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A wide-sense stationary random time series $zeta(t)$ is characterized by its mean value and its autocovariance function, which in the Wiener–Khinchin theorem is equivalent to the Fourier transform of the power spectral density $$langle[zeta-langlezetarangle][zeta’-langlezetarangle]rangle = sigma^2 C(tau)-langlezetarangle^2 = frac 1{2pi}int_{-infty}^infty S(omega)exp(-omegatau)domega$$ where $sigma^2 C(tau)$ is the autocovariance function with rms amplitude $sigma$, $tau = t – t’$ is the stationary variable for the time series, and $langlezetarangle$ is the mean value of $zeta$. The inverse Fourier transform of the autocovariance function is $$S(omega) + 2pi langlezetarangle^2delta(omega) = int_{-infty}^infty C(tau)exp(omegatau)dtau$$ and consists of the absolutely continuous power spectral density and the discontinuous line spectral component $2pi langlezetarangle^2delta(omega)$ at $omega = 0$ that only appears when the mean is non-zero.

Conjecture: For a zero-mean random process, i.e. $langlezetarangle = 0$, it is conjectured that the power spectral density vanishes when the frequency is zero (i.e. $S(0) = 0$). The conjecture rests on the physical interpretation of the zero frequency ($omega = 0$) component of the spectrum. The zero frequency term corresponds to the DC term in a time series, and the DC term represents the mean value. If the random variable has a zero-mean then the DC term is zero and thus both the mean value $langlezetarangle$ and the spectrum must be zero at $omega = 0$. In fact, since the mean value is always contained in the discontinuous line spectral component $2pi langlezetarangle^2delta(omega)$, the continuous portion of the power spectral density $S(omega)$ must always be zero at $omega = 0$ whether the mean is zero or not.

Question: Is this conjecture true or false?

This result proves the conjecture is true. Consider a stationary complex random function $zeta(t)$. Stationarity requires that the autocorrelation function depend on the difference $t –t’$ only and that the mean value must be a constant (not a member of a distribution function nor depend on $t$). Assume the power spectral density (hereafter called the spectrum) is absolutely continuous everywhere. The Fourier transform of $zeta(t)$ itself does not exist in general, because stationary random functions are generally neither absolutely integrable nor square integrable, i.e. they do not satisfy Dirichlet’s condition. The correct representation of the random variable $zeta(t)$ is either as a Fourier series or as a Fourier-Stieltjes integral [see, for example, Akira Ishimaru, Wave Propagation and Scattering in Random Media, (IEEE Press, 1997) Appendix A] Here, the Fourier-Stieltjes integral will be used
$$zeta(t) = int_{-infty}^{+infty} e^{iomega t}dA(omega)$$
where $dA(omega)$ is the random amplitude. Typically, the mean value and the autocorrelation function are found using ensemble averages, but this method provides little insight into the behavior of $dA(omega)$. Hence, the average value and autocorrelation function for $zeta(t)$ will be found using temporal averages. The ergodic hypothesis states that for a stationary process the temporal average and the ensemble average are equivalent.

$mathbf Temporal Averages -$ The temporal average $overline{zeta(t)}$ is calculated over a finite interval $T$, and then the limit is taken as $T$ goes to infinity
$$overline {zeta(t)} = lim_{Tto infty} frac 1T int_{-T/2}^{+T/2}zeta(t) dt$$
Substitute the Fourier-Stieltjes integral into the equation and perform the $t$ integration to get
$$overline {zeta(t)} = lim_{Tto infty} frac 1T int_{-infty}^{+infty} dA(omega) int_{-T/2}^{+T/2} e^{iomega t}dt = lim_{Tto infty} frac 1T int_{-infty}^{+infty} dA(omega) frac {2sin(omega T/2)} {omega} $$
Use the following expressions to take the limit

$$ lim_{Tto infty} frac {2sin(omega T/2)}{omega} rightarrow 2pidelta(omega) quad mathrm{and} quad lim_{Tto infty} frac 1T = lim_{Tto infty} frac {Deltaomega} {2pi}rightarrow frac {domega}{2pi} $$
which gives for the temporal average
$$overline {zeta(t)} = int_{-infty}^{+infty} dA(omega) delta(omega) domega = dA(0) $$
Finally, for a zero-mean process

$$overline {zeta(t)} = langle zeta rangle = dA(0) = 0 $$
Thus, the mean value of $zeta(t)$ obtained via temporal averaging is $dA(0)$ which is the random amplitude evaluated at the origin.

