My question has to do with the definition of auto correlation/cross-correlation for random processes.
Oppenheim/Schafer (Discrete time Signal Processing, Pg. 815 (Appendix A.2),2nd ed.) define auto correlation in the following way:
$\phi_{xx}[n,m]=E[\mathbf{x_nx^*_m}]=\int_\infty^\infty x_nx^*_m p_{x_n,x_m}(x_n,n,x_m,m)dx_ndx_m$

{$\mathbf{x_n}$} is a random process and $p_{x_n,x_m}(x_n,n,x_m,m)$ is it's joint probability density function.
What is the significance of using the conjugate of $x_m$ in the above formula? Is there a physical interpretation for it?

  • $\begingroup$ Cojugation is done because random variable can be complex also. Even in convolution expression also conjugation is used if signals are complex, Hence "Conjugation" it is used in all generalized formulas. $\endgroup$
    – spectre
    Mar 30, 2016 at 4:55
  • $\begingroup$ @spectre No, conjugation is not used when convolving complex-valued signals; it is used when correlating them. $\endgroup$ Mar 30, 2016 at 15:49

1 Answer 1


it's funny, but i have an old copy of O&S and i would have expected them to be consistent and use "$x[n]$" notation instead of "$x_n$" notation for discrete-time signals.

The expression above

$$ \phi_{xx}[n,m] \ = \ E[\mathbf{x_n \ x^*_m}] \ = \ \int_\infty^\infty \int_\infty^\infty \ x_n \ x^*_m \ p_{x_n,x_m}(x_n,n,x_m,m) \ dx_n dx_m $$

is the average value of the pair $\mathbf{x_n \ x^*_m}$ where the average is computed from the joint probability density function $ p_{x_n,x_m}(x_n,n,x_m,m) $

If the process is stationary and ergodic (and i think that any process that is fully stationary is also ergodic) i think this can be simplified a little

$$ \phi_{xx}[n-m] \ = \ E[\mathbf{x_n \ x^*_m}] \ = \ \int_\infty^\infty \int_\infty^\infty \ x_n \ x^*_m \ p_{x_n,x_m}(x_n,x_m) \ dx_n dx_m $$

now the root meaning of "ergodic" is that probabilistic averages can be replaced with time averages and vise-versa:

$$ \langle \mathbf{x_n \ x^*_{n+k}} \rangle = \lim_{L \to \infty} \frac{1}{2L+1} \sum\limits_{n=-L}^L \mathbf{x_n \ x^*_{n+k}} $$

the process is ergodic if all averages, including the one expressed above, are the same, whether they are probabilistic or computed from the data

$$ \langle \mathbf{x_n \ x^*_{n+k}} \rangle = \phi_{xx}[k] = E[\mathbf{x_n \ x^*_{n+k}}] $$

  • $\begingroup$ Thank for your comment, Robert. However, that still does not answer my actual question - what is the need for taking the conjugate of $\mathbf{x_m}$, and not just $\mathbf{x_m}$? $\endgroup$
    – Saqib Shah
    Mar 30, 2016 at 3:32
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    $\begingroup$ that's a good point, now that you make me look at it. it appears to me that O&S are not sticking to the convention of using the angle bracket operator notation for the inner product. if you have a finite dimension vector space (this one is not because $L \to \infty$) then the inner product of two vectors (with possibly complex elements) is $$ \langle \mathbf{x,y}\rangle = \sum\limits_{n=1}^{L} x_n (y_n)^* $$ the $\mathbf{y}$ is not shown as conjugated in the $\langle \ . \ \rangle$ expression. $\endgroup$ Mar 30, 2016 at 4:46
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    $\begingroup$ the reason why the elements are conjugated on the right-hand side is $$ ||\mathbf{x}|| = \sqrt{\langle \mathbf{x,x} \rangle} $$. O&S's convention (which is bad, in my opinion) is $$ ||\mathbf{x}|| = \sqrt{ \langle \mathbf{x,x}^* \rangle } $$. $\endgroup$ Mar 30, 2016 at 4:47

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