Can two weak stationary time series $X_t, Y_t$ have covariance $Cov(X_t,Y_t)$ that changes over time? No solution is needed, please give me some hints. I think it can but I find it is difficult to support my idea.

  • $\begingroup$ If you cannot prove your idea, just sharing your idea is also good. I think Cov(Xt, Yt) may change over time. But I can not find the way to support my idea. $\endgroup$
    – Wayne W
    Nov 4 '20 at 12:13
  • $\begingroup$ Is that a homework exercise? If so, please add the self-study tag and read its Wiki. $\endgroup$ Nov 4 '20 at 12:27
  • $\begingroup$ It a question that teacher left us, not a necessary homework. Thanks your comment ! $\endgroup$
    – Wayne W
    Nov 4 '20 at 12:35
  • $\begingroup$ Hint: let each series consist of iid variables. Does this imply the variables in one series must be independent of those in the other? $\endgroup$
    – whuber
    Nov 5 '20 at 15:05
  • $\begingroup$ Thanks your hint. Yes! It must be independent. But because of independence, the covariance must be 0 which will not change with t. $\endgroup$
    – Wayne W
    Nov 6 '20 at 8:28

Let $(X_t),$ with "times" $t=0,1,2,\ldots,$ be variables with independent and identical distributions having finite variance $\sigma^2$ -- which implies they form a stationary (whence weakly stationary) stochastic process. Suppose further that this common distribution is symmetric about $0$: this means the $-X_t$ all have the same distribution as the $X_t.$ For instance, the standard Normal distribution is symmetric about $0.$


$$Y_t = (-1)^t X_t$$

and notice that

$$\operatorname{Cov}(X_t,Y_t) = \operatorname{Cov}(X_t,(-1)^tX_t) = (-1)^t \operatorname{Cov}(X_t,X_t) = (-1)^t\sigma^2$$

alternates between $\sigma^2$ and $-\sigma^2:$ that is, it changes over time. Nevertheless, the symmetry of the distribution implies the $(Y_t)$ are identically distributed and, since the $(X_t)$ are independent, the $(Y_t)$ are independent too. Thus $(Y_t)$ is a (weakly) stationary process, too. In this example the covariance changes over time.

  • 2
    $\begingroup$ Thanks!!! This is really ingenious! $\endgroup$
    – Wayne W
    Nov 6 '20 at 13:06

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