If two wide-sense stationary processes $X(t)$ and $Y(t)$ are uncorrelated, then the cross correlation is

$R_{XY}(t_1,t_2) = E\{X(t_1)Y(t_2)\} = E\{X(t_1)\}E\{Y(t_2)\}$,

which will be a constant, since $E\{X(t_1)\}$ and $E\{Y(t_2)\}$ are constant. But my question is what conclusion about $R_{XY}(t_1,t_2)$ can be made if the wide-sense stationary processes $X(t)$ and $Y(t)$ are correlated? Is it still a constant or loosely $R_{XY}(t_1,t_2) = R_{XY}(t_1-t_2)$?

Similarly, if $X(t)$ and $Y(t)$ are strict stationary process, and they are independent, then we have the joint probability distribution

$\text{pr}\{X(t_1),Y(t_2)\}= \text{pr}\{X(t_1)\}\text{pr}\{Y(t_2)\}$

which is independent of time, since $\text{pr}\{X(t_1)\}$ and $\text{pr}\{Y(t_2)\}$ do not depend on time. Again, what's conclusion about $\text{pr}\{X(t_1),Y(t_2)\}$ if the strict stationary processes $X(t)$ and $Y(t)$ are not independent? Is it still independent of time or $\text{pr}\{X(t_1),Y(t_1+\tau)\}= \text{pr}\{X(t_2),Y(t_2+\tau)\}$?

  • $\begingroup$ The last paragraph does not make much sense. What is meant by the notation $\mathrm{pr}(X(t_1))$? When you say "it is constant", you apparently mean in time, correct? But, still, you should clarify what this notation is intended to mean. $\endgroup$
    – cardinal
    Jul 20, 2012 at 1:21
  • 1
    $\begingroup$ @cardinal you are right. I fixed it. $\endgroup$
    – chaohuang
    Jul 20, 2012 at 1:30

2 Answers 2


It might help to think about it in terms of copulae. $X$ and $Y$ can be converted to grades and joined together by a copula. This copula could potentially be constant or a time varying DCC copula. As a result, the joint distribution could be potentially multivariate constant in the first and second moments or time-varying. In that sense, they would be univariate stationary, but not necessarily multivariate stationary unless you make some assumptions about the dependence structure.


Say $X(n)$ is either all $1$'s or all $-1$'s with 50% probability, constant in time n.

Say $Y(0)=X(0)$ and then $Y(n) = -Y(n-1)$ so its always flipping up and down.

Both processes are stationary.

$E\{X(0)Y(0)\}=1$ but $E\{X(1)Y(1)\}=-1$.


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