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.
- Is there a proof for covariance-only stationarity → mean stationarity?
- 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
Best Answer
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 $t$.