Traditionally a weak stationary process is also called covariance stationary, but those 3 properties are exposed:

$$E[Xt] = μ , \forall t$$ $$var(Xt) = \sigma^2, \forall t$$ $$cov(Xt, Xt−j) = \gamma_j, \forall t$$

that are respectively mean, variance and covariance stationarity. We can merge 2 and 3 by extending to $j=0$ and $\gamma_0 = \sigma^2$, and let's call it covariance-only stationarity

But seeing the name covariance stationarity and also trying to figure out examples where the variance is stationary and not the mean, or where covariances (j=0 too) are stationary but not the mean, I was wondering if maybe covariance-only stationarity could lead to mean stationarity.

  1. Is there a proof for covariance-only stationarity → mean stationarity?
  2. or are both covariance-only stationarity and mean stationarity required for having a covariance stationary process?

note: if 2., in my opinion 'covariance stationary' is a very misleading term, weak stationary would be better to use

  • $\begingroup$ A standard example in signal-processing circles is a process consisting of a time-varying deterministic signal plus zero-mean weakly stationary noise. This process is not weakly stationary since the mean function of the process is the signal but it is covariance stationary. The examples in Karl Oskar's answer are of this type, with noise process being strictly stationary instead of only weakly stationary. $\endgroup$ Commented Mar 31, 2014 at 12:51

1 Answer 1


No, the requirement on the covariance function is not enough to ensure weak stationarity. The following is a counterexample to such a claim:

Let $\xi_t $ be iid random variables with zero mean and unit variance. Next, let $X_t=\xi_t + 1$ for $t$ even and $X_t=\xi_t$ for $t$ odd. Consider the process $\{X_t\}_{t=0}^\infty$. Then $Cov(X_t,X_s)=0$ if $t\neq s$ and 1 otherwise. So the covariance function satisfies the condition for weak stationarity. The mean does not, however, since $\mathbb EX_t=1$ when $t$ even and 0 otherwise.

If you want an even simpler example, take $X_0=\xi_0+1$ and $X_t=\xi_t$ for all other $t$.

  • $\begingroup$ For cov(Xt,Xs) t≠s, the 2 variables are independant, so it's 0 . For var(Xt)=E[(X-E[X])²], X-E[X]=$\epsilon_t$ and $E[\epsilon_t^2]=1$ from the chi-squared law with 1 degree of freedom? $\endgroup$
    – caub
    Commented Mar 31, 2014 at 13:51
  • $\begingroup$ No, not quite. The important thing to note is that adding or subtracting a constant to a random variable does not affect its variance. That is, $Var(X_t)=Var(\xi_t)=1$ for all t. $\endgroup$
    – KOE
    Commented Mar 31, 2014 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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