So I am using the Pavliotis (you can find it here ) and in propostion 3.5 they prove the correlation of an Ornstein-Uhblenbeck process. It goes like this:
For the Ornstein-Uhlenbeck $dX(t) =-AX(t)dt+ \sigma dW(t)$, the autocorrelation matrix of the process $X(t)$ is given by $$R(t,s)=e^{-At}\left(R(0)+\int_0^{min(s,t)}e^{Ar}\Sigma e^{A^T r}dr\right)e^{-A^Ts}, \hspace{0.5cm} \Sigma=\sigma \sigma^T, \; R(0)=E[x x^T].$$
Then, on to the proof (literally the first line :( ).. they say:
Define $S_t=e^{-At}$, and $B(t)=\Sigma W(t)$, then $X_t = S_t x + \int_0^t S_{t-s}dB(s)$. Consequently, using the properties of the stochastic integral \begin{align} R(s,t)=E[X(t)X(s)^T]=E[S_t x (x S_s)^T] +E[ \int_0^t\int_0^s (S_{t-l} dB(l))(S_{s-r} dB(r))^T] \hspace{0.5cm} (1) \end{align}
So, I tried computing it directly and I got that
\begin{align} X(t)X(s)^T &= (S_t x + \int_0^t S_{t-l}dB(l))(S_s x + \int_0^s S_{s-r}dB(r))^T\\ &=(S_t x)(S_s x)^T + (S_t x)(\int_0^s S_{s-r}dB(r))^T+(\int_0^t S_{t-l}dB(l))(S_s x)^T+(\int_0^t S_{t-l}dB(l))(\int_0^s S_{s-r}dB(r))^T \end{align}
But I don't know how to go from here to (1). A wild guess is that they use Ito's multiplication table where $dt dt=0, dB_i dB_j =0, dB_i dt = 0 , dB_i dB_i = dt$ but I am not sure how to use them here. Any further explanation is very appreciated.
Also, I think that the terms in the middle can be $0$ once you find their expected value, since it is $0$ for an Itô integral. However, for this shouldn't I asume that $x$ is independent? (since $S_t$,$S_s$ are constants).