I've been try to either prove or find a counter-example to the idea of jointly-wss being transitive.
In other words: does ($x$ and $y$ are jointly wss) $\wedge$ ($y$ and $z$ are jointly wss) imply that $x$ and $z$ are jointly wss?
Obviously it boils down to only asking about the cross-correlation condition, so one can ask:
Does $\forall t, \Delta t : C_x,_y(t, t+\Delta t)=C_x,_y(0, \Delta t) \ and \ C_y,_z(t, t+\Delta t)=C_y,_z(0, \Delta t)$ imply that $\forall t, \Delta t : C_x,_z(t, t+\Delta t)=C_x,_z(0, \Delta t)\ $?
My basic intuition is that this is not the case, but I cannot find an appropriately pathological set of signals...Nor have I managed to prove transitivity.
Any help would be very much appreciated!
Thank you.


The simplest counter example I can think of (and it is certainly pathological) is $y(t)=0$ for any $x,z$ which are not jointly WSS.

| improve this answer | |
  • $\begingroup$ Good point... * slightly embarrased*...I wasn't thinking about "trivial" signals. This does answer the question, but what if we don't include 0-signals? $\endgroup$ – ShlomiF Jun 18 '15 at 17:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.