# Second order weakly stationary process

I was wondering, whether a stochastic process with a covariance, that is not depending on time, imply constant variances. Since I read a textbook, where they call a process, which is second order weakly stationary also covariance stationary.

A weakly stationary process is defined as one that has two properties:

• the mean is a constant: $$E[X_t] = \mu$$ for all choices of $$t$$
• the autocorrelation function $$R_X[s,t]=E[X_sX_t]$$ has finite value for all choices on $$s$$ and $$t$$,and the value of $$R_X[s,t]$$ depends only on $$s-t$$ and not on the individual values of $$s$$ and $$t$$. In particular, for all $$t$$, $$R_x(t,t) = E[X_t^2]= E[X_0^2]$$ has finite constant value (sometimes referred to as the power of the process, especially by engineers) as does $$\operatorname{Var}(X_t) = E[X_t^2] - \mu^2 = E[X_0^2]- \mu^2$$.

Since $$\operatorname{Cov}(X_s, X_t) = E[X_sX_t] - E[X_s]E[X_t] = R_X[s,t]-\mu^2$$, we see that $$\operatorname{Cov}(X_s, X_t)$$ also depends only on the difference $$s-t$$ and not on the individual values of $$s$$ and $$t$$, that is, weak stationarity implies covariance stationarity.. The reverse implication -- that covariance stationarity implies weak stationarity -- is not true since the mean might not be constant. A standard example of a covariance-stationary process that is not weakly stationary is the sum $$X_t+y_t$$ of a weakly stationary process $$X_t$$ and a deterministic (time-varying) signal $$y_t$$.

See the latter half of this answer of mine over on dsp.SE for more than what you probably want to know about this matter.

• A technical point that is insisted on by almost all authorities is that the variances be finite. Otherwise, these calculations cannot be carried out.
– whuber
Jul 16, 2022 at 13:58
• @whuber Thanks. I have incorporated this point into my answer. Jul 16, 2022 at 15:19

Actually, it does. If you have a stochastic process $$(X_t)$$, with $$Cov(X_t,X_s)$$ that does not depend on time, then $$V(X_t)=Cov(X_t,X_t)$$ does not depend on $$t$$ either. Of course, the question remains whether or not $$V(X_t)$$ is finite. Usually, some parameter restrictions need to be satisfied to guarantee that.

For example, an AR(1) process $$X_t=\phi_0+\phi_1X_{t-1}+\epsilon_t$$, $$\epsilon \sim \mathcal N(0,\sigma^2)$$ has autocorrelation function $$\gamma_k=\phi_1\gamma_{k-1}$$ that does not depend on $$t$$ and the variance is given by $$\gamma_0=\frac{\sigma^2}{1-\phi_1^2}$$ which is also constant and finite as long as $$\vert \phi_1 \vert <1$$.