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?

Answer

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.

Attribution
Source : Link , Question Author : clotilde , Answer Author : javlacalle

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