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Suppose that we are given a wide-sense stationary random process $X$ with autocorrelation function $R_X(t)$. Power spectral density $S_X(f)$ of $X$ is then given by the Fourier transform of $R_X(t)$, i.e. $S_X(f)=\mathcal{F}(R_X(t))$.

I am wondering if there is a valid power spectral density function $S_X(f)$ such that, for a positive integer $n$, the integral over the entire frequency domain of the absolute value of $S_X(f)$ taken to the $n$-th power is not a finite constant. Formally, is there $S_X(f)$ such that:

$$\int_{-\infty}^{\infty} |S_X(f)|^n df=\infty$$

I know that this is impossible for $n=1$, as $\int_{-\infty}^{\infty} S_X(f) df=R_X(0)=E[X^2]<\infty$, however, I haven't found any result for $n>1$. Perhaps it's very obvious one way or the other (though it seems to me that such $S_X(f)$ does not exist, but I can't find a formal proof). In any case, I would appreciate elucidation.

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I'm confused, are you asking us to find a definite integral that does not evaluate to a constant? There doesn't appear to be additional variables of which this integral could be a function. –  prpl.mnky.dshwshr Mar 26 '12 at 18:41
    
Sorry for the confusion. I am asking if it's possible that the integral evaluates to infinity. I'll clarify the question. –  M.B.M. Mar 26 '12 at 19:10
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Consider the function $\frac{1}{\sqrt{|f|}}$. The singularity is just barely integrable, but if you square it, then it's no longer integrable. Really, you're looking for function that are in $L^{k}(\mathbb{R})$ for some $k<n$ but not for $L^{n}(\mathbb{R})$. I'm sure some of the basic examples from functional analysis textbooks would satisfy what you're looking for. –  prpl.mnky.dshwshr Mar 26 '12 at 19:48
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As an aside, I used to work as a radar engineer, and once an analyst kept grumbling about why his radar simulation was giving him nonsense $\infty$ values. I looked at his code and it was because he was using a non-integrable function for the antenna gain pattern (hence infinite energy radar). When I tried to explain this, he got angry that I was using the word "integrable" (a concept he apparently thought he could forget about after college). –  prpl.mnky.dshwshr Mar 26 '12 at 19:51
    
The thing about your example $\frac{1}{\sqrt{|f|}}$ is that I don't think that it's a valid PSD, as it's not continuous on $[-1/2,1/2]$ and its derivative at $f=0$ is not zero (I think it's $-\infty$ in the limit as $f\rightarrow 0$, but the derivative at $f=0$ is undefined, as the derivative of absolute value is undefined at 0). I like your story from your radar engineering days. –  M.B.M. Mar 26 '12 at 20:23

2 Answers 2

up vote 1 down vote accepted

This is impossible just by definition.

$S_X(f)$, if it exists, is the Radon-Nikodym derivative of the spectral measure $\mu$ with respect to the Lebesgue measure and, being nonnegative, necessarily $L^1$ as you noted. By definition, the Fourier transform of $S_X(f)$ (or $\mu$ in general) is $R_X(t)$. If one insists that $R_X(t)$ lies in the domain of the (inverse) Fourier transform, then Fourier inversion theorem implies that $S_X(f)$ is in fact continuous almost everywhere. A continuous function in $L^1$ lies also in $L^p$ for any $p > 1$.

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I like your answer, thanks! It matches the intuition I originally had. However, could you please let me (and the world) know where can one find the proof (or the name of the theorem if it's sufficiently well-known) of the statement in the last sentence: "A continuous function in $L^1$ lies also in $L^p$ for any $p>1$." I don't think I've encountered that before... –  M.B.M. May 28 '13 at 4:51
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Well, a continuous $f \in L^1$ must lie in $L^{\infty}$ since Lebesgue measure of $\mathbb{R}$ is not finite. Take $p > 1$, $|f|^p \leq \|f\|^{p-1}_{\infty} \cdot |f|$. So $\|f\|_p \leq \|f\|^{\frac{p-1}{p}}_{\infty} \cdot \|f\|_1 ^{\frac{1}{p}}$. Informally, continuity means that the blowup that might take $f$ out of $L^p$ can only occur at $\infty$ but boundedness precludes that. Same estimate goes through if $\infty$ is replaced by a finite $q > 1$, if we use Holder's inequality. –  Michael May 28 '13 at 14:09

A non-negative function is not necessarily $L^1$ on the real axis, although that seems to be what one of the answers (the other one) is saying. First of all, one must assume the process is not only stationary, but ergodic (a much stronger assumption) so that its power spectrum at least goes to zero at infinity. Secondly, even if the power spectrum is, e.g., compactly supported, e.g., a narrow band-limited but flat "pink noise", its Fourier transform need not be $L^1$ and so Fourier inversion will not be true. In fact the auto-correlation of pink noise is the sinc function $\sin \omega t \over t$ which is not absolutely integrable. So it is not true to say the the power spectral density is the Fourier transform of the sinc function, it is not true that the power spectral density and the auto-correlation function form a Fourier transform pair (except sometimes in the sense of distributions, e.g., when the process is ergodic).

so we cannot "insist that $R_X(t)$ lies in the domain of the (inverse) Fourier transform". there are a lot of mistakes floating around the internet, copied from electrical engineering textbooks or handouts which have mistakes in them.

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