The mean of Langevin equation I have a very basic question regarding the mean of the Langevin equation. 
So we have an equation of the form:
$$\dot{v}(t)=-\beta  v(t)+ \xi (t)$$
Where $\xi (t)$ is a Gaussian white noise with an average zero and a $\delta$ correlation in time.
As known, this equation has the following solution:
$$v(t)=v(0) e^{-\beta t}+\int_0^t dt' e^{-\beta (t-t')} \xi(t') $$
and I want to take the mean of this equation, i.e., $\langle v(t)\rangle$
The second term goes is zero since $\langle\xi (t)\rangle=0$, which leaves the first term. This, according to some books I have been reading, should be: 
$$\langle v(t)\rangle=v(0) e^{-\beta t},$$
 but I really don't get how we got this result? It's a bit confusing to me.
 A: To see why the first is left when taking the mean, re-write the SDE as,
$$
\mathrm dv=-\beta v\mathop{}\!\mathrm dt+g(x,\,t)\mathop{}\!\mathrm dW,
$$
where $g(x,\,t)$ is some function and $\mathrm dW$ the stochastic noise. Then we can obtain a differential equations for the mean of $v$ by taking the mean of both sides:
$$
\mathrm d\langle\mathrm v\rangle=\langle\mathrm d v\rangle =\langle -\beta v\rangle\mathop{}\!\mathrm dt
$$
since $\beta$ is a constant and $\langle\mathrm dW\rangle=0$. This can be rearranged to,
$$
\frac{\mathrm d\langle v\rangle}{\mathrm dt}=-\beta\langle v\rangle,
$$
the solution of this differential equation should be immediately seen:
$$
\langle v(t)\rangle=e^{-\beta t}\langle v(0)\rangle
$$
which matches the sources you've seen, aside from the lack of $\langle \cdot\rangle$ wrapping $v(0)$, which I suspect is some level of your confusion (the other bit being the exponential being dependent on $t$, though this confusion should be cleared up from the derivation above).
A: Let me answer the original question as I understood it literally. It has a methodological value.
I understand taking a mean value as some integration over a limited period $T$: $$\langle f(t)\rangle=\frac{1}{T}\int_t^{t+T}f(\tau)d\tau$$ Only doing this we may "eliminate" the fluctuating force $\xi(t)$. Now, applying this averaging to the solution $v(t)$ we obtain: $$\langle v(t)\rangle=\frac{1}{T}\int_t^{t+T}v(0)\text{e}^{-\beta \tau}d\tau + \frac{1}{T}\int_t^{t+T}\text{e}^{-\beta \tau}d\tau\int_0^{\tau} dt' \text{e}^{t'} \xi(t')$$ Now let us take the first integral: $$\frac{1}{T}\int_t^{t+T}v(0)\text{e}^{-\beta \tau}d\tau=v(0)\frac{\text{e}^{-\beta t}(1-\text{e}^{-\beta T})}{\beta T}$$ This is an exact result of our averaging over some period $T$. Now, if inequality $\beta T\ll 1$ holds, in the first approximation the difference in the nominator is equal to $\beta T$ with a good precision, thus you obtain the answer to the original question.
In order to make sure that the second term in the exact solution vanishes after averaging, you must respect - in addition! - another inequality, namely $T/\delta\gg 1$. Otherwise some "long-time" fluctuations will still be present in the mean solution. I leave the proof of this as an exercise to those who downvoted my answer without explanation.
Finally, let me note that although the noise force mean value is zero, it does not mean that this force does not displace the particle in the space. Originally still particle ($v(0)=0$) and without friction ($\beta=0$) may be found elsewhere: $x(t)=\int_0^t dt'\int_0^{t'}\xi(t'')dt''\ne x(0)$, remember the Brownian motion, for example. Another (funny) example: a periodic external force $F(t)=F_0\cdot \sin(\omega t)$. On average this force is zero, but it displaces the particle with time far far away ;-)  One cannot obtain this result by just dropping out this force from the exact equation of motion.
EDIT: As pointed out Alexander in his comment, "Langevin equation is about stochastic processes. Averages of stochastic processes are (usually) with respect to different noise realizations (all taken from same distribution). In different language - tracing out the ensemble of stochastic environment." In other words, averaging is done in fact over some parameter independent of time. It is like averaging over a random initial phase $\varphi_0$ in my last example with a periodic external force $F_0\cdot\sin(\omega t + \varphi_0)$. Thus, it is not averaging the equation or its solution over $t\le\tau\le (t+T)$. Although mathematically it is possible, I do not fully understand how it may naturally arise in physical calculations and what meaningful conclusion one can draw from such averaged things.
