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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.

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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.

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  • $\begingroup$ 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

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