How to prove that the Fourier Transform of white noise is flat? 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?
 A: 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.
A: 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).
A: 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
