I know that 3 conditions must be met in order a pair of stochastic processes $X(t)$ and $Y(t)$ to be characterized as jointly WSS:

1. $X(t)\;\; WSS$

2. $Y(t)\;\; WSS$

3. $R_{xy}(t_1,t_2) = R_{xy}(t_1 - t_2) = R_{xy}(t_2 - t_1)$ which means that their cross-correlation function should depend only on the difference $T = t_1 - t_2$.

My question is: if the first 2 conditions are met but the cross-correlation function does not depend on any variable (I mean that it's a constant e.g. $R_{xy}(t_1,t_2) = 0$) can we assume that the two stochastic processes $X(t)$ and $Y(t)$ are jointly WSS?


1 Answer 1


Yes, when $R_{X,Y}(t_1, t_2) = E[X(t_1)Y(t_2)] = 0~ \forall ~t_1, t_2$, $\{X(t)\}$ and $\{Y(t)\}$ are often called uncorrelated WSS processes, and they are considered to be jointly WSS processes. More nitpicky folks agree that the processes are indeed jointly WSS when $R_{X,Y}(t_1, t_2) = 0~ \forall ~t_1, t_2$, but insist on reserving the adjective uncorrelated for processes for which the cross-covariance function
$$C_{X,Y}(t_1,t_2) = E\left[\left(X(t_1)-E[X(t_1)]\right)\left(Y(t_2)-E[Y(t_2]\right)\right] =E[X(t_1)Y(t_2)]-E[X(t_1)]E[Y(t_2)] = R_{X,Y}(t_1,t_2)-E[X(t_1)]E[Y(t_2)]$$ equals $0$ for all $t_1, t_2$, that is, every pair of random variables $X(t_1)$, $Y(t_2)$ from the processes is a pair of uncorrelated random variables. Of course, when at least one of the processes has zero mean (as is usually the case in signal processing applications), then $C_{X,Y}(t_1,t_2) = R_{X,Y}(t_1,t_2)$ and so the nit gets picked without requiring any further consideration.


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.