I have the following stochastic process:
$X(t)=e^{-t}W(e^{2t})$
Where $W(t)$ is a Brownian motion and $t\in[0,\infty)$. Now consider the following process:
$Y(t)=\int_{0}^{t}X(s)ds$
I now want to find the following expectation:
$\mathbb{E}[X(t)Y(s)]=\mathbb{E}[e^{-t}W(e^{2t})\int_{0}^{s}e^{-y}W(e^{2y})dy]$
I am not sure what to do next, surely $X(t)$ and $Y(s)$ are not independent so i can't write:
$\mathbb{E}[X(t)Y(s)]=\mathbb{E}[X(t)]\mathbb{E}[Y(s)]$
Any ideas?
$E [\xi \int_{0}^a \eta(t) dt] = \int_{0}^a E \xi \eta(t) dt$ hence $$ \mathbb{E}[X(t)Y(s)]=\mathbb{E}[e^{-t}W(e^{2t})\int_{0}^{s}e^{-y}W(e^{2y})dy] = $$ $$=\mathbb{E}[\int_{0}^{s}e^{-y}e^{-t}W(e^{2t})W(e^{2y})dy] = \int_{0}^{s}e^{-y-t}\min(e^{2t},e^{2y})dy = $$ $$\int_{0}^{s}e^{-y-t}e^{2\min(t,y)}dy=\int_{0}^{s}e^{-\min(t,y)-\max(t,y)}e^{2\min(t,y)}dy = $$ $$=\int_{0}^{s}e^{\min(t,y)-\max(t,y)}dy = \int_{0}^{s}e^{-|t-y|}dy$$
Remark: here is an intuitive argument for $E [\xi \int_{0}^a \eta(t) dt] = \int_{0}^a E \xi \eta(t) dt$. We have $E [\xi \int_{0}^a \eta(t) dt] \approx E [\xi \sum_j \eta(t_j) \Delta_j]= \sum_j E [\xi \eta(t_j)] \Delta_j \approx \int_{0}^a E \xi \eta(t) dt$