Calculating the ensemble average of a known function multiplied with a random term in a Langevin equation? Let's say I have a Langevin equation that reads
$$\frac{dx(t)}{dt} = A(t) + B(t) + \eta(t)$$
Where $A(t)$ and $B(t)$ are some non-random functions that we know the functional form of, and $\eta(t)$ is a random term. For the sake of this question, lets say it has Gaussian properties such that it has zero mean, a finite standard deviation and $\langle\eta(t_1)\eta(t_2)\rangle = D\delta(t_1-t_2)$.
If $A(t) = B(t) = 0$, the equation just describes a random walk, and we can calculate
$$\langle x(t_1)x(t_2)\rangle = \int_{0}^{t_1}\int_{0}^{t_2}\langle\eta(t'_1)\eta(t'_2)\rangle dt'_1dt'_2$$
$$\langle x(t_1)x(t_2)\rangle = \int_{0}^{t_1}\int_{0}^{t_2}D\delta(t'_1-t'_2) dt'_1dt'_2$$
$$ \langle x(t_1)x(t_2)\rangle= Dt_1$$
But now lets say we know that $A(t)\neq 0$ and $B(t)\neq 0$. If we now calculate $\langle x(t_1)x(t_2)\rangle$ we have
$$\langle x(t_1)x(t_2)\rangle = \int_{0}^{t_1}\int_{0}^{t_2}\Big\langle(A(t'_1) + B(t'_1) + \eta(t'_1))(A(t'_2) + B(t'_2) + \eta(t'_2))\Big\rangle dt'_1dt'_2$$
And we find we have ensemble average terms where we mix random and non-random terms for example
$$\langle A(t'_1)\eta(t'_2)\rangle$$
My question is: how do you deal with these terms? In the case of only random terms, only one term $\langle\eta(t_1)\eta(t_2)\rangle = D\delta(t_1-t_2)$ appears under the integral which we've defined before. But how do we deal with the ensemble average terms with both random and non-random terms in it?
 A: A non-random function behaves as a constant in respect to statistical averaging, that is
$$
\langle A(t_1)\eta(t_2)\rangle=A(t_1)\langle \eta(t_2)\rangle=0
$$
(the second equality is because the OP assumes that the random process has the zero average.)
Remarks

*

*A word of caution - why are there explicitly two different non-random functions in the equation? It might be that one of them is a function of $x(t)$, in which case it is no more a non-random term, and the problem is more complex.

*As linear differential equations are exactly solvable, the corresponding Langevin equations can be formally solved and the solutions used for calculating all the necessary averages.

A: Given a deterministic function, $f(t)$, then it follows from Ito isometry & independent increments that
$$ \int_0^tf(u)\,\mathrm{d}W_u\sim\mathcal{N}\left(0,\,\int_0^tf^2(u)\,\mathrm{d}u\right).\tag{1}$$
where $W_t$ is the Wiener process (such that one could at least loosely say $\mathrm{d}W_t=\eta(t)\,\mathrm{d}t$) and $\mathcal{N}(\mu,\,\sigma)$ is the normal distribution. This is to say, the integral in Eq (1) is a Gaussian process with zero mean--see, for instance, these lecture notes. So one can safely ignore these terms when we average them out,
$$\left\langle\int f(u)\,\mathrm{d}W_u\right\rangle\equiv0.$$
In the cases we don't take an average, we could replace the integral with an arbitrary term, e.g.
$$Z(t)\sim\int f(u)\,\mathrm{d}W,$$
and not really miss anything when solving stochastic differential equation (though a final solution might need to be solved numerically or through an approximation such as being small compared to another term).
