if $X(t)$ is the Ornstein-Uhlenbeck process and $Y(t)$ the time integrated OU process I am trying to calculate the autocovariance $cov(Y_t, Y_s)$.
I have a bunch of results but I don't know how to connect them.
First in this post they treat the OU process and it's also the result of the variance:
$$ Var(Y_t) = \mathbb{E} Y_t^2 = \mathbb{E} \int_0^t \int_0^t X_s X_u ds du = \int_0^t \int_0^t Cov(X_s, X_u) ds du = 2 \int_0^t \int_0^u Cov(X_s, X_u) ds du.$$
and they use $Cov(X_s, X_u) = \frac{\sigma^2}{2\theta}\left( e^{-\theta(u-s)} – e^{-\theta(u+s)} \right)$ for $s\leq u$. And in that case the result is
$$
Var(Y_t)=\frac{\sigma^2 t}{\theta^2}-\frac{3\sigma^2}{2 \theta^3}+\frac{\sigma^2}{2 \theta^3}(4 e^{-\theta t}-e^{-2 \theta t})
$$
The thing is that (if I am not wrong) the way to calculate the autocovariance is using a similar integral but with different limits.
I am not sure about this but I was thinking in $
Cov(Y_t,Y_v)=\mathbb{E} \int_0^t \int_0^v X_s X_u ds du
$
but I am not completely sure.
And in that case I don't know how to solve the integral because I have to respect the validity of the formula of the covariance for $s \leq u$
On the other hand I also found the following result[Bhattacharya] :
$$
Cov(Y_t, Y_{t+v}-Y_t)=\frac{\sigma^2}{2 \theta³}(1-e^{-\theta v})
(1-e^{-\theta t})^2
$$
And also I was thinking in using that
$$
Cov(Y_t, Y_{t+v}-Y_t) = Cov(Y_t, Y_{v+t}) – Var(Y_t)
$$
But again, I am not completely sure about this, and if this is true, there is a way to solve the integral and have the same result.
Any help would be really appreciated.
REF:
Bhattacharya, Stochastic Processes with Applications
Best Answer
It's ultimately a simple calculation but it can be hard to see if you haven't done it before.
Since everything in sight has mean zero, the covariance is given by $\mathbb{E}[Y_t Y_v]$. Without loss of generality, assume $t \ge v$. Then $$\begin{align*} \mathbb{E}[Y_t Y_v] &= \mathbb{E}\left[\int_0^t X_s\,ds \int_0^v X_u\,du\right] \\ &= \mathbb{E}\left[\int_0^v \int_0^t X_s X_u\,ds\,du\right] \\ &= \int_0^v \int_0^t \mathbb{E}[X_s X_u]\,ds\,du && \text{(Fubini)} \\ &= \int_0^v \left(\int_0^u \mathbb{E}[X_s X_u]\,ds + \int_u^t \mathbb{E}[X_s X_u]\,ds\right)\,du. \end{align*}$$ Now use the formula you know for the covariance of $X_s$ and $X_u$. In the first integral you have $s \le u$, so use your formula as is, and in the second integral $u \ge s$, so use your formula with $u$ and $s$ reversed. The calculus is then an exercise.
Justifying the use of Fubini's theorem is also an exercise. The Cauchy–Schwarz inequality might be a convenient way to go.