In the following analysis the random variable $zeta(t)$ will be decomposed into the sum of zero-mean random variable $zeta_0(t)$ and the mean $langlezetarangle$ so that the mean value is everywhere explicit. Now the temporal average for the autocorrelation function of $overline{zeta(t)zeta^*(t’)}$ is given by
$$ overline{zeta(t)zeta^*(t+tau)} = overline{zeta_0(t)zeta_0^*(t+tau)} + langle zeta rangle^2 = sigma^2C(tau) + langle zeta rangle^2 $$
$$ lim_{Tto infty} frac 1T iint_{-infty}^{+infty} e^{-iomegatau}dA(omega)dA(omega’) int_{-T/2}^{+T/2} e^{i(omega-omega’)t}dt $$
Performing the $t$ integration yields
$$ sigma^2C(tau) + langle zeta rangle^2 = iint_{-infty}^{+infty} e^{-iomegatau}dA(omega)dA(omega’) lim_{Tto infty} frac 1T frac {2sin[(omega-omega’) T/2]} {(omega-omega’)} $$
The following expressions are used in the limit process

$$ lim_{Tto infty} frac {2sin[(omega-omega’) T/2]}{(omega-omega’)} rightarrow 2pidelta(omega-omega’) quad mathrm{and} quad lim_{Tto infty} frac 1T = lim_{Tto infty} frac {Deltaomega} {2pi}rightarrow frac {domega}{2pi} $$
Substituting these expressions into the autocorrelation gives
$$ sigma^2C(tau) + langle zeta rangle^2 = iint_{-infty}^{+infty} e^{-iomegatau}dA(omega)dA(omega’) delta(omega-omega’) domega $$
Finally, performing the $omega$ integration yields
$$ sigma^2C(tau) + langle zeta rangle^2 = int_{-infty}^{+infty} e^{-iomega'tau}|dA(omega’)|^2 qquad mathrm {Equation (1)}$$
The quantity $dA(omega’)$ can be related to the mean value $langlezetarangle$ and spectrum $S(omega)$ by taking the inverse Fourier transform of the autocorrelation function

$$ int_{-infty}^{+infty}sigma^2C(tau)e^{iomegatau} dtau + int_{-infty}^{+infty} langlezeta rangle^2 e^{iomegatau} dtau = int_{-infty}^{+infty}|dA(omega’)|^2int_{-infty}^{+infty}e^{i(omega-omega’)tau} dtau$$
Performing the $tau$ integration yields
$$ S(omega)+2pi langle zeta rangle^2 delta(omega)= 2pi int_{-infty}^{+infty} |dA(omega’)|^2 delta(omega-omega’) $$
Consider the case of $omega = 0$ for a zero-mean process, i.e. $langlezetarangle = 0$; both sides of the expression reduce to
$$ S(0)+0 = 2pi int_{-infty}^{+infty} |dA(omega’)|^2 delta(omega’) $$
The Dirac delta function forces the terms on the right side to be evaluated at $omega’ = 0$ to produce

$$ S(0) = 2pi |dA(0)|^2delta(omega’) = 2pi langle zeta rangle ^2 delta(omega’) =0 $$
where the previous result $dA(0) = langlezetarangle$ has been used. Thus, it is shown that the spectrum for a zero-mean process is zero when the frequency is zero, i.e. $S(0) = 0$. It can be seen in Equation (1) for the general case that the required value of $|dA(omega’)|^2$ must be given by
$$|dA(omega’)|^2 = frac 1{2pi}S(omega’) domega’+|dA(0)|^2 delta(omega’) domega’ $$
$mathbf Discussion -$ The vanishing of the spectrum at $omega = 0$ establishes some new requirements on the autocorrelation function and spectrum. For a zero-mean random variable, when $tau = 0$ the autocorrelation form of the Wiener-Khinchin theorem reduces to

$$ sigma^2 = frac 1{2pi}int_{-infty}^{+infty}S(omega) domega $$
This expression has proven useful and has wide application. Now consider the spectrum form of the Wiener-Khinchin theorem when $omega = 0$. This expression becomes

$$ S(0) = int_{-infty}^{+infty}sigma^2C(tau) dtau =0 $$
Besides the information on the behavior of the spectrum, this gives useful information on the autocorrelation function. For example, the autocovariance function must have both positive and negative excursions, and the area under these excursions must be equal otherwise the integral cannot be zero. This means that the simple Gaussian or exponential forms often used for the autocorrelation function are not valid for a zero mean process – they have no negative excursions anywhere.