I can show that a process $X(t)$ is Wide Sense stationary (WSS) by showing that $E[X(t)]$ is constant and that its autocorrelation function is in function of $\tau=t_1-t_2$, that is, $R_X(t+\tau,t)=R_X(\tau)$.

My question is:

If $E[X(t)]$ is constant and $R_X(t+\tau,t)=0$ can I say the process is WSS? What does $R_X(t+\tau,t)=0$ mean? Can I say $R_X(t+\tau,t)=0=R_X(\tau)$ and therefore is WSS?

  • $\begingroup$ by the way, almost certain that you also need the restriction that either a) the ACF is bounded or b) the distribution of any $X(t)$ has a bounded variance. That's mathematical nitpicking, though. (Imagine a process that is uncorrelated, zero-mean, cauchy-distributed: it's not WSS.) $\endgroup$ Jun 17, 2018 at 15:38

1 Answer 1


If E[X(t)] is constant and RX(t+τ,t)=0 can I say the process is WSS?

Can I say RX(t+τ,t)=0=RX(τ) and therefore is WSS?

Two times the same question. It fulfills the definition (as you noticed yourself), so why even ask? yes.

What does RX(t+τ,t)=0 mean?

It means the process is uncorrelated ($R_X(\tau \ne 0)=0$), but also that its variance is 0 ($R_X(0) = \sigma_X^2 =0$). There's only one process that fulfills that:

$X(t) = 0$

not a really exciting process, is it?


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