Why Fourier Conventions Need to Be Fixed
A discrete-time covariance sequence is indexed by integer lags, but its frequency description uses a continuous frequency variable. The same spectrum can therefore be written using angular frequency, measured in radians per time step, or cycle frequency, measured in cycles per time step. The formulas are equivalent, but their frequency intervals and density normalizations differ. A missing factor of \(2\pi\), or a sign change made in only one of the paired formulas, changes the answer.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((X_n)_{n\in\mathbb{Z}}\) be a real-valued, weakly stationary process with state space \(S=\mathbb{R}\) and discrete time index \(n\in\mathbb{Z}\). Write its autocovariance sequence as \(\gamma(h)=\operatorname{Cov}(X_{n+h},X_n)\), for \(h\in\mathbb{Z}\). Weak stationarity makes this independent of \(n\), and for a real-valued process \(\gamma(-h)=\gamma(h)\). The variance is \(\gamma(0)\). In the formulas below, an ordinary spectral density is defined when \(\sum_{h\in\mathbb{Z}}|\gamma(h)|<\infty\); this condition guarantees convergence of the Fourier series. Not every covariance sequence satisfies it.
The minus sign in each forward series and the plus sign in each reconstruction formula are a pair: multiplying by the positive-sign exponential selects the coefficient at lag \(h\). For a real covariance sequence, evenness also means the forward series can be written as a cosine series, so changing the sign in the exponential happens not to change its value. That symmetry is not a reason to mix conventions: the paired formulas specify how the lag and frequency variables correspond.
Coefficient Recovery and Normalization
Proof. For every \(\omega\), the absolute value of the tail of the series is bounded by the corresponding tail of \(\sum_h|\gamma(h)|\). Since that numerical series converges, its tails tend to zero independently of \(\omega\), so the Fourier series converges uniformly. For a finite partial sum, multiply by \(e^{ik\omega}\) and integrate. The elementary identity \(\int_{-\pi}^{\pi}e^{i(k-h)\omega}\,d\omega=2\pi\) when \(h=k\), and \(0\) when \(h\ne k\), leaves precisely the coefficient \(\gamma(k)\) after the factor \(1/(2\pi)\) is included. Uniform convergence permits taking the limit of the partial sums inside the integral, since the interval has finite length and the integral of the uniform error is bounded by \(2\pi\) times that error. Thus \(\int_{-\pi}^{\pi}e^{ik\omega}f_{\mathrm{ang}}(\omega)\,d\omega=\gamma(k)\). Taking \(k=0\) gives the stated normalization. \(\square\)
The normalization has a useful interpretation: the total density over the angular-frequency interval is the process variance, not a probability density that must integrate to one. The cycle-frequency density has the same total variance over its own interval. Its scale changes because the unit of frequency changes.
Proof. Substitute \(\omega=2\pi\nu\) into the angular series. Directly from the definitions, \(2\pi f_{\mathrm{ang}}(2\pi\nu)=\sum_{h\in\mathbb{Z}}\gamma(h)e^{-2\pi ih\nu}=f_{\mathrm{cyc}}(\nu)\), which proves the pointwise conversion. For the normalization, set \(\omega=2\pi\nu\), so \(d\omega=2\pi\,d\nu\). Then \(\int_{-1/2}^{1/2}f_{\mathrm{cyc}}(\nu)\,d\nu=\int_{-1/2}^{1/2}2\pi f_{\mathrm{ang}}(2\pi\nu)\,d\nu=\int_{-\pi}^{\pi}f_{\mathrm{ang}}(\omega)\,d\omega=\gamma(0)\), where the last equality is the coefficient check at lag zero. \(\square\)
The factor \(2\pi\) in the conversion is required by the change of frequency coordinate. It does not mean that one convention has more total spectral mass: the frequency interval is \(2\pi\) times as long in angular units, while the density per unit frequency is correspondingly rescaled. Both densities are periodic with their respective periods, \(2\pi\) and \(1\); the displayed half-open intervals choose one representative period.
