Given a probability distribution P(x,t), we can take the average of a time-dependent quantity x(t) as
\begin{equation} \overline{x(t)}=\int dx x(t) P(x,t) \end{equation}
My question is: what if I have to take the average of $\dot{x}(t)=-ax(t)+b\eta(t)$, where $\overline{\eta(t)}=0$ and $\overline{\eta(t)\eta(t')}=\delta(t-t')$?
Is it correct to assume that then:
\begin{equation} \int dx \dot{x}(t) P(x,t)=\overline{\dot{x}(t)} \end{equation}
I think the claim is correct, but the notation can lead to difficulties in computing time derivatives. I would change your first expression for \begin{equation} \overline{x(t)}=\int dx P(x,t) x. \end{equation} Indeed, the whole time dependence is encoded in the probability distribution $P(x,t):=P[x(t)=x]$. The variable $x$ inside the integral is a dummy integration variable that runs over all possible positions of the walker, and is not one particular realization or trajectory, which is noted as $x(t)$. Maybe it is more clear using another dummy variable $y$: \begin{equation} \overline{x(t)}=\int dy P(y,t) y=\int dy P[x(t)=y] y. \end{equation}
For an Itô stochastic differential equation of the type $$ \dot{x}(t)=a(x,t)+b(x)\eta (t),$$ we can compute the first moments of the position and velocities as
\begin{equation} \overline{x(t)}=\int_0^t a(x,s) ds +x(0), \end{equation} and \begin{equation} \overline{\dot{x}(t)}=a(x,t), \end{equation} respectively.
Therefore, it is true that \begin{equation} \overline{\dot{x}(t)}=\frac{d}{dt}\overline{x(t)}=\frac{d}{dt}\int dy P(y,t) y. \end{equation}
I don't know if the above equality is true for an arbitrary time-dependent diffusion though [$b(x,t)$ instead of $b(x)$].
