Given a random signal $ Z \left( t \right) $ which is addition of two independent signals $ X \left( t \right) $ and $ Y \left( t \right) $ with constant parameters $ a $ and $ b $:

$$ Z (t) = aX(t) + bY (t) $$

If the auto correlation function of $ R_{XX} \left( \tau \right) $ and $ R_{YY} \left( \tau \right) $ is known (Assume both are Stationary Signals).
What would be the Auto correlation function of $ Z \left( t \right) $?

Is it given by (I skipped the steps):

$$ {R}_{ZZ} \left( \tau \right) = a^2 R_{XX} \left( \tau \right) + b^2 R_{YY}\left( \tau \right) $$

I neglected the $ {R}_{XY} \left( \tau \right) $ contribution because it is zero as $ X \left( t \right) \perp Y \left( t \right) $ (Independent Random Signals).


You're correct as the Cross Correlation function vanishes.
This has the implicit assumption the process has zero mean (Actually, at least one of them).

Namely, in order to have $ {R}_{XY} \left( \tau \right) = 0 $ having $ X \left( t \right) \perp Y \left( t \right) $ isn't enough but at least of them has zero mean (Namely, $ \mathbb{E} \left[ X \left( t \right) \right] = 0 $, $ \mathbb{E} \left[ Y \left( t \right) \right] = 0 $ or both).

Pay attention that in practice for a single realization the empirical cross correlation won't vanish.

  • 1
    $\begingroup$ just a minor point. If you're going to use $\tau$ to denote Z's correlation lag, then you should use it for x and y also. otherwise, the assumption is usually that $\tau = 0$ which can possibly confuse readers. $\endgroup$
    – mark leeds
    May 31 '18 at 21:32
  • $\begingroup$ I disagree, E(x y)=E(x)E(y) when x and y are independent. $\endgroup$
    – user28715
    Jun 2 '18 at 0:16
  • $\begingroup$ What here not to agree? He states the variables are independent. I implicitly assume each is centered as well. $\endgroup$
    – Royi
    Jun 2 '18 at 8:33
  • $\begingroup$ @Royi You implicitly assumed that it was centered. Assumptions should be explicit. Neither you or the OP state that it is "centered." If the processes are not "centered," do you contend that your answer is correct of not? So what's not to agree, It's wrong as is. $\endgroup$
    – user28715
    Jun 2 '18 at 21:50
  • $\begingroup$ @StanleyPawlukiewicz, I updated the question to make the implicit assumption explicit. Yet I'd say that no one would use the Auto Correlation Function for non centered signals (What's the point of using the ACF for analysis of the data in that case?). So think the above is all about the Autocovariance Function. $\endgroup$
    – Royi
    Jun 2 '18 at 22:50

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