Worked Conversions
Worked Example: White Noise and the Density Scale
Let \((X_n)_{n\in\mathbb{Z}}\) be mean-zero white noise with variance \(\sigma^2=4\), so \(\gamma(0)=4\) and \(\gamma(h)=0\) for every nonzero integer \(h\). Compute both densities and check their integrals.
Only the zero-lag term remains in either series. Thus \(f_{\mathrm{ang}}(\omega)=\frac{1}{2\pi}\cdot4=\frac{2}{\pi}\), and \(f_{\mathrm{cyc}}(\nu)=4\). The angular interval has length \(2\pi\), so \(\int_{-\pi}^{\pi}\frac{2}{\pi}\,d\omega=\frac{2}{\pi}(2\pi)=4\). The cycle interval has length \(1\), so \(\int_{-1/2}^{1/2}4\,d\nu=4(1)=4\).
Check: The conversion gives \(2\pi f_{\mathrm{ang}}(2\pi\nu)=2\pi\cdot\frac{2}{\pi}=4=f_{\mathrm{cyc}}(\nu)\), and both integrals equal the variance \(4\).
Worked Example: A Geometric Covariance Sequence
Consider the covariance sequence \(\gamma(h)=(\tfrac12)^{|h|}\), for \(h\in\mathbb{Z}\). It is absolutely summable because its positive-lag tail is geometric. Compute its angular and cycle-frequency densities, then evaluate both at the endpoints \(0\) and the half-period.
Pairing the positive and negative lags gives \(\sum_{h\in\mathbb{Z}}(\tfrac12)^{|h|}e^{-ih\omega}=1+2\sum_{h\geq1}(\tfrac12)^h\cos(h\omega)\). Summing the two geometric series with ratios \(\tfrac12e^{-i\omega}\) and \(\tfrac12e^{i\omega}\) gives \(\frac{1-(\tfrac12)^2}{1-2(\tfrac12)\cos\omega+(\tfrac12)^2}=\frac{3/4}{5/4-\cos\omega}\). Including the angular factor \(1/(2\pi)\) yields \(f_{\mathrm{ang}}(\omega)=\frac{3}{8\pi(\tfrac54-\cos\omega)}\). Conversion then gives \(f_{\mathrm{cyc}}(\nu)=\frac{3}{4(\tfrac54-\cos(2\pi\nu))}\).
At \(\omega=0\), the denominator factor \(\tfrac54-\cos(0)=\tfrac14\), so \(f_{\mathrm{ang}}(0)=\frac{3}{8\pi(\tfrac14)}=\frac{3}{2\pi}\). At \(\nu=0\), the cycle formula gives \(f_{\mathrm{cyc}}(0)=\frac{3}{4(\tfrac14)}=3\). At \(\omega=\pi\), \(\cos(\pi)=-1\), so \(\tfrac54-\cos(\pi)=\tfrac94\) and \(f_{\mathrm{ang}}(\pi)=\frac{3}{8\pi(\tfrac94)}=\frac{1}{6\pi}\). At \(\nu=\tfrac12\), \(\cos(2\pi\cdot\tfrac12)=\cos(\pi)=-1\), so \(f_{\mathrm{cyc}}(\tfrac12)=\frac{3}{4(\tfrac94)}=\frac13\).
Check: The conversion holds at zero since \(2\pi\cdot\frac{3}{2\pi}=3\), and at the other endpoint since \(2\pi\cdot\frac{1}{6\pi}=\frac13\). The coefficient check at lag zero also requires both full-interval integrals to equal \(\gamma(0)=1\).
Worked Example: A Finite Moving Average
Let \(X_n=\varepsilon_n+\tfrac12\varepsilon_{n-1}\), where \((\varepsilon_n)_{n\in\mathbb{Z}}\) is mean-zero white noise with variance \(1\). Compute the covariance sequence and its two density formulas.
