Is a white noise always zero mean and uncorrelated?

Question

It seems to be silly, but is it possible to deduce mathematically that a white noise is necessarily zero mean and uncorrelated?

I have seen some people defining a white process as a zero-mean and constant-variance process with uncorrelated samples. However, the corollary that I saw on books is that, if $x(n)$ is a discrete-time white noise, then $R_x (\tau) = \frac{N_0}{2} \delta (\tau)$ and $S_x (f) = \frac{N_0}{2}$. Nothing more. It is called "white" because of the analogy with the white light that has all frequencies. I can infer the following statement from it:

1. Since $$ \begin{align} R_x(\tau) = E[x(n) x(n+\tau)] = 0\text{ for }\tau \neq 0 \label{1} \tag{1} \end{align} $$ we have that $x(n)$ and $x(n+\tau)$ are orthogonal.

To be an uncorrelated sequence, the covariance shall be $$ \begin{align} C_x(\tau)=R_x(\tau)-m_x ^2 =0\text{ for }\tau \neq 0. \label{2} \tag{2} \end{align} $$

If I guarantee that it is true, then $$ \begin{align} m_x=E[x(n)] = 0. \label{3} \tag{3} \end{align} $$

How to prove that?

Answer 1

I've found a straightforward prove for my question on this book, in PDF's page 538, or book's page 523 :)

One can observe that "$R_x (\tau)$ approaches of the square of the mean of $x(n)$ as $\tau \rightarrow \infty$". Mathematically, $$ \begin{align} \lim_{\tau \rightarrow \infty} R_x (\tau) = E[x(n)]^2 = m_x^2 \end{align} $$ then $$ \begin{align} m_x = 0. \label{4} \tag{4} \end{align} $$

With it and my question's introduction in mind, it becomes very easy. Substituting the Equation $(4)$ into $(2)$, we have that $$ \begin{align} C_x (\tau) = 0, \text{ for } \tau \neq 0 \label{5} \tag{5} \end{align} $$

Therefore a white noise is always zero mean and uncorrelated.

edit: here, we are considering a WSS (wide-sense stationary) white noise, since a white noise is not necessarily WSS. However, such statement is commonly assumed.

Answer 2

A white noise is a random signal having a constant power spectral density (PSD): $$S_x(f) = \frac{N_0}{2}.$$ From that, you can deduce everything else.

The autocorrelation and the PSD are linked through the Wiener-Khinchin theorem:
  • The PSD is the Fourier transform of the autocorrelation.
  • The autocorrelation is the inverse Fourier transform of the PSD.
The inverse Fourier transform of a constant is a Dirac, therefore the autocorrelation of a white noise is: $$R_{xx}(\tau) = \mathcal{F}^{-1}(S_x(f)) = \mathcal{F}^{-1}\left(\frac{N_0}{2}\right) = \frac{N_0}{2} \delta(\tau).$$

This means that the autocorrelation is zero for any delay different than zero. Then, think of the meaning of the autocorrelation. The autocorrelation is a measure of similarity of a signal with itself in the future, or a measure of how the future is related to the present. An autocorrelation of zero means that there is no similarity, hence the value in the future $x(t+\tau)$ is uncorrelated with the present value $x(t)$.

For the zero mean, we can look at the autocorrelation of a signal that is not zero mean. Let's decompose a signal as $x(t) = x_\mathrm{DC} + x_\mathrm{AC}(t)$, where $x_\mathrm{DC}$ corresponds to the DC part of $x(t)$, i.e. it is a constant value corresponding to the mean of $x(t)$, and $x_\mathrm{AC}(t)$ corresponds to the AC part of $x(t)$, i.e. it contains only the fluctuations of $x(t)$ and has zero mean. Then, the autocorrelation of $x(t)$ is $$ \begin{align} R_{xx}(\tau) &= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} x(t) x(t+\tau) \mathrm{d}t \\ &= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} (x_\mathrm{DC} + x_\mathrm{AC}(t)) (x_\mathrm{DC} + x_\mathrm{AC}(t+\tau)) \mathrm{d}t \\ &= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} x_\mathrm{DC}^2 + x_\mathrm{DC} x_\mathrm{AC}(t) + x_\mathrm{DC} x_\mathrm{AC}(t+\tau) + x_\mathrm{AC}(t) x_\mathrm{AC}(t+\tau) \mathrm{d}t \\ &= \lim_{T \to \infty} \left( \frac{1}{T} \int_{-T/2}^{T/2} x_\mathrm{DC}^2 \mathrm{d}t + x_\mathrm{DC} \int_{-T/2}^{T/2} x_\mathrm{AC}(t) \mathrm{d}t + x_\mathrm{DC} \int_{-T/2}^{T/2} x_\mathrm{AC}(t+\tau) \mathrm{d}t + \int_{-T/2}^{T/2} x_\mathrm{AC}(t) x_\mathrm{AC}(t+\tau) \mathrm{d}t \right). \end{align} $$ Since $x_\mathrm{AC}(t)$ has zero mean, we have $$ \lim_{T \to \infty} \int_{-T/2}^{T/2} x_\mathrm{AC}(t) = 0. $$ Therefore, the autocorrelation is $$ \begin{align} R_{xx}(\tau) &= \lim_{T \to \infty} \left( \frac{1}{T} \int_{-T/2}^{T/2} x_\mathrm{DC}^2 \mathrm{d}t + \int_{-T/2}^{T/2} x_\mathrm{AC}(t) x_\mathrm{AC}(t+\tau) \mathrm{d}t \right) \\ &= x_\mathrm{DC}^2 + \lim_{T \to \infty} \int_{-T/2}^{T/2} x_\mathrm{AC}(t) x_\mathrm{AC}(t+\tau) \mathrm{d}t. \end{align} $$ In order that the autocorrelation of $x(t)$ is zero for all $\tau \neq 0$, it requires that the autocorrelation of $x_\mathrm{AC}(t)$ is $-x_\mathrm{DC}^2$, i.e. a negative constant. Having a negative constant is not possible, because it means that for any $\tau_1$ and $\tau_2$ different than 0, you have $x_\mathrm{AC}(t+\tau_1) = - x_\mathrm{AC}(t+\tau_2)$, which is not possible (it can be the case for some delays, but not all).
Therefore, if a white noise autocorrelation is zero everywhere except for a zero delay, it necessarily means that the white noise has zero mean.

The zero mean property can also be deduced directly from the PSD. If the mean was nonzero, then there would be a Dirac at 0 Hz in the PSD, which is not the case. Indeed, it is not because the Fourier transform or the PSD has a nonzero value at 0 Hz that it means there is a DC component. For instance, an infinite sequence of rectangular pulses of value +1 or −1 randomly has zero mean, but its PSD is shaped as a sinc² with a nonzero value at 0 Hz. DC component means Dirac at 0 Hz, and vice-versa.

To conclude, when we see sentence as "$x(t)$ is a zero mean additive white Gaussian noise", the zero mean is redundant because it is already implied by the white noise.