# 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?

• Why differentiate between $A$ and $B$? You can reformulate the question in terms of $C(t)=A(t)+B(t)$ without loss of generality. The functions do not depend on an stochastic process (e.g. $A(\eta)$ or $A(x)$)?
– Javi
Commented Sep 13, 2022 at 12:34

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.)
• 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.
• Thank you for the answer! I added two non-random functions because I also was wondering about the case for $\langle A(t_1)B(t_2)\rangle$, though this case should be answered by your response, and it just comes out to be $\langle A(t_1)B(t_2)\rangle =A(t_1)B(t_2)$? If there was another term, $C(x(t))$, how would this be dealt with in calculating $\langle C(x(t_1))\eta(t_2)\rangle$ and also $\langle C(x(t_1))A(t_2)\rangle$? Commented Sep 13, 2022 at 18:40
• @physics_fan_123 if there were a term dependent on $x(t)$, the problem would be a lot more complex, since $x(t)$ depends on $\eta (t)$. However, if the equation is linear, it would be still tractable - see my remarks. Otherwise, one would have to use more special techniques... Or work with the Fokker Planck equation. Commented Sep 13, 2022 at 18:58
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).