Covariance of Ornstein - Uhlenbeck Process

Question

I'm considering the Ornstein - Uhlenbeck process $ X(t)=x_{\infty}+e^{-at}(x_{0}-x_{\infty})+b \int_{0}^{t} e^{-a(t-s)} dW(s)$ where $a, b > 0 $ are given constants. I used the Itô Isometry to compute the variance but I did not figure out how to use it to compute the covariance. What is the most general context in which I can use Itô Isometry to compute the covariance?

Thank you for your help.

Answer 1

Let $Y_t=\displaystyle\int_0^t\mathrm e^{au}\mathrm dW_u$ then, for every $t$, $X_t=b\mathrm e^{-at}Y_t+z(t)$ where $z(\ )$ is deterministic hence $$\mathrm{cov}(X_t,X_s)=(b\mathrm e^{-at})(b\mathrm e^{-as})\mathrm{cov}(Y_t,Y_s),$$ and, for every $t$, $Y_t=\displaystyle\int_0^\infty g_t(u)dW_u$ where $g_t$ is deterministic since $g_t(u)=\mathrm e^{au}\mathbf 1_{u\lt t}$ hence $$\mathrm{cov}(Y_t,Y_s)=\int_0^\infty g_t(u)g_s(u)du=\int_0^\infty \mathrm e^{2au}\mathbf 1_{u\lt\min\{t,s\}}du=\int_0^{\min\{t,s\}}\mathrm e^{2au}du=\ldots$$

Answer 2

Complementing the answer above to make explicit use of Ito's isometry as you requested. The appropriate version of ito's isometry to use in this case is the following: $$ \mathbb{E} \int_0^T f(r)dW(r) \int_0^T g(r) dW(r) = \mathbb{E} \int_0^T f(s)g(s)ds,$$ where in your case $f(r) = e^{ar} \mathbf{1}_{(0,s)}(r)$, and $f(r) = e^{ar} \mathbf{1}_{(0,t)}(r)$. I followed all the steps and got that the covariance in your case is $$ \frac{ b^2 }{a} e^{-a ( t \vee s ) } { \rm cosh } ( s \wedge t ),$$ where for any pair of numbers, $ x,y \in \mathbb{R} $, $x \vee y = \max \{ x,y \}$, and $x \wedge y = \min \{ x,y \}$.