Independence across distinct noise times makes the variance \(\gamma(0)=1+(\tfrac12)^2=\tfrac54\). At lag \(1\), the only shared noise term is \(\varepsilon_{n}\) in \(X_{n+1}\) and \(X_n\), with coefficients \(\tfrac12\) and \(1\); hence \(\gamma(1)=\tfrac12\). By symmetry \(\gamma(-1)=\tfrac12\), and for \(|h|>1\) there is no shared noise term, so \(\gamma(h)=0\).
Substitution into the angular convention gives \(f_{\mathrm{ang}}(\omega)=\frac{1}{2\pi}(\tfrac54+\tfrac12e^{-i\omega}+\tfrac12e^{i\omega})=\frac{1}{2\pi}(\tfrac54+\cos\omega)\). The cycle formula is \(f_{\mathrm{cyc}}(\nu)=\tfrac54+\cos(2\pi\nu)\). The angular density is nonnegative because its minimum is \(\frac{1}{2\pi}(\tfrac54-1)=\frac{1}{8\pi}>0\); the cycle density has minimum \(\tfrac54-1=\tfrac14>0\).
Check: The angular integral is \(\frac{1}{2\pi}\int_{-\pi}^{\pi}(\tfrac54+\cos\omega)\,d\omega=\frac{1}{2\pi}(\tfrac54\cdot2\pi+0)=\tfrac54\). The cycle integral is \(\int_{-1/2}^{1/2}(\tfrac54+\cos(2\pi\nu))\,d\nu=\tfrac54+0=\tfrac54\). These match the variance. The conversion also gives \(2\pi\cdot\frac{1}{2\pi}(\tfrac54+\cos(2\pi\nu))=f_{\mathrm{cyc}}(\nu)\).
Conventions, Limits, and a Common Error
The formulas above apply directly when the covariance sequence is absolutely summable. For some stationary processes, covariances do not decay fast enough for the ordinary Fourier series to converge to an integrable density. In that situation the spectral description may require a measure rather than a density. The existence and structure of that representation are separate questions; the convention formulas here do not assert that every covariance sequence has an ordinary density. In particular, a non-decaying periodic covariance should not be inserted into an infinite Fourier series and treated as though absolute convergence had been established.
Common mistake: losing the normalization factor. For the geometric example, one might correctly obtain the Fourier sum \(\frac{3/4}{5/4-\cos\omega}\) and then write \(f_{\mathrm{ang}}(\omega)=\frac{3}{4\pi(5/4-\cos\omega)}\). That calculation has multiplied the sum by \(1/\pi\), rather than by the required \(1/(2\pi)\). At \(\omega=0\), it gives \(3/\pi\). But substituting \(\omega=0\) into the defining series gives \(\frac{1}{2\pi}[1+2\sum_{h\geq1}(1/2)^h]=\frac{1}{2\pi}[1+2]=\frac{3}{2\pi}\). The corrected formula is \(\frac{3}{8\pi(5/4-\cos\omega)}\), which gives exactly \(\frac{3}{2\pi}\) there. Its cycle-frequency counterpart is \(\frac{3}{4(5/4-\cos(2\pi\nu))}\), not twice that value.
When using a spectrum, record three choices together: the frequency interval, the forward-series sign, and the density normalization. Then use the paired reconstruction sign and the appropriate integration interval. This makes a proposed formula checkable: its integral must recover \(\gamma(0)\), and its Fourier coefficient at lag \(h\) must recover \(\gamma(h)\). For an even real covariance sequence, a cosine expression is often the shortest route to the formula, but it does not remove the normalization check.
Check Your Understanding
Use the stated angular and cycle-frequency conventions in each question, and show the normalization or coefficient check that supports your answer.
- If \(\gamma(0)=2\) and all other covariance lags are zero, what are the two constant densities, and what does each integrate to over its stated interval?
- For an absolutely summable real covariance sequence, why does the angular forward series have the same value if its exponential sign is reversed?
- State the exact relation between \(f_{\mathrm{ang}}\) and \(f_{\mathrm{cyc}}\), including the argument substitution.
- For the moving average in the third worked example, recover \(\gamma(1)\) from the angular density by integrating against the paired exponential.
- What condition used here guarantees an ordinary Fourier-series density, and what should you avoid claiming when that condition is not known?