White noise has the ACF: $R_{WW}[\kappa] = c_0 \delta [\kappa]$ and zero mean $m_W[\kappa] = 0$.

The first and second order moments of a WSS process depend only upon the time difference $\kappa$. With that being said, since $m_W $ is always zero, it does not depend on time and the ACF doesn't either which is (according to my book) sufficient to prove the WSS property of a process.

Am I right with this? Solutions to an exam I'm doing says that white noise must not always be WSS?

  • 1
    $\begingroup$ White noise is stationary in every sense.because the joint distribution of any set of $\epsilon_i$ does not depend on time at all. Observations are independent and identically distributed. $\endgroup$ – Michael Chernick Mar 21 '18 at 19:55
  • 2
    $\begingroup$ @MichaelChernick I don't know how widely accepted this is, but I generally refer to IID noise as independent noise, and white noise as just uncorrelated, which is weaker when second moments exist $\endgroup$ – Taylor Mar 21 '18 at 23:27
  • $\begingroup$ I can accept that but the argument still holds as shown in Winkelreid's answer. Strong stationarity also holds I believe. $\endgroup$ – Michael Chernick Mar 21 '18 at 23:33

Well, this depends on your definition of white noise. This question asks for that definition.

One answer gives:

A white noise process is a random process of random variables that are uncorrelated, have mean zero, and a finite variance. Formally, $X(t)$ is a white noise process if $E(X(t))=0,E(X(t)^2)=S^2$, and $E(X(t)X(h))=0$ for $t≠h$. A slightly stronger condition is that they are independent from one another; this is an "independent white noise process."

Under this definition (the first of the two, the weaker one), which is I presume the same definition you have, your reasoning is perfectly correct and white noise is always wide-sense stationary. Note that this definition asks for the variances be finite, what I think you also do since you probably mean a finite number when you write $c_0$.

For the stronger version of the definition given in the same quoted answer, the same applies.

In contrast, the definition that Dilip Sarwate gave in his answer doesn't require the variances to be finite and hence allows for a white noise not to be wide-sense stationary as he explained.

There are probably other definitions for white noise out there. Possibly, in the context of your exam another definition of white noise is assumed than the one in your book and therefore the apparent contradiction.


White noise has the properties that you state, but those properties are not the properties that define white noise. As Michael Chernick's comment points out, a (discrete-time) white noise process is a collection of independent identically distributed zero-mean random variables, one for each time instant under consideration. If the random variables have finite variance $\sigma^2$, then the autocorrelation function of the process is $\sigma^2\delta[n]$ where $\delta[n]$ is the Kronecker delta function defined by $$\delta[n] = \begin{cases}1, &n=0,\\0, &n \neq 0.\end{cases}$$ Now, if the common distribution function of random variables does not have a variance, e.g. Cauchy random variables, then white noise is not a wide-sense-stationary process (even though it is a strictly stationary process).

  • $\begingroup$ I thought that being strict sense stationary implied that the process is also wide sense stationary, that is what I learned from this pdf page 2 $\endgroup$ – Aditya Feb 7 at 8:25
  • $\begingroup$ @Aditya Have you checked the pdf to see if somewhere it says that all the random variables have the same finite second moment, that is, $E\left[(X[n])^2\right] = \sigma^2$ for all $n$? (This implies that the first moment exists and is finite too). Without this, we cannot even begin to define wide-sense-stationarity for a process consisting of iid Cauchy random variables. We cannot say that $E[X[n]] = 0$ because the mean does not exist! So, read your pdf some more to see if it says something about finite second moments. See also here. $\endgroup$ – Dilip Sarwate Feb 7 at 12:58
  • $\begingroup$ Thanks for the clarification. I think I understand, The Cauchy distribution is a "pathological" distribution since both its expected value and its variance are undefined. Since the mean does not exist, we cannot even begin to say that it is time invariant. So such processes are in that sense not wide sense stationary. Like the same way an alphabet A is not an odd number? It is not even a number? $\endgroup$ – Aditya Feb 7 at 13:19
  • $\begingroup$ @Aditya A process consisting of iid random variables is strictly stationary_ (and even more because all sets of $n$ random variables have the same joint distribution, not just $X(a) X(b),X(c),\ldots$ having the same joint distribution as $X(a+t),X(b+t),X(c+t),\ldots$). So the process is time-invariant; it looks the same no matter where you start. But it is not necessarily wide-sense stationary unless we further insist that the strictly stationary process has finite second moments too (which will of course all have the same value since the process is stationary). $\endgroup$ – Dilip Sarwate Feb 7 at 13:39
  • $\begingroup$ So, you are saying the problem is because the the Cauchy distribution does not have finite moments of any order, by definition it can't be WSS since it talks about the mean and autocorrelation function. While the process was SSS if for any finite value of $n$ like you said, those distributions just need to be time invariant. In case of cauchy if we observe the analytical form of the first order pdf, there is no "t" in it, so the process is time invariant. So it is SSS. I think I understand better now, but it seems like a technicality. $\endgroup$ – Aditya Feb 7 at 14:37

Wise sense stationarity implies weak or covariance stationarity, i.e only the first two moments(mean and variance) are time-invariant or constant. A white noise time series in its simplest form has 0 mean, constant variance and is serially uncorrelated. Hence white noise implies wide-sense stationarity


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.