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: 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 $\vert \phi_1 \vert <1$.
A: 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.
