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Suppose $f(t)$ is a continuous-valued, zero-mean stochastic signal with Gaussian autocorrelation (with variance $\sigma^2$). Suppose I then pass this signal through a step function, producing a new $\pm$1 signal $g(t)$ that has value +1 wherever $f(t) \ge 0$, and -1 wherever $f(t) < 0$. This signal is similar to a random telegraph signal, though perhaps with different statistics. My question is: does the autocorrelation of $g(t)$ have a simple form? If it were a random telegraph signal where the switching was described by a Poisson process, then I take it the autocorrelation would decay exponentially with the lag. But the switches between the two states (+1 and -1) don't seem to be independent here. Any guidance or references would be helpful.

More generally, suppose f is composed with a "soft" step function, such as the error function $\mathrm{erf}(u)$. For instance, let $g(t)=\mathrm{erf}(\alpha f(t))$. For large $\alpha$, this approaches the case above. In general, what is the autocorrelation of $g$?

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  • $\begingroup$ Why do you make the statement or the claim that the switches are not independent here? Obviously, switches between $+1$ alternate with $-1$, so in one sense of sequence alone with disregard to the time between the sign changes there is a predictable component: $+1$ follows $-1$ follows $+1$... However, the time interval between the sign changes is still a random variable, isn't it? Also, the $\LaTeX$ command you want for $\pm$ is \pm, not \plusminus $\endgroup$
    – sleepless in beantown
    Commented Jan 5, 2011 at 6:28
  • $\begingroup$ I could be wrong about the independence (and thanks for thinking about this). I was thinking that the switches are not independent in the sense of a Poisson process. Thank you, Didier, for the answer below. Is there are reference you could point me to? Also, is it possible to generalize this in the case where f is composed with not the sgn function, but the function $h(u) = \erf(\alpha u)$ (i.e., $g(t) = h(f(t))$). For large \alpha, this approaches the case above. Is there a simple form for the autocorrelation of g in general? Thank you. $\endgroup$
    – James Hsieh
    Commented Jan 5, 2011 at 13:32
  • $\begingroup$ Hi Didier -- thank you very much for your reply. I've edited my original question to include the more general one. $\endgroup$
    – James Hsieh
    Commented Jan 5, 2011 at 17:14

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Assume $(f(t))_t$ is Gaussian and centered with variance $\sigma^2$ and autocorrelation $E[f(t)f(s)]=c(t,s)$ and let $g(t)=\mathrm{sgn}(f(t))$. Then $(g(t))_t$ is $\pm1$ and centered with autocorrelation $$ E[g(t)g(s)]=\frac{2}{\pi}\arcsin\left(\frac{c(t,s)}{\sigma^2}\right).$$

Now I answer the OP's more general question.

The computation in the sign function case is based on the fact that $(f(t),f(s))$ is a Gaussian vector distributed like $(N,aN+bN')$ with $a=c(t,s)/\sigma^2$ and $b=\sqrt{1-a^2}$, for two independent centered Gaussian $N$ and $N'$ with variance $\sigma^2$. One computes $$ E[g(t)g(s)]=2P[N>0,N'\ge -uN] $$ as a two-dimensional Gaussian integral for the suitable value $u=a/b$ and this yields the result.

Likewise, if $g(t)=h(f(t))$ for a given odd function $h$, $E[g(t)g(s)]=E[h(N)h(aN+bN')]$ and it remains to compute this two-dimensional Gaussian integral.

If $h(u)=\mathrm{erf}(\alpha u)$ as suggested by the OP, (I believe that) one gets $$ E[g(t)g(s)]=\frac{2}{\pi}\arcsin\left(\frac{\alpha^2c(t,s)}{1+\alpha^2\sigma^2}\right). $$

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