Autocorrelation of OU process How do you derive $\mathbb{E}\left[\left(\int_0^{t+\tau} e^{-\frac{t+\tau-s}{T_{\tilde{\chi}}}} \mathrm{d} \mathcal{W}_s\right) \left( \int_0^t e^{-\frac{t-s}{T_{\tilde{\chi}}}} \mathrm{d} \mathcal{W}_s\right)\right]$ ? 
By decomposing from 0 to t and from t to t+$\tau$ and using Itô isometry, one has 
$\mathbb{E}\left[\left(\int_0^{t+\tau} e^{-\frac{t+\tau-s}{T_{\tilde{\chi}}}} \mathrm{d} \mathcal{W}_s\right) \left( \int_0^t e^{-\frac{t-s}{T_{\tilde{\chi}}}} \mathrm{d} \mathcal{W}_s\right)\right] = \int_0^{t} e^{-\frac{2t+\tau-2s}{T_{\tilde{\chi}}}} \mathrm{d}s + \mathbb{E}\left[\left(\int_t^{t+\tau} e^{-\frac{t+\tau-s}{T_{\tilde{\chi}}}} \mathrm{d} \mathcal{W}_s\right) \left( \int_0^t e^{-\frac{t-s}{T_{\tilde{\chi}}}} \mathrm{d} \mathcal{W}_s\right)\right]$
but I can't see what to do with the second term...
Thanks for the help !
 A: Following your result. Without loss of generality, take $T_{\tilde{\chi}}=1$.
Provided that
$$
e^s{\rm d}W_s
$$
is an Ito integral, for which
$$
\int_0^te^s{\rm d}W_s
$$
is a martingale. Thanks to this fact,
$$
\int_0^te^s{\rm d}W_s\quad\text{and}\quad\int_t^{t+\tau}e^s{\rm d}W_s
$$
are independent from each other. Therefore,
$$
\mathbb{E}\left(C_1\int_t^{t+\tau}e^s{\rm d}W_s\right)\left(C_2\int_0^te^s{\rm d}W_s\right)=\mathbb{E}\left(C_1\int_t^{t+\tau}e^s{\rm d}W_s\right)\mathbb{E}\left(C_2\int_0^te^s{\rm d}W_s\right)=C_1C_2\mathbb{E}\left(\int_t^{t+\tau}e^s{\rm d}W_s\right)\mathbb{E}\left(\int_0^te^s{\rm d}W_s\right).
$$
Finally, note that
$$
\int_0^te^s{\rm d}W_s\quad\text{and}\quad\int_t^{t+\tau}e^s{\rm d}W_s
$$
are both a sum of independent zero-mean normal random variables, for which their expectations are both zero, i.e.,
$$
\mathbb{E}\left(\int_t^{t+\tau}e^s{\rm d}W_s\right)=\mathbb{E}\left(\int_0^te^s{\rm d}W_s\right)=0.
$$
Thus we conclude that
$$
\mathbb{E}\left(C_1\int_t^{t+\tau}e^s{\rm d}W_s\right)\left(C_2\int_0^te^s{\rm d}W_s\right)=0,
$$
meaning that your second term vanishes, and thus could be neglected.
