2
$\begingroup$

The Bussgang theorem states that if a stationary Gaussian process $X(t)$ with covariance function $R(\tau)$ is passing any non-linear function, affecting only the amplitude, $g(x)$, then the covariance of $y(t)=g(X(t))$ is $CR(\tau)$, for some C.

Now, let the process be instead cyclo-stationary with covariance $R(\tau_1,\tau_2)=R(\tau_1-\tau_2,\tau_2)=R(\tau_1-\tau_2,\tau_2+T)$, where $T$ is the period. Is it true that the covariance of $Y(t)=g(X(t))$ is $CR(\tau_1,\tau_2)$, possibly with another constant $C$?

Improve this question
$\endgroup$
1
  • $\begingroup$ Do you really mean what you write? what does ``affecting only the amplitude'' mean? The statement can't be true as written for stationary processes - take $g(x)=x^2$ to see that, as in that case $Ey^2(t)y^2(s)=R(0)^2+2R(|t-s|)^2$. Further, the wikipedia page linked is also not clear, and in any case does not deal with the covariance of $y$ but rather with the cross-correlation of the $x$ and $y$ processes. In short, this question needs cleanup. $\endgroup$
    – ofer zeitouni
    Commented Nov 12, 2013 at 5:57

1 Answer 1

Reset to default
2
$\begingroup$

The correct statement is the following: $$E(Y(t)X(s))=F(t) E(X(t)X(s))$$ for some function $F(t)$ which is periodic. To see that, follow verbatim the proof in the appendix of http://www3.alcatel-lucent.com/bstj/vol61-1982/articles/bstj61-7-1519.pdf

Improve this answer
$\endgroup$
2
  • $\begingroup$ I think the link is broken. Could you please update the link @ofer? Thanks in advance! $\endgroup$
    – Shanks
    Commented Apr 11, 2019 at 16:24
  • $\begingroup$ This is the bell systems technical journal, volume 61, number 7, page 1519. I don't have open access to it but it can probably be accessed through library (it is a wiley journal). You can try archive.org/details/… if you have the patience to sort through their user-unfriendly interface. $\endgroup$
    – ofer zeitouni
    Commented Apr 12, 2019 at 18:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .