# Is a white noise always zero mean and uncorrelated?

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?

• @V.V.T corrected. Jun 6 at 1:59
• Have you seen this question? Do the answers answer your question? Jun 6 at 10:37
• @MattL. No, I haven't. But this question doesn't answer my question directly. I just found the answer the posted here :) Jun 6 at 19:07
• 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. Jun 6 at 22:45
• @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. Jun 6 at 22:52

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}