What a Spectral Measure Measures
For a weakly stationary sequence, the covariance depends on the time lag rather than on the particular times being compared. Herglotz’s theorem expresses that covariance sequence as the Fourier coefficients of a finite, nonnegative measure on the frequency circle. Interpreting this measure starts with three distinct questions: how much mass there is altogether, how much lies below a given frequency, and how much is concentrated at one frequency. These quantities describe different features and should not be interchanged.
Throughout, let \((X_n)_{n\ge0}\) be a real-valued weakly stationary process on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Write its constant mean as \(m_X=\mathbb{E}[X_n]\), and its autocovariance as \(\gamma(h)=\operatorname{Cov}(X_{n+h},X_n)\) for integer lags \(h\). We use the Fourier convention from the earlier material: frequencies are represented in \((-\pi,\pi]\), with \(-\pi\) and \(\pi\) denoting the same point of the frequency circle, and \(\gamma(h)=\int_{(-\pi,\pi]}e^{ih\omega}\,F(d\omega)\). Here \(F\) is the spectral measure. In particular, it is a finite, nonnegative measure; its total mass is not required to be one.
A frequency band is an arc, or a finite union of arcs, on that circle. Its mass is obtained by applying \(F\) to the band, with the endpoints included or excluded according to the band being asked about. When an endpoint carries no atom, including or excluding it makes no difference; when it carries an atom, it does. Because a measure is additive on disjoint sets, the masses of a disjoint partition of the frequency circle sum to the total mass.
In this sense, the spectral measure assigns variance across frequency bands. This phrase means that the variance equals the total mass and that the measure allocates this total among frequency sets. It does not, by itself, construct a separate random variable for each band. Such a process-level decomposition belongs to the spectral representation of stationary sequences, treated later.
Total Mass, Cumulative Mass, and Atoms
Proof. Set \(h=0\) in the defining Fourier relation for \(F\). The integrand is \(e^{i0\omega}=1\) at every frequency, so \(\gamma(0)=\int 1\,F(d\omega)=F((-\pi,\pi])\). By the definition of autocovariance at lag zero, \(\gamma(0)=\operatorname{Var}(X_n)\). For the jump formula, the disjoint union \((-\pi,x]=(-\pi,x)\cup\{x\}\) gives \(H(x)=F((-\pi,x))+F(\{x\})\), by additivity of the measure. Rearranging gives the stated equality. At the identified endpoint, the same disjoint-union argument uses the representative \(\pi\) and frequencies approaching it from below. \(\square\)
The cumulative value \(H(x)\) counts everything from just above \(-\pi\) through \(x\), not just the mass at \(x\). A point mass is a jump, not the height of the cumulative distribution. Also, the total mass is the variance, not the variance divided by \(2\pi\); the factor \(2\pi\) appears only in particular measures, depending on the Fourier convention and the process.
Worked Example: White Noise and a Band’s Mass
Let \((X_n)_{n\ge0}\) be mean-zero white noise with variance \(\sigma^2>0\): its autocovariance is \(\gamma(0)=\sigma^2\) and \(\gamma(h)=0\) for every nonzero integer \(h\). Its spectral measure is \(F(d\omega)=\frac{\sigma^2}{2\pi}\,d\omega\) on \((-\pi,\pi]\). Compute the total mass, the mass of the band \([-\pi/2,\pi/2]\), and the cumulative distribution.
The total mass is the constant \(\frac{\sigma^2}{2\pi}\) times the length \(2\pi\) of the frequency interval: \(\frac{\sigma^2}{2\pi}(2\pi)=\sigma^2\). Thus it agrees with \(\gamma(0)\). The band \([-\pi/2,\pi/2]\) has length \(\pi\), so its mass is \(\frac{\sigma^2}{2\pi}\pi=\frac{\sigma^2}{2}\). Endpoints do not change this calculation because a constant multiple of Lebesgue measure assigns zero mass to a single point.
For \(-\pi<x\le\pi\), the interval \((-\pi,x]\) has length \(x+\pi\). Therefore \(H(x)=\frac{\sigma^2}{2\pi}(x+\pi)\). The limiting value as \(x\) decreases to \(-\pi\) is zero, and \(H(\pi)=\frac{\sigma^2}{2\pi}(2\pi)=\sigma^2\). Every singleton has mass zero, so this measure has no jumps. Check: the band and its complement each have mass \(\sigma^2/2\); their sum is \(\sigma^2\), the total variance.
The cumulative distribution is useful when a question asks for all the mass up to a cutoff. If the cutoff moves across a region with no atoms, the cumulative mass changes continuously for the examples at hand; crossing an atom adds its entire mass at once. On the circle, a band crossing the endpoint must be treated as one arc that wraps from frequencies near \(\pi\) to frequencies near \(-\pi\), or as the union of two intervals. The representative interval is only a way to write frequencies; \(-\pi\) and \(\pi\) are not two distinct frequencies.
Worked Examples with Concentrated Mass
Worked Example: A Constant-in-Time Random Process
Let \(Z\) be a random variable on \((\Omega,\mathcal{F},\mathbb{P})\) with \(\mathbb{E}[Z]=0\) and \(\operatorname{Var}(Z)=\tau^2>0\), and set \(X_n=Z\) for every \(n\ge0\). Find the spectral measure and distinguish its total mass, cumulative mass, and atom.
For any integer lag \(h\), \(X_{n+h}=Z=X_n\), so \(\gamma(h)=\operatorname{Cov}(Z,Z)=\tau^2\). The measure \(F=\tau^2\delta_0\), where \(\delta_0\) assigns mass one to the frequency \(0\), has Fourier coefficients \(\int e^{ih\omega}\,\tau^2\delta_0(d\omega)=\tau^2 e^{ih0}=\tau^2\). It therefore gives the required covariance sequence.
