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. Any ideas?
Thanks you very much for your help!
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. Any ideas?
Thanks you very much for your help!
$\mathbb E(X(t)X(s))=\int_0^t\int_0^s\mathbb E(B(u)B(v))\,\mathrm dv\,\mathrm du=\int_0^t\int_0^s\min\{u,v\}\,\mathrm dv\,\mathrm du$ 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.
$E[x(s)x(t)] = \int_0^t \int_0^s E\bigg(B(u)B(v)\bigg)dvdu$ 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.