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:
$Y(t)=\int_{0}^{t}X(s)ds$
Can we then calculate the expectation as follows?
$\mathbb{E}[Y(t)]=\int_{0}^{t}\mathbb{E}[e^{-s}W(e^{2s})]ds=\int_{0}^{t}e^{-s}\mathbb{E}[W(e^{2s})]ds=\int_{0}^{t}e^{-s}0ds$=0
Or have I made a mistake somewhere?