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?

  • $\begingroup$ You sure you can exchange the limit and the integral like that? $\endgroup$
    – ACuriousMind
    Commented Apr 15, 2015 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. $\endgroup$
    – user73762
    Commented Apr 15, 2015 at 17:03
  • 2
    $\begingroup$ @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. $\endgroup$ Commented Apr 15, 2015 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? $\endgroup$ Commented Apr 15, 2015 at 17:21
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Apr 15, 2015 at 21:21

1 Answer 1


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 $\xi(t) \to \xi(t) -\xi_0$.

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Apr 15, 2015 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
    Commented Apr 15, 2015 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). $\endgroup$ Commented Apr 15, 2015 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
    Commented Apr 16, 2015 at 9:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.