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?

  • $\begingroup$ @V.V.T corrected. $\endgroup$ Jun 6 at 1:59
  • 1
    $\begingroup$ Have you seen this question? Do the answers answer your question? $\endgroup$
    – Matt L.
    Jun 6 at 10:37
  • $\begingroup$ @MattL. No, I haven't. But this question doesn't answer my question directly. I just found the answer the posted here :) $\endgroup$ Jun 6 at 19:07
  • $\begingroup$ For a discrete-time white noise process, the delta in the autocorrelation function is generally expressed as $\delta[n]$, the Kronecker delta, and not as $\delta(t)$ which denotes the Dirac delta. $\endgroup$ Jun 6 at 22:45
  • $\begingroup$ @DilipSarwate depends on which reference you are basing. For Oppenheim, you are right, Kronecker delta must be denoted as $\delta[n]$, whereas $\delta(t)$ indicates the Dirac delta. However, for many other authors (Proakis, Diniz,... etc), both discrete- and continuous-time are denoted in parentheses. But again, there is nothing wrong with that, it is just a matter of notation. $\endgroup$ Jun 6 at 22:52

1 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 $$ \begin{align} C_x (\tau) = 0. \label{5} \tag{5} \end{align} $$

Therefore a white noise is always zero mean and uncorrelated.


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