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I try to solve a Langevin equation in the Fourier space. My understanding of the white noise in the Fourier space seems to be wrong.

Suppose I have a particle with its time evolution of the position given by the stochastic differential equation.

$$ \frac{d f(t)}{dt}=g(f)+\xi(t) $$

$f$ is the position, $\xi$ is the white noise, $g$ some function of the position, and $t$ the time.

The particle will move according to the differential equation but because of the noise, every time I redo the experiment I will observe a different trajectory. What is of interest is not the position of the particle at time $t_0$ but the average over a large number of realizations, all evaluated at time $t_0$.

I write this "average over all realization" with $<>$.

The noise, at any time $t$, is zero in average: $<\xi(t)>=0 \quad \forall ~t$.

This means: $$\lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum\limits_{j=1}^{N}\xi_j(t)=0 \quad \forall ~t $$

with "$j$" the different realizations.

I want to Fourier transform the noise. Given one particular realization "$j$" of the noise $\xi_j(t)$ I can compute its Fourier transform:

$ \hat\xi_j(\omega)=\int \xi_j(t) e^{i\omega x}dt $

Now, I want to compute the average over all realization:

$$ <\hat\xi(\omega)>=\lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum\limits_{j=1}^{N} \hat\xi_j(\omega) \\ \quad \quad \quad \quad ~= \lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum\limits_{j=1}^{N} \int \xi_j(t) e^{i\omega t}dt \\ \quad \quad \quad \quad ~= \int \left( \lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum\limits_{j=1}^{N} \xi_j(t) \right) e^{i\omega t}dt \\ \quad \quad \quad \quad ~= \int <\xi(t)> e^{i\omega x}dt \\ \quad \quad \quad \quad ~= 0 $$

Which is wrong since the Fourier spectrum of a white noise is a constant function. What am I missing?

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  • $\begingroup$ You sure you can exchange the limit and the integral like that? $\endgroup$
    – ACuriousMind
    Apr 15 '15 at 16:55
  • $\begingroup$ I am sure you cannot; hence you need another way to calculate $\left\langle \xi(t) e^{i\omega t} \right\rangle$. That is the reason why the answer by Mark Mitchison, focussing on $\left\langle \xi(t) \xi(t^\prime) \right\rangle$, is relevant. Note that, unless your wording and initial approach, it applies to the average of all processes ($\xi(t)$ rather than $\xi_j(t)$), but that is really the interesting aspect. $\endgroup$
    – pyramids
    Apr 15 '15 at 17:03
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    $\begingroup$ @pyramids Nevertheless, I think the conclusion, that $\langle \int\mathrm{d}t\, \mathrm{e}^{\mathrm{i}\omega t} \xi(t) \rangle = 0$, should be correct, no? Since we are considring a bunch of delta-correlated random functions, I would expect the Fourier transform of each one of these functions, evaluated at some frequency $\omega$, to be distributed equally between positive and negative values. $\endgroup$ Apr 15 '15 at 17:09
  • $\begingroup$ @pyramids What do you mean the Fourier transform conserves power? To consider the power, aren't we then talking about the second moment anyway? $\endgroup$ Apr 15 '15 at 17:21
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    $\begingroup$ there's a difference between the Fourier Transform of something and the Power Spectrum of the same something. the Fourier Transform of a stochastic function is, itself, also random. but, given certain conditions (like ergodicity), the expectation value of the Power Spectrum of white noise is a constant. $\endgroup$ Apr 15 '15 at 21:21
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The OP is correct in stating that the Fourier transform $$\xi(\omega) = \int\mathrm{d}t\, \mathrm{e}^{\mathrm{i}\omega t} \xi(t), $$ vanishes upon averaging over realisations, $\langle \xi(\omega)\rangle = 0$, so long as we assume that the noise is also zero on average in the time domain, $\langle \xi(t)\rangle = 0 $.

However, the noise is not only characterised by its first moment, but also by its auto-correlation function: $$ \langle \xi(t) \xi(t^\prime)\rangle = \eta\delta(t - t^\prime).$$ This last equation characterises the fluctuations of $\xi(t)$ in time; the presence of the delta function on the right-hand side is what actually defines white noise. The Fourier transform of the auto-correlation function gives the power spectrum: how noise power is distributed over different frequencies. For white noise this clearly takes the constant value $\eta$ in frequency space (up to a choice of normalisation for the Fourier transform). This means that the fluctuations contain equal contributions from all frequencies, i.e. fast and slow fluctuations contribute equally.


As a sidenote, it is worth mentioning that we could easily consider white noise with non-zero average $\langle \xi(t) \rangle = \xi_0.$ This simply means that the noise has a constant (i.e. non-random) component. In this case we have that $\langle\xi(\omega)\rangle = 2\pi\xi_0 \delta(\omega)$, and the white noise condition is $$\langle \xi(t) \xi(t^\prime)\rangle =\xi_0^2+ \eta\delta(t - t^\prime).$$ The choice $\xi_0 = 0$ is merely a convention that simplifies these expressions. We can always get zero-mean white noise by the shift $\xi(t) \to \xi(t) -\xi_0$.

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    $\begingroup$ You said "the Fourier spectrum of white noise is a constant function". This is true for the second moment, not the first. The average of $\xi(\omega)$ is zero. $\endgroup$ Apr 15 '15 at 16:59
  • $\begingroup$ Ok so here is my mistake... ? Not considering the moments, what would be Fourier spectrum of a white noise? $\endgroup$
    – David
    Apr 15 '15 at 17:03
  • $\begingroup$ What exactly do you mean "the Fourier spectrum of white noise"? Do you mean $\xi(\omega) = \int\mathrm{d}t\,\mathrm{e}^{\mathrm{i}\omega t} \xi(t)$? This is a random variable, the only thing you can say about it are its moments (or any other equivalent quantities such as the cumulants). $\endgroup$ Apr 15 '15 at 17:06
  • $\begingroup$ Yes this is what I mean. And yes I am actually trying to compute $\xi(\omega)$ first moment : $<\xi(\omega)>$ $\endgroup$
    – David
    Apr 16 '15 at 9:48

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