))\,\mathrm dv\,\mathrm du=\int_0^t\int_0^s\min\{u,v\}\,\mathrm dv\,\mathrm du$$ 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.
Thank you very much :) That's exactly what I did and I got $\frac{1}{2}s^2t$, but shouldn't it give me something "symmetric"? I mean, if I exchange $s$ by $t$ then get the same answer? Since covariance matrices are symmetric. Thanks :) –  martin Nov 20 '12 at 22:56
Once you assume that $s\lt t$, the setting is not symmetric with respect to $(s,t)$ anymore. I guess your formula is really $\frac12\min(s,t)^2\max(s,t)$, which is symmetric (but I did not check this was indeed the result). –  Did Nov 21 '12 at 6:18
@TheBridge Right, post modified. (Unrelated: beware of the space between @ and user's name.) –  Did Nov 21 '12 at 8:57