If we take $X_n$ a series a random vector with its components each having a probability distribution with zero mean and finite variance, and are statistically independent. How do we prove that the power spectrum of $X_n$ is flat?

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    $\begingroup$ Umm...isn't that the definition? "Constant power spectral density" means the Fourier transform is "flat". $\endgroup$
    – whuber
    Oct 24, 2014 at 20:44
  • $\begingroup$ Thanks for your comment. Ah yes this is the definition of a white noise. I didn't ask the question correctly. Actually I meant if you have a series of X_n following identically independent gaussian distributions, would that have a flat power spectrum? $\endgroup$
    – clotilde
    Oct 24, 2014 at 20:54
  • $\begingroup$ Please edit your question if you wish to change it--not everybody will read these comments and most people will answer only what is in the question itself. $\endgroup$
    – whuber
    Oct 24, 2014 at 21:21

3 Answers 3


The power spectrum at frequency $\lambda \in [-\pi,\pi]$ can be obtained by taking the Fourier transform of the autocovariances $\gamma(\tau)$ of orders $\tau=-\infty,...,-1,0,1,...\infty$:

$$ f(\lambda) = \frac{1}{2\pi} \sum_{\tau=-\infty}^\infty \gamma(\tau) e^{-i\lambda\tau} \,. $$

Using the facts that in a white noise process $\gamma(-\tau) = \gamma(\tau)$ and $e^{-i\lambda\tau} = \cos(\lambda\tau) - i \sin(\lambda\tau)$, $\cos(0)=1$ and that $\sum_{\tau=-\infty}^\infty \sin(\lambda\tau) = 0$ for a given $\lambda$, the above expression can be written as:

$$ f(\lambda) = \frac{1}{2\pi} \left( \gamma(0) + 2 \sum_{\tau=1}^\infty \gamma(\tau) \cos(\lambda\tau) \right) \,. $$

$\gamma(0)$ is the variance of the process, while the remaining covariances are zero in a white noise process, $\gamma(\tau)=0$ for $\tau\neq 0$. Thus, we are left with the constant:

$$ f(\lambda) = \frac{\gamma(0)}{2\pi} \,. $$

According to this view in the frequency-domain, a white noise process can be viewed as the sum of an infinite number of cycles with different frequencies where each cycle has the same weight.

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    $\begingroup$ What happened to $\gamma$ in the initial formula? $\endgroup$
    – whuber
    Oct 24, 2014 at 22:50
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    $\begingroup$ @whuber It was missing in the first two equations, thanks I fixed it. $\endgroup$
    – javlacalle
    Oct 24, 2014 at 22:54
  • $\begingroup$ Is it really true that $\sum_{\tau=-\infty}^\infty \sin(\lambda\tau) = 0$? For $\lambda = \pi/2$, the sum is of the form $$\cdots, 0, 1, 0, -1, 0, 1, 0, -1, \cdots$$ and thus does not converge. $\endgroup$ Oct 24, 2014 at 23:23
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    $\begingroup$ @Dilip Sarwate, what is required to converge is $\sum_{\tau=-\infty}^\infty \gamma(\tau)\sin(\lambda\tau)$; this will be the case if $\sum_{i=\infty}^\infty|\gamma(\tau)|$ converges, which is usually assumed (ergodicity assumptions). $\endgroup$
    – F. Tusell
    Oct 31, 2014 at 12:21

The question in the title is not the same as the question in the text or as described in the comments.

The Fourier transform of an infinitely long sequence is a discrete-time Fourier transform which is a (complex-valued) periodic function of the frequency variable $\omega$. See also, @javlacalle's answer. Thus, it cannot be "flat" except in the trivial case when the function is a constant, or one includes any complex number of magnitude $1$ in the notion of "flat". Furthermore, when the sequence is a realization of a white noise (normal) process (which is a sequence of i.i.d. (normal) random variables), then the Fourier transform of the sequence differs from realization to realization, and it boggles the mind that all of these Fourier transforms turn out to be "flat" in any sense of the word.

So, what is asked for in the title of the question is meaningless.

The question asked in the text of the question is, as pointed out by whuber, essentially the definition of white noise. It is better to approach the problem of defining white noise by starting with a sequence of i.i.d random variables of finite variance $\sigma^2$ and noting that the autocovariance function is a unit pulse function. To borrow notation from javlacalle, $\gamma(0) = \sigma^2$, and $\gamma(n) = 0$ for all other integers $n$. From this, it follows that the power spectral density (Fourier transform of the autocovariance as per the Wiener-Khinchine theorem) is a constant (which is why the noise process is called white noise, in mistaken analogy with white light which is a flat mixture of wavelengths, not frequencies).


Actually, your question is pretty legitimate. However, it needs to be asked in a slightly different manner. You need to specify the distribution of the periodogram ordinates and get a statistical test to examine whether your real data conform to that. One attempt is our recent paper


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