The total mass is \(F((-\pi,\pi])=\tau^2\), matching \(\operatorname{Var}(X_n)\). For \(x<0\), \(H(x)=0\), while for \(x\ge0\), \(H(x)=\tau^2\). In particular, the mass at zero is the jump \(\tau^2-0=\tau^2\), and every band containing zero has mass \(\tau^2\); a band excluding zero has mass zero. Check: at lag \(h=0\), the Fourier integral is \(\tau^2\); at any integer lag it is also \(\tau^2\), as required.
Worked Example: A Random-Phase Oscillation
Let \(A\) and \(B\) be independent, mean-zero random variables on \((\Omega,\mathcal{F},\mathbb{P})\), each with variance \(2\). For \(\omega_0=\pi/3\), define \(X_n=A\cos(n\omega_0)+B\sin(n\omega_0)\). Determine the covariance and the mass assigned to the positive-frequency band \([0,\pi/2]\).
For integers \(n,h\ge0\), independence and zero means imply \(\operatorname{Cov}(A,B)=0\). Hence \(\gamma(h)=2\cos((n+h)\omega_0)\cos(n\omega_0)+2\sin((n+h)\omega_0)\sin(n\omega_0)\). Using \(\cos(u-v)=\cos(u)\cos(v)+\sin(u)\sin(v)\), this becomes \(\gamma(h)=2\cos(h\omega_0)=2\cos(h\pi/3)\), which depends only on \(h\). The sequence is weakly stationary.
Consider \(F=\delta_{\pi/3}+\delta_{-\pi/3}\). Its Fourier coefficient at integer lag \(h\) is \(e^{ih\pi/3}+e^{-ih\pi/3}=2\cos(h\pi/3)\), so it gives the calculated covariance. Each of the two frequencies has mass \(1\), and the total mass is \(1+1=2\). The band \([0,\pi/2]\) contains \(\pi/3\) but not \(-\pi/3\), so its mass is \(1\); the endpoint frequencies \(0\) and \(\pi/2\) have no mass. The cumulative distribution is \(0\) below \(-\pi/3\), \(1\) from \(-\pi/3\) up to but not including \(\pi/3\), and \(2\) from \(\pi/3\) onward. Check: \(H(\pi)=2\), the total variance, and the jumps at \(-\pi/3\) and \(\pi/3\) are each \(1\).
These examples show why a band’s location matters. White noise spreads its variance uniformly across frequency, while the constant process places all its mass at frequency zero. The oscillating process places mass at two opposite frequencies. In each case the total remains the variance, but the cumulative distribution and the point masses reveal how that total is arranged.
Reading Jumps Correctly
Worked Example: Cumulative Mass Is Not Point Mass
Use the oscillating-process measure \(F=\delta_{\pi/3}+\delta_{-\pi/3}\). What is the mass at \(\pi/3\), and what would go wrong if the cumulative value there were treated as the point mass?
All mass at or below \(\pi/3\) is counted in \(H(\pi/3)\). The atom at \(-\pi/3\) contributes \(1\), and the atom at \(\pi/3\) contributes another \(1\), so \(H(\pi/3)=2\). Just to the left of \(\pi/3\), only the negative-frequency atom has been counted, so \(F((-\pi,\pi/3))=1\). The mass at the individual frequency is therefore \(H(\pi/3)-F((-\pi,\pi/3))=2-1=1\).
A mistaken calculation would call \(H(\pi/3)=2\) the mass at \(\pi/3\). That incorrectly includes the mass at \(-\pi/3\), which is already below the cutoff. The correction is to take the jump, not the cumulative height. Check: the two atoms have mass \(1\) each, so the total is \(2\); the jump at \(\pi/3\) is \(1\), not \(2\).
Common pitfall. Do not infer that a frequency has positive mass merely because the cumulative distribution is positive there. For example, in the oscillating-process example, \(H(0)=1\), but \(F(\{0\})=0\): the cumulative mass comes from \(-\pi/3\), and there is no jump at zero. Conversely, a jump at a cutoff is evidence of an atom exactly at that frequency. When a band endpoint is included, its atom must be counted; when excluded, it must not be counted.
The interpretation is therefore systematic: use \(F(B)\) for the mass of a specified band, use \(H(x)\) for all mass up to a cutoff in the chosen interval representation, and use the jump of \(H\) for the mass at a single frequency. First identify whether the question concerns a band, a cumulative total, or a point. Then account for the endpoint convention and, for a band crossing \(\pi\), remember that the frequency domain is a circle.
Check Your Understanding
Use the definitions and identities above. For questions involving a point mass, distinguish the cumulative value from its jump.
- A weakly stationary process has variance \(5\). What is the total mass of its spectral measure, and which covariance lag gives this identity?
- For \(F(d\omega)=\frac{3}{2\pi}\,d\omega\) on \((-\pi,\pi]\), find the mass of \([-\pi/3,\pi/3]\). Do its endpoints affect the answer?
- For \(F=2\delta_0\), give \(H(x)\) for \(x<0\) and for \(x\ge0\), and state the mass at zero.
- If \(H(x)=4\) but its left limit at \(x\) is \(3\), what is \(F(\{x\})\)? Explain why \(4\) is not the point mass.
- Why must a frequency band crossing \(\pi\) be interpreted as an arc on the frequency circle rather than as an ordinary interval extending past \(\pi\)?