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my ultimate goal is to compute the auto-covariance of the time-integrated Ornstein-Uhlenbeck process which has an initial value that is drawn from a Gaussian distribution with the same variance as the long-term variance of the OU process, however my following question is rather generic.

Let $Y_t = \int_0^t X_s dt$. Then:

\begin{equation} Cov(Y_t,Y_u) = Cov(\int_0^t X_s ds, \int_0^u X_s' ds')\\ = E[\int_0^t X_s ds \int_0^u X_s' ds'] - E[\int_0^t X_s ds] E[\int_0^u X_s' ds']\\ = E[\int_0^t \int_0^u X_s X_s' ds' ds] - E[\int_0^t X_s ds] E[\int_0^u X_s' ds'] \end{equation}

So far I believe that I did not make any noteworthy assumptions. Beforehand I computed the case where $E[X_t]=0$. In this case the second part of the sum is zero.

Making use of Fubini's theorem to put the expectation into the integral, I obtain \begin{equation} Cov(Y_t,Y_u) = \int_0^t \int_0^u E[X_s X_s'] ds' ds \end{equation}

using the definition of the covariance in reverse order, this can again be expanded to \begin{equation} Cov(Y_t,Y_u) = \int_0^t \int_0^u Cov(X_s,X_s') + E[X_s] E[X_s'] ds' ds = \int_0^t \int_0^u Cov(X_s,X_s') \end{equation}

Hence, if the expected value is zero, I can compute the auto-covariance of the time-integrated process by integration of the original auto-covariance.

Now I want to consider an expected value different from zero. From my current viewpoint the relationship \begin{equation} Cov(Y_t,Y_u) = E[\int_0^t \int_0^u X_s X_s' ds' ds] - E[\int_0^t X_s ds] E[\int_0^u X_s' ds'] \end{equation} should still hold.

What follows however cannot be correct and I would like to ask for the error in the computation and advice for the correct computation (possibly with respect to the Ornstein-Uhlenbeck process as $X_t$).

Please find the error in the following: Using Fubini, we obtain \begin{equation} Cov(Y_t,Y_u) = \int_0^t \int_0^u E[X_s X_s'] ds' ds - \int_0^t E[X_s] ds \int_0^u E[X_s'] ds'\\ = \int_0^t \int_0^u E[X_s X_s'] ds' ds - \int_0^t \int_0^u E[X_s] E[X_s'] ds'ds \end{equation}

Using the definition of the covariance results in \begin{equation} Cov(Y_t,Y_u) = \int_0^t \int_0^u Cov(X_s,X_s') + E[X_s] E[X_s'] ds' ds - \int_0^t \int_0^u E[X_s] E[X_s'] ds'ds \end{equation} Splitting the integral then leads to \begin{equation} Cov(Y_t,Y_u) = \int_0^t \int_0^u Cov(X_s,X_s') ds' ds + \int_0^t \int_0^u E[X_s] E[X_s'] ds' ds - \int_0^t \int_0^u E[X_s] E[X_s'] ds'ds \end{equation} which brings me to the same final relationship \begin{equation} Cov(Y_t,Y_u) = \int_0^t \int_0^u Cov(X_s,X_s') \end{equation}

Again, I strongly suspect that this relationship is only valid for a non-zero expected value, however I also obtain it for a non-zero expected value. Hence: which step is wrong?

If possible, I also would like to know how to tackle in a next step the varying $X_0$, if I draw if from a Gaussian distribution with the long-term variance of the OU process.

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