2
$\begingroup$

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.

$\endgroup$

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ A technical point that is insisted on by almost all authorities is that the variances be finite. Otherwise, these calculations cannot be carried out. $\endgroup$
    – whuber
    Commented Jul 16, 2022 at 13:58
  • 1
    $\begingroup$ @whuber Thanks. I have incorporated this point into my answer. $\endgroup$ Commented Jul 16, 2022 at 15:19
4
$\begingroup$

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 $\vert \phi_1 \vert <1$.

$\endgroup$

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.