# Null autocorrelation function and stationary

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?

• 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.) Jun 17, 2018 at 15:38

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?