I need to calculate the expected value $\mathbb{E}[\eta(s) \cdot \eta( \tilde{s} )]$, where $\eta$ is an Ornstein-Uhlenbeck process defined by the stochastic differential equation
\begin{equation*}
\tau_{\eta} \frac{d \eta(t)}{dt} = – \eta(t) + \sqrt{2(\tau + \tau_{\eta})} \sigma \xi(t).
\end{equation*}
$\tau, \tau_{\eta}, \sigma \in \mathbb{R}$ are constants and $\xi(t)$ is an Gaussian white noise process $ \xi(t) = \frac{dW(t)}{dt}$. \
The needed expected value is equal to the Covariance $Cov(\eta(s), \eta(\tilde{s}))$, if he Ornstein-Uhlenbeck processes have mean zero. I also calculated the explicit solution of the SDE
\begin{align*}
\eta(s) = \exp \left( – \frac{s}{\tau_{\eta}} \right) \left[ \eta_0 + \int_0^s \frac{1}{\tau_{\eta}} \exp \left( \frac{x}{\tau_{\eta}} \right)
\sqrt{2(\tau + \tau_{\eta}) } \sigma dW_x \right],
\end{align*}
where $W_x$ is a Brownian motion.
I inserted the solution of the stochastic differential equation in the expected value with two different time variables $s, \tilde{s}$. Because the increments of the Brownian motion have mean zero, the mixed terms are zero. This leads to
\begin{align*}
\mathbb{E} \left[ \eta(s) \cdot \eta(\tilde{s}) \right] &=
\mathbb{E} \left[ \exp \left( – \frac{s + \tilde{s}}{\tau_{\eta}} \right) \eta_0^2 \right]
+ \mathbb{E} \left[ n_0^2 + \int_0^s \int_0^{\tilde{s}} \frac{1}{\tau_{\eta}^2} \exp \left( \frac{x + y}{\tau_{\eta}} \right) \cdot 2(\tau + \tau_{\eta}) \sigma^2 dW_x dW_y \right]
\end{align*}
How can I calculate the expected value of the last double integral.
It probably has something to do with the Covariance of the 2 Brownian motions $dW_x, dW_y$.
Best Answer
The explicit solution of your SDE is of the form $$ \eta(s)=e^{-f(s)}\Big[\eta_0+\int_0^sae^{f(x)}\,dW_x\Big]\,. $$ The expectation $E[\eta(s)\,\eta(\tilde s)]$ is easily obtained from $E[\int_0^se^{f(x)}\,dW_x]=0$ and $$ E\Big[\int_0^se^{f(x)}\,dW_x\cdot\int_0^\tilde{s}e^{f(x)}\,dW_x\Big]=\int_0^{\min(s,\tilde s)}e^{2f(x)}\,dx\,. $$